General Organic and Biological Chemistry An Integrated Approach, 4th Edition By Raymond
Chapter 1 Science and Measurements Outline Notes I. Scientific Method (Section 1.1) A. Science and Chemistry Overview 1. Science greatly impacts our lives (i.e. DNA discovery) 2. Science covers a wide range of subjects including chemistry, biology, biochemistry, geology, astronomy, physics, etc. 3. Chemistry is the study of matter and its changes 4. Chemistry lays the groundwork for many other fields of science 5. Examine Figure 1.1 B. Scientific Method 1. Observation is the first step where information is gathered (i.e. Newton’s apple) 2. Laws are statements that describe things consistently and reproducibly observed (i.e. Newton’s Law of Gravity) 3. A hypothesis is a tentative and testable explanation based on observations or known facts (i.e. a clinician diagnosis a patient’s problem) 4. Experiments test the hypothesis (i.e. medical tests) a. Designed to be directly related to the problem or question at hand b. If the experiment invalidates the hypothesis then the hypothesis must be revised c. If a hypothesis survives repeated testing in can become part of a theory 5. A theory is an experimentally tested explanation of an observed behavior a. Must be consistent with existing experimental evidence b. Must accurately predict the results of future experiments c. Must explain future observations 6. Examine Figure 1.2 a. Scientist may not follow this order and they may not use all the steps b. Creativity and sufficient knowledge are required for the process c. Theories are continually reevaluated and possibly revised *See Sample Problem 1.1 and Practice Problem 1.1 DID YOU KNOW? Diabetes HEALTH LINK Science and Medicine (Examine Figure 1.4) II. Matter and Energy (Section 1.2) A. Matter Properties 1. Matter is anything that has mass and occupies space (i.e. body, air, book) 2. Physical properties are characteristics that can be determined without changing the chemical composition of matter (what it’s made of) a. Odor, color, malleability, taste, melting point, state/phases 3. Three physical states/phases a. Solids have fixed shapes and volumes b. Liquids have variable shapes and fixed volumes c. Gases have variable shapes and volumes 4. Examine Figure 1.5 5. Physical state is partially determined by the interactions of particles (atoms, molecules, etc.) 6. Examine Figure 1.6 B. Physical Changes 1. A physical change does not alter the chemical composition of matter (ice to water to steam) 1-1
2. Examine Figure 1.7 *See Sample Problem 1.2 and Practice Problem 1.2 C. Energy and Matter 1. Anytime matter is changed, work has been done 2. Energy is the ability to do work or transfer heat 3. Potential energy is stored energy and kinetic energy is the energy of motion 4. Potential energy (i.e. gasoline in a car) can be converted into kinetic energy (i.e. movement of a car’s piston), which produces work and heat *See Sample Problem 1.3 and Practice Problem 1.3 D. Heat and Changes 1. Heat plays a role in physical changes (i.e. heat added to water causing the motion of the particles/kinetic energy to increase turning it into steam) 2. Heat of fusion is energy put in during the melting process (solid to liquid) 3. Heat of vaporization is the energy put in during the boiling/vaporization process (liquid to gas) 4. These processes can be reversed with the removal of heat (i.e. condensation and freezing) 5. Examine Figure 1.8 6. Under certain conditions, some substances skip the liquid phase a. Sublimation is the conversion of a solid to a gas (i.e. dry ice) b. Deposition is the conversion of a gas to solid 7. Examine Figure 1.9 *See Sample Problem 1.4 and Practice Problem 1.4 III. Units of Measurement (Section 1.3) A. Measurement overview 1. Measurements consist of a number and a unit (i.e. 3 minutes, 3 hours, 3 miles) 2. The three systems/units of measurement are the metric units (used worldwide), the English units (common in US) and the SI units (international derived from metric system) 3. Examine Figure 1.10 B. Mass 1. Measure of the amount of matter in a sample 2. See Table 1.1 and 1.2 for common units and relationships 3. Weight (different from mass) is a measure of the force of gravity on an object’s mass C. Length 1. Measure of the distance between two points 2. One meter is equal to 3.281 feet therefore it is slightly longer than a yard D. Volume 1. Measure of the space occupied by an object 2. One liter is 1/1000 that of cubic meter 3. Examine Figure 1.11 DID YOU KNOW? Oil Spill E. Temperature 1. Three scales/units are degrees Fahrenheit (°F), degrees Celsius (°C), and Kelvin (K) a. °F is the English unit, water freezes at 32°F and boils at 212°F b. °C is the metric unit, water freezes at 0°C and boils at 100°C c. K is the SI unit based on absolute zero (coldest possible temperature: 0K/-273.15°C) 2. Examine Figure 1.12 F. Energy 1. The calorie (cal) is the metric and English unit for energy 2. 1 cal is the amount of energy required to raise 1 gram of water 1°C 3. The joule (J) is the SI unit; 4.184J = 1 cal 4. The food calorie (Cal) is used commonly; 1 Cal = 1000cal *See Sample Problem 1.5 and Practice Problem 1.5 1-2
IV. Scientific Notation, SI and Metric Prefixes (Section 1.4) A. Scientific Notation 1. Condenses very large or very small numbers 2. Expresses a number between 1 and 10 multiplied by a power of ten a. Positive exponents indicate how many times to multiply by 10 (shift decimal left) b. Negative exponents indicate how many times to divide by 10 (shift decimal right) 3. See Table 1.3 for scientific notation examples 4. Allows numbers to be compared without counting zeros *See Sample Problem 1.6 and Practice Problem 1.6 B. SI and Metric Prefixes 1. Prefixes are another way to express large or small numbers 2. The prefix indicates how the new unit relates to the original 3. See Table 1.4 for SI and metric prefixes and their relation to the original unit 4. Note there is more than one way to write equalities (1mL = 1x10-3L or 1000mL = 1L) 5. Prefixes can be applied to other units but it is uncommon (i.e. milliquarts) *See Sample Problem 1.7 and Practice Problem 1.7 V. Measurements and Significant Figures (Section 1.5) A. Accuracy and Precision 1. Accuracy is how close a measured value is to an accepted value (i.e. body temp is 98.6°F) 2. Precision is the reproducibility of a series of measurements (i.e. 98°F, 99°F, 97°F) 3. Measurements can be precise only, accurate only (on average), both precise and accurate, or neither precise and accurate 4. Examine Figure 1.14 B. Significant Figures 1. The quality of the equipment used to make measurements affects the accuracy and precision 2. Significant figures are defined as digits in a measurement that are reproducible when the measurement is repeated, plus the first uncertain digit (i.e. 5.671 g means the first three numbers 5.67 are certain while the last number 1 has some uncertainty) 3. Examine Figure 1.15 4. Reported zeros are sometimes significant and other times not (i.e. 2.00 has three significant figures while 0.009 only has one significant figure) 5. See Table 1.5 for rules involving significant figures 6. Significant figures only apply to measurements a. Exact numbers have an unlimited number of significant figures b. Come from exact counts (i.e. number of patients) and defined units (i.e. 1km = 1000m) *See Sample Problem 1.8 and Practice Problem 1.8 HEALTH LINK Body Mass Index (Examine Figure 1.16, Figure 1.17, and Table 1.6) C. Calculations with Significant Figures 1. Calculations should not change the degree of uncertainty in a value 2. When doing multiplication and division the answer should have the same number of significant figures as the quantity with the fewest (i.e. 5.3cm x 6.1cm should have only two: 32cm2) 3. The first digit to be removed is the digit immediately to right of the significant digit in a number *See Sample Problem 1.9 and Practice Problem 1.9 *See Sample Problem 1.10 and Practice Problem 1.10 4. When doing addition or subtractions the answer should have the same number of decimal places as the quantity with the fewest decimal places (i.e. 13.5g + 2.335g + 653g should have no places after the decimal: 669g) *See Sample Problem 1.11 and Practice Problem 1.11 HEALTH LINK Body Temperature (Examine Figure 1.18 and 1.19) VI. Conversion Factors and the Factor Label Method (Section 1.6) A. Unit Conversions 1-3
1. Many can be done in your head (i.e. 6 eggs are in a half dozen) 2. Others require a systematic approach called the factor label method B. Factor Label Method 1. Uses conversion factor to transform one unit into another 2. These are derived from the numerical relationship between two units (i.e. 2.205 lbs = 1kg) 3. All conversion factors are equal to 1 when divided 4. Review examples in text (What is the kilogram weight of 185 lb patient?) 5. Examine Figure 1.20 6. Make sure units cancel out, multiply the numerators, multiply the denominators, divide the two answers, and report final answer with proper units and significant figures DID YOU KNOW? Mars Climate Orbiter *See Sample Problem 1.12 and Practice Problem 1.12 *See Sample Problem 1.13 and Practice Problem 1.13 C. Temperature Conversions 1. To convert from °F to °C use one of the two equations (°F and °C are not the same size) a. °F = (1.8 x °C) + 32 b. °C = (°F – 32)/1.8 2. To convert from °C and K use one the two equations (°C and K are the same size and note there are no degrees in Kelvin) a. K = °C + 273.15 b. °C = K – 273.15 *See Sample Problem 1.14 and Practice Problem 1.14 VII. Density, Specific Gravity, and Specific Heat (Section 1.7) A. Density 1. The amount of mass contained in a given volume 2 Usually expressed in g/cm3 for solids, g/mL for liquids and g/L for gases 3. Density of water at 20°C is 1.00 g/mL 4. As temperature varies, density varies (i.e. how mercury in a thermometer works) 5. See Table 1.7 for the densities of common substances 6. For calculations density (D) = mass (m)/volume (V) 7. Review examples in text (What is the mass in grams of 2.15L of isopropyl alcohol?) *See Sample Problem 1.15 and Practice Problem 1.15 B. Specific Gravity 1. Relates the density of a substance to that of water 2. Specific Gravity = (Density of Substance)/(Density of Water) 3. Review example in text (What is the specific gravity of isopropyl alcohol?) 4. A substance with a specific gravity of less than 1 will float in water, while one that has a specific gravity of greater than 1 will sink 5. Specific gravity has a variety of applications (i.e. car batteries, alcohol content, urine analysis) 6. Examine Figure 1.21 HEALTH LINK Making Weight (Examine Figure 1.22) C. Specific Heat 1. The amount of heat required to raise the temperature of 1 gram of a substance by 1°C 2. Relates energy (cal), mass (g), and temperature (°C) 3. Review example in text (How many calories of heat are needed to increase the temperature of 2.5g of water by 5.0°C?) 4. See Table 1.8 for some specific heats *See Sample Problem 1.16 and Practice Problem 1.16 VIII. Measurements in General Chemistry, Organic Chemistry, and Biochemistry (Section 1.8) A. Sub-Disciplines of Chemistry 1-4
1. General chemistry is a study of the fundamental principles of chemistry (i.e. atoms, units, etc.) 2. Organic chemistry is a study of the chemistry of carbon 3. Biochemistry is a study of the chemistry of living things B. Examples of how chemistry relates to what goes on around you 1. A look at Cadmium (Examine Figure 1.23) *See Sample Problem 1.17 and Practice Problem 1.17 2. A look at Gasoline (Examine Figure 1.24 and 1.25) *See Sample Problem 1.18 and Practice Problem 1.18 3. A look at DNA (Examine Figure 1.26 and 1.27) *See Sample Problem 1.19 and Practice Problem 1.19
Chapter Summary Section 1.1 Science encompasses many subjects that greatly impact our lives. One such subject is chemistry, which is defined as the study of mater and its changes. Chemistry lays the groundwork for many other fields of study. Science, as a discipline, comes from the process known as the scientific method. The first step of the scientific method requires an observation. Next, a hypothesis is formed, which is a tentative and testable explanation of the observation. The hypothesis is then tested with an experiment. If the hypothesis survives repeated testing, then that hypothesis may become part of a theory. A theory should not be confused with a law, which is a statement that describes things consistently and reproducibly observed. Theories never become laws. Theories are an explanation and laws are a description. Section 1.2 Matter is anything that has mass and occupies space. The physical properties of matter can be determined without changing its chemical composition (what it’s made of). Some examples of physical properties include malleability, taste, melting point, and state (solid, liquid, or gas). A change in state is an example of a physical change (one that does not alter the chemical composition of matter). When matter is physically changed, energy is involved. Energy is the ability to do work or transfer heat. Energy can be found in two forms: potential energy (stored energy) and kinetic energy (energy of motion). Heat energy plays a major role in physical changes. For example, the heat of fusion is energy put in during the melting process (solid to liquid) and the heat of vaporization is the energy put in during the boiling/vaporization process (liquid to gas). The physical change of matter’s state can also sometimes skip the liquid phase. Sublimation is the conversion of a solid to a gas where deposition is the conversion of a gas to solid. Section 1.3 Scientific measurements consist of a number and a unit. The three systems of measurement units are the metric units, the English units, and the SI units. Five types of common measurements are mass, length, volume, temperature, and energy. Mass is measure of the amount of matter in a sample, which is different from weight (gravity dependent). Length is the measure of distance and volume is the measure of the space occupied by an object. There are three scales used to measure temperature: degrees Fahrenheit (English unit), degrees Celsius (metric unit), and Kelvin (SI unit based on absolute zero). The calorie is the metric and English unit for energy. One calorie is the amount of energy required to raise 1 gram of water 1°C. The joule (SI unit) and the food calorie are also used for energy measurements. Section 1.4 Scientific notation, which condenses very large or very small numbers, expresses a number between 1 and 10 multiplied by a power of ten. Positive exponents shift the decimal left (multiply by 10 “n” times); conversely, negative exponents shift the decimal right (divide by 10 “n” times). SI & metric prefixes are another way to express large or small numbers. The prefix indicates how the new unit relates to the original unit. Table 1.4 summarizes the common SI and metric prefixes. 1-5
Section 1.5 The accuracy of a measurement refers to how close a measured value is to a true value. Precision, on the other hand, is the measure of reproducibility. The quality of the equipment used to make measurements affects both accuracy and precision. Significant figures can be used to indicate the certainty of a measurement. Significant figures are defined as digits in a measurement that are reproducible when the measurement is repeated, plus the first uncertain digit. Significant figures only apply to measurements: exact numbers have an unlimited number of significant figures. Calculations should not change the degree of uncertainty in a value. When performing multiplication and division, the answer should have the same number of significant figures as the quantity with the fewest. When performing addition or subtractions, the answer should have the same number of decimal places as the quantity with the fewest decimal places. Section 1.6 When dealing with measurements it is often necessary to perform unit conversions. One approach to unit conversions is the factor label method, which uses conversion factors to transform one unit into another. Conversion factors are derived from the numerical relationship between two units, and when divided, they equal 1. Temperature conversions require equations that come from the relationships of the different temperature scales. For the conversion between degrees Fahrenheit and degrees Celsius the relationship is °F = (1.8 x °C) + 32 or °C = (°F – 32)/1.8. For the conversion between degrees Celsius and Kelvin the relationship is K = °C + 273.15 or °C = K – 273.15. Section 1.7 Another unit used to describe matter is density, which is the amount of mass contained in a given volume. Density is usually expressed in g/cm3 for solids, g/mL for liquids and g/L for gases. The density of water at 20°C is 1.00 g/mL(temperature affects density). Specific gravity relates the density of a substance to that of water. This relationship, which has many applications, is summarized by the equation Specific Gravity = (Density of Substance)/(Density of Water). A substance with a specific gravity of less than 1 will float in water, while one that has a specific gravity of greater than 1 will sink. Specific heat is yet another unit used to describe matter. Specific heat is the amount of heat required to raise the temperature of 1 gram of a substance by 1°C. It relates energy (cal), mass (g), and temperature (°C). Section 1.8 Chemistry has many sub-disciplines, three of which will be studied in this text. General chemistry is a study of the fundamental principles of chemistry. Organic chemistry is a study of the chemistry of carbon. Biochemistry is a study of the chemistry of living things. Some examples of how chemistry relates to what goes on around us include: a look at cadmium (General Chemistry), a look at gasoline (organic chemistry), and a look at DNA (biochemistry).
Lecture Suggestions Section 1.1 and 1.2 To introduce these section, ask students what makes a science textbook different from another subject’s textbook (history, English, etc.). Many students will say it is because a science book contains only facts. Explain that it is incorrect to describe science as simply a collection of facts. Describe how science is actually a process known as the scientific method, and the information in this textbook is the result of numerous scientists doing the scientific method over numerous generations. While the end results or “facts” are very important, it is the process (scientific method) that distinguishes science from other areas of study. Ask students to name ways in which science has directly affected their lives. 1-6
Section 1.3 and 1.4 To introduce these sections, an interesting approach is to say “I am doing three this weekend,” and ask students what this means. Obviously, they will have no clue what this means. Then say “what if I said 3 kilometers, or 3 hours, or 3 fraternity parties.” Explain that in science, numbers must have a unit. Explain how the metric and SI system prefixes are much easier to work with than the English system, and once they memorize a few prefixes, they can apply them to anything (a lot less memorization than the English system). For example, if you know giga is a billion greater than the base unit, then you can quickly figure out that a USB stick with 15 Gbytes had 15 billion bytes. Suggest to students that they can even make up their own units— “I have 2 kilofriends.” Finally, state how important it is in the health professions to know how units and their relationships work (mention the Figure 1.20 on math errors in medicine). Section 1.5 and 1.6 Significant figures are always difficult for students to grasp, especially since their whole lives they have never had to bother with them. A lot of students can recognize and even calculate them correctly, but they have no idea what they really mean. During this lecture, it’s helpful to use numerous visuals and have students make measurements (length of a pencil with three different rulers, volume of water with three different graduated cylinders, and temperature of the room with three different thermometers). For section 1.6, try teaching the factor label method by having students give the larger unit the 1 when writing metric/SI equalities. Tell students that if the larger unit always gets the 1, then the smaller unit will always have positive exponents (there will always be a bunch of little things in 1 large thing). For example, 1km =1000m (1x103m) or 1m = 1,000,000,000nm (1x109). This really helps the students understand the relationships and it prevents them from getting the equalities backwards. Section 1.7 and 1.8 At the beginning of the density lecture, show students the “burning water” demonstration. Add 1-2mL of lighter fluid to a 500mL Erlenmeyer flask before class. Then have a student open a brand new bottle of water (500mL size) and take a small drink to verify it is pure water. Pour the water into the flask and light the top of it on fire. Ask students to explain how you are able to burn water. Explain the concept of density and buoyancy. For section 1.8, talk about how in chemistry, “organic” refers to carbon and not living like the general public thinks. Originally, organic meant living because it got its name from organisms, but in chemistry, biochemistry is the discipline that deals with living things. From a chemistry standpoint, organic foods just mean they have carbon, which is actually every food.
Handouts for Students 1) For the Quantity Conversions handout (next page) have students memorize these common relationships. This will help them significantly when doing the factor label method.
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Handout Quantity Conversions (Memorize) English ↔ English Conversion Factors Distance 12 inches (“/in) = 1 foot (‘/ft) 3 feet (‘/ft) = 1 yard (yd) 1,760 yard (yds) = 1 mile (mi)
Mass 16 ounces (oz) = 1 pound (lb) 2,000 pounds (lbs) = 1 ton (t) 8 fluid ounces (fl oz) = 1 cup (c)
Time 60 seconds (sec) = 1 minute (min) 60 minutes (mins) = 1 hour (hr) 24 hours (hrs) = 1 day (d) 365.25 days (d) = 1 year (yr)
Volume 2 cups (c) = 1 pint (pt) 2 pints (pt) = 1 quart (qt) 4 quarts (qt) = 1 gallon (gal)
Metric ↔ Metric Conversion Factors 1 Giga-base unit (G) = 1 billion base units (1x109) 1 Mega-base unit (M) = 1 million base units (1x106) 1 kilo-base unit (k) = 1 thousand base units (1x103) 1 base unit = ten deci-base units (d) (1x101) 1 base unit = 1 hundred centi-base units (c) (1x102) 1 base unit = 1 thousand milli-base units (m) (1x103) 1 base unit = 1 million micro-base units (μ) (1x106) 1 base unit = 1 billion nano-base units (n) (1x109) 1 base unit = 1 trillion pico-base units (p) (1x1012)
Base Units distance – meter (m) mass – gram (g) time – second (s) energy – calorie (cal) substance – mole (mol) *volume – liter (l) Note: 1 cm3 (cc) = 1 mL
Metric ↔ English Conversion Factors Distance 2.54 centimeters (cm) = 1 inch (in) Mass 2.2 pounds (lbs) = 1 kilogram (kg) Volume 3.785 liters (l) = 1 gallon (gal)
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Chapter 2 Atoms and Elements Outline Notes I. Atoms (Section 2.1) A. Atoms Overview 1. Greek philosophers believed that all matter was made of indestructible building blocks called atoms (from Greek atomos meaning indivisible) 2. English scientist John Dalton described atoms based on measurements and experiments 3. Dalton’s Atomic Theory states that the atom is the basic unit from which matter is constructed a. A sample of iron can only be divided so many times—eventually an indivisible iron atom is all that remains b. Iron is different from a carbon, oxygen, or hydrogen atom c. The theory has been revised due to subsequent experiments (i.e. electricity and cathode rays) 4. Examine Figure 2.1 B. Subatomic Particles 1. Subatomic (smaller than an atom) particles make up atoms 2. The three subatomic particles are electrons (discovered in 1887), protons (discovered in 1919), and neutrons (discovered in 1932) 3. All matter consists of atoms, which are built from these three subatomic particles 4. Examine Figure 2.2 C. Properties of Atoms 1. Atoms are mostly empty space where electrons move 2. Atomic mass units (amu) or daltons are used to represent the mass of the subatomic particles a. A carbon atom with 6 protons and 6 neutrons has a mass of exactly 12 amu or daltons b. Protons are 1.0073 amu, neutrons are 1.0087 amu, and electrons are 5.486x10-4 amu c. The nucleus is the site of most of the atom’s mass and it is made up of protons and neutrons 3. All three subatomic particles have different electrical charges a. Protons are +1, neutrons are zero, and electrons are -1 b. The nucleus is positive and the electron clouds around the nucleus are negative c. Opposite electrical charges attract while similar charges repel d. A strong force holds the similar charges of protons (+) together 4. See Table 2.1 for subatomic particle information *See Sample Problem 2.1 and Practice Problem 2.1 II. Elements (Section 2.2) A. Elements Overview 1. Thousands of years ago the Greeks described matter as being combinations of the basic elements: earth, air, fire, and water 2. An element today is a substance that contains only one type of atom (i.e. carbon atom, graphite, and diamond are all examples of the element carbon) 3. Examine Figure 2.3 B. Element Names and Symbols 1. By the year 1800 only 31 elements had been discovered; however today there are 117 (92 are natural and 112 have been given official names) 2. Element names come from a variety of sources: i. Named after appearance (iodine comes from Greek iodes meaning violet) ii. Named after place of discovery (berkelium for Berkeley California) iii. Named after a person (einsteinium for Albert Einstein) 3. Examine Figure 2.4 2-1
4. Atomic symbols also identify elements i. They are one or two letter abbreviations for the name ii. Some are based on alternate names (sodium is Na after natron meaning soda) 5. See Table 2.2 for some element names and symbols DID YOU KNOW? Romans and Lead DID YOU KNOW? Element 112 III. Trace Elements (Section 2.3) A. Elements and the Body 1. The USDA publishes a set of dietary guidelines to help people have healthy diets 2. The current guidelines released in 2005 recommend a diet low in saturated and trans fats, cholesterol, salt, and sugar, but should include fruits, vegetables, whole grains, and low-fat milk 3. Examine Figure 2.5 4. Healthy diets should also include proteins, carbohydrates, fats, and trace elements 5. Trace elements are elements needed by the body in small quantities (less 100mg per day) 6. Oxygen, carbon, hydrogen, and nitrogen make up most of the mass of the human body 7. See Table 2.3 for biochemical roles of some elements B. NAS Guidelines 1. Recommended Dietary Allowance (RDAs) is for the daily intake of nutrients in a healthy diet 2. Dietary Reference Intakes (DRIs) help health professionals make nutritional recommendations 3. Adequate Intake (AI) is used when RDAs are not available 4. Tolerable Upper Intake Level (UL) is the highest intake that can be safely consumed (i.e. 40mg for zinc—taking too much can result in nausea, vomiting, and dizziness) 5. See Table 2.4 for DRIs 6. The FDA requires manufactures to provide a nutrition facts label a. Serving size should always be noted b. Nutrient facts and percent daily amount are based on a 2,000 Cal diet 7. Examine Figure 2.6 *See Sample Problem 2.2 and Practice Problem 2.2 IV. Atomic Number and Mass Number (Section 2.4) A. Atomic/Mass Number Overview 1. Atomic number is the number of protons an atom has in its nucleus (i.e. sulfur has 16 protons, thus its atomic number is 16) 2. Mass number is the number of protons and neutrons in the nucleus of an atom (i.e. an aluminum atom containing 13 protons and 14 neutrons has a mass number of 27) 3. Atomic notation expresses the mass number as a superscript and the atomic number as a subscript (i.e. 147N means the mass number is 14, the atomic number is 7, and it has 7 protons and 7 neutrons) *See Sample Problem 2.3 and Practice Problem 2.3 4. Mass number is not the mass of the atom i. It is just a count of protons and neutrons ii. Protons and neutrons do not have the same mass nor are their masses 1 amu (due to mass defect) B. Isotopes 1. Atoms of a particular element always have the same atomic number, but may have different mass numbers 2. Isotopes are atoms of an element that have different numbers of neutrons 3. Isotopes can be written in atomic notation (i.e. 11H, 21H, and 31H), or by following the element’s name with a hyphen and the mass number (i.e. hydrogen-1, hydrogen-2, and hydrogen-3) 4. Many elements have more than one isotope, and those are usually not present in equal amounts (i.e. 3517Cl is the most abundant of the two chlorine isotopes: chlorine-35 and chlorine-37) 5. Examine Figure 2.7 2-2
*See Sample Problem 2.4 and Practice Problem 2.4 C. Electrons 1. The number of electrons can be determined by the number of protons in a neutral atom 2. A neutral atom (charge = 0) contains an equal number of protons (+) and electrons (-) 3. A neutral oxygen atom (atomic number 8) has 8 protons and 8 electrons HEALTH LINK Stable Isotopes and Drug Testing (Examine Figure 2.8 and 2.9) V. Periodic Table (Section 2.5) A. Periodic Table Overview 1. The periodic table of the elements is a complete list of the elements arranged from smallest to largest atomic number 2. The table includes the atomic symbol, the atomic number, and the atomic weight 3. Atomic weight is the average mass of the atoms of an element as it is found in nature a. Bromine has two isotopes found in nature: 50.69% appears as bromine-79 and 49.31% appears as bromine-81, thus its atomic weight is 79.92 amu b. Some elements on the periodic table have their atomic weights in brackets noting that they are only found as unstable isotopes B. Metals, Nonmetals, and Semimetals 1. Elements can be grouped into metals, nonmetals, and semimetals (or metalloids) 2. Examine Figure 2.10 3. The heavy line on the periodic table separates the metals from the nonmetals a. Metals to the left and nonmetals are to the right b. Semimetals border the line 4. As solids, metals are conductors (electricity and heat), lustrous, malleable, and ductile 5. As solids, nonmetals are poor conductors and they are nonlustrous and brittle 6. Semimetals have physical properties that are intermediate of metals and nonmetals 7. Examine Figure 2.11 *See Sample Problem 2.5 and Practice Problem 2.5 C. Groups 1. Groups are vertical columns of elements on the periodic table 2. Groups are identified by numbers (1-8) and letters (A or B) 3. Some chemical groups have been given names (families) a. Alkali metals are group 1A (Li, Na, K, Rb, Cs, Fr) b. Alkaline earth metals are group 2A (Be, Mg, Ca, Sr, Ba, Ra) c. Halogens are group 7A (F, Cl, Br, I, At) d. Noble gases (or inert gases) are group 8A (He, Ne, Ar, Kr, Xe, Rn) 4. Groups 1A-8A are collectively called the representative elements 5. Groups 1B-8B are collectively called the transition metals 6. Elements 57 through 71 are called the lanthanide elements 7. Elements 89 through 103 are called the actinide elements D. Periods 1. Periods are horizontal rows of elements on the periodic table 2. The term “period” relates to the periodic (regular) changes that take place in the element properties as atomic number is increased 3. When Dmitri Mendeleev (1860s) arranged the known elements in a row by increasing atomic number, he noticed that the properties changed; when the properties began to repeat he started a new row (elements with similar properties ended up in the same group) E. Periodic Change 1. Elements on the far left side of a period are the most metallic while those on the far right are most nonmetallic (i.e. in period 3, Na, Mg, and Al are metals, Si is a semimetal, and P, S, Cl, and Ar are nonmetals) 2-3
2. Moving down a group, atoms become more metallic (i.e. in group 4A, C is a nonmetal, Si and Ge, are semimetals, and Sn and Pb are metals) 3. Atomic size also shows periodic behavior a. Moving across a period from left to right decreases atomic size b. Moving down group increases atomic size 4. Examine Figure 2.12 *See Sample Problem 2.6 and Practice Problem 2.6 HEALTH LINK Lead (Examine Figure 2.13 and 2.14) VI. The Mole (Section 2.6) A. Mole Overview 1. Atoms are too small to deal with one by one; therefore they are best represented by the mole 2. A small sample of an element contains a huge number of atoms (i.e. one cube of gold that is 1in3 contains 1x1024 gold atoms) 3. One mole of something contains 6.02x1023 items (Avogadro’s number) 4. Avogadro’s number is equal to the number of atoms in exactly 12 grams of pure carbon-12 5. Examine Figure 2.15 6. Even though a mole is a huge number of items, atoms are so small that one mole of a substance doesn’t occupy that much space 7. Examine Figure 2.16 8. Because 1 mole is equal to 6.02x1023 atoms, this can be used as a conversion factor 9. Review example in text (How many atoms are in 0.500 mol C?) *See Sample Problem 2.7 and Practice Problem 2.7 B. The Mole and Mass 1. The molar mass of an element is the mass in grams of one mole of its atoms 2. Numerically it is equal to the atomic weight of the element (i.e. Li has an atomic weight of 6.94 amu, which means it has a molar mass of 6.94 g/mol or 6.94 g = 1 mol) 3. The molar mass can be used as a conversion factor 4. Review examples in text (How many grams are in 0.770 mol C?) *See Sample Problem 2.8 and Practice Problem 2.8 VII. The Arrangement of Electrons (Section 2.7) A. Bohr Model 1. Chemistry requires an understanding of the electrons that are placed around an atom’s nucleus 2. When electricity is passed through a tube of hydrogen gas, the atoms emit energy as electromagnetic radiation, some of which may be observed as visible light 3. This light can be separated with a prism and viewed as an emission spectrum 4. Examine Figure 2.17 5. In the Bohr model, electrons circle the nucleus in specific orbits, which correspond to different energy levels a. The ground state (most stable electron arrangement) is when the electrons are in energy levels as near as possible to the nucleus b. If a ground state hydrogen atom absorbs energy, its electron is pushed to an excited state, which is an energy level farther from the nucleus 6. Examine Figure 2.18 7. When an atom returns its electron(s) from an excited state back to a ground state, electromagnetic radiation (energy that travels as waves) is given off 8. Types of electromagnetic radiation include radio waves, microwaves, infrared light, visible light, ultraviolet light, and x-rays B. Quantum Mechanics 1. Bohr’s model was only good for explaining the emission spectrum of hydrogen—it didn’t work for other elements 2. It was demonstrated that electrons behave as energy waves 2-4
3. Austrian physicist Erwin Schödinger devised a mathematical equation that described electrons in atomic orbitals—this approach is called quantum mechanics C. Atomic Orbitals 1. Atomic orbitals are 3D regions of space where there is a high probability of finding an electron 2. Examine Figure 2.19 3. Quantum mechanics allows us to calculate the maximum number of electrons in an energy level a. Maximum number of electrons per energy level = 2n2 b. n is the number of the energy level c. For example, the second energy level (n = 2) can hold up to 8 electrons (2 x 22 = 8) d. See Table 2.5 for maximum electrons per energy level 4. Ground state electron arrangements can be predicted from quantum mechanics 5. Examine Figure 2.20 6. See Table 2.6 for the ground state electron distributions of the first 20 elements *See Sample Problem 2.9 and Practice Problem 2.9 D. Valence Electrons 1. Elements in a given group are related based on the arrangement of their electrons 2. The valence shell is the highest numbered, occupied energy level 3. In groups 1A through 8A, the group number tells how many valence electrons each element has a. All group 1A atoms have one valence electron, group 2A have two, etc. b. Helium is an exception: it only has two (not eight). 4. The period of an element indicates what energy level its valance electrons are in (i.e. K is in period four, so its valence electron is in the fourth energy level) *See Sample Problem 2.10 and Practice Problem 2.10 E. Electron Dot Structures 1. American chemist Gilbert Lewis developed electron dot structures in the early 1900s 2. They show the number of valence electrons that an atom carries by representing valence electrons as dots (i.e. Li has one valence electron, so it has one dot by its symbol) 3. Examine Figure 2.21 *See Sample Problem 2.11 and Practice Problem 2.11 BIOCHEMISTRY LINK Bioluminescence (Examine Figure 2.22 and 2.23) VIII. Radioactive Isotopes (Section 2.8) A. Nuclear Change 1. Recall that in a physical change the composition of matter is not changed 2. When mater undergoes a chemical change, the arrangements of atoms is changed 3. A nuclear change involves the change of atomic nuclei (not a physical or chemical change) 4. There are over 300 isotopes for the natural 91 elements and over 1000 isotopes have been artificially produced 5. Some isotopes (radioactive isotopes or radioisotopes) have unstable nuclei that spontaneously disintegrate to become more stable and release high-energy particles called nuclear radiation B. Common Forms of Nuclear Radiation 1. An alpha particle (42α) is a helium-4 (42He) atom’s nucleus moving at 5-10% the speed of light a. Examine Figure 2.24a b. A nuclear equation shows the starting radioisotope (left), an arrow, and the products (right) c. In a balanced nuclear equation the sum of the mass number and the sum of the charges on atomic nuclei and subatomic particles are the same on both sides of the equation (i.e. 23092Th alpha decay) 2. See Table 2.7 for common forms of nuclear radiation *See Sample Problem 2.12 and Practice Problem 2.12 3. A beta particle (0-1β) is an electron that is ejected from the nucleus of a radioisotope at up to 90% the speed of light (i.e. 125B beta decay) a. Examine Figure 2.24b 2-5
b. In beta decay a neutron transforms into a proton and an electron, which is ejected (0-1β) 4. A positron (01β+) has the same mass as a beta particle but carries a 1+ charge a. It is also ejected from the nucleus of a radioisotope at up to 90% (i.e. 189F positron emission) b. Positron radiation is used in the medical procedure positron emission tomography c. Examine Figure 2.24c 5. A gamma ray (00γ) is a very high energy form of electromagnetic radiation a. Alpha, beta, and positron radiation are often accompanied by the release of gamma rays because the nucleus is often in an unstable state (i.e. 13153I gamma emission) b. Examine Figure 2.24d *See Sample Problem 2.13 and Practice Problem 2.13 DID YOU KNOW? Fission and Fusion IX. Radioisotopes in Medicine (Section 2.9) A. Effects of Radiation 1. Nuclear radiation can transfer kinetic energy to matter (living tissue) causing it to change a. The changes can alter water and important biochemical substances (i.e. proteins, DNA, etc.) b. These changes can disrupt normal cellular functions, which can cause health problems 2. Commonly used radiation units are used to understand radiation exposure a. Units are based on one of three properties i. How fast a sample of an unstable radioisotope disintegrates ii. The energy transferred when radiation strikes matter iii. The biological effect of radiation exposure b. One curie (Ci) is 3.7x1010 disintegrations per second c. One rad (for radiation absorbed dose) is a dose of 0.01 Joules per kilogram of the mass of an object (0.01 J/kg) d. The rem unit takes biological effect into account by multiplying the dose in rads by a quality factor (QF) that depends on the type of radiation i Dose (rad) x QF = dose (rem) ii. See Table 2.8 for quality factors iii. Review example in text (Lethal dose of radiation) e. SI units are used in medical applications: 1 gray (Gy) = 100 rads; 1 seivert (Sv) = 100 rem 3. The effects of short-term radiation exposure range from nausea to death a. See Table 2.9 for the health effects of short term radiation exposure b. Examine Figures 2.25 and 2.26 4. The effects of long-term exposure at low levels are not well understood a. Everyone is continually exposed to naturally occurring background radiation b. A significant portion of background radiation exposure comes from radon gas (Radon-222) c. X-rays are another common source of radiation exposure DID YOU KNOW? X-Ray Scanners *See Sample Problem 2.14 and Practice Problem 2.14 B. Controlling Exposure to Radiation 1. Background radiation exposure can’t be well controlled, but exposure from other sources can 2. Shielding is a common way to block radiation 3. Examine Figure 2.27 4. Paper can block alpha particles, a thin sheet of plastic or metal can block beta and positron particles, and a thick slab of concrete or lead is needed to block gamma rays C. Half-life 1. An important factor in controlling exposure is how long a radioisotope will remain hazardous 2. The half-life of a radioisotope is the time required for 1/2 of the atoms in a sample to decay 3. Examine Figure 2.28 4. Review examples in text (Radon-222 decay, Pultonium-239 decay, etc.) 5. See Table 2.10 for the half-life of some radioisotopes 2-6
6. Half-life also plays a role in determining whether a particular isotope is found in nature DID YOU KNOW? Half-life and Herbicides and Pesticides *See Sample Problem 2.15 and Practice Problem 2.15 D. Diagnosis and Therapy 1. Radioisotopes can be used to diagnose and treat disease 2. Radioisotope half-life is a major factor: too short and it decays before serving its purpose, too long and it will cause problems after its intended use 3. See Table 2.11 for some uses of radioisotopes in medicine 4. Positron Emission Tomography (PET) can be used to monitor glucose metabolism, oxygen usage, blood flow, and other biochemical processes 5. Review example in text (PET equation with Fluorine-18) 6. Examine Figure 2.30 7. Thyroid scan is another application where Iodine-131 emits radiation and a scanner records it 8. Examine Figure 2.29 9. Gamma knife uses gamma rays emitted by Cobalt-60 or Cesium-137 to kill cancerous cells 10. Examine Figure 2.31 *See Sample Problem 2.16 and Practice Problem 2.16 HEALTH LINK Radioisotopes for Sale (Examine Figure 2.32 and 2.33) HEALTH LINK CT and MRI Imaging (Examine Figure 2.34, 2.35, and 2.36)
Chapter Summary Section 2.1 The Greeks believed that all matter was made of indestructible building blocks, which they called atoms. In the 1800s, English scientist John Dalton described atoms based on measurements and experiments. In Dalton’s Atomic Theory, the atom is the basic unit from which matter is constructed. An atom is composed of three subatomic particles: the electron, the proton, and the neutron. Atomic mass units (amu) or daltons represent the mass of the subatomic particles. Atoms are made up of two regions: the nucleus, which is the site of most of the atom’s mass and it is made up of protons and neutrons, and the electron cloud, which is mostly empty space where electrons move. All three subatomic particles have different electrical charges. Protons are +1, neutrons are 0, and electrons are -1. Since opposite electrical charges attract while similar charges repel, a strong force holds the similar charges of protons (+) together in the nucleus. Section 2.2 Thousands of years ago, the Greeks described matter as being combinations of the basic elements: earth, air, fire, and water. Today, an element is a substance that contains only one type of atom. By the year 1800 only 31 elements had been discovered; however, today there are 117 (92 are natural). Elements names come from a variety of sources. Some are named after appearance, place of discovery, or a person. Atomic symbols also identify elements. They are one or two letter abbreviations for the name and some are based on alternate names. Section 2.3 Elements are the building blocks of matter, including the body. The USDA publishes a set of dietary guidelines to help people stay healthy. The current guidelines recommend a diet low in saturated and trans fats, cholesterol, salt, and sugar, but should include fruits, vegetables, whole grains, and low-fat milk. Healthy diets should also include proteins, carbohydrates, fats, and trace elements. Trace elements are elements needed by the body in small quantities. Oxygen, carbon, hydrogen, and nitrogen make up most of the mass of the human body. The NAS has created Recommended Dietary Allowances (RDA), Dietary Reference Intakes (DRI), Adequate Intake (AI), and Tolerable Upper Intake Level (UL) values. The FDA requires manufacturers to provide a nutrition facts label, which includes serving sizes and per percent daily amount are based on a 2,000 Cal diet. 2-7
Section 2.4 Atomic number is the number of protons an atom has in its nucleus, and mass number is the number of protons and neutrons in that atom’s nucleus. Atomic notion expresses the mass number as a superscript and the atomic number as a subscript. The mass number is not the mass of the atom; instead it is just a count of protons and neutrons. Atoms of a particular element always have the same number of protons; however, the number of neutrons can vary. Isotopes are atoms of an element that has different numbers of neutrons. Many elements have more than one isotope and those are usually not present in equal amounts. The number of electrons can be determined by the number of protons in an atom because a neutral atom (charge = 0) contains equal numbers of protons (+) and electrons (-). Section 2.5 The periodic table of the elements is a complete list of the elements arranged from smallest to largest atomic number. The table includes the atomic symbol, the atomic number, and the atomic weight (the average mass of the atoms of an element as it is found in nature). Elements can be grouped into metals, nonmetals, and semimetals. The heavy line on the periodic table separates the metals (left) from the nonmetals (right)—semimetals border the line. Groups are vertical columns of elements on the periodic table (identified by numbers 1-8 and letters A or B). Some chemical groups have been given names (families): alkali metals (1A), alkaline earth metals (2A), halogens (7A), and noble gases (8A). Groups 1A-8A are called the representative elements while groups 1B-8B are called the transition metals. The lanthanide elements (57-71) and the actinide elements (89-103) are at the bottom of the periodic table. Periods are horizontal rows of elements on the periodic table and relate to the periodic (regular) changes that take place in the element properties as atomic number is increased (first proposed by Dmitri Mendeleev in 1860s). Elements become more metallic as you move left and down on the periodic table. As you move right and up the periodic table, atomic size decreases. Section 2.6 Atoms are too small to deal with one by one, therefore they are best represented by the mole. One mole of something contains 6.02x1023 items, which is called Avogadro’s number. Avogadro’s number is equal to the number of atoms in exactly 12 grams of pure carbon-12. The molar mass of an element is the mass in grams of one mole of that element’s atoms. The molar mass is determined from the atomic weight of an element. The mole relationship can be summarized as a conversion factor. For example, 6.02x1023 atoms Li = 1 mole Li = 6.94 g Li. Section 2.7 In the Bohr model, electrons circle the nucleus in specific orbits, which correspond to different energy levels. The energy level closest to the nucleus is called the ground state and it is the most stable electron arrangement. If an atom absorbs energy, electron(s) are pushed to an excited state, which is an orbit farther from the nucleus. When an atom returns its electron(s) from an excited state back to a ground state, electromagnetic radiation is given off. It was determined that electrons can behave as energy waves rather than particles. Erwin Schrödinger devised a mathematical equation that described electrons in atomic orbitals (quantum mechanics approach). Atomic orbitals are 3D regions of space where there is a high probability of finding an electron. The maximum number of electrons per energy level is equal to 2n2, where n is the energy level. The valence shell is the highest numbered, occupied energy level. An element’s group number indicates how many valence electrons it has. The period of an element indicates what energy level its valance electrons are in. In electron dot structures, valence electrons are represented by dots around the atomic symbol. Section 2.8 A nuclear change involves the change of atomic nuclei. Some isotopes (radioactive isotopes or radioisotopes) have unstable nuclei that spontaneously disintegrate to become more stable. When this occurs, these radioisotopes release high-energy particles called nuclear radiation. The common forms 2-8
of nuclear radiation are alpha particle (42α), beta particle (0-1β), positron (01β+), and gamma ray (00γ). Nuclear change can be represented by a nuclear equation, which can then be balanced. Section 2.9 Nuclear radiation disrupts normal cellular functions, which can cause health problems. Commonly used radiation units such as the curie, the rad, and the rem are used to understand radiation exposure. The effects of short-term radiation exposure range from nausea to death, but the effects of long-term exposure at low levels are not well understood (background radiation). Shielding is a common way to block radiation (i.e. paper can block alpha particles, a thin sheet of plastic or metal can block beta and positron particles, and a thick slab of concrete or lead is needed to block gamma rays). Half-life, which is the time required for half of the atoms in a sample to decay, is another important property of nuclear radiation. Radioisotopes can also be used to diagnose and treat disease. Positron Emission Tomography (PET), thyroid scans, and gamma knifes are all medical applications of nuclear radiation.
Lecture Suggestions Sections 2.1 and 2.2 To introduce these sections, take a piece of paper and tear it in half. Then take one half and tear it in half again. Repeat this several times before posing the question: “assuming I had small enough fingers (forceps), could I tear this piece of paper in half infinite times or does there come a point where there is only one thing left?” Some students might say you could go on to infinity, but most will disagree and many might mention the atom. Then go through the brief history of the Greeks explaining how Democritus and Aristotle argued this same question thousands of years ago. Mention that the Greeks believed that all matter was made up of four elements, which is very similar to our concept of matter (everything is made up of a few elements). Sections 2.3 and 2.4 For these sections, begin by asking students how many possible words there are in the English language. Of course they won’t be able to say, but they’ll know the number is huge. Ask how many letters make up all those words. They will obviously know the answer is 26. Then ask how many physical things (matter) are there in the world. Again, they won’t be able to say, but they’ll know the number is huge. Then say 91 natural elements make up all matter. This leads into the fact that everything is made from elements and atoms, including us. Explain to students that they were not born with the mass and volume they have now, so ask them where all that matter (elements and atoms) in their bodies came from. Mention that all matter is made up of elements/atoms. Sections 2.5 and 2.6 To introduce these sections, begin by filling a small beaker with 18mL of water, and explain that this represents 1 mole of water. It is helpful to write Avogadro’s number out in decimal notation on the board, and then explain that this small amount of water (in the beaker) has this many things in it. To impress upon them how big the number is and how small atoms are, you can use the following scenario: ask if each water molecule was the size of a ping-pong ball, how much space would a mole take up? Most students might say something like the entire classroom while others might say something closer to the entire university. Explain that if you had a mole of ping-pong balls, it would be enough to cover the entire surface of the world. If there are that many things in the 18mL of water, what does that tell you about the size of atoms? This should help students understand how big Avogadro’s number is and how small atoms are. Sections 2.7, 2.8, and 2.9 For these sections, you can explain the difference between chemical and nuclear change by saying traditional chemistry is based on electrons and orbitals, where nuclear chemistry deals with protons/neutrons and the nucleus. It was once believed that an atom could not be divided, but that is in fact, not accurate (briefly discuss fission and the atomic bomb). Mention that alchemists obsessed over 2-9
how to turn one element into another, which was later believed to be impossible. To explain that this is not completely accurate, you can mention fusion in the sun and the creation of man-made elements. While discussing the atomic bomb, explain some of the biological effects of radiation.
Handouts for Students 1) For the Elements and Symbols handout (next page) have students memorize enough to be able to match these common elements with their symbols. This will be directly tied to their success in chemistry.
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Handout Elements and Symbols (Memorize) Nonmetals Hydrogen Helium Boron Carbon Nitrogen Oxygen Fluorine Neon Silicon Phosphorus
H He B C N O F Ne Si P
Sulfur Chlorine Argon Arsenic Selenium Bromine Krypton Tellurium Iodine Xenon
S Cl Ar As Se Br Kr Te I Xe
Metals Lithium Beryllium Sodium (Natrium) Magnesium Aluminum Potassium (Kalium) Calcium Titanium Chromium Manganese Iron (Ferrum) Cobalt Nickel Copper (Cuprum) Zinc Gallium Germanium
Li Be Na Mg Al K Ca Ti Cr Mn Fe Co Ni Cu Zn Ga Ge
Rubidium Rb Strontium Sr Zirconium Zr Silver (Argentium) Ag Tin (Stannum) Sn Antimony (Stibium) Sb Cesium Cs Barium Ba Tungsten (Wolfram) W Iridium Ir Platinum Pt Gold (Aurum) Au Mercury (Hydrargyrum) Hg Lead (Plumbum) Pb Bismuth Bi Uranium U Plutonium Pu 2-11
Chapter 3 Compounds Outline Notes I. Ions (Section 3.1) A. Ions Overview 1. Recall the number of protons and electrons are equal in a neutral atom (i.e. 31H contains 1 proton, 1 electron, and 2 neutrons) 2. An ion results when an atom or group of atoms gain or lose valence electrons a. Ions have an unequal number of protons and electrons b. Monoatomic ions are formed from a single atom (i.e. Li+ has 3 protons and 2 electrons) c. In atomic notation, the charge on an ion is shown as a subscript to the right of the symbol 4. Cations are ions that carry a positive charge due to a loss of electron(s) (i.e.Ca2+) a. Monoatomic cations formed from representative elements are given the same name as the original elements plus the word ion (i.e. Ca = calcium and Ca2+ = calcium ion) b. Examine Figure 3.1a c. Some transition metals are able to form several different cations (i.e. Fe2+ or Fe3+) i. These transition metal ions are named using a roman numeral in parentheses to indicate the charge of the ion [i.e. Fe2+ = iron(II) ion] ii. An alternative naming system uses the suffix ous for the smaller charged ion and ic for the larger charged ion (i.e. Fe2+ = ferrous ion and Fe3+ = ferric ion) d. See Table 3.1 for some transition metal ion names *See Sample Problem 3.1 and Practice Problem 3.1 5. Anions are ions that carry a negative charge due to a gain of electron(s) (i.e. F-) a. Monoatomic anions are named by changing the ending on the element name to ide and adding the word ion (i.e. F = fluorine and F- = fluoride ion) b. Examine Figure 3.1b 6. Polyatomic ions are built from two or more atoms (i.e. NH4+) a. Review example in text (NH4+ protons, electrons, and total charge) b. See Table 3.2 for common polyatomic ions c. Many polyatomic ions are found in significant amounts in all living things (i.e. HCO3- is involved in the transport of carbon dioxide from tissue to the lungs) II. The Octet Rule (Section 3.2) A. Octet Rule Overview 1. Electron arrangements allow us to understand why ions form a. Recall electron dot structures show the number of valence electrons an atom/element has b. For representative elements the number of valence electrons corresponds to the group 2. Examine Figure 3.2 3. Group 8A inert gases get their name because they resist change and don’t lose/gain electrons a. This is due in part to them having full valence shells 4. The octet rule states atoms gain, lose, or share valence electrons in order to end up with eight valence electrons B. Ion Formation using the Octet Rule 1. Nonmetals gain electron(s) to complete their octet causing them to become anions (i.e. the addition of one electron to a fluorine atom with 7 valence electrons allows it to reach an octet— this forms the negative fluoride anion F-) 2. Metals lose electron(s) to complete their octet causing them to become cations (i.e. the removal of one electron from a sodium atom with 1 valence electron allows it to drop back an energy level to reach an octet—this forms the positive sodium cation Na+) 3-1
3. Atoms rarely gain or lose more than 3 electrons (i.e. carbon with four valence electrons will reach its octet by other means) 4. See Table 3.3 for atoms and ions of representative elements 5. The octet rule and electron dot structures cannot always predict the charge on transition metal atoms/ions (i.e. Cr with 24 electrons would have to lose 6 to complete its octet) BIOCHEMISTRY LINK Ionophores and Biological Ion Transport (Examine Figure 3.3 and 3.4) *See Sample Problem 3.2 and Practice Problem 3.2 III. Ionic Compounds (Section 3.3) A. Pure Substances 1. Elements have only one kind of atom and are considered pure substances 2. Compounds are matter constructed of two or more chemically combined elements 3. Compounds are also pure substances (i.e. table salt, which contains sodium ions and chloride ions, is not just a haphazard mixture of two elements) 4. A particular compound always has the same proportions of the same elements 5. The identity of a compound is distinct from the identities of the elements that went into making it 6. Examine Figure 3.5 DID YOU KNOW? Salt Consumption B. Chemical Changes 1. When the atoms in an element or compound combine to make new compounds, a chemical change has taken place (i.e. iron reacts with oxygen to form the new compound rust) 2. Examine Figure 3.6 3. Chemical properties are the chemical changes that an element or a compound undergoes 4. Compounds can be either ionic or covalent depending on how the elements combine 5. Note that pure substances (elements and compounds) differ from mixtures a. In the compounds (pure substance) elements are chemically combined b. In mixtures pure substances are physically combined 6. Examine Figure 3.7 C. Ionic Bonds 1. Ionic compounds are constructed of cations and anions 2. The simplest ionic compounds are called binary compounds because they contain just two elements (metal cations bonded to nonmetal anions) 3. Opposite charges on ions hold the compound together in what is called an ionic bond 4. Ionic compounds, in the solid state, consist of crystal lattice, which is an array of alternating cations and anions 5. Examine Figure 3.8 6. In an ionic formula the cation is listed first (anion is listed second) and subscripts indicate the ratio of ions (i.e. MgCl2 has one Mg2+ ions and two Cl- ions) 7. Review examples in text (Writing the formula for lithium sulfide and iron I/II sulfide) *See Sample Problem 3.3 and Practice Problem 3.3 D. Ionic Compounds Containing Polyatomic Ions 1. Polyatomic ions also form ionic compounds (i.e. the food preservative sodium sulfite) 2. When there is more than one polyatomic ion, the formula is surrounded by parentheses and the subscript follows [i.e. the laxative Mg(OH)2 and the expectorant (NH4)2CO3] 3. Polyatomic ions “act as one” in ionic compounds (i.e. NaNO2 consists of Na+ and NO2-) E. Naming Ionic Compounds 1. The cation name is always placed before the anion name (i.e. LiBr is lithium bromide) 2. Subscripts (the number of times an ion appears) are not specified in the name (i.e. BaCl2 is barium chloride not barium dichloride) 3. It is assumed that the formula can be determined from the name because charges on ions are known and they must cancel each other out making the compound neutral 4. Transition metal compounds use roman numerals [i.e. CuCl2 is copper (II) chloride] 3-2
5. Examine Figure3.9 6. Ionic compounds are widely used in medicine, industries, and households 7. See Table 3.5 for the uses of some ionic compounds *See Sample Problem 3.4 and Practice Problem 3.4 DID YOU KNOW? Statue of Liberty HEALTH LINK Pass the Salt (Examine Figures 3.10) IV. Covalent Bonds (Section 3.4) A. Covalent Bonding Overview 1. Nonmetals can share valence electrons to attain an octet (i.e. two F atoms with 7 VE each) 2. A covalent bond is a shared pair of electrons 3. The number of covalent bonds that a nonmetal atom forms is the same as the number of electrons that it needs to reach an octet (i.e. F forms 1, O forms 2, N forms 3, C forms 4, and H forms 1) 4. Examine Figure 3.11 B. Line-Bond Structures 1. In the line-bond method each pair of shared bonding electrons is represented by a line 2. The line-bond structures (also called Lewis structures) are used instead of electron dot structures 3. Nonbonding electrons are simply electrons not involved in bonds C. Multiple Bonds 1. In a single bond one pair of electrons are shared (i.e. H2 has one single bond) 2. In a double bond two pairs of electrons are shared (i.e. O2 has one double bond) 3. In a triple bond three pairs of electrons are shared (i.e. N2 has one triple bond) 4. Examine Figure 3.12 *See Sample Problem 3.5 and Practice Problem 3.5 *See Sample Problem 3.6 and Practice Problem 3.6 V. Molecules (Section 3.5) A. Molecules Overview 1. Molecules are uncharged groups of atoms connected by covalent bonds 2. Molecular formulas list the number of each type of atom that is present in a molecule (i.e. H2O) 3. Most molecules are compounds because they contain atoms of different elements (i.e. NH3) 4. Seven elements appear as diatomic (two atoms) molecules—H2, N2, O2, F2, Cl2, Br2, I2 B. Naming Binary Molecules 1. Binary molecules contain just two different elements (i.e. CO2) 2. Elements are listed in order of appearance in the molecular formula 3. The ending of the second element’s name is changed to ide (i.e. HCl is hydrogen chloride) 4. Binary molecules can sometimes combine in several different ways (i.e. SO2 and SO3) *Recall ionic compounds always combine in fixed ratios to form neutral compounds (i.e. CaCl2) 5. Prefixes are added to distinguish molecules a. See Table 3.5 for prefixes 1-10 b. First names should not begin with “mono” (i.e. NO2 is nitrogen dioxide not mononitrogen) c. Usually when two vowels are placed together the “a” or “o” on the prefix is dropped (i.e. NO is nitrogen monoxide not nitrogen monooxide) d. Sometimes binary molecules are known better by other names (i.e. H2O is water not dihydrogen monoxide and H2S is hydrogen sulfide not dihydrogen sulfide) *See Sample Problem 3.7 and Practice Problem 3.7 HEALTH LINK Dental Fillings (Review Figure 3.13) VI. Formula Weight, Molecular Weight, and Molar Mass (Section 3.6) A. Formula Weight is the sum of the atomic weights of the elements in a formula 1. Formula weights are most commonly used for ionic compounds (i.e. NaCl is 58.44 amu) 2. Review examples in text [NaCl & Cu(NO3)2 formula weight calculations] 3. Molar mass (the mass in grams of one mole) is numerically equivalent to formula weight [i.e. Cu(NO3)2 has a formula weight of 187.52 amu so its molar mass is 187.52 g/mol] 3-3
4. Review example in text (AgNO3 mole calculation) *See Sample Problem 3.8 and Practice Problem 3.8 B. Molecular Weight is the sum of the atomic weights of the elements in molecular formula 1. Molecular weights are used to form covalent (molecular) compounds (i.e. H2O is 18.02 amu) 2. Review examples in text (H2O & SO3 molecular weight calculations) 3. Molar mass is numerically equivalent to molecular weight (i.e. SO3 has a molecular weight of 80.07 amu so its molar mass is 80.07 g/mol) 4. Review example in text (SO3 mole calculation) *See Sample Problem 3.9 and Practice Problem 3.9 HEALTH LINK Nitric Oxide (Review Figure 3.14)
Chapter Summary Section 3.1 An ion results when an atom or group of atoms gain or lose electrons; therefore, ions have an unequal number of protons and electrons. Monoatomic ions are formed from a single atom while polyatomic ions are built from two or more atoms. Positively charged ions (called cations) are formed from a loss of electrons; and negatively charged ions (called anions) are due to a gain of electrons. Monoatomic cations are given the same name as the original elements plus the word ion. Some transition metals are able to form several different cations, so roman numerals are used to distinguish them. Monoatomic anions are named by changing the ending on the element name to ide and adding the word ion. Polyatomic ions have specific names and many are found in living things. Section 3.2 Electron arrangements allow us to understand why ions form. The octet rule states atoms gain, lose, or share valence electrons in order to end up with eight. Nonmetals gain electron(s) to complete their octet causing them to become anions; and metals lose electron(s) to complete their octet causing them to become cations. Atoms rarely gain or lose more than 3 electrons and the octet rule and electron dot structures cannot always predict the charge on transition metal. Section 3.3 Elements and compounds (two or more chemically combined elements) are considered pure substances. A particular compound always has the same proportions of the same elements, and its identity is distinct from the identities of the elements that went into making it. When the atoms in an element or compound combine to make new compounds, a chemical change has taken place. Compounds can be either ionic or covalent depending on how the elements combine. Ionic compounds are constructed of cations and anions and the simplest ionic compounds are called binary compounds. Ionic compounds are held together with ionic bonds in crystal lattice structure. When naming ionic compounds, the word ion is dropped and the cation is written first, the anion second. Section 3.4 Nonmetals can share valence electrons in a covalent bond to attain an octet. The number of covalent bonds that a nonmetal atom forms is the same as the number of electrons that it needs to reach an octet. In the line-bond method, each pair of shared bonding electrons is represented by a line. Nonbonding electrons are simply electrons not involved in bonds. Some elements can form single bonds (one shared electron pair), double bonds (two shared electron pairs), and even triple bonds (three shared electron pairs). Section 3.5 Molecules are uncharged groups of atoms connected by covalent bonds. Molecules are described using molecular formulas, which list the number of each type of atom that is present in a molecule. Most molecules are compounds because they contain atoms of different elements. Seven elements appear as 3-4
diatomic (two atoms) molecules. Binary molecules contain just two different elements. When naming binary molecules, the elements are listed in order of appearance in the molecular formula. The ending of the second element’s name is changed to ide and prefixes are used to indicate the number of each element. Section 3.6 The formula weight of a compound is the sum of the atomic weights of the elements in its formula. Formula weights are most commonly used for ionic compounds. Likewise, molecular weight is the sum of the atomic weights of the elements in a molecular formula. Molecular weights are used for covalent (molecular) compounds. Molar mass is numerically equivalent to formula weight or molecular weight.
Lecture Suggestions Sections 3.1 and 3.2 When lecturing on these sections, compare the periodic table to the social order at a high school. You can say that the noble gases are the most popular kids that everyone wants to be like. Mention that these kids (elements) are stuck-up and they don’t care to interact (react) with the other kids (elements). Then mention that students (elements) will change their appearance on the outside (their valence shell/electrons) so they can appear like the popular kids (noble gases). While these kids (elements) may change their appearance (valence shell), they are still the same person on the inside (they still have the same number of protons). Explain that this is how ions are formed. While the analogy is juvenile, the students will like it. Sections 3.3 and 3.4 Compare ionic and covalent compound formation and nomenclature to two people (two elements) getting married. The man’s name is always written first and his name (first element’s name) doesn’t change (sodium stays sodium). The woman’s name is always written second (second element’s name), and once she gets married, her last name (ending) changes (chlorine becomes chloride). You can also compare some of the transition metals to twins: while they are closely related, they are actually different; therefore, they must have roman numerals to differentiate them. This analogy will help students remember which element gets changed to ide, and in ionic compound, the metal name never changes. Sections 3.5 and 3.6 To introduce these sections, start by asking how much a dozen weighs. Some students might ask: “a dozen of what?” Say a dozen oranges. They should then ask how much one weighs. Explain that the mole is just like the dozen—it is a counting number, not a weight. Then ask the students how much a mole weighs. They should catch on and ask “a mole of what?” Explain how the periodic table can be used to calculate the weight in grams of a mole of any substance (assuming you know its formula).
Handouts for Students 1) For the Ions handout (next page) teach students how to use the periodic table to get the ionic states for the first two sections (Metal Cations and Nonmetal Anions). Have students memorize the last two sections (Transition Metal and Polyatomic Ions). Tell students that they are responsible for the whole sheet, but if they know how to use the periodic table then they should only have to memorize the bottom half. Recommend that students make flash cards to help them study/learn these.
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Handout Ions (Memorize) Main Group Metals (Cations) Group 1: Positive 1 (1+) Hydrogen Ion H+ (NM) Lithium Ion Li+ Sodium Ion Na+ Potassium Ion K+ Rubidium Ion Rb+ Cesium Ion Cs+ Francium Ion Fr+
Group 2: Positive 2 (2+) Beryllium Ion Be2+ Magnesium Mg2+ Calcium Ion Ca2+ Strontium Ion Sr2+ Barium Ion Ba2+ Radium Ion Ra2+
Group 3: Positive 3 (3+) Aluminum Ion Al3+ Gallium Ion Ga3+
Main Group Nonmetals (Anions) Group 5: Negative 3 (3–) Nitride Ion N3– Phosphide Ion P3–
Group 6: Negative 2 (2–) Oxide Ion O2– Sulfide Ion S2– Selenide Ion Se2–
Group 7: Negative 1 (1–) Fluoride Ion F– Chloride Ion Cl– Bromide Ion Br – Iodide Ion I–
Transition Metals and Sn/Pb (Cations) +
Copper (I) Ion Copper (II) Ion Iron (II) Ion Iron (III) Ion
Cu Cu2+ Fe2+ Fe3+
Ammonium Ion Hydronium Ion Hydroxide Ion Carbonate Ion Nitrite Ion Nitrate Ion
+
Lead (II) Ion Lead (IV) Ion Tin (II) Ion Tin (IV) Ion
Pb2+ Pb4+ Sn2+ Sn4+
Zinc Ion Zn2+ Silver Ion Ag+ Chromium (II) Ion Cr2+ Chromium (III) Ion Cr3+
Polyatomic Ions NH4 H3O+ OH– CO32– NO2– NO3–
Sulfite Ion Sulfate Ion Cyanide Ion Acetate Ion Chromate Ion Dichromate Ion
SO32– SO42– CN– C2H3O2– CrO42– Cr2O72–
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Permanganate Ion MnO4– Hydrogen Carbonate HCO3– Hydrogen Sulfate Ion HSO4– Phosphate Ion PO43– Hydrogen Phosphate Ion HPO42– Dihydrogen Phosphate Ion H2PO4–
Chapter 4 An Introduction to Organic Compounds Outline Notes I. Structural Formulas (Section 4.1) A. Structural Formula Overview 1. Molecular formulas tell nothing about how atoms are attached to each other (C3H8O) 2. Electron dot structures show all valence electrons in a structure 3. Line-Bond structures are basically electron dot structures with the covalent bonds (shared electrons) shown with lines rather than dots 4. Nonmetals that typically covalently bond a. H forms 1 bond b. C forms 4 bonds c. N (&P) forms 3 bonds d. O (&S) forms 2 bonds e. F (& halogens) forms 1 bond 5. Examine Figure 4.1 B. Drawing Line-Bond Structures 1. Count total number of valence electrons 2. Use single bonds to connect atoms to one another 3. Add remaining electrons as lone pairs to outer atoms and then move to central atom 4. If needed, complete all atoms’ octets by moving electrons to form double or triple bonds *See Sample Problem 4.1 and Practice Problem 4.1 5. For polyatomic ions, add electrons for each negative charge and subtract electrons for each positive charge. Place the line-bond structure in brackets with the charge written outside as superscript. *See Sample Problem 4.2 and Practice Problem 4.2 C. Condensed Structural Formulas and Skeletal Structures 1. Condensed structural formulas describe the attachment of atoms to one another without showing all the bonds (CH3CH2COOH) 2. In skeletal structures covalent bonds are represented by lines, C atoms are represented by line connections (i.e. angles) and line ends, and H atoms (not shown) are understood to complete each carbon atom’s octet (4 covalent bonds around C) *See Sample Problem 4.3 and Practice Problem 4.3 3. Examine Figure 4.2 II. Polar Covalent Bonds, Shape, and Polarity (Section 4.2) A. Electronegativity and Polar Bonds 1. Electronegativity is the ability of an atom to attract bonding electrons a. Periodic trend: up a group and right in a period increases electronegativity (i.e. N > P; F > O) b. F is the most electronegative (all elements are compared to F) *Examine Figure 4.3 2. Polar covalent bonds result from elements with different electronegativites bonding a. H-F is a polar covalent bond (F is 1.9 more electronegative than H) b. More electronegative element pulls the electrons closer to itself creating a partial negative charge (δ-) c. Less electronegative has electrons pulled away from it creating a partial positive charge (δ+) d. Examine Figure 4.4 3. Table 4.1 Bond Types: Nonpolar Covalent, Polar Covalent, Ionic *See Sample Problem 4.4 and Practice Problem 4.4 4-1
B. Shapes of Molecules 1. Line-bond structures are not intended to represent the actual 3D shape of a molecule 2. Electrons (because they are negative) want to be as far apart as possible 3. Examine Figure 4.5 DID YOU KNOW? Some organic chemistry history 4. Table 4.2 Common Molecular Shapes a. Tetrahedral: 4 bonding groups b. Pyramidal: 3 bonding groups and 1 lone pair of electrons c. Trigonal Planar: 3 bonding groups d. Bent: 2 bonding groups and 1 lone pair of electrons or two bonding groups and two lone pairs of electrons e. Linear: 2 bonding groups 5. Complex Shapes: Examine Figure 4.6 C. Polarity 1. Molecules as a whole can be either polar or nonpolar 2. If a molecule has no polar covalent bonds then it is nonpolar (F2) 3. If a molecule has only C and H covalent bonds then it is nonpolar (CH4) 4. If a molecule has polar bonds but they cancel out then it is nonpolar (CO2) 5. If a molecule has polar bonds and they do not cancel out then it is polar (CH2Cl2) 6. Examine Figure 4.8 DID YOU KNOW? Foldit HEALTH LINK Prion diseases (Examine Figure 4.7) *See Sample Problem 4.5 and Practice Problem 4.5 III. Noncovalent Interactions (Section 4.3) A. Noncovalent Interactions Overview 1. Molecular forces such as polarity allow us to determine the properties of molecules (i.e. why some vitamins are soluble in water and others are not) 2. Noncovalent interactions do not involve the sharing of electrons 3. These interactions can be due to a permanent or temporary charge B. Noncovalent Interactions due to Permanent Charge 1. Hydrogen bonds result from the interactions of N, O, or F with an H atom that is covalently bonded to a different N, O, or F (δ+ H is attracted to δ- N, O, or F in another molecule) 2. Dipole-Dipole forces result from the attraction between oppositely charged ends of neighboring polar molecules 3. Examine Figure 4.9 C. Noncovalent Interactions due to Temporary Charge 1. London forces are when nonpolar molecules temporarily attract each other because of a dynamic temporary induced dipole caused by the continuous motion of electrons 2. The bigger the molecule (more surface area) the greater the London forces 3. Examine Figure 4.10 *See Sample Problem 4.6 and Practice Problem 4.6 IV. Families of Organic Compounds (Section 4.4) A. Organic Chemistry Overview 1. Recall the chemistry branch that is based on carbon compounds is called organic chemistry 2. Carbon has its own branch because it is present in more than 10 million compounds, which is many more than inorganic chemistry (compounds without carbon) 3. Organic families are determined by functional groups (atoms that give rise to chemical properties) B. Hydrocarbons 1. Hydrocarbons contain only carbon and hydrogen atoms 2. Alkanes have only single-bonded carbon atoms 3. Alkenes have carbon-carbon double bond(s) 4-2
4. Alkynes have carbon-carbon triple bond(s) 5. Aromatic compounds contain a ring of six carbon atoms with alternating double and single bonds; this bonding arrangement gives aromatics a special stability 6. Examine Figure 4.11 7. Hydrocarbons come in all sizes and all of them are nonpolar (attracted only by London Forces) 8. Examine Figure 4.12 *See Sample Problem 4.7 and Practice Problem 4.7 BIOCHEMISTRY LINK Ethylene, a plant hormone C. Alcohol, Phenols, and Ethers 1. Heteroatoms are any atom other than C or H (all organic families except hydrocarbons) 2. Alcohols have an -OH group attached to an alkane-like C atom 3. Phenols have an -OH group attached to an aromatic C atom 4. Ethers have a C-O-C linkage D. Thiols, Sulfides, and Disulfides 1. Thiols have an -SH group attached to a C atom 2. Sulfides have a C-S-C linkage 3. Disulfides have a C-S-S-C linkage 4. Examine Figure 4.14 E. Amines 1. Amines have an N atom directly attached to one or more C atoms 2. Amines are found in many biochemical molecules (amino acids) 3. N, O, and S atoms have lone pair(s) of electrons, which cause bent shapes DID YOU KNOW? Origin of organic family names F. Alkyl Halides 1. Alkyl halides have a halogen atom (F, Cl, Br, I) attached to an alkane-like C atom 2. Alcohols, phenols, ethers, amines, and some alkyl halides (F and Cl) have polar bonds G. Ketones and Aldehydes 1. A carbonyl group is an O atom double bonded to a C atom (C=O) 2. Ketones have the carbonyl C atom attached to two other C atoms 3. Aldehydes have the carbonyl C atom attached to one or two H atoms 4. Examine Figure 4.15 H. Carboxylic Acids, Esters, and Amides 1. Carboxylic acids contain a carboxyl group which is an -OH attached to a carbonyl C atom (-CO2H or COOH); the carbonyl C atom is also attached to an H or C atom 2. Esters contain the C-O-C linkage like ethers but one of the C belongs to a carbonyl 3. Amides contain an N atom like amines but the N is also attached to a carbonyl C atom 4. Examine Figure 4.16 5. Table 4.3 Important families of organic compounds HEALTH LINK Sunscreens *See Sample Problem 4.8 and Practice Problem 4.8
Chapter Summary Section 4.1 Structural formulas, such as electron-dot structures and line-bond structures, are more descriptive than molecular formulas because they show how the atoms are attached to each other. Both electron-dot and line-bond structures show how the valence electrons of nonmetals are shared to form covalent bonds. To draw structural formals, first count the total valence electrons, then connect the atoms with covalent bonds (two shared electrons). Then, add electrons and/or form double/triple bonds to complete each element’s octet. Condensed formulas (CH3CH2COOH) are abbreviated versions of line-bond structures 4-3
that do not show every bond. Skeletal structures are even more abbreviated in that carbon is represented by line-ends and line-connections and hydrogens are assumed to satisfy each carbon atom’s octet. Section 4.2 Electronegativity is the measure of an atom’s ability to attract bonding electrons. On the periodic table, electronegativity increases up each group and right across each period (F had the greatest). Differences in electronegativity determine if a bond is ionic, polar covalent, or nonpolar covalent (Table 4.1). In polar covalent bonds, the more electronegative element has a δ- charge while the other element has a δ+ charge. In molecular compounds the molecules have distinct shapes. The five basic shapes are tetrahedral, pyramidal, bent, trigonal planar, and linear. The shape of the molecule along with the polarity of each covalent bond determines if a molecule is polar or nonpolar. Section 4.3 Molecular forces (i.e. polarity) can explain many of the properties of molecular compounds (i.e. solubility). Molecular forces are caused by permanent or temporary noncovalent interactions (ones that do not involve the sharing of electrons). Hydrogen bonds are the strongest of these forces and are formed when an H (δ+) that is attached to a N, O, or F is attracted to a different N, O, or F (δ-). DipoleDipole forces arise when polar molecules attract. London forces are the weakest of these forces and are caused by temporary dipole (due to electrons moving); however, London forces are the strongest noncovalent interaction observed in nonpolar molecules. London forces are more significant in larger molecules due to more surface area. Section 4.4 Organic chemistry is based on carbon compounds (there are many more organic/carbon compounds than inorganic/non-carbon compounds). Organic chemistry is organized into families (Table 4.3) based on functional groups (atoms that give rise to chemical properties). Hydrocarbons contain only C and H atoms. The four families of hydrocarbons are alkanes (single-bonded C atoms), alkenes (double-bonded C atoms), alkynes (triple-bonded C atoms), and aromatics (ring of six carbon atoms with alternating double and single bonds). Other families are considered heteroatomic because they have additional elements (N, O, etc.) attached to the C atoms. Alcohols have an -OH group attached to an alkane-like C atom. Phenols have an -OH group attached to an aromatic C atom. Ethers have a C-O-C linkage. Amines have an N atom directly attached to one or more C atoms. Alkyl halides have a halogen atom (F, Cl, Br, I) attached to an alkane-like C atom. Ketones have a carbonyl group (C=O) attached to two other C atoms where aldehydes have the carbonyl attached to one or two H atoms. Carboxylic acids contain a carboxyl group (-CO2H or COOH). Esters contain the C-O-C linkage like ethers, but one of the Cs belongs to a carbonyl. Amides contain an N atom like amines, but the N is also attached to carbonyl.
Lecture Suggestions Section 4.1 To introduce this section, ask students to describe what makes each person in the room distinguishable from everyone else. Of course they’ll say “looks” are what they use to tell each other apart. That is correct, and in this chapter, we will distinguish compounds by their structure or “looks.” Then, tell students a molecular formula is very generic because it is like saying a student is a male or a female—it only tells us what type of compound (or person) it is. A structural formula, on the other hand, tells us exactly how each molecule looks (persons’ hair color, height, etc.). When drawing line-bond formulas it is helpful to have students draw out the electron-dot structure for each individual element. This allows them to easily count the valence electrons, and it also allows them to see how many bonds each element usually makes. Tell them to memorize that if carbon is present, then it is the center atom and hydrogen and halogens are never the center atom because they only form one bond. 4-4
Section 4.2 To introduce polarity, ask for a male (usually larger and stronger) and a female volunteer. Have them hold hands to represent a covalent bond (each student’s arm represents one electron). Then, tell them that each person or atom wants their hands as close to their bodies as possible. Of course the stronger male pulls his hand closer to his body, and even though the stronger person/atom pulls the arm/electrons closer, the weaker student still hangs on; therefore it is not like an ionic bond where one person/atom completely takes the arm/electron, but instead, they are just sharing them unequally. Explain that this is polarity because the male is partially negative while the female is partially positive. For whole molecule polarity, use the example of a Fraternity guy at a party pulling on girls, or girls pushing him away. If four girls are all pulling away from one guy in opposite directions like in CF4, then all the forces will cancel out and he won’t be able get close to any of their arms/hands (nonpolar are when forces cancel out). This may seem kind of silly, but students will find it entertaining and relatable. Section 4.3 To introduce noncovalent interactions, ask students what is easier to throw in the air: a baseball (CH4) or a bowling ball (C10H22)— they should reply the baseball because it is lighter. Then ask them which is easier to throw in the air: a little-league baseball (CH4) or a major-league baseball (one H2O)—and they should reply that it is about the same because they weigh about the same. Ask what would happen if you super glued a thousand major-league baseballs (many H2O hydrogen bonding) together—obviously they may now say the major-league baseballs. Explain that this is why water is liquid at room temperature while methane is a gas. You can tell them that hydrogen bonds are like super glue holding molecules together where dipole-dipole forces are like Elmer’s glue, and London forces are like chewing gum— they don’t hold as long. Section 4.4 When introducing organic chemistry, point to carbon on the periodic table and ask why this element is so special that it gets its own branch of chemistry while all the other combined elements get another branch called inorganic? There are about 10 million known organic compounds, but only approximately 300,000 known inorganic compounds. Carbon is special because it is the only true nonmetal in group 4A and it can form four covalent bonds with other elements including itself. During lecture, explain that organic chemistry is organized into families that are based on functional groups. You can tell students that if they want to master any organic concept, they must first learn the functional groups and their names. Compare functional groups/families to letters in the alphabet—you can’t read/write a word (compound) let alone a sentence (equation/reaction) without first learning the letters (functional groups/families). Encourage students to memorize (possibly with flash cards) the first and third column of Table 4.3—this foundation can make or break a student for the rest of the semester.
Handouts for Students 1) For the Molecular Geometry handout (next page) have students draw molecules in the squares as you work them out on the board/screen 2) For the Families and Functional Groups handout tell students they should memorize the first and third columns
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Handout Molecular Geometry *If there are only two atoms in the compound bonded together then it linear (H-F) Number of atoms (not bonds) or lone pairs of electrons (L.P.) around the central atom
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3
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All Atoms Linear
All Atoms Trigonal Planar
All Atoms Tetrahedral
2 Atoms & 1 L.P. Bent
3 Atoms & 1 L.P. Pyramidal
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Handout
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Chapter 5 Reactions Outline Notes I. Chemical Equations (Section 5.1) A. Reactions and Equations Overview 1. In chemical reactions covalent and ionic bonds are broken and new bonds are formed (i.e. the atoms in N2 and H2 can be rearranged to form NH3) 2. Examine Figure 5.1 3. The reaction between nitrogen and hydrogen is very important for living things a. Nitrogen is present proteins and DNA b. For organisms to use nitrogen, N2 must be “fixed” into compounds like NH3 c. Most fixed nitrogen comes from microorganisms converting atmospheric N2 d. Chemist Fritz Haber developed a process to manufacture NH3 from N2 (Haber process) e. The process is useful in making fertilizers, pharmaceuticals, explosives, and cleaners 4. Chemical equations describe chemical reactions a. An arrow separates reactants (the elements or compounds present at the start) from the products (the new substances formed) b. The numbers placed in front of the formulas are called coefficients i. They tell how much of a product or reactant is present ii. The coefficient 1 is never written c. The physical state of reactants/products can be indicated: s for solid, l for liquid, g for gas, and aq for aqueous (dissolved in water) DID YOU KNOW? Lightning Fixation *See Sample Problem 5.1 and Practice Problem 5.1 B. Balancing Chemical Equations 1. Matter (atoms) are not destroyed or created during a chemical reaction 2. Therefore, a chemical equation must be balanced (has the same number of atoms of each element on both sides of the reaction arrow) 3. Review example in text (How to make a cheese sandwich) 4. Balanced chemical equations make sure that there is enough of each reactant for the reaction 5. Review example in text (Nitrogen gas reacts with oxygen gas to produce nitrous oxide) *See Sample Problem 5.2 and Practice Problem 5.2 C. Avoiding Common Balancing Errors 1. Do not change the formula of a reactant or product (don’t change the subscripts) 2. Do not add new reactant or products 3. Do not use multiples of the coefficients when writing the balanced equation 4. Review examples in text (Common errors made balancing N2 + O2 → N2O) *See Sample Problem 5.3 and Practice Problem 5.3 II. Reaction Types (Section 5.2) A. Reaction Types Overview 1. The number of possible chemical reactions is immense 2. One classification system uses four broad reaction types (below) B. Synthesis 1. Two or more elements or compounds combine to form one more complex compound 2. The general reaction is A + B → AB (i.e. CO2 + H2O → H2CO3) C. Decomposition 1. The breakdown of one compound to form elements or simpler compounds (reverse of synthesis) 2. The general reaction is AB → A + B (i.e. 2H2O → 2H2 + O2) 5-1
D. Single Replacement 1. One element trades places with a different element in a compound 2. The general reaction is A + BC → AC + B (i.e. Fe + CuSO4 → FeSO4 + Cu) E. Double Replacement 1. Parts of two compounds switching places 2. The general reaction is AB + CD → AD + CB (i.e. NaCl + AgNO3 → NaNO3 + AgCl) *See Sample Problem 5.4 and Practice Problem 5.4 III. Reactions Involving Water (Section 5.3) A. Water Reactions Overview 1. Water is a reactant or product in many reactions important to organic and biochemistry 2. Alkenes, alcohols, carboxylic acids, and esters are all involved in water reactions 3. Recall alkenes contain a double bond (C=C), alcohols contain a hydroxyl (-OH), carboxylic acids contain a carboxyl (—CO2H), which is the combination a hydroxyl and a carbonyl (C=O), and esters contain a contain a C—O—C bond with a carbonyl 4. Examine Figure 5.2 DID YOU KNOW? Local Anesthetic Procaine B. Hydrolysis 1. Hydrolysis is a double replacement reaction where water (hydro) is used to split (lyse) a molecule 2. Esters undergo hydrolysis when treated water and an acid catalyst (H+) a. Ester hydrolysis yields a carboxylic acid and an alcohol b. A catalyst is a substance that speeds up a reaction without itself being altered 3. Examine Figure 5.3 4. Hydrolysis is a factor in the length of time that some drugs remain active a. Local anesthetics procaine (Novocain) and chloroprocaine are short acting (about 1 hour) because they are hydrolyzed b. A particular enzyme (biological catalyst) deactivates these anesthetics by catalyzing hydrolysis of their ester groups to form carboxylic acids and alcohols 5. Examine Figure 5.4 C. Hydration 1. Hydration is a synthesis reaction where water is added to a double bond 2. An alkene can by hydrated with water and an acid catalyst to produce an alcohol 3. Examine Figure 5.5 D. Dehydration 1. Dehydration is a decomposition reaction where an alcohol is heated with an acid catalyst to form an alkene and water (reverse of hydration) 2. Examine Figure 5.6 3. Dehydration and hydration play important roles in biochemistry (i.e. citric acid conversion) 4. Examine Figure 5.7 *See Sample Problem 5.5 and Practice Problem 5.5 IV. Oxidation and Reduction (Section 5.4) A. Oxidation and Reduction Overview 1. Reactions can be grouped as oxidation and reduction (different classification systems) 2. When atoms lose electron(s) they are oxidized (i.e. Na → Na+) 3. When atoms gain electron(s) they are reduced (i.e. Cl → Cl-) 4. The oxidizing agent removes electron(s) (i.e. Cl2) 5. The reducing agent gives electron(s) (i.e. Na) 6. Review examples in text (i.e. Copper and silver nitrate single replacement reaction) 7. Examine Figure 5.8 8. Oxidation and reduction always occur together because electrons move from the substance being oxidized to the one being reduced *See Sample Problem 5.6 and Practice Problem 5.6 5-2
B. Combustion 1. Combustion is a type of oxidation reaction where O2 is always a reactant and heat is released 2. When organic molecules are combusted, carbon dioxide and water are produced 3. Examine Figure 5.9 4. Review example in text (Combustion of methane) 5. When balancing combustion reactions, it is helpful to balance carbon atoms first, hydrogen atoms second, and oxygen atoms last 6. In the combustion of methane, C is oxidized because it gains a partial positive charge (δ+) a. Methane has nonpolar bonds where carbon dioxide has polar bonds b. In CO2, C carries a δ+ and O carries a δ- due to differences in electronegativity 7. Examine Figure 5.10 8. Instead of analyzing each covalent bond in organic reactions, use the following simple rules a. Atoms are oxidized if they gain oxygen or lose hydrogen b. Atoms are reduced if they lose oxygen or gain hydrogen 9. In the combustion of methane, C is oxidized because it loses H and gains oxygen in CO2 *See Sample Problem 5.7 and Practice Problem 5.7 C. Oxidizing Alcohols and Aldehydes 1. Some alcohols can be oxidized to form aldehydes, other alcohols can be oxidized to form ketones 2. Aldehydes can be oxidized to form carboxylic acids 3. Review reactions in text (Alcohol and aldehyde oxidation) D. Hydrogenation 1. Hydrogenation is a synthesis reaction that alkenes and other unsaturated hydrocarbons undergo 2. Catalytic hydrogenation uses H2 as a reducing agent and Pt as a catalyst 3. Review reaction in text (Unsaturated hydrocarbon is hydrogenated) 4. Aldehydes and ketones can also be hydrogenated 5. Review reactions in text (Aldehyde and ketone hydrogenation) 6. Hydrogenation takes place in living things with biochemical compounds (i.e. fatty acids) 7. Review reaction in text (Enzyme-catalyzed reaction with NADPH) HEALTH LINK Antiseptics and Oxidation (Examine Figure 5.11) *See Sample Problem 5.8 and Practice Problem 5.8 V. Mole and Mass Relationships (Section 5.5) A. Mole and Mass Relationships Overview 1. Balanced equations allow us to predict which reactant will run out first 2. They also allow us to predict how much product can be expected to form B. Mole Example Problems 1. Review example in text (Cheese sandwich making) 2. Review example in text (Moles of CO2 calculation) *See Sample Problem 5.9 and Practice Problem 5.9 C. Moles and Grams Example Problems 1. Review example in text (Cheese calculation in ounces) 2. Review examples in text (Grams of KO2 and grams of Ag calculations) *See Sample Problem 5.10 and Practice Problem 5.10 *See Sample Problem 5.11 and Practice Problem 5.11 3. Note that a calculation that involves a balanced equation must use the number of moles of reactant and products, not their mass 4. Examine Figure 5.12 VI. Calculating the Yield of a Reaction (Section 5.6) A. Limiting Reactant 1. The limiting reactant determines (limits) the amount of product that can be formed in a reaction 2. Examine Figure 5.13 3. The limiting reactant is the reactant that gets totally consumed in the reaction 5-3
4. Review example in text (Limiting reactant determination of Haber reaction) *See Sample Problem 5.12 and Practice Problem 5.12 B. Theoretical Yield 1. The reaction stops when the limiting reactant has been used up (even though other non-limiting reactants still remain) 2. The theoretical yield (the maximum amount of product that can be obtained) is determined by the amount of the limiting reactant that is initially present 3. Review example in text using the following steps (Propane combustion theoretical yield of CO2) a. Determine the limiting reactant b. Determine the theoretical yield of CO2 C. Percent Yield 1. The theoretical yield of a product is rarely obtained due to various reasons: a. The reactants might not completely react b. An unexpected or unwanted product may be produced c. The product is difficult to separate from the reaction mixture 2. The actual yield is the amount of product obtained from a reaction 3. The percent yield indicates how well the actual yield agrees with the theoretical yield; this can be calculated by the equation: percent yield = (actual yield)/(theoretical yield) x 100 4. Review example in text (Percent yield calculation) *See Sample Problem 5.13 and Practice Problem 5.13 VII. Free Energy and Reaction Rate (Section 5.7) A. Spontaneous and Nonspontaneous Reactions 1. A spontaneous reaction continues by itself once it has started 2. A nonspontaneous process will not take place unless something starts it and keeps it going 3. Energy is a key factor in determining the spontaneity of a reaction a. Energy (ability to do work and transfer heat) is involved when matter undergoes a change b. In a spontaneous process energy is released (i.e. rock rolls down a hill) c. In a nonspontaneous process energy must be continually put in (i.e. rock pushed up a hill) d. The breaking of bonds requires energy while the making of bonds releases energy 4. Review example in text (Combustion of propane) B. Free Energy 1. Free energy (G) is the energy released or gained in a reaction a. The symbol Δ stands for change, so ΔG is the charge in free energy for a reaction b. A negative ΔG indicates that energy is released and the reaction is spontaneous c. A positive ΔG indicates that energy must be input and the reaction is nonspontaneous 2. A reaction that releases heat is called exothermic (i.e. combustion of propane) 3. A reaction that requires heat to be added is called endothermic C. Energy Diagrams 1. Examine Figure 5.14 2. An energy diagram represents the energy change that takes place during a chemical reaction a. Energy is shown on the y-axis (bottom is lowest energy) b. The progress of the reaction is plotted on the x-axis (reactants on left and products on right) c. In a spontaneous reaction the products contain less energy (more stable) than the reactants d. In a nonspontaneous reaction the products contain more energy than the reactants 3. Examine Figure 5.15 4. Spontaneous (-ΔG) reactions don’t necessarily take place within a reasonable amount of time (i.e. the rusting of the metal in an automobile) 5. Reaction rate is the measure of how quickly products form 6. Activation energy (Eact) is the energy barrier that must be crossed for the reaction to occur (i.e. a rock sitting at the top of the hill will not roll down unless it is given a push) 5-4
a. Activation energy is related to the energy that reactant molecules must have to collide and the energy associated with bond breaking or making b. The lower the activation energy the faster a reaction will take place c. Two reactions with the same ΔG will take place at very different rates if their Eact is different 7. Examine Figure 5.16 8. Temperature change can alter the rate of a chemical reaction a. The higher the temperature the faster the reaction will go b. This is because the reactants have greater kinetic energy causing more collisions 9. Examine Figure 5.17 10. A change in the reactant concentrations can alter the rate of a chemical reaction a. The higher the concentration of reactants, the faster the reaction occurs b. This is because crowded reaction mixtures result in more collisions 11. Examine Figure 5.18 12. A catalyst increases the rate of a reaction by lowering the activation energy a. A majority of the chemical reactions that take place in living things are catalyzed by enzymes b. Most enzymes increase the reaction rates by a factor of 1,000 to 100,000 13. Examine Figure 5.19 *See Sample Problem 5.14 and Practice Problem 5.14 HEALTH LINK Carbonic Anhydrase (Examine Figure Examine Figures 5.20 and 5.21)
Chapter Summary Section 5.1 In chemical reactions, covalent and ionic bonds are broken and new bonds are formed. The reaction between nitrogen and hydrogen (Haber process) is very important for living things because nitrogen is present proteins and DNA; furthermore, nitrogen must be “fixed” into compounds like NH3 for most organisms to use it. Chemical reactions like this Haber reaction can be described with chemical equations. In a chemical equation, an arrow separates reactants from products and numbers called coefficients, which are placed in front of the formulas. Equations are balanced when each side of the arrow has the same number of atoms of each element. There are many common errors when balancing equations which should be avoided. Section 5.2 One classification system uses four broad reaction types to classify reactions. In a synthesis reaction, two or more elements or compounds combine to form one more complex compound. The reverse of a synthesis reaction is a decomposition reaction, which is the breakdown of one compound to form elements or simpler compounds. In a single replacement reaction, one element trades places with a different element in a compound, while in a double replacement reaction, both parts of two compounds switch places. Section 5.3 Water is involved in many reactions important to organic and biochemistry. The first reaction is hydrolysis, which is when water is used to split a molecule. Esters undergo hydrolysis (when treated with water and an acid catalyst) to produce carboxylic acids and alcohols. The next water reaction is hydration, which is when water is added to a double bond. Alkenes can by hydrated with water and an acid catalyst to produce alcohols. Lastly, dehydration (reverse of hydration) is when an alcohol is heated with an acid catalyst to form an alkene and water. Section 5.4 A different classification system for grouping reactions uses the terms oxidation and reduction. When atoms lose electron(s) they are oxidized, and when they gain electron(s) they are reduced. Similarly, an oxidizing agent removes electron(s) while a reducing agent gives electron(s). A specific type of 5-5
oxidation that produces heat and where O2 is a reactant is called combustion. When organic molecules are combusted, carbon dioxide and water are produced. Instead of analyzing each covalent bond in organic reactions, the following simple rules allow you to determine what is oxidized and what is reduced. Atoms are oxidized if they gain oxygen or lose hydrogen, conversely atoms are reduced if they lose oxygen or gain hydrogen. Some alcohols can be oxidized to form aldehydes, other alcohols can be oxidized to form ketones, and aldehydes can be oxidized to form carboxylic acids. A specific type of reduction that saturates hydrocarbons is called hydrogenation. In catalytic hydrogenation, H2 is added as a reducing agent with Pt as a catalyst. Section 5.5 Balanced equations allow us to predict which reactant will run out first and how much product can be expected to form. Moles of each reaction species (reactants and products) can be compared using coefficients from the balanced equation. Moles and mass relationships can be determined using both the balanced equation and molar masses of the reaction species. It is important to note that a calculation involving a balanced equation must use the number of mole reactants and products, not their masses. Section 5.6 The limiting reactant determines the amount of product that can be formed in a reaction. In reactions, the limiting reactant gets totally consumed and thus the reaction stops when it has been used up (even though other non-limiting reactants still remain). The limiting reactant determines the theoretical yield of a reaction, which is the maximum amount of product that can be obtained. The theoretical yield of a product is rarely obtained due to various chemical reasons; therefore, the actual yield is the amount of product(s) that is actually obtained. Percent yield indicates how well the actual yield agrees with the theoretical yield. Section 5.7 Reactions can be spontaneous (continues by itself once it has started) or nonspontaneous (does not take place unless something starts it and keeps it going). Energy is involved when matter undergoes a change and thus it is a key factor in determining the spontaneity of a reaction. In reactions, the breaking of chemical bonds requires energy (nonspontaneous) while the making of chemical bonds releases energy (spontaneous). Free energy (G) is the energy released or gained in a reaction. A reaction that releases heat is called exothermic while a reaction that requires heat is called endothermic. Energy diagrams are a good way to represent the energy change that takes place during a chemical reaction. Another important reaction factor is reaction rate, which is the measure of how quickly products form. The reaction rate is affected by the activation energy (Eact), which is the energy barrier that must be crossed for the reaction to occur. An increase in temperature or reactant concentration, or the addition of a catalyst all speed up the rate of a reaction.
Lecture Suggestions Sections 5.1, 5.2, 5.3, and 5.4 The following analogy can be used when lecturing on these sections. Compare the types of reactions to students going to a dance. In a synthesis reaction, two people (elements/compounds) go to the dance by themselves. They meet and decide they are happier (more stable) together as a couple. In a decomposition reaction a couple (compound) goes to the dance together and they decide to break up (elements/compounds). Mention that there is often something/someone that instigates this break up (heat, electrolysis, etc.). For a single replacement reaction, one guy (element) goes by himself and another guy goes with his girl as a couple (compound). The single guy “cuts in” on another guy’s date leaving him by himself. Mention that the girl (element) is more attracted (reactive) to the new guy (element) than the original one. In a double replacement reaction, two couples (compounds) go to the dance. While the girls (elements) are in the restroom together, they decide they like each other’s date better (elements), and so they switch dates. Also mention that many reactions need a certain 5-6
environment to take place in—here you can compare the dance floor to an aqueous environment. Students will like this analogy; it will liven up a chemistry lecture. Section 5.5 For this section, you can give students some helpful hints for balancing equations—a lot of these are in the text. Write the list on the board and tell students to go through them each time they balance an equation until they no longer need them. • Balance atoms that appear the least first • Balance any single elements last (O2 or Na) • Balance water last (second to last if single elements are present) • Balance 1 atom at a time unless polyatomic ions are present • Treat polyatomic ions as one unit and balance their parts at the same time • If a half (1/2) of coefficient is needed, temporarily add it and then multiply the entire equation by two • Never adjust the subscripts to balance • Check work by counting atoms to make sure they are equal on both sides • Add phase labels and catalyst if applicable Section 5.6 For this section, you can use “making s’mores” as an example (similar to the author’s “making cheese sandwiches” example). To make a perfect s’more (encourage students to join in), you need 2 graham crackers, 1 marshmallow, and 3 pieces of chocolate (small Hershey® sized pieces), just as an example Then ask them to help you write an equation for this process, and then balance it with the previous mentioned ratios or coefficients (2Gc + 1Mm + 3Ch → 1Gc2MmCh3 ). Ask them a series of questions and then do the calculations together. 1. How many s’mores can we make from 27 pieces of chocolate? 2. How many graham crackers are needed to produce 4 s’mores? 3. If one marshmallow weighs 7.0g, how many grams of marshmallows are needed to make 8 s’mores? 4. If you want to make 25g of s’mores, how many grams of chocolate do you need (1 chocolate piece weighs 3.0g)? 5. If you have 24g of graham crackers and one graham cracker weighs 6.0g, how many grams of s’mores can you make? 6. If you have 15g of each ingredient/reactant, which is the limiting reagent? 7. Using the previous limiting reagent, calculate the theoretical yield of s’mores. Section 5.7 After introducing endothermic and exothermic reactions, there are two chemical demonstrations you can do. The first, “beaker freeze”, is an endothermic reaction that is cold enough to freeze a beaker to a board. The second, “methanol cannon”, is an exothermic reaction where methanol is combusted to produce heat. The instructions for both of these demonstrations are below. Begin with the beaker freeze demo and then set it aside while you do the methanol cannon one. Ask students what kind of change is taking place when the methanol liquid vaporizes and when it is combusted. You can also ask students why there is water in the jug after the combustion reaction (section 5.4). Then pass around the plastic water jug and let students feel the heat and see the water. By this time, the beaker has frozen to the board, so you can pick it up to show students. If you have a small class, pass this around so they can feel the cold beaker, but always warn students not to touch the contents of the beaker nor directly inhale the mixture (ammonia is a product). Beaker Freeze: Type: Endothermic Reaction: Ba(OH)2(s) + 2NH4SCN(s) → 2NH3(g) + H2O(l) + Ba(SCN)2(aq) Materials: barium hydroxide, ammonium thiocyanate, beaker, board 5-7
Procedure:
Put beaker on board, add equal parts of Ba(OH)2 solid and NH4SCNsolid in beaker. Stir and a liquid should form. Add one drop of water to the board and place the beaker on top of it. Within minutes the beaker will freeze to the board (pick it up).
Methanol Cannon: Type: Combustion: CH3OH(l) + O2(g) spark→ CO2(g) + H2O(l) Materials: Huge plastic water jug, methanol, matches, tongs Procedure: Place about 25ml of methanol in the water jug. Move the jug around so the methanol coats the inside and then dump out the liquid methanol. Light a match and hold it by the tongs. Place it over the mouth of the jug to ignite the methanol vapors. Turn off lights to make it more dramatic. *Note: do not take too long between dumping out liquid and lighting it.
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Chapter 6 Gases, Solutions, Colloids, & Suspensions Outline Notes I. Gases and Pressure (Section 6.1) A. Gases and Pressure Overview 1. Gas particles do not interact with one another and are free to move about their container 2. The pressure of a gas is the force of the collisions that take place between the particles and an object (i.e. the walls of container, your chemistry book, your skin, etc.) 3. Atmospheric pressure is caused by the mass of all the air above us in the atmosphere a. A column of air 1 meter in diameter that extends upwards 32 km has a mass of 10,000kg b. Examine Figure 6.1 c. At sea level the atmospheric pressure is 14.7 pounds per square inch (psi) or 1 atmosphere (atm) d. Examine Figure 6.2 e. The less air above you in the atmosphere, the lower the atmospheric pressure i. In Denver, CO (elevation is 1.6 km) the pressure is 0.74 atm or 11 psi ii. At Mt. Everest’s summit (elevation 9.3 km) the pressure is 0.26 atm or 3.8 psi B. Barometers and Torr 1. A barometer is a device used to measure pressure a. Barometer’s are constructed by filling a long glass tube with mercury and inverting the tube in a mercury-filled dish (the mercury in the dish does not drain out because the atmosphere pushes down on it with enough force) b. Examine Figure 6.3 c. At 1 atm and 0°C the mercury column has a height of 760 mm i. The greater the atmospheric pressure the higher the mercury level in the column ii. The lesser the atmospheric pressure the lower the mercury level in the column 2. Torr a. The height of the mercury column in a barometer is the source of the pressure unit torr b. It is named after the Italian scientist Torricelli (1608-1647) who invented the barometer c. Each millimeter of mercury column height equals one torr d. Note that 760 mmHg = 760 torr = 1atm = 14.7 psi *See Sample Problem 6.1 and Practice Problem 6.1 *See Sample Problem 6.2 and Practice Problem 6.2 DID YOU KNOW? Weather Reports C. Standard Temperature and Pressure 1. Pressure and temperature have an effect on the physical state of a substance (i.e. at 90°C and 1 atm water is a liquid but at 90°C and 0.6 atm water is a gas) 2. Standard temperature and pressure (STP) is 0°C and 1 atm 3. Eleven elements exist as gases at STP (H2, N2, O2, F2, Cl2, and the inert gases-group 8A) 4. Many covalent compounds such as CO2, NH3, and N2O also exist as gases at STP D. Manometers 1. A manometer is used to measure the pressure of enclosed gases (i.e. in bottles) 2. A manometer consists of a curved tube filled with mercury attached to a flask 3. Examine Figure 6.4 4. Each millimeter of height difference between the two mercury columns corresponds to a pressure difference of 1 torr between the atmosphere pressure and the flask pressure 5. A sphygmomanometer is composed of an inflatable cuff that is attached to a manometer and is used to measure blood pressure E. Vapor Pressure 6-1
1. Particles in liquids constantly undergo collisions that cause the particles at the surface to continually evaporate (bounce off into the gas phase) 2. At the same time, gas phase molecules are being trapped and converted to liquid 3. Examine Figure 6.5 4. Vapor pressure is the maximum level of pressure a liquid (i.e. water) exerts causing the gas above it (i.e. air) to become completely saturated (maximum amount possible) with vapor 5. At 20ºC when air in a container is saturated with water vapor, the vapor pressure is 17.54 torr 6. See Table 6.1 for how vapor pressure varies with temperature F. Boiling Point 1. The boiling point of a liquid is the temperature at which the vapor pressure of the liquid equals the atmospheric pressure 2. At this temperature vapor forms in bubbles throughout the liquid 3. Water boils at 100ºC because its vapor pressure equals the atmospheric pressure of 760 torr (i.e. in Denver CO water boils at 92ºC because the air pressure is only 560 torr) 4. Increasing the pressure above 760 torr will push the boiling point of water to temperature greater than 100ºC (i.e. a pressure cooker increases the temperature at which water boils) 5. The high pressure reached in autoclaves used for sterilization allow water to boil at temperatures high enough to kill most infectious agents (i.e. bacteria) 6. Examine Figure 6.6 *See Sample Problem 6.3 and Practice Problem 6.3 HEALTH LINK Blood Pressure (Examine Figure 6.7 and Table 6.2) II. The Gas Laws (Section 6.2) A. Pressure and Volume 1. Examine Figure 6.8 2. Robert Boyle (late 1600’s) determined that as volume increases, pressure decreases 3. Boyle’ law states that pressure and volume are inversely proportional for a gas at fixed temperature 4. Equation: P1V1 = P2V2 where P is pressure, V is volume, 1 is initial, and 2 is final 5. Review example in text (Finger and syringe barrel calculation) *See Sample Problem 6.4 and Practice Problem 6.4 B. Pressure and Temperature 1. Joseph Gay-Lussac (1778-1850) determined that as temperature increases, pressure increases (i.e. as you drive your car the tire pressure increases due to an increase in temperature) 2. Gay-Lussac’s law states that pressure and temperature are directly related for a gas at fixed volume 3. Equation: P1/T1 = P2/T2 where P is pressure, T is temperature in K, 1 is initial, and 2 is final 4. Examine Figure 6.9 5. Review example in text (Steel tank calculation) C. Volume and Temperature 1. Jacques Charles (1746-1823) determined that as temperature increases, volume increases 2. Charles’ law states that volume and temperature are directly related for a gas at fixed pressure 3. Examine Figure 6.10 4. Equation: V1/T1 = V2/T2 where V is volume, T is temperature in K, 1 is initial, and 2 is final 5. Review example in text (Balloon in freezer calculation) *See Sample Problem 6.5 and Practice Problem 6.5 D. Moles and Volume 1. Amedeo Avogadro (early 1800s) determined that as amount increases, volume increases 2. Avogadro’s law states that volume and the number of moles are directly related for a gas at fixed temperature and pressure 3. Equation: V1/n1 = V2/n2 where V is volume, n is moles, 1 is initial, and 2 is final 4. Review example in text (Blowing up a balloon calculation) 6-2
E. Combined Gas Law 1. When the number of moles of a particular gas is held constant, Boyle’s law, Gay-Lussac’s law, and Charles’ law can be merged into one called the combined gas law 2. Equation: P1V1/T1 = P2V2/T2 where P is pressure, V is volume, T is temperature in K, 1 is initial, and 2 is final 3. This law is more useful because it allows the properties of a gas to be predicted over a wider range of conditions 4. Examine Figure 6.11 5. Review example in text (Balloon release calculation) 6. The combined gas law can also be used if the number of moles of a gas is allowed to change 7. Equation: P1V1/n1T1 = P2V2/n2T2 where P is pressure, V is volume, T is temperature in K, n is moles, 1 is initial, and 2 is final F. Ideal Gas Law 1. For a given gas, pressure multiplied by volume divided by number of moles multiplied by temperature is constant 2. This constant is called the gas constant (R), which is represented by PV/nT =R 3. Ideal gases are those that completely obey all of the previous gas laws 4. Ideal gases have an R value of 0.0821 L atm/mol K 5. Except at very high pressure or very low temperatures, most real gases come fairly close to ideal behavior (all gases will be treated as ideal in this text) 6. Equation: PV=nRT where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in K 7. Review example in text (Helium-filled calculation) *See Sample Problem 6.6 and Practice Problem 6.6 III. Partial Pressure (Section 6.3) A. Partial Pressure Overview 1. Gas laws are useful for predicting the behavior of a mixture of gases 2. When ideal gases are mixed, each exerts the same pressure that it would if it were alone 3. Examine Figure 6.12 4. Review example in text (He and Ne in cylinders calculation) HEALTH LINK Breathing (Examine Figure 6.13) B. Dalton’s Law of Partial Pressure 1. Partial pressure is the pressure that each gas in a mixture would exert if it were alone 2. Dalton’s law of partial pressure states that the total pressure of a mixture of gases is the sum of the partial pressures of its components 3. Equation: Ptotal = PA + PB + PC where Ptotal is the total pressure, PA is the partial pressure of the first gas, PB is the partial pressure of the second gas, and PC is the partial pressure of the third gas *See Sample Problem 6.7 and Practice Problem 6.7 IV. Solutions (Section 6.4) A. Classification of Matter 1. Matter can be classified as a pure substance or a mixture 2. Pure substances consist of one element or compound (i.e. Hg, Ag, O2, water, salt, sucrose, etc.) 3. Mixtures are a physical combination of two or more pure substances a. Homogeneous mixtures are uniformly distributed (i.e. Kool Aid) b. Heterogeneous mixtures are not evenly distributed (i.e. solid that sink in liquid) B. Solutions 1. Solutions are homogenous mixtures where all parts are in a single phase 2. Examine Figure 6.14 3. Solutions are made of a solvent (the solution component present in the greater amount) and solute(s) (component dissolved in the solvent) 4. In aqueous solutions water is the solvent 6-3
5. Examine Figure 6.15 6. Solutes and solvents that make up solutions can be solids, liquids, or gases (i.e. carbonated water) C. Solubility 1. In a solution, solutes and solvents stay uniformly mixed for a period of time 2. A precipitate is a solid reaction product that falls out of a solution 3. For liquid solutions, a solute particle must be about the same size as the solvent molecules and they must be able to interact with the solvent molecules through noncovalent interactions 4. “Like dissolves like” refers to a substance being soluble (dissolves) or insoluble (doesn’t dissolve) a. For example NaCl (table salt) is soluble in water b. This is because Na+ and Cl- ions are roughly the same size as the water molecules and the charges allow them to interact with the polar water molecules 5. Solubility is the amount of solute that will dissolve in a solvent at a given temperature a. The solubility of gas solutes in water decreases as temperature increases b. Most liquid and solid solutes become more soluble in water as temperature increases c. When a solute has reached its solubility limit, no additional solute will dissolve 6. Examine Figure 6.16 *See Sample Problem 6.8 and Practice Problem 6.8 V. Precipitation Reactions (Section 6.5) A. Ionic Compound Solubility 1. The strength of the ionic crystal lattice determines if an ionic compound is water soluble 2. Some ionic bonds hold the crystal lattice together so strong that the compound will not dissolve in water even if the individual ions would otherwise be soluble (review Figure 6.15) 3. See Table 6.3 for the water solubility behavior of some ionic compounds *See Sample Problem 6.9 and Practice Problem 6.9 B. Ionic Compound Reactions in Water 1. The information in Table 6.3 can be used to decide whether or not a chemical reaction will take place when a given pair of ionic compounds is mixed in water 2. Examine Figure 6.17 3. An (aq) is added for water soluble compounds and an (s) is written for insoluble compounds, which precipitate out of solution (two homogeneous mixtures become heterogeneous) 4. In an ionic equation the formulas of electrolytes are written as individual ions (all soluble/ aqueous compounds are broken apart into ions and all insoluble/solid compounds remain intact) 5. Spectator ions are ions that appear identically on both sides of the equation 6. The net ionic equation results when spectator ions are removed 7. Review examples in text [CoCl2 reacts with Ca(OH)2 & Na2CO3 reacts with Mg(NO3)2] *See Sample Problem 6.10 and Practice Problem 6.10 DID YOU KNOW? Hard Water VI. Solubility of Gases in Water (Section 6.6) A. Henry’s Law 1. Henry’s law states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the liquid (i.e. doubling pressure of a gas over a liquid, doubles its solubility in that liquid) 2. Carbonated beverages are a good example of Henry’s Law (i.e. when a bottle is opened the pressure of the gas above the beverage is decreased and the dissolved gas escapes “fizzes”) 3. Examine Figure 6.18 4. Some gases have a very high solubility in water because they react with the solvent and are converted to water-soluble products 5. Review examples in text (CO2 and NH3 solubility in water reactions) B. Solubility of Gases in Health Care 1. Blood can take up about 66 times more O2 because of its attraction to hemoglobin 2. Examine Figure 6.19 3. Hyperbaric (high pressure) oxygen treatment causes more oxygen to dissolve in the blood 6-4
a. Used to treat gangrene and tetanus (anaerobic bacteria) b. Used to speed the healing of some skin grafts and wood c. Slow decompression from hyperbaric conditions treats sickness experience by divers 4. Examine Figure 6.20 5. Some anesthetics are gases administered by having a patient breathe them 6. Examine Figure 6.21 7. The solubility of these inhaled anesthetics in the blood effects: a. How quickly the effects are felt b. How easily the dept of anesthesia can be changed c. How quickly the recovery from anesthesia takes 8. Dentists commonly use nitrous oxide (N2O) as an inhaled anesthetic 9. The solubility of N2O in the blood is low so its effects rapidly wear off *See Sample Problem 6.11 and Practice Problem 6.11 VII. Organic and Biochemical Compounds (Section 6.7) A. Organic Compound Solubility 1. Alkanes readily mix with one another due to London forces (i.e. pentane is soluble in hexane) 2. Examine Figure 6.22a 3. Hydrocarbons do not dissolve in water because they are nonpolar a. When mixed, the hydrocarbon molecules associated through London forces and the water molecules associate through hydrogen bonding b. Although hydrocarbons are not soluble in water, many organic compounds are 4. Ethyl alcohol is soluble in water because the molecules can hydrogen bond with each other 5. Examine Figure 6.22b 6. The more carbon atoms an alcohol has the less water soluble it becomes a. Long chains make the alcohol more hydrocarbon-like and less water-like b. Large alcohols tend to be more soluble in nonpolar solvents than in water 7. This relationship can be applied to other organic compounds 8. Review examples in text (Acetic acid vs. octanoic acid solubility) B. Biochemical Compound Solubility 1. Hydrophilic compounds are soluble in water (water loving) 2. Hydrophobic compounds are insoluble in water (water fearing) 3. Amphipathic compounds have both hydrophilic and hydrophobic parts 4. For large biochemical molecules to be hydrophilic, they must have enough polar groups a. The polar groups (N and O atoms) allow for hydrogen bonding with water b. There must be enough polar groups to “overcome” the insoluble carbon chains 5. Examine Figure 6.23 C. Soaps 1. Fatty acids (i.e. oleic acid) are hydrophobic even though they have a polar carboxyl group 2. When fatty acids are reacted with NaOH, they form ionic compounds called soaps 3. Soap is amphipathic because of the negative charge on the O atom (i.e. anions are hydrophilic) 4. Soaps form micelles, which are tiny spheres in which hydrophilic ends make up the outer surface, and hydrophobic ends cluster in the center (this gives soap its cleaning ability) 5. Examine Figure 6.24 *See Sample Problem 6.12 and Practice Problem 6.12 HEALTH LINK Prodrugs (Examine Figure 6.25) VIII. Concentration (Section 6.8) A. Concentration Overview 1. An important characteristic of a solution is how much solute is present (i.e. blood serum solutes) 2. Concentration is the amount of solute that is dissolved in a solvent (exact amount) 3. A saturated solution holds the maximum amount of a solute that can be dissolved, while unsaturated solution holds less than the saturating amount 6-5
4. A dilute solution contains less solute than a concentrated solution 5. A number of different units are used to describe solution concentration B. Percent, PPT, PPM, and PPB 1. Percent is the same thing as parts per hundred (solute present in every 100 parts of solution) a. Weight/volume percent, % (w/v), indicates mass of solute per volume of solution b. Equation: % (w/v) = (g of solute)/(mL of solution) x 100 c. Review example in text (Potassium bitartrate calculation) 2. Very dilute solutions use the units: a. Parts per thousand (ppt); Equation: ppt = (g of solute)/(mL of solution) x 1000 b. Parts per million (ppm); Equation: ppm = (g of solute)/(mL of solution) x 1,000,000 c. Parts per billion (ppb); Equation: ppb = (g of solute)/(mL of solution) x 1,000,000,000 d. Examine Figure 6.26 e. Review example in text (Sodium fluoride calculation) *See Sample Problem 6.13 and Practice Problem 6.13 C. Molarity 1. Molarity (M) is the number of moles of solute present in each liter of solution 2. Equation: M = (moles of solute)/(liters of solution) 3. Examine Figure 2.27 *See Sample Problem 6.14 and Practice Problem 6.14 D. Concentration Units used in Medicine 1. See Table 6.4 for the concentration of six blood serum solutes 2. Review example in text (mg/dL calculation) 3. An equivalent (Eq) is the number of moles of charges that a solute contributes to a solution (i.e. 1 mole Na+ = 1 Eq of Na+ and 1 mole of Ca2+ = 2 Eq of Ca2+) 4. Review example in text (Phosphate ion concentration in mEq calculation) 5. Concentration allows us to calculate the amount of solute present in a solution 6. Review examples in text (Moles of glucose calculation and fibrinogen concentration calculation) *See Sample Problem 6.15 and Practice Problem 6.15 *See Sample Problem 6.16 and Practice Problem 6.16 IX. Dilution (Section 6.9) A. Dilution Overview 1. A solution can be diluted by the addition of more solvent 2. Dilution is an easy way to reduce the concentration of a solution 3. Dilution is important in health care because some drugs must be diluted to the proper concentration before they can be administered to patients 4. Examine Figure 6.28 B. Dilution Equation 1. Voriginal x Coriginal = Vfinal x Cfinal where V is volume and C is concentration 2. Any units for volume and concentration can be used as long as they are the same on both sides of the equation (i.e. if the original volume is in mL then the final volume must also be in mL) 3. Review example in text (Glucose dilution calculation) *See Sample Problem 6.17 and Practice Problem 6.17 X. Colloids and Suspensions (Section 6.10) A. Solution-like Mixtures 1. Solutions are typically clear (you can’t distinguish the solutes from the solvents) 2. In solutions, the solutes do not settle out because of their interactions with the solvent 3. Colloids and suspensions are both mixtures that resemble solutions B. Suspensions 1. Suspensions contain large particles suspended in a liquid 2. The particles are too large to dissolve in the liquid and many are large enough to be seen 3. The particles will settle due to gravity unless the suspension is stirred or mixed 6-6
4. Due to their different sizes, the particles can be separated out by filter or centrifuge 5. Examine Figure 6.29 6. Suspensions have a cloudy appearance (i.e. milk of magnesia, Pepto Bismol, etc.) C. Colloids 1. Colloids contain particles larger than those of solutions but smaller than those of suspensions 2. Colloids usually cannot be distinguished with the naked eye or light microscope 3. The particles in colloids do not separate out like in suspensions but they do not form a solution 4. Colloid particles can be separated by special filtration or centrifugation techniques 5. Some examples of colloids include soapy water, fog, shaving cream, mayonnaise, and milk 6. See Table 6.5 for some solutions, colloids, and suspensions *See Sample Problem 6.18 and Practice Problem 6.18 HEALTH LINK Saliva (Examine Figure 6.30) XI. Diffusion and Osmosis (Section 6.11) A. Diffusion 1. Diffusion is the process where substances move from areas of higher concentration to lower concentration (i.e. chopped onion odor) 2. Examine Figure 6.31 3. Semipermeable membranes are barriers to diffusion that allow solvents but not all solutes to pass 4. Solvents moving from low to high solute concentration is the same principle as solutes moving from high to low solute concentration (i.e. when solute cannot move across the membrane but solvents can) B. Osmosis 1. Osmosis is the movement of water across a membrane from a solution of lower concentration to one of higher concentration (i.e. seawater and freshwater separated by a semipermeable membrane) 2. Osmotic pressure is the pressure exerted by water as it flows through the membrane (measured as the pressure necessary to stop osmosis) 3. Examine Figure 6.32 4. Sufficient pressure can be applied to counteract osmotic pressure and thus prevent osmosis C. Cells and Osmosis 1. Cell membranes are semipermeable and can be affected by osmosis 2. Isotonic is when there is no net movement of water because the concentration inside and outside the cell are the same 3. Hypertonic is when water moves out of the cell because the concentration is greater on the outside 4. In hypertonic solutions, red blood cells shrink in a process called crenation 5. Hypotonic is when water moves into the cell because the concentration is greater on the inside 6. In hypotonic solutions, red blood cells expand and may burst in a process called hemolysis 7. Examine Figure 6.33 8. If the blood serum is hypertonic or hypotonic, a health problem may exist *See Sample Problem 6.19 and Practice Problem 6.19 D. Colligative Properties 1. A colligative property is one that depends only on the concentration of solute (i.e. osmosis) 2. Freezing point depression is a colligative property where the freezing point can be lowered (depressed) by the addition of a solute (i.e. salt, antifreeze, etc.) 3. Boiling point elevation is a colligative property where the addition of a solute raises (elevates) the boiling point HEALTH LINK Diffusion and Kidneys (Examine Figure 6.34)
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Chapter Summary Section 6.1 The pressure of a gas is the force of the collisions that take place between the particles and an object. At sea level the atmospheric pressure is 14.7 pounds per square inch (psi) or 1 atmosphere (atm). Atmospheric pressure can be measured with a device called a barometer, which reads 760 mmHg or torr at 1 atm. One atm (760 torr) and 0°C is called standard temperature and pressure (STP). A manometer can be used to measure the pressure of enclosed gases, and a special type that measures blood pressure is called sphygmomanometer. The boiling point of a liquid is the temperature at which the vapor pressure of a liquid equals the atmospheric pressure. Section 6.2 The physical properties of gases can be described with various gas laws. Boyles’ law states that pressure and volume are inversely proportional for a gas at fixed temperature. In Gay-Lussac’s law, pressure and temperature are directly related for a gas at fixed volume. Charles’ law states that volume and temperature are directly related for a gas at fixed pressure. And, Avogadro’s law states that volume and the number of moles are directly related for a gas at fixed temperature and pressure. These four laws can be merged in the combined gas law. Pressure multiplied by volume divided by number of moles multiplied by temperature is constant for a given gas. This gas constant (R) can be used for ideal gases in what is known as the ideal gas law. Section 6.3 When ideal gases are mixed, each exerts the same pressure that it would if it were alone. Partial pressure is the pressure that each gas in a mixture would exert if it were alone. Dalton’s law of partial pressure states that the total pressure of a mixture of gases is the sum of the partial pressures of its components. Section 6.4 Matter can be classified as a pure substance (element or compound) or a mixture (combination of pure substances). Homogeneous mixtures are uniformly distributed while heterogeneous mixtures are not. Solutions are homogenous mixtures composed of a solvent and solute(s). In a solution, solutes and solvents stay uniformly mixed for any period of time. A precipitate is a solid reaction product that falls out of solution. This is determined from the “like dissolves like” guidelines, which refers to a substance being soluble or insoluble. Solubility is the amount of solute that will dissolve in a solvent at a given temperature. Section 6.5 The strength of the ionic crystal lattice determines if an ionic compound is water soluble. Table 6.3 can be used to determine the solubility of ionic compounds as well as the reactions that form them. An (aq) is added for water soluble compounds and an (s) is written for insoluble compounds, which precipitate out of solution. In an ionic equation the formulas of electrolytes are written as individual ions. In a net ionic equation, spectator ions (exist as ions on both sides of the chemical equation) are removed. Section 6.6 Henry’s law states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the liquid. While pressure affects solubility, some gases have a very high solubility in water because they react with the solvent and are converted to water-soluble products. There are many health care applications relating to the solubility of gases (i.e. hyperbaric oxygen treatment and inhaled anesthetics). Section 6.7 Many organic and biochemical compounds are soluble in water, however, many are not. Hydrocarbons do not dissolve in water because they are nonpolar, while small alcohols do because they are polar. Biochemical compounds can be classified as hydrophilic (water loving), hydrophobic (water fearing), or amphipathic (both hydrophilic and hydrophobic). For large biochemical molecules to be hydrophilic, 6-8
they must have enough polar groups to “overcome” the insoluble carbon chains. Soap is an excellent example of an amphipathic compound; it has a long hydrocarbon chain that is hydrophobic and it has an O atom with a negative charge that his hydrophilic. Because of this, soaps form micelles, which give them their cleaning ability. Section 6.8 Concentration is the amount of solute that is dissolved in a solvent. Solutions can be saturated or unsaturated and a dilute solution contains less solute than a concentrated solution. Many units are used for solution concentration including weight/volume percent (%w/v), parts per thousand (ppt), parts per million (ppm), parts per billion (ppb), and molarity (M). Some examples of concentration units used in medicine include milligrams per deciliter (mg/dL) and equivalents (Eq). Section 6.9 Dilution is an easy way to reduce the concentration of a solution by the addition of more solvent. Dilution is important in health care because some drugs must be diluted to the proper concentration before they can be administered to patients. Section 6.10 Colloids and suspensions are both mixtures that resemble solutions. Suspensions contain large particles suspended in a liquid. The suspended particles will settle due to gravity unless the suspension is stirred or mixed. Suspension particles can be separated by size using filtration or centrifugation. Colloids contain particles larger than those of solutions, but smaller than those of suspensions. The particles in colloids do not separate out like in suspensions; however, they do not form a solution. Section 6.11 The process where substances move from areas of higher to lower concentrations is called diffusion. Semipermeable membranes act as barriers to diffusion in that they allow solvents and only some solutes to pass. In the process of osmosis, water moves across a membrane from a solution of lower concentration to one of higher concentration. The pressure exerted by water as it flows through the membrane is known as osmotic pressure. Cell membranes are semipermeable and can be affected by osmosis. Cells can be isotonic, hypertonic, or hypotonic, which affects the direction of osmosis. In hypertonic solutions, red blood cells shrink in a process called crenation; conversely, in hypotonic solutions, red blood cells expand and burst in a process called hemolysis. Osmosis is a colligative property, which is one that depends only on the concentration of solute. Freezing point depression and boiling point elevation are also examples of colligative properties.
Lecture Suggestions Sections 6.1, 6.2, and 6.3 Pick a willing student to come to the front of the class and have them draw (with a pen/marker) a square on their hand that is approximately 1 inch by 1 inch. Then give them a 15lb weight and tell them to hold it as long as they can with their arm extended straight. Explain to students that this is how much weight is pressing on each little square inch of their bodies. Explain that 14.7psi is the atmospheric pressure caused by weight of the air above them. Mention that they are used to this pressure so they don’t notice it; however, if it was significantly decreased they would blow up like a balloon, and if it was significantly increased they would be crushed. Perform the boiling water demo for students. Hook a vacuum up to a stoppered flask with water in it. Turn on the vacuum and the water will begin to boil at room temperature. Discuss how this demonstration relates to vapor pressure, boiling point, and the gas laws.
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Sections 6.4 and 6.5 Have students use Table 6.3 to predict the products between the reaction of aqueous potassium iodide and aqueous lead (II) nitrate. After they balance the equation, have students write the total ionic equation, circle the spectator ions, and then write the net ionic equation. After they are down you can work it on the board and have them check their answers. Perform the “yellow precipitate” demonstration (below) and discuss it. Yellow Precipitant: Type: Double Rep.: KI(aq) + Pb(NO3)2(aq) → PbI2(s) + KNO3(aq) Materials: KI(aq), Pb(NO3)2(aq), beaker Procedure: Fill a beaker ¾ full with distilled water. Add 10-15mL of aqueous potassium iodide to the beaker. Then add 10-15mL of aqueous lead (II) nitrate to the beaker a yellow precipitant will form instantly. Dispose of the lead product appropriately. Sections 6.6 and 6.7 To introduce these sections, ask students to write down one way in which the solubility of either a gas, liquid, or solid applies to something their future health-related careers. Have students volunteer their ideas and then briefly discuss each. You can mention applications in the text such as hyperbaric oxygen treatment, inhaled anesthetics, drug solubility, etc. Later in the lecture you can discuss how soap has affected medicine and societies in general over the last two thousand years. Sections 6.8, 6.9, 6.10, and 6.11 Demonstrate the difference between solutions, colloids, and suspensions. Place three beakers of water on your desk. Add a small amount of table salt to the first, 1-2mL of milk to the second, and some sand to the third. Stir them all up and ask students to identify them as a solution, colloid, or suspension. Turn out the lights and shine a laser pointer at the solution and explain how the light is not scattered. Repeat with the colloid and explain why it scatters the light (Tyndall Effect). You can then draw the following on the board to help students compare these mixtures
Solution (Homogeneous Mixture)
Colloid (Homogeneous Mixture)
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Suspension (Heterogeneous Mixture)
Chapter 7 Acids, Bases & Equilibrium Outline Notes I. Acids and Bases (Section 7.1) A. Acids vs. Bases 1. Acids have a sour taste (i.e. citric acid in citrus fruits) while bases have a bitter taste (i.e. caffeine) 2. Acids dissolve some metals while bases feel slippery 3. Acids turn litmus (plant pigment) to a pink color while bases turn litmus blue 4. Examine Figure 7.1 B. Common Acids and Bases 1. Many acids and bases will cause extensive tissue damage and can be fatal; therefore they should not be identified by taste 2. See Table 7.1 for common acids and bases II. Bronsted-Lowry Acids and Bases (Section 7.2) A. Arrhenius Acids and Bases 1. Svante Arrhenius (1880’s) identified acids as substances that produce H+ in water 2. He also determined that bases produce OH3. Review example in text (HCl and NaOH equations) B. Bronsted-Lowry Acids and Bases 1. Johannes Bronsted and Thomas Lowry (1920’s) determined that acids release H+ and bases accept H+ 2. Review example in text (HCN in water equation) 3. Double arrows indicate that a reaction is reversible 4. Review example in text (NH3 in water equation) *See Sample Problem 7.1 and Practice Problem 7.1 C. Conjugate Acids and Bases 1. Compounds that differ only in the presence or absence of H+ are called conjugates 2. Review example in text (Acetic acid conjugate equation) 3. Water is amphoteric, which means that it can act as either an acid or a base *See Sample Problem 7.2 and Practice Problem 7.2 III. Equilibrium (Section 7.3) A. Equilibrium Overview 1. The terms reactant and product are somewhat unclear in a reversible reaction (the product of one direction is the reactant of the other); therefore we will define reactants as the compounds that appear on the left side of the reaction arrow and products on the right side 2. Review example in text (Decomposition of N2O2 equation) 3. Examine Figure 7.2 4. Equilibrium is the point where the rate of the forward reaction equals the rate of the reverse reaction 5. Reactants and products are produced as rapidly as they are consumed at equilibrium, thus there is no change in their concentrations B. Equilibrium Constants 1. In chemical equation: aA + bB → cC + dD, A and B are reactants, C and D are products, and a, b, c, and d are coefficients 2. The equilibrium constant (Keq) for this chemical equation can be calculated using the following equation: Keq =[C]c[D]d / [A]a[B]b 3. In this equation the brackets represent concentration in molarity (M) 4. Review example in text (Formation of ammonia) 7-1
*See Sample Problem 7.3 and Practice Problem 7.3 5. Only concentrations that can change are written in an equilibrium constant expression a. The concentration of solids at a given temperature cannot change, thus they are omitted b. Solvents are also omitted because their concentration is very high and does not change significantly during a reaction (i.e. H2O(l)) 6. Review example in text (PbI2 equilibrium constant expression) 7. Examine Figure 7.3 8. Review example in text (HCN + H2O equilibrium constant expression) C. Acidity Constant 1. The acidity constant (Ka) is the equilibrium constant for an acid reaction 2. When an equilibrium constant has a value of less than 1, the denominator is greater than the numerator (more reactants than products) 3. When an equilibrium constant has a value of greater than 1, the numerator is greater than the denominator (more products than reactants) 4. See Table 7.2 for interpreting equilibrium constants *See Sample Problem 7.4 and Practice Problem 7.4 IV. LeChatelier’s Principle (Section 7.4) A. LeChatelier’s Principle Overview 1. Equilibrium conditions can easily be disturbed by changes in conditions 2. LeChatelier’s principle states that when a reversible reaction is pushed out of equilibrium, the reaction responds to reestablish the equilibrium 3. Examine Figure 7.4 B. Varying Concentration 1. Varying the concentration of a reactant or product disturbs equilibrium 2. Review example in text (Carbon dioxide/carbonic acids equilibrium) 3. Increasing the concentration of a reactant causes a net forward reaction to take place 4. Decreasing the concentration of a reactant causes a net reverse reaction to take place 5. Increasing the concentration of a product causes a net reverse reaction to take place C. Varying Pressure 1. For reactions that involve gases, varying the pressure may disturb equilibrium 2. If the pressure is increased, then the system will shift to decrease the total number of moles of gas that are present (to reduce the pressure) 3. Review example in text (Ammonia gas formation equilibrium) 4. If the pressure is decreased, then the reaction will shift to increase the pressure 5. Review example in text (N2O2/NO2 equilibrium) D. Varying Temperature 1. Changes in temperature can also affect an equilibrium 2. Heat can be treated as a reactant or product 3. Review example in text (Ammonia gas with heat formation equilibrium) *See Sample Problem 7.5 and Practice Problem 7.5 E. Catalysts 1. When a catalyst is used in a reversible reaction, the lowered activation energy speeds up the forward and reverse reactions to the same extent 2. A catalyst has no effect on the equilibrium or on the value of Keq BIOCHEMISTRY LINK Diving Mammals, Oxygen, and Myoglobin (Examine Figure 7.5) V. Ionization of Water (Section 7.5) A. Water Overview 1. Water is amphoteric because it can act as an acid or base depending on what it reacts with 2. Review example in text (Water acid-base equations) 3. Water molecules can react with other molecules in an acid-base reaction 4. Examine Figure 7.6 7-2
B. Ionization and Equilibrium 1. Ionization is the formation ions 2. In pure water ionization is spontaneous 3. Ionization of water: 2H2O(l) → H3O+(aq) + OH-(aq) 4. The equilibrium constant Kw is equal to 1.0 x10-14 5. Expression: Kw = [H3O+][ OH-] = 1.0x10-14 6. If the concentration of H3O+ or OH- is known then the other can be calculated 7. Review example in text (Calculation of H3O+ concentration) *See Sample Problem 7.6 and Practice Problem 7.6 VI. The pH Scale (Section 7.6) A. pH Scale Overview 1. The concentration of H3O+ determines if a solution is acidic, basic, or neutral 2. An acidic solution has a concentration of H3O+ greater than 1x10-7 M; the higher the concentration, the more acidic it is 3. A basic solution has a concentration of H3O+ less than 1x10-7 M; the lower the concentration, the more basic it is 4. A solution with the concentration of H3O+ equal to 1x10-7 M is neutral 5. pH is a logarithmic scale in which a one-unit change in pH is equivalent to a ten-fold change in [H3O+] (condenses a wide range of concentrations down to an easier scale) B. Calculating pH 1. pH is calculated with the equation: pH = -log[ H3O+] 2. A solution is acidic when pH is less than 7, basic when pH is greater than 7, and neutral when the pH equals 7 3. Examine Figure 7.7 4. See Table 7.3 for pH values of common solutions 5. Review example in text (Calculating the pH of 8x10-4M H3O+) *See Sample Problem 7.7 and Practice Problem 7.7 6. H3O+ concentration can be calculated from pH (i.e. reverse of previous calculation) 7. Review example in text (Calculation of H3O+ concentration of pH 2.0) 8. pH can be measured with a pH meter or an indicator 9. Examine Figure 7.8 *See Sample Problem 7.8 and Practice Problem 7.8 MATH SUPPORT Logs and Antilogs VII. Acid and Base Strength (Section 7.7) A. Acid Strength 1. The stronger the acid the more H3O+ it produces and the lower its pH 2. Strong acids almost completely ionize in water (i.e. HCl) 3. Review example in text (HCl pH calculation) 4. Weak acids ionize much less in water (i.e. HF) 5. Review example in text (HF pH calculation) 6. The larger the Ka for an acid, the stronger an acid it is 7. pKa can also be used to indicate the strength of an acid: pKa = -log Ka 8. The lower the pKa, the stronger the acid 9. Review example in text (HCl & HF pKa calculation) 10. See Table 7.4 Ka and pKa values of some acids B. Base Strength 1. The stronger the base the more OH- it produces and the higher its pH 2. Strong bases dissociate (fall apart) completely in water (i.e. NaOH) 3. Review example in text (NaOH pH calculation) 4. Some weak bases have extremely low solubility in water thus [OH-] remains low [i.e. Mg(OH)2] 5. Other weak bases are poor H+ acceptors (i.e. NH3) 7-3
6. Review example in text (NH3 in water) 7. The stronger an acid, the weaker its conjugate base *See Sample Problem 7.9 and Practice Problem 7.9 8. See Table 7.5 for strengths of acids and their conjugates bases BIOCHEMISTRY LINK Plants as pH Indicators VIII. Neutralizing Acids and Bases (Section 7.8) A. Neutralization 1. Neutralization takes place when an acid and base react to form water and a salt (ionic compound) 2. Review example in text (HCl and NaOH neutralization reaction) 3. If proper amounts of an acid and base are combined then they will cancel each other out to produce a neutral solution 4. Review example in text (HCl and NaOH neutralization calculation) DID YOU KNOW? Fish and Amines B. Titration 1. A titration is a technique used to determine the concentration of an acid or base solution 2. Examine Figure 7.10 3. Review example in text (Titration calculation of NaOH and HCl) 4. Examine Figure 7.11 *See Sample Problem 7.10 and Practice Problem 7.10 IX. Effect of pH on Acid and Conjugate Base Concentration (Section 7.9) A. Acid and Conjugate Base Overview 1. Some compounds important to living things are acids 2. pH determines if they are found in acid or conjugate base form 3. Ka and pKa provide information about the relative concentrations of reactants and products 4. Review example (HF in water pKa calculation) 5. LeChatelier’s Principle can be used to predict how changing the pH affects the equilibrium B. pH and Concentration Relationship 1. When the pH = pKa the concentration of acid equals the concentration of its conjugate base 2. When the pH < pKa the concentration of the acid is higher than that of its conjugate base 3. When the pH > pKa the concentration of the conjugate base is higher than that of its acid 4. Fatty acids (long-chained carboxylic acids) are a biochemical example of the effect pH has 5. Review example in text (Oleic acid and oleate ion equation) 6. See Table 7.6 for acid and conjugate base concentration as a function of pH *See Sample Problem 7.11 and Practice Problem 7.11 X. Buffers (Section 7.10) A. Buffers Overview 1. A buffer is a solution that resists changes in pH when small amounts of acid or base are added 2. Buffers are prepared from weak acids and their conjugate bases 3. When small amounts of acid/base are added, the pH remains mostly unchanged because of LeChatelier’s Principle a. The presence of a weak acid is capable of neutralizing small amounts of added base b. The presence of the conjugate base is capable of neutralizing small amounts of added acid 4. Review example in text (Acetic acid and acetate ion buffer system) B. Buffers and pH 1. Buffers are most resistant to pH changes when the pH equals the pKa of the weak acid 2. Buffers are effective when the pH is within one unit of the pKa (i.e. pH = pKa ± 1) *See Sample Problem 7.12 and Practice Problem 7.12 BIOCHEMISTRY LINK The Henderson-Hasselbalch Equation XI. Maintaining the pH of Blood Serum (Section 7.11) A. Blood Serum Overview 1. pH of the blood serum normally falls between 7.35 and 7.45 7-4
2. 3. 4. 5.
If the pH moves out of this range then enzymes (biochemical catalysts) will not work correctly Enzymes are proteins that can be affected by pH Examine Figure 7.12 Maintaining the appropriate blood serum pH involves acids produced by metabolism (i.e. fatty acids, lactic acid, phosphoric acid, ketone bodies, etc.) 6. CO2 produced by cells is another significant source of acid B. Maintaining Blood Serum pH Mechanisms 1. CO2 is carried in the blood as dissolved gaseous CO2 or H2CO3 or HCO32. Review equations in text (H2O + CO2 & H2CO3 + H2O) 3. The body deals with these acids in three ways: a. Carbonic acid/bicarbonate buffer system b. Respiration c. Action of the kidneys 4. Review examples in text (Carbonic acid and dihydrogen phosphate buffer systems) 5. If the pH of the blood serum becomes too acidic, the brain signals for faster and deeper breathing a. This leads to more CO2 being exhaled and a drop in its partial pressure b. A loss in CO2 decreases the formation of H3O+ making the blood serum less acidic 6. Review example in text (Exercise and the carbonic acid/bicarbonate buffer system) 7. If the pH of the blood serum becomes too basic, the brain signals for slower and shallower breathing a. This leads to the production of more H2CO3 through an increase in CO2 b. More H2CO3 causes more H3O+ making the blood serum less basic 8. The kidneys release HCO3- and remove H3O+ if the serum is too acidic 9. Examine Figure 7.13 C. Acid/Base Disorders 1. Even with multiple controls (i.e. buffers systems, breathing, and kidneys), it is still possible for the pH of the blood serum to fall outside the normal range 2. Acidosis is the condition in which the blood serum has a low pH a. Mild acidosis can result in light-headedness b. Severe acidosis can lead to coma or death c. Review examples in text (Respiratory and metabolic acidosis) 3. Alkalosis is the condition in which the blood serum has a high pH a. Alkalosis symptoms range from headaches, nervousness, cramps, to convulsions and death b. Review examples in text (Respiratory and metabolic alkalosis) 4. See Table 7.7 for acid/base disorders DID YOU KNOW? Altitude alkalosis *See Sample Problem 7.13 and Practice Problem 7.13
Chapter Summary Section 7.1 Some compounds can be classified as acids or bases. Acids have a sour taste while bases taste bitter. Acids also dissolve some metals and turn litmus pink. On the other hand, bases feel slippery and turn litmus blue. Many acids and bases will cause extensive tissue damage and can be fatal if ingested. Section 7.2 Svante Arrhenius identified acids as substances that produce H+ in water while bases produce OH- in water. In the Bronsted-Lowry definition, acids release H+ and bases accept H+. Compounds that differ only in the presence or absence of H+ are called conjugates. Water is amphoteric, which means it can act as an acid or base depending on what it reacts with.
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Section 7.3 Reactions can be reversible, which means the reactant(s) can be converted into product(s) (forward reaction) and the product(s) can be converted back into reactant(s) (reverse reaction). Equilibrium is the point where the rate of the forward reaction and the rate of the reverse reaction are equal. For the chemical equation: aA + bB → cC + dD, an equilibrium constants (Keq) can be calculated from the the expression: Keq =[C]c[D]d/[A]a[B]b. The brackets represent concentration, and solids and solvents are not included in the expression. The acidity constant (Ka) is the equilibrium constant for an acid reaction. When Ka>1, the concentration of the reactant(s) is greater than the product(s). When Ka<1, the concentration of the product(s) is greater than the reactant(s). The concentrations of the reactant(s) and products are the same when Ka=1. Section 7.4 Equilibrium conditions can easily be disturbed by changes in conditions. LeChatelier’s principle states that when a reversible reaction is pushed out of equilibrium, the reaction responds to reestablish the equilibrium. When the concentration of a reactant or product is varied, the reaction shifts to reestablish equilibrium. In reactions that involve gases, if the pressure is increased, then the system will shift to decrease the total number of moles of gas that are present. Changes in temperature can also affect equilibrium. Heat can conveniently be treated as a reactant or product. Catalysts have no effect on equilibrium. Section 7.5 Water molecules are amphoteric and thus can react with other molecules in an acid-base reaction. Water spontaneously undergoes ionization according to the equation: 2H2O(l) → H3O+(aq) + OH-(aq). The equilibrium constant (Kw) for this reaction is equal to 1.0 x10-14. The concentration of H3O+ or OHcan be calculated using Kw. Section 7.6 The concentration of H3O+ determines if a solution is acidic, basic, or neutral. Scientist use pH, which is a logarithmic scale in which a one-unit change in pH is equivalent to a ten-fold change in [H3O+]. A solution is acidic when pH is less than 7, basic when pH is greater than 7, and neutral when the pH equals 7. pH can be measured with a pH meter or chemicals called indicators. Section 7.7 Strong acids almost completely ionize in water; and the stronger the acid the more H3O+ it produces, thus the lower its pH is. On the other hand, weak acids ionize much less in water. pKa (-log Ka) can be used to indicate the strength of an acid: the lower the pKa, the stronger the acid. Strong bases dissociate completely in water; and the stronger the base the more OH- it produces, thus the higher its pH is. Some weak bases have extremely low solubility in water therefore the [OH-] remains low. Other weak bases are simply poor H+ acceptors. Acid strength affects the strength of its conjugate base: the stronger an acid, the weaker its conjugate base is. Section 7.8 A neutralization reaction is when an acid and base react to form water and a salt. If proper amounts of an acid and base are combined, then they will cancel each other out to produce a neutral solution. A titration is a technique that can be used to determine the concentration of an acid or base solution. Section 7.9 Some compounds important to living things are acids, and pH determines if they are found in acid or conjugate base form. Ka or pKa and LeChatelier’s Principle can be used to predict how changing the pH affects the equilibrium. When the pH = pKa, the concentration of acid equals the concentration of its conjugate base. When the pH < pKa, the concentration of the acid is higher than that of its conjugate base. Likewise, when the pH > pKa the concentration of the conjugate base is higher than that of its acid. 7-6
Section 7.10 A solution that resists changes in pH when small amounts of acid or base are added is called a buffer. Buffers are prepared from weak acids and their conjugate bases. When small amounts of acid/base are added, the pH remains mostly unchanged. Buffers are most resistant to pH changes when the pH equals the pKa of the weak acid, but they are effective when the pH is within one unit of the pKa. Section 7.11 If the pH of the blood serum moves out of the range 7.35 to 7.45, then enzymes (protein catalysts) will not work correctly. Metabolism and CO2 are two significant sources of acid in the body. The body maintains pH by a carbonic acid/bicarbonate buffer system, respiration, and the action of the kidneys. Even with these multiple controls, it is still possible for the pH of the blood serum to fall outside the normal range. Acidosis is the condition in which the blood serum has a low pH while alkalosis is the condition in which the blood serum has a high pH.
Lecture Suggestions Sections 7.1 and 7.2 Start class by asking students what are some acids they know of. As students respond, briefly mention something about each acid (common use, source, etc.). Then see if students know of any bases. If they don’t, give examples of common basic substances (ammonia, antacids, etc.) Next ask them what they know about acids and bases (i.e. properties). Identify any misconceptions that arise (i.e. many students think all acids are poisonous and many don’t know that some bases can be just as caustic as acids). Sum up the discussion by mentioning some general properties of acids and bases (taste, litmus, etc.). You may also want to write the formula of a common acid such acetic acid (vinegar) on the board and circle the H that is released as H+. You can also do this with a base such as acid magnesium hydroxide (antacid) and circle the OH that is released as OH-. Sections 7.3 and 7.4 For these sections use the teeter totter (see saw) example. Explain that if four children are on each side of the teeter totter and the weight is the same, then it will be balanced in the air. Relate this to a reaction that is in equilibrium. Next, say the two children get off the right side (product side). Ask them what the other children will have to do to rebalance the teeter totter. They will of course say one child must move to the other side. Relate this to LeChatelier’s principle: a shift in equilibrium due to a decrease in product concentration. Then ask students what would need to be done if four children jumped on the left side (reactants side). Hopefully they will say two children must move sides to keep it balanced. Again, relate this to LeChatelier’s principle: a shift in equilibrium due to an increase in reactant concentration. Sections 7.5, 7.6, 7.7, and 7.8 For this lecture demonstrate with pH paper (litmus, hydrion, etc.) that distilled water in a beaker is neutral (freshly boiled distilled water may be required). Next, add half of the beaker of water to another beaker and then add some drops of HCl. Test the solution with the paper to show it is acidic. Write the reactants on the board and have students predict the products, balance it, and identify the acid, base, conjugate acid, and conjugate base. Next add a very small piece (about half the size of a small pea) of elemental sodium to the original beaker. As the sodium piece vigorously moves around the water, review the concept that sodium has 1 valence electron and is reacting (losing that electron) to achieve an octet. Test the solution with the paper to show it is basic (you may want to mention that this is how group 1A got the name alkali metals). Again write the reactants on the board and have students predict the products, balance it, and identify the acid, base, conjugate acid, and conjugate base. You can also mention that hypothetically, if concentrated HCl(aq) (caustic acid) was mixed with concentrated NaOH(aq) (caustic base) in the correct proportions, then they would neutralize each other and be harmless. Tell students it could be touched or even drank—though salt water doesn’t taste very good (remind students that this is only hypothetically and should never be attempted). 7-7
Sections 7.9, 7.10, and 7.11 Demonstrate in this lecture that when CO2 is dissolved in water, the solution becomes acidic. In a test tube, add distilled water and the indicator bromothymol blue. Tell students that the solution is neutral (you can verify this for them with pH paper). Ask for a student volunteer, and have him/her blow through a straw into the solution in the test tube. As his/her CO2 becomes dissolved, the neutral solution will become acidic due to the formation of carbonic acid. This acid will react with the indicator turning it from blue to yellow (you can again verify it is acidic with pH paper). Relate this to how the body uses the carbonic acid/bicarbonate buffer system, respiration, and kidney action to control the pH of the blood serum.
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Chapter 8 Organic Reactions 1: Hydrocarbons, Carboxylic Acid, Amines, and Related Compounds Outline Notes I. Alkanes (Section 8.1) A. Hydrocarbons 1. Recall alkanes consist only of C and H atoms connected by single bonds (i.e. methane: CH4) 2. Normal alkanes have C chains that are unbranched 3. Examine Figure 8.1 4. Hydrocarbons contain only C and H atoms and include the following organic groups i. Alkanes have only single bonds ii. Alkenes contain at least one carbon to carbon double bond iii. Alkynes contain at least one carbon to carbon triple bond iv. Aromatics contain a ring of alternating single and double bonds 5. Each C atom in an alkane molecule has four covalent bonds in a tetrahedral arrangement 6. Alkanes contain only nonpolar covalent bonds, therefore they are nonpolar molecules that are attracted to one another only by London forces 7. The more carbon atoms in a normal alkane, the higher its boiling point 8. See Table 8.1 for properties of selected hydrocarbons 9. Boiling point increases with the length of the carbon skeleton because there is more surface area, thus the London forces holding the molecules together are much stronger i. Boiling involves pulling molecules in the liquid phase apart to move them to the gas phase ii. An increase in London forces is also associated with a rise in melting point iii. At room temperature the first four alkanes exist as gases (boiling points are less than 25°C) iv. Alkanes that are 5 to 17 C atoms are liquids at room temperature v. Alkanes that are more than 17 C atoms are solids at room temperature B. IUPAC Nomenclature 1. The International Union of Pure and Applied Chemistry (IUPAC) has devised a set of rules that are used to name organic compounds 2. The parent chain is the longest continuous chain of C atoms in a molecule 3. Substituents are atoms or groups of atoms attached to the parent chain 4. In alkanes the substituents are called alkyl groups, which only contain C and H atoms 5. IUPAC Rules for naming alkanes follow: a. Name the parent chain by combing a numbering prefix (C atoms) with “ane” for alkane b. See Table 8.2 for IUPAC prefixes c. Name alkyl groups attached to the parent chain d. See Table 8.3 for formulas and names of alkyl groups e. Determine the point of attachment of alkyl groups to the parent chain; the parent chain is numbered from the end nearest the first substituent alkyl group f. Examine Figure 8.2 g. Construct the name by placing the alkyl groups in alphabetical order and specifying their position numbers (locants), followed by the name of the parent chain i. The prefixes di, tri, tetra, etc. are used if two or more identical substituents are present ii. The labels di, tri, tetra as well as sec and tert are ignored when alphabetizing substituents iii. Each substituent is denoted by both a name and a number (locant) *See Sample Problem 8.1 and Practice Problem 8.1 C. Representing Structures 8-1
1. Molecules can be drawn using a combination of line-bond and condensed structural formulas 2. Review examples in text (Alkane condensed formulas) 3. Molecular structures can also be represented using skeletal structures (C and H atoms aren’t shown and covalent bonds are lines) 4. Review examples in text (Skeletal structures of alkanes in Figure 8.2) II. Constitutional Isomers (Section 8.2) A. Constitutional Isomers Overview 1. Constitutional isomers are molecules that have the same molecular formula but different atomic connections (i.e. butane and 2-methylpropane) 2. As the number of atoms in a molecular formula increases, so does the number of possible constitutional isomers (i.e. C8H18 has 18 and C30H62 has 4,111,846,763) 3. Examine Figure 8.3 4. Structures can appear to be different but they may represent the same molecule 5. Review example in text (2-methylhexane structures) 6. Identical molecules have the same name where constitutional isomers have different names *See Sample Problem 8.2 and Practice Problem 8.2 B. Sources and Applications of Alkanes 1. Natural gas used to heat homes is mostly methane 2. Propane is used as a fuel for camp stoves and barbecues 3. Butane is used in disposable lighters 4. Petroleum (crude oil) is a mixture of hydrocarbons and is the source of most other alkanes 5. Examine Figure 8.4 6. Gasoline is a mixture of alkanes having 5-12 C atoms (normal alkanes as well as branched alkanes—constitutional isomers) 7. Diesel fuel is a mixture of alkanes having 12-18 C atoms 8. Motor oil is a mixture of alkanes with more than 15 C atoms 9. Asphalt is a mixture of alkanes with more than 35 C atoms III. Conformations (Section 8.3) A. Conformations Overview 1. The 3D shape of a molecule can affect many biochemical processes (i.e. the transport of certain compounds across cell membranes and proper functioning of proteins) 2. Rotation about single bonds allows most molecules to assume a number of different 3D shapes 3. Conformations are the shapes that a molecule can take because of bond rotations B. Conformation Properties 1. Conformations have the same molecular formula and the same atomic connections 2. Conformations have different 3D shapes and can be interchanged by the rotation of single bonds 3. Switching from one conformation to another only involves bond rotation, never bond breaking 4. Examine Figure 8.5 IV. Cycloalkanes (Section 8.4) A. Cycloalkanes Overview 1. Cycloalkanes are carbon atoms joined into rings 2. Examine Figure 8.6 3. Cycloalkanes are nonpolar molecules only attracted to one another by London forces 4. For simplicity, cycloalkanes are usually drawn using skeletal structures, although side views are useful when considering the orientation of their substituents 5. Cycloalkanes are named using the following guidelines a. The ring is usually designated as the parent and “cyclo” is attached to the alkane name b. When only one group is attached to the ring, no locant is necessary in naming the compound c. When the ring has more than one group attached, it is numbered to give the lowest possible number scheme 6. Examine Figure 8.7 8-2
B. Geometric Isomers 1. Limited rotation of C-C single bonds in cycloalkanes allows for the existence of stereoisomers 2. Stereoisomers have the same molecular formula and same atomic connections 3. Stereoisomers do not have the same 3D shape and are only interchanged by breaking bonds 4. Thus, stereoisomers (different molecules) are not conformations (same molecules) 5. Geometric isomers are stereoisomers that have restricted bond rotation 6. Examine Figure 8.8 7. Review examples in text (Butane and cyclohexane structures) 8. Geometric isomers come in pairs: cis and trans a. Cis have both groups on the same face of the ring b. Trans have groups on opposite faces of the ring *See Sample Problem 8.3 and Practice Problem 8.3 V. Alkenes, Alkynes, and Aromatic Compounds (Section 8.5) A. Unsaturated Hydrocarbons 1. Alkanes and cycloalkanes are saturated hydrocarbons (only contain single bonds) 2. Unsaturated hydrocarbons have a carbon to carbon double or triple bond functional group 3. Alkenes (double bond) have a trigonal planar shape around in each C atom 4. Alkynes (triple bond) have a linear shape around each C atom 5. Review example in text (Ethylene and acetylene structures) 6. Examine Figure 8.9 7. Aromatic hydrocarbons have a six-carbon atom ring structure called benzene 8. Review example in text (Benzene structures) 9. Aromatics compounds do not behave as alkenes, therefore they are often drawn with a circle to replace the double bonds 10. Unsaturated hydrocarbons (alkenes, alkynes, and aromatics) are nonpolar molecules only attracted to one another by London forces B. Naming Alkenes, Alkynes, and Aromatics 1. In alkenes/alkynes, the parent chain is the longest chain that contains the double/triple bond 2. The position of the double/triple bond is indicated with a number (locant), which is the lowest possible numbered carbon involved in the double or triple bond 3. Alkenes end with “ene” and alkynes end with “yne” 4. The same rules for alkanes apply for substituent groups 5. Examine Figure 8.10 6. Some organic compounds are known by common names that do not follow IUPAC rules 7. Benzene is an accepted IUPAC name so it is used as the parent ring 8. When only one group is attached to the benzene ring, the number is not specified 9. When there is more than one group present, the ring is numbered to give the groups the lowest number scheme possible 10. When there are only two groups on the ring the following designations can be used: a. Ortho (o) has a 1,2 arrangement b. Meta (m) has a 1,3 arrangement c. Para (p) has a 1,4 arrangement 11. Polycyclic aromatic hydrocarbons (PAHs) contain benzene rings that are fused together 12. Examine Figure 8.11 13. Many PAHs are formed from combustion and many are known to cause cancer (i.e. tobacco smoke) C. Geometric Isomers 1. Double bonds cannot rotate freely like single bonds 2. Because of this restricted rotation, alkenes can exist as cis and trans stereoisomers 3. Examine Figure 8.12 4. Review example in text (cis alkene vs. trans alkene structures) 8-3
5. Not all alkenes are found as cis and trans isomers (geometric isomers) 6. If either double bonded C atom carries two identical groups, then no geometric isomerism exists 7. Alkynes have a linear shape, therefore they do not exist as cis and trans isomers *See Sample Problem 8.4 and Practice Problem 8.4 VI. Reactions of Hydrocarbons (Section 8.6) A. Alkane Reactions 1. In alkane combustion, the alkane reacts with O2(g) to produce CO2(g) and water a. Combustion is accompanied by the rapid release of heat (i.e. gas furnace) b. Review examples in text (Methane and ethane combustion) c. Many other organic compounds can also be combusted (i.e. acetylene torch) d. Review examples in text (Acetylene and ethanol combustion) 2. Halogenation reactions occur when a Cl or Br atom replaces a H atom in an alkane a. Light is needed to initiate halogenation reactions b. Cl2 or Br2 are used as reactants to produce alkyl halides and either HCl or HBr c. Halogenation is a substitution reaction in which one group replaces another d. Review examples in text (Halogenation reaction equations) e. Halogenation may yield more than one constitutional isomer product f. Review examples in text (Halogenation reaction equations producing isomers) g. Only H atoms can be replaced in halogenation reactions h. Review examples in text (Halogenation reactions) i. Halogenation of alkanes is important for many reasons: i. Alkyl halides are used as insecticides, refrigerants, and anesthetics ii. Alkyl halides can be used in the production of many other compounds iii. Some halogenated compounds are found in nature j. Examine Figure 8.13 *See Sample Problem 8.5 and Practice Problem 8.5 B. Alkene and Alkyne Reactions 1. In catalytic hydrogenation, alkenes can be reduced into alkanes by H2(g) and Pt a. Review example in text (Alkene hydrogenation reaction equation) b. Alkynes can be hydrogenated but they require two moles of H2(g) c. Review example in text (Alkyne hydrogenation reaction equation) 2. Hydration of alkenes produces alcohols a. Water is the reactant and an acid (H+) catalyst is required b. Review example in text (Alkene hydration reaction equation) c. If hydration is carried out on asymmetric alkenes, two different products are possible d. Examine Figure 8.14 e. Markovnikov’s rule states that the major product (greater amount) is the one in which the H atom has been added to the C atom that already had more H atoms *See Sample Problem 8.6 and Practice Problem 8.6 C. Aromatic Compound Reactions 1. Aromatic substitution reactions involve a group replacing an H atom attached to the benzene ring a. In aromatic halogenation, a halogen replaces an H atom b. An Fe catalyst is required rather than light c. Review examples in text (Benzene halogenation reaction equations) d. Depending on the structure, more than one product may be obtained e. Review example in text (Toluene halogenation products) *See Sample Problem 8.7 and Practice Problem 8.7 2. Halogenated aromatic compound have a wide range of uses a. Examine Figure 8.15 b. Examine Figure 8.16 8-4
VII. Carboxylic Acids (Section 8.7) A. Carboxylic Acid Overview 1. Recall carboxylic acids contain a carboxyl group (-OH attached to C=O) 2. To name carboxylic acids, the parent chain is the longest chain containing the carboxyl group a. Numbering begins at the carboxyl group and substituent groups are numbered accordingly b. The final “e” of the corresponding hydrocarbon is dropped and the suffix “oic acid” is added 3. Examine Figure 8.17 4. Some carboxylic acid have common names: methanoic acid is commonly called formic acid and ethanoic acid is commonly called acetic acid 5. In biochemistry, some carboxylic acids contain a ketone group at the alpha (α) or beta (β) C atoms a. An α carbon atom is directly attached to the C atom in the carboxyl group b. A β carbon atom is two C atoms removed from the C atom in the carboxyl group c. Review example in text (α and β keto acids) *See Sample Problem 8.8 and Practice Problem 8.8 B. Carboxylic Acid Properties 1. Carboxylic acids have high boiling points compared to other organic compounds with similar molecular weights 2. This is caused by their ability to form hydrogen bonds with one another 3. Review example in text (Carboxylic acids forming hydrogen bonds) 4. Like all other organic compounds, the more carbons, the higher the boiling point (greater surface area leads to increased London forces) 5. Small carboxylic acids are soluble in water because they can hydrogen bond with water; however, increasing the length of the carbon skeleton leads to a reduction in water solubility 6. See Table 8.4 for physical properties of some carboxylic acids 7. Carboxylic acid that contain 3-10 C atoms have unpleasant odors (i.e. butanoic acid) VIII. Phenols (Section 8.8) A. Phenol Overview 1. Members of the phenol family have an –OH attached to an aromatic ring 2. When substituents are added the carbon carrying the –OH is designated carbon 1 3. The prefixes ortho (o), meta (m), and para (p) are used when other groups are attached to the ring 4. Examine Figure 8.18a B. Phenols in Biomolecules 1. Catechol, resorcinol, and hydroquinone are phenols that are common in biomolecules 2. See Table 8.5 for physical properties of selected phenols 3. Urushiol is present in poison ivy/oak and contributes to the itching and burning symptoms 4. Catechin is used for dyeing/tanning and is present in mahogany 5. Gentistic acid is used as an anti-rheumatism drug 6. Examine Figure 8.18b 7. Phenols have relatively high boiling points and they are somewhat water soluble because of their ability to form hydrogen bonds HEALTH LINK Chili Pepper Painkiller (Examine Figure 8.19) *See Sample Problem 8.9 and Practice Problem 8.9 IX. Carboxylic Acids and Phenols as Weak Organic Acids (Section 8.9) A. Carboxylic Acids as Weak Acids 1. Carboxylic acids can donate H+, therefore they are acids 2. Review example in text (Acetic acid in water) 3. Due to their small Ka, carboxylic acids are weak acids 4. The pKa for most carboxylic acids is about 5; therefore at a physiological pH of 7, they exist as carboxylate ions rather than carboxylic acids 5. Fatty acids are hydrophobic, however at physiological pH they exist as amphipathic ions 6. Review example in text (Palmitic acid and palmitate ion structures) 8-5
7. In general, carboxylate ions with less than 12 C atoms are hydrophilic while those with 12 or more C atoms are amphipathic 8. Carboxylate ions are named by removing the “ic acid” and replacing it with “ate ion” 9. Review examples in text (Propanoic acid and 2-methylbutanoic acid) B. Phenols as Weak Acids 1. Phenols are considerably weaker acids than carboxylic acids (much lower Ka’s) 2. Review example in text (Phenol in water) 3. The pKa for most phenols is about 10; therefore at a physiological pH of 7, they exist in their acidic form (phenol form) rather than their conjugate base form (phenoxide ion form) 4. The conjugate base of a phenol is named by dropping the “ol” and adding “oxide ion” 5. While both carboxylic acids and phenols contain an –OH, alcohols, which also contain an –OH, are not acidic (when –OH is attached to an alkane-type carbon it is not acidic) C. Carboxylate Salts and Phenoxide Salts 1. Strong bases are more effective than water at removing H+ from carboxylic acids 2. Examine Figure 8.20 3. When carboxylic acids react with strong bases, carboxylate salts are formed 4. Some carboxylate salts are commonly used as food preservatives (i.e. sodium benzoate) 5. Examine Figure 8.21 6. Carboxylate salts are usually more water-soluble than their carboxylic acid counterparts 7. When phenols react with strong bases, phenoxide salts are formed (more water soluble) 8. Review example in text (2.4-dinitrophenol solubility) *See Sample Problem 8.10 and Practice Problem 8.10 X. Preparing Esters (Section 8.10) A. Ester Synthesis 1. Esters are composed of a carboxylic acid residue and an alcohol residue 2. Residues are part of a reactant molecule that remains incorporated into a product 3. An ester is formed when a carboxylic acid and an alcohol react in the presence of an H+ catalyst 4. This double replacement reaction also produces a water molecule 5. Examine Figure 8.22 6. In living things, esters are prepared from molecules other than carboxylic acids 7. Esters are names by first listing the name of the alkyl group from the alcohol 8. The second part of the name comes from the carboxylic acid name: the “ic” acid is dropped and replaced with “ate” 9. Examine Figure 8.23 DID YOU KNOW? Juicy Fruit *See Sample Problem 8.11 and Practice Problem 8.11 10. Hydrogen bonds do not form between esters, therefore only dipole-dipole and London forces hold the molecules together 11. Esters have a lower boiling point than carboxylic acids of similar size B. Ester Hydrolysis 1. Ester formation is a reversible process 2. In the presence of H+, esters can react with water to produce carboxylic acids and alcohols 3. Le Chatelier’s principle can be applied to favor the forward reaction (ester synthesis) or the reverse reaction (ester hydrolysis) 4. Excess carboxylic acid and alcohol is used to favor the forward reaction while excess ester and water is used to favor the reverse reaction 5. Examine Figure 8.24 6. Ester hydrolysis can take place under acidic or basic conditions; basic conditions convert the carboxylic acid product into a carboxylate salt HEALTH LINK Alpha Hydroxy Acids (See Table 8.6) XI. Amines (Section 8.11) 8-6
A. Amine Classification and Nomenclature 1. Recall amines have an N atom directly attached to one or more C atoms 2. Primary (1°) amines have only one C atom attached to the N atom (plus two H atoms) 3. Secondary (2°) amines have two C atoms attached to the N atom (plus one H atom) 4. Tertiary (3°) amines have three C atoms attached to the N atom (no H atoms) 5. Quaternary (4°) ammonium ions have four carbon atoms attached to the N atom and are aminerelated derivatives of ammonium ion (NH4+) 6. Examine Figure 8.25 7. To name amines, the longest chain of C atoms attached to the N atom is the parent chain 8. The parent chain is numbered from the end nearest the N atom 9. The hydrocarbon name is used but the “e” is dropped and replaced with “amine” 10. Numbers are used to designate the location of the parent chain amine 11. N- is used as a locant to designate substituent groups attached to the N atom 12. Review examples in text (Naming 1°, 2°, and 3° amines) 13. Small amines can be named with common names: the alkyl groups attached to the nitrogen are alphabetized and the word amine is added to the end (i.e. methylethylamine) 14. Quaternary ammonium ions are named the same way but “ammonium ion” is used instead *See Sample Problem 8.12 and Practice Problem 8.12 B. Heterocyclic Amines 1. In heterocyclic amines, one or more N atoms are part of the ring 2. Pyridine, pyrimidine, and purine are very important heterocyclic amines (i.e. DNA) 3. Examine Figure 8.26 4. Review examples in text (Nicotine and melamine cases) C. Physical Properties of Amines 1. Amines have polar covalent bonds (C-N) and a pyramidal shape about the N atom 2. Only 1° and 2° amines can hydrogen bond with like amines (3° cannot since they do not have a H attached to the N atom) 3. Review example in text (1° and 2° amines hydrogen bonding) 4. 1° and 2° amines have higher boiling points than 3° amines of similar molecular weights 5. See Table 8.7 for physical properties of some amines 6. Review example in text (Trimethylamine vs. propylamine boiling points) 7. Small amines are highly water soluble because they can hydrogen bond with water 8. Many amines have a strong unpleasant odor (i.e. putrescine and cadaverine) HEALTH LINK Adrenaline and Related Compounds (Examine Figure 8.27) XII. Amines as Weak Organic Bases (Section 8.12) A. Amines as Weak Bases 1. Amines are weak bases meaning that they accept H+ 2. When amines are added to water, an H+ from water is transferred to the amine producing the conjugate acid of the amine (ammonium-type ion) and a hydroxide ion (makes solution basic) 2. Review example in text (Methylamine in water) 3. Examine Figure 8.28 4. Amines are weak bases with pKa values between 9-11; therefore, at pH less than 9, amines are found as their conjugate acids while at pH above 11 they exist as amines 5. Amine conjugate acids are named like quaternary ammonium cations (i.e. CH3NH3+ is named methylammonium ion) B. Ammonium Salts 1. Amines can accept H+ from acid (i.e. methylamine reacts with HCl to produce a salt and water) 2. Salts formed from amines are important to the drug industry because they are more water soluble than the original amine, making them easier to administer 3. Review example in text (Pseudoephedrine reaction with HCl and with H2SO4) 8-7
4. The salts of amine-containing drugs are named by combining the name of the drug with the name of the anion (i.e. pseudoephedrine hydrochloride and pseudoephedrine sulfate) 5. Quaternary ammonium ions are not basic but they can exist as salts (i.e. muscarine iodide) *See Sample Problem 8.13 and Practice Problem 8.13 XIII. Amides (Section 8.13) A. Amides Overview 1. A salt is formed when carboxylic acids react with ammonia or with a 1° or 2° amines 2. If this salt is heated, an amide is produced 3. Amides have a nitrogen atom directly attached to a carbonyl carbon, thus they contain a carboxylic acid residue and an ammonia/amine residue (review structures in text) 4. Examine Figure 8.29 5. There are alternative ways of producing amides (i.e. proteins synthesis) B. Amide Nomenclature and Properties 1. Amide names are based on the parent carboxylic acid: the “oic acid” is replaced by “amide” 2. When an amine residue is present, the substituents attached to the nitrogen atom are identified using the N locant like in amine nomenclature 3. Examine Figure 8.30 4. Amides tend to have high boiling points because of their ability to hydrogen bond with one another 5. Amides are not bases like amines (the acidic carboxylic acids has neutralized the basic amine) 6. Like esters, amides can be hydrolyzed with water and an H+ catalyst back into carboxylic acids and conjugate acids of ammonia/amines *See Sample Problem 8.14 and Practice Problem 8.14 DID YOU KNOW? Aspartame HEALTH LINK Biofilms (Examine Figure 8.31 and 8.32) BIOCHEMISTRY LINK A Cure for Fleas (Examine Figure 8.33)
Chapter Summary Section 8.1 Alkanes, alkenes, alkynes, and aromatics are collectively classified as hydrocarbons. Alkanes can have normal carbon chains or branched carbon chains. Alkanes contain nonpolar molecules that are attracted to one another only by London forces. As the number of carbon atoms in a normal alkane is increased, the boiling point is also increased. The International Union of Pure and Applied Chemistry (IUPAC) has devised a set of rules that can be used to name organic compounds. In IUPAC nomenclature, substituents such as alkyl groups are attached to a parent chain. In addition to names, organic molecules can be represented by skeletal structures or using a combination of line-bond and condensed structural formulas. Section 8.2 Constitutional isomers are molecules that have the same molecular formula, but different atomic connections. As the number of atoms in a molecular formula increases, so does the number of possible constitutional isomers. Even though structures can appear to be different, they may represent the same molecule. Identical molecules have the same name where constitutional isomers have different names. Alkanes come from various sources and have many applications. For example, methane is used to heat homes, propane is used as a fuel for camp stoves and barbecues, and butane is used in disposable lighters. Section 8.3 The 3D shape of a molecule can affect many biochemical processes. The shape of a molecule can be caused by rotation about single bonds. Conformations are the 3D shapes that molecules can take because of rotations. Conformations have the same molecular formula and the same atomic connections; 8-8
however, they have different 3D shapes and can be interchanged by the rotation of single bond. Switching from one conformation to another only involves bond rotation, never bond breaking. Section 8.4 When carbon atoms are joined into rings they form cycloalkanes. Cycloalkanes are nonpolar molecules only attracted to one another by London forces. Cycloalkanes are usually drawn using skeletal structures, and they are named by attaching “cyclo” to the alkane name. Limited rotation of C-C single bonds in cycloalkanes allows for the existence of stereoisomers, which have the same molecular formula and same atomic connection, but have different 3D shapes. Geometric isomers are stereoisomers that have restricted bond rotation. The two geometric isomers for cycloalkanes are cis (same ring face) and trans (opposite ring face). Section 8.5 Hydrocarbons can be saturated (alkanes and cycloalkanes) or unsaturated (alkenes, alkynes, and aromatics). Unsaturated hydrocarbons are nonpolar molecules that are only attracted to one another by London forces. In IUPAC nomenclature, alkenes are indicated by the suffix “ene,” alkynes by “yne,” and aromatics by “benzene.” Some smaller hydrocarbons are also known by common names, which do not follow IUPAC naming rules. Some compounds contain benzene rings that are fused together called polycyclic aromatic hydrocarbons (PAHs). Many PAHs are formed from combustion and many are known to cause cancer. Some alkenes can exist as geometric isomers because double bonds cannot rotate freely. Section 8.6 Hydrocarbons can undergo a number of chemical reactions. In alkane combustion, an alkane reacts with O2(g) to produce CO2(g), water, and heat. Alkanes can also be halogenated, which is when they react with Cl2 or Br2 to produce alkyl halides (substitution reaction). In catalytic hydrogenation, alkenes or alkynes are reduced into alkanes by H2(g) and Pt. Alkenes can also react with water to produce alcohols in a reaction called hydration. If hydration is carried out on an asymmetric alkene, two different products are possible. According to Markovnikov’s rule, the major product is the one in which the H atom has been added to the C atom that already had more H atoms. In aromatic halogenation, a halogen replaces an H atom. Halogenated aromatic compounds have a wide range of uses such as disinfectants. Section 8.7 Carboxylic acids contain a carboxyl group. In IUPAC nomenclature, carboxylic acids are indicated by the suffix “oic acid.” Some carboxylic acids have common names such as formic acid and acetic acid. In biochemistry, some carboxylic acids contain a ketone group at the alpha (α) or beta (β) C atoms. Carboxylic acids have high boiling points compared to other organic compounds with similar molecular weights because of their ability to form hydrogen bonds with one another. Small carboxylic acids are soluble in water and have unpleasant odors. Section 8.8 Members of the phenol family have an –OH attached to an aromatic ring. Phenols such as catechol, resorcinol, and hydroquinone are present in many biomolecules. Phenols have relatively high boiling points and they are somewhat water soluble because of their ability to form hydrogen bonds. Section 8.9 Carboxylic acids and phenols are weak acids because they can donate H+ and they have small Kas. The pKa for most carboxylic acids is about 5; therefore at a physiological pH of 7, they exist as carboxylate ions rather than carboxylic acids. In general, carboxylate ions with less than 12 C atoms are hydrophilic while those with 12 or more C atoms are amphipathic. In IUPAC nomenclature, carboxylate ions are indicated with the suffix “ate ion.” Phenols are considerably weaker acids than carboxylic acids. The pKa for most phenols is about 10; therefore, at a physiological pH of 7, they exist in their acidic form (phenol form) rather than their conjugate base form (phenoxide ion form). Strong bases are most 8-9
effective at removing H+ from carboxylic acids and phenols. When carboxylic acids react with strong bases, carboxylate salts are formed; moreover, when phenols react with strong bases, phenoxide salts are formed. Section 8.10 Esters are composed of a carboxylic acid residue and an alcohol residue. An ester is formed when a carboxylic acid and an alcohol react in the presence of an H+ catalyst. In IUPAC nomenclature, esters are identified with the suffix “ate.” Esters cannot hydrogen bond with like molecules, therefore their dipole-dipole and London forces cause them to have lower boiling points than carboxylic acids of similar size. In the presence of H+ and water, esters can be hydrolyzed to produce carboxylic acids and alcohols. Section 8.11 Amines have an N atom directly attached to one or more C atoms. Amines can be classified as primary (1°), secondary (2°), or tertiary (3°). Quaternary (4°) ammonium ions have four carbon atoms attached to the N atom and are amine-related derivatives of ammonium ion. In IUPAC nomenclature, amines are identified by the suffix “amine” and ammonium ions are named the same way but with the suffix “ammonium ion.” In heterocyclic amines, one or more N atoms are part of a ring. Amines are polar molecules and 1° and 2° amines can hydrogen bond with like amines. Small amines are highly water soluble because of their ability to hydrogen bond with water, and many have a strong unpleasant odor. Section 8.12 Amines are weak bases because they can accept H+ and they have pKa values between 9 and11. At pH less than 9, amines are found as their conjugate acids while at pH above 11, they exist as amines. Salts formed from amines are important to the drug industry. Amine salts are more water soluble than the original amine, which makes them easier to administer orally. The salts of amine-containing drugs are named by combining the name of the drug with the name of the anion. Section 8.13 A salt is formed when carboxylic acids react with ammonia or with a 1° or 2° amines. If this salt is heated, an amide is produced. Amides have a nitrogen atom directly attached to a carbonyl carbon. In IUPAC nomenclature, amides are identified with the suffix “amide.” Amides tend to have high boiling points because of their ability to hydrogen bond with one another. Amides are not bases like amines, but they can be hydrolyzed back into carboxylic acids and amine/ammonia conjugate acids.
Lecture Suggestions Sections 8.1, 8.2, 8.3, 8.4, 8.5, and 8.6 Using a molecular model set for organic chemistry, demonstrate to students the difference between conformations, constitutional isomers, and stereoisomers (geometric isomers). This activity will also allow you to practice IUPAC nomenclature. (Demonstrate the following.) • Construct a model of pentane (C5H12) and draw its expanded structural formula. Rearrange the carbon atoms to form all the possible chain isomers of C5H12. Draw the structural formulas for all of these constitutional isomers on the board. Name all the compounds using IUPAC rules. • Construct a model of 1,2-dibromocyclopentane (C5H8Br2) and draw its structural formula on the board. Rearrange the spatial geometry to get the cis isomer of the compound and draw the structural formula on the board. Rearrange the spatial geometry to get the trans isomer of the compound and draw the structural formula on the board. Name the compounds using IUPAC rules. Ask students why 1,1,2-tribromocyclopentane does not have geometric isomers. • Construct a model of 2-butene and draw its structural formula on the board. Rearrange the spatial geometry to get the cis isomer of the compound and draw the structural formula on the board. 8-10
Rearrange the spatial geometry to get the trans isomer of the compound and draw the structural formula on the board. Name the compounds using IUPAC nomenclature. Sections 8.7, 8.8, 8.9, and 8.10 For these sections, demonstrate the difference between a carboxylic acid and an ester. Draw the structure of butanoic acid on the board and ask students to name it using IUPAC rules. Pass around a sample of butanoic acid in a stoppered test tube. Instruct students to carefully remove the stopper and waft over the test tube to smell it. Students will obviously be disgusted by its odor as it smells like vomit, rancid food, and foot odor. Next, on the board add the reactant of ethanol to the butanoic acid. Have students predict the product and then name it using IUPAC rules. Check their work by writing the correct answer on the board. Next pass around the ester ethyl butanoate in a stoppered test tube. Again, instruct students to carefully remove the stopper and waft over the test tube to smell it. Students should like its odor as it smells like pineapples. When doing this, it is a good time to review the fact that when a chemical change occurs, the products do not have the same properties as the original reactants. You can also use the following carboxylic acid-ester examples: ethanoic acid + 3-methyl-1-butanol → 3-methyl-1-butyl ethanoate (Banana) propanoic acid + ethanol → ethyl propanoate (Butterscotch) ethanoic acid + 1-octanol → octyl ethanoate (Orange) *Safety Note: Use caution when working with carboxylic acids Sections 8.11, 8.12, and 8.13 For this lecture, demonstrate amide formation through the synthesis of nylon. Write the following on the board and explain (you will need to introduce the concept of polymerization): H2N(CH2)xNH2 + HOOC(CH2)yCOOH → [—NH(CH2)xNHCO(CH2)yCO—]n A Diamine A Dicarboxylic Acid A Polyamide Next, explain that when synthesizing amides (and ester), carboxylic acid are often replaced with other compounds that react better (if you desire, you can go into further detail using Le Chatelier’s principle). Write the reaction between 1,6-hexanediamine and sebacoyl chloride on the board: H2N(CH2)6NH2 + ClCO(CH2)8COCl → [—NH(CH2)6NHCO(CH2)8CO—]n 1,6-Hexanediamine Sebacoyl Chloride Nylon Next, in a well ventilated room, fill a 100mL beaker with approximately 40mL of 6% 1,6hexanediamine solution (food coloring can be added for a more dramatic effect). Then carefully pour 40mL of the 4% sebacoyl chloride down the side of the 100mL beaker making sure it does not mix with the 6% 1,6-hexanediamine solution. The clear sebacoyl chloride solution should sit on top of the colored 1,6-hexanediamine solution. At the interface, use a pair of forceps to grab the polyamide and gently pull. Pull the nylon slowly and wrap it around a stirring rod. You can then thoroughly rinse the strand of nylon with water and pass it around for students to touch. *Safety Note: A product of this reaction is HCl so the nylon should be thoroughly rinsed before handling
Handouts for Students 1) The IUPAC Nomenclature handout (next page) can be used as a reference sheet for students. However, students should learn/memorize the prefixes, suffixes, and substituent groups.
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Handout IUPAC Nomenclature Common Alkyl Group IUPAC Prefixes Number of Carbon Atoms 1 2 3 4 5 6 7 8 9 10
Prefix
Meth Eth Prop But Pent Hex Hept Oct Non Dec
Formula —CH3 —CH2CH3 —CH2CH2CH3 —CHCH3 ׀ CH3 —CH2CH2CH2CH3 —CH2CHCH3 ׀ CH3 —CHCH2CH3 ׀ CH3 CH3 ׀ —CHCH3 ׀ CH3
Name methyl ethyl propyl isopropyl butyl isobutyl sec-butyl (s-butyl) tert-butyl (t-butyl)
Common Functional Group Suffixes
Common Nonalkyl Group -F -Cl -Br -I -NO2
fluoro chloro bromo iodo nitro
alkane alkene alkyne aromatic carboxylic acid ester & carboxylate ion alcohol aldehyde ketone amine amide thiol 8-12
-ane -ene -yne -benzene -oic acid -ate -ol -al -one -amine -amide -thiol
Chapter 9 Organic Reactions 2: Alcohols, Ethers, Aldehydes and Ketones Outline Notes I. Alcohols, Ethers, and Related Compounds (Section 9.1) A. Functional Groups 1. Recall alcohols have an –OH attached to an alkane-type carbon (C–OH), and ethers have an O atom attached to two C atoms (C–O–C) 2. There are three sulfur-containing families related to alcohols and ethers (S and O have some similar properties because they are in the same group on the periodic table) a. Thiols contain an –SH attached to a C atom (C–SH) b. Sulfides contain an S atom attached to two C atoms (C–S–C) c. Disulfides contain two S atoms attached to two C atoms (C–S–S–C) B. Nomenclature 1. In alcohols, the parent chain is the longest chain that contains the –OH functional group 2. The position of the –OH is indicated with a number (locant), which is the lowest possible 3. The “e” is dropped on the corresponding hydrocarbon and “ol” is added 4. The same rules for alkanes apply for substituent groups 5. Examine Figure 9.1 6. Smaller alcohols are known by common names, which are obtained by combining the name of the alkyl group with the word alcohol (i.e. CH3CH2OH is ethyl alcohol) 7. Thiols are named the same as alcohols, however the ending “thiol” is added to the hydrocarbon name (i.e. CH3CH2SH is ethanethiol) 8. Common names of thiols consist of the name of the alkyl group plus the word mercaptan 9. In this text only, common names are used for ethers, sulfides, and disulfides 10. Review examples in text (Figure 9.1 names and structures) C. Alcohol Classification 1. Alcohols are categorized by the nature of the C atom that is bonded directly to the –OH group 2. In primary (1º) alcohols, the hydroxyl-carrying C is directly attached to only one other C atom 3. In secondary (2º) alcohols, the hydroxyl-carrying C is attached to two other C atoms 4. In tertiary (3º) alcohols, the hydroxyl-carrying C is attached to three other C atoms 5. Review examples in text (Classification of alcohols in Figure 9.1) D. Alcohol Properties 1. Alcohols have relatively high boiling points compared to hydrocarbons with similar weight 2. See Table 9.1 for physical properties of alcohols and related compounds 3. Review example in text (Methanol and ethane boiling points) 4. Alcohols have higher boiling points because they can hydrogen bond with similar molecules 5. Examine Figure 9.2 6. As with other organic families, the longer the alcohol C chain, the higher the boiling point (increased weight and London force interactions) 7. The boiling points of ethers, thiols, sulfides, and disulfides are lower than those of alcohols with similar molecular weights because they cannot hydrogen bond with like molecules 8. Ethers are slightly polar, but the dipole-dipole attractions are not strong enough to raise the boiling points much above those of similarly sized hydrocarbons *See Sample Problem 9.1 and Practice Problem 9.1 9. Small alcohol molecules are soluble in water because of the “like dissolves like” rule 9-1
10. Alcohols are able to hydrogen bond with water; however, large alcohols have a lower water solubility due to the nonpolar hydrocarbon portion 11. Ethers are less polar than alcohols and they form fewer hydrogen bonds with water; therefore, they are less soluble in water than alcohols 12. Many thiols, sulfides, and disulfides have unpleasant odors and have low water solubility (i.e. skunk spray smells unpleasant and does not wash off easily with water) DID YOU KNOW? Skunk Spray *See Sample Problem 9.2 and Practice Problem 9.2 II. Preparation (Section 9.2) A. Nucleophilic Substitution Overview 1. A nucleophilic substitution reaction involves a nucleophile, which is an electron-rich group, replacing a leaving group 2. A leaving group is an atom or group of atoms that is held to a C atom by a relatively weak bond 3. Chlorine and bromine are common leaving groups 4. Review example in text (Nucleophilic substitution of methyl chloride) 5. The key to understanding nucleophilic substitutions is recognizing the leaving group is always replaced by the nucleophile B. Alcohol, Ether, Thiol, and Disulfide Preparation 1. The following nucleophiles are used to produced these various functional groups a. OH- forms alcohols c. SH- forms thiols b. C-O form ethers d. C-S- forms disulfides 2. Examine Figure 9.3 3. Halogenation of alkanes (producing alkyl halides) can be coupled with nucleophilic substitution reactions 4. This allows alkanes to be converted into alcohols, thiols, etc. 5. Review examples in text (Preparation of methanol and ethanethiol) 6. It should be noted that when living things manufacture these compounds, they tend to use enzyme-catalyzed versions of the same reactions *See Sample Problem 9.3 and Practice Problem 9.3 III. Reactions (Section 9.3) A. Oxidation of Alcohols and Thiols 1. Recall an organic molecule has been oxidized if C atoms gain O and/or lose H 2. A common oxidizing agent in organic chemistry is potassium dichromate (K2Cr2O7) 3. When an alcohol is oxidized, a carbonyl group is formed because one H atom is removed from the –OH and one H atom is removed from the C atom carrying the –OH 4. Examine Figure 9.4 5. 1° alcohols (two H atoms attached to the alcohol carrying C atom) can be oxidized into aldehydes, which are immediately oxidized into carboxylic acids by K2Cr2O7 6. 2° alcohols (one H atom attached to the alcohol carrying C atom) can be oxidized into ketones 7. 3° alcohols (no H atoms attached to the alcohol carrying C atom) cannot be oxidized *See Sample Problem 9.4 and Practice Problem 9.4 8. NAD+ is an oxidizing agent used by living things and works in conjunction with certain enzymes 9. NAD+ assists in the oxidation of an alcohol by accepting one of its H atoms to become NADH 10. Examine Figure 9.5 11. The oxidation of thiols with the oxidizing agent I2 produces a disulfide 12. Review example in text (Ethanethiol oxidation with I2) B. Dehydration of Alcohols 1. Alcohols undergo dehydration in the presence of an H+ catalyst and heat 2. The –OH and –H from an adjacent C atom are lost to produce an alkene 3. When more than one alkene can be formed from an alcohol, the major product is the one produced by the removal of the H from the neighboring C atom that carries fewer H atoms 9-2
4. Examine Figure 9.6 *See Sample Problem 9.5 and Practice Problem 9.5 IV. Aldehydes and Ketones (Section 9.4) A. Functional Group 1. Recall ketones and aldehydes contain a carbonyl (C=O) group 2. In ketones the carbonyl C atom is attached to two other C atoms (2º) 3. In aldehydes the carbonyl C atom is attached to one other C atom and an H atom or to two H atoms B. Aldehyde and Ketone Nomenclature 1. In IUPAC nomenclature, the carbonyl group must be part of the parent chain, which is numbered from the end nearer this carbonyl group 2. The position of the carbonyl group is not specified in aldehydes because it is always in position 1 3. The position of the carbonyl group is specified in ketones unless the molecule is small enough that there is no question where the carbonyl is placed 4. The suffix “al” is used for aldehydes and the suffix “one” is used for ketones 5. Examine Figure 9.7 6. The common names of ketones are formed by placing the word ketone after the names of the alkyl groups attached to the carbonyl C atom (i.e. 2-propanone is dimethyl ketone or acetone) 7. The common names of aldehydes include the name aldehyde (i.e. methanol is formaldehyde) C. Aldehyde and Ketone Physical Properties 1. Aldehydes and ketones have much lower boiling points than alcohols with similar weight 2. Review example in text (Ethanol vs. ethanol boiling points) 3. See Table 9.2 for physical properties of aldehydes and ketones 4. The lower boiling points are due to the fact that alcohols can hydrogen bond with each other while aldehydes and ketones cannot (no H atom attached to the O atom) 5. Aldehydes and ketones can interact through London forces as well as dipole-dipole forces 6. The polarity of the carbonyl group allows aldehydes and ketones to hydrogen bond with water; therefore, small aldehydes and ketones are soluble in water 7. As with other organic families, aldehydes and ketones become less water soluble the bigger they get (more nonpolar CH2 groups) *See Sample Problem 9.6 and Practice Problem 9.6 V. Oxidation of Aldehydes (Section 9.5) A. Oxidation of Aldehydes Overview 1. Recall 1° alcohols can be oxidized into aldehydes, which are further oxidized in carboxylic acids 2. Examine Figure 9.8 3. Special reagents have been developed for oxidizing aldehydes without affecting alcohols 4. One of these, called Benedict’s reagent, uses Cu2+ as the oxidizing agent B. Benedict’s Reagent 1. Aldehydes can be oxidized into carboxylic acids by Benedict’s reagent 2. As the blue colored Benedict’s reagent oxidizes an aldehyde, it changes colors due to Cu2+ being reduced to Cu+ 3. The final color of the solution depends on the amount of aldehyde initially present 4. Examine Figure 9.9 5. Benedict’s reagent has been used to monitor diabetes by testing urine for the presence of glucose which is an aldehyde-containing sugar DID YOU KNOW? Treating Antifreeze Poisoning *See Sample Problem 9.7 and Practice Problem 9.7 HEALTH LINK Aldehyde Dehydrogenase (Examine Figure 9.10) VI. Reduction of Aldehydes (Section 9.6) A. Catalytic Hydrogenation 1. Recall alkenes can be hydrogenated into alkanes with H2 and a Pt catalyst 9-3
2. Hydrogenation is a reduction reaction because two H atoms are added across the double bond 3. Aldehydes and ketones can also be hydrogenated (reduced) into alcohols 4. When aldehydes/ketones are treated with H2 and a Pt catalyst, the carbon-oxygen double bond of the carbonyl is converted to a single bond forming an –OH group 5. Examine Figure 9.11 6. Aldehydes are hydrogenated into 1° alcohols while ketones are hydrogenated into 2° alcohols *See Sample Problem 9.8 and Practice Problem 9.8 B. Biochemical Hydrogenation 1. In the presence of the biochemical reducing agent NADH, enzymes in the body catalyze the reduction of aldehydes and ketones into alcohols 2. Review examples in text (HSDH catalysis of steroids) 3. Examine Figure 9.12 HEALTH LINK Protective Enzymes (Examine Figure 9.13) VII. Reactions of Alcohols with Aldehydes and Ketones (Section 9.7) A. Hemiacetals and Hemiketals 1. Alcohols (with H+) can undergo an addition reaction with the carbonyl of an aldehyde or ketone 2. Here the alcohol’s H atom attaches to the carbonyl O atom and the remainder of the alcohol attaches to the carbonyl C atom 3. The product that is formed from an aldehyde is called a hemiacetal, which contains a C atom attached to an –OH, –OC, and –H 4. The product that is formed from a ketone is called a hemiketal, which contains a C atom attached to an –OH, –OC, and –C 5. Molecules that contain both an alcohol and an aldehyde/ketone can form cyclic hemiacetals/hemiketals when the –OH reacts with a C=O in the same molecule 6. Examine Figure 9.14 B. Acetals and Ketals 1. When two alcohol molecules react with an aldehyde/ketone an acetal/ketal is formed 2. An acetal consists of a C atom that is attached to two –OC and an –H 3. A ketal consists of a C atom that is attached to two –OC and a –C 4. Examine Figure 9.15 5. Varying the alcohol used allows for the formation of a wide range of acetals/ketals from a given aldehyde/ketone 6. Acetals/ketals can also be directly produced from hemiacetals/hemiketals *See Sample Problem 9.9 and Practice Problem 9.9 HEALTH LINK Drugs in the Environment (Examine Figure 9.16)
Chapter Summary Section 9.1 Alcohols, ethers, thiols, sulfides, and disulfides are all organic functional groups containing either oxygen or sulfur atoms. In IUPAC nomenclature, alcohols are indicated by the suffix “ol.” Alcohols can be classified as primary (1º), secondary (2º), or tertiary (3º) depending on the number of C atoms attached to the alcohol carrying C atom. Alcohols have relatively high boiling points compared to hydrocarbons with similar weight because they can hydrogen bond with similar molecules. The boiling points of ethers, thiols, sulfides, and disulfides are lower because they cannot hydrogen bond with like molecules. Small alcohol molecules are soluble in water; and ethers are less soluble in water than alcohols. Thiols, sulfides, and disulfides have low water solubility and many have unpleasant odors. Section 9.2 A nucleophilic substitution reaction involves a nucleophile, which is an electron-rich group, replacing a leaving group. Chlorine and bromine are common leaving groups in organic chemistry. The OHnucleophile forms alcohols, C-O- forms ethers, SH- forms thiols, and C-S- forms disulfides. Alkanes can 9-4
be converted into alcohols, thiols, etc. when halogenation is coupled with nucleophilic substitution reactions. Section 9.3 When an alcohol is oxidized with an oxidizing agent such as K2Cr2O7, a carbonyl group is formed. Primary alcohols can be oxidized into aldehydes, which can further be oxidized into carboxylic acids. Secondary alcohols can be oxidized into ketones but tertiary alcohols cannot be oxidized. NAD+ is an oxidizing agent that assists enzymes in living things. Thiols can also be oxidized; however they are oxidized with I2 to produce disulfides. Alcohols can be dehydrated into alkenes. When more than one alkene can be formed from an alcohol, the major product is the one produced by the removal of the H atom from the neighboring C atom that carries fewer H atoms. Section 9.4 Aldehdyes and ketones contain a carbonyl (C=O) group. In IUPAC nomenclature, aldehydes are indicated by the suffix “al” and ketones by the suffix “one.” Aldehydes and ketones have much lower boiling points than alcohols with similar weight because they cannot hydrogen bond with similar molecules. The polarity of the carbonyl group allows aldehydes and ketones to hydrogen bond with water; therefore, small aldehydes and ketones are soluble in water. Section 9.5 Special reagents have been developed for oxidizing aldehydes without affecting alcohols. One of these reagents, Benedict’s reagent, uses Cu2+ as the oxidizing agent. As the blue colored Benedict’s reagent oxidizes an aldehyde solution, the solution changes colors. The final color of the solution depends on the amount of aldehyde initially present. Benedict’s reagent can be used to monitor diabetes by testing urine for the presence of glucose (an aldehyde-containing sugar). Section 9.6 Aldehydes and ketones can be hydrogenated (reduced) into alcohols. When aldehydes/ketones are treated with H2 and a Pt catalyst, the carbon-oxygen double bond is reduced to form an alcohol. Aldehydes can be hydrogenated into primary alcohols while ketones can be hydrogenated into secondary alcohols. In the presences of the biochemical reducing agent NADH, enzymes in the body catalyze the reduction of aldehydes and ketones into alcohols. Section 9.7 When alcohols are reacted with aldehydes or ketones in the presence of an H+ catalyst, the alcohol’s H atom attaches to the carbonyl O atom and the remainder of the alcohol attaches to the carbonyl C atom. The product formed from an aldehyde is called a hemiacetal while the product formed from a ketone is called a hemiketal. When two alcohol molecules react with an aldehyde or ketone, an acetal or ketal is formed. These reactions are very important in the chemistry of carbohydrates.
Lecture Suggestions Sections 9.1 and 9.2 Begin section 9.1 by asking students to draw the structure of propane. Have them calculate its molecular weight and list its physical state/phase at room temperature. Next, ask them to do the same for ethyl alcohol. If students don’t know the physical state of these two compounds, you can hint to them that propane is used in gas grills and ethyl alcohol is in alcoholic beverages. After you give the answers on the board, ask students why ethyl alcohol is a liquid at room temperature while propane is gas (especially since they have the similar molecular weights). Hopefully they will say something about the alcohol sticking together due to hydrogen bonding. You can then re-discuss the difference between London forces and hydrogen bonding and their effect on physical state. You can also re-explain the ball analogy/example mentioned in the lecture suggestions for section 4.3. 9-5
When discussing sulfur compounds, you can mention that while some have unpleasant odors (i.e. skunk spray), others have desirable ones (i.e. the odor/flavor of onions and garlic). You can also discuss why natural gas smells bad when methane itself has no odor (due to the addition of methanethiol and ethanethiol as a safety precaution). When discussing nucleophilic substitutions in section 9.2, you can re-explain the dance analogy/example mentioned in the lecture suggestions for section 5.2. Here the nucleophile “cuts in” on the leaving group’s “date.” Sections 9.3 and 9.4 Demonstrate how aldehydes can be oxidized and ketones cannot use the potassium dichromate test. Prepare a 15mL sample of a reagent by adding 10mL of 0.2M K2Cr2O7 to 5mL of concentrated H2SO4. Add 10 mL of acetone to one test tube and 10mL of formaldehyde to another (you can also use benzaldehyde; however due to low water solubility, a solvent such as acetone or ethanol will be required). Add 10mL of the potassium dichromate reagent to each test tube and swirl. The acetone will remain orange for a negative oxidation test result. The formaldehyde will change to green for a positive oxidation test (benzaldehyde will turn gray/white). You can then ask students to write out these reactions as you go through them on the board. If you have time, you may want to demonstrate Benedict’s test (oxidation of an aldehyde carbohydrate) or even Tollen’s test (silver the inside of a bottle as an aldehyde reduces the reagent to elemental silver). Sections 9.5, 9.6 and 9.7 To lead into lecture, you can discuss with students the DID YOU KNOW? Antifreeze section and the HEALTH LINK Aldehyde Dehydrogenase (some college students are very interested in the chemistry of the hangover-acetaldehyde). A hint for distinguishing hemiacetals from acetals is to tell students the “hemi’s have hydroxy’s.”
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Chapter 10 Carbohydrates Outline Notes I. Monosaccharides (Section 10.1) A. Carbohydrates Overview 1. More than half of the earth’s carbon atoms are found in carbohydrates 2. Plants produce most of these carbohydrates through the process of photosynthesis 3. Examine Figure 10.1 4. Carbohydrates play many important biological roles (i.e. energy storage and use) 5. Chemically, carbohydrate molecules are polyhydroxy aldehydes or ketones that containing three or more C atoms B. Carbohydrate Classification 1. Carbohydrates can be grouped into one of three categories: a. Monosaccharides are single residues that are the building blocks for larger carbohydrates b. Oligosaccharides contain from 2-10 monosaccharides residues c. Polysaccharides contain more than 10 monosaccharides residues 2. The term saccharide comes from the Greek word for sugar 3. Each monosaccharide is either an aldehyde or ketone that carries hydroxyl groups on other C atoms 4. Examine Figure 10.2 5. Monosaccharides can be classified as aldoses for aldehydes and ketoses for ketones 6. Monosaccharides can also be classified by the number of C atoms they have: a. Trioses have 3 C atoms c. Pentoses have 5 C atoms b. Tetroses have 4 C atoms d. Hexoses have 6 C atoms 7. An aldohexose is an aldehyde sugar with 6 C atoms; “ose” indicates it is a carbohydrate 8. See Table 10.1for monosaccharide classification 9. Monosaccharides have a sweet taste, high melting points, and are hydrophilic (due to the ability to form numerous hydrogen bonds) *See Sample Problem 10.1 and Practice Problem 10.1 II. Stereoisomers (Section 10.2) A. Chirality and Enantiomers 1. Recall stereoiosmers have the same molecular formulas and atomic connections, but they have different 3D shapes and are only interchanged by the breaking of bonds (i.e. cis and trans) 2. Examine Figure 10.3 3. Enantiomers are a type of stereoisomer that are nonsuperimposable mirror images of each other 4. Review examples in text (Baseball gloves vs. coffee cups) 5. Objects that cannot be superimposed on their mirror images are called chiral (i.e. your hands) 6. Chiral carbon atoms are attached to four different atoms or groups of atoms 7. A molecule is chiral if it contains one or more chiral carbons 8. Examine Figure 10.4 9. Review example in text (Chiral molecules and enatiomers) *See Sample Problem 10.2 and Practice Problem 10.2 B. Properties of Enantiomers 1. Pairs of enantiomers are identical in most ways (i.e. boiling points, melting points, and solubilities) 2. One difference is they rotate plane-polarized light in opposite directions 3. Examine Figure 10.5 10-1
4. When plane-polarized light is passed through a solution of one enantiomers, the light is rotated in a clockwise direction; this is called the dextrorotatory or (+) isomer 5. The solution of the other enantiomer rotates the plane-polarized light in a clockwise direction; this is called the levorotatory or (-) isomer 6. Living things typically produce/use either the (+) or (-) enantiomer, but not both 7. When chiral molecules are produced in laboratories, a 50:50 mixture of enantiomers called a racemic mixture is produced 8. Enatiomers also interact differently with other chiral molecules (i.e. taste buds) 9. Review examples in text (Baseball gloves, coffee cups, and your hands) 10. Examine Figure 10.6 11. Review examples in text (Taste of carvone) 12. Drug effectiveness can also be a function of enantiomers 13. Review examples in text (Chloramphenicol, estrone, propranolol, and ethambutol) C. Molecules with More than One Chiral Carbon 1. As the number of chiral carbon atoms increases, so does the number of stereoisomers that exist 2. Equation: Maximum Number of stereoiosmers = 2n where n is the number of chiral carbons 3. Examine Figure 10.7 4. Review examples in text (Stereoisomers of 2-bromo-3-chlorobutane) 5. Stereoiosmers of molecules that are not enantiomers are called diastereomers; diastereomers are nonmirror image stereoisomers *See Sample Problem 10.3 and Practice Problem 10.3 6. Carbohydrates also contain chiral carbon atoms, thus they exist as stereoisomers 7. Examine Figure 10.8 D. Fischer Projections 1. A Fischer projection is a simple way to show the 3D structure of a molecule 2. In Fischer projections, chiral carbons are represented by the intersection of two lines 3. Horizontal lines represent bonds pointing toward the viewer and vertical lines represent bonds pointing away from the viewer 4. For monosaccharides, the carbonyl is placed at or near the top 5. In D sugars, the –OH attached to the chiral carbon atom farthest from the carbonyl points to the right; conversely, in L sugars it points to the left 6. Monosaccharide enantiomers have the same name except for D or L designations (i.e. D-glyceraldehyde and L- glyceraldehyde) 7. In nature most monosaccharides are D sugars 8. Examine Figure 10.9 9. Review example in text (Stereoisomers of glucose) 10. Epimers are diastereomers that differ at only one chiral carbon *See Sample Problem 10.4 and Practice Problem 10.4 III. Important Monosaccharides and Monosaccharide Derivatives (Section 10.3) A. Monosaccharides 1. The most abundant monosaccharides in nature are pentoses and hexoses 2. D-Ribose and D-2-deoxyribose are aldopentoses that are incorporated larger biomolecules (i.e. NADPH, NAD+, RNA, & DNA) 3. Review examples in text (D-Ribose and D-2-deoxyribose structures) 4. D-glucose (dextrose or blood sugar) plays many important roles in human biochemistry (i.e. it is a key reactant that drives many nonspontaneous biochemical reactions) 5. D-galactose is combined with glucose to produce the disaccharide lactose (milk sugar) 6. When galactose is digested, it is transformed into a glucose derivative used for energy production or storage (glucose and galactose are diastereomers) 7. Galactosemias is a group of genetic disorders related to the inability to metabolize galactose; the buildup of galactose can lead to liver problems, mental retardation, and nervous system damage) 10-2
8. D-fructose (fruit sugar) is a ketose that is present in fruit and makes up 40% of honey 9. Review example in text (D-fructose structure) 10. There are many monosaccharide derivatives incorporated into oligosaccharides and polysaccharides; the four major classes are deoxy, amino, alcohol, and carboxylic acid sugars B. Monosaccharide Derivatives 1. In deoxy sugars, an H atom replaces one or more of the –OH groups in a monosaccharide 2. Review examples in text (L-fucose and L-rhamnose) 3. In amino sugars, an –OH group of a monosaccharide has been replaced by an –NH2 (amino) group 4. Review examples in text (D-glucosamine and N-acetyl-D-glucosamine) 5. In alcohol sugars the carbonyl group of a monosaccharide has been reduced to an alcohol group 6. Examine Figure 10.10 7. In carboxylic acid sugars an aldehyde and/or alcohol group of a monosaccharide has been oxidized to form a carboxyl group a. In aldonic acids the aldehyde group has been oxidized b. In uronic acids the alcohol group at the opposite end of the aldehyde has been oxidized 8. Review examples in text (D-gluconic acid and D-glucuronic acid) *See Sample Problem 10.5 and Practice Problem 10.5 IV. Reactions of Monosaccharides (Section 10.4) A. Reduction 1. When the carbon-oxygen double bond of an aldehyde or ketone is treated with H2 and a Pt or enzyme catalyst, it is reduced to an alcohol 2. Alcohol sugars formed by reduction of aldoses and ketoses are important in nature as well as the food industry (sweeteners) 3. Review example in text (Sorbitol use) B. Oxidation 1. The oxidation of an aldehyde/alcohol sugar into a carboxylic acid sugar is catalyzed by enzymes in nature 2. Sugars that can be oxidized and thus give a positive Benedict’s test are called reducing sugars 3. Aldoses and some ketoses are reducing sugars (ketoses can rearrange to become aldoses) 4. Review example in text (D-fructose conversion) 5. Examine Figure 10.11 6. Benedict’s test can be used to monitor diabetes, though it gives false positives 7. A new glucose-detection system uses an enzyme-based method in test strips 8. Examine Figure 10.12 C. Hemiacetal Formation 1. Recall aldehydes/ketones can react with alcohols to form hemiacetals/hemiketals 2. Monosaccharides contain both an aldehyde/ketone group and an alcohol group *See Sample Problem 10.6 and Practice Problem 10.6 V. Monosaccharides in their Cyclic Form (Section 10.5) A. Cyclic Hemiacetals 1. When the carbonyl group of an aldehyde or ketone reacts with the hydroxyl group of an alcohol, a hemiacetal forms (C atom attached to –OH and –OC) 2. Review example in text (Hemiacetal formation reaction) 3. Monosaccharides containing both an aldehyde/ketone and an alcohol can react with themselves to form cyclic hemiacetals 4. Review example in text (D-glucose cyclic hemiacetal formation) 5. Typically rings are drawn with the O atom at the back and the hemiacetal carbon on the right 6. Examine Figure 10.13 B. Naming Cyclic Monosaccharides 1. The different hemiacetal forms of cyclic monosaccharide are called anomers 10-3
In alpha (α) anomers the –OH group points down while in the beta (β) anomers it points up The complete name for a monosaccharide specifies the anomers as α or β Furan and pyran are two cyclic ethers that are used to indicate the ring size of a monosaccharide A furanose has a five-membered ring while a pyranose has a six-membered ring Review examples in text (Furan and pyran structure and glucose’s full name) When drawing cyclic monosaccharides, the O atom is at the back and the group on C atom 5 always points up for D sugars and down for L sugars 8. Review example in text (α and β-D-galactopyranose structures) 9. Amino sugars incorporate the word amino in the name 10. Review example in text (α and β-D-glucosaminopyranose structures) 11. When drawing furanose rings, the –CH2OH on the left side always points up for D sugars 12. Examine Figure 10.14 13. Deoxy sugars incorporate the word deoxy in the name 14. Review example in text (α and β-D-2-deoxyribofuranose structures) 15. Ketoses (i.e. D-fructose) follow the same conventions except the hemiketal (hemiacetal) carbon is at position number 2 16. Examine Figure 10.15 C. Mutarotation 1. Cyclic hemiacetal groups of α and β anomers undergo continuous change in solution 2. When a monosaccharide is dissolved in water, all three forms (α, β, and open) will appear 3. Mutarotation is the process of converting back and forth from α anomers to the open form, to the β form (changes the rotation of plane-polarized light) 4. In solution, monosaccharides spend most of their time in cyclic form; however, the open form (aldehyde or ketone) is still available to react so they are still reducing sugars *See Sample Problem 10.7 and Practice Problem 10.7 VI. Oligosaccarides (Section 10.6) A. Glycoside Formation 1. Reacting a hemiacetal with an alcohol produces an acetal and water 2. Review example in text (Acetal formation reaction) 3. Acetals formed between α or β anomers and alcohols are called glycosides 4. A glycosidic bond connects the acetal carbon to the newly added –OC group 5. Review example in text (Glycoside formation reaction) 6. Some examples of natural glycosides include those found in vanilla beans and willow bark 7. Examine Figure 10.16 8. Glycosides can be hydrolyzed back into monosaccharides and alcohols 9. Review examples in text (Vanillin and salicin hydrolysis) B. Disaccharides 1. Two monosaccharides that are connected by a glycosidic bond is called a disaccharide 2. Examine Figure 10.17 3. Review example in text (Maltose formation) 4. An α(1→4) glycosidic bond connects C atom 1 of one monosaccharide and C atom 4 of another (the arrow points from the hemiacetal residue C to the alcohol residue C) 5. Maltose contains an α(1→4) glycosidic bond that connects two glucose units a. Starch and glycogen can be digested into maltose, which can then be hydrolyzed into glucose b. Maltose is a reducing sugar which means it can form an aldehyde that is able to oxidized 6. Examine Figure 10.18 C. Oligosaccarides 1. Oligosaccarides are a class of carbohydrates that contain from 2-10 monosaccharide units 2. Disaccharides are the most common in nature (i.e. maltose, cellobiose, lactose, and sucrose) 3. Cellobiose contains two glucose residues connected by a β(1→4) glycosidic bond a. Review example in text (Cellobiose structure) 2. 3. 4. 5. 6. 7.
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b. Cellobiose is formed during the breakdown of cellulose c. When hydrolyzed, cellobiose gives two molecules of glucose, which are the same products that are produced by the hydrolysis of maltose d. Humans lack the enzyme that hydrolyzes cellobiose, therefore cellobiose cannot be used as fuel source e. Cellobiose is a reducing sugar 4. Lactose contains a galactose and glucose residue connected by an α(1→4) glycosidic bond a. Review example in text (Lactose structure) b. Lactose is found in milk (i.e 5% in cow’s milk and 7% in human milk) c. Lactose is an important fuel source for infants d. The enzyme lactase hydrolyzes lactose in the intestine to galactose and glucose molecules e. Lactose intolerance is the inability to hydrolyze lactose f. Lactose is a reducing sugar 5. Sucrose contains a glucose and fructose residue connected by an α1→β4 glycosidic bond a. Review example in text (Sucrose structure) b. Sucrose is not a reducing sugar because it contains no hemiacetal carbon (the glycosidic bond is between both anomeric carbons producing an acetal and ketal) c. Sucrose is commonly known as table sugar and is an energy storage molecule for many plants (i.e. sugar beets and sugar cane) d. When digested, sucrose is hydrolyzed into fructose and glucose molecules DID YOU KNOW? Trehalose 6. Oligosaccharides with 3 or more units are found in nature a. Trisaccharides (i.e. raffinose), tetersaccharides (i.e. stachyose), and pentasaccharides (i.e. verbascose) are present in beans and peas b. Examine Figure 10.19 7. Oligosaccharide digestion involves the hydrolysis of glycosidic bonds a. Review example in text (Lactose intolerance) b. Review example in text (α-galactosidase and oligosaccharide hydrolysis) c. Examine Figure 10.20 8. Many biologically important compounds are part of oligosaccharides a. Blood type is determined by an oligosaccharide attached to red blood cells b. Review example in text (ABO antigens) c. Examine Figure 10.21 *See Sample Problem 10.8 and Practice Problem 10.8 HEALTH LINK Natural and Artificial Sweeteners (See Table 10.2) HEALTH LINK Stevia (Examine Figure 10.22 and 10.23) DID YOU KNOW? The Sweetest Compounds VII. Polysaccharides (Section 10.7) A. Polysaccharides Overview 1. Carbohydrates containing more than 10 monosaccharide units are called polysaccharides 2. Polysaccharides provide structure and energy storage in living things 3. Homopolysaccharides are composed of just one type of monosaccharide 4. Heteropolysaccharides are built from more than one type of monosaccharide B. Homopolysaccharides 1. Cellulose provides structure for plants (50% of wood is cellulose) a. It is a water-insoluble protective barrier that provides support to stems and stalks b. It is composed of long chains of glucose molecules joined by β(1→4) glycosidic bonds c. The strength of cellulose comes from hydrogen bonds between durable sheets d. Examine Figure 10.24 e. Animals do not produce the enzyme required to hydrolyze cellulose; however, cows and horses contain a symbiotic bacteria in their digestive tract that can catalyze its hydrolysis 10-5
2. Starch is produced by plants as a way to store energy (i.e. potatoes store a lot of starch) a. It consists of two different homopolysaccharides: amylose and amylopectin b. Amylose is composed of glucose molecules joined by α(1→4) glycosidic bonds c. Cellulose and amylose are stereoisomers; amylose coils into flexible helical shapes unlike the hydrogen-bonded sheets in cellulose d. Examine Figure 10.25 e. Amylopectin is composed of glucose molecules joined by α(1→4) glycosidic bonds, with an α(1→6) glycosidic bond (branch) about every 25 or 30 glucose units f. The α(1→6) linkages produce branching in the chain g. Animals and plants can hydrolyze the α(1→4) glycosidic bonds in starch; this hydrolysis yields glucose and maltose, which is hydrolyzed by maltase 3. Glycogen is an energy storage molecule for animals (“animal starch”) a. It is composed of glucose molecules joined by α(1→4) glycosidic bonds, with α(1→6) branch points every 8-12 glucose units b. A variety of enzymes are responsible for catalyzing the hydrolysis of glycogen c. Examine Figure 10.26 4. Chitin makes up the exoskeleton of crustaceans and insects and is present in the cell walls of some algae and fungi a. It is composed of N-acety-D-glucosamine residues joined by β(1→4) glycosidic bonds b. Chitin is rigid due to hydrogen bonds between separate strands c. Review example in text (Chitin structure) DID YOU KNOW? Lunfenuron and Chitin C. Heteropolysaccharides 1. Hyaluronic acid is found in the lubricating fluid that surrounds joints and in the vitreous humor (clear gel) present inside the eye a. It is composed of alternating residues of N-acetyl-D-glucosamine and D-glucuronate b. It is connected by β(1→4) and β(1→3) glycosidic bonds 2. Chondroitin 4-sulfate is present in connective tissue a. It is composed of alternating N-acetyl-D-galactosamine-4-sulfate and D-glucuronate units b. It is connected by β(1→4) and β(1→3) glycosidic bonds c. It can be taken without a prescription to assist in the repair of damaged cartilage d. Review example in text (Chondroitin 4-sulfate structure) *See Sample Problem 10.9 and Practice Problem 10.9
Chapter Summary Section 10.1 Carbohydrates, which are polyhydroxy aldehydes or ketones (or derivatives), play many important roles in the biochemistry of living things. Carbohydrates can be classified as monosaccharides (single residue), oligosaccharides (2-10 monosaccharides residues), or polysaccharides (more than 10 monosaccharides residues). Monosaccharides can be further classified as aldoses, ketoses, trioses, tetroses, pentoses, etc. An aldohexose is an aldehyde sugar with 6 carbon atoms. Monosaccharides have a sweet taste, high melting points, and are hydrophilic. Section 10.2 Enantiomers are a type of stereoisomer that are nonsuperimposable mirror images of each other. This stereoisomerism is caused by chiral carbon atoms, which are carbon atoms attached to four different atoms or groups of atoms. Pairs of enantiomers are identical in most ways; however, they rotate planepolarized light in opposite directions. Dextrorotatory or (+) isomers rotate light right while levorotatory or (-) isomers rotate light left; and a racemic mixture contains equal parts of both isomers. As the number of chiral carbon atoms increases, so does the number of stereoisomers that exist. Stereoiosmers of molecules that are not enantiomers are called diastereomers, and diastereomers that 10-6
differ at only one chiral carbon are called epimers. A Fischer projection is a simple way to show the 3D structure of a molecule. In D sugars, the –OH attached to the chiral carbon atom farthest from the carbonyl in the Fischer projection points to the right; conversely, in L sugars it points to the left. Most monosaccharides in nature are D sugars. Section 10.3 The most abundant monosaccharides in nature are pentoses and hexoses. D-Ribose and D-2-deoxyribose are aldopentoses that are incorporated larger biomolecules. D-glucose plays many important roles in human biochemistry. D-galactose is combined with glucose to produce the disaccharide lactose. Dfructose is a ketose that is present in fruit and makes up 40% of honey. There are many monosaccharide derivatives incorporated into oligosaccharides and polysaccharides. In deoxy sugars an H atom replaces one or more of the –OH groups in a monosaccharide. In amino sugars an –OH group of monosaccharide has been replaced by an –NH2 group. In alcohol sugars the carbonyl group of a monosaccharide has been reduced to an alcohol group. Lastly, in carboxylic acid sugars an aldehyde and/or alcohol group of a monosaccharide has been oxidized to form a carboxyl group (i.e. in aldonic acids the aldehyde group has been oxidized while in uronic acids the alcohol group at the opposite end of the aldehyde has been oxidized). Section 10.4 When the carbon-oxygen double bond of an aldehyde or ketone is treated with H2 and a Pt or enzyme catalyst, it is reduced to an alcohol. Alcohol sugars formed by reduction of aldoses and ketoses are important in nature as well as the food industry. When monosaccharides are oxidized, the aldehyde/alcohol group is converted into a carboxylic acid. Sugars that can be oxidized, and thus, give a positive Benedict’s test, are called reducing sugars. Monosaccharides can also react with themselves because they contain both an aldehyde/ketone group and an alcohol group. When this occurs, hemiacetals/hemiketals are produced. Section 10.5 Monosaccharides containing both an aldehyde/ketone and an alcohol can react with themselves to form cyclic hemiacetals/hemiketals. The different hemiacetal forms of cyclic monosaccharide are called anomers. In alpha (α) anomers the –OH group points down while in the beta (β) anomers it points up. The ring size of a monosaccharide is indicated by the name furanose (five-membered ring) or pyranose (six-membered ring). Amino sugars and deoxy sugars incorporate the word amino and deoxy, respectively, in the monosaccharide name. Mutarotation is the process of converting back and forth from an α anomeric form, to the open form, to the β form. In a solution, monosaccharides spend most of their time in cyclic form; however, the open form is still available to react, thus they are still reducing sugars. Section 10.6 Acetals formed between α or β anomers and alcohols are called glycosides. Glycosides are connected via a glycosidic bond, which can be broken as glycosides are hydrolyzed back into monosaccharides and alcohols. Two monosaccharides that are connected by a glycosidic bond are called a disaccharide. Some important disaccharides are maltose, cellobiose, lactose, and sucrose. Many biologically important compounds are contain oligosaccharides (i.e blood type antigens). Section 10.7 Polysaccharides provide structure and energy storage in living things. Homopolysaccharides are composed of just one type of monosaccharide. Important homopolysaccharides include cellulose, starch (amylose and amylopectin), glycogen, and chitin. Heteropolysaccharides are built from more than one type of monosaccharide. Some notable heteropolysaccharides include hyaluronic acid and chondroitin 4-sulfate.
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Lecture Suggestions Sections 10.1, 10.2, and 10.3 Demonstrate that carbohydrates are nature’s way of storing, transferring, and utilizing chemical energy. Write the equations for photosynthesis and cellular respiration on the board. Ask to students to compare and contrast these equations/processes. Explain to students that the main purpose of these reactions is to transfer energy from the sun to all living things in the form of chemical carbohydrates. You may also want to mention that the atoms (C, H, and O) are continuously recycled, however, energy is not (organisms do not give energy back to the sun). Perform the “sugar combustion” demonstration below to reemphasize the chemical energy stored in carbohydrates. Sugar Combustion: Type: Decomposition and Combustion: C12H22O11(s) H2SO4→12C(s) + 11H2O(l) + Heat KClO3(s) Heat→ KCl(s) + O2(g) C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l) Materials: Evaporating Dish, C12H22O11, KClO3, Concentrated H2SO4 Procedure: Place equal parts of sucrose and potassium chlorate in an evaporating dish. Add 2-3 drops of concentrated sulfuric acid and stand back. The compounds will combust within 2-3 seconds. Sections 10.4, 10.5, and 10.6 Demonstrate Benedict’s test. Place a small amount (lest than 0.3g) of glucose into a test tube and add 5mL of water (stir/shake). Repeat with lactose and sucrose. Add 5mL of Benedict’s reagent to each test tube and place them in a hot water bath. While the reaction is proceeding (takes 5-10 minutes) explain the reaction with the class. On the board, show the reaction of glucose with Benedict’s reagent. Make sure you show the following: -how the α and β anomers open up into the straight chain form -how the hemiacteal reacts back into a carbonyl and hydroxyl -how the aldehyde is oxidized into a carboxylic acid (carboxylate ion) -how copper is reduced This is also a good time to reemphasize LeChatelier’s Principle. Explain why Benedict’s test takes 5-10 minutes to react (the cyclic forms must open up as the straight chain forms are consumed in the reaction). Once the reaction is complete, discuss why maltose is a reducing sugar while sucrose is not (you may want to draw the structures on the board). For further discussion, you can ask why Benedict’s reagent is not used anymore to test urine for glucose. Sections 10.7 To lead into lecture, ask students why humans cannot eat grass for energy like cows. Hopefully someone will know that they have symbiotic bacteria that actually break down the grass. Then show students the structure of cellulose. Explain that even though cellulose is made up of entirely glucose units, that we can’t break the β(1→4) glycosidic bonds and therefore we can use it as an energy source. Then show them the structure of starch and glycogen. Explain that we can break α(1→4) and α(1→6) glycosidic bonds, thus these carbohydrates are an important source of energy. For further discussion, you can ask students what the benefit of dietary fiber is and how it relates to these polysaccharides.
Handouts for Students 1) The Isomerism handout (next page) can help students distinguish between the different types of isomers encountered in this text.
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Handout Isomerism
Isomers Structural Isomers
Stereoisomers
Enantiomers Position Isomers CH3CH2CH2OH
Functional Group Isomers
&
CH3CH2OH
OH
&
&
Diasteromers
CH3OCH3 CH3CHCH3
Chain Isomers
Geometric Isomers
2 or More Chiral Carbon Isomers
&
&
&
Chapter 11 Lipids and Membranes Outline Notes I. Fatty Acids (Section 11.1) A. Fatty Acid Characteristics 1. Fatty acids are carboxylic acids that typically contain between 12 and 20 C atoms 2. In describing lipid structure, the term “head” refers to a polar group while the term “tail” refers to a nonpolar hydrocarbon chain 3. In fatty acids, the head is the carboxyl group and the tail is the long chain of C atoms 4. Fatty acids usually have an even number of C atoms 5. Review example in text (Structure of a fatty acids) B. Fatty Acid Classification 1. Fatty acids differ from one another in the number of C atoms and double bonds they have 2. See Table 11.1 for common fatty acids 3. Saturated fatty acids have only single bonds in their hydrocarbon tail 4. Monounsaturated fatty acids have one double bond in their hydrocarbon tail 5. Polyunsaturated fatty acids have two or more double bonds in their hydrocarbon tail 6. Examine Figure 11.1 C. Fatty Acid Properties 1. Melting points and boiling points are determined by saturation and carbon chain length a. The hydrocarbon tails of saturated fatty acids pack close together through London forces b. The longer the hydrocarbon tail, the stronger the interaction between molecules (more surface area increase London forces) c. Examine Figure 11.2a d. The melting points and points of saturated fatty acids rise as the molecule gets larger and most are solids at room temperature e. The double bonds in unsaturated fatty acids are usually in the cis form causing a kink in the tail f. The cis double bond kink causes the molecules of unsaturated fatty acids to be farther apart g. This reduction in contact between chains weakens the London forces h. The greater the amount of unsaturation, the lower the melting point i. Unsaturated fatty acids tend to be liquids at room temperature j. Examine Figure 11.2b 2. Fatty acids are water insoluble (hydrophobic) a. Recall in carboxylic acids, the more nonpolar hydrocarbons, the lower the water solubility b. Review example in text (Solubility of acetic acid, hexanoic acid, and myristic acid) c. Recall that the water solubility of carboxylic acids can be improved by reacting it with a base d. Review example in text (Benzoic acid reacting with NaOH to produce sodium benzoate) e. The negative charge on the carboxylate ion (conjugate base) makes the molecule more water soluble than the original carboxylic acid f. Examine figure 11.3 g. pH affects the solubility of fatty acids; low pH favors the carboxylic acid form which is hydrophobic; high pH favors the carboxylate ion form which is amphipathic h. At physiological pH (blood pH is 7.4), fatty acids exist in the carboxylate ion form D. Fatty Acids in the Body 1. Fatty acids are components of many other lipids 2. They are also significant sources of energy in living things 3. Polyunsaturated fatty acids play key roles in cellular control (i.e. omega-3 fatty acids) *See Sample Problem 11.1 and Practice Problem 11.1 11-1
*See Sample Problem 11.2 and Practice Problem 11.2 HEALTH LINK Omega 3 Fatty Acids (Examine Figure 11.4 and 11.5) II. Waxes (Section 11.2) A. Wax Formation Reaction 1. Recall that a carboxylic acid can be reacted with an alcohol to produce an ester 2. Review example in text (Formation of methyl propanoate) 3. Fatty acids can participate in the same reaction to form esters (i.e. biodiesel) 4. Review example in text (Formation of methyl linoleate) 5. When a fatty acid reacts with large alcohol, the product is a type of lipid found in waxes B. Wax Properties 1. Waxes are a mixture of water insoluble compounds 2. Waxes can contain esters (major component), alcohols, and alkanes 3. Wax esters are produced by combining fatty acids with long-chain alcohols 4. Typically, wax esters have fatty acids with 14-36 C atoms and alcohols with 16-30 C atoms 5. Examine Figure 11.6 C. Wax Function 1. Waxes often function to protect organisms by keeping water either in or out of them (i.e. wax on plant leaves prevent water loss from evaporation) 2. Examine Figure 11.7 3. Waxes protect skin and hair by keeping it soft and waterproof and some waxes are used as energy storage molecules in certain microorganisms 4. See Table 11.2 for key esters found in some waxes *See Sample Problem 11.3 and Practice Problem 11.3 III. Triglycerides (Section 11.3) A. Triglyceride Overview 1. Triglycerides (triacylglycerides) are triesters formed from glycerol and carboxylic (fatty) acids 2. Review example in text (Formation of triacetin from glycerol and acetic acid) 3. Animal fats and vegetable oils are triglycerides that are composed of three fatty acids and glycerol joined by ester bonds 4. Review example in text (Structure of a triglyceride) 5. Triglycerides often contain two or three different fatty acid residues 6. Examine Figure 11.8 7. Triglycerides like lard and beef fat are considered saturated due to their saturated fatty acids, where vegetable oils are considered unsaturated due to their unsaturated fatty acids 8. Review examples in text (Fatty acids in fats and oils) *See Sample Problem 11.4 and Practice Problem 11.4 B. Triglyceride Properties 1. The cis configuration in the hydrocarbon tails of unsaturated fatty acids does not allow the triglyceride molecules to pack together 2. Therefore, the more unsaturated fatty acids a triglyceride has, the lower its melting points 3. Fats are solids because they contain more saturated fatty acids, where oils are liquids because they contain more unsaturated fatty acids 4. Coconut oil is an exception: it is mostly saturated however its short residues (lauric acid) allow it to remain a liquid at room temperature 5. The shorter the hydrocarbon tail, the weaker the London forces and the lower the melting point 6. One of the primary biological roles of triglycerides is to provide energy a. Triglycerides provides twice as many calories per gram as carbohydrates and proteins b. In animals, triglycerides are stored in adipose tissue for future use 7. Examine Figure 11.9 8. Triglycerides also protect internal organs by acting as a cushion/padding against injury B. Important Reactions of Triglycerides 11-2
1. Triglycerides contain alkene and ester functional groups; thus two important reactions are catalytic reduction and ester hydrolysis 2. Catalytic reduction is when C=C double bonds are hydrogenated a. Recall that alkenes can be reduced with H2 and a Pt catalyst into alkanes b. In the commercial application of partial hydrogenation, vegetable oil is reduced with H2/Pt, however, the reaction is halted before all the double bonds are broken c. Examine Figure 11.10 d. This converts the liquid oil into a semisolid product e. This product is good for cooking and it does not contain the cholesterol that is present in fats f. Examine Figure 11.11 3. Autooxidation takes place when an alkene is exposed to air a. O2 in the air removes H atoms from the C atoms next to the C=C double bond b. C-C single bonds are then broken to form small organic molecules with unpleasant odors c. When this happens the triglyceride has spoiled or gone rancid d. Examine Figure 11.12 e. Vegetable oils spoil more quickly than fats because they have more unsaturated fatty acids f. Spoiling can be slowed by refrigeration (slower oxidation reaction), capping (less O2 exposure), partial hydrogenation (less double bonds), and the addition of antioxidants DID YOU KNOW? Phospholipase A2 *See Sample Problem 11.5 and Practice Problem 11.5 4. Saponification (soap making) is when triglycerides are hydrolyzed under basic conditions (i.e. with NaOH) into fatty acid salts and glycerol a. Fatty acid salts, commonly known as soaps, are amphipathic compounds that form micelles b. NaOH produces solid soaps while KOH produces liquid soaps c. The greater amounts of unsaturation also result in softer or liquid soaps d. The other saponification product is glycerol, which is used in lotions and creams e. Examine Figure 11.13 *See Sample Problem 11.6 and Practice Problem 11.6 HEALTH LINK Trans Fats (Examine Figure 11.14) HEALTH LINK Olestra (Examine Figure 11.15) IV. Phospholipids and Glycolipids (Section 11.4) A. Cell Membranes 1. Cell membranes are a double layer of amphipathic lipids that prevent the movement of polar molecules and ions 2. The hydrophilic head of each lipid is on the membrane surface and the hydrophobic tail points in towards the interior of the membrane 3. Examine Figure 11.16 4. Phospholipids and glycolipids are the two amphipathic lipids commonly found in membranes B. Phospholipids 1. Phospholipids contain “phosphate,” which refers to phosphoric acids and its three derivatives 2. The phosphate group is pH dependent (two forms present a physiological pH) 3. Review example in text (Phosphate structures) 4. Phosphates can react like carboxylic acids with alcohols to form esters 5. Review example in text (Formation of a phosphate monoester) 6. Phosphates can also react with two alcohol molecules to form a phosphate diester 7. The two classes of phospholipids are glycerophospholipids and sphingolipids 8. Glycerophospholipids are made by combing glycerol, two fatty acids, one phosphate, and one alcohol-containing compound a. Review example in text (Structure of a glycerophospholipid) b. Fatty acids are attached to glycerol via ester bonds and phosphate ester bonds connect the phosphate group to both glycerol and the alcohol compound 11-3
c. Examine Figure 11.17 d. Phosphatidlylcholine (lecithin) is the most common phospholipid found in plants and animals e. Phosphatidylethanolamine (cephalin) is also a major contributor to membrane structure f. Phosphatidylserine is an important part of the myelin of brain tissue g. Because glycerophospholipids are amphipathic, they are great emulsifying agents h. Emulsions are colloids formed from two liquids (i.e. lecithin in egg whites) 9. Sphingolipids are made by combining sphingosine with one fatty acid, one phosphate, and one alcohol-containing compound a. Review example in text (Structure of a sphingolipid) b. The sphingomyelin family of sphingolipids are prevalent in the myelin sheath C. Glycolipids 1. Glycolipids are made by combing sphingosine with one fatty acid and one sugar residue 2. Review example in text (Structure of a glycolipid) 3. Cerebrosides are glycolipids produced from monosaccharides and they are found at nerve synapses in the brain 4. Gangliosides are glycolipids produced from oligosaccharides and they act as cell surface receptors for hormones and drugs 5. Examine Figure 11.18 *See Sample Problem 11.7 and Practice Problem 11.7 V. Steroids (Section 11.5) A. Steroid Overview 1. Steroids are lipids that have the same fused ring structure consisting of three 6-C atom ring and one 5-C atom ring 2. Review example in text (Ring structure of a steroid) 3. Three important types of steroids are cholesterol, steroid hormones, and bile salts B. Cholesterol 1. Cholesterol is the most common steroid found in humans/animals and it is not found in plants 2. Some cholesterol comes from the diet and some is manufactured in the liver 3. Review example in text (Structure of cholesterol) 4. Cholesterol is somewhat amphipathic due to the hydrophilic –OH group; this makes it possible to form components of cell membranes 5. It is the starring material for the synthesis of other steroids (this is its primary biological function) 6. Cholesterol is not water soluble so it and other lipids are transported through the blood as suspensions by lipid-protein complexes called lipoproteins a. Lipoproteins called chylomicrons and VLDLs carry triglycerides, phospholipids, and cholesterol b. LDLs transport cholesterol and phospholipids from the liver to the cells where HDLs transport them from the cells back to the liver c. Examine Figure 11.19 7. Low HDL and high LDL are warning signs of atherosclerosis, which is the buildup of cholesterol deposits in the arteries a. Examine Figure 11.20 b. Deposits reduces the supply of blood flow to the heart and brain, which can cause heart attacks and strokes c. Review examples in text (Cholesterol screening tests) C. Steroid Hormones 1. Hormones are molecules that regulate the function of organs and tissues 2. Some sex hormones and adrenocorticoid hormones are steroids 3. Examine Figure 11.21 4. The manufacture of steroid hormones begins with cholesterol 11-4
5. Next, the female sex hormone progesterone is used to make other sex hormones and adrenocorticoid hormones 6. The male sex hormone testosterone is responsible for male traits while the female sex hormones progesterone and estradiol are responsible for female traits 7. Adrenocorticoid hormones are produced in the adrenal gland and some are potent antiinflammatory agents (i.e. cortisol and cortisone) DID YOU KNOW? Human Growth Hormone D. Bile Salts 1. Bile salts (i.e. glycocholate and taurocholate) are produced from cholesterol and they are amphipathic 2. Bile salts are released from the gallbladder into the small intestine where they aid digestion by forming emulsions with dietary lipids *See Sample Problem 11.8 and Practice Problem 11.8 HEALTH LINK Anabolic Steroids (Examine Figure 11.22) VI. Eicosanoids (Section 11.6) A. Eicosanoids Overview 1. Eicosanoids are hormones that are derived from arachidonic acid and other 20-C atom fatty acids 2. Examine Figure 11.23 3. Arachidonic acid undergoes reactions that transform it into the various eicosanoids— prostaglandins, thromboxanes, and leukotrienes B. Prostaglandins 1. Prostaglandins cause pain, inflammation, and fever 2. They can also affect blood pressure and the prostaglandin PGE2 induces labor C. Thromboxanes and Leukotrienes 1. Thromboxane A2 is involved in blood clotting 2. Leukotriene A4 induces muscle contractions in the lungs and is linked to asthma attacks 3. Examine Figure 11.24 D. Nonsteroidal Anti-Inflammatory Drugs (NSAIDs) 1. NSAIDs such as aspirin and ibuprofen reduce pain, fever, and inflammation by blocking the action of an enzyme involved in the conversion of arachidonic acid into prostaglandins and thromboxanes 2. The two forms of the enzyme (COX-1 and COX-2) are activated during injury or illness 3. NSAIDs alleviate pain symptoms by blocking the COX enzymes; however, they may cause ulcers and kidney damage (recall prostaglandins affect blood pressure and thromboxanes are involved in blood clotting) 4. Review examples in text (Pain relievers and side effects) *See Sample Problem 11.9 and Practice Problem 11.9 VII. Membranes (Section 11.7) A. Membranes Overview 1. Membranes are barriers that surround cells and they are composed of a bilayer of lipids 2. These amphipathic lipids are usually phospholipids, glycolipids, and cholesterol 3. Examine Figure 11.25 4. Many of the fatty acid residues in membrane lipids are unsaturated; cis double bonds do not allow for the tight packing of the membrane structure 5. However, the ring structure of cholesterol does add to the rigidity of the membrane 6. Proteins are another important component of membranes; they can be attached to the surface of the membrane or they can stretch from one side to the other 7. According to the fluid mosaic model, membranes are dynamic structures composed of mobile phospholipids and embedded proteins B. Selective Permeability 11-5
1. Membranes are selectively permeable in that they allow some substances to cross while they prevent the passage of other substances 2. Water, small nonpolar molecules (i.e. O2, N2, CO2), and amphipathic steroids are all able to diffuse through membranes 3. Membranes do not allow ions, small polar molecules (except water), and large biomolecules (i.e. DNA and RNA) to diffuse through them 4. Therefore, glucose and ions (i.e. Na+, K+, Cl-) must be assisted by proteins, which prevent them from coming in contact with the hydrophobic interior of the membrane 5. In facilitated diffusion, proteins aid solutes as they move from high to low concentrations 6. Movement against diffusion, from low to high concentrations, is called active transport 7. Active transport requires an input of energy into the proteins that move the substances across 8. Examine Figure 11.26 *See Sample Problem 11.10 and Practice Problem 11.10
Chapter Summary Section 11.1 Fatty acids are long carboxylic acids that are the building blocks for many types of lipids. Saturated fatty acids have only single bonds in their hydrocarbon tail while monounsaturated and polyunsaturated have one or more double bonds, respectively. Saturated fatty acids have higher melting and boiling points than unsaturated fatty acids: the cis double bond(s) in unsaturated fatty acids causes a kink, which reduces the London forces between molecules. Therefore, saturated fatty acids are usually solids at room temperature while unsaturated fatty acids are liquids. Fatty acids are hydrophobic, however, at physiological pH, they exist in the carboxylate ion form which is amphipathic. A primary biological function of fatty acids is to provide energy to living things Section 11.2 When a fatty acid reacts with large alcohol, the product is a type of lipid found in waxes. Waxes are mixtures of esters, alcohols, and alkanes. Waxes often function to protect organisms by keeping water either in or out of them. They also protect skin and hair, and some are even used as energy storage molecules in certain microorganisms. Section 11.3 Triglycerides (triacylglycerides) are triesters formed from glycerol and fatty acids. Animal fats, which have more saturated fatty acids, and vegetable oils, which have more unsaturated fatty acids, are both examples of triglycerides. Because of London forces, fats are usually solids at room temperature and oils are usually liquids. One of the primary biological roles of triglycerides is to provide energy (they provide twice as many calories per gram as carbohydrates and proteins). Three important reactions that triglycerides undergo are catalytic reduction, autooxidation, and saponification. In catalytic reduction the C=C double bonds are hydrogenated. This converts liquid oils into semisolid products. Autooxidation takes place when triglyceride is exposed to O2 in the air. When this happens the triglyceride has spoiled or gone rancid. Saponification (soap making) is when triglycerides are hydrolyzed under basic conditions into fatty acid salts (soaps) and glycerol. Section 11.4 Cell membranes are a double layer of amphipathic lipids that prevent the movement of polar molecules and ions. The hydrophilic head of each lipid is on the membrane surface and the hydrophobic tail points in towards the interior of the membrane. Phospholipids, which contain phosphate, and glycolipids, which contain a sugar residue, are the two amphipathic lipids commonly found in membranes. The two classes of phospholipids are glycerophospholipids and sphingolipids. Due to their amphipathic nature, phospholipids are great emulsifying agents (emulsions are colloids formed from two liquids). Glycolipids produced from monosaccharides and found at nerve synapses in the brain 11-6
are called cerebrosides. Gangliosides are glycolipids produced from oligosaccharides and they act as cell surface receptors for hormones and drugs. Section 11.5 All steroids have the same fused ring structure consisting of three 6-C atom ring and one 5-C atom ring. Three important types of steroids are cholesterol, steroid hormones, and bile salts. Cholesterol is the most common steroid found in humans/animals it is the starring material for the synthesis of other steroids. Cholesterol is not water soluble, so it and other lipids are transported through the blood as suspensions by lipid-protein complexes called lipoproteins (chylomicrons, VLDLs, LDLs, and HDLs). Hormones are molecules that regulate the function of organs and tissues. Some sex hormones and adrenocorticoid hormones are steroids. Bile salts are released from the gallbladder into the small intestine where they aid digestion by forming emulsions with dietary lipids. Section 11.6 Eicosanoids are hormones that are derived from arachidonic acid and other 20-C atom fatty acids. Arachidonic acid undergoes reactions that transform it into the various eicosanoids—prostaglandins, thromboxanes, and leukotrienes. Prostaglandins can cause pain, inflammation, and fever. They can also affect blood pressure and even induce labor. Thromboxanes are involved in blood clotting while leukotrienes induce muscle contractions in the lungs (i.e. asthma attacks). Nonsteroidal antiinflammatory drugs (NSAIDs) such as aspirin and ibuprofen reduce pain, fever, and inflammation by blocking the action of an enzyme involved in the conversion of arachidonic acid into prostaglandins and thromoboxanes. Section 11.7 Membranes are barriers that surround cells and they are composed of a bilayer of lipids (usually phospholipids, glycolipids, and cholesterol) and proteins. According to the fluid mosaic model, membranes are made of mobile phospholipids and embedded proteins. Membranes are selectively permeable: water, small nonpolar molecules, and amphipathic steroids are all able to diffuse through membranes. Membranes do not allow the diffusion ions, small polar molecules (except water), and large Biomolecules. In facilitated diffusion, proteins aid solutes as they move from high to low concentrations. Movement against diffusion, from low to high concentrations, is called active transport because it requires an input of energy.
Lecture Suggestions Sections 11.1, 11.2, and 11.3 Begin lecture by discussing the misconceptions of the word “fat.” First explain the negative connotations: why many people think fat should be completely avoided (obesity, heart disease, stroke etc.). Then explain to students why fats are very important (cellular components, energy supply, protection, etc.). Explain that while Americans consume too much fat (over 40%), a healthful diet should include 30% fat (10% saturated, 10% monounsaturated, and 10% polyunsaturated). Also mention that people who say they have 0% body fat are crazy as every cell and tissue in the body contains fats (i.e. membrane lipids, etc.). Before you begin section 11.3, continue the previous discussion of fat by asking students what they think is considered “bad” fat. They should mention both saturated fat and trans fat. Using the previously discussed physical properties of fatty acids, explain why saturated fats (solids) are considered less healthy than unsaturated fats (liquids). You can also discuss how these fats relate to cholesterol, lipoproteins, and atherosclerosis (section 11.5). Next, discuss the issues with trans fats. Explain that most unsaturated fats in nature are cis; therefore, trans usually only come about from the commercial process of catalytic hydrogenation. Ask students what they think the affect would be on a cell if its membrane’s cis unsaturated fatty acids were replaced with trans ones. Be sure to explain how fluidity of 11-7
the cell would be affected by London forces (you may want to wait until section 11.7 for this discussion). Sections 11.4, 11.5, and 11.6 For these sections, discuss how lipids related to medicine. When discussing membranes in section 11.4, mention how some vitamins and medications are “fat soluble” and thus require lipids to carry them throughout the body. Show students the structures of a fat soluble vitamin (i.e. vitamin A) and a medication (i.e. naproxen). Have students identify the hydrophobic and hydrophilic regions on the structures. You may also want to reemphasize how hydrophobic medications can be made more soluble by reacting them into amphipathic salts. When lecturing on section 11.5, briefly discus the misconceptions of the words “steroid” and “cholesterol.” Talk about the importance of steroids, specifically cholesterol, in the body. Next discuss how anabolic steroids (i.e. Dianabol) work and the dangers associated with their use. For section 11.6, discuss how the NSAIDs aspirin and ibuprofen work. Mention that they prevent the formation of eicosanoids by blocking COX enzymes. If time permits, you can also discuss the concerns with COX-2 inhibitors. Sections 11.7 In this lecture, compare a cell to the classroom. Ask students what the walls, celling, and floor represent. They should answer the cell membrane. Ask them if the inside of the classroom (cell) is hydrophilic or hydrophobic. They usually answer hydrophilic. Then ask them if the outside of the classroom (cell) is hydrophilic or hydrophobic. They usually answer incorrectly and say hydrophobic. Explain to students that while the membrane lipid bilayer has both hydrophilic and hydrophobic; only the hydrophilic regions are exposed on the surface. Go up to the doorway and show them that only the inside of the wall is hydrophobic. Next, review the principle of equilibrium and diffusion (solutes moving from high to low concentrations). Have them pretend students in the classroom are solutes. Ask them which way they will naturally flow if there are hundreds of students crammed in the hall, but there are only a few students in the classroom. Then ask them the reverse scenario. Emphasize that no energy is required because the students (solutes) are going where they want to go (diffusion). Then ask them how they could get more students in a classroom that is already full (cramped) with students. Explain that someone (protein pump) would have to push or pull them in and thus energy is required (active transport).
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Chapter 12 Peptides, Proteins, and Enzymes Outline Notes I. Amino Acids (Section 12.1) A. Amino Acid Overview 1. Amino acids are molecules that contain a carboxyl group (–CO2H) and an amino group (–NH2) 2. 20 amino acids are used to make peptides and proteins 3. All 20 are α-amino acids due to the fact that the amino group is attached to the C atom next to the carboxyl group (α-C) 4. The only differences between the amino acids is their side chain or R-group 5. Review example in text (Structure of an α-amino acid) 6. See Table 12.1 for the α-amino acids present in proteins B. Amino Acids and pH 1. Because amino acids contain carboxyl (acidic) and amino (basic) groups, their structure is dependent on pH 2. The pKa values for carboxyl groups in amino acids range from 1.8 to 4.3; therefore at pH 7 the carboxylic acid (–CO2H) portion is found in its conjugate base form (–CO2-) 3. The pKa values for the amino groups in amino acids range from 9.1 to 12.5; therefore at pH 7 the amine (-NH2) portion is found in its conjugate acid form (–NH3+) 4. This produces a 1+ and 1- charge on the amino acid (i.e. glycine) 5. Examine Figure 12.1 6. A zwitterion has one positive and one negative charge 7. As pH changes, so does the net charge on an amino acid a. At pH 1, glycine has a net charge of 1+ b. At pH 14, glycine has a net charge of 18. Slight differences in pKa values make each amino acid different from the others 9. The isoelectric point (pI) is the pH at which an amino acid has a net charge of zero *See Sample Problem 12.1 and Practice Problem 12.1 *See Sample Problem 12.2 and Practice Problem 12.2 C. Classifying Amino Acids 1. All amino acid are hydrophilic due to their polar covalent bonds (recall N, O, and H atoms allow for the formation of hydrogen bonds with water) 2. The water solubility of the different amino acids depends on the makeup of the side chain groups 3. The side chains of amino acids can be classified into four categories: a. Nonpolar (9) usually contain an alkyl group, aromatic ring, or a nonpolar collection of atoms b. Polar-acidic (2) contain a carboxyl group, which at pH 7 is found in the conjugate base form; therefore, polar-acid side chains carry a 1- charge at pH 7 c. Polar-basic (3) contain an amino group, which at pH is found in the conjugate acid form; therefore, polar-basic side chains carry a 1+ charge at pH 7 d. Polar-neutral (6) usually contain an alcohol, a phenol, or an amide D. Stereoisomers 1. 19 out of 20 α-amino acids are chiral (the α-C has four different groups attached to it) 2. Glycine as a side chain of an H atom, therefore it is not chiral 3. Due to chirality, these 19 amino acids exist as a pair of enatiomers 4. In the Fischer projection, the carboxyl group points up and the side chain points down a. D-amino acids have the amino group on the right side b. L-amino acids have the amino group on the left side 5. Examine Figure 12.2 12-1
6. Only L-amino acids are used in the proteins of living things 7. Some organisms can produce an enzyme that catalyzes the conversion of an L-amino acid into a D-amino acid once a protein has been assembled (i.e. bacterial cell walls and antibiotics) II. The Peptide Bond (Section 12.2) A. Peptide Bond Formation 1. Recall a carboxylic acid can react with an amine to form an amide 2. The carboxyl and amino groups of amino acids can react to form amides; however, in biochemistry, these amides (amide bonds) are called peptides (peptide bonds) 3. Review examples in text (Amide and peptide bond formation reactions) B. Peptide Classification 1. A dipeptide consists of two amino acids, a tripeptide consists of three, a tetrapeptide consist of four and the trend continues similar to the system used to describe carbohydrates 2. Oligopeptides contain 2-10 amino acids and polypeptides contain more than 10 amino acids 3. Oligopeptides and polypeptides are collectively referred to as peptides 4. Examine Figure 12.3 5. Proteins are polypeptide chains having more than 50 amino acids C. Peptide Nomenclature 1. Each amino acid can take part in two peptide bonds 2. Review examples in text (Peptide formations in figure 12.3) 3. The N-terminus is the end of the peptide with the unreacted amino group while the C-terminus is the end with the free carboxyl group 4. It is customary to draw peptides and proteins with the N-terminus on the left and the C-terminus on the right 5. Peptides and proteins are named by listing amino acids in order form the N- to C-terminus; however due to length, most large peptides and proteins are known by common names 6. Examine Figure 12.4 D. Peptide Hydrolysis and Size 1. Hydrolysis of the peptide bond restores amino acids’ amino and carboxyl groups to their original state (peptide hydrolysis is identical to amide hydrolysis) 2. Examine Figure 12.5 3. Biochemically active peptides and proteins come in all sizes 4. Review examples in text (Thyrotropin and chymotrypsinogen) *See Sample Problem 12.3 and Practice Problem 12.3 *See Sample Problem 12.4 and Practice Problem 12.4 III. Peptides, Proteins, and pH (Section 12.3) A. Peptide/Protein Structure and pH 1. Peptide and protein structure is just as sensitive to pH as is amino acid structure 2. In acidic solutions, amino and carboxyl groups appear in their acidic forms (–NH3+ and –CO2H) 3. In neutral solutions, amino groups remain in their acidic form (–NH3+), but carboxyl groups are found in their basic form (–CO2-) 4. In basic solutions, amino and carboxyl groups appear in their basic forms (–NH2 and –CO2-) 5. Peptide bonds are not acidic or basic and their structure does not vary with changes in pH B. Charge and pH 1. Examine Figure 12.6 2. Review examples in text (Change in net charge for the tripeptide in figure 12.6) 3. When peptides carry either a net positive or a net negative charge, the liked-charged molecules repel one another and stay dissolved in water 4. When the net charge is zero, the molecules are more likely to precipitate out of aqueous solution *See Sample Problem 12.5 and Practice Problem 12.5 IV. Protein Structure (Section 12.4) A. Protein Structure Overview 12-2
1. Proteins can be classified in general terms as either fibrous or globular 2. Fibrous proteins exist as long fibers or strings and are usually tough and water insoluble (i.e. collagen and keratin) 3. Globular proteins are spherical in shape, are highly folded, and tend to be water soluble 4. Protein structure is understood in four levels of organization: primary, secondary, tertiary, and quaternary B. Primary (1°) Structure 1. Primary structure refers to the order of amino acid residues in a peptide or protein 2. Review examples in text (Ala-Gly-Val tripeptide) 3. Examine Figure 12.7 4. Primary structure is analogous to the arrangements of letters in a word (i.e. diet, tide, tied) 5. Rearranging the amino acid residues in peptide or protein can change its function 6. Review examples in text (Artificial sweetener aspartame) C. Secondary (2°) Structure 1. The two types of secondary structure commonly observed are the α-helix and the β-sheet 2. Secondary structure results from hydrogen bonding that takes place between the amide N-H and C=O groups along the polypeptide backbone (series of alternating C atoms and amide groups) 3. Review example in text (Peptide backbone structure) 4. In the α-helix, the polypeptide folds into a shape like a coiled spring a. This is caused by hydrogen bonds between every fourth amino acid, which stabilizes the structure b. Examine Figure 12.8 c. The spring-like coils are flexible and stretchy (i.e. keratins in wool) 5. In the β-sheet, the polypeptide assumes a sheet-like arrangement that forms when different segments of a polypeptide chain align side by side a. In parallel sheets all of the interacting strands run in the same N-terminal to C-terminal direction b. In antiparallel sheets adjacent segments run in opposite directions c. Examine Figure 12.9 d. Many hydrogen bonds stabilize the β-sheet (i.e. silk is strong but not stretchy) D. Tertiary (3°) Structure 1. Tertiary structure refers to the overall three-dimensional shape of a protein 2. Examine Figure 12.10 3. There is usually only one shape that leads to a native (biologically active) molecule 4. The sequence of amino acids (primary structure) determines which folding patterns are selected, therefore it ultimately determines secondary and tertiary structures 5. In an aqueous environment, the nonpolar side chains are folded into the hydrophobic interior and the polar side chains are on the hydrophilic surface (“nonpolar in, polar out”) 6. The surface of a globular protein usually has binding sites (i.e. indentions or clefts) that allow other substances to attach/bind to the protein 7. The tertiary structure is maintained by interactions between the amino acid side chains E. Noncovalent Interactions Important to Tertiary Structure 1. Hydrogen bonds make the most important contributions to stabilizing tertiary structure a. Side chains must carry an N-H or O-H and another N or O atom to hydrogen bond b. The numerous polar amino acids allows for many opportunities for hydrogen bonding 2. The hydrophobic effect is caused by nonpolar groups that are drawn to one another to avoid water a. Nonpolar compounds can interact through London forces b. Nonpolar side chains become folded into the interior to the proteins to avoid water exposure 3. An ionic bond (salt bridge) is formed when positively charged side chains are attracted to negatively charged side chains a. The polar-basic and polar-acidic side chains are responsible for ionic bonds b. The ionic interactions usually take place on the protein surface 12-3
F. Covalent Interactions Important to Tertiary Structure 1. Disulfide bonds (bridges) are formed when the –CH2SH side chains of two cysteine residues react (recall the oxidation of thiols produces disulfides) a. Examine Figure 12.11 b. Disulfide bonds are easily reduced to re-form thiol groups; c. Because cells have reducing agents (i.e. NAHD), disulfide bonds are less common in proteins within a cell than in ones outside of a cell d. Disulfide bonds are strong (covalent) and often help proteins maintain their native structure 2. Prosthetic groups are nonpeptide components that are tightly bound to the polypeptide chain a. Examine Figure 12.12 b. Proteins that require a prosthetic group to be biological activity are called conjugated c. Proteins that function without a prosthetic group are called simple G. Quaternary (4°) Structure 1. Many native proteins are a combination of more than one polypeptide chain 2. This arrangement of chains in a functioning protein is called quaternary structure 3. Examine Figure 12.13 4. Quaternary structure is largely maintained through hydrogen bonding, the hydrophobic effect, ionic bonds, and disulfide bonds 5. Cooperation between the subunit chains is often a factor in the regulation of protein activity (i.e. the binding of a compound to one subunit can alter the tertiary structure of another subunit, which greatly affects that subunits ability to bind) *See Sample Problem 12.6 and Practice Problem 12.6 BIOCHEMISTRY LINK Hemoglobin and Collagen (Examine Figure 12.14 and 12.15) HEALTH LINK Immunotherapy (Examine Figure 12.16) V. Denaturation (Section 12.5) A. Denaturation Overview 1. Recall the native shape is the structure of a protein that is biologically active 2. Denaturation is any change in protein structure caused by disruption of the non-covalent forces and disulfide bonds that are responsible for maintaining secondary, tertiary, and quaternary structure 3. A loss of biological activity normally accompanies denaturation 4. Examine Figure 12.17 5. Denaturation is not reversible unless the changes that take place are minor DID YOU KNOW? Miraculin B. Denaturing Agents 1. An increase in temperature increases the kinetic energy; this increase in motion can disrupt hydrogen bonds and other noncovalent interactions 2. Varying pH affects protein shape because the charges on amino acid side chains involved in salt bridges may disappear 3. Detergents and soaps which are amphipathic interfere with hydrophobic interactions *See Sample Problem 12.7 and Practice Problem 12.7 VI. Enzymes (Section 12.6) A. Enzymes and Substrates 1. Most of the thousands of chemical reactions taking place within your body at any given time would not occur without the help of biological catalysts called enzymes 2. Most enzymes are globular proteins (a few newly discovered ones are RNA molecules) 3. Enzymes catalyze reactants known as substrates 4. Enzyme nomenclature involves the specification of the reaction or the substrate being catalyzed 5. Review examples in text (Aldehyde reductase and chymotrypsin nomenclature) 6. Examine Figure 12.18 B. Specificity 12-4
1. There are different types of enzyme specificity (selectivity) 2. In absolute specificity, the enzyme accepts only one specific substrate (i.e. urease) 3. In relative specificity, the enzyme will catalyze a wide range of molecules having the same functional groups or similar structures (i.e. alcohol dehydrogenase) 4. In stereospecificity, the enzyme will only react with or produce a particular stereoisomer: geometric isomers cis/trans, enantiomers D/L (i.e. stearoyl-CoA 9-desaturase) 5. Examine Figure 12.19 *See Sample Problem 12.8 and Practice Problem 12.8 C. Catalysis 1. Recall temperature, concentration, and catalysts can all influence the rate of a chemical reaction a. An increase in temperature increases the kinetic energy of the reactants, thus allowing them to have more frequent collisions; this results in an increased reaction rate b. Increasing the concentration of reactants causes more reactants to be packed into a given volume, which improves the chances they will collide properly; this results in an increased reaction rate c. Catalysts increase reaction rates by lowering the activation energy 2. Activation energy does not determine whether a reaction is spontaneous or nonspontaneous (an uncatalyzed reaction that is nonspontaneous is still nonspontaneous after the addition of a catalyst) 3. Enzyme catalysis is a direct result of the interactions that take place between it and the substrate a. Substrates attach to specific binding sites called active sites b. The enzyme (E) and substrate (S) bind to form the enzyme-substrate (ES) complex c. Amino acid side chains that line the enzyme’s active site interact with the substrate and hold it in place d. The substrate is held in the proper orientation near other atoms that are involved in the bond making or breaking processes e. Once the product (P) has been formed, it leaves the enzyme in its original form ready for another cycle of catalysis f. Examine Figure 12.20 4. Enzyme catalysis varies depending on the enzyme and the reaction being catalyzed 5. Review examples in text (Papain, acetylcholinesterase, and catalase rates) DID YOU KNOW? Orotidine 5’-Monophoshpate Decarboxylase D. Cofactors and Coenzymes 1. Some enzymes require the presence of cofactors for their catalytic action 2. Cofactors are inorganic ions (i.e. Fe2+, Cu2+, Na+) or organic compounds called coenzymes 3. Most coenzymes are derived from water-soluble vitamins 4. See Table 12.2 for selected enzyme cofactors 5. In about 1/3 of known enzymes, the metal ions are tightly bound to the enzyme, thus these cofactors are prosthetic groups 6. For other enzymes, the cofactors are not considered to be prosthetic groups because they are loosely held or they come and go during the reaction E. The Effect of pH and Temperature 1. Because enzymes are globular proteins, their ability to act as catalysts depends on maintaining their native state 2. Proteins are sensitive to pH because pH influences the charges on the amino acids side chains that maintain tertiary and quaternary structure a. The pH optimum is the pH at which an enzyme is most active b. Most human enzymes are most efficient around pH 7 (physiological pH) and they denature at pH extremes c. The lethal consequences of acidosis and alkalosis are mainly due to pH-induced changes in the structures of enzymes and proteins 12-5
d. Some enzymes have pH optimum far outside the usually physiological range (i.e. pepsin has an optimum near pH 1.5 because it functions in the acidic environment of the stomach) 3. Examine Figure 12.21 4. A rise in temperature leads to an increase in enzyme-catalyzed reaction rates until a temperature is reached where the protein begins to denature a. Deaths due to high fever are typically a result of the denaturation of enzymes/proteins b. The temperature optimum is the temperature at which an enzyme is most active c. Most human’s enzymes are most efficient around 37°C and denature before they reach 90°C d. Some enzymes in bacteria living in hot springs have temperature optimums at 100°C e. Examine Figure 12.22 VII. Control of Enzyme-Catalyzed Reactions (Section 12.7) A. Overview of Enzyme Controlled Reactions 1. Reaction rate is the measure of how quickly reactants are consumed and products are formed 2. Raising the reactant concentration of an enzyme-catalyzed reaction increases the reaction rate; however, once the enzyme is working at top speed, adding more substrate (reactant) will not increase the rate any further 3. The two well-studied groups of enzymes with respect to varying substrate concentrations are Michaelis-Menten enzymes and allosteric enzymes B. Michaelis-Menten Enzymes 1. Michaelis-Menten enzymes (named after two scientists) show how enzymes are controlled through a two step process: a. Step 1: Enzyme (E) binds to substrate (S) to form an enzyme-substrate complex (ES) b. Step 2: Substrate is transformed into product (P) and then released (ES → E + P) 2. KM (Michaelis constant) is the measure of the strength of the attraction between the enzyme and the substrate (describes step 1) 3. Vmax (maximum velocity) is the reaction rate that given concentration of enzyme can produce 4. Each enzyme has a unique KM and Vmax under particular reaction conditions 5. Review example in text (Alcohol dehydrogenase KM and Vmax) 6. Examine Figure 12.23 7. Review example in text (Orange peeling in figure 12.23) C. Inhibition 1. An inhibitor is any compound that reduces an enzymes ability to act as a catalyst 2. Irreversible inhibitors react with enzymes to produce a protein that is not a catalyst (i.e. aspirin blocking COX enzymes responsible for prostaglandins production—pain) a. Examine Figure 12.24 b. Irreversible inhibitors permanently render enzymes useless 3. Reversible inhibitors interfere with the enzymes’ catalytic ability, but the inhibition is not permanent because the inhibitors are loosely bound to the enzyme a. For Michaelis-Menten enzymes the two types of reversible inhibition are competitive and noncompetitive b. Competitive inhibitors compete with the substrate for the active site of the enzyme i. They usually have structure similar to that of the substrate ii. Some drugs act as competitive inhibitors (i.e. allopurinol, sulfanilamide) iii. Examine Figure 12.25 iv. Review example in text (Orange ball and orange peeling in figure 12.23) v. Competitive inhibitors affect KM but not Vmax c. Noncompetitive inhibitors attach to another site other than the active site of the enzyme i. They usually have a very different structure than the substrate ii. Heavy metal ions (Pb2+ and Hg2+) are examples of noncompetitive inhibitors because they react with the sulfur groups in enzymes iii. Review example in text (Mitten and orange peeling in figure 12.23) 12-6
iv. Noncompetitive inhibitors affect Vmax but not KM *See Sample Problem 12.9 and Practice Problem 12.9 HEALTH LINK Tamiflu and Relenza (Examine Figure 12.26 & Figure 12.27) D. Allosteric Enzymes 1. Allosteric enzymes are enzymes that display cooperativity, which is when the binding of substrate to the active site on one subunit affects the binding of the substrate at other subunits 2. Allosteric enzymes are regulated by ions or molecules called allosteric effectors a. Effectors bind at sites other than active site and they usually have very different structures from the substrates b. The attachment of the effector on the enzyme changes the 3D shape of the enzyme subunits and thus influences the enzyme’s ability to bind and catalyze the substrate c. Positive effectors enhance substrate binding and increase the speed of the catalyzed reaction d. Negative effectors reduce binding and slow the reaction e. Allosteric control is an important form of self-regulation for an enzyme (i.e. it acts as an “on/off” switch) f. KM and Vmax are terms reserved for Michaelis-Menten enzymes—they do not translate well to allosteric enzymes 3. Feedback inhibition is where the product of a metabolic pathway prevents its own production by controlling one or more of the enzymes responsible for its synthesis a. The product usually functions as a negative effector b. The control point is usually early in the metabolic pathway c. Examine Figure 12.28 E. Covalent Modification 1. In covalent modification, bonds are formed or broken within the enzyme 2. Phosphorylation (formation of phosphate esters) and dephosphorylation (phosphate ester hydrolysis) are both examples 3. Examine Figure 12.29 4 Review example in text (Glycogen phosphorylase) 5. Another type of covalent modification involves zymogens, which are inactive enzyme precursors that are activated by the hydrolysis of a peptide bond 6. Trypsin and chymotrypsin are both examples of enzymes that begin as zymogens 7. Examine Figure 12.30 8. Review example in text (Activation of trypsinogen and chymotrypsinogen) HEALTH LINK Proteins in Medicine (Examine Figure 12.31)
Chapter Summary Section 12.1 Amino acids are molecules that contain a carboxyl group (–CO2H) and an amino group (–NH2). There are twenty α-amino acids that are used to make peptides and proteins. Amino acid structure is dependent on pH; and at physiological pH, many amino acids exist as zwitterions, which are ions with one positive and one negative charge. The pH at which the net charge of an amino acid is zero is known as the isoelectric point (pI). All amino acids are hydrophilic due to their polar covalent bonds; however their side changes can be classified into four categories: nonpolar, polar-acidic, polar-basic, and polar-neutral. All the amino acids except glycine are chiral and thus they exist as enantiomers. In Fischer projections, D-amino acids have the amino group on the right side while L-amino acids have the amino group on the left side. Only L-amino acids are used in the proteins of living things. Section 12.2 The carboxyl and amino groups of amino acids can react to form peptide bonds (amide bonds). Amides formed by combing amino acids are called peptides. Oligopeptides contain 2-10 amino acids, and include dipeptides (two amino acids), tripeptides (three amino acids), and tetrapeptides (four amino 12-7
acids). Peptides with 10-50 amino acids are classified as polypeptides while those with more than 50 amino acids are considered proteins. It is customary to draw peptides and proteins with the N-terminus (unreacted amino group) on the left and the C-terminus (unreacted carboxyl group) on the right. Peptides and proteins are named by listing amino acids in order from the N- to C-terminus; however due to length, most large peptides and proteins are known by common names. Peptides and proteins can be hydrolyzed back into amino acids via an amide hydrolysis reaction. Section 12.3 Peptide and protein structure is just as sensitive to pH as is amino acid structure. In acidic solutions, amino and carboxyl groups appear in their acidic forms (–NH3+ and –CO2H); in neutral solutions, amino groups remain in their acidic form (–NH3+), but carboxyl groups are found in their basic form (–CO2-); and in basic solutions, amino and carboxyl groups appear in their basic forms (–NH2 and –CO2-). The peptide bonds are not acidic or basic, and therefore their structure does not vary with changes in pH. When peptides carry either a net positive or a net negative charge, the liked-charged molecules repel one another and stay dissolved in water. Conversely, when the net charge is zero, the molecules are more likely to precipitate out of aqueous solution. Section 12.4 Proteins can be classified in general terms as either fibrous, which exist as water insoluble fibers, or globular, which are water soluble spheres. Protein structure is understood in four levels of organization: primary, secondary, tertiary, and quaternary. Primary structure refers to the order of amino acid residues in a peptide or protein. Secondary structure, such as an α-helix or a β-sheet, results from hydrogen bonding that takes place between groups on the polypeptide backbone. Tertiary structure refers to the overall three-dimensional shape of a protein. There is usually only one shape that leads to a native (biologically active) molecule. The native state of globular protein usually has binding sites that allow other substances to attach/bind to the protein. Side-chain interactions such as hydrogen bonding, the hydrophobic effect, ionic bonds (salt bridges), and disulfide bonds (bridges) determine the tertiary structure of a protein. Some proteins have prosthetic groups, which are nonpeptide components tightly bound to the polypeptide chain. Conjugated proteins require a prosthetic group to be biological activity while simple proteins do not. Many native proteins are a combination of more than one polypeptide chains. This arrangement of chains in a functioning protein is called quaternary structure. Section 12.5 Denaturation is any change in protein structure caused by disruption of the non-covalent forces and disulfide bonds that are responsible for maintaining secondary, tertiary, and quaternary structure. A loss of biological activity normally accompanies denaturation, and it is not reversible unless the changes that take place are minor. Temperature and pH changes as well as detergents/soaps can all cause denaturation. Section 12.6 Globular proteins called enzymes are responsible for catalyzing nearly all biological reactions. Enzymes catalyze reactants known as substrates into products. Three major types of enzyme specificity (selectivity) include absolute specificity (the enzyme accepts only one specific substrate), relative specificity, (the enzyme will catalyze a wide range of molecules having the same functional groups or similar structures), and stereospecificity (the enzyme will only react with or produce a particular stereoisomer). Enzyme catalysis is a direct result of the interactions that take place between the enzyme and the substrate. Substrates (S) attach to specific binding sites called active sites on the enzyme (E) to form the enzyme-substrate (ES) complex. Once the product (P) has been formed, it leaves the enzyme in its original form which is then ready for another cycle of catalysis. Some enzymes require presence of cofactors, which are inorganic ions or organic compounds called coenzymes, for their catalytic ability. Because enzymes are proteins, their ability to act as catalysts depends on maintaining their native state. The pH optimum is the pH at which an enzyme is most active. Likewise temperature optimum is the temperature at which an enzyme functions best. 12-8
Section 12.7 Michaelis-Menten enzymes show how enzymes are controlled through two steps: 1) the enzyme (E) binds to substrate (S) to form an enzyme-substrate complex (ES), and 2) the substrate is transformed into product (P) and then released (ES → E + P). KM (Michaelis constant), which describes step 1, is the measure of the strength of the attraction between the enzyme and the substrate. Vmax (maximum velocity), which describes step 2, is the reaction rate that a given concentration of enzyme can produce. Compounds known as inhibitors reduce an enzyme’s ability to act as a catalyst. Irreversible inhibitors permanently render enzymes useless while reversible inhibitors only temporarily interfere with the enzyme’s catalytic ability. The two types of reversible inhibitors are competitive inhibitors, which compete with the substrate for the active site thus affecting KM, and noncompetitive inhibitors, which attach to another site on the enzyme thus affecting Vmax. Another type of enzymes, called allosteric enzymes, display cooperativity, which when the binding of substrate to the active site on one subunit affects the binding of the substrate at other subunits. Allosteric enzymes are regulated by allosteric effectors: positive effectors speed up catalyzed reactions while negative effectors slow them down. Another type of control, called feedback inhibition, involves the product of a metabolic pathway preventing its own production by controlling one or more of the enzymes responsible for its synthesis. In yet another type of control, called covalent modification, bonds are formed or broken within the enzyme, or zymogens (inactive enzyme precursors are activated by the hydrolysis of a peptide bond).
Lecture Suggestions Sections 12.1, 12.2, and 12.3 Begin lecture by drawing four nonionized amino acids (Phe, Ser, Asp, Lys) on the board. Show students the carboxyl group and the amino group of Phe and ask them what organic reaction these two functional groups are involved in. They should reply amide formation, which you can clarify this as peptide formation in biochemistry. Tell students that like polysaccharides, proteins are polymers made up of monomers (amino acids). Explain how each amino acid monomer can react with the next (due to the presence of both functional groups on each molecule) thus forming a polymer. Next show students the three parts to each amino acid (N, C, and side chains). Explain that all four amino acids on the board are polar; however, each is classified based on its side chain. Point to each side chain and have students tell if they are polar or nonpolar; then do the same having them tell if they are acidic, basic, or neutral (they should remember this from organic chemistry). Next, explain how pH affects each amino acid (Figure 12.1). For pH 1, draw a bunch of H+ ions and have students help you predict the form the amino acid will take. For pH 7, explain how the carboxylic acid loses an H+ and the amine gains an H+ to form the zwitterion. For pH 14, draw a bunch of OH- ions and have students help you predict the form the amino acid will take. For section 12.2, show the actual peptide formation between each of the four amino acids. Explain after you form the first peptide bond that the product is called a dipeptide. Repeat with the tripeptide and tetrapeptide. Show students how the tetrapeptide is named (N-terminus to C-terminus). For section 12.3, explain how tetrapeptide on the board is affected by pH using the same techniques as before (Figure 12.6). Sections 12.4 and 12.5 To explain the four levels of protein structure, bring in two coiled phone cords. Have two student volunteers pull one of them until it uncoils. Tell the class that this represents primary structure of a protein. For a better affect, you can use different colored permanent makers to color 2-inch sections (before class) to represent the different amino acids. Explain various aspects of primary structure (peptide bonds, etc.). Next, have the two students move together and let the phone cord coil back up. Tell the class that this represents an α-helix, which is a type of secondary structure; then explain secondary structure (α-helix vs. β-sheets, polypeptide backbone interactions, etc.). Next, have one student tangle the phone cord up by twisting and turning it on itself. Tell the class that this represents 12-9
tertiary structure and then explain it (side-chain interactions, noncovalent vs. covalent, etc.). Lastly, have the two students tangle both phone cords together. Tell the class that this represents quaternary structure and then explain it (multiple polypeptide chains, side-chain interactions, etc.). To take this demonstration even further, you could bring in a small object and have students tangle the phone cord around it to represent a prosthetic group. For section 12.5, you can have two to four students pull apart the tangled (quaternary structure) protein to show it being denatured by pH, temperature, or detergents. Sections 12.6 and 12.7 For specificity in section 12.6, compare enzymes to locks and substrates to keys. For absolute specificity explain that the key must be a perfect fit into the lock (no other shaped key will work). Compare relative specificity to a lock that has more than one key (i.e. generic key, or master key, or old-time skeleton key). For stereospecificity, ask students what the affect would be if you flipped one of the teeth upside down on the standard key. You can also ask why humans can digest starch with amylase but not cellulose. For section 12.7, you can actually demonstrate Figure 12.23. All you need is a few oranges, a few orange balls and a heat mitten. As a student volunteer acts as the enzyme, you can explain MichaelisMenten enzymes, competitive inhibitors, and noncompetitive inhibitors.
Handouts for Students 1) The Common Amino Acids handout (next page) can help students distinguish between the different types of classes of amino acids encountered in this chapter.
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Handout Common Amino Acids
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Chapter 13 Nucleic Acids Outline Notes I. Nucleic Acid Building Blocks (Section 13.1) A. Phosphate 1. Nucleotides, which make up nucleic acids, are composed of three building blocks: a phosphate, a monosaccharide, and an organic base 2. Recall phosphoric acid has three forms depending on pH; and at physiological pH the two major forms present are H2PO4- and HPO423. The phosphate group allows for the formation of phosphate monoesters and diesters 4. Examine Figure 13.1 B. Monosaccharides 1. D-Ribose and D-2-deoxyribose are the two monosaccharides incorporated into nucleotides 2. Each appears in its β-furanose form in nucleotides 3. In nucleotides the monosaccharide reacts to form phosphate esters and N-glycosides 4. Examine Figure 13.2 5. Review example in text (Phosphate esters and N-glycoside formation) C. Organic Bases 1. Amine bases incorporated into nucleotides resemble either purine or pyrimidine 2. The five major bases found in nucleotides include the purines adenine (A) and guanine (G) and the pyrimidines cytosine (C), thymine (T), and uracil (U) 3. Examine Figure 13.3 4. There are more than 50 other bases, which are called minor bases; these bases are formed by enzyme-catalyzed modifications of the major bases 5. Examine Figure 13.4 D. Nucleosides 1. A nucleoside is composed of a monosaccharide and purine or pyrimidine base 2. Nucleosides containing ribose are called ribonucleosides while those containing 2-deoxribose are called deoxyribonucleosides 3. The monosaccharides are connected to the organic bases via a β N-glycosidic bond 4. Examine Figure 13.5 5. Enzymes insure that nucleosides consisting of ribose are only paired with A, G, C, or U while those consisting of 2-deoxyribose are only paired with A, G, C, or T E. Nucleotides 1. Nucleotides are produced from nucleosides reacting with phosphate to form a phosphoester bond 2. Examine Figure 13.6 3. Ribonucleosides react to form ribonucleotides and deoxyribonucleosides react to form deoxyribonucleotides 4. Review examples in text (CMP, dAMP, etc.) *See Sample Problem 13.1 and Practice Problem 13.1 *See Sample Problem 13.2 and Practice Problem 13.2 II. Nucleoside Di- and Triphosphates, Cyclic Nucleotides (Section 13.2) A. Nucleoside Diphosphates 1. Attaching a second phosphate to the phosphate of a nucleotide generates a nucleoside diphosphate 2. The phosphoester bond attaching the two phosphates is called a phosphoranhydride bond 3. Examine Figure 13.7a B. Nucleoside Triphosphates 1. Attaching a third phosphate to the nucleotide generates a nucleoside triphosphate 13-1
2. Nucleoside triphosphates (i.e. ATP) are the immediate source of energy for most of the energyrequiring processes that take place in living things 3. Nucleoside triphosphates are also used in the synthesis of polynucleotides and cyclic nucleotides C. Cyclic Nucleotides 1. The phosphate residue can attach at two different positions on the same nucleoside 2. Cyclic nucleotides (i.e. cyclic AMP or cAMP) are formed from one phosphate residue attaching twice to the same monosaccharide residue (enzymes called cyclases form them) 3. Examine Figure 13.7b 4. Cyclic nucleotides help regulate many biochemical processes: many act as effectors for allosteric enzymes *See Sample Problem 13.3 and Practice Problem 13.3 III. Polynucleotides (Section 13.3) A. Oligonucleotides and Polynucleotides 1. Oligonucleotides contain 2-10 nucleotide residues connected to one another while polynucleotides contain more than 10 nucleotide residues 2. The nucleotide residues are joined to one another by 3′5′-phosphodiester bonds 3. Examine Figure 13.8 4. The sequence of nucleotides in an oligonucleotide/polynucleotide is listed from the 5′-terminus to the 3′-terminus 5. Typically the 5′ end is a phosphate monoester and the 3′ end is a free –OH group B. Sugar-Phosphate Backbone 1. The sugar-phosphate backbone consists of repeating units of phosphate attached to the monosaccharide attached to phosphate, and so on 2. One of the organic bases is attached at C-1′ of each residue, and the phosphate connecting the residues carries a negative charge (physiological pH) 3. Deoxyribonucleic acid (DNA) contains the sugar 2-deoxyribose and the bases A, G, C, T while ribonucleic acid (RNA) contains the sugar ribose and the base A, G, C, U (U replaces T) *See Sample Problem 13.4 and Practice Problem 13.4 4. Examine Figure 13.9 *See Sample Problem 13.5 and Practice Problem 13.5 IV. DNA Structure (Section 13.4) A. DNA Structure Overview 1. Every cell in particular living thing contains the exact same DNA 2. DNA is found in the cell nucleus of plant and animal cells 3. Human DNA consists of 3 billion deoxyribonucleotides and carries an estimated 25,000 genes a. Genes are stretches of DNA that carry the codes for protein production b. Only 5% of human DNA comes from genes and the rest is considered “noncoding” B. Primary Structure 1. The primary structure of nucleic acids refers to the sequence of its nucleotide residues 2. Review examples in text (Primary structure of DNA/RNA in Figure 13.8/13.9) 3. In dTGCA, only the order of base residues are noted because the sugar-phosphate backbone remains constant, and the “d” indicates it is DNA (contains 2-deoxyribose) 4. Alterations in primary structure can affect the function of a nucleic acid C. Secondary Structure 1. The secondary structure of nucleic acids pertains to the helix formed by the interaction of two DNA strands 2. Examine Figure 13.10 3. The two single strands lie side by side in an antiparallel arrangement (one running 5′ to 3′ and the other running 3′ to 5′) 4. The two strands are held together by base pairing, which involves hydrogen bonding between the bases attached to the sugar-phosphate backbone 13-2
5. The base pairing is complementary in that A bonds with T through 2 hydrogen bonds and G bonds with C through 3 hydrogen bonds 6. Examine Figure 13.11 7. The double helix (i.e. resembles a spiral staircase) has the negatively charged sugar-phosphate backbone on the outside and the bases on the inside 8. Hydrophobic interactions between neighboring bases on the same DNA strand help stabilize the double helix D. Tertiary Structure 1. The tertiary structure of nucleic acids refers to the overall 3D shape 2. Supercoiling involves the double helices being twisted into compact and complex shapes 3. Review example in text (Supercoiling in bacterial plasmids) 4. Examine Figure 13.12 5. In plants and animals, the DNA found in the nucleus is supercoiled by proteins called histones 6. Nucleosomes arise when DNA wraps around a group of histones 7. Examine Figure 13.13 8. Nucleosomes then twist further to produce chromatin, which makes up a chromosome 9. Human cells normally contain 23 pairs of chromosomes with one member of each pair contributed by each parent *See Sample Problem 13.6 and Practice Problem 13.6 V. Denaturation (Section 13.5) A. Denaturation Overview 1. Recall the double helix is held together by hydrogen bonding between complementary bases 2. When this hydrogen bonding is disrupted, the DNA strand has been denatured B. Denaturing Conditions 1. DNA can be denatured by the same conditions that denature proteins (i.e. heat, pH, etc.) 2. Depending on the amount of heat added, the DNA strand may unwind or separate entirely; however, DNA can “renature” when cooled 3. Examine Figure 13.14 4. Denaturation by heat has many applications *See Sample Problem 13.7 and Practice Problem 13.7 VI. Nucleic Acids and Information Flow (Section 13.6) A. DNA Replication 1. DNA contains all the information necessary for making the proteins required for life 2. In DNA replication, the information stored in DNA is passed to a new generation of cells 3. Cells use the existing DNA as a template for making new DNA molecules before it divides B. Transcription 1. Transcription is the process of mapping portions of the primary structure of DNA (templates) onto RNA molecules 2. Some viruses (i.e. HIV) are capable of reverse transcriptions, which is when viral RNA is read to produce viral DNA that used to make new viruses C. Translation 1. Translation is the process where the primary structure of RNA is read to generate the correct primary structure of a protein 2. Examine Figure 13.15 *See Sample Problem 13.8 and Practice Problem 13.8 VII. DNA Replication (Section 13.7) A. DNA Replication Process 1. In semiconservative replication, each strand of the double helix serves as a template for the manufacture of a new DNA strand 2. Replication begins at the origin which is where the separated DNA strands are exposed a. In bacteria only one origin is formed while in plants and animals many origins are opened 13-3
b. Single-strand binding proteins (SSBs) protect the DNA from endonucleases, which are protective enzymes that destroy foreign DNA c. Each end of the origin contains replication fork where replication occurs 3. Examine Figure 13.16 4. DNA polymerase is the enzyme that produces new DNA a. It attaches dATP, dGTP, dCTP, and dTTP with its proper complement on the parent strand b. Each deoxyribonucleotide residue is adding to the growing strand as DNA polymerase moves along the parent strand c. DNA polymerase moves in a 3′ to 5′ direction on the parent strand, thus the newly synthesized DNA strand runs from 5′ to 3′ 5. More than one DNA polymerase is used to speed up the process; however, this creates breaks in the newly formed DNA molecule 6. Enzymes called ligases seal the breaks by forming phosphoester bonds B. Proofreading and Repair 1. DNA polymerase inserts the wrong residues about once every 100,000 times 2. The DNA polymerase enzymes proofreads its work and tries to fix any mistakes 3. Repair enzymes fix any errors not caught by the DNA polymerase proofreading 4. Proofreading and repair lessens the error rate to less than one in 1 billion *See Sample Problem 13.9 and Practice Problem 13.9 VIII. Transcription and RNA (Section 13.8) A. Transcription 1. RNA differs from DNA in three major ways: a. RNA is composed of the sugar ribose while DNA is composed of the sugar 2-deoxyribose b. RNA uses the pyrimidine U instead of T c. RNA usually exists as a single strand rather than double strand 2. Transcription begins when RNA polymerase attaches to a promoter site on the DNA 3. RNA polymerase disrupts the double helix so the template strand can be read 4. Examine Figure 13.17 5. RNA polymerase attaches ATP, GTP, CTP, and UTP to growing RNA strand according to complementary base pairing a. C in DNA is attached to G in RNA and vice versa b. T in DNA is attached to A in RNA but A in DNA is attached to U in RNA 6. RNA polymerase moves down the DNA templates in 3′ to 5′ direction until it reaches a termination sequence a. The DNA double helix reforms after RNA polymerase b. The terminations sequence is a series of bases that signals the end of the DNA segment 7. In plants and animals transcription occurs in the nucleus a. The RNA then leaves the nucleus and enters the cytoplasm or endoplasmic reticulum b. The RNA is then used to synthesize proteins 8. Not all DNA is transcribed—most of the DNA transcribed is genes B. Types of RNA 1. The three types of RNA used in protein synthesis are transfer RNA (tRNA), messenger RNA (mRNA), and ribosomal RNA (rRNA) 2. tRNAs (the smallest type) carries the correct amino acid to the site of protein synthesis 3. Examine Figure 13.18 4. mRNAs carry the information that specifies which protein should be made 5. rRNAs (the longest type) combine with proteins to form ribosomes which are the multisubunit complexes where proteins synthesis occurs 6. RNA undergoes post-transcriptional modification (except in bacteria) where RNA is shortened and some major bases (A, G, C, U) are converted into minor bases HEALTH LINK Lupus (Examine Figure 13.19) 13-4
C. The Genetic Code 1. Codons are series of 3 mRNA bases that are read to produce a primary structure of a protein 2. There are 64 codons that code for the 20 amino acids used in proteins 3. Genetic code (triplet code) show which codon(s) specifies each amino acid 4. See Table 13.1 for codons in the 5′ to 3′ sequence of mRNA 5. The genetic code is universal because nearly all organisms use the same codons to specify the same amino acids 6. AUG is called the start codon because it codes for methionine as well as indicates the place to start making a protein (fMet is removed before the protein is completed) 7. The three codons UAA, UAG, and UGA single the place to stop protein synthesis 8. Each tRNA carries an anticodon, which is three bases that are complementary to a codon 9. Review example in text (AGC codon and UCG anticodon) 10. Base pairing between the codon and anticodon ensures that the correct amino acid is added to the growing chain at the right time 11. Synthetase enzymes are responsible for matching amino acids with their tRNA carriers *See Sample Problem 13.10 and Practice Problem 13.10 IX. Translation (Section 13.9) A. Translation Overview 1. Translation (protein synthesis) occurs in three steps: initiation, elongation, and termination 2. Initiation is where a ribosome, mRNA, and tRNA come together to form a complex 3. Elongation is where amino acids are joined to the growing polypeptide chain 4. Termination is when the protein has been synthesized and the complex dissociates B. Initiation 1. mRNA and tRNA combine with ribosome subunits near the 5′ end of the start codon (AUG) 2. The tRNA with the anticodon UAC carries the amino acid fMet 3. The fMet becomes the N-terminal amino acid residue in the protein being synthesized 4. Examine Figure 13.20 DID YOU KNOW? Viruses C. Elongation 1. A second tRNA-amino acid binds beside the first 2. The ribosome catalyzes a reaction between the two amino acids, forming a new peptide bond and breaking the bond between the first amino acid and its tRNA 3. The first tRNA leaves the ribosome and the translocation occurs, which is when the ribosome shifts down the mRNA three nucleotides to the next codon 4. A third tRNA-amino acid attaches and the process repeats causing the polypeptide chain to grow D. Termination 1. Protein synthesis is halted when the ribosome has a stop codon 2. An enzyme removes the tRNA from the last amino acid in the polypeptide chain 3. The mRNA and tRNA dissociates from the ribosome which then separates E. Post-Translational Modification of Proteins 1. After translation, proteins are processed before becoming active 2. Post-translational modifications include: a. Removal of fMet at the N-terminus b. Trimming and cutting of the polypeptide chain c. Folding the protein into its proper 3D shape d. Addition of prosthetic groups (i.e. hem, metal ions, etc.) e. Formation of disulfide bridges f. Modification of amino acid residues (i.e. converting L into D amino acids) g. Association with other polypeptide strands (i.e. quaternary structure) *See Sample Problem 13.11 and Practice Problem 13.11 X. Control of Gene Expression (Section 13.10) 13-5
A. Gene Expression 1. The DNA of organisms contain thousands of genes which are not continually expressed (read to make proteins) 2. This is because the production of unneeded proteins would be an inefficient use of resources 3. There are a variety of techniques that organisms use to control gene expression: a. Switching transcription on and off b. Controlling the rate of post-transcriptional modification of RNA c. Controlling the rate of post-translational modification of proteins B. Escherichia coli and Operons 1. Escherichia coli (E. coli) are symbiotic bacteria that live in the intestine of humans and animals 2. E. coli contains an enzyme β-galactosidase that catalyzes the hydrolysis of lactose into glucose and galactose 3. When lactose is present the manufacture of β-galactosidase is turned on 4. E. coli uses an operon, which is a group of genes under the control of one promoter site, to control this switching on/off 5. Examine Figure 13.21 6. Review example in text (Lac operon and gene expression) *See Sample Problem 13.12 and Practice Problem 13.12 HEALTH LINK RNA Interference (Examine Figure 13.22) XI. Mutation (Section 13.11) A. Mutation Overview 1. A mutation is any permanent change in the primary structure of DNA 2. Mutations can involve the switching of one base pair for another or the addition or deletion of base pairs 3. The common causes of mutations are errors in replication and exposure to mutagens 4. Mutagens are mutation-causing agents such as x-rays, ultraviolet (UV) radiation, nuclear radiation and chemicals 5. Many errors in DNA are repaired (recall DNA polymerase proofreads during replication and repair enzymes continually inspect and repair damage) B. Mutations and Humans 1. Only 5% of the total DNA in humans contains genes, thus the reaming 95% has little or no effect on gene expression 2. Even mutations in genes may have no adverse affect on an individual because several different codons code for the same amino acid (i.e. GGU, GGC, GGA, and GGG all code for glycine) 3. Genetic diseases are diseases caused by inherited mutations (over 4,000 genetic diseases have been identified in humans) *See Sample Problem 13.13 and Practice Problem 13.13 DID YOU KNOW? Breast Cancer Genes XII. Recombinant DNA (Section 13.12) A. Making Recombinant DNA 1. Scientists can now make recombinant DNA which contains DNA from two or more sources 2. The first step is to identify a gene of interest and cut it from the DNA strand 3. Restriction enzymes (endonucleases), which are bacterial protective enzymes, are used to cut the DNA in such a way that it can be combined to DNA from another source 4. Restriction enzymes create sticky ends which are single strands of DNA that are complementary 5. Examine Figure 13.23 6. Sticky ends allow DNA fragments from different sources to attach to each other 7. Review example in text (Human growth hormone combined with a bacterial plasmid) 8. Bacterial plasmids (circular DNA) that are recombinant (i.e. contain human DNA) can be introduced into bacteria to copy the gene B. Applications of Recombinant DNA 13-6
1. Recombinant DNA has allowed for the production of biological molecules such as human growth hormone and insulin 2. Review examples in text (Insulin produced from recombinant DNA) 3. Many applications have appeared in agriculture through the development of: a. Herbicide-resistant soybeans b. Insect-resistant cotton and rice 4. Review example in text (Flavr-Savr tomatoes) 5. Examine Figure 12.24 BIOCHEMISTRY LINK Glowing Cats (Examine Figure 13.25) *See Sample Problem 13.14 and Practice Problem 13.14 XIII. DNA Fingerprinting (Section 13.13) A. DNA Fingerprinting Overview 1. DNA fingerprinting can allow police to match crime scene DNA (i.e. hair, saliva, blood, etc.) to the DNA of a suspect 2. Forensic scientists only look at a number of short DNA fragments a. Recall human DNA is about 5% coding and 95% noncoding b. Much of the noncoding DNA consists of repeating series of base pairs c. Short tandem repeats (STRs) are relatively small stretches of DNA that contain short repeating sequences of bases (i.e. the FBI uses 13 STRs which are all less than 500 bases) 3. DNA analysis requires a minimum amount of DNA; therefore, polymerase chain reaction (PCR) is used to make additional DNA copies (only copies the DNA parts needed) B. Polymerase Chain Reaction 1. PCR involves combing the DNA of interest with the building blocks of DNA (dATP, dGTP, dCTP and dTTP), DNA polymerase, and primers (short segments of DNA that are complementary to the 3′ end of the DNA being copied/amplified) 2. The mixture is heated to 98°C which causes the double strand to separate 3. The temperature is then lowered to 60°C which allows the primers to bind 4. DNA polymerase then attaches near the primers and begins replicating the DNA 5. This process can be repeated multiple times to produce millions of copies within a few hours 6. Examine Figure 13.26 7. DNA polymerase isolated from bacteria that live in hot springs are used to overcome the problem of the enzyme denaturing at high temperatures C. STR Analysis 1. Once the DNA sample is large enough, restriction enzymes are used to cut the DNA strands into STRs, which can then be used to determine the number of repeats 2. See Table 13.2 for STRs and the probability of their occurrence 3. Review example in text (Probability of two people with the same DNA fingerprint) 4. Because STRs come from noncoding regions of DNA, DNA fingerprinting does not pose the confidentially problems associated with analyzing genes *See Sample Problem 13.15 and Practice Problem 13.15
Chapter Summary Section 13.1 Nucleotides, which are the building blocks of nucleic acids, are composed of a phosphate, a monosaccharide, and an organic base. The phosphate group allows for the formation of phosphate monoesters and diesters. D-Ribose and D-2-deoxyribose are the two monosaccharides incorporated into nucleotides. Amine bases incorporated into nucleotides resemble either purine, such as adenine (A) and guanine (G), or pyrimidine, such as cytosine (C), thymine (T), or uracil (U). A, G, C, T, and U are all considered major bases. When the monosaccharide reacts with a major base, a type of N-glycoside called a nucleoside is formed. Ribonucleosides contain ribose while deoxyribonucleosides contain 13-7
2-deoxribose. Ribose is only paired with A, G, C, or U while 2-deoxyribose is only paired with A, G, C, or T. When nucleosides react with phosphate to form a phosphoester bond, nucleotides such as ribonucleotide and deoxyribonucleotide are formed. Section 13.2 Attaching a second or third phosphate to the phosphate of a nucleotide generates a nucleoside diphosphate or nucleoside triphosphate, respectively. The phosphoester bond attaching the two phosphates is called a phosphoranhydride bond. Nucleoside triphosphates (i.e. ATP) are the immediate source of energy for most energy-requiring processes in living things. Nucleoside triphosphates are also used in the synthesis of polynucleotides and cyclic nucleotides, which are formed from one phosphate residue attaching twice to the same monosaccharide residue. Section 13.3 Oligonucleotides contain 2-10 linked nucleotide residues while polynucleotides contain more than 10 linked nucleotide residues. Nucleotide residues are joined to one another by 3′, 5′-phosphodiester bonds. Typically the 5′terminus has a phosphate monoester and the 3′ terminus has a free –OH group. The sugar-phosphate backbone of a nucleic acid consists of repeating units of phosphate attached to the monosaccharide attached to phosphate, and so on. Deoxyribonucleic acid (DNA) contains the sugar 2deoxyribose and the bases A, G, C, T while ribonucleic acid (RNA) contains the sugar ribose and the base A, G, C, U (U replaces T). Section 13.4 Human DNA consists of 3 billion deoxyribonucleotides and carries an estimated 25,000 genes, which are stretches of DNA that carry the codes for protein production. The primary structure of nucleic acids refers to the sequence of its nucleotide residues, and alterations in primary structure can affect the function of a nucleic acid. The secondary structure of nucleic acids pertains to the helix formed by the interaction of two DNA strands. The two strands are held together by base pairing, which involves the hydrogen bonding between the bases attached to the sugar-phosphate backbone. The base paring is complementary in that A bonds with T through 2 hydrogen bonds and G bonds with C through 3 hydrogen bonds. The tertiary structure of nucleic acids refers to the overall 3D shape, which arises from supercoiling (double helices being twisted into compact and complex shapes). In plants and animals, the DNA found in the nucleus is supercoiled by proteins called histones. When DNA wraps around a group of histones, a nucleosome arises, which ultimately makes up a chromosome. Section 13.5 When the hydrogen bonding that holds the double helix together is disrupted, the DNA strand has been denatured. DNA can be denatured by the same conditions that denature proteins (i.e. heat, pH, etc.). Depending on the amount of heat added, the DNA strand may unwind or separate entirely; however, DNA can “renature” when cooled. Section 13.6 DNA contains all the information necessary for making the proteins required for life. In DNA replication, the information stored in DNA is passed to a new generation of cells. Portions of DNA can also be mapped onto RNA molecules through the process of transcription. Some viruses (i.e. HIV) are capable of reverse transcription, which is when viral RNA is read to produce viral DNA. Once DNA is transcribed into RNA, the RNA can be read to generate the correct primary structure of a protein through a process known as translation. Section 13.7 In semiconservative replication, each strand of the double helix serves as a template for the manufacture of a new DNA strand. Replication begins at the origin which is where the separated DNA strands are exposed. Each end of the origin contains a replication fork where replication occurs. The enzyme DNA polymerase produces new DNA along the original DNA strand. More than one DNA 13-8
polymerase is used to speed up the process; however, this creates breaks in the newly formed DNA molecule, which must be sealed by ligases. DNA polymerase proofreads its work and tries to fix any mistakes. Repair enzymes then fix any errors not caught by the DNA polymerase proofreading. This lessens the error rate to less than one in 1 billion nucleotides. Section 13.8 Transcription begins when RNA polymerase attaches to a promoter site on the DNA. RNA polymerase disrupts the DNA double helix so the template strand can be read. RNA polymerase attaches the correct bases (U replaces T) as it moves down the DNA template until it reaches a termination sequence. Three types of RNA are produced during transcription: transfer RNA (tRNA), messenger RNA (mRNA), and ribosomal RNA (rRNA). tRNAs carries the correct amino acid to the site of protein synthesis, mRNAs carry the information that specifies which protein should be made, and rRNAs combine with proteins to form ribosomes, which is where protein synthesis occurs. A codon is a series of 3 mRNA bases that code for an amino acid; and an anticodon found on a tRNA molecule is complementary to each codon. The genetic code (or triplet code) shows which codon(s) specify each amino acid. Section 13.9 Translation (protein synthesis) occurs in three steps: initiation, elongation, and termination. In the initiation step, a ribosome, mRNA, and tRNA come together to form a complex. During the elongation step, amino acids are joined by peptide bonds into a growing polypeptide chain. In the termination step, the complex dissociates after the protein has been synthesized. After translation, proteins are processed through post-translational modifications before becoming active. Section 13.10 The DNA of organisms contain thousands of genes that are not continually expressed (read to make proteins) because the production of unneeded proteins would be an inefficient use of resources. Organisms use a variety of techniques to control gene expression including switching transcription on and off, controlling the rate of post-transcriptional modification of RNA, and controlling the rate of post-translational modification of proteins. For example, the bacteria Escherichia coli (E. coli) use an operon, which is a group of genes under the control of one promoter site, to control the switching on/off of its genes. Section 13.11 A mutation is any permanent change in the primary structure of DNA. The common causes of mutations are errors in replication and exposure to mutagens, which are mutation-causing agents such as x-rays, UV radiation, nuclear radiation and chemicals. Many errors in DNA are repaired DNA polymerase proofreading and repair enzymes. Because only 5% of the total DNA in humans contains genes, many mutations have no adverse affect on an individual. Mutations in genes also can have no adverse affect on an individual because several different codons code for the same amino acid. However, many mutations do have adverse affects including inherited mutations, which are the cause of over 4,000 genetic diseases in humans. Section 13.12 Scientists are now able make recombinant DNA which contains DNA from two or more sources. They use restriction enzymes to cut pieces of DNA. The restriction enzymes create sticky ends, which allow DNA fragments from different sources to attach to each other. There are many applications of recombinant DNA including 1) the production of biological molecules such as human growth hormone and insulin, 2) herbicide-resistant and insect-resistant crops, and 3) foods such as Flavr-Savr tomatoes. Section 13.13 DNA fingerprinting can allow police to match crime scene DNA (i.e. hair, saliva, blood, etc.) to the DNA of a suspect. Forensic scientists examine short tandem repeats (STRs), which are relatively small 13-9
stretches of DNA that contain short repeating sequences of bases. DNA analysis requires a minimum amount of DNA; therefore, polymerase chain reaction (PCR) can be used to make additional DNA copies. Once the DNA sample is large enough, restriction enzymes are used to cut the DNA strands into STRs, which can then be used to determine the number of repeats. Because STRs come from noncoding regions of DNA, DNA fingerprinting does not pose the confidentially problems associated with analyzing genes.
Lecture Suggestions Sections 13.1, 13.2, 13.3, 13.4, and 13.5 For these sections, use a Venn diagram to compare DNA and RNA. Place the following Venn diagram and words on the board/screen. Have students place the following the words correctly in the diagram.
DNA Adenine (A) Thymine (T) Guanine (G) Cytosine (C) Uracil (U) Ribose 2-Deoxyribose
RNA
Polynucleotides Phosphate Double Helix Single Stranded Nucleotides Phosphodiester Bonds Sugar-Phosphate Backbone
Sections 13.6, 13.7, 13.8, 13.9, 13.10, and 13.11 Write the following DNA strand (gene) on the board TATCCCGTACATGCAGATCCTGGCCGTGAAGTTTAAACGACCCTTTCC Have students transcribe the sequence into mRNA (answer below). AUAGGGCAUGUACGUCUAGGACCG GCACUUCAAAUUUGCUGGGAAAGG Next have students translate the mRNA sequence into a polypeptide using the genetic code (answer below). Ile-Gly-His-Val-Arg-Leu-Gly-Pro-Ala-Leu-Gln-Ile-Cys-Trp-Glu-Arg For a fun activity have students use the decoder sheet below (project on screen) to convert the mRNA sequence into a message (answer below). I LOVE CHEMISTRY
Decoder Chart First Position
Second Position U
U
UUU UUC UUA UUG
C
A
UCU UCC UCA UCG
Third Position
A
U
UAU UAC UAA UAG
} }
13-10
G J K
UGU UGC UGA UGG
}S }T
U C A G
C
A
G
CUU CUC CUA CUG
AUU AUC AUA AUG
GUU GUC GUA GUG
E
CCU CCC CCA CCG
} B } C
CAU CAC CAA CAG
I
ACU ACC ACA ACG
} D } F
AAU AAC AAA AAG
GCU GCC GCA GCG
} G } H
GAU GAC GAA GAG
O
} L } M } } } }
N P
Q R
CGU CGC CGA CGG
}V }W
U C A G
AGU AGC AGA AGG
} X } Y
U C A G
GGU GGC GGA GGG
} Z } Space
U C A G
Next, tell students to delete the first T in the DNA strand and repeat the previous steps (transcribe, translate, and decode). Once students realize the message make no sense, discuss how if mutations such as this deletion occurs in a gene, it can have a major adverse affect on the organism. You may also want to ask how a substitution of the second T with a G would affect the polypeptide and message (this will reemphasize that not all mutations cause adverse affects). You can also ask students where in the gene a deletion would have the least affect. Note 5′ and 3′ designations have been ignored—they can be added for a more challenging activity. Sections 13.12 and 13.13 Begin these sections with a brief discussion about the pros and cons of genetic technology. Have students try to come up with the benefits of some of the various applications of recombinant DNA and DNA fingerprinting. List them on the board (they should be able to come up with a few). You can then add any other major ones to the list (below are some categories you may want to include. Recombinant DNA DNA Fingerprinting -Medicine -Law Enforcement -Agriculture -Ancestry/Family Relations -Environmental -Paleontology/Archeology Next, discuss the cons associated with each. Depending on time, you may or may not want your discussion to move into the realm of ethics. Also, explain during section 13.13 how forensic science (i.e. PCR, STRs analysis, etc.) is very different than what is portrayed on TV (i.e. analysis and results are not instantaneous).
Handouts for Students 1) The Genetic Code handout (next page) will allow students to quickly translate mRNA codons into amino acids.
13-11
Handout Genetic Code First Position (5′ End)
Second Position
U
U
C
A
G
C
Third Position (3′ End)
A
G
UUU UUC UUA UUG
UCU UCC Ser UCA UCG
UAU UAC UAA UAG
UGU Cys UGC UGA STOP UGG Trp
U C A G
CUU CUC CUA CUG
CCU CCC CCA Pro CCG
CAU CAC CAA CAG
CGU CGC CGA CGG
Arg
U C A G
ACU ACC Thr ACA ACG
AAU AAC AAA AAG
AGU AGC AGA AGG
} Ser } Arg
U C A G
GCU GCC Ala GCA GCG
GAU GAC GAA GAG
GGU GGC GGA Gly GGG
U C A G
} Phe } Leu
AUU AUC AUA AUG
Leu
}
Ile
Met (Start)
GUU GUC Val GUA GUG
} Tyr } STOP } His } Gln } Asn } Lys } Asp } Glu
}
20 Amino Acid Abbreviations Alanine Arginine Asparagine Aspartic Acid Cysteine Glutamine Glutamic Acid Glycine Histidine Isoleucine
Ala Arg Asn Asp Cys Gln Glu Gly His Ile
Leucine Lysine Methionine Phenylalanine Proline Serine Threonine Tryptophan Tyrosine Valine 13-12
Leu Lys Met Phe Pro Ser Thr Trp Tyr Val
Chapter 14 Metabolism Outline Notes I. Metabolic Pathways, Energy, and Coupled Reactions (Section 14.1) A. Metabolic Pathways 1. The manufacture or break down of biomolecules (i.e. carbohydrates, lipids, proteins, etc.) involves groups of reactions called metabolic pathways 2. The three major types of metabolic pathways are: a. Linear, which is a continuous series of reactions in which the product of one reaction is the reactant in the next b. Circular, which is a series of reactions where the final product is an initial reactant c. Spiral, which is a series of repeated reactions used to break down or build up a molecule 3. Examine Figure 14.1 B. Energy and Coupled Reactions 1. Recall reactions are either spontaneous (-ΔG) or nonspontaneous (+ΔG) 2. In coupled reactions spontaneous reactions provide the energy needed by nonspontaneous reactions 3. Review example in text (ATP formation from ADP) 4. The production of ATP requires energy; therefore it is coupled with reactions that release enough energy to make its formation spontaneous 5. Review example in text (Glycolysis production of ATP reaction/calculation) 6. An entire metabolic pathway can be either spontaneous or nonspontaneous a. This is determined by the sum of the ΔG values for all steps of the pathway b. Review example in text (Hypothetical pathway calculation) *See Sample Problem 14.1 and Practice Problem 14.1 II. Overview of Metabolism (Section 14.2) A. Basics of Metabolism 1. Metabolism is the sum of all reactions that take place in a living thing 2. There are two parts of metabolism: a. Catabolism, in which compounds are broken down into smaller ones in processes that usually release energy b. Anabolism, in which larger compounds are biosynthesized from smaller ones in processes that usually require energy 3. Examine Figure 14.2 4. ATP plays a key role in metabolism: it is formed during catabolism and used during anabolism 5. Examine Figure 14.3 6. Review example in text (ATP hydrolysis reactions coupled to anabolism) 7. GTP and UTP are two other nucleoside triphosphates that carry energy in metabolism; and the energy from their hydrolysis is roughly equal to the energy supplied by ATP 8. Coenzymes such as NAD+, NADP+, and FAD also play important roles in metabolism a. They are reduced to NADH, NADPH, and FADH2 (respectively) during oxidation reactions b. The reduced forms can be used to produce ATP or the H atoms can be used in anabolism 9. Coenzyme A (CoA) and acetyl-CoA are also central to metabolism a. The attachment of CoA to reactants can raise the reactants’ potential energy b. When CoA is removed, energy is released that can be used to drive other reactions c. Acetyl-CoA is a product common to the break down of carbohydrates, fatty acids, and some amino acids d. Acetyl-CoA is also the starting point for the citric acid cycle that generates ATP B. Catabolism 14-1
1. The first step of catabolism is the digestion of food in which large biomolecules are broken down into their building blocks a. Carbohydrate polysaccharides are broken down into monosaccharides b. Triglycerides (lipids) are broken down into fatty acids and glycerol c. Proteins and broken down into amino acids 2. Monosaccharides enter glycolysis, which is a pathway that converts hexoses (6 C atoms) into two pyruvates (3 C atoms) while producing NADH and ATP 3. Pyruvate can be converted into acetyl-CoA (2 C atoms) which can then enter the citric acid cycle 4. Fatty acids undergo oxidation that removes two C atoms at a time and results in the production of NADH and FADH2 5. Amino acids lose their amino groups before being catabolized further 6. NADH and FADH2 can enter the electron transport chain where they are recycled to their oxidized forms, and energy is released during oxidative phosphorylation 7. Examine Figure 14.4 8. Glycolysis occurs in the cytoplasm of the cell while the other major catabolic pathways/ processes occur in the mitochondria (“powerhouses of the cell”) a. Mitochondria are found in plant and animal cells and they are where most ATP is produced b. Mitochondria have an outer membrane that is separated from a folded inner membrane by an intermembrane space c. The fluid inside the membrane is called the matrix d. Examine Figure 14.5 DID YOU KNOW? Mitochondria C. Anabolism 1. During anabolism, small molecules (i.e. pyruvate, acetyl-CoA, etc.) are used to make monosaccharides, fatty acids, and amino acids 2. These building blocks are then incorporated into polysaccharides, lipids, and proteins 3. Anabolic pathways are not the exact reverse of catabolic pathways *See Sample Problem 14.2 and Practice Problem 14.2 III. Digestion (Section 14.3) A. Digestion Overview 1. During digestion, large molecules in food are broken down into their respective building blocks 2. Digestion occurs in the mouth, stomach, and/or small intestine 3. Carbohydrate, lipid (triglyceride), and protein digestion all differ B. Carbohydrate Digestion 1. Starch in food begins to be digested in the mouth 2. Salivary amylase catalyzes the hydrolysis of some of the α(1→4) glycosidic bonds in starch 3. Amylase does not act on any other oligosaccharides or polysaccharides in food, and it stops working soon after food is swallowed due to stomach acid denaturing it 4. Any remaining starch/polysaccharides are split into monosaccharides in the small intestine 5. Debranching enzymes break α(1→6) glycosidic bonds in amylopectin while amylase continues to work on the α(1→4) glycosidic bonds 6. β-galactosidase hydrolyzes lactose while sucrase hydrolyzes sucrose 7. Monosaccharides are then absorbed by the intestinal lining cells where they are then released in the blood for distribution throughout the body C. Triglyceride Digestion 1. Triglyceride digestion begins in the stomach 2. The enzymes lingual lipase and gastric lipase catalyze the hydrolysis of triglycerides into diacylglycerides and fatty acids a. Lingual lipase is present in salvia but activated by acid in the stomach once it is swallowed b. Gastric lipase is secreted in the stomach 3. Most of the digestion of triglycerides occurs in the small intestine 14-2
4. The bile salts act as emulsifiers to form lipid-water colloids and assist the enzymes in breaking down diacylglycerides and remaining triglycerides into monoacylglycerides, fatty acids, and glycerol 5. These products then cross the small intestine cells and are converted back into triglycerides 6. The triglycerides then carried by the lymph and then the blood by lipoproteins D. Protein Digestion 1. Protein digestion begins in the stomach where the enzyme pepsin catalyzes the hydrolysis of some of the peptide bonds 2. The peptides then leave the stomach and enter the small intestine where any unhydrolyzed peptides bonds are split by trypsin, chymotrypsin, and other protease enzymes 3. The free amino acids are absorbed and move into the blood 4. Examine Figure 14.6 5. Examine Figure 14.7 *See Sample Problem 14.3 and Practice Problem 14.3 IV. Glycolysis (Section 14.4) A. Glycolysis Overview 1. Glycolysis is a 10-step linear pathway that occurs in the cytoplasm of a cell 2. One glucose molecule is converted into two pyruvates; 2 ATP and 2 NADH are also produced 3. Review example in text (Pyruvate in the body and its structures) 4. The net reaction of glycolysis is Glucose + 2NAD+ + 2ADP + 2Pi → 2Pyruvate + 2NADH + 2ATP + Energy 5. Examine Figure 14.8 6. Review examples in text (Steps of glycolysis—figure 14.8) B. Control of Glycolysis 1. The rate of glycolysis is controlled at steps 1, 3, and 10 2. The allosteric enzyme phosphofructokinase catalyzes step 3 which is the main control point a. ATP and citrate are negative effectors while ADP and AMP are positive effectors b. When a cell is energy-rich, the concentration of ATP and citrate are high which inhibits the further production of ATP by glycolysis c. When a cell is energy-poor, the concentration of ADP and AMP are high which activates phosphofructokinase and glycolysis resumes 3. The allosteric enzyme hexokinase catalyzes step 1 a. Glucose 6-phoshpate is a negative effector b. A rise in glucose 6-phoshpate can be caused by the i. inhibition of step 3 ii. manufacture of glucose iii. breakdown of glycogen c. These all cause step 1 of glycolysis to slow 4. The allosteric enzyme pyruvate kinase catalyzes step 10 a. ATP, fructose 1,6 bisphosphate, and alanine are all negative effectors b. Alanine is an amino acid that can be produced when pyruvate concentrations are high *See Sample Problem 14.4 and Practice Problem 14.4 DID YOU KNOW? Glycolysis and Cancer C. Pyruvate 1. Pyruvate can go a number of different directions during metabolism 2. In yeast cells, pyruvate undergoes alcoholic fermentation a. Pyruvate is split into acetaldehyde plus CO2 and the acetaldehyde is reduced to ethanol b. This recycles NADH back into NAD+ which allows glycolysis to continue 3. In humans, pyruvate is reduced to lactate under anaerobic conditions (O2 deficient) a. This occurs during vigorous exercise b. This converts the NADH back into NAD+ so glycolysis can continue 14-3
c. Lactate is send to the liver where it can be used to synthesize glucose 4. Under aerobic conditions (O2 sufficient), pyruvate is converted into acetyl-CoA which then goes into the citric acid cycle a. Pyruvate moves from the cytoplasm (glycolysis) to the mitochondria (citric acid cycle) b. It crosses the outer mitochondrial membrane by diffusion and then the inner mitochondrial membrane with the help of a transport protein c. In the mitochondrial matrix, pyruvate is then converted into acetyl-CoA via the pyruvate dehydrogenase complex d. The pyruvate dehydrogenase complex has ATP, NADH, and acetyl-CoA as negative effectors; therefore, when the cell is energy rich, this process is slowed 5. Examine Figure 14.9 D. Energy and Glycolysis 1. Under anaerobic conditions (in humans) one mole of glucose produces two moles of lactate and two moles of ATP through the spontaneous reaction: Glucose + 2ADP + 2Pi → 2Lactate + 2ATP ΔG -29.4 kcal 2. Review example in text (Energy summation of anaerobic glycolysis) 3. Only 33.2 % is used to make ATP, the remaining energy is released as heat E. Entry of Other Monosaccharides into Glycolysis 1. All common monosaccharides can enter glycolysis at some point 2. Review examples in text (Galactose, mannose, fructose conversion) 3. The net gain of ATP and NADH from glycolysis is usually the same for other monosaccharides as it is for glucose 4. Some intermediates of glycolysis are also the starting point for the pentose phosphate pathway, which is responsible for the production of ribose, 2-deoxyribose, other monosaccharides (3-7 C atoms big), and NADPH (used in the synthesis of fatty acid) 5. Review example in text (Diagram of the pentose phosphate pathway) V. Gluconeogenesis (Section 14.5) A. Gluconeogenesis Overview 1. The linear path involved in making glucose from noncarbohydrates (i.e. lactate, glycerol, amino acids) is called gluconeogenesis 2. One important role of gluconeogenesis is the conversion of lactate produced during anaerobic catabolism back into glucose 3. Gluconeogenesis also provides glucose during fasting or in the early stages of starvation a. This results when glucose and glycogen stores have been depleted b. This is very important because glucose is the only fuel for brain cells B. Gluconeogenesis Pathway 1. Gluconeogenesis is not the reverse of glycolysis even though they share many reactions/enzymes 2. Examine Figure 14.10 3. Review examples in text (Comparison of steps 1 and 3 of glycolysis and gluconeogenesis) 4. The reversal of step 10 of glycolysis requires several steps: pyruvate is converted into oxaloacetate which is then transformed into phosphoenolpyruvate 5. Compounds can enter the gluconeogenesis pathway at many different steps 6. Review examples in text (Conversion of lactate, glycerol, and amino acids) *See Sample Problem 14.5 and Practice Problem 14.5 VI. Glycogen Metabolism (Section 14.6) A. Glycogenesis and Glycogenolysis 1. Recall glycogen is highly branched and composed of glucose linked by α(1→4) and α(1→6) bonds 2. Glycogen is mainly found in the liver and muscle cells where it acts a storage molecule for glucose 3. Glycogenesis is the synthesis of glycogen a. Glucose 6-phosphate (formed in glycolysis or gluconeogenesis) is first isomerized into glucose 1-phosphate which is then reacted with UTP to form glucose-UDP 14-4
b. This activated glucose-UDP molecule can then be used to add a glucose residue to the end of a growing glycogen chain 4. Glycogenolysis is the breakdown of glycogen a. Glucose residues are removed from the ends of glycogen b. These glucose 1-phosphate residues are then isomerized into glucose 6-phosphate 5. Examine Figure 14.11 B. Glycogen Metabolism Regulation 1. When glucose concentrations are high, glycogenesis converts excess glucose into glycogen 2. When glucose concentrations are low, glycogenolysis begins a. In liver cells, glucose 6-phosphate is hydrolyzed to glucose which then enters the blood b. In muscle cells, glucose 6-phosphate enter glycolysis 3. The key enzyme in glycogenesis is glycogen synthase while the key enzyme in glycogenolysis is glycogen phophorylase a. Both enzymes are controlled by covalent modification b. When one enzyme is switched on, the other is switched off 4. The pancreatic hormones insulin and glucagon control which enzyme is active a. When blood glucose levels rise, insulin is released to reduce the amount of glucose present i. Insulin actives glycogen synthase and deactivates glycogen phophorylase ii. Insulin also causes an increase in the uptake of blood glucose by cells b. When blood glucose levels drop, glucagon is released to increase glucose levels i. Glycogenolysis is switch on and glycogenesis is switch off by covalent modification ii. The movement of glucose from the blood into cells slows iii. The breakdown of proteins and lipids is accelerated which provides amino acids and glycerol that can be used by gluconeogenesis *See Sample Problem 14.6 and Practice Problem 14.6 VII. Citric Acid Cycle (Section 14.7) A. Citric Acid Cycle Overview 1. Under aerobic conditions, the two pyruvate molecules from glycolysis can be catabolized further 2. Pyruvate is converted into acetyl-CoA which then enters the citric acids cycle in the mitochondria 3. The citric acid cycle is a circular pathway with 8 steps 4. Examine Figure 14.12 5. The net reaction for the citric acid cycle is Acetyl-CoA + 3NAD+ + FAD + GDP + Pi → 2CO2 + CoA + 3NADH + FADH2 + GTP 6. Review examples in text (Steps of the citric acid cycle) 7. GTP is equivalent to ATP, and NADH and FADH2 can enter the electron transport chain where their energy is used to generate more ATP B. Control of the Citric Acid Cycle 1. Recall, the conversion of pyruvate to acetyl-CoA is regulated by the allosteric enzyme complex; therefore, it regulates the citric acid cycle as acetyl-CoA is what enters the cycle 2. In the cycle itself, isocitrate dehydrogenase in step 3 is the main control point 3. Isocitrate dehydrogenase has ATP and NADH as negative effectors and ADP and NAD+ as positive effectors *See Sample Problem 14.7 and Practice Problem 14.7 VIII. Electron Transport Chain and Oxidative Phosphorylation (Section 14.8) A. Electron Transport Chain 1. The electron transport chain (ETC) is a group of proteins and other molecules embedded into the inner mitochondrial membrane 2. Examine Figure 14.13 3. There are multiple steps involved in the ETC a. First, complex I accepts two electrons as it oxidizes NADH to NAD+ b. FADH2 then donates two electrons at complex II 14-5
c. Pairs of electrons supplied by NADH and FADH2 are passed from one molecule in the chain to the next as they release energy d. The final acceptor of electrons is O2 which produces water through the reaction: O2 + 4H+ + 4e → 2H2O i. If this reaction is incomplete it can form the toxic superoxide ions (O2-) or H2O2 ii. These toxic compounds are broken down by protective enzymes B. Oxidative Phosphorylation 1. The energy released by the ETC is used for active transport to pump H+ across the inner mitochondrial membrane into the intermembrane space a. H+ concentration become higher in the intermembrane space than in the matrix b. H+ then diffuses through a channel from high to low concentration 2. The enzyme ATP synthase, which facilitates the H+ diffusion, converts the released energy into ATP through a process known as oxidative phosphorylation 3. On average each NADH molecule forms 2.5 ATPs, and each FADH2 forms 1.5 ATPs 4. No more than 35% of the potential energy of these molecules is converted into ATP, the rest is released as heat 5. Examine Figure 14.14 6. Review example in text (Shuttle systems in figure 14.14) *See Sample Problem 14.8 and Practice Problem 14.8 C. Summary of ATP Yield from Glucose 1. One glucose molecule produces 2 ATPs, 2 NADH, and 2 pyruvates from glycolysis 2. The 2 pyruvates yield 2 NADH as they are converted into acetyl-CoA 3. During the citric acid cycle, 2 acetyl-CoAs yield 6 NADH, 2 FADH2, and 2 GTP (ATP) 4. The net gain in ATP is 30-32 ATP per molecule of glucose 5. Examine Figure 14.15 HEALTH LINK Brown Fat (Examine Figure 14.16) IX. Lipid Metabolism (Section 14.9) A. Fatty Acid Catabolism 1. When triglycerides are hydrolyzed they produce glycerol and fatty acids 2. Glycerol can be used for catabolism or anabolism, depending on the needs of the cell 3. Fatty acids can be catabolized to produce energy a. 2 C atoms at a time are oxidized to release NADH, FADH2, and acetyl-CoA b. NADH and FADH2 go to the ETC to produce ATP c. Acetyl-CoA enters the citric acid cycle to produce GTP, NADH, and FADH2 4. Examine Figure 14.17 5. Review examples in text (Steps of fatty acid catabolism and figure 14.17) 6. β oxidation spiral is a series of reactions that is repeated on increasingly shorter reactants 7. Examine Figure 14.18 8. Review examples in text (β oxidation spiral and figure 14.18) *See Sample Problem 14.9 and Practice Problem 14.9 B. Ketone Bodies 1. Excess acetyl-CoA (more than the citric acid cycle can use) is converted into ketone bodies a. Acetoacetate forms when two acetyl-CoA molecules are combined b. When acetoacetate loses CO2 it forms acetone c. 3-hydroxybutyrate forms when acetoacetate is enzymatically reduced d. Examine Figure 14.19 2. Ketone bodies are easily transported (water soluble/hydrophilic) from the liver (where they are made) to other cells 3. Many cells use acetoacetate and 3-hydroxybutyrate as reactants in metabolism 4. Under starvation or diabetes, glucose concentrations drop and acetyl-CoA levels rise 5. Review examples in text (Reasons acetyl-CoA levels rise) 14-6
6. Examine Figure 14.10 7. Ketosis is the overproduction of ketone bodies a. It can be detected by the odor of acetone on the breath and the presence of ketone bodies in the urine and blood b. High concentrations of ketone bodies can lead to ketoacidosis, which is a potentially fatal drop in blood pH C. Fatty Acid Anabolism 1. Lipogenesis, biosynthesis of fatty acids, involves a spiral pathway that adds 2 C atoms at a time 2. Examine Figure 14.20 3. Lipogenesis occurs in the cytoplasm and uses different enzymes/coenzymes than β oxidation a. NADPH is the coenzyme used in lipogenesis b. Acyl carrier protein (ACP) activates the grown fatty acid for addition of the next two C atoms 4. Palmitic acid (16 C atoms) is the largest fatty acid that can be produced in humans 5. Unsaturated fatty acids are formed from saturated ones by desaturase enzymes a. Human desaturase enzymes will not place a C=C beyond C 10 in a fatty acid b. This is why linoleic and linolenic acid cannot be produced in human c. These two essential fatty acids must come from the diet X. Amino Acid Metabolism (Section 14.10) A. Amino Acid Metabolism Overview 1. Digestion of dietary proteins releases amino acids that can be transported throughout the body 2. Some amino acids are used for the synthesis of new peptides/proteins 3. Most amino acids are converted into compounds that take part in the citric acid cycle, gluconeogenesis, and other metabolic pathways 4. Amino acid catabolism requires the removal of the amino group through transamination or oxidative deamination B. Transamination and Oxidative Deamination 1. In transamination an amino group is transferred from an amino acid to an α-keto acid 2. In oxidative deamination an amino group is replaced by a carbonyl group 3. Examine Figure 14.21 4. Review examples in text (Transamination and oxidative deamination in figure 14.21) 5. Oxidative deamination produces NH4+ which is toxic; therefore it must be removed from the cell as rapidly as possible 6. In mammals the NH4+ is converted into urea by the urea cycle which a 7. Examine Figure 14.22 8. Birds and reptiles excrete nitrogen in white paste that contains uric acid (structure in text), while fish eliminate NH4+ directly into the water through their gills 9. All of the amino acids can be converted into citric acid cycle intermediates or into compounds that can enter the citric acid cycle 10. Examine Figure 14.23 C. Amino Acid Anabolism 1. Humans can only make 10 of the 20 amino acids used to synthesize proteins 2. The ones that cannot be made are called essential amino acids because they must be obtained in the diet 3. See Table 14.1 for essential amino acids 4. In humans, synthesis of many of the nonessential amino acids involves the reversal of the steps of amino acid catabolism *See Sample Problem 14.10 and Practice Problem 14.10
14-7
Chapter Summary Section 14.1 The manufacture or break down of biomolecules involves groups of reactions called metabolic pathways. The three major types of metabolic pathways are linear (a continuous series of reactions in which the product of one reaction is the reactant in the next), circular (a series of reactions where the final produce is an initial reactant), and spiral (a series of repeated reactions used to break down or build up a molecule). Many pathways involve coupled reactions which is when spontaneous reactions provide the energy needed by nonspontaneous ones. Entire metabolic pathways can be either spontaneous or nonspontaneous, which can be determined by the sum of the ΔG values for all steps of the pathway. Section 14.2 Metabolism is the sum of all reactions that take place in a living thing. The two parts of metabolism are catabolism, in which compounds are broken down into smaller ones in processes that usually release energy, and anabolism, in which larger compounds are biosynthesized from smaller ones in processes that usually require energy. ATP plays a key role in both parts of metabolism: it is formed during catabolism and used during anabolism. In catabolism, digested food that contains large biomolecules are broken down into their building blocks. Catabolism involves processes such as glycolysis, the citric acid cycle, the electron transport, and oxidative phosphorylation. During anabolism, small molecules are used to make monosaccharides, fatty acids, and amino acids which are then incorporated into polysaccharides, lipids, and proteins, respectively. Section 14.3 During digestion, large molecules in food are broken down into their respective building blocks. In carbohydrate digestion, polysaccharides such as starch are hydrolyzed into monosaccharides by enzymes. The monosaccharides are then absorbed by the small intestine where they are then released in the blood for distribution throughout the body. In lipid digestion, enzymes breakdown triglycerides into monoacylglycerides, fatty acids, and glycerol. Bile salts aid this digestion by acting as emulsifiers. These products cross the small intestine cells and are converted back into triglycerides. They are then carried by the lymph and blood throughout the body. In protein digestion, enzymes catalyze the hydrolysis of peptide bonds which results in free amino acids. These amino acids are then absorbed and move into the blood. Section 14.4 Glycolysis is a 10-step linear pathway that occurs in the cytoplasm of a cell. In glycolysis, one glucose molecule is converted into two pyruvates, 2 ATP, and 2 NADH. Glycolysis is controlled at steps 1, 3, and 10 by allosteric enzymes. When the cell does not need ATP, these enzymes slow the glycolysis pathway. Pyruvate produced from glycolysis can go a number of different directions during metabolism. In yeast cells, pyruvate undergoes alcoholic fermentation in which it reduced to ethanol. In humans, pyruvate is reduced to lactate under anaerobic conditions. Under aerobic conditions, pyruvate is converted into acetyl-CoA which then goes into the citric acid cycle. All common monosaccharides can enter glycolysis at some point. Some intermediates of glycolysis are also the starting point for the pentose phosphate pathway, which is responsible for the production of ribose, 2-deoxyribose, other monosaccharides, and NADPH. Section 14.5 The linear path involved in making glucose from noncarbohydrates is called gluconeogenesis. One important role of gluconeogenesis is the conversion of lactate produced during anaerobic catabolism back into glucose. Gluconeogenesis also provides glucose during fasting or in the early stages of starvation. The gluconeogenesis pathway is not the reverse of the glycolysis pathway even though they share many reactions and enzymes. Compounds can enter the gluconeogenesis pathway at many different steps. 14-8
Section 14.6 Glycogenesis is the synthesis of glycogen while glycogenolysis is the breakdown of glycogen. When glucose concentrations are high, glycogenesis converts excess glucose into glycogen. When glucose concentrations are low, glycogenolysis begins in liver and/or muscle cells. The key enzyme in glycogenesis is glycogen synthase while the key enzyme in glycogenolysis is glycogen phophorylase. When one enzyme is switched on, the other is switched off. The pancreatic hormones insulin and glucagon control which enzyme is active. Section 14.7 Under aerobic conditions, the two pyruvate molecules from glycolysis can be catabolized further via the 8-step circular pathway called the citric acid cycle. The net reaction for the citric acid cycle is AcetylCoA + 3NAD+ + FAD + GDP + Pi → 2CO2 + CoA + 3NADH + FADH2 + GTP. The citric acid cycle is controlled by the conversion of pyruvate into acetyl-CoA (entry point) and isocitrate dehydrogenase (step 3 in cycle). Isocitrate dehydrogenase has ATP and NADH as negative effectors and ADP and NAD+ as positive effectors. Section 14.8 The electron transport chain (ETC) is a group of proteins and other molecules embedded into the inner mitochondrial membrane. Electrons from NADH and FADH2 are passed from one molecule in the chain to the next as they release energy. O2 is the final acceptor of electrons as it is converted into water. The energy released by the ETC is used for active transport to pump H+ across the mitochondrial inner membrane. The enzyme ATP synthase, which facilitates this movement of H+, converts released energy into ATP through a process known as oxidative phosphorylation. The net gain is 30-32 ATP per molecule of glucose. Section 14.9 When triglycerides are hydrolyzed they produce glycerol and fatty acids. Depending on the needs of the cell, glycerol can be used for catabolism or anabolism. The fatty acids can be catabolized to produce energy through the β oxidation spiral, which is a series of reactions that is repeated on increasingly shorter reactants. Acetyl-CoA that is produced by pyruvate or β oxidation of fatty acids can accumulate. When excess acetyl-CoA builds up, it is converted into ketone bodies. The overproduction of ketone bodies, or ketosis, can lead to a potentially fatal drop in blood pH known as ketoacidosis. Fatty acids can also be biosynthesized (anabolism) in the process known as lipogenesis. Unsaturated fatty acids are formed from saturated ones by desaturase enzymes. Because human desaturase enzymes will not place a C=C beyond carbon 10 in a fatty acid, the essential fatty acids linoleic and linolenic acid must come from the diet. Section 14.10 Digestion of dietary proteins releases amino acids that can be transported throughout the body. Some amino acids are used for the synthesis of new peptides/proteins, but most are converted into compounds that take part in the citric acid cycle, gluconeogenesis, and other metabolic pathways. Amino acid catabolism requires the removal of the amino group through transamination, where an amino group is transferred from an amino acid to an α-keto acid, or oxidative deamination, where an amino group is replaced by a carbonyl group. Oxidative deamination produces NH4+ which is toxic; therefore, it must be removed from the cells. In mammals the NH4+ is converted into urea by the urea cycle. All of the amino acids can be converted into citric acid cycle intermediates or into compounds that can enter the citric acid cycle. Amino acids can also be biosynthesized (anabolism); however humans can only make 10 of the 20 required for proteins. The ones that cannot be made are called essential amino acids because they must be obtained in the diet.
14-9
Lecture Suggestions Sections 14.1, 14.2, and 14.3 It is helpful for students to understand the “big picture” of metabolism before they begin their study of these sections. Ask students if they know what the name of the process that plants use to manufacture their own food. They should answer photosynthesis. Ask them what three things plants need for this process. As they mention, water, CO2, and sunlight, write these reactants on the board. See if students can predict the products of this reaction (glucose/carbohydrates and O2). Next repeat this entire process with animals and cellular respiration. Once you have both equations on the board, ask students: “What is the point of doing these processes if they are just the “reverse” of each other?” Hopefully they will notice that energy is being transferred from the sun, to the plants, to the animals via chemical carriers (carbohydrates). Mention that these processes also allow for the biosynthesis of cellular components. Explain that while matter/atoms are recycled in these process, energy is not (animals do not give energy back to the sun). Lastly, lead into a discussion of catabolism and anabolism using these processes. This drives home the idea that these two parts of metabolism are “two sides of the same coin”. Sections 14.4, 14.5, 14.6, 14.7 For these sections explain how beer and wine is made through the fermentation process (carbohydrates from a plants are converted by yeast into ethanol). Ask students if they know why beer and wine are usually never over 14% ethanol (why the fermentation process stops). You can explain that the ethanol product kills the yeast that made it. You can then explain the liquors and spirits come from distillation (separation of a mixture based on a physical change). For fun, you can ask students if they know what plants (carbohydrate source) are used to make some of the various alcoholic beverages (i.e. vodka from potatoes, wine from grapes, etc.). Also ask students what the difference is between wine and champagne (sparkling) and how it relates to biochemical metabolism process. They may figure it out but if not, explain that it is the difference between letting the CO2 escape versus trapping it. Lastly, ask students how alcoholic fermentation in yeast differs from what humans do under anaerobic conditions. (As a joke you can mention that if humans did alcoholic fermentation rather than lactic acid under anaerobic conditions, there might be a lot more people who workout). Sections 14.7, 14.8, 14.9, 14.10 For these sections, explain that while you need protein for a healthy diet, you do not use other organisms’ proteins as your own. Explain to students that while protein catabolism does provide energy, it is the essential amino acids in the proteins that you actually need. Explain that if you eat a cow’s protein (i.e. hemoglobin) in a hamburger, you do not use that protein your body (i.e. you don’t use the cow’s hemoglobin to carry oxygen in your blood). Tell students that you break down (catabolism) the protein for energy and amino acids and then use the amino acids (or their atoms) to biosynthesize (anabolism) your own proteins according to your DNA. This reiterates the material in chapter 13 on protein synthesis.
14-10
Chapter 1 Science and Measurements SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
1.1
Which drawing correctly shows the relationship between pounds and kilograms?
1 kilogram is heavier than 1 pound (1 kilogram = 2.205 lb).
1.3
Is the statement “What goes up must come down” a scientific law or scientific theory? Explain. A law. It describes what is observed but does not explain why it happens.
1.5
How is a theory different from a hypothesis? A hypothesis is a tentative explanation based on presently known facts while a theory is an experimentally tested explanation that is consistent with existing experimental evidence and accurately predicts the results of future experiments.
1.7
Define the terms “matter” and “energy.”
1-1
Matter is anything that has mass and occupies space. Energy is the ability to do work or heat up something.
1.9
On a hot day, a glass of iced tea is placed on a table. a. What are some of the physical properties of the ice? Ice is water in the solid state. It is clear and colorless, hard, and feels cold to the touch. b. What change in physical state would you expect to take place if the iced tea sits in the sun for a while? The ice would slowly melt and turn into liquid water if the iced tea is left to sit in the sun for a while (melting).
1.11
Give an example of a physical change that involves starting with a liquid and ending up with a gas. Some examples of physical changes are: a puddle of water evaporating from the ground, rubbing alcohol evaporating from the skin, spilled gasoline evaporating from the ground.
1.13
What is potential energy? Potential energy is stored energy.
1.15
Describe a situation where an object’s potential energy varies as a result of changes in its position. As a Ferris wheel goes around, the potential energy of a rider changes relative to their position. When the rider is at the very top of the wheel, the potential energy is highest. As the wheel goes around, the potential energy of the rider decreases as the bottom of the wheel is approached. At the lowest point, the potential energy is lowest.
1.17
A battery-powered remote control toy car sits at the bottom of a hill. The car begins to move and is steered up the hill. a. Describe the changes to the car’s kinetic energy.
1-2
Kinetic energy is the energy associated with moving objects while potential energy is stored energy. Before the car starts to move up the hill, its kinetic energy is zero. As it begins its motion, the car gains kinetic energy. b. Describe the changes to the car’s potential energy that are related to its position. At the bottom of the hill, the car has lower potential energy. As the car moves uphill, the car gains potential energy.
1.19
Suppose that you are camping in the winter. To obtain drinking water, you use a propane-fueled camp stove to melt snow. a. Is the melting of snow a physical change? Explain. Melting snow is a physical change because the chemical composition does not change. b. When propane burns, the gases carbon dioxide and water vapor are formed. Is the burning of propane a physical change? Explain. Burning propane is a chemical change because new substances are produced. c. Describe the potential energy change that takes place for propane as it burns in the stove. The potential energy stored in the propane is lowered as the propane burns. d. Describe the kinetic energy change that takes place for water as the snow melts. As the snow melts, the water molecules move faster and so their kinetic energy increases.
1.21
a. What is heat of fusion? Heat of fusion is the energy required for a substance to melt or the energy absorbed as a solid is converted to a liquid. b. What is heat of vaporization? Heat of vaporization is the energy required for a substance to evaporate or the energy absorbed as a liquid is converted to a gas.
1-3
1.23
If you immerse your arm in a bucket of ice water, your arm gets cold. Where does the heat energy from your arm go and what process is the energy used for? The heat energy from the arm goes to the ice water and is used to melt the ice.
1.25
True or false? IF heat is continually added to a pan of boiling water, the temperature of water continually rises until all of the water has boiled away. False. The temperature of the water remains constant until all of the liquid water has boiled away.
1.27
Based on your experience or the information in Table 1.1, which is larger? a. 1 yd or 1 m 1 m. One meter is slightly larger than 1 yard because 1 yard is 3 feet and 1 meter is 3.281 feet. Therefore, 1 meter is 1.094 yard. b. 1 lb or 1 g 1 lb. One kilogram is 2.205 pounds. Therefore one pound is 453.5 grams. c. 1 cup or 1 mL 1 cup. One cup is 8 fluid ounces. One fluid ounce is 29.6 milliliter. Therefore, 1 cup is 237 mL.
1.29
Based on your experience or the information in Table 1.2, which is larger? a. 1 mg or 1 g? 1 mg. One mg equals 1000 g. b. 1 grain or 1 mg? 1 gr. One grain equals 65 mg. c. 1 T or 1 tsp? 1 T. One T equals 15 mL which equals 3 tsp. Therefore, 1T = 3 tsp. d. 1 T or 1 fl oz?
1-4
1fl oz. One fl oz equals 2T.
1.31
1.33
1.35
Convert each number into scientific notation. a. 1,300
1.3 x 103
b. 6,901,000
6.901 x 106
c. 0.00013
1.3 x 10-4
d. 0.0000006901
6.901 x 10-7
Convert each number into ordinary notation. a. 7 x 10-2
0.07
b. 7 x 102
700
c. 8.3 x 108
830,000,000
d. 8.3 x 10-8
0.000000083
a. In the year 2000, the world population is estimated to have been about 6 x 109. Convert this value into ordinary notation. 6,000,000,000 b. In the year 1000, the world population is estimated to have been about 3 x 106. Convert this value into ordinary notation. 3,000,000
1.37
Express each value using an appropriate metric prefix. a. one-thousandth of a meter
1 millimeter
b. one million meters
1 megameter
c. one billion meters
1 gigameter
1-5
1.39
1.41
Express each distance in scientific notation and ordinary (decimal) notation, without using metric prefixes (example: 6.2 cm = 6.2 x 10-2 m = 0.062 m) a. 1.5 km
= 1.5 x 103 m
= 1,500 m
b. 5.67 mm
= 5.67 x 10-3 m
= 0.00567 m
c. 5.67 nm
= 5.67 x 10-9 m
= 0.00000000567 m
d. 0.3 cm
= 3 x 10-3 m
= 0.003 m
Which is the greater amount of energy? a. 1 kcal or 1 kJ?
1 kcal. One calorie is larger than one joule (1 cal = 4.184 J), so 1 kcal (1000 cal) is larger than 1 kJ (1000 J).
b. 4.184 cal or 1 J? 4.184 cal. One calorie is larger than one joule (1 cal = 4.184 J), so 4.184 cal is larger than 1 J.
1.43
a. How many meters are in 1 km? The prefix kilo (k) stands for 103. Therefore, 1 km = 1 x 103 m. b. How many meters are in 5 km? The prefix kilo (k) stands for 103. Therefore, 5 km = 5 x 103 m. c. How many meters are in 10 km? The prefix kilo (k) stands for 103. Therefore, 10 km = 10 x 103 m or 1 x 104 m.
1.45
You are at the state fair and pay a dollar for the chance to throw three baseballs in an attempt to knock over a pyramid of bowling pins. After your three tosses, the pins remain standing. Which of the following statements about your throws might be correct? a. They were precise and accurate. b. They were neither precise nor accurate. c. They were precise but not accurate.
1-6
Either b or c could be correct since the pins remain standing. Your throws may or may not have been precise. 1.47
How many significant figures does each number have? Assume that each is a measured value. a. 1000000.5
All the zeroes count because they are between nonzero digits. 8
b. 887.60
The ending zero counts because there is a decimal point. 5
c. 0.668
Zeroes at the beginning of a number are not significant. 3
d. 45
All non-zero digits are significant. 2
e. 0.00045
Zeroes at the beginning of a number are not significant. 2
f. 70.
The ending zero counts because there is a decimal point. 2
1.49
Solve each calculation, reporting each answer with the correct number of significant figures. Assume that each value is a measured value. a. 14 x 3.6 In multiplication and division, the answer is rounded to have the same number of significant figures as the measurement with the fewest significant figures. Since both measured values have two significant figures, the calculator answer (50.4) rounds to 50. or 5.0 x 101 . The decimal point is necessary to indicate that the zero is significant. 5.0 x 101 b. 0.0027 ÷ 6.7784
1-7
The same rule applies as in part a. Dividing 0.0027 (2 significant figures) by 6.7784 (5 significant figures) gives 0.000398324, which rounds to 0.00040 or 4.0 x 10-4 (2 significant figures). 4.0 x 10-4 c. 12.567 + 34 When adding or subtracting, the answer should have the same number of decimal places as the quantity with the fewest decimal places.
12. 567 34
Three decimal places Zero decimal places
46.567
Rounds to 47 (0 decimal places)
47 d. (1.2 x 103 x 0.66) + 1.0 When more than one operation is involved, calculate the part in parentheses first and round it to the appropriate significant figures and then perform the next part of the calculation. In the first calculation, both numbers have two significant figures, so the answer (1.2 x 103 x 0.66 = 792) is rounded to two significant figures (790).
790 1.0
10s place is significant 10ths place is significant
791.0
Rounds to 790 with the 10s place significant
7.9 x 102
1.51
A microbiologist wants to know the circumference of a cell being viewed through a microscope. Estimating the diameter of the cell to be 11 m and knowing that circumference = π x diameter (we will assume that the cell is round, even though that is usually not the case), the microbiologist uses a calculator and gets the answer 34.55751919 m. Taking significant figures into account, what answer should actually be reported? (π = 3.141592654…..). In the calculation x diameter, the diameter (11 m) has the fewest number of significant figures (2). The answer (34.55751414… m) is reported with two significant figures.
1-8
35 m
1.53
Give the two conversion factors that are based on each equality. 12 eggs 1 dozen
or
1 dozen 12 eggs
3
b. 1 x 10 m = 1 km
1 x 103 m 1 km
or
1 km 1 x 103 m
c. 0.946 L = 1 qt
0.946 L 1 qt
or
1 qt 0.946 L
a. 12 eggs = 1 dozen
1.55
Convert a. 48 eggs into dozen 12 eggs = 1 dozen
48 eggs x
1 dozen = 4 dozens 12 eggs
b. 250 m into kilometers 1 x 103 m = 1 km 250 m x
1 km = 0.25 km 1 x 103 m
c. 2.7 L into quarts 0.946 L = 1 qt 2.7 L x
1.57
1 qt = 2.9 qt 0.946 L
Convert a. 92 µg into grams.
1-9
1 g = 1 x 10-6 g
92 g x
1 x 10-6 g = 9.2 x 10-5 g 1 g
b. 27.2 ng into milligrams. 1 ng = 1 x 10-9 g; 1 mg = 1 x 10-3 g.
27.2 ng x
1 x 10-9 g 1 mg x = 2.72 x 10-5 mg 1 ng 1 x 10-3 g
c. 0.33 kg into milligrams. 1 kg = 1 x 103 g; 1 mg = 1 x 10 -3 g
0.33 kg x
1 x 103 g 1 mg x = 3.3 x 105 mg 1 kg 1 x 10-3 g
d. 7.27 mg into micrograms. 1 mg = 1 x 10-3 g; 1g = 1 x 10-6 g
7.27 mg x
1.59
1 x 10−3 g 1 g x = 7.27 x 103 g -6 1 mg 1 x 10 g
Convert a. 1 cm into kilometers 1 cm = 1 x 10-2 m; 1 km = 1 x 103 m
1 cm x
1 x 10−2 m 1 km x = 1 x 10-5 km 3 1 cm 1 x 10 m
b. 25 pm into micrometers 1 pm = 1 x 10-12 m; 1 m = 1 x 10-6 m
25 pm x
1 x 10-12 m 1 m x 1 pm 1 x 10-6 m
= 2.5 x 10-5 m
1 - 10
c. 3.0 x 10-4 mm into decimeters 1 mm = 1 x 10-3 m; 1 dm = 1 x 10-1 m
1 x 10-3 m 1 dm x 1 mm 1 x 10-1 m
3.0 x 10-4 mm x
= 3.0 x 10-6 dm
d. 8.5 x 10-3 mm into nanometers 1 mm = 1 x 10-3 m; 1 nm = 1 x 10-9 m
8.5 x 10
1.61
-3
1 x 10-3 m 1 nm mm x x 1 mm 1 x 10-9 m
= 8.5 x 103 nm
Convert your weight from pounds to kilograms. Answers will vary depending on your weight. Below is a sample setup. If your weight is 175 lb then 175 lb x
1 kg 2.205 lb
=
79.4 kg
In general, answer: Your pound weight
1.63
x
1 kg . 2.205 lb
Convert a. 91°F into degrees Celsius C =
F - 32 91 - 32 = = 33°C 1.8 1.8
b. 53°C into degrees Fahrenheit °F = (1.8 x °C) + 32 = (1.8 x 53°C) + 32 = 127°F c. 0°C into kelvins K = °C + 273 = 0°C + 273 = 273 K d. 309 K into degrees Celsius
1 - 11
°C = K –273 = 309 K – 273 = 36°C
1.65
In 2011, 51,303 people completed the 12 km Bloomsday race in Spokane, WA. What is the distance of this race in miles? 12 km = distance of the race 1 mi = 1.609 km
12 km
1.67
x
1 mi 1.609 km
= 7.5 mi
Stavudine is an antiviral drug that has been tested as a treatment for AIDS. The daily recommended dosage of stavudine is 1.0 mg/kg of body weight. How many grams of this drug should be administered to a 150 lb patient? First, convert 150 lb to kg using the equivalence 2.205 lb = 1 kg. Then, use 1.0 mg/kg as a conversion factor to calculate the number of milligrams required by the patient. To convert the final answer to grams, use 1000 mg = 1 g.
150 lb
1.69
1 kg 2.205 lb
x
1.0 mg 1 kg
x
1g = 0.068 g 1000 mg
Ivermectin is used to treat dogs that have intestinal parasites. The effective dosage of this drug is 10.5 g/kg of body weight. How much ivermectin should be given to a 9.0 kg dog?
9.0 kg
1.71
x
x
10.5 g 1 kg
= 95 μg or 9.5 x 10-5 g
The tranquilizer Valium is sold in 2.0 mL syringes that contain 50.0 mg of drug per 1.0 mL of liquid (50.0 mg/1.0 mL). If a physician prescribes 25 mg of this drug, how many milliliters should be administered? Note that the size of the syringes does not have anything to do with the solution of the problem, since it asks for milliliters needed and not the number of syringes.
1 - 12
25 mg
1.73
x
1.0 mL 50.0 mg
= 0.50 mL
a. A vial contains 25 mg/mL of a particular drug. To administer 15 mg of the drug, how many milliliters should be drawn from the vial? Use the drug concentration in the vial 25 mg/mL (that is, 25 mg of the active drug is contained in every 1.0 mL of the vial content) as a conversion factor. To calculate how many milliliters of the drug should be dispensed to administer a dose of 15 mg:
15 mg drug x
1 mL 25 mg drug
= 0.60 mL
b. A patient is to receive 50 cc of a drug mixture intravenously over a 1 hr time period. What is the appropriate IV drip rate in gtt/min? This problem is multi-step and requires more than one conversion factor to complete. Use the conversion factor 15 drops (gtt) = 1 milliliter (mL). To use this relationship, we first have to convert 50 cc to mL using the equality 1 cc = 1 mL. Because the problem asks for gtt/min, we also have to convert hour to minute. Combining all of these steps results in the following calculation: 50 cc 1 mL 15 gtt 1 hr x x x = 13 gtt/min 1 hr 1 cc 1 mL 60 min
1.75
A dose of 3 mg/kg/day (3 mg of drug per kilogram of body weight per day) of Phenobarbital is to be given to a 24 kg patient once a day. Phenobarbital is sold in 35 mg tablets. How many tablets (rounded to the nearest one tablet) should be given to the patient per day? The recommended dosage in mg of drug for a patient that has 24 kg of body weight per day is: 24 kg ×
3 mg 1 kg
=
72 mg of drug or 70 mg (to 1sig. fig.)
1 - 13
Using the ratio 35 mg of drug per one tablet, calculate the number of tablets required to deliver 70 mg of the drug: 70 mg ×
1.77
1 tablet 35 mg
=
2 tablets
The antipsychotic drug thioridazine is administered at 0.5 mg/kg/day in three divided doses. The drug is sold in 10 mg tablets. How many tablets should be given per dose to a 180 lb patient? First, convert the weight of the patient from lb to kg: 180 lb ×
1 kg 2.205 lb
=
81.6 kg body weight
Per day, the recommended dosage (mg of drug) for this kg of body weight is: 81.6 kg ×
0.5 mg 1 kg
=
41 mg of drug
Using the ratio 10 mg of drug per one tablet, calculate the number of tablets required to deliver 41 mg of the drug: 41 mg ×
1 tablet 10 mg
=
4 tablets
Per dose, the recommended number of tablets is: 4 tablets 1 day × 1 day 3 doses
1.79
=
1.3 tablets 1 dose
At 20 °C, what is the mass in grams of (See Table 1.7) a. 2.0 mL of water? b. 2.0 mL of whole blood? c. 15.3 cm3 of salt? d. 71.2 cm3 of lead? To solve for the mass in grams, use the density as a conversion factor: a. 2.0 mL of water? 2.0 mL x
1.00 g = 2.0 g water 1 mL
1 - 14
b. 2.0 mL of whole blood? 1.06 g = 2.1 g whole blood 1 mL
2.0 mL x
c. 15.3 cm3 of salt? 15.3 cm3 x
2.17 g = 33.2 g salt 1 cm3
d. 71.2 cm3 of lead? 71.2 cm3 x
1.81
11.35 g = 808 g lead 1 cm3
At 20 °C what is the volume in milliliters occupied by (See Table 1.7) a. 15.2 g of water ? b. 2.0 kg of kerosene ? c. 9.2 x 10-2 g of isopropyl alcohol d. 75 g chloroform To solve for the volume in milliliters, use the density as a conversion factor: a. 15.2 g of water ?
15.2 g x
1 mL = 15.2 mL water 1.00 g
b. 2.0 kg of kerosene ?
2.0 kg x
1 x 103 g 1 mL x = 2.4 x 103 mL kerosene 1 kg 0.82 g
c. 9.2 x 10-2 g of isopropyl alcohol
9.2 x 10-2 g x
1 mL = 0.12 mL isopropyl alcohol 0.785 g
d. 75 g chloroform
1 - 15
75 g x
1.83
1 mL = 5.0 x 101 mL chloroform 1.49 g
A patient has 25.0 mL of blood drawn and this volume of blood has a mass of 26.5 g. What is the density of the blood? Density is expressed in g/mL. Therefore, the density is found by dividing the mass of the blood by its volume. Density =
1.85
mass 26.5 g = = 1.06 g/mL volume 25.0 mL
What is the specific gravity of whole blood at 20oC? (See Table 1.7.) The specific gravity relates the density of a substance to that of water at the same temperature:
specific gravity =
density of substance density of water
At 20C, the density of water is 1.00 g/mL and the density of whole blood is 1.06 g/mL.
specific gravity of whole blood =
1.87
1.06 g/mL = 1.06 1.00 g/mL
Calculate the number of calories of heat energy required for each (See Table 1.8) a. to warm 35.0 g of water from 21.0 °C to 29.0 °C Use the specific heat of water, 1.000 cal/g○C (Table 1.8), to convert mass and temperature change into calories. The temperature change is 8.0 °C (29.0 °C– 21.0 °C)
35.0 g x 8.0 C x
1.000 cal = 280 cal g C
b. to warm 17.5 g of water from 18.0 °C to 54.0 °C
1 - 16
17.5 g x 36.0 C x
1.89
1.000 cal = 6.30 x 102 cal g C
Calculate the number of calories of heat energy required for each (See Table 1.8) a. to warm 35.0 mL of water from 21.0 °C to 29.0 °C First, use the density of water to convert mL of water to g of water. 35.0 mL x
1.00 g = 35.0 g water 1 mL
Then, use the specific heat of water, 1.000 cal/g○C (Table 1.8), to convert mass and temperature change into calories. The temperature change is 8.0 °C (29.0 °C– 21.0 °C)
35.0 g x 8.0 C x
1.000 cal = 280 cal g C
b. to warm 17.5 mL of water from 18.0 °C to 54.0 °C
17.5 mL x
1.00 g = 17.5 g water 1 mL
17.5 g x 36.0 C x
1.91
1.000 cal = 6.30 x 102 cal g C
How much will the temperature change when 750 g of each of the following materials absorbs 1.25 x 104 cal of heat energy? To do these calculations, substitute all the known values into the conversion equation using specific heat and solve for the unknown (temperature change) a. iron (specific heat = 0.11 cal/g°C)
750 g x temperature change x
temperature change =
0.11 cal = 1.25 x 104 cal g C
1.25 x 104 cal g C x = 150 C 750 g 0.11 cal 1 - 17
b. stainless steel (specific heat = 0.12 cal/g°C)
750 g x temperature change x
temperature change =
0.12 cal = 1.25 x 104 cal g C
1.25 x 104 cal g C x = 140 C 750 g 0.12 cal
c. aluminum (specific heat = 0.89 J/g°C) First, convert 1.25 x 104 cal to J using the relationship given in Table 1.1. 1.25 x 104 cal x
4.184 J = 52,300 J 1 cal
750 g x temperature change x
temperature change =
1.93
0.89 J = 52,300 J g C
52,300 J g C x = 78 C 750 g 0.89 J
Cadmium has a density of 8.65 g/cm3 and beryllium, the lightest metal, has a density of 1.85 g/cm3. a. What volume (in cubic centimeters) is occupied by 25.0 g of cadmium? Use the density of the metal as a conversion factor to convert mass to volume. 25.0 g ×
1 cm3 8.65 g
=
2.89 cm3
b. What volume (in cubic centimeters) is occupied by 25.0 g of beryllium? 25.0 g ×
1 cm3 1.85 g
=
13.5 cm3
c. What is the mass (in pounds) of a cube of cadmium with a dimension of 5 inches on a side? First, convert the dimension 5 inches to centimeters:
1 - 18
5 in ×
1m 1 cm × 39.37 in 10-2 m
=
12.7 cm (without rounding to 1 sig. fig.)
Then, calculate the volume of the cube in cm3 volume of the cube = length x length x length = 12.7 cm x 12.7 cm x 12.7 cm volume of the cube = 2050 cm3 Now, use the density as a conversion factor to convert the volume to mass in g: 2050 cm3 ×
8.65 g 1 cm3
1.77 x 104 g or 2 x 104 g (rounded off to 1 sig. fig.)
=
Now, convert g to lb: 1.77 x 104 g ×
1 lb 454 g
=
40 lb (rounded off to 1 sig. fig.)
d. What is the mass (in pounds) of a cube of beryllium with a dimension of 5 inches on a side? The calculations are similar to part c. above. Because the dimensions of the cube are the same, the beryllium cube will have the same volume: volume of the cube = 2050 cm3 2050 cm3 ×
1.85 g 1 cm3
3.79 x 103 g ×
1.95
1 lb 454 g
3.79 x 103 g or 4 x 103 g (rounded off to 1 sig. fig.)
= =
8 lb (rounded off to 1 sig. fig.)
At 20 °C, what is the volume, in milliliters, occupied by a. 14.5 g of isooctane? Use the density of isooctane given in the text, 0.69 g/mL at 20 °C, as a conversion factor to convert the mass given to the volume of isooctane. 14.5 g ×
1 mL 0.69 g
=
21 mL
b. 5.55 x 105 g of isooctane?
1 - 19
Again, use the density of isooctane given in the text, 0.69 g/mL at 20 °C, as a conversion factor to convert the mass given to the volume of isooctane. 5.55 x 105 g ×
1 mL 0.69 g
=
8.0 x 105 mL
c. 3.99 lb of isooctane? First, convert the mass in lb to g. Then, use the density of isooctane given in the text, 0.69 g/mL at 20 °C, as a conversion factor to convert the mass given to the volume of isooctane.
1.97
3.99 lb ×
454 g 1 lb
=
1810 g
1810 g ×
1 mL 0.69 g
=
2600 mL
At 20 °C, what is the volume, in gallons, occupied by a. 3.11 lb of gasoline? First, convert the mass in lb to g. Then, use the density of gasoline given in the text, 0.73 g/mL at 20 °C, as a conversion factor to convert the mass given to the volume of gasoline in mL. Then, convert mL to gallon. 3.11 lb ×
454 g 1 lb
=
1410 g
1410 g ×
1 mL 0.73 g
=
1900 mL
10-3 L 1 qt 1 gallon 1900 mL × × × 1 mL 0.946 L 4 qt
b. 172 lb of gasoline? Similar calculations as part a.:
1 - 20
=
0.50 gallon
172 lb × 78100 g ×
454 g 1 lb
78100 g or 7.81 x 104 g
=
1 mL 0.73 g
=
107000 mL or 1.1 x 105 mL
10-3 L 1 qt 1 gallon 1.1 x 10 mL × × × 1 mL 0.946 L 4 qt 5
=
29 gallon
c. 5.43 kg of gasoline? First, convert the mass in kg to g. Then, use the density of gasoline given in the text, 0.73 g/mL at 20 °C, as a conversion factor to convert the mass given to the volume of gasoline. Convert mL to gallon. 103 g 5.43 kg × 1 kg
=
5430 g
1 mL 0.73 g
=
7400 mL or 7.4 x 103 mL
5430 g ×
10-3 L 1 qt 1 gallon 7.4 x 10 mL × × × 1 mL 0.946 L 4 qt 3
1.99
=
2.0 gallon
Assume that the building blocks used to make DNA are pennies, each of which is 1.55 mm thick. If 3 billion pennies are stacked on one another, as happens in DNA with its building blocks, how tall would the stack be (in meters)? 1.55 mm 10-3 m 3 x 10 pennies × × 1 penny 1 mm 9
=
5 x 106 m
1.101 In the past 200 years, in what ways have scientific discoveries led to changes in the treatment of diabetes? Insulin became available, the purity of insulin was improved, genetically engineered human insulin was put on the market, and oral drugs were developed. 1.103 a. A 6’ 2” tall adult weighs 180 lbs. What is his BMI? Based on this value, what is his status: underweight, normal, overweight, or obese?
1 - 21
The formula for BMI is given below, where weight is given in pounds and height is given in inches. First, calculate the total inches in height. weight BMI = 703 x height 2 12 in height = 6 ft 2 in = 6 ft x + 2 in = 74 in 1 ft 180 lbs BMI = 703 x (74) 2 BMI = 23 According to Table 1.6, this falls within the “recommended weight” status.
b. Answer part a, but using your height and weight. Answers will vary. c. A woman stands 1.65 m tall and weighs 72.7 kg. What is her BMI and what is her status? First, convert m to in and then kg to lb. Then use the same formula as above. 1.65 m x
3.281 ft 12 in x = 65.0 in 1 m 1 ft
2.205 lb = 1.60 x 102 lb 1 kg weight BMI = 703 x height 2
72.7 kg x
BMI
= 703 x
BMI
= 26.6
1.60 x 102 lbs (65.0) 2
This BMI places her in the overweight status.
1.105 A patient has a temperature of 31°C. Should her clinician be concerned? If a patient has a temperature of 31°C, her physician should be concerned because this is considerably lower than average normal temperature (37°C), even after factoring temperature measurement errors (up to 2°C) due to variation in body location from which measurements are made.
1 - 22
1.107 One of the rule changes that the NCAA made to discourage rapid weight loss was to shorten the time between weigh in and competition from 24 hours to just 2 hours. Why would this discourage athletes from trying to make weight? Two hours might not be enough time to rehydrate and to recover from the effects of dehydration.
1.109 a. Use the density of water (1.00 g/mL) to derive a conversion factor for water that has the units lb/cup.
1.00 g 1 lb 237 mL 0.522 lb x x = 1.00 mL 454 g 1 cup 1 cup b. If an athlete reduces her body’s water volume by 5.5 cups through restricting fluid intake and sweating in a sauna, how much weight has she lost? Is this a good idea? Explain.
0.522 lb = 2.9 lbs 1 cup No. Losing weight through dehydration can adversely affect endurance, strength, energy, and motivation. Extreme dehydration can lead to kidney failure, heart attack, and death. 5.5 cup x
1.111 a. Write the two conversion factors that are based on the equality 1 grain = 325 milligrams.
325 mg 1 grain and 325 mg 1 grain b. Which conversion factor in your answer to part a would be used to convert grains to milligrams?
325 mg 1 grain c. One aspirin tablet contains 5.0 grains of aspirin. How many milligrams of aspirin are in two tablets?
2 tablets x
5.0 grains 325 mg x = 3300 mg 1 tablet 1 grain
1 - 23
d. How many grams of aspirin are in two tablets?
1 x 10-3 g 3300 mg x = 3.3 g 1 mg e. How many micrograms of aspirin are in two tablets?
3.3 g x
1 g = 3.3 x 106 g -6 1 x 10 g
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 1.2 Which drawing correctly shows the relationship between decigrams and grams?
1 dg
1g
10 dg
1g
10 dg
1g
1 dg
10 g
Answer:
10 dg = 1 g 1 dg = 10-1 g. Therefore, there are 10 dg in 1 g.
SECTION 1.1
1.4
THE SCIENTIFIC METHOD
Centuries before any experiments had been carried out, philosophers had proposed the existence of the atom. Why are the proposals of these philosophers not considered theories?
1 - 24
Answer: Theories are explanations of observed phenomena based on experimental observations. The philosophers’ proposals on the existence of the atom are not theories because they were not based on experimental observations.
1.6
How is a law different from a hypothesis?
Answer: A law describes events that are consistently observed and are reproducible. A hypothesis is simply a tentative explanation of an observed phenomenon.
SECTION 1.2
1.8
MATTER AND ENERGY
What are the three states of matter?
Answer: Solid, liquid, and gas.
1.10
a. List a few of the physical properties of a piece of copper wire. b. Give examples of some of the physical changes that a piece of copper wire could undergo.
Answer: a. A piece of copper is wire is brown-gold in color, shiny, malleable, ductile, and can conduct electricity. b. A piece of copper wire can be cut into shorter pieces, coiled into different shapes, and melted using heat.
1.12
Give an example of a chemical change (change in chemical composition) that involves starting with a liquid and ending up with a gas.
1 - 25
Answer: Burning gasoline (combustion) is an example. In the combustion of gasoline (or any liquid fuel like propane or butane), the liquid fuel is converted into gaseous carbon dioxide and water vapor.
1.14
What is kinetic energy?
Answer: Kinetic energy is the energy of motion.
1.16
Describe a situation where an object’s potential energy varies as a result of changes due to its chemical composition.
Answer: An example is the burning of wood. When wood is burned, it undergoes a change in chemical position and produces heat in the process. Once the wood has completely burned, the remaining solids and the gases expelled can no longer burn to provide heat.
1.18
In the autumn, a leaf falls from a tree. a. Describe the change to the leaf’s kinetic energy. b. Describe changes to the leaf’s potential energy that are related to its position.
Answer: a. The leaf has kinetic energy as it falls but when it hits the ground, its kinetic energy goes to zero. b. The potential energy of the leaf is highest just as it starts to fall. Its potential energy is lowest once it hits the ground.
1.20
Rather than melting snow on a camp stove as described in the previous problem, you decide to eat handfuls of snow. a. Describe the change in the potential energy of your body as the snow that you have swallowed is melted. b. If you are stranded in the woods during the winter, why is it better to obtain water by melting snow than eating it?
1 - 26
Answer: a. Some of the potential (stored) energy in the body provides the heat used to melt the snow. b. The body’s energy reserves can be put to other uses, like keeping warm.
1.22
a. What change in physical state takes place during sublimation? b. Give an example of a substance that sublimes.
Answer: a. In sublimation, a substance undergoes a change from the solid state to the gas state. b. Dry ice, which is frozen carbon dioxide, is an example of a substance that sublimes. Iodine crystals also undergo sublimation. 1.24
Some over-the-counter (nonprescription) wart removers contain ether. When a few drops are placed on a wart, it feels cold as the ether rapidly evaporates. Where does the heat energy from the wart go and what process is the energy used for?
Answer: The heat energy from the wart is absorbed by the ether molecules causing their average kinetic energy to increase. This enables the ether molecules to separate from each other. The energy is used for the process of evaporation, when liquid ether changes to gaseous ether.
1.26
True or false? A container holding a mixture of ice and water at a temperature of 0 °C (the freezing temperature of water) is placed in a -30 °C freezer. The ice-water mixture stays at a temperature of 0 °C until all the water has frozen.
Answer: True. The temperature of the water remains constant until all of the liquid water has been converted to ice.
1 - 27
SECTION 1.3
1.28
UNITS OF MEASUREMENT
Based on your experience or the information in Table 1.1, which is larger? a. 1 pint or 1 L? b. 1°F or 1°C c. 1 gal or 1 L?
Answer: a. 1 L. One quart is slightly smaller than one liter, and one pint is only onehalf quart. b. 1°C. One degree Celsius is 1.8 times larger than one degree Fahrenheit. c. 1 gal. One gallon is four quarts and one quart is just a bit smaller than one liter.
1.30
Based on your experience or the information in Table 1.2, which is larger? a. 1 gtt or 1 mL b. 1 T or 1 mL c. 1 tsp or 1 fl oz d. 1 mg or 1 gr
Answer: Use Table 1.2 for the relationships between the units for comparison. a. 1 mL. One milliliter (mL) contains 15 drops (gtt) or another way to look at it, 1 gtt is 1/15 of one mL. Therefore, 1 mL is greater than 1 gtt. b. 1 T. One tablespoon (T) contains 15 mL or another way to look at it, 1 mL = 1/15 T. Therefore, 1 T is greater than 1 mL. c. 1 fl oz. One teaspoon (tsp) is 5 mL. One tablespoon is 15 mL. Therefore, there are 3 teaspoons for every 1 tablespoon. Two tablespoons are equal to 1 fluid ounce (fl oz). Therefore, there are 6 teaspoons in every 1 fluid ounce, and 1 teaspoon is equal to 1/6 of 1 fluid ounce. d. 1 gr. One grain (gr) is 65 milligrams (mg).
1 - 28
SECTION 1.4
1.32
SCIENTIFIC NOTATION, SI, AND METRIC PREFIXES
Convert each number into scientific notation. a. 2,000,000,000 b. 850 c. 0.2 d. 0.0085
Answer: Values written in scientific notation are written as a number between 1 and 10 multiplied by a power of ten. As discussed in the chapter, to convert to scientific notation, the decimal point is shifted to the left for numbers equal to or greater than 10 and shifted to the right for numbers less than 1.
1.34
a. 2,000,000,000
2 x 109
b. 850
8.5 x 102
c. 0.2
2 x 10-1
d. 0.0085
8.5 x 10-3
Convert each number into ordinary notation. a. 4.23 x 10-4 b. 4.23 x 101 c. 9.66 x 10-6 d. 9.66 x 106
Answer: To convert from scientific notation to ordinary notation, use the exponent to determine how many times to multiply by 10 (for positive exponents) or how many times to divide by 10 (for negative exponents) a. 4.23 x 10-4 =
4.23 = 0.000423 10 x 10 x 10 x 10
b. 4.23 x 101 = 4.23 x 10 = 42.3 c. 9.66 x 10-6 =
9.66 = 0.00000966 10 x 10 x 10 x 10 x 10 x 10
1 - 29
d. 9.66 x 106 = 9.66 x 10 x 10 x 10 x 10 x 10 x 10 = 9,660,000
1.36
a. Light microscopes can see objects as small as 0.0000002 m in diameter. Convert this value into scientific notation. b. Electron microscopes can see objects as small as 0.0000000002 m in diameter. Convert this value into scientific notation.
Answer: a. 2 x 10-7 m b. 2 x 10-10 m
1.38
Express each value using an appropriate metric prefix. a. one thousand grams b. one-millionth of a gram c. one-billionth of a gram
Answer: Refer to the Table of SI and Metric Prefixes to match the multiplier with the appropriate prefix. a. one thousand grams = 1,000 g = 1 x 103 g = 1 kilogram or 1 kg
b.
one-millionth of a gram =
1 gram = 0.000001 gram 1,000,000 = 10-6 gram = 1 microgram or 1 g
c.
one-billionth of a gram =
1 gram = 0.000000001 gram 1,000,000,000 = 10-9 gram = 1 nanogram or 1 ng
1.40
Express each mass in scientific notation and ordinary (decimal) notation, without using metric prefixes (example: 5 g = 5 x 10-6 g = 0.000005 g). a. 24 g b. 0.716 mg c. 15 dg d. 412 kg
1 - 30
Answer: Refer to the Table of SI and Metric Prefixes to match the multiplier with the appropriate prefix. a. 24 g = 24 x 10-6 g = 2.4 x 10-5 g = 0.000024 g b. 0.716 mg = 0.716 x 10-3 g = 7.16 x 10-4 g = 0.000716 g c. 15 dg = 15 x 10-1 g = 1.5 x 100 g = 1.5 g d. 412 kg = 412 x 103 g = 4.12 x 105 g = 412,000 g
1.42
Which is the greater amount of energy? a. 100 cal or 100 J b. 100 Cal or 100 kJ
Answer: Use the relationships between calorie and joule (1 calorie = 4.184 joules) and between Calorie and calorie (1 Calorie = 1,000 calories). a. 100 cal is greater than 100 J because 1 cal = 4.184 J b. 100 Cal is greater than 100 kJ because 1 Cal = 1,000 cal = 4,184 J = 4.184 kJ
1.44
a. How many grams are in 1 kg? b. How many grams are in 2 kg? c. How many grams are in 2.5 kg?
Answer: Kilo means 103, milli means 10-3, and micro means 10-6. Therefore, 1 kilogram = 103 grams = 1000 grams; 1 milligram = 10-3 gram = 0.001 gram; and 1 microgram = 10-6 gram = 0.000001 gram. Use these equivalences to convert the measurements given.
a. 1 kg = 1 x 103 g b. 2 kg = 2 x 103 g c. 2.5 kg = 2.5 x 103 g 1 - 31
SECTION 1.5
1.46 a
MEASUREMENTS AND SIGNIFICANT FIGURES
A scientist buys a thermometer and tests it by measuring the melting point of particular substance. a. What results would show that the thermometer is not precise? b. What results would show that the thermometer is not accurate?
Answer: a. The thermometer is not precise if several measurements of the melting point are made and the resulting values are not close to each other. b. The thermometer is not accurate if the measured values of the melting point are not close to the true value of the melting point.
1.48
How many significant figures does each number have? Assume that each is a measured value. a. 1.466 b. 3.5895 c. 600.2 3
d. 4.55 x 10 e. 0.001 f. 2 x 101 g. 2.0 x 101
Answer: Use the rules given in Table 1.5 for determining the number of significant figures in a given measurement. The significant figures are underlined for each number below. a. 1.466
4 sig. figs.
b. 3.5895
5 sig. figs.
c. 600.2
4 sig. figs.
d. 4.55 x 10
3 sig. figs.
e. 0.001
1 sig. fig.
3
1 - 32
1.50
f. 2 x 101
1 sig. fig.
g. 2.0 x 101
2 sig. figs.
Solve each calculation, reporting each answer with the correct number of significant figures. Assume that each value is a measured quantity. a. 0.114 x 5.2377 b. 3.11 x 14.5 c. 123.667 - 78.9 d. (6.21 + 0.04) x 16.72
Answer: Calculations should not change the degree of uncertainty in the given values. When multiplying or dividing numbers, the answer should be rounded off to the same number of significant figures as the measurement with the least number of significant figures. In addition or subtraction, the answer should carry the same number of decimal places as the measurement with the fewest decimal places. (sf = significant figures) a. 0.114 (3 sf) x 5.2377 (5 sf)
= 0.597 (3 sf)
b. 3.11 (3 sf) x 14.5 (3 sf) c. 123.667 - 78.9 44.8
= 45.1
(3 sf)
(3 decimal places) (1 decimal place) (1 decimal place)
d. (6.21 + 0.04) x 16.72 = (6.25) x 16.72 = 105 (3 sf)
1.52
Given that area = x radius2 and radius = diameter/2, what is the area of the cell described in Problem 1.51? Report your answer with the correct number of significant figures.
Answer: area = x radius2 = 3.14 radius = diameter/2 = 11 μm/2 = 5.5 μm
1 - 33
area = 3.14 x (5.5 μm)2 = 95 μm2
SECTION 1.6
1.54
CONVERSION FACTORS AND THE FACTOR LABEL METHOD
Give the two conversion factors that are based on each equality. a. 2 T = 1 oz b. 15 gtt = 1 mL c. 1 mg = 1000 g
Answer: To obtain the first conversion factor, we divide each side of the equality by one of the terms. We divide each side by the other term to obtain the second conversion factor.
1.56
a.
2T 1 oz and 1 oz 2T
b.
15 gtt 1 mL and 1 mL 15 gtt
c.
1 mg 1000 g and 1000 g 1 mg
Convert a. 15 T into fluid ounces b. 45 gtt into milliliters c. 0.33 mg into micrograms
Answer: Determine the equality that relates the unit of the known value and the unit that is required. Create a conversion factor based on the equality. Use the factor-label method to perform the unit conversion. a. 15 T = ? fl oz
15 T
b. 45 gtt = ? ml
2 T = 1 fl oz
1 fl oz 2 T
= 7.5 fl oz
15 gtt = 1 ml 1 - 34
45 gtt
1 mL 15 gtt
c. 0.33 mg = ? g
1.58
10-6 g = 1 g
1 mg = 10-3 g
0.33 mg
= 3.0 mL
10-3 g 1 mg
1 g 10 -6 g
= 330 g
Convert a. 81.2 g into kilograms b. 81.2 kg into grams c. 29 g into milligrams d. 47.66 g into decigrams
Answer: Determine the equality that relates the unit of the known value and the unit that is required. Create a conversion factor based on the equality. Use the factor-label method to perform the unit conversion. a. 81.2 g = ? kg
81.2 g
1000 g = 1 kg
1 kg 1000 g
b. 81.2 kg = ? g
1000 g = 1 kg
81.2 kg c. 29 g = ? mg 29 g
= 0.0812 kg
1000 g 1 kg
= 8.12 x 104 g
1 g = 10-6 g
10-6 g 1 g
10-3 g = 1 mg
1 mg = 0.029 mg or 2.9 x 10-2 mg 10-3 g
d. 47.66 g = ? dg; 1 g = 10-6 g; 10-1 g = 1 dg 47.66 g
10-6 g 1 g
1 dg = 4.766 x 10-4 dg -1 10 g
1 - 35
1.60
Convert a. 0.33 L into pints b. 0.33 gal into liters c. 1.0 cup into deciliters d. 1.0 mL into quarts
Answer: Determine the equality that relates the unit of the known value and the unit that is required. Create a conversion factor based on the equality. Use the factor-label method to perform the unit conversion. a. 0.33 L = ? pints
0.33 L
0.946 L = 1 qt
1 qt 0.946 L
2 pt 1 qt
1 qt = 2 pt (pints)
= 0.70 pt
b. 0.33 gal = ? liters 1 gal = 4 qt
0.33 gal
4 qt 1 gal
1 qt = 0.946 L
0.946 L 1 qt
= 1.2 L
c. 1.0 cup = ? deciliters 4 cup = 1 qt 1 qt = 0.946 L = 1 dL 1 qt 0.946 L 1 dL 1.0 cup = 2.4 dL 4 cup 1 qt 10-1 L d. 1.0 mL into quarts 10−3 L 1.0 mL × 1 mL 1.62
1 mL = 10-3 L
×
1 qt 0.946 L
10-1 L
0.946 L = 1 qt
=
1.1 𝑥 10−3 qt
Convert your height from feet and inches into meters.
Answer: Sample calculation for a person whose height is 5’ 7” (5 feet and 7 inches):
1 ft height = 5’ 7” = 5 ft + 7 in = 5 ft + 7 in = 5 ft + 0.6 ft = 5.6 ft 12 in 1m 5.6 ft = 1.7 m 3.281 ft 1 - 36
1.64
Convert a. 103° F into degrees Celsius b. 25 °C to Fahrenheit c. 35 °C into kelvins d. 405 K into degrees Fahrenheit
Answer: Use the following conversion formulas:
C =
F - 32 1.8
°F = (1.8 x °C) + 32
K = °C + 273.15
a. 103° F = ? °C
C =
103 - 32 = 39°C 1.8
b. 25 °C = ? °F °F = (1.8 x °C) + 32 = (1.8 x 25) + 32 = 45 + 32 = 77 °F c. 35 °C = ? kelvins K = 35 + 273.15 = 308 d. 405 K = ? °F First, convert K to °C: 405 = °C + 273.15
°C = 405 - 273.15 = 132 °C
Then, convert to °F: °F = (1.8 x 132) + 32 = 270 °F
1.66
It is estimated that an accordion player expends 9.2 kJ of energy per minute of playing time. Convert this value into Calories (1 food Calorie = 1000 cal).
Answer: Since 1 kJ = 1 x 103 J and 1 cal = 4.184 J
1 - 37
9.2 kJ
1.68
103 J 1 kJ
1 cal 4.184 J
1 Cal 1000 cal
= 2.2 Cal
As an alternative to ear tags and lip tattoos, tetracycline (an antibiotic) is used to mark polar bears. The advantages of using tetracycline in this fashion are that it leaves a detectable deposit on the bears’ teeth, it can be administered remotely, and using it doesn’t require that the animal be sedated. If 25 mg/kg is an effective dose, how much tetracycline is needed (in grams) to mark a 1000 kg polar bear?
Answer: First, determine the number of milligrams of tetracycline needed and then convert that to grams. Note that 25 mg/kg (25 mg of tetracycline per kilogram of body weight) is used as a conversion factor.
1000 kg polar bear
25 mg tetracycline 1 kg polar bear
25, 000 mg tetracycline
1.70
10-3 g 1 mg
= 25, 000 mg tetracycline
= 25 g tetracycline
Chloroquine is used to treat malaria. Studies have shown that an effective dose for children is 3.5 mg per kilogram (3.5 mg/kg) of body weight, every 6 hours. If a child weighs 12 kg, how many milligrams of this drug should be given in a 24 hour period?
Answer: First determine the number of doses required for a 24-hour period using the information that 1 dose is required every 6 hours.
24 hr
1 dose 6 hr
= 4 doses for a 24-hr period
Next, determine the required mg of drug for 4 doses using the information that 1 dose is equivalent to 3.5 mg drug/kg body weight.
4 doses
3.5 mg drug/kg body weight 1 dose
1 - 38
= 14 mg drug/kg body weight for a 24-hr period
If the child weighs 12 kg, then the required mg drug for a 24-hr period is:
12 kg body weight
1.72
14 mg drug 1 kg body weight
= 1.7 x 102 mg drug
An antibiotic is sold in 3.0 mL ampoules that contain 60.0 mg of drug (60.0 mg/3.0 mL). How many milliliters of the antibiotic should be withdrawn from the ampoule if 45 mg are to be administered to a patient?
Answer: 45 mg antibiotic = ? mL
45 mg
1.74
3.0 mL 60.0 mg
60.0 mg of antibiotic = 3.0 mL ampoule
= 2.3 mL
a. A prescription of antibiotics for a 30 lb child says to give 100 mg three times daily. Is this dosage safe if the proper pediatric dosage range for this drug is 10-30 mg per kilogram of body weight per day? b. A patient’s cough syrup prescription comes in a 250 mL bottle. For how long will the cough syrup last if he takes two teaspoons three times a day?
Answer: a. First, convert the weight of the child from lb to kg:
30 lb
1 kg = 14 kg body weight 2.2 lb
The recommended total daily dose in mg is:
100 mg 3 doses 300 mg drug = 1 dose 1 day 1 day Calculate the mg per kg of body weight equivalent to administering the total daily dose to a 14-kg child.
300 mg drug 22 mg drug 14 kg body weight= 1 day kg body weight/day which is between 10-30 mg per kilogram of body weight per day so it is a safe dose.
1 - 39
b. Convert 250 mL into teaspoons and then use the conversion factor 6 teaspoons/day (equivalent to two teaspoons three times a day).
250 mL
1.76
1 tsp 5 mL
1 day = 8.3 days 6 tsp
To treat migraines, valproic acid can be given at a dosage of 15 mg/kg/day (15 mg of drug per kilogram of body weight per day). Valproic acid is sold in 250 mg capsules. How many capsules per day should a 115 lb patient be prescribed?
Answer: First, convert the weight of the patient from lb to kg: 1 kg 2.2 lb
115 lb ×
=
52.3 kg body weight
Per day, the recommended dosage (mg of drug) for this kg of body weight is: 52.3 kg ×
15 mg 1 kg
=
784 mg of drug
Using the ratio 250 mg of drug per one capsule, calculate the number of capsules required to deliver 784 mg of the drug: 784 mg ×
1.78
1 capsule 250 mg
=
3.1 capsules or 3 capsules
A prescription calls for giving a 95 lb patient 5 mg/kg/day (5 mg of drug per kilogram of body weight per day) of an anticonvulsant drug, with the half the dose given in the morning and the other half at night. The drug is sold in 100 mg tablets. How many tablets (rounded to the nearest one tablet) should be given at any given time.
Answer: First, convert the weight of the patient from lb to kg: 95 lb ×
1 kg 2.2 lb
=
43 kg body weight
Per day, the recommended dosage (mg of drug) for this kg of body weight is:
1 - 40
43 kg ×
5 mg 1 kg
=
215 mg of drug
Using the ratio 100 mg of drug per one tablet, calculate the number of tablets required to deliver 215 mg of the drug: 215 mg ×
1 tablet 100 mg
=
2.15 tablets or 2 tablets per day
Therefore, 1 tablet should be given in the morning and 1 tablet should be given at night.
SECTION 1.7
1.80
DENSITY, SPECIFIC GRAVITY, AND SPECIFIC HEAT
At 20 °C what is the mass in grams of (See Table 1.7) a. 1.0 L of kerosene? b. 1.0 gal of isopropyl alcohol? c. 75.2 cm3 of cork? d. 5.05 cm3 of lead?
Answer: Density can be used as a conversion factor that relates mass and volume. Look up the density for each substance and express it as a conversion factor. a. 1.0 L of kerosene? The density of kerosene from Table 1.7 is 0.82 g/mL which can be expressed as a conversion factor in two different ways:
0.82 g 1 mL or 1 mL 0.82 g Because the volume is given, use the first conversion factor. But first, convert L to mL to be able to use the density conversion factor.
1.0 L kerosene
1 mL 10-3 L
0.82 g kerosene = 820 g kerosene 1 mL
b. 1.0 gal of isopropyl alcohol? The density of isopropyl alcohol from Table 1.7 is 0.785 g/mL. First, convert gal to L to mL.
1 - 41
1.0 gal isopropyl alcohol
4 qt 1 gal
3784 mL isopropyl alcohol
0.785 g isopropyl alcohol = 3.0 x 103 g isopropyl alcohol 1 mL
0.946 L 1 mL = 3784 mL isopropyl alcohol 1 qt 10-3 L
c. 75.2 cm3 of cork? The density of cork from Table 1.7 is 0.25 g/cm3.
75.2 cm3 cork
0.25 g cork = 19 g cork 1 cm3
d. 5.05 cm3 of lead? The density of lead from Table 1.7 is 11.35 g/cm3.
5.05 cm3 lead
1.82
11.35 g lead = 57.3 g lead 1 cm3
At 20 °C what is the volume in milliliters occupied by (see Table 1.7) a. 1.50 kg of water? b. 77.2 g of kerosene? c. 5.0 mg of isopropyl alcohol? d. 1.0 lb of water?
Answer: In each of these cases, the mass is given. Use the density conversion factor in which the mass is in the denominator to solve for the volume. a. 1.50 kg of water? The density of water from Table 1.7 is 1.00 g/mL. First, convert the mass in kilogram to gram then use the density conversion factor to calculate the volume. 103 g 1.50 kg water 1 kg
1 mL = 1.50 x 103 mL water 1.00 g
b. 77.2 g of kerosene?
1 - 42
The density of kerosene from Table 1.7 is 0.82 g/mL.
77.2 g kerosene
1 mL = 94.1 mL kerosene 0.82 g
c. 5.0 mg of isopropyl alcohol? The density of water from Table 1.7 is 1.00 g/mL. First, convert the mass in mg to g then use the density conversion factor to calculate the volume. 10-3 g 1 mg
5.0 mg isopropyl alcohol
1 mL = 6.4 x 10-3 mL isopropyl alcohol 0.785 g
d. 1.0 lb of water? The density of water from Table 1.7 is 1.00 g/mL. First, convert the mass in lb to g then use the density conversion factor to calculate the volume. 1.0 lb water
1.84
1 kg 2.205 lb
103 g 1 kg
1 mL = 450 mL water 1.00 g
A patient has 0.050 L of blood drawn and this volume of blood has a mass of 55.0 g. What is the density of the blood?
Answer: To calculate the density of a substance, divide its mass by its volume.
density =
mass 55.0 g = = 1100 g / L volume 0.050 L
1100 g 10-3 L = 1.1 g/mL 1 L 1 mL
1.86
What is the specific gravity of kerosene at 20○C? (See Table 1.7.)
Answer: Specific gravity is the density of a substance divided by the density of water at the same temperature.
1 - 43
1.88
specific gravity of kerosene =
density of kerosene density of water
specific gravity of kerosene =
0.82 g/mL = 0.82 1.00 g/mL
Calculate the number of calories of heat energy required for each (See Table 1.8) a. to warm 2.60 g of isopropyl alcohol from 15.0 oC to 35.0 oC b. to warm 17.5 g of isopropyl alcohol from 32.0 oC to 87.0 oC
Answer: a. Use the specific heat of isopropyl alcohol, 0.612 cal/g○C, to convert mass and temperature change into calories. The temperature change is 20.0 °C (35.0 °C – 15.0 °C).
2.60 g x 20.0 C x
0.612 cal = 31.8 cal g C
b. The temperature change is 55.0 °C (32.0 °C – 87.0 °C).
17.5 g x 55.0 C x
1.90
0.612 cal = 589 cal g C
Calculate the number of calories of heat energy required for each. (See Table 1.8) a. to warm 2.60 mL of isopropyl alcohol from 21.0°F to 29.0°F b. to warm 17.5 mL of isopropyl alcohol from 18.0°F to 54. 0°F
Answer: a. First, convert the volume to mass using density. Use the specific heat to convert mass and temperature change into calories. For isopropyl alcohol, the density is 0.785 g/mL and the specific heat is 0.612 cal/g°C.
2.60 mL isopropyl alcohol
0.785 g = 2.04 g isopropyl alcohol 1 mL
Convert the temperatures to °C then calculate the change in temperature.
1 - 44
C =
29.0 - 32 = -1.7C 1.8
C =
21.0 - 32 = -6.1C 1.8
The temperature change is -1.7°C – (-6.1°C) = 4.4°C Use these values to calculate the energy required: 2.04 g 4.4 C
0.612 cal = 5.5 cal g C
b. to warm 17.5 mL of isopropyl alcohol from 18.0°F to 54. 0°F
17.5 mL isopropyl alcohol
0.785 g = 13.7 g isopropyl alcohol 1 mL
Convert the temperatures to °C then calculate the change in temperature.
C =
54.0 - 32 = 12.2C 1.8
C =
18.0 - 32 = -7.8C 1.8
The temperature change is 12.2°C – (-7.8°C) = 20.0°C Use these values to calculate the energy required: 13.7 g 20.0 C
1.92
0.612 cal = 168 cal g C
How much will the temperature change when 55.0 g of each of the following material absorbs 125 cal of heat energy? a. silver (specific heat = 0.056 cal/g ○C) b. olive oil (specific heat = 0.47 cal/g ○C) c. table salt (specific heat = 0.86 J/g ○C)
Answer: a. silver (specific heat = 0.056 cal/g ○C)
55.0 g x temperature change x temperature change =
0.056 cal = 125 cal g C
125 cal g C x = 41 C 55.0 g 0.056 cal
b. olive oil (specific heat = 0.47 cal/g ○C) 1 - 45
55.0 g x temperature change x
0.47 cal = 125 cal g C
125 cal g C x = 4.8 C 55.0 g 0.47 cal
temperature change =
c. table salt (specific heat = 0.86 J/g ○C)
0.86 cal = 125 cal g C 125 cal g C x = 2.6 C 55.0 g 0.86 cal
55.0 g x temperature change x temperature change =
SECTION 1.8
1.94
MEASUREMENTS IN GENERAL CHEMISTRY, ORGANIC CHEMISTRY, AND BIOCHEMISTRY
Cadmium has a density of 8.65 g/cm3, lead a density of 11.35 g/cm3, and zinc a density of 7.14 g/cm3. e. A 15.0 cm3 sample of each metal is obtained. What is the mass of each sample, in grams, milligrams, and pounds? f. A 20.0 in3 sample of each metal is obtained. What is the mass of each sample, in grams, milligrams, and pounds? g. A 50.0 g sample of each metal is obtained. What is the volume of each sample, in cubic centimeters, cubic millimeters, and cubic inches? h. A 1.50 lb sample of each metal is obtained. What is the volume of each sample, in cubic centimeters, cubic millimeters, and cubic inches?
Answer: a. Use the density of each metal as a conversion factor to convert volume to mass. Cadmium 15.0 cm3 ×
8.65 g 1 cm3
1.30 x 102 g ×
=
1 mg 10−3 g
1.30 x 102 g =
1.30 x 105 mg
1 - 46
1.30 x 102 g ×
1 kg 2.205 lb × 3 10 g 1kg
=
0.287 lb
Lead 15.0 cm3 ×
11.35 g 1 cm3
1.70 x 102 g
=
1.70 x 102 g ×
1 mg 10−3 g
1.70 x 102 g ×
1 kg 2.205 lb × 3 10 g 1kg
=
1.70 x 105 mg
=
0.375 lb
Zinc 15.0 cm3 ×
7.14 g 1 cm3
=
1.07 x 102 g
1.07 x 102 g ×
1 mg 10−3 g
1.07 x 102 g ×
1 kg 2.205 lb × 3 10 g 1kg
=
1.07 x 105 mg
=
0.236 lb
b. First, convert the volume from in3 to cm3 because the density is given with cm3 as the volume unit. To convert in3 to cm3, the conversion factor needs to be raised to the power of 3 to change inch to in3 and centimeter to cm3. 20.0 in
3
2.54 cm 3 ) × ( 1 in
=
328 cm3
Then, use the density of each metal as a conversion factor to convert volume to the different mass units. Cadmium 328 cm3 ×
8.65 g 1 cm3
2.83 x 103 g ×
=
1 mg 10−3 g
2.83 x 103 g =
2.83 x 106 mg
1 - 47
2.83 x 103 g ×
1 kg 2.205 lb × 3 10 g 1kg
=
6.24 lb
Lead 328 cm3 ×
11.35 g 1 cm3
3.72 x 103 g
=
3.72 x 103 g ×
1 mg 10−3 g
3.72 x 103 g ×
1 kg 2.205 lb × 3 10 g 1kg
=
3.72 x 106 mg
=
8.20 lb
Zinc 328 cm3 ×
7.14 g 1 cm3
2.34 x 103 g
=
2.34 x 103 g ×
1 mg 10−3 g
2.34 x 103 g ×
1 kg 2.205 lb × 3 10 g 1kg
=
2.34 x 106 mg
=
5.16 lb
c. Use the density of each metal as a conversion factor to convert mass to volume. To convert a unit that is cubed to another unit that is cubed, raise the conversion factor(s) to the third power. Cadmium 50.0 g ×
1 cm3 8.65 g
5.78 cm3
=
3
1 mm 3 × ( -3 ) 10 m
3
10-2 m × ( ) 1 cm
3
1 in 3 ( ) = × 2.54 cm
5.78 cm
5.78 cm
0.353 in3
1 - 48
=
5.78 x 103 mm3
Lead 50.0 g ×
1 cm3 11.35 g
4.41 cm3
= 3
× (
1 mm 3 ) 10-3 m
1 in 3 ) = 4.41 cm3 × ( 2.54 cm
0.269 in3
3
4.41 cm
10-2 m × ( ) 1 cm
=
4.41 x 103 mm3
=
7.00 x 103 mm3
Zinc 50.0 g ×
1 cm3 7.14 g
7.00 cm3
=
3
× (
1 mm 3 ) 10-3 m
1 in 3 ) = 7.00 cm3 × ( 2.54 cm
0.427 in3
3
7.00 cm
10-2 m × ( ) 1 cm
d. First, convert the mass from lb to g. Then, use the density of each metal as a conversion factor to convert mass to volume. Cadmium 1.50 lb ×
454 g 1 lb
=
681g
681 g ×
1 cm3 8.65 g
=
78.7 cm3 3
× (
1 mm 3 ) 10-3 m
1 in 3 ) = 78.7 cm3 × ( 2.54 cm
4.80 in3
3
78.7 cm
10-2 m × ( ) 1 cm
1 - 49
=
7.87 x 104 mm3
Lead 1.50 lb ×
454 g 1 lb
=
681g
681 g ×
1 cm3 11.35 g
=
60.0 cm3 3
× (
1 mm 3 ) 10-3 m
1 in 3 ) = 60.0 cm3 × ( 2.54 cm
3.66 in3
3
60.0 cm
10-2 m × ( ) 1 cm
=
6.00 x 104 mm3
=
9.54 x 104 mm3
Zinc 454 g 1 lb
=
681g
1 cm3 681 g × 7.14 g
=
95.4 cm3
1.50 lb ×
3
10-2 m × ( ) 1 cm
3
1 in 3 ) = × ( 2.54 cm
95.4 cm
95.4 cm
1.96
1 mm 3 × ( -3 ) 10 m
3
5.82 in3
At 20 °C, what is the volume, in milliliters, occupied by d. 8.49 g of isooctane? e. 4.71 x 102 g of isooctane? f. 12.0 lb of isooctane?
Answer: a. Use the density of isooctane given in the text, 0.69 g/mL at 20 °C, as a conversion factor to convert the mass given to the volume of isooctane. 8.49 g ×
1 mL 0.69 g
=
12 mL
1 - 50
b. Again, use the density of isooctane given in the text, 0.69 g/mL at 20 °C, as a conversion factor to convert the mass given to the volume of isooctane. 8.49 x 102 g ×
1 mL 0.69 g
=
1200 mL
c. First, convert the mass in lb to g. Then, use the density of isooctane given in the text, 0.69 g/mL at 20 °C, as a conversion factor to convert the mass given to the volume of isooctane.
1.98
12.0 lb ×
454 g 1 lb
=
5450 g
5450 g ×
1 mL 0.69 g
=
7900 mL
At 20 °C, what is the volume, in gallons, occupied by d. 8.99 lb of gasoline? e. 114 lb of gasoline? f. 22.6 kg of gasoline?
Answer: a. First, convert the mass in lb to g. Then, use the density of gasoline given in the text, 0.73 g/mL at 20 °C, as a conversion factor to convert the mass given to the volume of gasoline.
8.99 lb ×
454 g 1 lb
=
4080 g
4080 g ×
1 mL 0.73 g
=
5600 mL
b. Similar calculations as part a.:
114 lb × 51800 g ×
454 g 1 lb 1 mL 0.73 g
51800 g or 5.18 x 104 g
= =
71000 mL or 7.1 x 104 mL
1 - 51
c. First, convert the mass in kg to g. Then, use the density of gasoline given in the text, 0.73 g/mL at 20 °C, as a conversion factor to convert the mass given to the volume of gasoline. 103 g 22.6 kg × 1 kg
=
22600 g
1 mL 0.73 g
=
31000 mL or 3.1 x 104 mL
22600 g ×
1.100 When it is compacted, DNA fits into a cell nucleus that is 80,000 times smaller in diameter than the DNA is long. If the strand in Problem 1.99 were compacted by the same amount, what size package would it fit in? Answer: length of stack of pennies from Problem 1.99
=
5 x 106 m
diameter of compacted DNA = 5 x 106 m / 80,000 = 60 m
HEALTH LINK
SCIENCE AND MEDICINE
1.102 Until the late 1980s, what was the source of the insulin used to treat diabetes? Answer: Insulin was isolated from cattle and pigs until 1982 when genetically engineered human insulin became the source of insulin for treatment of diabetes.
HEALTH LINK
BODY MASS INDEX
1.104 September 2006 was the first time that models were banned from a top-level fashion show for being too thin. The organizers of the Madrid Fashion Week defined “too thin” as having a BMI of less than 18. How much would a 5’2” model weigh if she had a BMI of 16?
1 - 52
Answer:
BMI = 703
x
weight height 2
height = 5 ft 2 in = 5 ft x
12 in + 2 in = 62 in 1 ft
So, solving for weight:
weight =
BMI x height 2 703
weight =
16 x (62) 2 = 87 lbs 703
HEALTH LINK
BODY TEMPERATURE
1.106 Suppose that you take your temperature orally and see that it is 99.1°F. Does this necessarily mean that you are running a fever? Explain. Answer: No. A normal body temperature of approximately 98.6oF is based on the average of temperatures in healthy people. A difference of only 0.5 °F could easily be normal for you.
HEALTH LINK
MAKING WEIGHT
1.108 Why does a high urine specific gravity indicate dehydration? Answer: The high specific gravity indicates that there is more dissolved material per volume of water in the urine. This is the result of dehydration. 1.110 a. Use the density of water (1.00 g/mL) to derive a conversion factor for water that has the units kg/L.
1 - 53
b. If an athlete reduces his body’s water volume by 0.75 L through restricting fluid intake and sweating in a sauna, how much weight (in kilograms) has he lost? Answer: a.
1.00 g 1 mL
b. 0.75 L
1 kg 103 g
1 mL = 1.00 kg/L 10−3 L
1.00 kg = 0.75 kg 1 L
LEARNING GROUP PROBLEMS
1.112 a. Write the two conversion factors that are based on the equality 15 drops = 1 milliliter. b. Which conversion factor in your answer to part a would be used to convert milliliters to drops? c. 65 drops of water is how many milliliters? d. 65 drops of water is how many microliters? e. 65 drops of water is how many tablespoons? (See Table 1.2) Answer: a.
15 drops 1 mL and 15 drops 1 mL
b.
15 drops 1 mL
c. 65 drops
1 mL = 4.3 mL 15 drops
d. From part c., 65 drops is equivalent to 4.3 mL. Convert 4.3 mL to L
4.3 mL
10-3 L 1 mL
1 L = 4.3 x 103 L -6 10 L
e. 65 drops = 4.3 mL as in part c. From Table 1.2, 15 mL = 1 T
1 - 54
4.3 mL
1T = 0.29 T 15 mL
1 - 55
Chapter 2 Atoms and Elements SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
2.1
Identify the two nuclei and the type of nuclear radiation that this released.
To identify the two nuclei, use the number of protons, which is the same as the atomic number. Refer to the periodic table to determine the element. The first nucleus belongs to a Be atom because it contains 4 protons. Its mass number is 10 (4 protons + 6 neutrons). The product nucleus belongs to a He atom because it contains 2 protons. Its mass number is 6 (2 protons + 4 neutrons). The difference between the two nuclei corresponds to a loss of 2 protons and 2 neutrons released as an alpha particle radiation. The equation is: 10 4
2.3
Be →
6 2
He +
4 2
Why does the nucleus of an atom have a positive charge? It consists of protons (positive charge) and neutrons (no charge).
2.5
a. A hydrogen atom has a diameter of 5.3 x 10-11 m. Express this diameter in nanometers and picometers. b. Placed side-by-side, how many hydrogen atoms would it take to equal a distance of 1 inch (1 in = 2.54 cm)? Set-up the following conversion steps to solve these two problems.
2 - 25
a. A hydrogen atom has a diameter of 5.3 x 10-11 m. Express this diameter in nanometers and picometers (pico=10-12). 5.3 x 10-11 m x
1 nm = 5.3 x 10-2 nm or 0.053 nm -9 1 x 10 m
5.3 x 10-11 m x
1 pm = 5.3 x 101 pm or 53 pm -12 1 x 10 m
b. Placed side-by-side, how many hydrogen atoms would it take to equal a distance of 1 inch (1 in = 2.54 cm)? First, convert the diameter of one hydrogen atom to cm: 5.3 x 10-11 m x
1 cm = 5.3 x 10-9 cm 1 x 10-2 m
Then, set-up a conversion to determine how many hydrogen atoms would “fit” into a distance of 1 inch or 2.54 cm using the equivalence of 1 hydrogen atom occupying 5.3 x 10-9 cm: 2.54 cm x
2.7
1 hydrogen atom = 4.8 x 108 hydrogen atoms 5.3 x 10-9 cm
A tin atom has a diameter of 145 pm. If the nucleus of an atom is 100,000 times smaller than the atom, what is the diameter of the nucleus of a tin atom in picometers? According to the problem, the diameter of the nucleus of a tin atom is 1/100,000 that of the diameter of the whole atom:
1 x 145 pm = 1.45 x 10-3 pm 100,000
2.9
What is the atomic symbol for each element? a. b. c. d.
iron iodine helium hydrogen
Fe I He H
2 - 26
2.11
The elements carbon, calcium, chlorine, copper, and cobalt are present in the human body. Match each element to its correct atomic symbol: Co, Cu, Ca, C, Cl. Co is cobalt; Cu is copper; Ca is calcium; C is carbon; Cl is chlorine
2.13
Give the name and atomic symbol for the four most abundant elements in the human body. oxygen (O), carbon (C), hydrogen (H), nitrogen (N)
2.15
Table 2.4 lists the RDA for some elements and the AI for others. Why? Adequate Intake or AI is used when an RDA is not available for a given nutrient.
2.17
In NAS dietary tables the dietary reference intakes for selenium is reported in micrograms per day, rather than the milligrams per day given in Table 2.4. For this element, convert the values in Table 2.4 into micrograms per day. The RDA for selenium is 0.055 mg/day. In micrograms per day: 0.055 mg 10-3 g μg × × day 1 mg 10-6 g
=
55 μg day
The UL for selenium is 40 mg/day. In micrograms per day: 40 mg 10-3 g μg × × day 1 mg 10-6 g
2.19
=
40,000 μg day
The nutrition facts label on a box of macaroni and cheese says that one serving contains 20% of the daily recommended amount (Adequate Intake) of calcium. To how many milligrams of calcium does this correspond? (See Table 2.4) The Adequate Intake for calcium given in Table 2.4 is 1300 mg/day. 20% of 1300 mg is: 0.20 × 1300 mg = 260 mg Therefore, one serving of macaroni and cheese contains about 260 mg of calcium.
2 - 27
2.21
How many protons and neutrons are present in the nucleus of each? The number of protons is equal to the atomic number. The number of neutrons is equal to the mass number minus the number of protons. a. 9 F
19
9 protons
19 - 9 = 10 neutrons
b. 11 Na
23
11 protons
23 - 11 = 12 neutrons
238
92 protons
238 - 92 = 146 neutrons
c. 92 U
2.23
Give the atomic number and mass number of a. a helium atom with 2 neutrons b. a lithium atom with 3 neutrons c. a neon atom with 10 neutrons Use the periodic table to look up the atomic number for each element. The mass number can be calculated by adding the number of protons (equal to the atomic number) and the number of neutrons. Atomic number
2.25
Mass number
a.
a helium atom with 2 neutrons
2
2 protons + 2 neutrons = 4
b.
a lithium atom with 3 neutrons
3
3 protons + 3 neutrons = 6
c.
a neon atom with 10 neutrons
10
10 protons + 10 neutrons = 20
Give the atomic notation for each atom described in Problem 2.23. A
The general symbol for atomic notation is Z X , where A is the mass number, Z is the atomic number, and X is the symbol of the element. 4
a. 2 He
6
b. 3 Li
20
c. 10 Ne
2 - 28
2.27
Give the number of neutrons in an atom of 40 a. 18 Ar 42
b. 20 Ca 50
c. 23 V To calculate the number of neutrons, subtract the atomic number from the mass number: a. 18 Ar
40
number of neutrons = 40 – 18 = 22 neutrons
b. 20 Ca
42
number of neutrons = 42 – 20 = 22 neutrons
50
number of neutrons = 50 – 23 = 27 neutrons
c. 23 V
2.29
Give the number of electrons present in a neutral atom of each element. a. copper b. calcium c. chlorine For a neutral atom of an element: number of electrons = number of protons (equal to the atomic number)
2.31
a. copper
Cu
29 electrons
b. calcium
Ca
20 electrons
c. chlorine
Cl
17 electrons
Complete the table. Answers are in bold: Name
Calcium
Carbon
Copper
Atomic notation
40 20
13 6
63 29
Number of protons
20
6
29
Number of neutrons
20
7
34
Ca
2 - 29
C
Cu
2.33
Atomic number
20
6
29
Mass number
40
13
63
Which of the following statements do not accurately describe isotopes of an element? a. same number of protons b. same mass number c. same atomic number Choice b, the same mass number, would not accurately describe isotopes of an element. Isotopes are atoms of the same element so they have the same number of protons but different number of neutrons. Since the number of protons and atomic number are the same, then answers a and c are correct. The mass number would not be the same for isotopes of an element.
2.35
Which are isotopes? 64 65 65 64 28 Ni , 30 Zn , 29 Cu , 30 Zn Isotopes are atoms of the same element that have different number of neutrons. 64 30
2.37
Zn and 3065Zn are isotopes. 85
87
Rubidium, which appears in nature as 37 Rb (mass = 84.91 amu) and 37 Rb (mass = 86.91 amu) has an atomic weight of 85.47 amu. Which isotope predominates? 85 37
Rb . The mass of this isotope (84.91 amu) is closest to the atomic weight given
in the periodic table (85.47 amu) and therefore is the more abundant one.
2.39
List some of the physical properties of metals. Metals are good conductors of heat and electricity, shiny, malleable, and ductile.
2 - 30
2.41
Identify two elements that belong to each group. a. halogen Possible answers are: fluorine, chlorine, bromine, and iodine. b. inert gas Possible answers are: helium, neon, argon, krypton, xenon, and radon.
2.43
Identify two representative nonmetallic elements that are in the third period. Possible answers are: phosphorus, sulfur, chlorine, and argon.
2.45
Which of the two elements is more metallic? Refer to the periodic table and find where the elements are located. The farther they are to the left of the metal-nonmetal transition line (“zigzag” line), the more metallic they are. Elements to the right of this line are nonmetals. Another way of thinking of this is that the farther left in a period or the farther down a group, the more metallic the element is. a. Na and Cl Na. Na is found in the leftmost column of the periodic table and is more metallic than Cl (a nonmetal) which is to the right of the transition line. b. O and Te Te. Te is found closer to the right of the transition line than O and therefore is more metallic.
2.47
Which would you expect to be larger, an atom of cesium or an atom of francium? Francium. Francium is below cesium in Group 1A of the periodic table. Generally, atomic size increases moving down a group.
2 - 31
2.49
List the two conversion factors that relate number of lithium atoms and moles of lithium. Given that 1 mole of atoms is equal to 6.02 x 1023 atoms, the two conversion factors are: 6.02 x 1023 Li atoms 1 mol Li and 1 mol Li 6.02 x 1023 Li atoms
2.51
a. How many moles of lithium is 3.01 x 1023 lithium atoms? (Refer to Problem 2.49) To convert number of Li atoms to mol Li atoms, use the conversion factor that has mol Li atoms in the numerator: 3.01 x 1023 Li atoms x
1 mol Li = 0.500 mol Li 6.02 x 1023 Li atoms
b. How many lithium atoms is 0.525 mol of lithium? To convert mol Li atoms to number of Li atoms, use the conversion factor that has number of Li atoms in the numerator:
0.525 mol Li x
2.53
6.02 x 1023 Li atoms = 3.16 x 1023 Li atoms 1 mol Li
List the two conversion factors that relate grams of calcium and moles of calcium. Use the molar mass of Ca which is numerically equal to its atomic weight to setup the two conversion factors. From the periodic table, 1 mole of Ca weighs 40.08 g.
40.08 g Ca 1 mol Ca and 1 mol Ca 40.08 g Ca
2.55
a. How many moles of calcium is 126 g of Ca? (Refer to Problem 2.53) To convert grams of Ca to mol Ca, use the conversion factor that has mol Ca in the numerator:
2 - 32
126 g Ca x
1 mol Ca = 3.14 mol Ca 40.08 g Ca
b. How many grams of calcium is 2.25 mol of Ca? To convert mol Ca to grams of Ca, use the conversion factor that has grams Ca in the numerator: 2.25 mol Ca x
2.57
40.08 g Ca = 90.2 g Ca 1 mol Ca
How many atoms are present in 2.00 mol of aluminum? One mole of aluminum contains 6.02 x 10 23 atoms. P
2.00 mol Al x
2.59
P
6.02 x 1023 Al atoms = 1.20 x 1024 Al atoms 1 mol Al
7.25 x 1012 sodium atoms is how many moles? One mole of sodium contains 6.02 x 10 23 atoms. P
7.25 x 1012 Na atoms x
2.61
P
1 mol Na = 1.20 x 10-11 mol Na 6.02 x 1023 Na atoms
What is the a. atomic weight of helium (He)? 4.00 amu (rounded to two decimal places) b. molar mass of helium? 4.00 g/mol The molar mass of an element is numerically equal to its atomic weight but expressed in g/mol. c. mass (in grams) of 5.00 mol of helium? One mole of He has a mass of 4.00 g.
2 - 33
5.00 mol He x
4.00 g He = 20.0 g He 1 mol He
d. number of helium atoms in 8.85x10-5 mol of helium? One mole of He contains 6.02 x 1023 He atoms.
8.85 x 10
-5
6.02 x 1023 He atoms mol He x = 5.33 x 1019 He atoms 1 mol He
e. mass (in grams) of 3.39 x 1020 helium atoms? One mole of He has a mass of 4.00 g and contains 6.02 x 1023He atoms.
3.39 x 1020 He atoms x
2.63
1 mol He 4.00 g He x = 2.25 x 10−3 g He 23 6.02 x 10 He atoms 1 mol He
In 3.45 mg of boron there are how many of the following?
1 x 10-3 g = 3.45 x 10-3 g B 1 mg
a. grams of boron
3.45 mg x
b. moles of boron
3.45 x 10-3 g B x
1 mol B = 3.19 x 10-4 mol B 10.81 g B
c. boron atoms
3.19 x 10-4 mol B x
2.65
6.02 x 1023 B atoms = 1.92 x 1020 B atoms 1 mol B
Helium has a different emission spectrum than hydrogen. Account for this difference. The energy levels of a helium atom have different energies than those of a hydrogen atom. This produces a different emission spectrum when helium atoms move from various excited states back to more stable ones.
2 - 34
2.67
Using a periodic table determine the number of electrons held in energy levels 13 of each atom. The electrons in an atom are arranged according to energy levels. Each energy level is specified by a number n. The maximum number of electrons that each energy level can hold is 2n2. When assigning the electrons of a given element to these energy levels, we start filling the n = 1 level completely first and make our way up until all of the electrons have been assigned to an energy level.
2.69
Total # of electrons
n=1 (max. 2 e-)
n=2 (max. 8 e-)
n=3 (max. 18 e-)
a. Be
4
2
2
0
b. N
7
2
5
0
c. Cl
17
2
8
7
Specify the number of valence electrons for each atom. Valence electrons are the electrons found in the outermost shell of an atom. The valence shell is the highest occupied energy level in an atom. Atoms in the same group have the same number of valence electrons. For representative elements, the number of valence electrons corresponds to the group number to which the element belongs. a. H b. Be c. C d. Br e. Ne
2.71
1 valence electron 2 valence electrons 4 valence electrons 7 valence electrons 8 valence electrons
For each, give the total number of electrons, the number of valence electrons, and the number of the energy level that holds the valence electrons. The energy level is correlated with the period number for the element. Total # of electrons
Number of valence electrons
Energy level of valence electrons
a. He
2
2
1
b. Xe
54
8
5
c. Te
52
6
5
2 - 35
d. Pb
2.73
82
4
6
For a neutral atom of element 112 a. how many total electrons are present? 112. In a neutral atom, the number of electrons is equal to the number of protons which is the same as the atomic number of the element. b. how many valence electrons are present? 2. Element 112 is predicted to have 2 valence electrons like the element zinc which belongs to the same group. c. which energy level holds the valence electrons? Energy level n = 7. The valence electrons are held in the highest occupied energy level which is the same as the period number for that element. Element 112 belongs to Period 7 of the periodic table. d. is the valence energy level full? No, the n = 7 energy level can contain 98 electrons. There are only 2 electrons in the n = 7 energy level of element 112.
2.75
Draw the electron dot structure of each atom. The electron dot structure shows the number of valence electrons, drawn as dots around the symbol for the element.
2.77
a. Na
Na
c. Ar
Ar
b. Cl
Cl
d. S
S
a. What is an alpha particle? An alpha particle is a type of radiation that can be emitted by radioisotopes. b. How is an alpha particle similar to a helium nucleus? Like a helium nucleus, an alpha particle has 2 protons, 2 neutrons, and 2+ charge.
2 - 36
c. How is an alpha particle different than a helium nucleus? The alpha particle has a much greater energy than a helium nucleus.
2.79
a. What is a positron? A positron is a subatomic particle that can be emitted by radioisotopes as radiation. b. How is a positron similar to a beta particle? Positrons have the same mass as beta particles and are emitted at speeds of up to 90% of the speed of light like beta particles. c. How is a positron different than a beta particle? A positron has a 1+ charge but a beta particle has 1- charge.
2.81
Identify the missing product in each nuclear equation. In balancing nuclear equations, the sum of the mass numbers and the sum of the charges on atomic nuclei and subatomic particles must be the same on both sides of the equation.
→ ? + 0 a. 32 15 P 1 32 15
P →
32 14
Si +
+
0 + 1
→ ? + 0 b. 40 19 K -1 40 19
K →
c.
40 19 K
40 19
K →
40 20
Ca +
0 −1
→ 40 Ar + ? 18
40 18
Ar +
0 1
β+
2 - 37
2.83
a. Write the balanced nuclear equation for the loss of an alpha particle from 203 83 Bi. The loss of an alpha particle results in the loss of 2 neutrons and 2 protons: 203 83
Bi →
Tl +
199 81
4 2
b. Write the balanced nuclear equation for the loss of a positron from 179 F. The loss of a positron results in the loss of 1 proton but no change in the mass number: 17 9
2.85
F →
17 8O
+
0 + 1
β
Write a balanced nuclear equation for each process. To find the atomic number of an isotope produced in the nuclear reaction, the atomic number of the emitted particle is subtracted from the atomic number of the original radioisotope. Looking this number up on the periodic table gives the atomic symbol of the newly produced isotope. The mass number of the emitted particle is subtracted from the mass number of the original radioisotope to give the new mass number. 187
a. 80 Hg emits an alpha particle 187 80
4 Hg → 183 78 Pt + 2
b. 226 88Ra emits an alpha particle 226 222 88Ra → 86Ra
+ 42α
238
c. 92 U emits an alpha particle 238 92
U
→
234 90
Th
+
4 2
2 - 38
2.87
197 80 Hg, a beta and gamma emitter, is used for brain scans.
Write a balanced equation for the loss of 1 beta particle and 1 gamma ray from this radioisotope. The loss of a beta particle results in the loss of 1 neutron and the gain of 1 proton. The loss of a gamma ray results in no change in the mass number and charge of the original isotope. 197 80
2.89
2.93
+
197 81Tl
+
0 −1
0 0
Write the balanced nuclear equation for the loss of a positron from iodine-128. 128 53
2.91
Hg →
I →
128 52Te
0 + 1
+
In a radioactive decay series, one radioisotope decays into another, which decays into another, and so on. For example, in fourteen steps uranium-238 is converted to lead-206. Starting with uranium-238, the first decay in this series releases an alpha particle, the second decay releases a beta particle, and the third releases a beta particle. Write balanced nuclear equations for these three reactions. 238 92 U
→
234 90Th
+
234 90Th
→
234 91 Pa
+
234 91 Pa
→
234 92 U
+
4 2
0 −1
0 −1
Smoke detectors contain an alpha emitter. Considering the type of radiation released and the usual placement of a smoke detector, do these detectors pose a radiation risk? Explain. No. Since alpha particles travel only 4 -5 cm in air and smoke detectors are usually on the ceiling, the risk of exposure to alpha radiation is small.
2.95
Radioisotopes used for diagnosis are beta, gamma, or positron emitters. Why are alpha emitters not used for diagnostic purposes? Alpha particles are relatively large and do not penetrate tissue very deeply. Radiation emissions must be able to pass through the body and reach a detector to be useful for diagnostic purposes.
2 - 39
2.97
Exposure to 50 rem of radiation can cause nausea and a temporary drop in the white blood cell count. a. Convert this value into seiverts. Use the equivalence 1 seivert = 100 rem 50 rem ×
1 seivert = 0.5 seivert 100 rem
b. Convert this value into rads, assuming that the source of radioactivity is an alpha emitter. Use the equation dose (rad) x QF = dose (rem) where QF is the quality factor that depends on the type of radiation. Table 2.8 in the text gives a QF = 20 for alpha particles. Given 50 rem, we can solve for the value in rads by rearranging the equation above: dose (rad) =
dose (rem) 50 rem = = 2.5 rad or 3 rad QF 20
c. Convert this value into rads, assuming that the source of radioactivity is a beta emitter. For beta particles, the QF is 1: dose (rad) =
2.99
dose (rem) 50 rem = = 50 rad QF 1
Your body contains about 0.1 Ci of naturally-occurring potassium-40. This number of microcuries corresponds to how many disintegration per a. second One curie is defined as 3.7 x 1010 disintegrations per second or 3.7 x 1010 dps. 0.1 Ci ×
10-6 Ci 3.7 x 1010 dps × = 3700 dps or 4000 dps 1 Ci 1 Ci
2 - 40
1 sec ×
4000 disintegrations = 4000 disintegrations 1 sec
b. minute One curie is defined as 3.7 x 1010 disintegrations per second or 3.7 x 1010 dps. 4000 disintegrations 60 seconds × = 2.4 x 105 disintegrations or 2 x second 1 minute 105 disintegrations
2.101
59 26 Fe,
a beta emitter with a half-life of 45 days, is used to monitor iron metabolism. a. Write a balanced nuclear equation for this radioactive decay. 59 26
Fe →
59 27 Co
+
β
0 −1
b. How much time must elapse before a patient contains just 25% of an administered dose of 59 26 Fe, assuming that this radioisotope is eliminated from the body only by radioactive decay? 90 days. The half-life of the iron radioisotope is 45 days. After 45 days, only 50% of the administered dose is left in the patient. After another 45 days, only half of the 50% remaining dose is left. Half of the 50% dose left is 25% of the original. Therefore, it takes 90 days for the amount of administered dose to go down to 25% of the original.
2.103 Phosphorus-32 has a half-life of 14.3 days. Beginning with a 2.00 µg sample of this radioisotope, a. how many micrograms remain after 2 half-lives? Original amount
2.00 µg
After 1 half-life (14.3 days)
1.00 µg remain
After 2 half-lives (28.6 days)
0.500 µg remain
b. how many micrograms remain after 42.9 days? 2 - 41
number of half-lives = 42.9 days x
1 half-life = 3 half-lives 14.3 days
Continuing the series from part a: Original amount
2.00 µg
After 1 half-life (14.3 days)
1.00 µg remain
After 2 half-lives (28.6 days)
0.500 µg remain
After 3 half-lives (42.9 days)
0.250 µg remain
c. how many days will it take for the mass of phosphorus-32 to drop to 0.125 µg? 57.2 days 0.125 µg is what remains after 4 half-lives (42.9 days + 14.3 days = 57.2 days). Refer to the table above.
2.105 a. Neon-19 is a positron emitter with a half-life of about 20 seconds. Write a balanced nuclear equation for the loss of a positron from this radioisotope. 19 10
Ne →
19 9
F +
β
0 + 1
b. Neon-24 is a beta emitter with a half-life of about 200 seconds. Write a balanced nuclear equation for the loss of a beta particle from this radioisotope. 24 10
Ne →
24 11
Na +
0 −1
c. Beginning with 2 µg of neon-24, how many micrograms would remain after 200 seconds? 1 µg of neon-24 would remain. 200 seconds is equivalent to 1 half-life. d. How many half–lives of neon-19 will pass in 200 seconds? number of half-lives = 200 seconds x
1 half-life = 10 half-lives 20 seconds
2.107 a. In chemical terms, why can exposure to nuclear radiation be harmful? Exposure to nuclear radiation is harmful to living tissues because of the kinetic energy that radioactive emissions impart to surrounding atoms. This transfer of energy can change the structure of water and important biochemical substances
2 - 42
such as proteins, DNA, lipids, and others found within cells, disrupting normal biochemical functions. b. What are some of the short-term effects of being exposed to a high dose of radiation? The short-term effects of being exposed to a high dose of radiation include nausea to death within a few weeks depending on the dose. c. What might be the source of a high dose of radiation? Accidents in nuclear power plants may cause release of high doses of radiation.
2.109 a. Which is more easily shielded against, alpha particles, beta particles, or gamma rays? alpha particles b. Which of the types of radiation in part a is the most difficult to shield against? gamma particles
2.111
52 26
Fe , a positron emitter with a half-life of 8.2 hours, is used for PET bone
marrow scans. a. Write a balanced nuclear equation for this radioactive decay. 52 26
Fe →
52 25 Mn
+
β
0 + 1
b. The detector used for this scan measures gamma rays. How is the release of positrons connected to the formation of gamma rays? When a positron is emitted by the 52 26 Fe radioisotope, it collides with an electron from a nearby atom. This collision destroys both the positron and the electron, resulting in the release of two gamma rays. c. Assuming that this radioisotope is eliminated from the body only by radioactive decay, how much time must elapse before a patient contains just 52 25% of an administered dose of 26 Fe ? 16.4 hours. The half-life of the 52 26 Fe radioisotope is 8.2 hours. After 1 halflife (8.2 hours), 50% of the administered dose remains. After 2 half-lives
2 - 43
(16.4 hours), half of the remaining 50% will decay, bringing the quantity of the radioisotope inside the patient down to 25% of the administered dose. 52
d. The product of 26 Fe positron decay is a radioisotope that emits positrons. Write a balanced nuclear equation for the decay of this product. 52 The product of the 52 26 Fe decay is the 25 Mn radioisotope. The decay by positron emission of this radioisotope is described by:
52 25 Mn
→
52 24 Cr
+
β
0 + 1
2.113 What is brachytherapy? Brachytherapy is a type of cancer treatment in which a radiation source is placed internally near a tumor. The radiation source delivers a high dose of radiation over time killing cancer cells in the process.
2.115 a. What are the two stable isotopes of carbon, and which predominates in nature? 12
13
12
The two stable isotopes of carbon are 6 C and 6 C . The 6 C isotope is more predominant in nature.
b. Compared to most of the carbon found in nature, plants and animals contain slightly more of the lighter isotope of carbon and slightly less of the heavier isotope. Explain why. 12
The lighter 6 C isotope reacts slightly faster and is therefore incorporated in plants and animals in higher amounts.
2.117 In what ways was lead used in Ancient Rome, and how is one of those uses related to the word “plumbing”? In ancient Rome, lead was used to line containers in which to boil wine. Lead was also used for water pipes, thus the use of the term “plumbing” derived from the Latin name of lead “plumbum”.
2.119 How are electrons involved in the production of light by luciferin?
2 - 44
When luciferin is acted on by a particular enzyme in the presence of oxygen gas and ATP, electrons in luciferin are raised to an excited state and, upon returning to the ground state, light is emitted.
2.121 Why were thorium and radium present in Tho-Radia face creams and powders? People back then believed that radioactive substances would improve one’s beauty by energizing and revitalizing the skin. 2.123 a. One of two techniques described in this Health Link is considered “invasive”. What do you suppose is meant by this term? Invasive refers to the fact that the imaging technique requires something (x rays in this case) to penetrate the body. b. To which technique does this term apply? CT
2.125 In late 2006, a former Russian spy was poisoned when a small amount of polonium – 210 was put in his food. He died a few weeks later. a. What is the atomic symbol of polonium? Po b. How many protons and neutrons does an atom of polonium-210 have? 84 protons and 126 neutrons c. Describe the structure of an alpha particle. An alpha particle has 2 protons, 2 neutrons, and a 2+ charge. d. Polonium-210 emits alpha particles. Write a balanced nuclear reaction for this process. 210 84 Po
→
4 2
+
206 82 Pb
2 - 45
e. Polonium-210 has a half-life of 138 days. Of a 2 g sample, how much would remain after 5 half-lives? See table below: 210 84 Po
(half-life = 138 days)
Original amount
2 µg
After the 1st half-life
1 µg remain
After the 2nd half-life
0.5 µg remain
After the 3rd half-life
0.25 µg remain
After the 4th half-life
0.125 µg remain
After the 5th half-life
0.06 µg remain
f. If alpha particles can be blocked by a sheet of paper (Figure 2.27), why is polonium – 210 so poisonous? Since polonium-210 was ingested by the victim, any released alpha particles were readily absorbed by cells and their harmful effects were felt. The alpha particles are capable of breaking certain chemical bonds within the body which could affect important biochemical processes.
ANSWERS TO EVEN NUMBERED END OF CHAPTER PROBLEMS 2.2
Identify the two nuclei and the type of nuclear radiation that this released.
Answer: To identify the two nuclei, use the number of protons, which is the same as the atomic number. Refer to the periodic table to determine the element. 2 - 46
The first nucleus belongs to a Be atom because it contains 4 protons. Its mass number is 10 (4 protons + 6 neutrons). The product nucleus belongs to a B atom because it contains 5 protons. Its mass number is 10 (5 protons + 5 neutrons). The difference between the two nuclei corresponds to a gain of 1 proton and loss of 1 neutron. Therefore, the missing particle is a beta-particle. The equation is: 10 4
SECTION 2.1
2.4
Be→105B+ −10
ATOMS
Why is most of the mass of an atom located in the nucleus?
Answer: The nucleus of an atom is composed of protons and neutrons. These two subatomic particles are much more massive than the electrons. Therefore, most of the mass of an atom is in the nucleus.
2.6
a. A cesium atom has a diameter of 2.98 x 10-10 m. Express this diameter in nanometers and picometers. b. Placed side-by-side, how many cesium atoms would it take to equal a distance of 1 inch (1 in = 2.54 cm)?
Answer: a.
2.98 x 10-10 m x
2.98 x 10-10 m x b.
1 nm = 2.98 x 10-1 nm or 0.298 nm -9 10 m
1 pm = 2.98 x 102 pm or 298 pm -12 10 m
First, convert the diameter of one cesium atom to cm:
2.98 x 10-10 m x
1 cm = 2.98 x 10-8 cm 10-2 m
2 - 47
Then, set-up a conversion to determine how many cesium atoms would “fit” into a distance of 1 inch or 2.54 cm using the equivalence of 1 cesium atom occupying 2.98 x 10-8 cm:
2.54 cm x
2.8
1 cesium atom = 8.52 x 107 cesium atoms -8 2.98 x 10 cm
A carbon atom has a diameter of 67 pm. If the nucleus of an atom is 100,000 times smaller than the atom, what is the diameter of the nucleus of a carbon atom in picometers?
Answer: According to the problem, the diameter of the nucleus of a carbon atom is 1/100,000 that of the diameter of the whole atom:
1 x 67 pm = 6.7 x 10-4 pm 100,000
SECTION 2.2
2.10
ELEMENTS
What is the atomic symbol for each element? a. nickel b. nitrogen c. neon d. neptunium
Answer: a. nickel b. nitrogen c. neon d. neptunium
2.12
Ni N Ne Np
The elements nitrogen, nickel, potassium, phosphorus, sodium, and sulfur are present in the human body. Match each element to its correct atomic symbol: N, Na, Ni, P, K, S.
2 - 48
Answer: nitrogen
N
nickel
Ni
potassium
K
phosphorus
P
sodium
Na
sulfur
S
SECTION 2.3
2.14
TRACE ELEMENTS AND VITAMINS
Give names and atomic symbols for four trace elements.
Answer: The trace elements in the human body listed in the text are:
2.16
iodine (I)
iron (Fe)
zinc (Zn)
copper (Cu)
nickel (Ni)
cobalt (Co)
selenium (Se)
manganese (Mn)
Which is a better measure of the amount of nutrient needed to meet a person’s dietary needs, RDA or AI? Explain.
Answer: The RDA’s (Recommended Dietary Allowance) recommended daily intake of nutrients is validated to meet the needs of 97-98% of healthy people. AIs (Adequate Intake), however, are based on less reliable experimental data on the percentage of people covered by a particular adequate intake recommendation.
2.18
In NAS dietary tables the dietary reference intakes for iodine is reported in micrograms per day, rather than milligrams per day given in Table 2.4. For this element, convert the values in Table 2.4 into micrograms per day.
Answer: The RDA for iodine given is 0.15 mg/day. In micrograms per day: 0.15 mg 10-3 g μg × × day 1 mg 10-6 g
=
150 μg day
The UL for iodine given is 1.1 mg/day. In micrograms per day:
2 - 49
1.1 mg 10-3 g μg × × day 1 mg 10-6 g
2.20
=
1100 μg day
The nutrition facts label on a package of cookies states that one serving contains 1% of the daily recommended amount of iron. To how many milligrams of iron does this correspond? (See Table 2.4)
Answer: The RDA for iron given in Table 2.4 is 18 mg/day. 1% of 18 mg is: 0.01 × 18 mg =
0.18 mg = 0.2 mg
Therefore, one serving of the cookies contains about 0.2 mg of iron.
SECTION 2.4
2.22
ATOMIC NUMBER AND MASS NUMBER
How many protons and neutrons are present in the nucleus of each? a. 133 54 Xe (used to detect lung malfunctions) b. 75 34 Se (used in pancreas scans) c. 84 37 Rb (used to measure cardiac output)
Answer: In the atomic notation: number
superscript = mass number
mass number
=
number of protons + number of neutrons
atomic number
=
number of protons
number of neutrons
=
mass number - number of protons
a. mass number = 133
atomic number = 54
2 - 50
subscript = atomic
number of protons
=
atomic number
number of neutrons
=
mass number – number of protons
=
133 – 54
b. mass number = 75
54 protons
=
79 neutrons
=
41 neutrons
=
47 neutrons
atomic number = 34
number of protons
=
34 protons
number of neutrons
=
75 – 34
c. mass number = 84
2.24
=
atomic number = 37
number of protons
=
37 protons
number of neutrons
=
84 – 37
Give the atomic number and mass number of each atom. a. a boron atom with 6 neutrons b. a magnesium atom with 12 neutrons c. a phosphorus atom with 16 neutrons
Answer: Use the periodic table to look up the atomic number for each element. The mass number can be calculated by adding the number of protons (equal to the atomic number) and the number of neutrons. Atomic number
2.26
Mass number
a.
a boron atom with 6 neutrons
5
5 protons + 6 neutrons = 11
b.
a magnesium atom with 12 neutrons
12
12 protons + 12 neutrons = 24
c.
a phosphorus atom with 16 neutrons
15
15 protons + 16 neutrons = 31
Give the atomic notation for each atom described in Problem 2.24.
Answer: In general, the atomic notation is given by:
2 - 51
mass number atomic number
atomic symbol
a. 115 B
2.28
b. 24 12 Mg
31 c. 15 P
Give the number of neutrons in an atom of 28 a. 14 Si 35 b. 17 Cl c. 55 25 Mn
Answer: To calculate the number of neutrons, subtract the atomic number from the mass number:
2.30
28 a. 14 Si
number of neutrons = 28 – 14 = 14 neutrons
35 b. 17 Cl
number of neutrons = 35 – 17 = 18 neutrons
c. 55 25 Mn
number of neutrons = 55 – 25 = 30 neutrons
Give the number of electrons present in a neutral atom of each element. a. cadmium b. californium c. cobalt
Answer: For a neutral atom of an element: number of electrons = number of protons (equal to the atomic number) a. cadmium
Cd
48 electrons
b. californium
Cf
98 electrons
c. cobalt
Co
27 electrons
2 - 52
2.32
Complete the table:
Answer: The answers are in bold italics: Name
Hydrogen
Potassium
1 1
H
41 19
Number of protons
1
19
26
Number of neutrons
0
22
30
Atomic number
1
19
26
Mass number
1
41
56
Atomic notation
2.34
K
Which of the following statements do not accurately describe the isotopes of an element? a. same atomic symbol b. same number of neutrons c. same number of protons and neutrons
Answer: a. is an accurate statement because isotopes are atoms of the same element that differ only in the numbers of neutrons. b. and c. are not accurate statements because isotopes differ in numbers of neutrons. Therefore, their mass numbers are different and they do not have the same number of protons and neutrons.
2.36
Which are isotopes? 35 17
35 37 Cl, 37 17 Cl, 18 Ar, 19 K
Answer: Isotopes are atoms of the same element (therefore, same symbol) that have different numbers of neutrons (and therefore, different mass numbers). 35 17
Cl and 37 17 Cl are isotopes.
2 - 53
Iron 56 26
Fe
2.38
Oxygen, which appears in nature as 168 O (mass = 16.00 amu), 178 O (mass = 17.00 amu), and 188 O (mass = 18.00 amu) has an atomic weight of 16.00 amu. Which isotope predominates?
Answer: 16 8
O. The mass of this isotope (16.00 amu) is closest to the atomic weight given in the periodic table (16.00 amu) and therefore is the most abundant one.
SECTION 2.5
2.40
PERIODIC TABLE
List some of the physical properties of nonmetals.
Answer: They are poor conductors of heat and electricity. As solids they are not lustrous and are brittle.
2.42
Identify two elements that belong to each group. a. alkali metal b. alkaline earth metal
Answer: a. Alkali metals are the elements in Group 1A (excluding hydrogen): lithium, sodium, potassium, rubidium, cesium, and francium b. Alkaline earth metals are the elements in Group 2A: berrylium, magnesium, calcium, strontium, barium, and radium.
2.44
Identify two representative metallic elements that are in the fourth period.
Answer: Potassium, calcium, and gallium.
2 - 54
2.46
Which of the two elements is more metallic?
Answer: The metallic character of elements increases as you move to the left across a period and down a group on the periodic table. a. Al. Al and Si are in the same period but Al is farther to the left than Si. b. Ca. Ca and Mg are in the same group but Ca is farther down the column than Mg.
2.48
Which would you expect to be larger, an atom of francium or an atom of radium?
Answer: Atom of francium. Atomic size decreases going right across the same row. Francium is to the left of radium and therefore is the larger atom.
SECTION 2.6 THE MOLE
2.50
List the two conversion factors that relate number of nitrogen atoms and moles of nitrogen.
Answer: Given that 1 mole of atoms is equal to 6.02 x 1023 atoms, the two conversion factors are:
6.02 x 1023 N atoms 1 mol N and 1 mol N 6.02 x 1023 N atoms 2. 52 a. How many moles of nitrogen is 1.03 x 1018 nitrogen atoms? (Refer to Problem 2.50) b. How many nitrogen atoms is 1.53 mol of nitrogen? Answer: a. To convert number of N atoms to mol N, use the conversion factor that has mol N in the numerator:
2 - 55
1.03 x 1018 N atoms b.
1 mol N = 1.71 x 10-6 mol N 23 6.02 x 10 N atoms
To convert mol N to number of N atoms, use the conversion factor that has number of N atoms in the numerator:
6.02 x 1023 N atoms 1.53 mol N = 9.21 x 1023 N atoms 1 mol N
2.54
List the two conversion factors that relate grams of iron and moles of iron.
Answer: Use the molar mass of Fe which is numerically equal to its atomic weight in amu to set-up the two conversion factors. From the periodic table, 1 mole of Fe weighs 55.85 g.
55.85 g Fe 1 mol Fe and 1 mol Fe 55.85 g Fe
2.56
a. How many moles of iron is 90.7 g of Fe? (Refer to Problem 2.54) b. How many grams of iron is 6.88 x 10-2 mol Fe?
Answer: a. To convert grams of Fe to mol Fe, use the conversion factor that has mol Fe in the numerator:
90.7 g Fe
1 mol Fe = 1.62 mol Fe 55.85 g Fe
b. To convert mol Fe to g Fe, use the conversion factor that has g Fe in the numerator:
6.88 x 10-2 mol Fe
55.85 g Fe = 3.84 g Fe 1 mol Fe
2 - 56
2.58
How many atoms are present in 0.470 mol of uranium?
Answer: Use the equivalence 1 mol U = 6.02 x 1023 U atoms:
0.470 mol U
2.60
6.02 x 1023 U atoms 1 mol U
= 2.83 x 1023 U atoms
4.82 x 106 potassium atoms is how many moles?
Answer: One mole of potassium contains 6.02 x 10 23 atoms. P
4.82 x 106 K atoms x
2.62
P
1 mol K = 8.01 x 10-18 mol K 6.02 x 1023 K atoms
What is the a. atomic weight of magnesium (Mg)? b. molar mass of magnesium? c. mass (in grams) of 5.00 mol of magnesium? d. number of magnesium atoms in 8.85 x 10-5 mol of magnesium? e. mass (in grams) of 3.39 x 1020 magnesium atoms?
Answer: Refer to the periodic table for the atomic weight of magnesium and use this number to calculate the other quantities. a. 24.31 amu b. 24.31 g/mol c. To calculate the mass in grams of 5.00 mol of magnesium, use the equivalence 1 mol Mg = 24.31 g:
5.00 mol Mg
24.31 g 1 mol Mg
= 122 g
d. To calculate the number of magnesium atoms in 8.85 x 10-5 mol Mg, use the equivalence 6.02 x 1023 Mg atoms = 1 mol Mg.
2 - 57
8.85 x 10-5 mol Mg
6.02 x 1023 Mg atoms 1 mol Mg
= 5.33 x 1019 Mg atoms
e. Use the equivalences 6.02 x 1023 Mg atoms = 1 mol Mg and 1 mol Mg = 24.31 g to calculate the mass of 3.39 x 1020 magnesium atoms:
3.39 x 1020 Mg atoms
2.64
1 mol Mg 6.02 x 1023 Mg atoms
24.31 g 1 mol Mg
= 0.0137 g
In 3.45 ng of silicon there are how many of the following? a. grams of silicon? b. moles of silicon? c. silicon atoms?
Answer:
3.45 ng
e.
3.45 x 10-9 g Si
f.
1.23 x 10-10 mol Si
SECTION 2.7
2.66
1 x 10-9 g = 3.45 x 10-9 g Si 1 ng
d.
1 mol Si = 1.23 x 10-10 mol Si 28.09 g Si 6.02 x 1023 Si atoms = 7.40 x 1013 Si atoms 1 mol Si
THE ARRANGEMENT OF ELECTRONS
Neon bulbs give off a reddish-orange light, while sodium bulbs emit a yellow light. In terms of emission spectra, account for this difference.
Answer: Neon emits light in a different part of the spectrum from sodium because Neon’s ground state electron arrangement is different from that of sodium.
2.68
Using a periodic table determine the number of electrons held in energy levels 1-3 of each atom.
2 - 58
Answer: The electrons in an atom are arranged according to energy levels. Each energy level is specified by a number n. The maximum number of electrons that each energy level can hold is 2n2. When assigning the electrons of a given element to these energy levels, we start filling the n=1 level completely first and make our way up until all of the electrons have been assigned to an energy level.
2.70
Total # of electrons
n=1 (max. 2 e-)
n=2 (max. 8 e-)
n=3 (max. 18 e-)
a. B
5
2
3
0
b. C
6
2
4
0
c. Mg
12
2
8
2
Specify the number of valence electrons for each atom. a. Li b. Si c. Al d. Kr
e. P
Answer: The number of valence electrons is the same as the number of the group to which the element belongs in the periodic table.
2.72
a.
Li is in group 1A
1 valence electron
b.
Si is in group 4A
4 valence electrons
c.
Al is in group 3A
3 valence electrons
d.
Kr is in group 8A
8 valence electrons
e.
P is in group 5A
5 valence electrons
For each, give the total number of electrons, the number of valence electrons, and the number of the energy level that holds the valence electrons. a. Br b. Kr c. As d. I
Answer: The total number of electrons is equal to the atomic number, the number of valence electrons is the same as the group number, and the energy level that holds the valence electrons is the same as the row of the periodic table (period) that the element is in.
2 - 59
2.74
a.
Br
35 total electrons
7 valence electrons
level 4
b.
Kr
36 total electrons
8 valence electrons
level 4
c.
As
33 total electrons
5 valence electrons
level 4
d.
I
53 total electrons
7 valence electrons
level 5
For a neutral atom of element 114 a. how many total electrons are present? b. how many valence electrons are present? c. which energy level holds the valence electrons? d. is the valence energy level full?
Answer:
2.76
a.
114 electrons. The total number of electrons in a neutral atom is equal to the total number of protons which is the same as the atomic number of the element.
b.
4 valence electrons. When placed in the periodic table, element 114 would fall under Group 4A.
c.
n = 7. When placed in the periodic table, element 114 would be in the 7th period.
d.
No, it contains only 4 electrons.
Draw the electron dot structure of each atom. a. K b. Se c. Ca d. O
Answer: The electron dot structure shows the number of valence electrons, drawn as dots around the symbol for the element. a.
K
b.
Se
c.
2 - 60
Ca
d.
O
SECTION 2.8
2.78
RADIOACTIVE ISOTOPES
a. What is a beta particle? b. How is a beta particle different than an electron?
Answer: a. A beta particle is a negatively charged particle ejected from the nucleus of a radioisotope. It has the same mass and charge as an electron. b. Unlike an electron, a beta particle is ejected from the nucleus of a radioisotope at up to 90% of the speed of light.
2.80
a. What is a gamma ray? b. How is a gamma ray similar to a radio wave? c. How is a gamma ray different than a radio wave?
Answer:
2.82
a.
A gamma ray is a high energy form of electromagnetic radiation.
b.
Both gamma rays and radio waves are forms of electromagnetic radiation.
c.
A gamma ray carries a lot more energy than a radio wave.
Identify the missing product in each nuclear equation. a. 148 O → ? + 01+ c. 146 C → ? + −01 b. 31 H →
3 2
He
+ ?
Answer: To find the atomic number of the particle produced in the nuclear reaction, the atomic number of the emitted particle is subtracted from the atomic number of the original radioisotope. Look up the atomic number of the product particle in the periodic table to determine the elemental identity of the newly produced isotope. The mass number of the emitted particle is subtracted from the mass number of the original radioisotope to give the new mass number. Remember that for some types of nuclear radiation, including , , + and , charge can be treated as an atomic number. P
2 - 61
P
2.84
a.
14 8
b.
3 1
c.
14 6
O → 147 N + 01+
H → 23 He
+
+ −01
→ 147 N
C
0 −1
35 a. Write the balanced nuclear equation for the loss of an alpha particle from 16 S.
b. Write the balanced nuclear equation for the loss of a beta particle from 27 12 Mg .
Answer: a.
To find the atomic number of the particle produced in the nuclear reaction, the atomic number of the emitted particle is subtracted from the atomic number of the original radioisotope. Look up the atomic number of the product particle in the periodic table to determine the elemental identity of the newly produced isotope. The mass number of the emitted particle is subtracted from the mass number of the original radioisotope to give the new mass number. 35 16
b.
S →
Si
+
4 2
For a beta particle, the charge is treated as an atomic number. 27 12
2.86
31 14
Mg
→ 27 13 Al
+ −01
a. Write the balanced nuclear equation for the loss of a positron from 142 62 Sm. b. Write the balanced nuclear equation for the loss of an alpha particle from 148 62 Sm. c. Write the balanced nuclear equation for the loss of a beta particle from 157 62 Sm.
Answer: To find the atomic number of the particle produced in the nuclear reaction, the atomic number of the emitted particle is subtracted from the atomic number of the original radioisotope. Look up the atomic number of the product particle in the periodic table to determine the elemental identity of the newly produced isotope. The mass number of the emitted particle is subtracted from the mass number of the original radioisotope to give the new mass number. Remember that for some
2 - 62
types of nuclear radiation, including , , + and , charge can be treated as an atomic number. P
a.
The loss of a positron results in the loss of 1 proton but no change in the mass number: 142 62
b.
Sm →
Pm +
0 + 1
β
Sm →
144 60
Nd +
4 2
The loss of a beta particle results in the gain of 1 proton but no change in the mass number: 157 62
2.88
142 61
The loss of an alpha particle results in the loss of 2 neutrons and 2 protons: 148 62
c.
P
Sm →
157 63
Eu +
β
0 −1
Write the balanced nuclear equation for the loss of 1 beta particle and 1 gamma ray from 170 72 Hf.
Answer: The loss of a beta particle results in the gain of 1 proton but no change in the mass number. The loss of a gamma ray has no effect the atomic number and the mass number. 170 72
2.90
Hf →
170 73
Eu +
+
β
0 −1
0 0
a. In the type of nuclear decay called electron capture, an electron from an atom’s electron cloud falls into the nucleus. Iodine-128 undergoes electron capture to produce an x- ray and another product. Write a balanced nuclear reaction for this process. 128 53
I
+
0 -1
e
→ ? +
0 0
x ray
b. Looking only at the new atom formed from the reactions in part a of this question and Problem 2.89, is it possible to tell electron capture from positron emission.
2 - 63
Answer: 128 53
a.
I
+
0 − −1
e
→ 12852Te +
0 0
x ray
b. No. Because 12852Te is formed in each case, it is not possible to tell electron capture from positron emission by looking only at the new atom formed.
2.92
In its radioactive decay series, in eleven steps uranium-235 is converted into lead207. Starting with uranium-235, the first decay in this series releases an alpha particle, the second decay releases a beta particle, and the third releases an alpha particle. Write the balanced nuclear equations for these three reactions.
Answer: 235 92 U
→
231 90Th
+
4 2
231 90Th
→
231 91 Pa
+
0 −1β
231 91 Pa
→
227 89 Ac
+
4 2
SECTION 2.9
2.94
RADIOISOTOPES IN MEDICINE
Because of an increasing number of illnesses causes by bacterial contamination, in 2008 the U.S. Food and Drug Administration approved the irradiation of iceberg lettuce and spinach. In the irradiation process, bacteria is killed when food is exposed to high energy x rays or to gamma rays. Will irradiated spinach be radioactive? Explain.
Answer: No. The irradiated spinach is exposed only to the x-ray and gamma ray emissions but not the radioisotopes used to produce them. The high energy radiation is dissipated after irradiation.
2.96
A nurse is assisting a patient who has just undergone radiation therapy in which a radioisotope was administered intravenously. To ensure her own safety, what information might the nurse want to have regarding the radioisotope?
2 - 64
Answer: To minimize the risks of exposure to radiation, the nurse might want to know what type of radiation was used in the patient’s radiation therapy. Different types of radiation involve different levels of energy and have different half-lives. These factors determine how far the radiation can travel in the air, its ability to penetrate surrounding matter, and how long the radioisotopes will remain hazardous. Knowing these will help the nurse decide how best to protect herself to ensure her safety.
2.98
Exposure to 100 rem of radiation can cause radiation sickness, which includes headaches and an increased risk of infection. a. Convert this value into seiverts. b. Convert this value into rads, assuming that the source of radioactivity is an alpha emitter. c. Convert this value into rads, assuming that the source of radioactivity is a beta emitter.
Answer: a. Convert this value into seiverts. Use the equivalence 1 seivert = 100 rem 1 seivert
100rem × 100 rem = 1 seivert b. Convert this value into rads, assuming that the source of radioactivity is an alpha emitter. Use the equation dose (rad) x QF = dose (rem) where QF is the quality factor that depends on the type of radiation. Table 2.8 in the text gives a QF = 20 for alpha particles. Given 100 rem, we can solve for the value in rads by rearranging the equation above: dose (rad) =
dose (rem) 100 rem = = 5 rad QF 20
2 - 65
c. Convert this value into rads, assuming that the source of radioactivity is a beta emitter. For beta particles, the QF is 1: dose (rad) =
dose (rem) 100 rem = = 100 rad QF 1
2.100 The Becquerel (Bq), an S.I. unit used to measure radiation, is defined as 1 disintegration per second. One Becquerel is how many curies?
Answer: 1 Bq ×
2.102
1 disintegration/second 1 curie × = 2.7 x 10-11 curie 10 1 Bq 3.7 x 10 disintegrations/second
197 80
Hg , a radioisotope used in brain scans, has a half-life of 66 hours. a. Beginning with a 1.00 mg sample of this isotope, how many milligrams of the isotope will remain after 264 hours? b. This isotope decays by emitting 1 neutron ( 01 n ) and 1 gamma ray ( 00 ). Write a balanced nuclear reaction for this decay process.
Answer: a.
0.0625 mg will remain after 264 hours.
After one half-life (66 hours), 1.00 mg will be reduced to 0.500 mg. After a second half-life (66 + 66 = 132 hours), the 0.500 mg will have decayed to 0.250 mg. After a third half-life (66 + 66 + 66 = 198 hours), the 0.250 mg will be reduced to 0.125 mg. After a fourth half-life ( 66 + 66 + 66+ 66 =264 hours), the 0.125 mg will be reduced to 0.0625 mg.
b. To find the atomic number of the isotope produced in the nuclear process, the atomic numbers of the emitted radiation are subtracted from the atomic number of the original radioisotope. Looking this number up on the periodic table gives the atomic symbol of the newly produced radioisotope. The mass numbers of the emitted radiation are subtracted from the mass number of the original radioisotope.
2 - 66
197 80
196 80
Hg
Hg +
1 0
n +
0 0
2.104 Bismuth-213 has a half-life of 47 min. Beginning with a 1.00 μg sample of this radioisotope, a. how many micrograms remain after 3 half-lives? b. how many micrograms remain after 94 min? c. how many minutes will it take for the mass of bismuth-213 to drop to 0.125 μg? Answer: a.
b.
After every one half-life, the amount of radioactive bismuth is cut down by a half: Original amount
1.00 µg
After 1 half-life (47 min)
0.500 µg remains
After 2 half-lives (2 x 47 min = 94 min)
0.250 µg remains
After 3 half-lives (3 x 47 min = 141 min)
0.125 µg remains
94 minutes correspond to 2 half-lives as shown below:
number of half-lives = 94 min
1 half-life = 2 half-lives 47 min
From the table in part a, 0.250 µg remains after 2 half-lives.
c.
141 minutes or 3 half-lives as shown in the table from part a.
2.106 a. Oxygen-15 is a positron emitter with a half-life of about 120 seconds. Write a balanced nuclear equation for the loss of a positron from this radioisotope. b. Oxygen-19 is a beta emitter with a half-life of about 30 seconds. Write a balanced nuclear equation for the loss of a beta particle from this radioisotope. c. If, after a wait of 120 sec, a sample contains 1 μg of oxygen-15 and 1 μg of oxygen-19, how many micrograms of each were initially present? Answer: a.
15 8O
→
15 7N
+
0 + 1β
b.
19 8O
→
19 9F
+
0 −1β
2 - 67
c.
For oxygen-15, 0.5 μg remains because 120 sec is equal to 1 half-life.
For oxygen-19: Original amount
1µg
After 1 half-life (30 sec)
0.5 µg remains
After 2 half-lives (60 sec)
0.25 µg remains
After 3 half-lives (90 sec)
0.125 µg remains
After 4 half-lives (120 sec)
0.06 µg remains
2.108 a. Everyone is exposed to low levels of radiation. What are two sources of this background radiation? b. Name one adverse health effect that is linked to exposure to background radiation. Answer: a. Cosmic rays and radon gas are two sources or background radiation. b. Lung cancer is an adverse health effect linked to radon exposure.
2.110 Alpha emitters are the most harmful form of radioisotope to take internally. Why do you suppose that this is the case? Answer: Alpha particles are the most massive of the particles emitted as nuclear radiation. They are not able to travel far but once ingested, these massive particles carry a lot of energy that can cause a lot of damage to surrounding tissue nearby.
2.112 A compound containing iodine-125 is being tested for use in PET scans of myelin, a substance that surrounds and insulates nerve cells. a. Which type of radiation does iodine-125 release? b. Which type of radiation does a PET scanner detect coming from the body, and how is this radiation formed? c. Would you expect iodine-125 to have a short half-life or a long one? Why? Answer: a. Positron. Iodine-125 to be used for positron emission tomography (PET) sca is positron emitter. 2 - 68
b. Gamma rays are detected by PET scanner. These gamma rays are formed when the positron emitted collides with an electron producing two gamma rays. c. Iodine-125 is expected to have a relatively short-life corresponding to a more rapid decay rate necessary for PET scans.
2.114 What is a gamma knife? Answer: A gamma knife is a beam of gamma rays that can be used to target a specific point to destroy tumors.
HEALTH LINK
STABLE ISOTOPES AND DRUG TESTING
2.116 How are the relative amounts of stable carbon isotopes used to distinguish natural from synthetic testosterone? Answer: A smaller percentage of the carbon-12 isotope can be detected in testosterone naturally produced from substances present in food in a Western diet than that measured in synthetic testosterone produced from a substance found in yams and soy plants.
HEALTH LINK
LEAD
2.118 a. Why should you be concerned about being exposed to lead? b. What steps can you take to reduce your exposure to lead? Answer: a. Lead is a toxic metal that can cause some severe health problems: abdominal cramps, vomiting, convulsions, and brain and nerve damage. b. To reduce exposure to lead, especially by children, avoid contact with material known to contain lead like paint chips and dust from paint used in 2 - 69
homes built before 1980. It has also been found in toys and toy jewelry, furniture, food, and clothing that have been the subject of recall by the US Consumer Product Safety Commission.
BIOCHEMISTRY LINK
BIOLUMINESCENCE
2.120 What are some of the biological functions of bioluminescence? Answer: Some of the biological functions of bioluminescence include attracting mates and prey and, in the case of squids, hiding from prey.
HEALTH LINK
RADIOISOTOPES FOR SALE
2.122 While searching online, you come across someone selling a Gilbert U-238 Atomic Energy Lab science kit. While these lab kits were designed for use by children, it is difficult to imagine any current parent allowing their child to play with it. What has changed since the 1950s when kits were first sold? Answer: Currently, more is known and documented about the range of health risks upon exposure to radioactivity based on dosage received.
HEALTH LINK
CT and MRI Imaging
2.124 In some cases, the images produced by CT and MRI are quite similar. If given a choice, which would you rather have run?
2 - 70
Answer: This is a matter of personal choice. There are potential risks associated with exposure to x rays, so some people may not want a CT scan run. Many MRI devises are quite confining – some may not prefer that experience.
LEARNING GROUP PROBLEMS
2.126 a. What is the atomic symbol of holmium? b. How many protons and neutrons does an atom of homium-151 have? c. Holmium-151 is a radioisotope that decays by emitting either a positron or an alpha particle. Write a balanced nuclear equation for the loss of a positron from holmium-151. d. Write a balanced nuclear equation for the loss of an alpha particle from holmium-151. e. Holmium-151 is formed from a particular radioisotope when it emits a positron. Write a balanced equation for this nuclear reaction. f. Holmium-151 is formed from a particular radioisotope when it emits an alpha particle. Write a balanced equation for this nuclear reaction. g. Holmium-151 has a half-life of 36 seconds. What percent of an initial sample of holmium-151 will remain after 360 seconds have passed? Answer: a. Ho number of neutrons = 151 – 67 = 84
b. number of protons = 67 c. 151 67 Ho →
151 67 Dy
+
0 + 1
d. 151 67 Ho →
147 65Tb
+
4 2
e. 151 68 Er →
151 67
f. 15569Tm →
151 67
β
Ho +
0 + 1
Ho +
4 2
β
1 half-life = 10 half-lives 36 sec 10 half-lives means that the amount of the initial sample will be multiplied by ½ ten times to determine the remaining amount after this time period has passed. g. number of half-lives = 360 sec
10
1 % remaining = 2
x 100% = 0.098 %
2 - 71
Chapter 3 Compounds SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
3.1
Which picture represents molecules and which one represents an ionic compound?
Ionic compounds consist of a crystal lattice or array of alternating cations and anions, like the drawing below.
ionic compound Molecules consist of distinct groups of atoms that are covalently bonded to each other, like the particles pictured below.
molecules
3.3
Give the total number of protons and electrons in each ion. The number of protons is equal to the atomic number of the element (refer to a periodic table). Neutral atoms contain the same number of protons and electrons. Cations contain one fewer electron for each positive charge. Anions contain one more electron for each negative charge. The charge on the ion indicates the number of electrons lost or gained. For positive ions, subtract this charge value from the atomic number to calculate the number of electrons. For negative ions, add this number to the atomic number to calculate the number of electrons. a. K + P
P
19 protons
19 – 1 = 18 electrons 3-1
b. Mg 2+
12 protons
12 – 2 = 10 electrons
c. P 3-
15 protons
15 + 3 = 18 electrons
P
P
3.5
P
P
Give the total number of protons and electrons in each ion. 26 protons
26 – 2 = 24 electrons
b. Fe 3+
26 protons
26 – 3 = 23 electrons
c. Cu +
29 protons
29 – 1 = 28 electrons
a. Fe 2+ P
P
P
P
P
P
d. Cu 2+ P
3.7
29 – 2 = 27 electrons
Give the total number of protons, neutrons, and electrons in each ion. 63
a. 29 Cu
+
29 protons 19
b. 9 F
37
63 – 29 = 34 neutrons
29 – 1 = 28 electrons
19 – 9 = 10 neutrons
9 + 1 = 10 electrons
37 – 17 = 20 neutrons
17 + 1 = 18 electrons
-
9 protons c. 17 Cl
−
17 protons
3.9
29 protons
P
Identify the following monoatomic ions. Use the number of protons (the same as the atomic number) to identify the element. Use the difference between the number of protons and the number of electrons to determine the net charge on the ion. a. 15 protons and 18 total electrons. P3-
15 protons
15 + 3 = 18 total electrons
b. 20 protons and 18 total electrons. Ca2+
20 – 2 = 18 total electrons
20 protons 3-2
c. 7 protons and 8 valence electrons. N3-
7 protons, 5 valence electrons
5 + 3 = 8 valence electrons
d. 25 protons and 22 total electrons. Mn3+
3.11
25 – 3 = 22 total electrons
25 protons
Give the name of each ion. a. F -
fluoride ion
The element is fluorine. As an anion, “ine” is replaced with “ide” b. O2-
oxide ion
The element is oxygen. As an anion, “ygen” is replaced with “ide” c. Cl -
chloride ion
The element is chloride. As an anion, “ine” is replaced with “ide”. d. Br -
bromide ion
The element is bromine. As an anion, “ine” is replaced with “ide”.
3.13
Give a name for each ion. All of these elements are transition metal elements. When naming transition metal ions, the charge of the ion is indicated as a Roman numeral enclosed in parentheses. a. Co2+
cobalt(II) ion
b. Pb2+
lead(II) ion
c. Cr3+ d. Cu+
B
chromium(III) ion
B
copper(I) ion
3-3
3.15
Give the name of each ion. The names of polyatomic ions cannot be predicted in the same way as monatomic ions. Refer to the Table of Common Polyatomic Ions in the text. a. CO 3 2B
carbonate ion
B
b. NO 3 B
nitrate ion
B
c. SO 3 2B
sulfite ion
B
d. CH 3 CO 2 B
3.17
B
B
B
acetate ion
Write the formula of each ion. These are polyatomic ions. Formulas cannot be predicted in the same way as monatomic ions. Refer to the Table of Common Polyatomic Ions in the text.
3.19
a. hydrogen carbonate ion
HCO 3 -
b. nitrite ion
NO 2 -
c. sulfate ion
SO 4 2-
B
B
B
B
B
B
Draw the electron dot structure of a hydrogen cation. A hydrogen cation is a hydrogen atom from which the lone electron has been removed: H+
3.21
How many valence electrons must each atom gain to reach an octet? Add as many electrons to the number of valence electrons each atom already has to determine the number required to acquire 8 valence electrons (an octet). a. O already has 6 valence electrons:
8 – 6 = 2 valence electrons needed
b. F already has 7 valence electrons:
8 – 7 = 1 valence electron needed
c. Se already has 6 valence electrons:
8 – 6 = 2 valence electrons needed
d. Br already has 7 valence electrons:
8 – 7 = 1 valence electron needed
3-4
3.23
Draw the electron dot structure for the ion expected to be formed from each of the atoms in Problem 3.21. Be sure to indicate the sign (positive or negative) and magnitude of the charge on the ion. Electron dot structures show the number of valence electrons that a given atom has. According to the octet rule, an atom will gain, lose, or share electrons to get a complete set of eight valence electrons. Each atom in Problem 3.21 needs to gain valence electrons to reach an octet. The charge is negative and the magnitude is equal to the number of valence electrons required.
2a.
3.25
b.
1-
2c.
1d.
When a nitrogen atom is converted into an ion, a. what is the name of the ion? nitride ion b. how many electrons does nitrogen gain? 3 electrons. Nitrogen has 5 valence electrons and gains 3 valence electrons to reach an octet. c. the ion ends up with what charge? The charge on the nitride ion is 3-.
3.27
Draw the electron dot structure of each atom and of the ion that it is expected to form. a. Na
Na+
Losing one electron leaves a positively charged Na atom with an octet of valence electrons. ..
b. : Cl ..
.. − : Cl : ..
Gaining one electron creates a chloride ion with a 1- charge and an octet of valence electrons. .. : c. : Ar ..
no ion formed
An argon atom already has an octet of electron and therefore does not need to gain or lose any electrons. 3-5
3.29
Name each ionic compound. a. MgO
magnesium oxide
MgO is combination of magnesium ions (Mg2+) and oxide ions (O2-). b. Na 2 SO 4 B
B
B
B
sodium sulfate
Na 2 SO 4 is a combination of sodium ions (Na+) and sulfate ions (SO 4 2-). The number of times that each ion appears in the formula is not specified in the name of the compound. B
B
B
c. CaF 2 B
B
B
B
calcium fluoride
B
CaF 2 is a combination of calcium ions (Ca2+) and fluoride ions (F-). B
B
d. Na 2 S B
sodium sulfide
B
Na 2 S is a combination of sodium ions (Na+) and sulfide ions (S2-). B
3.31
B
Name each ionic compound. When naming ionic compounds that contain transition metal cations, the charge of the cation is indicated in the name as a Roman numeral enclosed in parentheses. (Note: There are some elements that are not transition metals that are able to form different cations.) a.
FeCl 2 B
B
b. CoS
cobalt(II) sulfide
c. CoCl 3 B
d. Al 2 S 3 B
3.33
B
iron(II) chloride
B
cobalt(III) chloride
B
aluminum sulfide (aluminum is not a transition metal)
B
Name each ionic compound. a.
Na2SO4
sodium sulfate
b. Ca(NO3)2
calcium nitrate
c. K3PO4
potassium phosphate
B
3-6
d. NH4OH
3.35
Name each ionic compound. a.
3.37
ammonium hydroxide
B
Mg(HSO4)2
magnesium hydrogen sulfate
b. Na2Cr2O7
sodium dichromate
c. MgSO3
magnesium sulfite
d. Ca(HCO3)2
calcium hydrogen carbonate
Write the formula of each ionic compound. a. calcium hydrogen phosphate CaHPO 4 Calcium ion is Ca2+ and the polyatomic ion hydrogen phosphate is HPO 4 2-. For the compound to be neutral, an equal number of Ca2+ and HPO 4 2- are required. B
B
B
B
B
B
b. copper(II) bromide CuBr 2 2+ Copper(II) ion is Cu and bromide ion is Br-. For the compound to be neutral, twice as many Br- ions as Cu2+ are required. B
B
c. copper(II) sulfate CuSO 4 Copper(II) ion is Cu2+ and sulfate ion is SO 4 2-. For the compound to be neutral, an equal number of Cu2+ and SO 4 2- are required. B
B
B
B
B
B
d. sodium hydrogen sulfate NaHSO 4 Sodium ion is Na+ and the polyatomic ion hydrogen sulfate is HSO 4 -. For the compound to be neutral, an equal number of Na+ and HSO 4 - are required. B
B
B
B
3.39
B
B
Give the name of each ionic compound. a. Li 2 SO 4 (an antidepressant) B
B
B
B
B
lithium sulfate Li+ is the lithium ion and SO 4 2- is the sulfate ion. B
b. Ca(H 2 PO 4 ) 2 B
B
B
B
B
B
B
(used in foods as a mineral supplement)
calcium dihydrogen phosphate Ca2+ is the calcium ion and H 2 PO 4 - is the dihydrogen phosphate ion. B
B
B
B
3-7
c. BaCO 3 B
B
barium carbonate Ba2+ is the barium ion and CO 3 2- is the carbonate ion. B
3.41
B
Write the formula of the ionic compound that forms between each pair. a. magnesium ions and fluoride ions MgF 2 Twice as many (F ) ions as magnesium ions (Mg2+) are required to give a neutral compound. B
B
b. potassium ions and bromide ions KBr + An equal number of potassium ions (K ) and bromide ions (Br-) give a neutral compound. c. potassium ions and sulfide ions K2S Twice as many potassium ions (K+) as sulfide ions (S2-) are required to form a neutral compound. B
B
d. aluminum ions and sulfide ions Al 2 S 3 To obtain a neutral compound, for every two aluminum ions (Al3+), three sulfide ions (S2-) are required. B
3.43
3.45
B
B
B
In addition to sugars, citric acid, and other ingredients, the sports drink called “Powerade” contains potassium phosphate and potassium dihydrogen phosphate. Write the formula of each of these ionic compounds. potassium phosphate
K 3 PO 4
potassium dihydrogen phosphate
KH 2 PO 4
B
B
B
B
B
B
B
B
Predict the number of covalent bonds formed by each nonmetal atom. The number of covalent bonds that a nonmetal atom usually forms is the same as the number of electrons it requires to achieve an octet. a. N 3 covalent bonds A nitrogen atom (group 5A) has five valence electrons and needs 3 more for an octet. b. Cl
1 covalent bond
3-8
A chlorine atom (group 7A) has seven valence electrons and needs 1 more for an octet. c. P 3 covalent bonds A phosphorus atom (group 5A) has five valence electrons and needs 3 more for an octet.
3.47
Draw the electron dot structure of each molecule. In the molecular drawings given, replace each single bond with two dots (:) and each double bond with four dots (::).
H a.
H
H
H C O C H H
H
H
H
H O
H C
C
C C H
H
C
H O
H
C
H
H
O
b.
H H N
H
H
H
H C
C
C C
H
H H N
H
3.49
H
H
H
On the structure shown in Problem 3.48a, point out a pair of bonding electrons.
H O H C C Cl H This is an example of a bonding pair. Other examples include the C-Cl bond, all C-H bonds, and each of the 2 C-O bonds.
3-9
3.51
Draw the line-bond structure of pyruvic acid, a compound formed during the breakdown of sugars by the body.
H O O H C C C O H H Pyruvic acid Substitute a line for each pair of dots that represent a bond.
3.53
Name each binary molecule. When naming binary molecules, the relative number of each atom is specified using a prefix. (mono = 1, di = 2, tri = 3, tetra =4, penta = 5, and hexa = 6) Note that the mono prefix is not used on the first element written when only one atom is present. a. NCl 3 B
b. PCl 3 B
c. PCl 5 B
3.55
nitrogen trichloride
B
phosphorus trichloride
B
phosphorus pentachloride
B
Phosphine (PH 3 ) is a poisonous gas that has the odor of decaying fish. Give another name for this binary molecule. B
B
phosphorus trihydride
3.57
Draw the electron dot structure of the molecule formed when sufficient H atoms are added to give each atom an octet of valence electrons. Atoms will form as many covalent bonds as are required to acquire eight valence electrons. 3 - 10
H H
C H H
a. C H
N
H
H
b. N
H O H
c. O
3.59
Which structure best fits the formula CO2?
is the structure that best fits the formula CO2. This structure represents the most common bonding patterns expected for both C and O.
3.61
Which structure best fits the formula CH2O?
is the structure that best fits the formula CH2O. This structure represents the most common bonding patterns expected for H, C, and O.
3.63
Indicate whether each is an ionic compound or a binary molecule. Ionic compounds contain at least one metallic element. A binary molecule contains only nonmetallic elements. a. BaCl 2 B
b. OCl 2 molecule B
ionic
d. HgO
binary molecule
e. N 2 O 3
B
B
B
3 - 11
B
B
ionic B
binary
c. CS 2 B
3.65
binary molecule
B
B
ionic
B
Name each of the ionic compounds or binary molecules in Problem 3.63. a. BaCl 2
barium chloride
d. HgO
b. OCl 2 trioxide
oxygen dichloride
e. N 2 O 3
c. CS 2
carbon disulfide
f. Cu 2 O
B
B
B
3.67
f. Cu 2 O
B
B
B
B
B
B
B
mercury(II) oxide dinitrogen
B
B
copper(I) oxide
Are all diatomic molecules compounds? Explain No. A diatomic molecule consists of two atoms bonded together. If the atoms are of different elements (such as CO), then the diatomic molecule is a compound. If the atoms are the same (such as N 2 ), then the diatomic molecule is an elemental substance. B
3.69
B
a. What is the formula weight of ammonium hydroxide? The formula weight of a compound can be calculated by adding the atomic weights of all the atoms of each element in the formula. Ammonium hydroxide has the formula NH 4 OH. B
B
Formula weight of NH 4 OH = B
B
(1 x 14.01 amu) + (5 x 1.01 amu) + (1 x 16.00 amu) = 35.06 amu b. What is the mass of 0.950 mol of ammonium hydroxide? Use the molar mass of NH 4 OH (35.06 g/mol) as a conversion factor to calculate the mass of 0.950 mol of NH 4 OH. B
B
B
0.950 mol NH 4OH x
B
35.06 g NH 4OH = 33.3 g NH 4OH 1 mol NH 4OH
c. How many moles of ammonium hydroxide are present in 0.475 g? To calculate the moles of NH4OH present in 0.475 g of NH4OH, use the molar mass as a conversion factor as shown below.
3 - 12
0.475 g NH 4OH x
3.71
1 mol NH 4OH 35.06 g NH 4OH
= 0.0135 mol NH 4OH
a. What is the formula weight of sodium oxide? Sodium oxide has the formula Na 2 O. B
B
Formula weight of Na 2 O = (2 x 23.00 amu) + (1 x 16.00 amu) = 62.00 amu B
B
b. How many oxide ions are present in 0.25 mol of sodium oxide? Use the ratio given by the chemical formula of the compound to determine the amount of a component in a compound. In every 1 mole of Na 2 O, there is 1 mole of O2- ions. B
0.25 mol Na 2O
1 mol O21 mol Na 2O
x
x
B
6.02 x 1023 O2- ions = 1.5 x 1023 O2- ions 1 mol O2-
c. How many sodium ions are present in 0.25 mol of sodium oxide? In every 1 mole of Na 2 O, there are 2 moles of Na+ ions. B
0.25 mol Na 2O
x
B
2 mol Na + 1 mol Na 2O
x
6.02 x 1023 Na + ions = 3.0 x 1023 Na + ions 1 mol Na +
d. How many oxide ions are present in 2.30 g of sodium oxide? First, convert the mass of the sample to moles of the sample. Then, proceed as above.
2.30 g Na 2O
x
1 mol Na 2O 62.00 g Na 2O
x
1 mol O21 mol Na 2O
x
6.02 x 1023 O2- ions = 2.23 x 1022 O2- ions 21 mol O
e. How many sodium ions are present in 2.30 g of sodium oxide?
2.30 g Na 2O
x
1 mol Na 2O 62.00 g Na 2O
x
2 mol Na + 1 mol Na 2O
3 - 13
x
6.02 x 1023 Na + ions = 4.47 x 1022 Na + ions + 1 mol Na
3.73
The food additive potassium sorbate, K(C 6 H 7 O 2 ), is a mold and yeast inhibitor. B
B
B
B
B
B
a. What is the charge on the sorbate ion? 1–. The charge on the potassium ion is 1+. Because the formula indicates a oneto-one ratio between potassium and sorbate, the charge on the sorbate ion must be 1-. b. How many C atoms are present in 0.0150 g of potassium sorbate? First, calculate the formula weight in grams of the potassium sorbate compound. Then use this to convert the mass of the sample to moles of the sample. Use the ratio given by the formula and Avogadro’s number to determine the number of C atoms as shown below:
0.0150 g KC6 H 7 O 2
1 mol KC6 H7 O2 150.22 g KC6 H 7 O 2
x
x
3.75
x
6 mol C 1 mol KC6 H 7 O 2
6.02 x 1023 C atoms = 3.61 x 1020 C atoms 1 mol C
a. What is the molecular weight of CCl 4 ? B
B
molecular weight of CCl 4 = (1 x 12.01 amu) + (4 x 35.45 amu) = 153.81 amu B
B
b. What is the mass of 61.3 mol of CCl 4 ? B
61.3 mol CCl4 x
153.81 g CCl4 1 mol CCl4
B
= 9.43 x 103 g CCl 4
c. How many moles of CCl 4 are present in 0.465 g of CCl 4 ? B
0.465 g CCl4 x
B
B
1 mol CCl4 153.81 g CCl4
B
= 3.02 x 10-3 mol CCl4
d. How many molecules of CCl 4 are present in 5.50 x 10-3 g of CCl 4 ? B
5.50 x 10-3 g CCl4
1 mol CCl4 x 153.81 g CCl4
B
B
x
B
6.02 x 1023 CCl4 molecules = 2.15 x 1019 CCl4 molecules 1 mol CCl4
3 - 14
3.77
One tablet of a particular analgesic contains 250 mg acetaminophen (C 8 H 9 NO 2 ). How many acetaminophen molecules are contained in the tablet? B
B
B
B
B
B
First convert milligrams into grams, then proceed as in previous problems.
250 mg x
3.79
1 x 10-3 g 1 mol 6.02 x 1023 molecules x x = 1.0 x 1021 molecules 1 mg 151.17 g 1 mol
Isovaleric acid (C 5 H 10 O 2 ) is the molecule responsible for foot odor. To smell this compound, it must be present in the air at a minimum of 250 parts per billion. This means that a 0.50 L volume of air would hold 0.00013 g of isovaleric acid. B
B
B
B
B
B
a. To how many moles of isovaleric acid does this correspond?
0.00013 g C5H10O2 x
1 mol C5H10O2 102.13 g C5H10O2
= 1.3 x 10-6 mol C5H10O2
b. To how many molecules of isovaleric acid does this correspond?
1.3 x 10-6 mol C5 H10O2
3.81
x
6.02 x 1023 molecules C5H10O 2 1 mol C5 H10O2
= 7.8 x 1017 molecules
Explain how ionophores act as antibiotics. Ionophores are molecules that transport ions across cell membranes. Ionophores can be used to kill bacteria (as antibiotics) by forcing the passage of ions across the cell membrane of the bacteria and disrupting the balance of ions which hampers the normal biological functions of the bacterial cell.
3.83
Name the compounds that, in addition to NaCl, are present in Hawaiian sea salt and Black Indian salt. Hawaiian sea salt also contains iron(III) oxide (Fe2O3). Black Indian salt also contains small amounts of sodium sulfate (Na2SO4), iron(II) sulfide FeS, and hydrogen sulfide (H2S).
3.85
Sometimes a filling can make a tooth sensitive to temperature changes. Given that amalgam is made from metals and that composites are made from nonmetals, which type of filling would you expect is most likely to cause thermal sensitivity? The amalgam would likely result in increased thermal sensitivity because metals conduct heat more readily than the nonmetals in composites. 3 - 15
3.87
Viagra, one of the drugs used to treat erectile dysfunction, works by enhancing the effects of nitric oxide produced by the body. Men who take nitroglycerin for angina are advised against taking Viagra. Why? Nitroglycerin reduces constriction of arteries (angina) by producing nitric oxide which causes blood vessels to dilate. Viagra enhances the effects of nitric oxide. Taking Viagra with nitroglycerin may result in a precipitous, and possibly lifethreatening, drop in blood pressure when blood vessel dilation is prolonged or severe.
3.89
Some ionic compounds can form hydrates, in which water molecules are incorporated into their crystal structures. One example is iron(II) nitrate hexahydrate, Fe(NO 3 ) 2 ∙6H 2 O. The dot in the formula indicates that there are 6 water molecules associated with each formula unit of iron(II) nitrate. Because water molecules are neutral, they contribute no charges to the compound. B
B
B
B
B
B
a. What is the charge on the cation in this compound? 2+. The charge on each of the nitrate ions is 1-. Two nitrate ions to make a neutral compound means that the charge on the iron must be 2+. b. Give an alternate name for the cation. ferrous ion c. What is the charge on the anion in this compound? 1-. The anion is the polyatomic ion NO 3 -. B
B
d. Does the compound contain any ionic bonds? Explain. Yes. The bonds between 1 Fe2+ ion and 2 NO 3 - ions are ionic. B
B
e. Does the compound contain any covalent bonds? Explain. Yes. The bonds between 1 N and 3 O atoms in NO 3 - and the bonds between 2 H atoms and 1 O atom in each H 2 O molecule are all covalent. B
B
B
B
f. Give an alternate name for water, using the rules for naming binary compounds. dihydrogen monoxide
3 - 16
g. What is the formula weight of iron(II) nitrate hexahydrate? Formula weight of Fe(NO 3 ) 2 ·6H 2 O = B
B
B
B
B
B
(1 x 55.85 amu) + (2 x 14.01 amu) + (12 x 16.00 amu) + (12 x 1.01 amu) = 287.99 amu h. How many moles is 4.93 g of this compound?
4.93 g x
1 mol 287.99 g
= 1.71 x 10-2 mol
i. How many grams is 0.639 mol of this compound? 0.639 mol x
287.99 g = 184 g 1 mol
j. How many water molecules are in 8.58 x 10-6 mol of this compound?
8.58 x 10-6 mol Fe ( NO3 )2 ·6H 2 O x
x
6 mol H 2 O 1 mol Fe ( NO3 )2 ·6H 2 O
6.02 x 1023 molecules H 2 O 1 mol H 2 O
= 3.10 x 1019 molecules H 2 O
k. How many iron(II) ions are in 43.8 g of this compound? 1 mol Fe ( NO3 )2 ·6H 2O 43.8 g Fe ( NO3 )2 ·6H 2O x 287.99 g Fe ( NO3 )2 ·6H 2 O
x
1 mol Fe2+ 1 mol Fe ( NO3 )2 ·6H 2 O
x
6.02 x 1023 Fe2+ ions 1 mol Fe2+
3 - 17
= 9.16 x 1022 Fe 2+ ions
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 3.2
Atoms of two nonmetal elements can combine to form the three compounds shown. Are these compounds ionic or are they binary molecules?
Answer: These compounds are binary molecules because they are formed by the combination of atoms of two nonmetal elements. Binary ionic compounds form from the combination of a metal and a nonmetal element.
SECTION 3.1
IONS
3.4 Give the total number of protons and electrons in each ion. + a. Li b. Al3+ c. S2Answer: Neutral atoms contain an equal number of electrons and protons, which is equal to the atomic number of the element. When ions are formed, the number of protons stays the same but the number of electrons changes. Thus, ions have unequal numbers of protons and electrons. Positive ions are created by the loss of 1 or more electrons and negative ions are created by the gain of 1 or more electrons. Recall that electrons have a negative charge. The charge on the ion indicates the number of electrons lost or gained. For positive ions, subtract this charge value from the atomic number to calculate the number of electrons. For negative ions, add this number to the atomic number to calculate the number of electrons. a. Li+, atomic number = 3
3 protons
3-1
b. Al3+, atomic number = 13
13 protons
13 - 3 = 10 electrons
c. S2-, atomic number = 16
16 protons
16 + 2 = 18 electrons
3.6
Give the total number of protons and electrons in each ion. a. Cr2+
b. Cr3+
c. Sn2+ 3 - 18
d. Sn4+
= 2 electrons
Answer: Neutral atoms contain an equal number of electrons and protons, which is equal to the atomic number of the element. When ions are formed, the number of protons stays the same but the number of electrons changes. Thus, ions have unequal numbers of protons and electrons. Positive ions are created by the loss of 1 or more electrons and negative ions are created by the gain of 1 or more electrons. Recall that electrons have a negative charge. The charge on the ion indicates the number of electrons lost or gained. For positive ions, subtract this charge value from the atomic number to calculate the number of electrons. For negative ions, add this number to the atomic number to calculate the number of electrons. a. Cr2+, atomic number = 24
24 protons
24 - 2 = 22 electrons
b. Cr3+, atomic number = 24
24 protons
24 - 3 = 21 electrons
c. Sn2+, atomic number = 50
50 protons
50 - 2 = 48 electrons
c. Sn4+, atomic number = 50
50 protons
50 - 4 = 46 electrons
3.8
Give the total number of protons, neutrons, and electrons in each ion.
23
16
a. 11Na+
35
b. 8O2-
c. 17Cl-
Answer: In an atomic notation, the top number gives the mass number and the bottom number is the atomic number. + a. 23 11 Na
11 protons
23 – 11 = 12 neutrons
11 – 1 = 10 electrons
b. 168 O 2 −
8 protons
16 – 8 = 8 neutrons
8 + 2 = 10 electrons
17 protons
35 – 17 = 18 neutrons
17 + 1 = 18 electrons
c.
35 17
3.10
Cl
−
Identify the monatomic ion that has a. 3 protons and 2 total electrons b. 35 protons and 8 valence electrons c. 47 protons and 46 total electrons d. 28 protons and 26 total electrons
3 - 19
Answer: Use the number of protons (the same as the atomic number) to identify the element. Use the difference between the number of protons and the number of electrons to determine the net charge on the ion. a. 3 protons and 2 total electrons Li1+ The element is Li (atomic number = 3). A neutral Li atom has 3 protons and 3 electrons. Because the ion has 1 fewer electron than the neutral atom, it has a net 1+ charge. b. 35 protons and 8 valence electrons Br1The element is Br (atomic number = 35). A neutral atom of Br has 7 valence electrons. Because the ion has 1 more valence electron than the neutral, it has a net charge of 1-. c. 47 protons and 46 total electrons Ag1+ The element is Ag (atomic number = 47). A neutral atom of Ag has 47 electrons. Because the ion has 1 fewer electron than the neutral, it has a net charge of 1+. d. 28 protons and 26 total electrons Ni2+ The element is Ni (atomic number = 28). A neutral atom of Ni has 28 electrons. Because the ion has 2 fewer electrons than the neutral, it has a charge of 2+.
3.12 Give the name of each ion. a. Sr2+ c. S2b. Cl- d. Se2-
Answer: Monatomic positive ions are named by using the name of the element and adding the word “ion”. Monatomic negative ions are named by using the name of the element but changing the ending to “-ide” and adding the word ion. With some transition metals that are able to form several different cations, the charge of the ion is indicated as a Roman numeral enclosed in parentheses. a. strontium ion b. chloride ion 3 - 20
c. sulfide ion d. selenide ion
3.14
Give a name for each ion. a. Co3+
b. Pb4+
c. Cr2+
d. Cu2+
Answer: Monatomic positive ions are named by using the name of the element and adding the word “ion”. With some transition metals that are able to form several different cations, the charge of the ion is indicated as a Roman numeral enclosed in parentheses. a. cobalt(III) ion b. lead(IV) ion c. chromium(II) ion d. copper(II) ion
3.16 Give the name of each ion. a. H3O+ c. HPO42b. OH d. H2PO4-
Answer: Table 3.2 lists the names of some common polyatomic ions. a. hydronium ion b. hydroxide ion c. hydrogen phosphate ion d. dihydrogen phosphate ion
3.18 Write the formula of each ion. a. hydrogen sulfate ion b. phosphate ion c. dichromate ion Answer: These are all polyatomic ions. Check Table 3.2 for the formulas. a. HSO4b. PO433 - 21
c. Cr2O72-
3.20
Hydride ion has a 1- charge. Draw the electron dot structure of a hydride ion.
Answer: A hydride ion is a negatively charged hydrogen atom that has gained one electron. -
H
hydride ion
SECTION 3.2
3.22
THE OCTET RULE
How many valence electrons must each atom lose to reach an octet? a. Li
b. Al
c. Cs
d. Sr
Answer: Each atom is expected to lose all of its valence electrons to leave an octet (or duet) number of electrons in the resulting ion. a. Li b. Al c. Cs d. Sr
1 valence electron 3 valence electrons 1 valence electrons 2 valence electrons
3.24 Draw the electron dot structure for the ion expected to be formed from each of the atoms in Problem 3.22. Be sure to indicate the sign (positive or negative) and magnitude of the charge on the ion. Answer: In the formation of cations, all of the valence electrons are lost from each atom of the elements listed. Therefore, the electron dot structure carries no electrons or “dots”.
a. Li
3.26
+
b. Al
3+
c. Cs
+
2+
d. Sr
When a potassium atom is converted into an ion, a. what is the name of the ion? 3 - 22
b. how many electrons does potassium lose? c. the ion ends up with what charge?
Answer: a. Potassium ion. b. 1 electron. A potassium atom loses its 1 valence electron to achieve an octet. c. 1+. Losing one electron produces an ion with one more proton than electrons.
3.28
Draw the electron dot structure of each atom and of the ion that it is expected to form. a. K b. I c. Ca
Answer: Electron dot structure is a way to represent valence electrons. For monoatomic anions, all valence electrons are shown, usually an octet of them. For monoatomic cations, no valence electrons are shown – the original valence electrons were removed when the ion formed.
. K
a.
K
b.
I
+
. I
c. Ca
SECTION 3.3
2+
Ca
IONIC COMPOUNDS
3.30 Name each ionic compound. a. Li2O c. Al2O3 b. BaCl2 d. MgSe
3 - 23
Answer: To name ionic compounds, first name the cation then the anion. For cations that have more than one type of charge (e.g. transition metals), the charge is written as a Roman numeral enclosed in parentheses. a. lithium oxide b. barium chloride c. aluminum oxide d. magnesium selenide
3.32
Name each ionic compound. a. Fe2O3
b. CrBr3
c. CrBr2
d. PbI2
Answer: To name ionic compounds, first name the cation then the anion. For cations that have more than one type of charge (e.g. transition metals), the charge is written as a Roman numeral enclosed in parentheses. To determine which cation is present (for those that can exist in several different ones), use the charge of the anion and the number of anions required to balance the charge on the cation. See the first one below as an example. a. iron(III) oxide (Oxide has a 2- charge. Because it requires 2 Fe ions to balance a total of 6- charges (3 x 2- = 6-), each Fe ion must have a charge of 3+ (2 x 3+ = 6+) b. chromium(III) bromide c. chromium(II) bromide d. lead(II) iodide
3.34
Name each ionic compound. a. K2SO3
b. Sr(NO2)2
c. K2HPO4
Answer: a. potassium sulfite b. strontium nitrite c. potassium hydrogen phosphate 3 - 24
d. (NH4)2CO3
d. ammonium carbonate
3.36
Name each ionic compound. a. Al2(CO3)3
b. NH4CN
c. BaSO3
d. CH3CO2Na
Answer: a. aluminum carbonate b. ammonium cyanide c. barium sulfite d. sodium acetate
3.38 Write the formula of each ionic compound. a. iron(III) hydroxide b. ammonium bromide c. copper(I) sulfate d. tin(II) oxide Answer: a. Fe(OH)3
Iron(III) is Fe3+ and hydroxide is OH-. For the compound to be neutral, 3 times as many OH- ions as Fe3+ ions are required.
b. NH4Br
Ammonium is NH4+ and bromide is Br-. For the compound to be neutral, equal numbers of NH4+ and Br- are required.
c. Cu2SO4
Copper(I) is Cu+ and sulfate is SO42-. For the compound to be neutral, twice as many Cu+ as SO42- are needed.
d. SnO
Tin(II) is Sn2+ and oxide is O2-. For the compound to be neutral, equal numbers of Sn2+ and O2- are needed.
3.40 Give the name of each ionic compound. a. AlPO4(used in some dental cements) b. MgHPO4 (a laxative) c. SrBr2 (an anticonvulsant)
3 - 25
Answer: a. aluminum phosphate b. magnesium hydrogen phosphate c. strontium bromide
3.42
Write the formula of the ionic compound that forms between a. copper(I) ions and chloride ions b. copper(II) ions and hydrogen sulfate ions c. lithium ions and iodide ions d. cobalt(III) ions and phosphate ions
Answer: Each ionic compound should be neutral: the total number of positive charges should equal the total number of negative charges. Use as many of each cation and each anion as needed to balance out the opposite charges.
3.44
a. CuCl
A chloride ion has a 1- charge so only one is needed to balance the charge on the copper(I).
b. Cu(HSO4)2
A hydrogen sulfate ion has a 1- charge; therefore 2 HSO4ions are needed to balance the charge on the copper(II) ion.
c. LiI
Each lithium ion carries a 1+ charge and an iodide ion carries a 1- charge and, therefore, equal numbers of ions are needed.
d.CoPO4
Cobalt(III) has a 3+ charge and a phosphate ion has a 3charge. Equal numbers of the two ions are needed.
To fight tooth decay, some toothpastes contain stannous fluoride, also known as tin(II) fluoride. Write the formula of this compound.
Answer: SnF2 The formula for tin(II) fluoride is SnF2. The charge on the tin ion is 2+ and the charge on a fluoride ion is 1–. Therefore, we need 2 fluoride ions to balance the 2+ charge on the tin ion.
3 - 26
SECTION 3.4
3.46 a. C b. O c. Br
COVALENT BONDS
Predict the number of covalent bonds formed by each nonmetal atom.
Answer: The number of covalent bonds formed by a nonmetal atom can generally be predicted from the number of additional valence electrons the atom needs to gain by sharing in order to achieve an octet of electrons. a. C has 4 valence electrons. Therefore, it is expected to form 4 covalent bonds. b. O has 6 valence electrons. Therefore, it is predicted to form 2 covalent bonds. c. Br has 7 valence electrons. Therefore, it would likely form 1 covalent bond.
3.48
Draw the electron dot structure of each molecule.
a.
H O H C C Cl H b.
H O H C C C C O H H Answer: In a line-bond structure, each line represents a pair of valence electrons shared as a covalent bond. Show each line-bond as a pair of electron dots in the electron dot structure. Retain the valence electrons that are not shared (nonbonding electrons) as electron dots. a.
H O H C C Cl H 3 - 27
b.
H O H C
C C C O H H
3.50 On the structure shown in Problem 3.48a, point out a pair of nonbonding electrons. Answer:
H O H C C Cl H
3.52
Draw the line-bond structure of lactic acid, a compound formed during anaerobic exercise. H
H O H C C
O C O H
H H Lactic Acid Answer: The line-bond structure for lactic acid is given below:
H H O H C C
O C O H
H H
3 - 28
SECTION 3.5
MOLECULES
3.54 Name each binary molecule. a. CS2 b. N2O3 c. NF3
Answer: To name binary molecules, name the elements in the order that they appear in the molecular formula. For the second element, change the ending of the name to ide. Use a prefix to indicate the number of atoms of each element. When there is only one atom of the first element in the formula, the prefix mono is not used. a. carbon disulfide b. dinitrogen trioxide c. nitrogen trifluoride
3.56 Dentists use nitrous oxide (N2O) as an anesthetic. Give another name for this binary molecule. Answer: dinitrogen monoxide
3.58 Draw the electron dot structure of the molecule formed when sufficient H atoms are added to give each atom an octet of valence electrons. Answer: First draw the electron dot structure for the atom and then use H· to finish pairing up the dots in the structure.
a.
.. . .F. .. .. .P. .
b.
.. ..
becomes H .. F ..
..
becomes H .. P .. H .. H
3 - 29
c.
3.60
.. . . Br .. .
.. ..
becomes H .. Br ..
Which structure best fits the formula C2H3Cl?
Answer:
is the structure that best fits the formula C2H3Cl. This structure represents the most common bonding patterns expected for H, C, and Cl based on the octet rule.
3.62
Which structure best fits the formula C2H5Cl?
3 - 30
Answer:
is the structure that best fits the formula C2H5Cl. This structure represents the most common bonding patterns expected for H, C, and Cl based on the octet rule.
3.64 Indicate whether each is an ionic compound or a binary molecule. a. SF2 b. MgF2 c. SnCl2 d. PF5 e. NO2 f. SnO2 Answer: Ionic compounds contain at least one metallic element. A binary molecule contains only nonmetallic elements. a. SF2 b. MgF2 c. SnCl2 d. PF5 e. NO2 f. SnO2
3.66
binary molecule ionic ionic binary molecule binary molecule ionic
Name each of the ionic compounds or binary molecules in Problem 3.64.
Answer: To name binary molecules, name the elements in the order that they appear in the molecular formula. For the second element, change the ending of the name to ide. Use a prefix to indicate the number of atoms of each element. When there is only one atom of the first element in the formula, the prefix mono is not used. To name ionic compounds, name the cation first then name the anion. When naming ionic compounds that contain transition metal cations, the charge of the cation is indicated in the name as a Roman numeral enclosed in parentheses. a. SF2
sulfur difluoride
3 - 31
b. MgF2
magnesium fluoride
c. SnCl2
tin(II) chloride
d. PF5
phosphorus pentafluoride
e. NO2
nitrogen dioxide
f. SnO2
tin(IV) oxide
3.68
Do any elements exist as binary molecules? Explain
Answer: No, from the definition, binary molecules contain atoms of two different elements.
SECTION 3.6 MASS
3.70
FORMULA WEIGHT, MOLECULAR WEIGHT, AND MOLAR
a. What is the formula weight of Li2CO3 ? b. What is the mass of 1.33 x 10-4 mol of Li2CO3 ? c. How many moles of CO32- ions are present in 73.5 g of Li2CO3?
Answer: a. What is the formula weight of Li2CO3 ? Count the number of atoms of each element in the formula, multiply that number by the atomic weight for the element as given on the periodic table (rounded to 2 decimal places), and then add the total weight contributed by each element to get the formula weight.
Li C O
# of atoms 2 1 3
total weight contributed atomic weight by each element 6.94 13.88 12.01 12.01 16.00 48.00 Answer: 73.89 amu
b. What is the mass of 1.33 x 10-4 mol of Li2CO3 ? Use the formula weight expressed as a molar as a conversion factor.
3 - 32
1.33 x 10-4 mol Li 2CO3 x
73.89 g Li 2CO3 1 mol Li 2CO3
= 9.83 x 10 -3 g Li 2CO3
c. How many moles of CO32- ions are present in 73.5 g of Li2CO3? Use the molar mass to convert grams of Li2CO3 to moles. Then use the fact that 1 mol Li2CO3 contains 1 mol CO32- as a conversion factor to calculate moles of CO32-.
73.5 g Li 2CO3 x
3.72
1 mol Li 2CO3 1 mol CO32− x 73.89 g Li 2CO3 1 mol Li 2CO3
= 0.995 mol CO32−
a. What is the formula weight of magnesium iodide? b. How many magnesium ions are present in 7.5 x 10-6 mol of magnesium iodide? c. How many iodide ions are present in 7.5 x 10-6 mol of magnesium iodide? d. How many magnesium ions are present in 4.5 mg of magnesium iodide? e. How many iodide ions are present in 4.5 mg of magnesium iodide?
Answer: a. What is the formula weight of magnesium iodide? First, write the correct formula, MgI2. Next, count the number of atoms of each element in the formula, multiply that number by the atomic weight for the element as given on the periodic table (keeping one decimal place is sufficient), and then add the total weight contributed by each element together to get the formula weight.
Mg I
# of atoms 1 2
atomic weight total weight contributed by each element 24.31 24.31 126.90 253.8 Answer: 278.11 amu
b. How many magnesium ions are present in 7.5 x 10-6 mol of magnesium iodide?
7.5 x 10-6 mol MgI 2 x
1 mol Mg 2+ 6.02 x 1023 Mg 2+ x 1 mol MgI 2 1 mol Mg 2+
= 4.5 x 1018 Mg 2+
c. How many iodide ions are present in 7.5 x 10-6 mol of magnesium iodide?
7.5 x 10-6 mol MgI2 x
2 mol I− 6.02 x 1023 I− x 1 mol MgI2 1 mol I−
3 - 33
= 9.0 x 1018 I −
d. How many magnesium ions are present in 4.5 mg of magnesium iodide? Convert milligrams of MgI2 to grams of MgI2 and then use the molar mass to calculate moles of MgI2. Then use the ratio that one mole of MgI2 contains one mole of Mg2+ ions (6.02 x 1023 ions).
4.5 mg MgI2 x
1 mol MgI2 10-3 g 1 mol Mg 2+ x x 1 mg 278.11 g MgI2 1 mol MgI2
x
6.02 x 1023 Mg 2+ 1 mol Mg 2+
= 9.7 x 1018 Mg 2+
e. How many iodide ions are present in 4.5 mg of magnesium iodide?
4.5 mg MgI 2 x
1 mol MgI2 10-3 g 2 mol I− x x 1 mg 278.11 g MgI 2 1 mol MgI 2
x
6.02 x 1023 I − 1 mol I−
= 1.9 x 1019 Mg 2+
3.74 The food additive sodium benzoate, C6H5CO2Na, extends the shelf life of food by preventing bacterial growth. a. What is the charge on benzoate ion? b. How many C atoms are present in 0.250 g of sodium benzoate? Answer: a. 1- . The benzoate ion has a charge because Na has a 1+ charge. b. First, convert 0.250 g to moles then use the molar ratio, 7 mol C for every 1 mole of C6H5CO2Na to calculate the moles of C atoms. Then use the equivalence 1 mol of C atoms = 6.02 x 1023 C atoms.
0.250 g C6 H5CO 2 Na x
1 mol C6 H5CO 2 Na 7 mol C x 144.11 g C6 H 5CO 2 Na 1 mol C6 H5CO 2 Na
x
3.76
6.02 x 1023 C = 7.31 x 1021 C atoms 1 mol C
a. What is the molecular weight of aspirin (C9H8O4)? 3 - 34
b. What is the mass of 0.00225 mol of aspirin? c. How many moles of aspirin are present in 500 mg of aspirin? d. How many molecules of aspirin are present in 1.00 g of aspirin? Answer: a. molecular weight of C9H8O4 = (9 x 12.01 amu) + (8 x 1.01 amu) + (4 x 16.00 amu) = 180.17 amu b. Use the molecular weight of aspirin to determine the mass of aspirin from the given moles.
0.00225 mol C9 H8O4
x
180.17 g C9 H8O4 1 mol C9 H8O4
= 0.405 g C9 H8O4
c. First, convert mg to g. Then, use the molar mass of aspirin to calculate the moles of aspirin from the given mass.
500 mg C9 H8O4
x
10-3 g 1 mg
x
1 mol C9 H8O4 180.17 g C9 H8O4
= 3 x 10-3 mol C9 H8O4
d. Follow the sequence of steps as in c. above and then use Avogadro’s number to determine the number of molecules.
1.00 g C9 H8O4
1 mol C9 H8O4 x 180.17 g C9 H8O4
6.02 x 1023 C9 H8O4 molecules x 1 mol C9 H8O4 = 3.34 x 1021 molecules of C9 H8O4
3.78
A vitamin tablet contains 500 mg of vitamin C (C6H8O6). How many vitamin C molecules are contained in the tablet?
Answer: The molar mass for a vitamin C molecule is 176.14 g. 500 mg C6 H8O6 x
x
1 mol C6 H8O6 10-3 g x 1 mg 176.14 g C6 H8O6 6.02 x 1023 C6 H8O6 molecules 1 mol C6 H8O6
3 - 35
= 2 x 1021 molecules of C6H8O6
3.80 2-Isobutyl-3-methoxypyriazine, C9H14N2O, has an “earthy” smell and is one of the compounds responsible for the aroma of coffee. a. If 1.00 L of brewed coffee contains 8.3 x 10-2 mg of 2-isobutyl-3menthoxypyriazine, how many molecules of this compound are present? b. How many molecules of the compound would be present in 1 cup of this coffee? Answer: a. First, calculate the molecular weight of 2-isobutyl-3-methoxypyriazine or C9H14N2O molecular weight of C9H14N2O = (9 x 12.01 amu) + (14 x 1.01 amu) + (2 x 14.01 amu) + (1 x 16.00 amu) = 166.25 amu Use this to calculate the number of moles and number of molecules after converting mg to g:
8.3 x 10-2 mg x
1 mol 10-3 g 6.02 x 1023 molecules x x 1 mg 166.25 g 1 mol = 3.0 x 1017 molecules of C9 H14 N 2 O
b. Use the relationship that 1.00 L contains 8.3 x 10-2 mg of the compound to calculate the mg in 1 cup:
1 cup x
1 gallon 3.785 L 8.3 x 10-2 mg x x = 0.02 mg of compound 16 cups 1 gallon 1.00 L
Then, use the same conversion set-up as part a to calculate the number of molecules:
2 x 10-2 mg x
1 mol 10-3 g 6.02 x 1023 molecules x x 1 mg 166.0 g 1 mol = 7 x 1016 molecules of C9 H14 N 2O
3 - 36
BIOCHEMISTRY LINK
3.82
IONOPHORES AND BIOLOGICAL ION TRANSPORT
K+ attaches more strongly to valinomycin than Na+ because of the size of the cavity (binding site) in the center of the compound. Is the cavity too large for Na+ or is it too small?
Answer: Since K+ ions are larger than Na+ ions, the cavity must be too large for the Na+ ion.
HEALTH LINK
3.84
PASS THE SALT, PLEASE
What is iodized salt and why is it sold?
Answer: Iodized salt is a form of table salt that has compounds containing iodine. It is sold to provide people with the recommended daily intake of iodine.
HEALTH LINK
3.86
DENTAL FILLINGS
Although amalgam is sometimes referred to as a compound, it does not fit the definition of this term. Explain.
Answer: A compound always has the same proportion of the same elements. In amalgam, the amounts of mercury, silver, tin, copper, and zinc vary depending on how it is prepared, so there is no one amalgam compound.
HEALTH LINK
NITRIC OXIDE
3.88 Why does the risk of heart attack and stroke increase when a person is given a blood transfusion?
3 - 37
Answer: Stored blood, used in blood transfusions, typically lack the normal concentration of nitric oxide. Nitric oxide is important in regulating blood pressure and for controlling muscles that dilate arteries and blood vessels.
LEARNING GROUP PROBLEMS
3.90
a. Name the hydrate Cu3(PO4)2·3H2O. (Refer to Problem 3.89.) b. What is the charge on the cation in this compound? c. Give a different name for the cation than used in your answer to part a of this question. d. What is the charge on the anion in this compound? e. What is the formula weight of Cu3(PO4)2·3H2O? f. How many moles is 1.22 g of Cu3(PO4)2·3H2O? g. How many grams is 5.73 mol of Cu3(PO4)2·3H2O? h. How many water molecules are in 6.39 x 10-2 mol of Cu3(PO4)2·3H2O? i. How many cupric ions are in 0.227 g of Cu3(PO4)2·3H2O?
Answer: a. Name the hydrate Cu3(PO4)2·3H2O. Copper(II) phosphate trihydrate b. What is the charge on the cation in this compound? 2+ c. Give a different name for the cation than used in your answer to part a of this question. Cupric ion. d. What is the charge on the anion in this compound? 3e. What is the formula weight of Cu3(PO4)2·3H2O? formula weight of Cu3(PO4)2·3H2O = (3 x 63.55 amu) + (2 x 30.97 amu) + (11 x 16.00 amu) + (6 x 1.01 amu) = 434.65 amu 3 - 38
f. How many moles is 1.22 g of Cu3(PO4)2·3H2O?
1.22 g Cu 3 (PO4 ) 2 • 3H 2O x
1 mol Cu 3 (PO4 )2 • 3H 2O = 2.81 x 10-3 mol Cu 3 (PO4 )2 • 3H2O 434.65 g Cu 3 (PO4 ) 2 • 3H 2O
g. How many grams is 5.73 mol of Cu3(PO4)2·3H2O?
5.73 mol Cu 3 (PO4 )2 • 3H 2O x
434.65 g Cu 3 (PO4 )2 • 3H 2O = 2.49 x 103 g Cu 3 (PO4 )2 • 3H 2O 1 mol Cu 3 (PO4 )2 • 3H 2O
h. How many water molecules are in 6.39 x 10-2 mol of Cu3(PO4)2·3H2O? 3 mol H 2O 6.02 x 1023 molecules H 2O 6.39 x 10 mol Cu 3 (PO4 )2 • 3H 2O x x 1 mol Cu 3 (PO4 )2 • 3H 2O 1 mol H 2O -2
= 1.15 x 1023 molecules H 2O
i. How many phosphate ions are in 0.227 g of Cu3(PO4)2·3H2O?
0.227 g Cu 3 (PO 4 ) 2 • 3H 2O x
1 mol Cu 3 (PO 4 ) 2 • 3H 2O 2 mol PO 43x 434.65 g Cu 3 (PO 4 ) 2 • 3H 2O 1 mol Cu 3 (PO 4 ) 2 • 3H 2O
x
3 - 39
6.02 x 1023 PO43- ions = 6.29 x 1020 PO43- ions 31 mol mol PO4
Chapter 4 An Introduction to Organic Compounds SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
4.1
One of the alkenes is nonpolar and the other is polar. Which is which?
The one on the left, with the two F atoms on the same side, is polar. The one on the right, with the two F atoms opposite each other, is nonpolar.
4.3
4.5
Indicate the number of covalent bonds that each nonmetal atom is expected to form. a. C
4 bonds. Carbon has 4 valence electrons and therefore needs to share 4 more electrons to complete its octet number of electrons.
b. O
2 bonds. Oxygen has 6 valence electrons and therefore needs to share 2 more electrons to complete its octet number of electrons.
c. P
3 bonds. Phosphorus has 5 valence electrons and therefore needs to share 3 more electrons to complete its octet number of electrons.
d. Br
1 bond. Bromine has 7 valence electrons and therefore needs to share only 1 more electron to complete its octet number of electrons.
Draw the structural formula of the molecule that contains a. one oxygen atom and two hydrogen atoms. b. one hydrogen atom and one iodine atom. c. one nitrogen atom and three hydrogen atoms.
4-1
Use the procedure for drawing line-bond structures given in the text, keeping in mind that a hydrogen atom should have a pair of electrons and all other atoms in the molecule should have an octet of electrons. a. one oxygen atom and two hydrogen atoms.
H O H b. one hydrogen atom and one iodine atom.
H
I
c. one nitrogen atom and three hydrogen atoms.
H
IN
H
H
4.7
Draw the line-bond structure of each molecule. Use the procedure for drawing line-bond structures given in the text, keeping in mind that each atom in the diatomic molecule should have an octet of electrons. a. F2
F
F
b. O2
O O
4.9
Draw the line-bond structure of each molecule. a. CH2S S H
C
H
F
N
F
b. NF3
F
4-2
4.11
Draw each molecule Using line-bond structures: H H a.
a. C2H6
b.
H H
H C C H
c. H C C H
b C2H4
H H H C C H
H H
H C C H
b.
H H 4.13
H H c. CC 2H2 C H H
c. H C C H
Draw each polyatomic ion. Each atom, except for hydrogen, should have an octet of valence electrons.
a. OH–
b. NH4+
c. CN
4.15
-
–
Draw each of the following. Each atom should have an octet of valence electrons.
O a. SO3
S
O
O 2-
O S O b. SO32-
O
4-3
4.17
Draw two different molecules that have the formula C2H6O.
H
H
H C O C H H
4.19
H H H C C O H
H
H H
Write a condensed structural formula for each molecule.
H
a.
H
H
H
H
C
C
C
C
H
H
H
H
H
CH3CH2CH2CH3
H H H
H H
b.
4.21
C C N H H
CH3CH2NH2
Draw a skeletal structure for each molecule in Problem 4.19.
H
a.
H
H
H
H
C
C
C
C
H
H
H
H
H
H H H
b.
4.23
4.25
C C N H
N
H H
H
H
H
Which atom in each pair is more electronegative? a. N and O
O. The electronegativity of O is 3.5 and N is 3.0.
b. O and S
O. The electronegativity of O is 3.5 and S is 2.5.
c. Na and Cl
Cl. The electronegativity of Cl is 3.0 and Na is 0.9.
Calculate the electronegativity difference between each pair of atoms. 4-4
4.27
a. N and O
3.5 – 3.0 = 0.5
b. C and P
2.5 – 2.1 = 0.4
c. H and O
3.5 – 2.1 = 1.4
d. Li and Cl
3.0 – 1.0 = 2.0
For each pair of atoms in Problem 4.25, identify the bond that would form between them as nonpolar covalent, polar covalent, or ionic. Use the following ranges of electronegativity differences to identify the type of bond:
4.29
Less than 0.5 Between 0.5 – 1.9 Greater than 1.9
nonpolar covalent polar covalent ionic
a. N and O
3.5 – 3.0 = 0.5
polar covalent
b. C and P
2.5 – 2.1 = 0.4
nonpolar covalent
c. H and O
3.5 – 2.1 = 1.4
polar covalent
d. Li and Cl
3.0 – 1.0 = 2.0
ionic
Identify each bond as being polar covalent or nonpolar covalent. Use the following ranges of electronegativity differences to identify the type of bond:
4.31
Less than 0.5 Between 0.5 – 1.9 Greater than 1.9
nonpolar covalent polar covalent ionic
a. C - H
2.5 – 2.1 = 0.4
nonpolar covalent
b. C - O
3.5 – 2.5 = 1.0
polar covalent
c. F - F
4.0 – 4.0 = 0.0
nonpolar covalent
Label any polar covalent bond(s) in each molecule or ion.
4-5
Use the following ranges of electronegativity differences to identify the type of bond: Less than 0.5 Between 0.5 – 1.9 Greater than 1.9
nonpolar covalent polar covalent ionic
a. None. The electronegativity difference between H and S is 0.4. b. Each C-O bond is polar covalent because the electronegativity difference between the atoms is 1.0. polar covalent bond 2-
O
C
O
O polar covalent bond c. None. The electronegativity difference between N and Cl is 0.
4.33
Specify the shape around each specified atom in Problem 4.31. a. The S atom in H2S b. The C atom in CO32c. The N atom in NCl3 a. Bent. The S atom has four groups of electrons around it: two single bonds to H atoms and two pairs of nonbonding electrons. b. Trigonal planar. The C atom has three groups of electrons around it: 2 single bonds and 1 double bond to three O atoms. c. Pyramidal. The N atom has four groups of electrons around it: 3 covalent bonds to Cl atoms and a nonbonding pair of electrons.
4.35
Identify the shape around each specified atom in Problem 4.9. a. The C atom in CH2S. Trigonal planar. The C atom has three groups of electrons around it: two single bonds to H atoms and a double bond to a S atom. 4-6
b. The N atom in NF3. Pyramidal. The N atom has four groups of electrons around it: 3 single bonds to 3 F atoms and a nonbonding pair.
4.37
Which of the molecules are polar?
H
H a. H C H
F
b.
F
C
F F
c.
H
F
C
F
F
a and b. To be polar the molecule must meet two basic criteria: it must have polar bonds and these bonds cannot be equally distributed in the molecule (the molecule must be unsymmetrical). Since all of the molecules have polar bonds, the only criterion to check is bond distribution. The shape around the C atom in each molecule is tetrahedral. Only in molecule c do the polar bonds cancel one another and make the molecule nonpolar.
4.39
Which, if any, of the molecules in Problem 4.9 are polar? a. CH2S Not polar. This molecule has no polar covalent bonds. b. NF3 Polar. The N-F bonds are polar covalent and the shape of the molecule leads to an unsymmetrical distribution of electrons.
4.41
True or false? A bent molecular shape always has a bond angle near 110°. False. Depending on the number of groups of electrons around an atom, a bent molecular shape can have a bond angle near 104.5° or near 115° .
4.43
Do hydrogen bonds form between formaldehyde molecules (Table 4.2)? No. In order to have hydrogen bonding, there must be at least one hydrogen atom attached to a nitrogen, oxygen, or fluorine atom. In examining the formaldehyde 4-7
structure, H2C=O, we see that the hydrogen atoms are not attached to the oxygen, so hydrogen bonds do not form between formaldehyde molecules.
4.45
Which pairs of molecules can form a hydrogen bond with one another? a.
CH 3 CH3 and
CH3 CH3
No. There are no hydrogen atoms bonded to an oxygen, nitrogen, or fluorine atom in either of the molecules. Therefore, no hydrogen bonds can form. O b.
CH C 3
O H
and
CH C 3
H
No. There are no hydrogen atoms attached to N, O, or F.
c.
CH3 CH 2 OH
and
CH3 CH2 OH
Yes. The two molecules can form a hydrogen bond between them because there is a hydrogen atom in each molecule attached to an oxygen atom. The hydrogen atom from one molecule can form a hydrogen bond with the oxygen atom from the other molecule.
O d. CH3 C
O OH
and CH3 C
OH
Yes. The two molecules can form a hydrogen bond between them because there is a hydrogen atom in each molecule attached to an oxygen atom. The hydrogen atom from one molecule can form a hydrogen bond with the oxygen atom from the other molecule.
4.47
Which of the molecules in Problem 4.45 can form a hydrogen bond with a water molecule? b, c, and d. A hydrogen bond is the interaction of a nitrogen, oxygen, or fluorine atom with a hydrogen atom that is covalently bonded to a different nitrogen, oxygen, or fluorine atom. This gives two criteria for hydrogen bonding to occur; first, at least one of the molecules must have a hydrogen atom that is attached to a nitrogen, oxygen, or fluorine atom, second, the other molecule must have a nitrogen, oxygen, or fluorine atom in its structure. Water meets either criterion, the 4-8
molecule in part b meets the second criterion, and the molecules in parts c and d meet either criterion.
4.49
Which share the stronger London force interactions, two CH3CH2CH2CH2CH3 molecules or two CH3CH2CH2CH2CH2CH2CH2CH3 molecules? Two CH3CH2CH2CH2CH2CH2CH2CH3 molecules. The larger the molecule, the more electrons there are that can participate in a London force interaction. Therefore, the bigger molecules will experience stronger London force interactions.
4.51
In ancient Greece, Socrates and others were executed by being forced to drink hemlock. The major poisonous ingredient in hemlock is an organic molecule called coniine.
N H Coniine Show two ways that a coniine molecule can form a hydrogen bond to a water molecule. The coniine molecule contains an N-H bond that is capable of hydrogen-bonding with a water molecule in two ways as shown below:
hydrogen-bonding
hydrogen-bonding
N H
O H
H O H
4.53
H
a. This molecule is largely responsible for the odor and flavor of bananas. Which organic family is present? 4-9
Ester. b. This molecule is used to give movie theater popcorn a buttery flavor. Which organic family is present?
Ketone.
4.55
a. Spearmint flavoring is due to carvone. Which organic families are present in this molecule?
carvone
Alkene and ketone. b. Which organic families are present in adrenaline?
adrenaline Phenol, alcohol, and amine.
4 - 10
4.57
a. Which organic families are present in the local anesthetic procaine?
procaine
Amine, aromatic, and ester. b. Which organic families are present in the local anesthetic lidocaine?
lidocaine
Aromatic, amide, amine.
4.59
a. Acetaminophen is the active ingredient in Tylenol and many other non-aspirin pain relievers. Name the organic families that are present in acetaminophen.
acetaminophen Phenol and amide. b. Some creams used to relieve muscle pain contain the molecule below. Which organic families are present?
4 - 11
Phenol and ester.
4.61
Some cough syrups contain the expectorant guiafenesin. Name the organic families that are present in this molecule.
Guiafenesin Ether, aromatic, and alcohol.
4.63
Name the organic families present in Vitamin E.
Vitamin E
Phenol and ether.
4.65
a. Draw two alcohols with the formula C4H10O.
b. Draw two ethers with the formula C4H10O
4.67
a. Draw two thiols with the formula C5H12S. 4 - 12
b. Draw two sulfides with the formula C5H12S.
c. Draw two disulfides with the formula C5H12S2.
4.69
a. Draw two carboxylic acids with the formula C6H12O2.
b. Draw two esters with the formula C6H12O2.
c. Draw two aldehydes with the formula C7H14O.
d. Draw two ketones with the formula C7H14O.
4 - 13
4.71
Draw two molecules with the formula C5H10O2 that a. are carboxylic acids
b. are esters
c. have both aldehyde and alcohol groups
d. have both ketone and ether groups
4.73
Label all polar covalent bonds in the molecules using the symbols + and -.
−
+ a.
b. There are no polar covalent bonds in this molecule because no two bonded atoms have electronegativity differences greater than 0.5. 4 - 14
− − + c.
4.75
+
+
Using the symbols + and -, label all polar covalent bonds for − + +
a. the molecule in problem 4.54b
−
− +
carvone
b. the molecule in problem 4.55a
4.77
Using the symbols + and -, label all polar covalent bonds for
− +
a. the molecule in problem 4.60b
almond flavor + +
b. the molecule in problem 4.61
4.79
− +
+
− +
− +
Guiafenesin
Predict the shape and approximate bond angle around 4 - 15
+
− +
a. the C=O carbon atom.
Trigonal planar with an approximate bond angle of 120°. The C atom has three groups of electrons around it: three covalent bonds to another C atom, an O atom, and a H atom.
b. the S atom.
Bent with an approximate bond angle of 104°. The S atom has four groups of electrons around it: two covalent bonds and two nonbonding pairs of electrons.
c. the C atom on the left.
Tetrahedral with an approximate bond angle of 109°. The C atom has four groups of electrons around it constituting four covalent bonds.
4.81
Will hydrogen bonds form between each pair of molecules? a. Yes. The two molecules can form a hydrogen bond between them because there is a hydrogen atom in each molecule attached to an oxygen atom. The hydrogen atom from one molecule can form a hydrogen bond with the oxygen atom from the other molecule.
4 - 16
b. Yes. The hydrogen atom from the molecule with an –OH group can form a weak hydrogen bond with the oxygen atom on the C=O group.
c. No. There are no hydrogen atoms bonded to oxygen, nitrogen, or fluorine in either of the molecules.
4.83
Will hydrogen bonds form between each pair of molecules? a. No. There are no hydrogen atoms bonded to oxygen, nitrogen, or fluorine in either of the molecules. b. No. There are no hydrogen atoms bonded to oxygen, nitrogen, or fluorine in either of the molecules.
c. Yes. The two molecules can form a hydrogen bond between them because there is a hydrogen atom in each molecule attached to an oxygen atom. The hydrogen atom from one molecule can form a hydrogen bond with the oxygen atom from the other molecule.
4.85
Will hydrogen bonds form between each pair of molecules? a.
4 - 17
Yes. The two molecules can form a hydrogen bond between them because there is a hydrogen atom in each molecule attached to a nitrogen atom. The hydrogen atom from one molecule can form a hydrogen bond with the nitrogen atom from the other molecule.
b. No. There are no hydrogen atoms bonded to oxygen, nitrogen, or fluorine in these molecules.
c. No. There are no hydrogen atoms bonded to oxygen, nitrogen, or fluorine in these molecules.
4.87
Of the pairs of molecules in Problem 4.83, which interact primarily through London forces? The molecules in 4.83a and 4.83b are nonpolar molecules that interact primarily through London forces:
4.89
Of the pairs of molecules in Problem 4.81, which can interact through dipoledipole forces, but not hydrogen bonds? The molecules in 4.81c can interact through dipole-dipole forces but not hydrogen bonds. These molecules are polar but have no hydrogen atoms bonded to oxygen, nitrogen, or fluorine.
4.91
Are covalent bonds broken when PrPC is converted into PrPSC? Explain.
4 - 18
No, covalent bonds are not broken when PrP is converted into PrPSC. The conversion occurs through rotations around single bonds which change the overall shape of the PrP molecule.
4.93
During ripening, bananas produce small amounts of ethylene. When bananas are shipped, why should they not be shipped in containers? The ethylene produced by bananas will build up inside the closed containers causing high levels of ethylene. This may cause the bananas to ripen rapidly and, possibly, spoil.
4.95
What properties are important for molecules used as sunscreens? The molecules should absorb UV-A and UV-B radiation and should not be toxic when applied to the skin.
4.97
a. To which organic family does the molecule belong? CH3CH2CH2CH2CH2OH The molecule belongs to the alcohol family. b. Give the molecular formula of the molecule in part a. C5H12O c. Can two of the molecules in part a interact through London forces? Yes. All organic molecules can interact through London forces. d. Can two of the molecules in part a interact through dipole-dipole forces? Yes. The alcohol molecule is polar and all polar molecules can interact through dipole-dipole forces. e. Can two of the molecules in part a interact through hydrogen bonds? Yes. The molecule contains a H atom attached to an oxygen atom. f. Draw a molecule that has the same molecular formula as the molecule in part a, but belongs to a different family of organic compounds. CH3CH2CH2CH2 – O – CH3
4 - 19
The molecule drawn above has the same molecular formula (C5H12O) as the alcohol in part a but belongs to the ether family. Other possible answers include placing the O atom between any other two C atoms. g. Can two of the molecules in part f interact through London forces? Yes. All organic molecules can interact through London forces. h. Can two of the molecules in part f interact through dipole-dipole forces? Yes. The molecule is polar and therefore can interact through dipole-dipole forces. i. Can two of the molecules in part f interact through hydrogen bonds? No. The molecule in part f does not have a H atom bonded to an O atom.
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 4.2
One of the aromatic compounds is nonpolar and the other is polar. Which is which?
Answer: The one on the left is polar; there is a non-symmetric distribution of the F atoms around the ring. The one on the right, with the three F atoms symmetrically distributed around the ring, is nonpolar.
4 - 20
4.4
Indicate the number of covalent bonds that each nonmetal atom is expected to form. a. Se
b. H
c. I
d. N
Answer: a. 2 bonds. Selenium has 6 valence electrons and therefore needs to share 2 more electrons to complete its octet number of electrons. b. 1 bond. Hydrogen has 1 valence electron and needs to share just 1 more electron to complete its valence shell (hydrogen does not follow the octet rule). c. 1 bond. Iodine has 7 valence electrons and therefore needs to share just 1 more electron to complete its octet number of electrons. d. 3 bonds. Nitrogen has 5 valence electrons and therefore needs to share 3 more electrons to complete its octet number of electrons.
4.6
Draw the structural formula of the molecule that contains the following atoms. a. one selenium atom and two fluorine atoms b. one phosphorus atom and three hydrogen atoms c. one hydrogen atom and one bromine atom
Answer: Use the procedure for drawing line-bond structures given in the text, keeping in mind that a hydrogen atom should have a pair of electrons and all other atoms in the molecule should have an octet of electrons. F
Se
F
H
IP
H
a.
H
b. H
Br
c.
4 - 21
4.8
Draw the line-bond structure of each molecule. a. I2 b. N2
Answer: Use the procedure for drawing line-bond structures given in the text, keeping in mind that each atom in the diatomic molecule should have an octet of electrons.
I
I
N
N
a. b.
4.10
Draw the line-bond structure of each molecule. a. OCl2 b. CS2
Answer:
a.
Cl
O Cl
b.
4.12
Draw each molecule a. C3H8
b. C3H6
c. C3H4
4 - 22
Answer:
H H H
H H H a. H C C C H
b. H C C C H H
H H H
H c.
H
H C C C H or
H
H C C C H
H
4.14
Draw each polyatomic. Each atom, except for hydrogen, should have an octet of valence electrons. a. PO43b. HPO42c. H2PO4-
Answer:
O
a.
P
O
O
b.
P
O
O H
O
4.16
2-
O
3-
O
H O c.
O
P O H
Draw each of the following. Each atom should have an octet of valence electrons. a. PO33b. SH-
Answer: a.
O P O
3b.
- S
O 4 - 23
H
O
4.18
Draw three different molecules that have the formula C3H9N.
Answer:
H
H H
H
C C N
C H
H H H
H
H H H
H
H
C C C
N H
H H H
H
H
H C
N
C H
H
H H C
H
H
4.20
Write a condensed structural formula for each molecule.
H a.
H H
H
C C
C O C H
H H
H
H H
b.
H
H H
H C
C C
H
H H
F
Answer: a. CH3CH2CH2OCH3
4.22
b. CH3CH2CH2F
Draw a skeletal structure for each molecule in Problem 4.20.
Answer:
a.
b.
4 - 24
4.24
Which atom in each pair is more electronegative? a. C and F
b. I and F
c. H and F
Answer: a. F. The electronegativity of F is 4.0 and C is 2.5. b. F. The electronegativity of F is 4.0 and I is 2.5. c. F. The electronegativity of F is 4.0 and H is 2.1.
4.26
Calculate the electronegativity difference between each pair of atoms. a. Mg and N
b. N and Br
c. C and F
d. C and C
Answer:
4.28
a. Mg and N
3.0 – 1.2 = 1.8
b. N and Br
3.0 – 2.8 = 0.2
c. C and F
4.0 – 2.5 = 1.5
d. C and C
2.5 – 2.5 = 0.0
For each pair of atoms in Problem 4.26, identify the bond that would form between them as nonpolar covalent, polar covalent, or ionic.
Answer: Use the following ranges of electronegativity differences to identify the type of bond: Less than 0.5 Between 0.5 – 1.9 Greater than 1.9
nonpolar covalent polar covalent ionic
a. Mg and N
3.0 – 1.2 = 1.8
polar covalent
b. N and Br
3.0 – 2.8 = 0.2
nonpolar covalent
4 - 25
4.30
c. C and F
4.0 – 2.5 = 1.5
polar covalent
d. C and C
2.5 – 2.5 = 0.0
nonpolar covalent
Identify each bond as being polar covalent or nonpolar covalent. a. C-N
b. C-C
c. C-Cl
Answer: Use the following ranges of electronegativity differences to identify the type of bond:
4.32
Less than 0.5 Between 0.5 – 1.9 Greater than 1.9
nonpolar covalent polar covalent ionic
a. C-N
3.0 – 2.5 = 0.5
polar covalent
b. C-C
2.5 – 2.5 = 0.0
nonpolar covalent
c. C-Cl
3.0 – 2.5 = 0.5
polar covalent
Label any polar covalent bond(s) in each molecule or ion.
H H a.
+
H C N H H H O
b. H C H
c. [ O Cl O ] Answer: Use the following ranges of electronegativity differences to identify the type of bond:
4 - 26
Less than 0.5 Between 0.5 – 1.9 Greater than 1.9
a.
nonpolar covalent polar covalent ionic 3.0 – 2.5 = 0.5 2.5 – 2.1 = 0.4 3.0 – 2.1 = 0.9
C-N C-H N-H
polar covalent nonpolar covalent polar covalent
polar covalent +
H H H
C N
H
H H b.
all N-H bonds are polar covalent 3.5 – 2.5 = 1.0 2.5 – 2.1 = 0.4
C-O C-H
polar covalent nonpolar covalent
polar covalent O H
c.
O-Cl
C
H
3.5 – 3.0 = 0.5
polar covalent
both O-Cl bonds are polar covalent
[
4.34
O Cl O ]
Specify the shape around each specified atom in Problem 4.32. a. The N atom in CH3NH3+. b. The C atom in CH2O. c. The Cl atom in ClO2 -.
Answer: a. Tetrahedral. The N atom has four groups of electrons around it: three single bonds to H atoms and one single bond to a C atom. 4 - 27
b. Trigonal planar. The C atom has three groups of electrons around it: two single bonds to two H atoms and 1 double bond one O atom. c. Bent. The Cl atom has four groups of electrons around it: two covalent bonds to two O atoms and two nonbonding pairs of electrons.
4.36
Identify the shape around each specified atom in Problem 4.10. a. The O atom in OCl2. b. The C atom in CS2.
Answer:
Cl
O Cl
a. Bent. The O atom has four groups of electrons around it: two single bonds to Cl atoms and two nonbonding pairs of electrons.
b. Linear. The C atom has two groups of electrons around it: two double bonds to two S atoms.
4.38
Which of the molecules are polar?
H a.
H C O H H
b.
Cl
C C Cl
H O H c.
H C C C H H
H
Answer: In a polar molecule, one side has a partial positive charge and the other side has a partial negative charge. Determine what kinds of bonds exist in the molecule. If 4 - 28
there are no polar covalent bonds or all of the polar covalent bonds oppose and cancel each other, then the molecule is nonpolar. If the polar covalent bonds do not oppose and cancel each other, then the molecule is polar. a. Polar molecule. Polar covalent bonds exist and they do not cancel each other out. The shape around the oxygen atom is bent. b. Nonpolar molecule. Polar covalent bonds exist but they cancel each other out because of the linear shape of the molecule. c. Polar molecule. Polar covalent bonds exist and they do not cancel each other out. The shape around the C=O is trigonal planar.
4.40
Which, if any, of the molecules in Problem 4.10 are polar?
Answer: In a polar molecule, one side has a partial positive charge and the other side has a partial negative charge. Determine what kinds of bonds exist in the molecule. If there are no polar covalent bonds or all of the polar covalent bonds oppose and cancel each other, then the molecule is nonpolar. If the polar covalent bonds do not oppose and cancel each other, then the molecule is polar.
Cl
O Cl
a. Polar molecule. Polar covalent bonds exist and they do not cancel each other out because the molecule has a bent shape.
b. Nonpolar molecule. No polar covalent bonds exist.
4.42
True or false? An atom surrounded by four groups of electrons always has a tetrahedral arrangement of atoms.
Answer: False. An atom surrounded by 4 groups of electrons has a tetrahedral arrangement of atoms only if all 4 groups each form a single bond to 4 attached atoms.
4 - 29
4.44
Do hydrogen bonds form between CH3OH molecules?
Answer: Hydrogen bonds form between a nitrogen, oxygen, or fluorine atom and a hydrogen attached to a different nitrogen, oxygen, or fluorine atom. Methanol has a hydrogen attached to an oxygen atom. Therefore, two methanol molecules can form a hydrogen bond between each other.
4.46
Which pairs of molecules can form a hydrogen bond with one another?
and
a.
CH3CH2SH
b. CH3CH2SH
and
c. CH OH
CH3OH
3
d. CH3OCH3
and
and
CH3OCH3
Answer: a. No. There are no hydrogen atoms bonded to oxygen, nitrogen, or fluorine in either of the molecules. Therefore, no hydrogen bonds can form. b. No. There no hydrogen atoms attached to either N, O, or F. c. Yes. The two molecules can form a hydrogen bond between them because there is a hydrogen atom in each molecule attached to an oxygen atom. The hydrogen atom from one molecule can form a hydrogen bond with the oxygen atom from the other molecule. d. No. There are no hydrogen atoms attached to either O, N, or F.
4.48
Which of the molecules in Problem 4.46 can form a hydrogen bond with a water molecule?
Answer: The molecules in 4.46 part c. can hydrogen bond with a water molecule. 4 - 30
The oxygen atom in the molecules in 4.64 part d. can form a weak hydrogen bond with a hydrogen atom in water.
4.50
Which share the stronger London force interactions, two CH3CH2CH2CH2CH3 molecules or two CH3CH(CH3)CH2CH3 molecules?
Answer: Two CH3CH2CH2CH2CH3 molecules share the stronger London force. London force interactions depend on surface area. As the surface area decreases, so do the London force interactions. Each of the molecules in this question has the formula C5H12, but branching in CH3CH(CH3)CH2CH3 reduces surface area and London force interactions.
4.52
Can two coniine molecule (Problem 4.51) interact by London Forces? Explain.
N H Coniine Answer: Yes. London forces can exist between any two molecules because this interaction is due only to the presence of electrons that can develop temporary dipoles.
4.54
a. The flavor of lemons comes from limonene. Which organic family is present in limonene?
limonene
b. The molecule below is largely responsible for the odor and flavor of pears. Which organic family is present?
4 - 31
Answer: a. Alkene. b. Ester.
4.56
a. Which organic families are present in the stimulant called amphetamine?
amphetamine b. The molecule below is the active ingredient in quick tanning lotions. Which organic families are present?
Answer: a. Aromatic and amine. b. Alcohol and ketone.
4.58
a. Name the organic families that are present in aspirin.
aspirin
4 - 32
b. Name the organic families that are present in adrenaline. O CH
C
OH
CH3 CH3 CH
CH2
CH3
ibuprofen
Answer: a. Aromatic, carboxylic acid, and ester. b. Aromatic and carboxylic acid.
4.60
a. Which organic families are present in the neurotransmitter dopamine?
dopamine b. The molecule below has an almond flavor. Which organic families are present?
almond flavor Answer: a. Phenol and amine. b. Aromatic and aldehyde.
4 - 33
4.62
Name the organic families present in the antibiotic amoxicillin.
amoxicillin
Answer: Phenol, amine, amide, sulfide, and carboxylic acid.
4.64
Name the organic families present in Vitamin K.
Vitamin K
Answer: Aromatic, alkene, and ketone.
4.66
a. Draw two alkenes with the formula C4H8. b. Draw two alkynes with the formula C6H10.
Answer:
a.
b. 4 - 34
4.68
a. Draw two phenols with the formula C8H10O. b. Draw two alcohols that are aromatic and have the formula C8H10O.
Answer:
a.
b.
4.70
a. Draw two amines with the formula C3H9N. b. Draw two amides with the formula C3H7NO.
Answer:
a. O
b. 4.72
CH3 CH2 C
O NH2
CH3 C
NH
CH3
Draw two molecules with the formula C9H10O2 that. a. have both aromatic and carboxylic acid groups. b. have both aromatic and ester groups. c. have both phenol and aldehyde groups. d. have both phenol and ketone groups.
Answer: O CH2 CH2 C
OH
a. 4 - 35
O CH2 C
O
CH3
b. O CH2 CH2 C
HO
H
c. O CH2 C
HO
CH3
d. 4.74
Label all polar covalent bonds in the molecules, using the symbols + and -. O
H
a. H
H
H
C
C
H
H
H
O
C
C
H O
C
H
H
OH
b. O
c.
H
H
Answer:
-
O H H
C
+
H
-
H
C
O
C
+
+ +
H
O
a.
H
H
H
b.
4 - 36
-
+ H
H
O
C
C
+ O
+
H
c.
4.76
H
+ H
Using the symbols + and -, label all polar covalent bonds for. a. the molecule in problem 4.58a b. the molecule in problem 4.59a
Answer:
O
+
+ O
H
+
O
+
O
aspirin
a.
-
+ H
- + O
+
N H
b.
4.78
acetaminophen
O C
+
CH3
+
For this problem, you may need to partially redraw the molecules in line-bond notation. Using the symbols + and -, label all polar covalent bonds for a. the molecule in problem 4.55b b. the molecule in problem 4.56b
4 - 37
Answer:
+
-
H
O
+
-
H
O
-
+
O
H
+ CH
+
-
+
CH2 N
CH3
H
+
adrenaline
a.
+ - +
H
O
b.
4.80
O
CH2 C
+
+
- +
CH2 O
H
Predict the shape and approximate bond angle around b. the O atom.
b. the C atom that holds the –OH group.
b. the O atom on the right.
4 - 38
Answer: a. Bent with an approximate bond angle of 104°. The O atom has four groups of electrons around it: two covalent bonds to two C atoms and two nonbonding pairs of electrons. c. Tetrahedral with an approximate bond angle of 109°. The C atom has four groups of electrons around it: 4 covalent bonds to 2 C atoms, a H atom, and the O atom. b. Bent with an approximate bond angle of 104°. The O atom has four groups of electrons around it: two covalent bonds and two nonbonding pairs of electrons.
4.82
Will hydrogen bonds form between each pair of molecules?
d.
e.
f. Answer: a. Yes. The two molecules can form a hydrogen bond between them because there is a hydrogen atom in each molecule attached to an oxygen atom. The hydrogen atom from one molecule can form a hydrogen bond with the oxygen atom from the other molecule. b. No. There are no hydrogen atoms bonded to oxygen, nitrogen, or fluorine in either of the molecules. Therefore, no hydrogen bonds can form.
4 - 39
c. Yes. The hydrogen atom from the molecule with an –OH group can form a weak hydrogen bond with the oxygen atom between the two C atoms of the other molecule.
4.84
Which of the molecules in Problem 4.83 can form a hydrogen bond with water?
Answer: The molecules with the –OH group can form a hydrogen bond with water.
4.86
Which of the molecules in Problem 4.85 can form a hydrogen bond with water?
Answer: The molecule with the –NH2 group can form a hydrogen bond with water.
The molecule with the C=O group can form a weak hydrogen bond with water.
4.88
Of the pairs of molecules in Problem 4.85, which interact primarily through London forces?
Answer: The molecules in Problem 4.85 c interact primarily through London forces because they are nonpolar.
4.90
Of the pairs of molecules in Problem 4.82, which can interact through dipoledipole forces, but not hydrogen bonds?
4 - 40
Answer: The molecules in Problem 4.82 b interact primarily through dipole-dipole forces because they are both polar but no hydrogen bonding capability.
HEALTH LINK
4.92
PRION DISEASES
Suggest a way to reduce the spread of mad cow disease between cattle.
Answer: Cattle feed should not contain parts from sheep, cattle, or other animals.
BIOCHEMISTRY LINK
4.94
ETHYLENE, A PLANT HORMONE
Ethylene gas can be produced from petroleum and then stored in metal cylinders. When food processors want to ripen bananas, they expose the fruit to this manufactured ethylene. Would you expect plants to react differently to ethylene made from petroleum than to ethylene that they have produced themselves?
Answer: No. As long as it is only the ethylene gas that is responsible for the natural ripening process, then the same reaction should occur with ethylene made from petroleum.
HEALTH LINK
4.96
SUNSCREENS
When applied to the skin of mice, forskolin, a compound present in an Asian plant, was shown to increase the production of melanin. Which, do you suppose, were the results of this scientific study? a. The mice tanned more quickly. b. The mice did not sunburn as easily. c. The mice were less susceptible to skin cancer
4 - 41
Answers: a., b., and c. Probably all of these were found since the presence of melanin is associated with each result.
4.98
a. To which organic family does the molecule belong?
b. Give the molecular formula of the molecule in part a. c. Can two of the molecules in part a interact through London forces? d. Can two of the molecules in part a interact through dipole-dipole forces? e. Can two of the molecules in part a interact through hydrogen bonds? f. Draw a molecule that has the same molecular formula as the molecule in part a but is an ester. g. Can two of the molecules in part f interact through London forces? h. Can two of the molecules in part f interact through dipole-dipole forces? i. Can two of the molecules in part f interact through hydrogen bonds? h. Draw a molecule that has the same molecular formula as the molecule in part a but is both an aldehyde and an ether. Answer: a. The molecule belongs to the carboxylic acid family. b. C5H10O2 c. Yes. All organic molecules can interact through London forces. d. Yes. The carboxylic acid molecule is polar and all polar molecules can interact through dipole-dipole forces. e. Yes. The molecule contains a H atom attached to an O atom.
f. The molecule drawn above has the same molecular formula (C5H10O2) as the carboxylic acid in part a but belongs to the ester family. Other possible answers include placing the COO group between any other two C atoms.
4 - 42
g.Yes. All organic molecules can interact through London forces. h.Yes. The molecule is polar and therefore can interact through dipole-dipole forces. i. No. The molecule in part f does not have a H atom bonded to an O atom.
j.
4 - 43
Chapter 5 Reactions SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
5.1
The drawing below represents a chemical reaction.
a. Write a balanced chemical equation for this reaction. A + 4B → 2C
b. If 12 As are reacted with 40 Bs, what is the limiting reactant and what is the theoretical yield of C? B is the limiting reactant and the theoretical yield is 20 C The limiting reactant is the reactant that would get used up first. To determine which of the two reactants the limiting reactant is, calculate the amount of A2 required to react with all of the B present using molar ratios from the balanced equation. If there is more than enough A2 to consume all of B, then B is the limiting reactant because not all of the A2 will be used up. If there is not enough A2 to react with all of the B present, then A2 is the limiting reactant because it will be used up first before all of the B present is used up. 40 B ×
1A = 10 A 4B
Because only 10 A are required to react with ALL of the B present, B will get used up first and there will be an excess of 2A. Therefore, B is the limiting reactant. To calculate the theoretical yield, use the amount of limiting reactant B and the appropriate molar ratio to determine amount of C: 40 B x
2C = 20 C 4B
5-1
5.3
Write the following sentence as a balanced chemical equation. Phosphorus reacts with chlorine (Cl2) to produce phosphorus trichloride. 2P + 3Cl2 → 2PCl3 Prefixes like “tri” tell you the subscript for the element in the compound’s formula. First translate the names to symbols: P + Cl2 → PCl3. Next, check to see if the equation is balanced. It is not balanced because there are 2 Cl atoms on the left and 3 Cl atoms on the right. Place the coefficient 2 in front of the PCl3 (2PCl3) which makes 6 Cl atoms on the right that can be balanced by placing a 3 in front of the Cl2 (3Cl2) to make 6 Cl atoms on the left. The 2PCl3 also means you have 2 P atoms on the right, so you need to place a 2 in front of the P on the left (2P) in order to have 2 P atoms on that side. This makes the equation balanced.
5.5
What would you observe if you carried out the following reactions? a. NH3(g) + HCl(g) → NH4Cl(s) Formation of a solid. The symbols given after the formulas indicate the physical state of the reactants and products: (s) means solid, (l) means liquid, (g) means gas, and (aq) means aqueous (dissolved in water). The (s) on the NH4Cl (s) indicates it is produced as a solid and, therefore, you would observe a solid being produced during the reaction of the two gases. The solid is in the form of microscopic particles suspended in space. b. HCO3-(aq) + H3O+(aq) →CO2(g) + 2H2O(l) Bubbling, indicating the formation of a gas. The CO2(g) would be observed as a bubbling gas formed after mixing two liquid solutions.
5.7
Balance the reaction equations. a. SO2 + O2 → SO3 2SO2 + O2 → 2SO3 You have 3 O atoms on the right, and 4 on the left. Place a 2 in front of SO3 (2SO3). This gives 6 O atoms and 2 S atoms on the right, so place a 2 in front of the SO2 on the left, (2SO2). This balances the S atoms and gives 4 O atoms to add to the O2, for a total of 6 O atoms on the left. 5-2
b. NO + O2 → NO2 2NO + O2 → 2NO2 There are 3 O atoms on the left and 2 on the right. Place a 2 in front of the NO2 (2NO2). This means you need 2 in front of the NO (2NO) so that the N atoms are balanced. This also gives 4 O atoms on both sides and makes the equation balanced.
5.9
Balance the reaction equations. a. K + Cl2 → KCl 2K + Cl2 → 2KCl There is 1 Cl atom on the right and 2 on the left. Place a 2 in front of the KCl (2KCl). This means that you need 2 in front of the K (2K) so that the K atoms are balanced. The equation is balanced. b. CH4 + Cl2 → CH2Cl2 + HCl CH4 + 2Cl2 → CH2Cl2 + 2HCl There are 3 H atoms on the right and 4 on the left. Place a 2 in front of the HCl (2HCl). This means you need 2 in front of the Cl2 (2Cl2) so that the Cl atoms are balanced. The equation is balanced.
5.11
Balance the reaction equations. Follow a similar process as in previous problems to balance these equations. In many cases, finding the balancing coefficients requires a trial-and-error method. a. Al + CuO → Al2O3 + Cu 2Al + 3CuO → Al2O3 + 3Cu b. Mg + P4 → Mg3P2 6Mg + P4 → 2Mg3P2
5.13
Balance the reaction equations. It may help to think in terms of nitrate ion, ammonium ion, and sulfate ion, rather than individual N, O, H, and S atoms. a. CaCl2 + AgNO3 → AgCl + Ca(NO3)2 5-3
CaCl2 + 2AgNO3 → 2AgCl + Ca(NO3)2 As stated in the problem, treating the polyatomic ion NO3- as an “individual particle” helps to simplify the balancing process: CaCl2 + AgNO3
→
AgCl + Ca(NO3)2
Ca
1
Ca
1
Cl
2
Cl
1
Ag
1
Ag
1
NO3- 1
NO3- 2
As is apparent from the table above, placing a 2 next to AgCl doubles the Cl on the product side to equal the number of Cl on the reactant side. This also doubles the Ag so a 2 needs to be placed next to AgNO3 on the reactant side. This makes the number of Ag atoms equal on both sides (2) and also makes the number of NO3- ions equal on both sides (2). Using these two balancing coefficients results in the balanced equation given above. b. (NH4)2SO4 + CaBr2 → NH4Br + CaSO4 (NH4)2SO4 + CaBr2 → 2NH4Br + CaSO4 Count each of the polyatomic ions as an individual particle (NH4+ and SO42-). Proceed with a trial and error method to determine the balancing coefficients as was done in the previous problems.
5.15
Balance the reaction equations. a. Cl2 + NaBr → NaCl + Br2 Cl2 + 2NaBr → 2NaCl + Br2 b. BiCl3 + H2O → Bi2O3 + HCl 2BiCl3 + 3H2O → Bi2O3 + 6HCl
5.17
One source of acid rain is the reaction between sulfur dioxide, oxygen, and water to form sulfuric acid (H2SO4). Balance the equation for this reaction. SO2(g) + O2(g) + H2O (g) → H2SO4(aq) 2SO2(g) + O2(g) + 2H2O (g) → 2H2SO4(aq) 5-4
5.19
The oxyacetylene torch used for welding burns acetylene gas (C2H2). Balance the equation for the incomplete combustion of acetylene. C2H2(g) + O2(g) → CO(g) + H2O(g) 2C2H2(g) + 3O2(g) → 4CO (g) + 2H2O(g)
5.21
Classify each reaction as involving synthesis, decomposition, single replacement, or double replacement. a. 2Na + Cl2 → 2NaCl Synthesis. Two elements combine to form a compound. b. 2K + 2H2O → H2 + 2KOH Single replacement. K replaces one hydrogen atom in H2O.
5.23
Classify each reaction as involving synthesis, decomposition, single replacement, or double replacement. a. HBr + NaOH → NaBr + H2O Double replacement. Na and H exchange places to form two different compounds. b. Cu + 2AgNO3 → Cu( NO3)2 + 2Ag Single replacement. Cu replaces Ag in the compound AgNO3. c. 2S + 3 O2 → 2SO3 Synthesis. S and O2 combine to form a compound. d. H2SO4 → H2O + SO3 Decomposition. A compound breaks down to form two simpler compounds.
5-5
5.25
Classify the reactions in Problem 5.6 as involving synthesis, decomposition, single replacement or double displacement. a. AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq ) Double replacement. Ag is replaced by K to form KNO3 and K is replaced by Ag to form AgCl. b. 2 Al(s) + Fe2O3(s) → 2 Fe(l) + Al2O3(s ) Single replacement. Al replaces Fe in Fe2O3 to form Al2O3.
5.27
Classify the reactions in Problem 5.11 as involving synthesis, decomposition, single replacement or double displacement. a. Al + CuO → Al2O3 + Cu Single replacement. Al replaces Cu in Cuo to form Al2O3. b. Mg + P4 → Mg3P2 Synthesis. Mg and P4 combine to form a compound, Mg3P2.
5.29
The reaction shown here is one step in the fermentation process. To which reaction type does the reaction belong?
Pyruvic acid
Acetaldehyde
Decomposition. Pyruvic acid decomposes into two simpler compounds, acetaldehyde and carbon dioxide.
5.31
As we will see in Section 8.10, an ester can be prepared in the following way. Does the reaction involve synthesis, decomposition, single replacement, or double displacement?
5-6
H+
Double replacement. –OCH3 in HOCH3 replaces the –OH in the first reactant molecule to form the first product molecule. The –OH replaces –OCH3 in HOCH3 to form HOH.
O
CH3 C O CH3
+ HOCH3
CH3 C OH
5.33
O
H+
+
HOH
Draw the products of each hydrolysis reaction. When an ester is hydrolyzed in the presence of H+, the two products are a carboxylic acid and an alcohol.
O
H+
H C
O CH2 CH3
a.
+ H2O
O
+
H C OH
HOCH2CH3
O
b.
CH3 C
H+ O CH2
+
H2O
O CH3 C OH
+ HOCH2 O
c.
H+
CH3 CH2 C O CH2 CH2 CH3
5-7
+ H2O
O CH3CH2
5.35
+
C OH
HOCH2CH2CH3
Draw the missing reactant for each hydrolysis reaction. In each case, the reaction is a hydrolysis reaction producing a carboxylic acid and an alcohol. The missing reactant is an ester molecule whose components can be derived from the product carboxylic acid and alcohol molecules as shown in part a.
? a.
+ H2O
H+
O CH3CH2
C OH
+
HOCH3
O
The missing reactant is
?
+
H2O
H+
CH3CH2
C O CH3
.
O H C OH +
HO CH2 CH CH3 CH3
b.
O H C O CH2 CH CH3 CH3
The missing reactant is
H+ ? c.
O C The missing reactant is 5-8
O CH3
5.37
Draw the hydration product formed when each alkene is reacted with H2O in the presence of H+.
a. Hydration of an alkene is the process of adding a water molecule across a double bond. The end result is that, to one of the C atoms in the double bond, a H atom is added and, to the other C atom, an –OH group is added. The product that forms is an alcohol molecule.
becomes
b. See directions in part a.
becomes
CH3 CH3 c. See directions given in part a.
CH3
OH
CH3 CH3
CH3 becomes
5-9
5.39
Draw the missing reactant for each hydration reaction. In each case, the missing reactant is an alkene. The hydration of an alkene produces an alcohol. In a hydration, a water molecule is split apart with a H atom making a bond to one of the C atoms in a C=C and the –OH group forming a bond with the other C atom in the C=C group as shown for part a below.
a.
?
+
H+
H2O
CH3CH2OH H C C H
The missing reactant is
H C C H
+
HOH
H H
H+ CH3CH2OH
H H
b.
?
+
H2O
H+
OH CH3 CH CH3
H C C CH3
The missing reactant is
c.
?
+
H2O
H+
H H
OH
The missing reactant is
5.41
Draw the organic dehydration product formed when each alcohol is heated in the presence of H+. Dehydration is the reverse of hydration. First, identify the location of an H atom and an OH group that are on adjacent carbons. The H and and OH groups are removed and the bond between the two carbon atoms becomes a double bond. The H and OH combine to give water as a second product.
5 - 10
a.
OH
CH3C=CH2
CH3CCH3 CH3
becomes
CH3
b. OH
becomes
c. OH CH3CH2CHCH2CH3
5.43
becomes
CH3CH2CH=CHCH3
Draw each missing reactant. In each case, determine the missing reactant based on the products.
a.
?
+
H2O
H+
HO
The missing reactant is This is a hydration reaction of an alkene because the product is an alcohol.
O
b.
?
+
H2O
C OH
H+
+
5 - 11
HOCH3
O C O CH3 The missing reactant is The reaction is the hydrolysis of an ester because the products are a carboxylic acid and an alcohol.
CH3 H+ ? c.
heat
H2O
+ CH3
The missing reactant is
OH
This is a dehydration reaction of an alcohol producing an alkene and water.
5.45
Aspirin (acetylsalicylic acid) is an ester of acetic acid.
a. Circle the ester group in aspirin.
b. Aspirin is sold with cotton placed in the neck of the bottle to help keep moisture out. When exposed to water, the ester group of aspirin slowly hydrolyzes. Draw the hydrolysis products obtained if aspirin is reacted with H2O in the presence of H+. 5 - 12
Hydrolysis of an ester in the presence of H+ results in the formation of the carboxylic acid molecule and an alcohol molecule. In this case, the acetyl group becomes part of the carboxylic acid molecule and the salicylic acid group becomes the alcohol molecule.
O O
C OH OH
5.47
+
HO C CH3
The ester shown below is reacted with H2O and H+ to form carboxylic acid A and alcohol B. Alcohol B is heated in the presence of H+ to produce molecule C. Draw A, B, and C. + O H H2O CH3CH2CH C O CH CH3 + A + B
CH3
CH3 +
H B
C heat
When an ester is reacted with H2O and H+, it undergoes a hydrolysis reaction producing a carboxylic acid (molecule A) and an alcohol (molecule B):
O
+
H
CH3CH2CH C O CH CH3 CH3
+
H2O
CH3 O CH3CH2CH C OH
+
HO CH CH3
CH3
CH3
molecule A
molecule B
When alcohol B is heated in the presence of H+, it undergoes a dehydration reaction to form an alkene (molecule C).
5 - 13
HO CH CH3
H+
CH3
heat
molecule B
5.49
H2C CH CH3 molecule C
Zinc reacts with copper(II) sulfate according to the equation Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) In an oxidation and reduction reaction, the atom or ion that loses electrons is oxidized and the atom or ion that gains electrons is reduced. The chemical that receives electrons is the oxidizing agent and the chemical that donates electrons is the reducing agent. a. Is zinc oxidized or is it reduced? Oxidized. Zinc is oxidized because it lost 2 electrons, going from neutral Zn(s) to Zn2+. b. Is copper(II) ion oxidized or is it reduced? Reduced. The copper(II) ion is reduced because it gained 2 electrons, going from a positively charged species, Cu2+, to neutral Cu. c. What is the oxidizing agent? Cu(II) ion is the oxidizing agent because it accepted electrons from zinc. d. What is the reducing agent? Zn is the reducing agent because it donated electrons to Cu2+.
5.51
Sodium metal reacts with oxygen gas (O2) to form sodium oxide. a. Write a balanced equation for this oxidation-reduction reaction. (In this problem you need not worry about the physical state of the reactants or product.) 4 Na + O2 → 2 Na2O b. Which reactant is oxidized? Na is oxidized because it lost electrons, going from neutral Na to Na+. c. Which reactant is reduced? 5 - 14
O2 is reduced because it gained electrons, going from neutral O to O2-. d. What is the oxidizing agent? O2 is the oxidizing agent because it removed electrons from Na. e. What is the reducing agent? Na is the reducing agent because it gave electrons to O2.
5.53
5.55
Classify the reactions in Problems 5.49 and 5.51 as involving synthesis, decomposition, single replacement, or double replacement. Problem 5.49:
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) Single replacement.
Problem 5.51:
4 Na + O2 → 2 Na2O Synthesis.
A reaction of iron metal with oxygen gas produces ferrous oxide (FeO). Fe + O2
→
FeO
a. Balance the equation. There is 1 O atom on the right and 2 on the left. Place a 2 in front of the FeO, (2FeO). This gives two O atoms on the right. Since this also makes two Fe atoms on the right, place a 2 in front of the Fe (2Fe) and the equation is balanced. 2Fe + O2
→
2FeO
b. When FeO forms, has Fe been oxidized or has it been reduced? Fe has been oxidized. Since oxidized means loss of electrons and reduced means gain of electrons, to determine which atom is oxidized or reduced, the charge on each atom or ion must be determined. Some of these can be predicted from the periodic table. When that is not possible (transition metals, for example) the charge is calculated using the other ions of known charge. charges:
0
0
2Fe + O2
2+ 2-
→
2FeO
Fe is changing from a 0 charge to a 2+ charge which means it has lost 2 electrons so it is oxidized. 5 - 15
c. When FeO forms, has O2 been oxidized or has it been reduced? O2 has been reduced. Using the charges calculated in part b, the O2 with 0 charge on each atom is changed to O2- which means it has gained 2 electrons. It is reduced.
5.57
Ethanol (CH3CH2OH) is mixed with gasoline to produce gasohol, a cleaner burning fuel than gasoline. Write the balanced chemical equation for the complete oxidation of ethanol by O2 to produce CO2 and H2O. The balanced equation is:
CH3CH2OH + 3O2 → 2CO2 + 3H2O
See the steps below for how to arrive at this answer. Write the formulas in equation format by placing the reactants on the left side and the products on the right side of the arrow. CH3CH2OH + O2 → CO2 + H2O To begin the balancing process, examine the equation carefully to see if it may already be balanced. In this case it is not. Next, note that there are more hydrogen atoms than any other element so this is probably a good element to start with. Since there are 6 H atoms on the left, place a 3 in front of H2O (3H2O). This makes six hydrogen atoms on both sides. Then, since there are 2 C atoms on the left, place a 2 in front of the CO2 (2CO2). This leaves only the O atoms to check. On the right there are 4 O atoms from the 2CO2 and 3 from the 3H2O for a total of 7 O atoms on the right. On the left there is 1 O atom from the CH3CH2OH and 2 O atoms from the O2. Complete the balancing by placing a 3 in front of the O2 (3O2). This makes 1 + 6 = 7 O atoms on the left which balances the equation. As recommended by the text, this equation can also be balanced by balancing C first, then H, then O.
5.59
Many city water departments use chlorine gas (Cl2) to disinfect their water supplies. When bubbled through water, Cl2 is converted into hypochlorite ion (OCl-), an oxidizing agent that kills harmful bacteria. In this reaction, chlorine atoms are both oxidized and reduced. Explain. Cl2(aq) + H2O(l) → OCl-(aq) + 2 H+(aq) + Cl-(aq) In this reaction, one Cl atom is oxidized by gaining oxygen atoms to become OCl. The other Cl atom is reduced by gaining an electron to become Cl-. 5 - 16
5.61
Draw the missing product of each reaction. Each of these reactions is a hydrogenation reaction as indicated by the presence of the H2 reactant and the Pt catalyst. Each of the 2 H atoms of H2 adds across the double bond converting the alkene to an alkane.
Pt a.
CH2=CHCH2CH3 +
?
H2
The missing product is CH3CH2CH2CH3 Pt H2
?
b. CH3 CH CH2 CH2 CH3 CH3
The missing product is
CH3
Pt +
?
H2
c.
CH3 The missing product is
5.63
Draw the skeletal structure of the product formed when each alkene is reacted with H2, in the presence of Pt.
a. Upon reaction with H2, 2 H atoms are added, one bonding to each C atom in the C=C bond. The double bond is converted to a single bond. Note that the hydrogen atoms added will not be shown since you are drawing a skeletal structure.
5 - 17
b. Upon reaction with H2, the double bond is converted to a single bond.
5.65
Draw the saturated product expected when 3 mol of H2, in the presence of Pt, are reacted with 1 mol of the termite trail marking pheromone shown below. H
H C
H C CH3CH2CH2
C CH2
C
CH2CH2OH C
H H
C H
CH3(CH2)10CH2OH is the saturated product expected. 3 mol of H2 will add to all of the double bonds in 1 mol of the pheromone.
5.67
For the reaction between carbon and oxygen to form carbon monoxide, beginning with 0.65 mol of C, 2C(s) + O2(g) → 2CO(g) a. how many moles of O2 are required to completely consume the C? Use the mole to mole ratio between C and O2 given by the equation as a conversion factor to convert moles of C to moles of O2. 5 - 18
0.65 mol C x
1 mol O2 = 0.33 mol O2 2 mol C
b. how many moles of CO are obtained when the C is completely reacted? Use the mole to mole ratio between C and CO given by the equation as a conversion factor to convert moles of C to moles of CO. 0.65 mol C x
5.69
2 mol CO = 0.65 mol CO 2 mol C
For the combustion of methane (CH4), beginning with 3.15 mol of CH4, CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) a. how many moles of O2 are required to totally consume the CH4? Use the mole to mole ratio for CH4 to O2 given by the equation as a conversion factor to convert moles of CH4 to moles of O2.
3.15 mol CH 4 x
2 mol O2 = 6.30 mol O2 1 mol CH4
b. how many moles of CO2 are obtained when the CH4 is totally consumed? Use the mole to mole ratio for CH4 to CO2 given by the equation as a conversion factor to convert from moles of CH4 to moles of CO2.
3.15 mol CH 4 x
1 mol CO2 = 3.15 mol CO2 1 mol CH4
c. how many moles of H2O are obtained when the CH4 is totally consumed? Use the mole to mole ratio for CH4 to H2O given by the equation as a conversion factor to convert from moles of CH4 to moles of H2O. 2 mol H 2O 3.15 mol CH 4 x = 6.30 mol H 2O 1 mol CH 4
5.71
For the reaction in Problem 5.55, a. how many grams of Fe are required to react completely with 4.22 x 10-3 mol of O2? 5 - 19
The balanced equation for the reaction in Problem 5.55 is: 2Fe + O2
→
2FeO
First, convert the mol of O2 to mol Fe using the molar ratio from the balanced equation. Then, use the molar mass of Fe to convert mol Fe to grams of Fe.
4.22 x 10−3 mol O2 x
2 mol Fe 55.85 g Fe x = 0.471 g Fe 1 mol O2 1 mol Fe
b. how many grams of FeO are produced from the complete reaction of 17.44g of Fe? First, convert the grams of Fe to mol Fe. Then, use the molar ratio between Fe and FeO to find mol FeO. Use the molar mass of FeO to convert to grams of FeO.
17.44 g Fe x
5.73
1 mol Fe 2 mol FeO 71.85 g FeO x x = 22.44 g FeO 55.85 g Fe 2 mol Fe 1 mol FeO
Isopropyl alcohol (rubbing alcohol) can be produced by the reaction: OH H+ CH3CH CH2 + H2O CH CH CH 3
propene
3
isopropyl alcohol
a. How many moles of H2O, are required to completely react with 55.7 mol of propene? Use the mole to mole ratio for C3H6 (propene) to H2O given by the equation as a conversion factor to convert from moles of C3H6 to moles of H2O.
55.7 mol C3H6 x
1 mol H 2O = 55.7 mol H 2O 1 mol C3H6
b. How many moles of H2O are required to completely react with 1.66 mol of propene? Use the mole to mole ratio for C3H6 to H2O given by the equation as a conversion factor to convert from moles of C3H6 to moles of H2O.
5 - 20
1.66 mol C3H6 x
1 mol H 2O = 1.66 mol H 2O 1 mol C3H6
c. How many moles of isopropyl alcohol are expected from the complete reaction of 47.2 g of propene? Use the molar mass of propene to convert from grams to moles. Then, use the mole to mole ratio for C3H6 to C3H7OH given by the equation as a conversion factor to convert from moles of C3H6 to moles of C3H7OH.
47.2 g C3H6 x
1 mol C3H6 1 mol C3H7 OH x = 1.12 mol C3H7OH 42.08 g C3H6 1 mol C3H6
d. How many grams of isopropyl alcohol are expected from the complete reaction of 125 g of propene? Use molar mass of propene to convert from grams to moles. Then, use the mole to mole ratio for C3H6 to C3H7OH given by the equation as a conversion factor to convert from moles of C3H6 to moles of C3H7OH. Finally, use the molar mass of 2-propanol to convert moles of 2-propanol to grams.
125 g C3H6 x
5.75
1 mol C3H6 1 mol C3H7 OH 60.10 g C3H7OH x x = 179 g C3H7OH 42.08 g C3H6 1 mol C3H6 1 mol C3H7 OH
Consider the reaction KOH(s) + CO2(g) →
KHCO3(s)
a. How many grams of KOH are required to react completely with of 5.00 mol of CO2? Use the mole to mole ratio for CO2 to KOH given by the equation as a conversion factor to convert from moles CO2 to moles of KOH. Then use the molar mass of KOH to convert moles of KOH to grams.
5.00 mol CO2 x
1 mol KOH 56.11 g KOH x = 281 g KOH 1 mol CO2 1 mol KOH
b. How many grams of KHCO3 are produced from the complete reaction of 75.9 g of KOH? Use the molar mass of KOH to convert from grams to moles. Then, use the mole to mole ratio for KOH to KHCO3 given by the equation as a conversion factor to 5 - 21
convert from moles of KOH to moles of KHCO3. Finally, use the molar mass of KHCO3 to convert moles of KHCO3 to grams. 75.9 g KOH x
5.77
1 mol KHCO3 100.12 g KHCO3 1 mol KOH x x = 135 g KHCO3 56.11 g KOH 1 mol KOH 1 mol KHCO3
One treatment for mercury poisoning is to administer chelating agents, compounds that are able to form several bonds to metal ions. The resulting compound is water soluble and can be excreted. Each molecule of the chelating agent dimercaprol (C3H5OS2) reacts with one mercury ion. What is the minimum number of micrograms of dimercaprol that are required to remove all of the mercury from the bloodstream of a child whose blood contains 5.0 g Hg/dL and whose total blood volume is 2.5 L? First, determine the moles of mercury ions in 2.5 L of the blood using the mercury concentration 5.0 g Hg/dL: 5 μg Hg 10-6 g 1 mol Hg ions 2.5 L × -1 × × × = 6.2 x 10-7 mol Hg ions 1 dL 1 μg 200.6 g Hg 10 L 1 dL
Then, use the 1 to 1 ratio of dimercaprol molecules to mercury ions (which is the same as the mole-to-mole ratio) given in the problem to determine the moles of dimercaprol required to chelate the number of mercury ions calculated above. Use the molar mass of dimercaprol to convert moles to grams of dimercaprol. Convert grams to micrograms. 6.2 x 10-7 mol Hg ions × 1 μg 10-6 g
5.79
1 mol dimercaprol 121.21 g dimercaprol × × 1 mol Hg ions 1 mol dimercaprol
= 75 μg dimercaprol
a. How many grams of water are required to completely react with 2.7 g of Procaine (Figure 5.4)? The balanced equation for the hydrolysis of procaine can be rewritten as C13H20N2O2 Procaine
2.7 g C13H20N2O2 ×
+
H2O
→
C7H7NO2 carboxylic acid
1 mol C13H20N2O2 1 mol H2O × × 236.3 g C13H20N2O2 1 mol C13H20N2O2
18.02 g H2O = 0.21 g H2O 1 mol H2O
5 - 22
+
C6H15NO alcohol
b. How many grams of the carboxylic acid hydrolysis product are obtained when 2.7 g of Procaine are completely hydrolyzed? 2.7 g C13H20N2O2 ×
1 mol C13H20N2O2 1 mol C7H7NO2 × × 236.3 g C13H20N2O2 1 mol C13H20N2O2
137.1 g C7H7NO2 = 1.6 g C7H7NO2 1 mol C7H7NO2
c. How many grams of the alcohol hydrolysis product are obtained when 2.7 g of Procaine are completely hydrolyzed? 2.7 g C13H20N2O2 ×
1 mol C13H20N2O2 1 mol C6H15NO × × 236.3 g C13H20N2O2 1 mol C13H20N2O2
117.2 g C6H15NO = 1.3 g C6H15NO 1 mol C6H15NO 5.81
One form of phosphorus, called white phosphorus, burns when exposed to air. P4(s) + O2(g)
→
P4O10(s)
a. Balance the reaction equation. There are 2 O atoms on the left and 10 O atoms on the right. Place a 5 in front of the O2 (5O2). This gives ten O atoms on the left. Since the P atoms are already balanced, the equation is balanced. P4(s) + 5O2(g) → P4O10(s) b. What is the theoretical yield (in grams) of P4O10 if 33.0 g of P4 are reacted with 40.0 g of O2? The theoretical yield is the amount of product expected when all of the reactant is used up. When it cannot be assumed that another reactant in the reaction is in excess, you must first determine which reactant produces the smaller amount of product when used up. Do this by converting each number of grams given to moles of product formed. Then use the smaller number of moles of product to determine the grams of product that would actually be produced. 33.0 g P4 x
1 mol P4O10 1 mol P4 x = 0.266 mol P4O10 123.90 g P4 1 mol P4
40.0 g O2 x
1 mol P4O10 1 mol O2 x 32.00 g O2 5 mol O2
5 - 23
= 0.250 mol P4O10
Notice that the O2 will produce the lesser amount of product when it is used up and it will therefore determine the amount of product formed. Convert the 0.250 mol P4O10 to grams using the molar mass of P4O10. 0.250 mol P4O10 x
283.89 g P4O10 1 mol P4O10
= 71.0 g P4O10
5 - 24
5.83
Magnesium and iodine react to form magnesium iodide. Mg + I2 → MgI2 a. What is the limiting reactant if 10.0 g Mg are reacted with 95.0 g of I2? Make sure the equation is balanced. It is in this case. The limiting reactant is the reactant that is completely consumed first and therefore limits the amount of product formed. To determine the limiting reactant in this reaction, first convert the amounts of reactants in grams to moles:
10.0 g Mg x
1 mol Mg = 0.411 mol Mg 24.31 g Mg
95.0 g I2 x
1 mol I2 = 0.374 mol I2 limiting reactant 253.81 g I2
Because the balanced equation requires 1 mole of Mg to react with 1 mole of I2, it is clear that I2 is the limiting reactant. When 0.374 mol of I2 has reacted with 0.374 mol Mg, there will be some Mg left unreacted because the original amount (0.411 mol) is greater than 0.374 mol. b. What is the theoretical yield of MgI2 in grams? Use the amount of the limiting reactant to calculate the theoretically yield in grams of MgI2: 0.374 mol I2 x
1 mol MgI2 278.11 g MgI2 x = 104 g MgI 2 1 mol I2 1 mol MgI2
c. If 25.7 g of MgI2 are obtained, what is the percent yield? The percent yield can be calculated the using the formula below. The actual yield is 25.7 g and the theoretical yield is 104 g.
percent yield =
actual yield x 100% theoretical yield
percent yield =
25.7 g x 100% = 24.7% 104 g
5 - 25
5.85
If a person drinks a glass of beer or wine, the first step in metabolizing the ethanol (CH3CH2OH) is its enzymatic conversion to acetaldehyde (CH3CHO). OH CH3 C H
O
enzyme
+
NAD+
CH3 C H
H ethanol
+
NADH
+
H+
acetaldehyde
a. In this reaction is the ethanol oxidized or reduced? Oxidized In the case of an organic molecule, it is not possible to simply look up the charges on the atoms to determine the charges. Oxidation has occurred when H atoms are removed, oxygen is added, or both. Reduction is typically evidenced by the addition of H atoms, removal of oxygen atoms from, or both. In this case H atoms are removed as the molecule is changed from ethanol to acetaldehyde. b. What theoretical yield of acetaldehyde (in grams) is expected from the reaction of 10.0 g of ethanol? (Assume that ethanol is the limiting reactant.) First, use the molar mass of ethanol to convert grams of ethanol to moles of ethanol. Then convert moles of ethanol to moles of acetaldehyde using the mole to mole ratio from the equation. Finally, convert moles of acetaldehyde to grams of acetaldehyde using the molar mass of acetaldehyde. The molecular formula of ethanol is C2H5OH and the molecular formula of acetaldehyde is C2H4O. 10.0 g C2 H5OH x
5.87
1 mol C2 H5OH 1 mol C2 H 4O 44.05 g C2 H 4O x x = 9.56 g C2 H 4O 46.07 g C2 H5OH 1 mol C2 H5OH 1 mol C2 H 4O
Carbon(s) and sulfur dioxide(l) react to form carbon disulfide(l) and carbon monoxide(g). a. Write a balanced chemical equation for this reaction. 5C(s) + 2SO2(l) → CS2(l) + 4CO(g) b. If 7.2 g of carbon and 10.5 grams of sulfur dioxide react, what is the limiting reactant? SO2 is the limiting reactant. The limiting reactant is the reactant that is completely consumed first and therefore limits the amount of product formed. To determine the limiting reactant in this reaction, first convert the amounts of reactants in grams to moles: 5 - 26
7.2 g C ×
1 mol C = 0.60 mol C 12.01 g C
10.5 g SO2 ×
1 mol SO2 = 0.16 mol SO2 64.07 g SO2
According to the balanced equation above, for every 2 mol SO2, 5 mol of C are required. Calculate the mol C required to react all of 0.16 mol SO2. 0.16 mol SO2 ×
5 mol C = 0.40 mol C 2 mol SO2
SO2 is the limiting reactant because there is more than enough mol C to completely react with all of the SO2 present. Carbon is in excess.
c. For the reaction in part b, what is the theoretical yield of carbon disulfide? Use the amount of the limiting reactant, SO2, to calculate the theoretical yield of carbon disulfide. 10.5 g SO2 ×
1 mol SO2 1 mol CS2 76.15 g CS2 × × = 6.24 g CS2 64.07 g SO2 2 mol SO2 1 mol CS2
d. If 2.1 g of carbon disulfide are obtained from the reaction in part b, what is the percent yield? percent yield =
5.89
2.1 g CS2 × 100% = 34% 6.24 g CS2
During photosynthesis, the energy of sunlight is used to convert carbon dioxide and water into glucose (C6H12O6) and O2. a. Write a balanced chemical equation for this process. 6CO2 + 6H2O → C6H12O6 + 6O2 b. If 6.10 g of glucose are produced through photosynthesis, what is the minimum number of moles of carbon dioxide and of water that were used? 6.10 g C6H12O6 ×
1 mol C6H12O6 6 mol CO2 × = 0.203 mol CO2 180.2 g C6H12O6 1 mol C6H12O6
5 - 27
6.10 g C6H12O6 ×
1 mol C6H12O6 6 mol H2O × = 0.203 mol H2O 180.2 g C6H12O6 1 mol C6H12O6
c. If the 6.10 g of glucose is an 83.2% yield, and assuming that carbon dioxide was the limiting reactant, how many moles of carbon dioxide were available for the reaction? Assuming 6.10 g is the actual yield and the percent yield is 83.2%, we can calculate the theoretical yield in grams of the glucose by setting up the following equation: percent yield =
6.10 g C6H12O6 × 100% = 83.2% theoretical yield in g C6H12O6
Solving for the theoretical yield gives 7.33 g C6H12O6. Use this mass of C6H12O6 to calculate the moles of carbon dioxide available for the reaction:
7.33 g C6H12O6 ×
5.91
1 mol C6H12O6 6 mol CO2 × = 0.244 mol CO2 180.2 g C6H12O6 1 mol C6H12O6
Which of the reactions are spontaneous? For a spontaneous reaction, ΔG must have a negative value. For a nonspontaneous reaction, ΔG has a positive value. a. CO(g) + H2O(g) → CO2(g) + H2(g)
ΔG = -4.1 kcal/mol
This is a spontaneous reaction because the ΔG value is negative. b. 2HI(g) → H2(g) + I2(g)
ΔG = 0.6 kcal/mol
This is a nonspontaneous reaction because the ΔG value is positive.
5.93
a. Define the term “spontaneous reaction”. A spontaneous reaction proceeds without outside intervention once started. b. Define the term “nonspontaneous reaction”. A nonspontaneous reaction requires outside intervention to get it started and to keep it going.
5 - 28
5.95
Draw an energy diagram for the reaction in Problem 5.91a. Label the x- and yaxis, reactants, products and ΔG. The reaction in 5.91a is: CO(g) + H2O(g) → CO2(g) + H2(g)
ΔG = -4.1 kcal/mol
Energy
ΔG = -4.1 kcal/mol CO(g) + H2O(g) CO2(g) + H2(g)
Progress of Reaction
5.97
Draw two reaction energy diagrams that illustrate that type of results expected for the reaction below, one diagram for the reaction in the presence of Pt catalyst, and one diagram in the absence of Pt catalyst. In drawing the energy diagrams, assume that the reactions are spontaneous. Pt CH3CH2CH3 CH3CH=CH2 + H2 Reaction energy diagram for the reaction without a catalyst. Activation Energy
Energy
CH3CH=CH2 + H2 CH3CH2CH3
5 - 29
Reaction energy diagram for the reaction with a catalyst. A catalyst lowers the activation energy for the reaction. In the diagram below, the height of the activation energy is less than that for the diagram of the uncatalyzed reaction.
Activation Energy Energy
CH3CH=CH2 + H2 CH3CH2CH3
Reaction Progress
5.99
What is a catalyst? A catalyst is a substance that increases the rate of a reaction without being changed or consumed itself.
5.101 When hydrogen peroxide is used as a disinfectant, it is the O2 gas produced from the decomposition of H2O2 that kills bacteria. Balance the reaction equation. H2O2(aq) → H2O(l) + O2(g) 2 H2O2(aq) → 2 H2O(l) + O2(g)
5.103 Explain the role that carbonic anhydrase has in the transport of CO2 in the bloodstream. Carbonic anhydrase is an enzyme that catalyzes the conversion of CO2 in red blood cells to H2CO3, carbonic acid, which breaks apart into the bicarbonate ion HCO3 – and H+. The bicarbonate ion is then released into the blood plasma. In the lungs, this process is reversed; carbonic anhydrase catalyzes the conversion of H2CO3 back to CO2 which is exhaled along with additional CO2 dissolved in blood serum and released from its attachment to hemoglobin.
5 - 30
5.105 The reaction that takes place in a car battery when it starts the car is Pb + PbO2 + H2SO4 → PbSO4 + H2O a. Balance the chemical equation. Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2 H2O b. Pb is the atomic symbol for which element? lead c. What is the charge on Pb (the first reactant in the reaction equation)? zero d. What is the charge on the Pb ion in PbO2? +4 e. What is the charge on the Pb ion in PbSO4? +2 f. In this reaction, what is oxidized? Pb g. What is reduced? Pb4+ h. What is the oxidizing agent? Pb4+ i. What is the reducing agent? Pb j. If 5.0g of Pb are reacted with 5.6 g PbO2, what is the theoretical yield of PbSO4? Assume that there is more than enough H2SO4 for the reaction to take place. First, determine the limiting reactant. Convert grams of each reactant to moles. 5 - 31
5.0 g Pb x
1 mol Pb = 0.024 mol Pb 207.20 g Pb 1 mol PbO2 = 0.023 mol PbO2 239.20 g PbO2
5.6 g PbO2 x
Use the molar ratio from the balanced equation to determine how much PbO2 is required if all of the Pb is consumed.
0.024 mol Pb x
1 mol PbO2 = 0.024 mol PbO2 1 mol Pb
Because the amount of PbO2 required is greater than what is present, PbO2 is the limiting reactant. Use the amount of PbO2 to calculate the theoretical yield of PbSO4
0.023 mol PbO2 x
2 mol PbSO4 303.26 g PbSO4 x = 14 g PbSO4 1 mol PbO2 1 mol PbSO4
k. If 4.5 g of PbSO4 are obtained in the reaction described in part j, what is the percent yield?
percent yield =
4.5 g x 100% = 32% 14 g
l. Would you expect ΔG for this reaction to have a negative or positive value? Negative. The reaction is spontaneous which is the basis for its use in batteries. m. When it is in use, a battery heats up. Is the reaction exothermic or endothermic? Exothermic. Heat is released as the reaction occurs. n. A running car engine recharges the battery by reversing the reaction above. Would you expect ΔG for this process to have a negative or positive value? Positive. The reverse reaction is nonspontaneous as it requires an external source of driving force (car engine) to occur.
5 - 32
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 5.2
The drawing below represents a chemical reaction.
a. Write a balanced chemical equation for the reaction. b. If 26 Cs are reacted with 31 Ds, what is the limiting reactant and what is the theoretical yield of F? Answer: a.
2C + 3D → 2E + 4F
b.
D is the limiting reactant and the theoretical yield is 41 F
If all 26 C react, 39 D are required (see calculation below).
26 C x
3D = 39 D 2C
Because there are only 31 D available, there will be an excess of C and D is the limiting reactant. The theoretical yield is calculated by using the amount of the limiting reactant and the molar ratio between that reactant and F.
31 D x
SECTION 5.1
5.4
4F = 41 F 3D CHEMICAL EQUATIONS
Write the following sentence as a balanced chemical equation. Aqueous sodium chloride reacts with aqueous silver nitrate to form aqueous sodium nitrate and solid silver chloride.
5 - 33
Answer: Balanced chemical equations should include correct formulas for each reactant or product, the phase or state of each product, and the correct balancing coefficients. In this case, the equation is balanced as written and no coefficients need to be changed. NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)
5.6
What would you observe if you carried out the following reactions? a. AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq ) b. 2Al(s) + Fe2O3(s) → 2Fe(l) + Al2O3(s )
Answer: Note the physical states of the reactants and the products and describe what changes they undergo. a. When aqueous silver nitrate reacts with aqueous potassium chloride, a solid precipitate of silver chloride forms. b. When solid aluminum reacts with solid iron (III) oxide, liquid iron and solid aluminum oxide will form.
5.8
Balance the reaction equations. a. Mg + O2 → MgO b. K + O2 → K2O
Answer: When balancing equations, count the number of atoms of each element on both sides of the equation. If the number of atoms of each element on the reactant side does not equal the number of atoms of the same element on the product side, change the coefficients by trial and error until all the elements are balanced. a. Unbalanced: Mg + O2 → MgO Because there are 2 oxygen atoms on the reactant side, add the coefficient 2 to MgO. Then, add the coefficient 2 to Mg. Balanced:
2 Mg + O2 → 2 MgO
b. Unbalanced: K + O2 → K2O Because there are 2 oxygen atoms on the reactant side, add the coefficient 2 to K2O. Then, add the coefficient 4 to K. 5 - 34
Balanced:
5.10
4 K + O2 → 2 K2O
Balance the reaction equations. a. C4H6 + H2 → C4H10 b. C3H8 + O2 → CO + H2O
Answer: When balancing equations, count the number of atoms of each element on both sides of the equation. If the number of atoms of each element on the reactant side does not equal the number of atoms of the same element on the product side, change the coefficients by trial and error until all the elements are balanced. a. C4H6 + 2 H2 → C4H10 b. 2 C3H8 + 7 O2 → 6 CO + 8 H2O
5.12
Balance the reaction equations. a. S8 + O2 → SO2 b. SiO2 + HF → SiF4 + H2O
Answer: When balancing equations, count the number of atoms of each element on both sides of the equation. If the number of atoms of each element on the reactant side does not equal the number of atoms of the same element on the product side, change the coefficients by trial and error until all the elements are balanced. a. S8 + 8O2 → 8SO2' b. SiO2 + 4 HF → SiF4 + 2H2O
5.14
Balance the reaction equations. It may help to think in terms of nitrate ion and sulfate ion, rather than individual N, O, H, and S atoms. a. Pb(NO3)2 + K2SO4 → KNO3 + PbSO4 b. BaI2 + Na2SO4 → BaSO4 + NaI
Answer: When balancing equations, count the number of atoms of each element on both sides of the equation. If the number of atoms of each element on the reactant side does not equal the number of atoms of the same element on the product side, change the coefficients by 5 - 35
trial and error until all the elements are balanced. In this case, treat the nitrate ion and the sulfate ion each as a particle that moves around as one unit. a. Unbalanced: Pb(NO3)2 + K2SO4 → KNO3 + PbSO4 Because there are 2 nitrate ions on the reactant side, place a 2 in front of KNO3. The equation is now balanced: on each side, there are 1 lead ion, 2 nitrate ions, 2 potassium ions, and 1 sulfate ion. Balanced:
Pb(NO3)2 + K2SO4 → 2 KNO3 + PbSO4
b. Unbalanced: BaI2 + Na2SO4 → BaSO4 + NaI Because there are 2 iodide ions on the reactant side, place a 2 in front of NaI. The equation is now balanced: on each side, there are 1 barium ion, 2 iodide ions, 2 sodium ions, and 1 sulfate ion. Balanced:
5.16
BaI2 + Na2SO4 → BaSO4 + 2NaI
Cave explorers can use lamps that contain calcium carbide (CaC2) to light their way. When moistened, calcium carbide reacts to form acetylene gas (C2H2), which immediately burns to produce light. Balance the equation for the reaction of calcium carbide with water. CaC2(s) + H2O(l) → Ca(OH)2(aq) + C2H2(g)
Answer: There is 1 O atom on the left and 2 on the right. Place a 2 in front of the H2O (2H2O). This gives two O atoms on both sides. It also gives 4 H atoms on both sides, so the equation balanced. CaC2(s) +2 H2O(l) → Ca(OH)2(aq) + C2H2(g)
5.18
One source of acid rain is the reaction between nitrogen dioxide, oxygen, and water to form nitric acid (HNO3). Balance the equation for this reaction. NO2(g) + O2 (g) + H2O (g) → HNO3(aq)
Answer: 4NO2(g) + O2 (g) + 2H2O (g) → 4HNO3(aq)
5.20
Octane (C8H18), one component of gasoline, burns to form carbon dioxide and water, Balance this reaction equation. 5 - 36
C8H18 (l) + O2 (g) → CO2 (g) + H2O (g) Answer: 2C8H18(l) + 25O2 (g) → 16CO2(g) + 18H2O (g)
SECTION 5.2
5.22
REACTION TYPES
Classify each reaction as involving synthesis, decomposition, single replacement, or double replacement. a. CaCO3 + 2HCl → CaCl2 + H2CO3 H+ CH3CH2OH CH2=CH2 + H2O heat b.
Answer: a. Double replacement. In this reaction, the Ca in CaCO3 and the H in HCl switch places.
CaCO3 + 2HCl
CaCl2 + H2 CO3
b. Decomposition. A single compound is broken down into two smaller compounds, CH2=CH2 and H2O.
5.24
Classify each reaction as involving synthesis, decomposition, single replacement, or double replacement. a. Cl2 + 2NaI → 2NaCl + I2 b. 2K + 2HCl → 2KCl + H2 c. N2 + 3H2 → 2NH3 d. 2KClO3 → 2KCl + 3O2
Answer: a. Single replacement. Cl replaces I in NaI. b. Single replacement. K replaces H in HCl. c. Synthesis. The two elements are chemically combined to form a new compound. 5 - 37
d. Decomposition. A bigger compound is broken down into a smaller compound and an element.
5.26 Classify the reactions in Problem 5.9 as involving synthesis, decomposition, single replacement, or double replacement. Answer: a. K + Cl2 → KCl Synthesis. The two elements are chemically combined to form a new compound. b. CH4 + Cl2 → CH2Cl2 + HCl Single replacement. H replaces Cl in Cl2.
5.28
Classify the reactions in Problem 5.15 as involving synthesis, decomposition, single replacement, or double replacement.
Answer: a. Cl2 + NaBr → NaCl + Br2 Single replacement. Cl replaces Br in NaBr. b. BiCl3 + H2O → Bi2O3 + HCl Double replacement. O in H2O and the Cl in BiCl3 switch places.
5.30
The reaction shown here is one step in the citric acid cycle (Section 14.7). To which reaction type does the reaction belong?
O
O O
C
CH CH C O
+ H2O
O
O
OH O
C
CH2 CH C O Malate
Fumarate
Answer: Synthesis. The reaction shows the combination of the ion fumarate and water to form the bigger ion, malate. 5 - 38
5.32
As we will see in Section 9.2, an alcohol can be produced in the following way. Does the reaction involve synthesis, decomposition, single replacement, or double replacement? OH
O CH3CH2CH
+
H2
Pt
CH3CH2CH2
Answer: Synthesis. The reaction given in this problem shows the starting compound combining with H2 to form an alcohol.
SECTION 5.3
5.34
REACTIONS INVOLVING WATER
Draw the products of each hydrolysis reaction. O a.
CH3 CH C
+
O CH3
+
H
H2O
CH3 O b.
CH3 C
+
O CH CH3
+
H
H2O
CH3
O c.
C
+
+
O CH2
H2O
H
Answer: In a hydrolysis reaction, water is used to split a molecule. In this problem, each reactant is an ester breaks apart into a carboxylic acid and an alcohol. O a.
CH3 CH C
OH
+
HO CH3
CH3 O b.
CH3 C
OH +
HO CH CH3 CH3
5 - 39
O C
c.
5.36
OH
+
HO CH2
Draw the missing reactant for each hydrolysis reaction. O
+
a.
+
?
H2O
H
CH3 CH C
OH
+
HOCH2CH3
OH
+
HO CH2
CH3
O
+
b.
+
?
H2O
H
CH3 CH2 C
O
+
c.
+
?
H2O
H
CH3 C
OH
+
HO CH2
Answer: The missing reactant is drawn by forming a bond between the two products to form an ester. O a.
CH3 CH C O CH2 CH3 CH3
O b.
CH3 CH2 C O CH2
O c.
CH3 C O CH2
5 - 40
5.38
Draw the hydration product formed when each alkene is reacted with H2O in the presence of H+. a. CH3CH2CH2CH=CHCH2CH2CH3
b.
c.
Answer: In the hydration of an alkene in the presence of H+, water is added across the double bond, producing an alcohol.
a.
b.
c.
5 - 41
5.40
For each reaction, draw two different alkenes that could form the product shown.
OH
+
a.
?
+
H
H2O
CH3 CH2 CH2 CH CH3 OH
+
b.
?
+
H
H2O
CH3 CH2 C CH3 CH3
CH3
+
c.
?
+
H
H2O
OH
Answer: The two different alkenes formed would differ by the position of the C=C. The alcohol formed can form either by the hydration of a C=C to the left or to the right of the C bonded to the OH.
a.
CH3 CH2 CH CH CH3
b.
CH3 CH C CH3
or or
CH3 CH2 CH2 CH CH2 CH3 CH2 C CH2 CH3
CH3
CH3
CH3 c.
or These are the same compounds.
5.42
Draw the organic dehydration product formed when each alcohol is heated in the presence of H+.
OH a.
CH3 CH2 C CH2 CH3 CH2 CH3
5 - 42
OH b.
c.
Answer: In a dehydration reaction, an alcohol is converted to an alkene by the removal of water formed from the OH group and a H atom from a neighboring C atom. This is the reverse of hydration.
a.
CH3 CH2 C CH CH3 CH2 CH3
b.
c.
5.44
H2C CH
Draw each missing product.
a.
+
HO
H heat
+ H2O
?
+
b. CH3 (CH2)3 CH CH
O c.
+
(CH2)3 CH3
CH3
CH3 CH2 O C CH2 C CH3 CH3
5 - 43
H2O
H
?
+
H
CH3CH2OH
+
?
Answer:
a. This is a dehydration reaction. OH b.
CH3 (CH2)3 CH CH2 (CH2)3 CH3
This is a hydration reaction. O
CH3
c. HO C CH2 C CH3 CH3
This is a hydrolysis reaction.
5.46
Triflusal is a drug used to inhibit blood clot formation. Draw the organic products obtained when triflusal is hydrolyzed in the presence of H+.
O C OH
O
O C CH3
CF3 Triflusal Answer: In the hydrolysis of trifusal, the trifusal molecule is split apart at the ester group to form a carboxylic acid and an alcohol.
O C OH O
OH +
CH3 C OH
CF3 5 - 44
5.48
The ester shown below is reacted with H2O and H+ to form a carboxylic acid A and alcohol B. Alcohol B is heated in the presence of H+ to produce molecule C. Draw A, B, and a possible structure for C.
O
+
H2O
+
CH3 CH2 CH O C CH CH3
H
+
A
B
CH3
CH3
+
H
C
B
heat
Answer: When an ester is reacted with H2O in the presence of H+, it undergoes a hydrolysis reaction producing a carboxylic acid (molecule A) and an alcohol (molecule B): +
O
H
CH3 CH2 CH O C CH CH3 CH3
+
H2O
CH3 O +
HO C CH CH3
CH3 CH2 CH OH
CH3
CH3
molecule A
molecule B
When alcohol B is heated in the presence of H+, it undergoes a dehydration reaction to form an alkene (molecule C).
H+ CH3 CH2 CH OH CH3
heat
5.50
CH3 CH2 CH
CH3
molecule B
SECTION 5.4
or
CH3 CH CH
molecule C
OXIDATION AND REDUCTION
Magnesium reacts with iron(II) chloride according to the equation 5 - 45
CH2
Mg(s) + FeCl2(aq) → MgCl2(aq) + Fe(s) a. Is magnesium oxidized or is it reduced? b. Is iron(II) ion oxidized or is it reduced? c. What is the oxidizing agent? d. What is the reducing agent? Answer: Oxidized means to lose electrons and reduced means to gain electrons. To determine which atom is oxidized or reduced, the charge on each atom or ion must be determined. Some of these can be predicted from the periodic table. When that is not possible (transition metals, for example), the charge is calculated using the other ions whose charges are known. charges:
0
2+ 1-
Mg(s) + FeCl2(aq) →
2+
1-
0
MgCl2(aq) + Fe(s)
a. Mg is oxidized. Mg changes from a 0 charge to a 2+ charge, so it loses electrons and is oxidized. b. Fe2+ is reduced. Fe2+ is changed to Fe(s) with 0 charge, so it gains electrons and it is reduced. c. Fe2+ is the oxidizing agent. By definition, the species that causes the oxidation to occur is the oxidizing agent. Since Fe2+ ion accepted electrons from Mg(s), Fe2+ is the oxidizing agent. d. Mg(s) is the reducing agent. By definition the species that causes the reduction to occur is the reducing agent. Since Mg(s) gave the electrons to Fe2+ ion, Mg(s) is the reducing agent.
5.52
Aluminum metal reacts with oxygen gas (O2) to form aluminum oxide. a. Write a balanced equation for this oxidation-reduction reaction. (In this problem you need not worry about the physical state of the reactants or product.) b. Which reactant is oxidized? c. Which reactant is reduced? d. What is the oxidizing agent? e. What is the reducing agent?
Answer: a. 4Al + 3O2 → 2Al2O3 5 - 46
b. Al is oxidized because its charge changes from 0 to 3+. charges:
0
0
Al + O2
3+
→
2-
Al2O3
c. O2 is reduced because its charge changes from 0 to 2-. d. O2 is the oxidizing agent because it was the acceptor of electrons (reduced). e. Al is the reducing agent because it donated the electrons to O2.
5.54
Classify the reactions in Problems 5.50 and 5.52 as involving synthesis, decomposition, single replacement or double replacement.
Answer: 5.50 Mg(s) + FeCl2(aq) → MgCl2(aq) + Fe(s) Single replacement. Mg replaces Fe in FeCl2. 5.52 4Al + 3O2 → 2Al2O3 Synthesis. Aluminum and oxygen combine to form the compound aluminum oxide.
5.56
A reaction of iron metal with oxygen gas produces ferric oxide (Fe2O3). Fe + O2 → Fe2O3 a. Balance the reaction equation. b. When Fe2O3 forms, has Fe been oxidized or has it been reduced? c. When Fe2O3 forms, has O2 been oxidized or has it been reduced?
Answer: a. 4 Fe + 3 O2 → 2 Fe2O3 b. Oxidized. When Fe2O3 forms, Fe has gone from neutral to positively charged, Fe3+. c. Reduced. When Fe2O3 forms, oxygen has gone from neutral to negatively charged, O2-.
5 - 47
5.58
1-Butene (CH2=CHCH2CH3) burns in the presence of O2 to produce CO2 and H2O. Write the balanced chemical equation for this reaction.
Answer: C4H8(g) + 6 O2(g) → 4 CO2(g) + 4 H2O(g) C4H8 is the condensed molecular formula for 1-butene, CH2=CHCH2CH3. A reaction that involves burning in the presence of O2 producing CO2 and H2O is a combustion reaction.
5.60
One of the reactions used to develop photographic film is shown below. Hydroquinone, a developer, reacts with silver ions to form silver metal, which gives the black color on a film negative.
OH
O +
2 Ag+
+
2 OH-
OH
+
2 Ag
O Benzoquinone
Hydroquinone
a. Is hydroquinone oxidized or is it reduced? b. Is Ag+ oxidized or is it reduced? c. What is the oxidizing agent? d. What is the reducing agent? Answer: a. Hydroquinone is oxidized because it loses hydrogens when it is converted to benzoquinone. b. Ag+ is reduced to neutral Ag because it gains an electron. c. Ag+ is the oxidizing agent because it removes electrons. d. Hydroquinone is the reducing agent because it donates electrons.
5 - 48
+
2 H2 O
5.62
Draw the missing product of each reaction. a.
CH3 CH2 CH C CH3
H2
+
Pt
CH3 CH3 b.
+ H2
CH3 CH2 C C CH2 CH3
Pt
CH2CH3
CH2 c.
+ H2
Pt
Answer: These are catalytic hydrogenation reactions. H2 acts as a reducing agent and adds across the C=C, forming a saturated compound. a.
CH3 CH2 CH2 CH CH3 CH3
CH3 b.
CH3 CH2 CH CH CH2 CH3 CH2CH3
CH3 c.
5.64
Draw the skeletal structure of the product formed when each alkene is reacted with H2, in the presence of Pt.
a.
b.
5 - 49
Answer:
a.
5.66
b.
Draw the saturated product expected when 4 mol of H2, in the presence of Pt, are reacted with -farnesene, a compound present in the natural wax coating of apples.
-Farnesene Answer: When a compound with a C=C or CC reacts with H2 in the presence of a catalyst (e.g. Pt), the double or triple bonds are hydrogenated forming a saturated (no double or triple bond) compound:
SECTION 5.5
MOLE AND MASS RELATIONSHIP IN REACTIONS
5.68 For the reaction between nitrogen and oxygen to form nitric oxide, beginning with 0.086 mol of N2, N2(g) + O2(g) → 2 NO(g) a. how many moles of O2 are required to completely consume the N2? b. how many moles of NO are obtained when the N2 is completely reacted? Answer: a. Use the mole to mole ratio between N2 and O2 given by the balanced equation as a conversion factor to convert moles of N2 to moles of O2. 0.086 mol N 2 x
1 mol O2 1 mol N 2
= 0.086 mol O2
5 - 50
b. Use the mole to mole ratio between N2 and NO given by the equation as a conversion factor to convert moles of N2 to moles of NO. 0.086 mol N 2 x
5.70
2 mol NO = 0.17 mol NO 1 mol N 2
For the combustion of ethane (CH3CH3), beginning with 0.27 mol of CH3CH3, 2CH3CH3(g) + 7O2(g) → 4CO2(g) + 6H2O(g) a. how many moles of O2 are required to totally consume the CH3CH3? b. how many moles of CO2 are obtained when the CH3CH3 is totally consumed? c. how many moles of H2O are obtained when the CH3CH3 is totally consumed?
Answer: Use the balancing coefficients given in the equation as conversion factors to determine the amount being asked for from the amount given. a. Convert from moles of ethane, CH3CH3, to moles of O2 required:
0.27 mol CH3CH3
7 mol O2 2 mol CH3CH3
= 0.95 mol O2
b. Convert from moles of CH3CH3 to moles of CO2:
0.27 mol CH3CH3
4 mol CO2 2 mol CH3CH3
= 0.54 mol CO2
c. Convert from moles CH3CH3 to moles of H2O:
0.27 mol CH3CH3
5.72
6 mol H 2O 2 mol CH3CH3
= 0.81 mol H 2O
For the reaction in Problem 5.56, a. how many moles of O2 are required to react completely with 39.2 mg of Fe? b. how many grams Fe2O3, are produced from the complete reaction of 14.8 g of
Fe?
5 - 51
Answer: The balanced equation for the reaction in Problem 5.56 is: 4Fe + 3O2 → 2Fe2O3 a. First, convert the mg of Fe to g of Fe. Then, convert g of Fe to mol Fe. Use the molar ratio from the balanced equation to determine the mol of O2 required.
39.2 mg Fe x
10−3 g 1 mol Fe x 1 mg 55.85 g Fe
x
3 mol O2 = 5.26 x 10-4 mol O 2 4 mol Fe
b. First, convert the grams of Fe to mol Fe and then, use the molar ratio between Fe and Fe2O3 to find mol Fe2O3. Use the molar mass of Fe2O3 to convert to grams of Fe2O3.
14.8 g Fe x
5.74
1 mol Fe 2 mol Fe2O3 159.70 g Fe2O3 x x = 21.2 g Fe 2O3 55.85 g Fe 4 mol Fe 1 mol Fe2O3
In the presence of H2 and Pt, propene is reduced to propane. Pt CH3CH2CH3 + CH3CH CH2 H2
Propene
Propane
a. How many moles of H2 are required to completely react with 14.3 mol of propene? b. How many moles of H2 are required to completely react with 0.88 mol of propene? c. How many moles of propane are expected from the complete reaction of 27.7 g of propene? d. How many grams of propane are expected from the complete reaction of 125 g of propene? Answer: a. Convert from moles of propene to moles of H2:
14.3 mol CH3CH=CH2
1 mol H 2 1 mol CH3CH=CH 2
b. Convert from moles of propene to moles of H2:
5 - 52
= 14.3 mol H 2
0.88 mol CH3 CH=CH 2
1 mol H 2 1 mol CH3 CH=CH 2
= 0.88 mol H 2
c. First, convert the mass of propene given to moles of propene using the molar mass of propene:
1 mol CH3CH=CH2 = 0.658 mol CH3CH = CH 2 42.09 g Then, convert moles of propene to moles of propane: 27.7 g CH3CH=CH 2
0.658 mol CH3CH=CH 2
1 mol CH3CH 2CH3 1 mol CH3CH=CH2
= 0.658 mol CH3CH 2CH3
d. First, convert mass of propene to moles of propene:
125 g CH3CH=CH 2
1 mol CH3CH=CH 2 42.09 g
= 2.97 mol CH3CH = CH 2
Then, convert moles of propene to moles of propane:
2.97 mol CH3CH=CH 2
1 mol CH3CH 2CH3 1 mol CH3CH=CH2
= 2.97 mol CH3CH 2CH3
As a last step, convert moles of propane to grams of propane using the molar mass of propane:
2.97 mol CH3CH 2CH3
5.76
44.11 g 1 mol CH3CH 2CH3
= 131 g CH3CH 2CH3
Consider the reaction 2KClO3(s) → 2KCl(s) + 3O2(g) a. How many grams of KCl are produced from the complete reaction of 6.60 mol of KClO3? b. How many grams of KCl are produced from the complete reaction of 47.5 g of KClO3?
5 - 53
Answer: a. First, convert moles of KClO3 to moles of KCl using the balanced equation:
2 mol KCl 2 mol KClO3
6.60 mol KClO3
= 6.60 mol KCl
Then, convert moles of KCl to grams of KCl:
6.60 mol KCl
74.55 g KCl = 492 g KCl 1 mol KCl
b. First, convert mass of KClO3 to moles of KClO3:
47.5 g KClO3
1 mol KClO3 122.55 g
= 0.388 mol KClO3
In the second step, convert moles of KClO3 to moles of KCl using the balanced equation:
0.388 mol KClO3
2 mol KCl 2 mol KClO3
= 0.388 mol KCl
Then, convert the moles of KCl to grams of KCl:
0.388 mol KCl
5.78
74.55 g KCl = 28.9 g KCl 1 mol KCl
An automobile air bag contains sodium azide (NaN3). When an electrical switch triggers a reaction of sodium azide, the rapidly formed nitrogen gas (N2) causes the air bag to inflate. Sodium azide decomposes to form sodium metal and nitrogen gas. The sodium metal then reacts with KNO3 (also present in the air bag) to form K2O(s), Na2O(s), and N2(g). a. Balance the equation for the first reaction. NaN3(s) → Na(s) + N2(g) b. Balance the equation for the second reaction. Na(s) + KNO3(s) → K2O(s) + Na2O(s) + N2(g) c. Identify each of the three products of the second reaction as being an element, an ionic compound, or a covalent compound. d. Name the compounds that are the products of the second reaction. e. If an airbag contains 135 g of sodium azide, how many moles of N2 will be produced by the two reactions that take place when the air bag is triggered?
5 - 54
Answer: a. 2NaN3(s) → 2Na(s) + 3N2(g) b. 10Na(s) + 2KNO3(s) → K2O(s) + 5Na2O(s) + N2(g) c. K2O and Na2O are ionic compounds.
N2 is an element.
d. K2O is potassium oxide and Na2O is sodium oxide. e. First, calculate the moles of N2 formed from the first reaction: 135 g NaN3 ×
1 mol NaN3 3 mol N2 × = 3.11 mol N2 65.02 g NaN3 2 mol NaN3
Calculate the moles of Na formed by the first reaction to determine the moles of N2 formed from the second reaction. 135 g NaN3 ×
1 mol NaN3 2 mol Na × = 2.08 moles Na 65.02 g NaN3 2 mol NaN3
Assuming Na is the limiting reactant in the second reaction, calculate the additional moles of N2 using the moles of Na.
2.08 moles Na ×
1 mol N2 = 0.208 mol N2 10 mol Na
The total moles of N2 formed from the 2 reactions is 3.11 mol N2 + 0.208 mol N2 = 3.32 mol N2
5.80
The local anesthetic Lidocaine can be hydrolyzed as shown in the reaction equation below.
5 - 55
a. How many grams of water are required to completely hydrolyze 0.54 g of lidocaine? b. How many grams of the carboxylic acid hydrolysis product are obtained when 0.54 g of lidocaine are completely hydrolyzed? c. How many grams of the amine hydrolysis product are obtained when 0.54 g of lidocaine are completely hydrolyzed? Answer: a. The balanced equation for the hydrolysis of lidocaine using chemical formulas is C14H22N2O Procaine 0.54 g C14H22N2O ×
+
H2O
→
C6H13NO2 carboxylic acid
1 mol C14H22N2O 1 mol H2O × × 234.4 g C14H22N2O 1 mol C14H22N2O
18.02 g H2O = 0.042 g H2O 1 mol H2O
b. Use the same balanced equation as above. 0.54 g C14H22N2O ×
1 mol C14H22N2O 1 mol C6H13NO2 × × 234.4 g C14H22N2O 1 mol C14H22N2O
131.2 g C6H13NO2 = 0.30 g C6H13NO2 1 mol C6H13NO2
c. Use the same balanced equation as above. 0.54 g C14H22N2O ×
1 mol C14H22N2O 1 mol C8H11N × × 234.4 g C14H22N2O 1 mol C14H22N2O
121.2 g C8H11N = 0.28 g C8H11N 1 mol C8H11N
5 - 56
+
C8H11N amine
SECTION 5.6
5.82
CALCULATING THE YIELD OF A REACTION
Mercury reacts with oxygen to produce mercury(II) oxide. Hg(l) + O2(g) → HgO(s) a. Balance the reaction equation. b. What is the theoretical yield (in grams) of HgO if 254 g of Hg are reacted with 254 g of O2?
Answer: a. 2 Hg(l) + O2(g) → 2 HgO(s) b. The theoretical yield in a chemical reaction is the maximum amount of product that can be obtained when the limiting reactant is completely consumed. It is calculated from the initial amount of the limiting reactant. The first step in this calculation is to determine which of the given reactants is the limiting one. First calculate the number of moles of each of the reactants:
254 g Hg
1 mol Hg 200.59 g
= 1.27 mol Hg
254 g O2
1 mol O 2 32.00 g
= 7.94 mol O 2
Determine the number of moles of O2 required to react completely with all of the Hg:
1.27 mol Hg
1 mol O2 2 mol Hg
= 0.635 mol O2
A total of 0.635 moles of O2 are required and there are 7.94 mol of O2 available. All of the Hg can react because there is sufficient O2. Therefore, Hg is the limiting reactant. Now, calculate the theoretical yield in grams of HgO based on the initial amount of Hg:
1.27 mol Hg
2 mol HgO = 1.27 mol HgO 2 mol Hg 5 - 57
1.27 mol HgO
5.84
216.59 g 1 mol HgO
= 275 g HgO
Aluminum and oxygen react to form aluminum oxide. 4 Al + 3 O2 → 2 Al2O3 a. What is the limiting reactant if 15.0 g Al are reacted with 15.0 g of O2? b. What is the theoretical yield of Al2O3, in grams? c. If 8.8 g of Al2O3 are obtained, what is the percent yield?
Answer: a. Al is the limiting reactant. To determine the limiting reactant, first calculate the number of moles of each of the reactants:
15.0 g Al
1 mol Al = 0.556 mol Al 26.98 g
15.0 g O2
1 mol O 2 32.00 g
= 0.469 mol O 2
Determine the number of moles of O2 required to react completely with all of the Al:
0.556 mol Al
3 mol O2 4 mol Al
= 0.417 mol O2
A total of 0.417 mole of O2 is required and there is 0.469 mol of O2 available. All of the Al can react because there is sufficient O2. Therefore, Al is the limiting reactant. b. To calculate the theoretical yield in grams of Al2O3 based on the initial amount of Al:
0.556 mol Al
2 mol Al2O3 4 mol Al
0.278 mol Al2O3
= 0.278 mol Al2O3
101.96 g 1 mol Al2O3
c. The percent yield is:
5 - 58
= 28.3 g Al2O3
% yield =
5.86
8.8 g Al2O3 28.3 g Al2O3
x 100% = 31%
The toxic effects of methanol (CH3OH) are due to its enzymatic conversion into formaldehyde (CH2O) in the liver.
O
OH H C H
+
NAD+
enzyme
H C H
+
NADH +
H Methanol
Formaldehyde
a. In this reaction is methanol oxidized or is it reduced? b. What theoretical yield of formaldehyde (in grams) is expected from the reaction of 10.0 g of methanol? (Assume that methanol is the limiting reactant.) Answer: a. Oxidized. Methanol is oxidized in this reaction because it loses hydrogen. O enzyme H C H CH3OH b.
NAD+
NADH
+
H+
b. The theoretical yield in grams of formaldehyde can be calculated as follows:
10.0 g CH3OH
5.88
1 mol CH3OH = 0.312 mol CH3OH 32.05 g
0.312 mol CH3OH
1 mol CH 2O 1 mol CH3OH
0.312 mol CH 2O
30.03 g = 9.37 g H 2CO 1 mol CH2O
= 0.312 mol CH 2O
Butyl alcohol (C4H10O) reacts with O2 to form carbon dioxide and water. a. Write a balanced chemical equation for this reaction. b. Butyl alcohol has a density of 0.810 g/mL at 20 °C. If 12.5 mL of butyl alcohol react with 1.2 mol of O2, what is the limiting reactant?
5 - 59
H+
c. For the reaction in part b, what is the theoretical yield (in moles) of carbon dioxide? d. If a 61% yield of carbon dioxide is obtained from the reaction in part b, how many moles of carbon dioxide were obtained? Answer: a. C4H10O + 6O2 → 4CO2 + 5H2O b. C4H10O is the limiting reactant. First, calculate the moles of butyl alcohol in 12.5 mL: 12.5 mL C4H10O ×
0.810 g C4H10O 1 mol C4H10O × = 0.137 mol C4H10O 1 mL C4H10O 74.14 g C4H10O
According to the balanced equation above, for every 1 mol C4H10O, 6 mol of O2 are required. Calculate the mol O2 required to react with all of the 0.137 mol C4H10O. 0.137 mol C4H10O ×
6 mol O2 = 0.822 mol O2 1 mol C4H10O
Since the amount of O2 available (1.2 mol) is more than that required to react with all of the C4H10O, O2 must be in excess and the C4H10O is the limiting reactant. c. Use the amount of the limiting reactant, C4H10O, to calculate the theoretical yield of carbon dioxide in moles. 0.137 mol C4H10O × d.
4 mol CO2 =0.548 mol CO2 1 mol C4H10O
0.344 mol CO2
percent yield =
actual yield × 100% = 61% 0.548 mol CO2
Solving for the actual yield gives 0.344 mol CO2. Or, actual yield = (0.61 x 0.548 mol CO2) = 0.344 mol CO2
5.90
The artificial sweetener Aspartame cannot be used for baking because, when heated, it undergoes hydrolysis to yield three products (two amino acids and one alcohol) that are not sweet. 5 - 60
a. Balance this reaction equation. b. If 83.5 mg of aspartame is reacted with an excess of water, what is the theoretical yield (in grams) of each product? c. A 35.2% yield of each product corresponds to how many grams of each product for the reaction in part b? Answer: a. The balanced equation for the hydrolysis of aspartame using chemical formulas is C14H18N2O5 + 2H2O → HOCH3 Aspartame alcohol b.
C4H7NO4 amine
+
C9H11NO2
+
amine
10-3 g C14H18N2O5 1 mol C14H18N2O5 83.5 mg C14H18N2O5 × × × 1 mg C14H18N2O5 294.3 g C14H18N2O5
5 - 61
1 mol C4H7NO4 133.1 g C4H7NO4 × = 0.0378 g C4H7NO4 1 mol C14H18N2O5 1 mol C4H7NO4 83.5 mg C14H18N2O5 ×
10-3 g C14H18N2O5 1 mol C14H18N2O5 × × 1 mg C14H18N2O5 294.3 g C14H18N2O5
1 mol C9H11NO2 165.2 g C9H11NO2 × = 0.0469 g C9H11NO2 1 mol C14H18N2O5 1 mol C9H11NO2 83.5 mg C14H18N2O5 ×
10-3 g C14H18N2O5 1 mol C14H18N2O5 × × 1 mg C14H18N2O5 294.3 g C14H18N2O5
1 mol CH4O 32.04 g CH4O × = 0.00909 g CH4O 1 mol C14H18N2O5 1 mol CH4O c. actual yield = (0.352 x 0.0378 g C4H7NO4) = 0.0133 g C4H7NO4 actual yield = (0.352 x 0.0469 g C9H11NO2) = 0.0165 g C9H11NO2 actual yield = (0.352 x 0.00909 g CH4O) = 0.00320 g CH4O
SECTION 5.7
FREE ENERGY AND REACTION RATE
5.92 Which of the reactions are spontaneous? a. 3C2H2(g) → C6H6(g) G = -119.0 kcal/mol b. fructose 1,6-diphosphate + ADP → fructose 6-phosphate + ATP G = 3.4 kcal/mol Answer: For a spontaneous reaction, G must have a negative value. For a nonspontaneous reaction, G has a positive value. a. This is a spontaneous reaction because the G value is negative. b. This is a nonspontaneous reaction because the G value is positive.
5.94
a. An exothermic reaction takes place in a beaker. What will you feel if you touch the beaker? b. An endothermic reaction takes place in a beaker. What will you feel if you touch the beaker?
Answer: 5 - 62
a. The beaker will start to feel warm as heat is released in an exothermic reaction. b. The beaker will start to feel cold as heat is absorbed from the surrounding by the endothermic reaction.
5.96
Draw an energy diagram for the reaction in Problem 5.91b. Label the x- and yaxis, reactance, products and ΔG.
Answer:
Energy
2HI(g)
H2(g) + I2(g)
ΔG = 0.6 kcal/mol
Progress of reaction
5.98
a. Draw an energy diagram for the spontaneous reaction A → B b. Draw an energy diagram for the nonspontaneous reaction C → D
Answer:
Energy
a. For a spontaneous reaction, G is negative:
A
B
Progress of reaction
5 - 63
b. For a nonspontaneous reaction, G is positive:
D
Energy
C Progress of reaction
5.100 What is the likely effect on the rate of a reaction if a. reactant concentration is decreased? b. temperature is decreased? c. a catalyst is added? Answer: a. Reaction rate decreases. Molecules must collide in order to react together. The fewer that are present, the less likely it is for collisions to occur and the reaction will be slower. Decreasing the concentration means fewer molecules in a given volume and, therefore, fewer collisions. b. Reaction rate decreases. Molecules must collide in order to react. The faster they are moving, the more likely it is for collisions to occur and the reaction will be faster. Also, activation energy is required for a reaction to occur and the higher the temperature the more energy that is available. Decreasing the temperature means fewer molecules colliding at one time and less energy available to ensure reaction. c. Reaction rate increases. A catalyst will lower the activation energy and therefore increase the likelihood of a reaction when molecules collide. This means more success whenever the molecules collide, which speeds up the reaction rate.
HEALTH LINK
ANTISEPTICS AND OXIDATION
5.102 When present in ointments and creams for treating acne, the oxidizing agent benzoyl peroxide (C14H10O4) is safely handled. Pure benzoyl peroxide is quite 5 - 64
hazardous to deal with, however. Any heat source can cause it to explode as the compound undergoes rapid oxidation. Complete the balancing of the equation for the combustion of benzoyl peroxide. 2 C14H10O4 + O2 → CO2
+ H2O
Answer: 2 C14H10O4 + 29 O2 → 28CO2
+ 10H2O
Place 28 (2 x 14) in front of the C on the right to make the number of C atoms equal. Next, place 10 in front of the H2O to make 20 H atoms on the left to equal the 2 x 10 = 20 H atoms on the right. Finally, count the O atoms on the right. 28 x 2 = 56 and 10 x 1 = 10 for a total of 66 O atoms on the right. Since there are only 8 (2 x 4) O atoms in the first compound on the left, the remaining O atoms must be balanced using O2 molecules. Place 29 in front of the O2 mole to make 58 O atoms [8 + 58 (on left) = 56 + 10 (on right)].
HEALTH LINK
CARBONIC ANHYDRASE
5.104 Only 5% of the CO2 that moves from cells into blood is carried as dissolved CO2. What happens to the other 95%? Answer: Approximately 1/3 of the CO2 attaches to hemoglobin and the remaining part of the 95% is converted into carbonic acid in a hydration reaction catalyzed by the enzyme carbonic anhydrase.
LEARNING GROUP PROBLEMS 5.106 a. When reacted with water and H+, the ester reacts to form a carboxylic acid. Write a balanced chemical equation for the reaction.
b. Which type of reaction is this: hydration, hydrogenation, hydrolysis, or elimination?
5 - 65
c. How many grams of water are required to completely react with 15.8 g of the ester? d. If 52.7 g of ester are reacted with 13.9 g of water, what is the theoretical yield of carboxylic acid? e. If a 76.2% yield is obtained from the reaction in part d, how many grams of carboxylic acid are obtained? Answer: a. 2
2
b. hydrolysis c.
15.8 g ester
1 mol ester 222.26 g ester
d. 52.7 g ester
2 mol H 2O 1 mol ester
1 mol ester 222.26 g ester
13.9 g H 2O
18.02 g H 2O = 2.56 g H 2O 1 mol H 2O
= 0.237 mol ester
1 mol H2O 18.02 g H2O
= 0.771 mol H 2O
Calculate moles of H2O required to react with the amount of ester present to determine the limiting reactant:
0.237 mol ester
2 mol H2O 1 mol ester
= 0.474 mol H2O
Because there is an excess of H2O to react with the ester, the ester is the limiting reactant. The theoretical yield is calculated based on the amount of the ester:
0.237 mol ester
1 mol carboxylic acid 1 mol ester
166.14 g carboxylic acid 1 mol carboxylic acid 5 - 66
= 39.4 g carboxylic acid
e. 76.2 % =
actual yield 100% 39.4 g
actual yield = 39.4 g
76.2 % = 30.0 g carboxylic acid 100%
5 - 67
Chapter 6 Gases, Solutions, Colloids, and Suspensions SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
6.1
a. For the gas shown below, has the volume increased, decreased, or remained the same? Decreased as indicated by the lower position of the piston. b. Has the pressure inside the container increased, decreased, or remained the same? Increased. The particles occupy a smaller volume and will collide with the inner walls with higher frequency (as long as the temperature remains the same). c. Have the number of moles increased, decreased, or remained the same? Remained the same. The number of moles of gas particles should stay the same as long as it can be assumed that the container is tightly sealed.
6.3
Define STP. STP or standard temperature and pressure is defined to be 1 atm (or 14.7 psi or 760 torr) and 0°C
6.5
A pressure of 9.2 psi is how many Use the following information to set-up the conversion calculations: 1 atm = 760 torr or mmHg = 14.7 psi a. atmospheres?
6-1
9.2 psi
1 atm 14.7 psi
= 0.63 atm
760 torr 14.7 psi
= 480 torr
b. torr?
9.2 psi
6.7
A pressure of 814 torr is how many Use the following information to set-up the conversion calculations: 1 atm = 760 torr or mmHg = 14.7 psi a. atmospheres?
1 atm = 1.07 atm 760 torr
814 torr x b. psi?
14.7 psi = 15.7 psi 760 torr
814 torr x
6.9
In the SI measurement system, the unit of pressure is called the kilopascal (kPa). One atmosphere equals 101.3 kPa. a. Convert 1.29 atm into kilopascals. 1.29 atm
x
101.3 kPa 1 atm
= 131 kPa
b. Convert 87.2 kPa into atmospheres. 87.2 kPa
x
1 atm 101.3 kPa
= 0.861 atm
c. Convert 612 torr into kilopascals. 612 torr
x
1 atm 760 torr
x
101.3 kPa 1 atm
6-2
= 81.6 kPa
6.11
Explain how a barometer works. A barometer is an instrument that measures atmospheric pressure. It is constructed by inverting a glass tube containing mercury into a dish that also contains mercury. The height of the mercury inside the tube will go up and down depending on how much atmospheric pressure is pushing down on the mercury in the dish. When the atmospheric pressure is equal to 1 atm, the height of the mercury in the tube will be 760 mm.
6.13
At an atmospheric pressure of 760 torr and a temperature of 0○C, the mercury level in the right arm of the manometer pictured in Figure 6.4b is 5 mm higher than the mercury in the left arm. What is the gas pressure inside the flask? 760 torr + 5 torr = 765 torr The first thing to note is that the manometer pictured in Figure 6.4b is an openended manometer. This means that the gas pressure in the gas bulb is pushing against the outside atmospheric pressure of 760 torr. Since the mercury moved up on the right arm this tells you that the pressure of the gas in the bulb is 5 mmHg greater than atmospheric pressure and this much pressure can be added to the atmospheric pressure to get the pressure of the gas. The second thing to note is that the atmospheric pressure is in torr, but the torr and mmHg are interchangeable units.
6.15
Use Table 6.1 to estimate the boiling point of water at a pressure of 920 torr. 105C The boiling point of a liquid is the temperature at which the vapor pressure equals the pressure of the atmosphere above it. Table 6.1 gives the vapor pressure of water at various temperatures. When the pressure of the atmosphere above it is 920 torr, water will boil at a temperature of about 105C.
6.17
The record for the highest barometric pressure ever measured on earth is 32.06 mmHg, which was measured in Mongolia in 2001. Estimate the boiling point of water at this pressure. 30C The boiling point of a liquid is the temperature at which the vapor pressure equals the pressure of the atmosphere above it. From Table 6.1, 32.06 mmHg or torr is closest to 31.82 torr which is the vapor pressure of water when its temperature is 30C. When the pressure of the atmosphere above it is 32.06 torr, water will boil at a temperature of about 30C.
6-3
6.19
At a pressure of 760 torr a balloon has a volume of 1.50 L. If the balloon is put into a container and the pressure is increased to 2500 torr (at constant temperature), what is the new volume of the balloon? Assign values to the variables. P1 = 760 torr P2 = 2500 torr V1 = 1.50 L V2 = ? (this is what you are asked to calculate) Select the gas law that has these variables ONLY. In this case it will be Boyle’s Law; P1V1 = P2V2 . Rearrange the equation so that the required variable is by itself. Replace the variables in the equation with the values you identified above and solve for the one missing.
6.21
V2 =
P1V1 P2
V2 =
760 torr x 1.50 L = 0.46 L 2500 torr
At a temperature of 30○C, a balloon has a volume of 1.50 L. If the temperature is increased to 60○C (at constant pressure), what is the new volume of the balloon? In gas law problems the temperature must be converted to Kelvin. K = oC + 273.15 = 30○C + 273 = 303 K K = oC + 273.15 = 60○C + 273 = 333 K Assign values to the variables. T1 = 303 K T2 = 333 K V1 = 1.50 L V2 = ? (this is what you are asked to calculate) Select the gas law that has these variables ONLY. In this case it will be Charles’ Law; V1/ T1 = V2 / T2. Rearrange the equation so that the required variable is by itself. Replace the variables in the equation with the values you identified above and \solve for the one missing.
V2 =
V1T2 T1
V2 =
333 K x 1.50 L = 1.65 L 303 K
6-4
6.23
At a temperature of 30○C, a gas inside a 1.50 L metal canister has a pressure of 760 torr. If the temperature is increased to 60○C (at constant volume), what is the new pressure of the gas? In gas law problems the temperature must be converted to kelvins. K = oC + 273.15 = 30○C + 273 = 303 K K = oC + 273.15 = 60○C + 273 = 333 K Assign values to the variables. T1 = 303 K T2 = 333 K P1 = 760 torr P2 = ? (this is what you are asked to calculate) Note: even though the volume was stated it is not included as a variable because the problem states that it remained constant. Select the gas law that has these variables ONLY. In this case it will be GayLussac’s Law; P1/ T1 = P2 / T2. Rearrange the equation so that the required variable is by itself. Replace the variables in the equation with the values you identified above and solve for the one missing.
6.25
P2 =
P1T2 T1
V2 =
760 torr x 333 K = 835 torr 303 K
A 2.0 L balloon contains 0.35 mol of Cl2(g). At constant pressure and temperature, what is the new volume of the balloon if 0.20mol of gas is removed? Identify the variables. Note that 0.20 mol is the amount removed NOT final number of moles. n2 = 0.35 – 0.20 = 0.15 mol n1 = 0.35 mol V1 = 2.0 L
n2 = 0.15 mol V2 = ? (this is what you are asked to calculate)
Select the gas law that has these variables ONLY. In this case it will be Avogadro’s Law; V1 / n1 = V2 / n2. Rearrange the equation so that the required variable is by itself. Replace the variables in the equation with the values you identified above and solve for the one missing.
6-5
6.27
V2 =
V1n 2 n1
V2 =
2.0 L x 0.15 mol = 0.86 L 0.35 mol
A balloon with a volume of 1.50 L is at a pressure of 760 torr and a temperature of 30○C. If the balloon is put into a container and the pressure is increased to 2500 torr and the temperature is raised to 60 ○C, what is the new volume of the balloon? First convert oC to K. K = oC + 273.15 = 30○C + 273 = 303 K K = oC + 273.15 = 60○C + 273 = 333 K Assign values to the variables. T1 = 303 K T2 = 333 K P1 = 760 torr P2 = 2500 V1 = 1.50 L V2 = ? (this is what you are asked to calculate) Select the gas law that has all these variables. In this case it will be the Combined Gas Law; P1V1/ n1T1 = P2V2 / n2T2. Rearrange the equation so that the required variable is by itself. Replace the variables in the equation with the values you identified above and solve for the one missing. Note that it is not necessary to include n1 and n2 since the number of moles is constant.
6.29
V2 =
P1V1T2 T1P2
V2 =
760 torr x 1.50 L x 333 K = 0.50 L 2500 torr x 303 K
A 575 mL metal can contains 2.50 x 10-2 mol of He at a temperature of 298K. What is the pressure (in atm) inside the can? Is this pressure greater than or less than standard atmospheric pressure? Assign values to the variables. T = 298 K V = 0.575 L
P= ? n = 2.50 x 10-2 mol
Select the gas law that has just one of each variable. In this case it will be the Ideal Gas Law; PV = n R T. R is the gas constant (0.0821 L atm /mol K). Rearrange the equation so that the required variable is by itself. Replace the
6-6
variables in the equation with the values you identified above and solve for the one missing. P =
nRT V
Note: since the R constant has the units atm and L in it, before solving for P the volume unit must be converted to L.
1 x 10-3 L 575 mL x = 0.575 L 1 mL P =
2.50 x 10-2 mol x (0.0821 L atm/mol K) x 298 K = 1.06 atm 0.575 L
1.06 atm is greater than standard atmospheric pressure.
6.31
A 250.0 mL flask contains 0.350 mol of O2 at 40°C. Use the ideal gas equation to calculate the pressure, which is the only unknown property of the gas. Use the conversion equivalences: 1 atm = 760 torr = 14.7 psi to convert the pressure from one unit to another. a. What is the pressure in atm? PV = nRT P=?
V = 250.0 mL = 2.500 x 10-1 L
n = 0.350 mol O2
T = 40C + 273.15 = 310 K
PV = nRT therefore P = P =
nRT V
0.350 mol x 0.0821 L atm/mol K x 310 K = 36 atm 2.500 x 10-1 L
b. What is the pressure in torr? 1 atm = 760 torr
36 atm
760 torr = 2.7 x 104 torr 1 atm
6-7
c. What is the pressure in psi? 1 atm = 14.7 psi
36 atm
6.33
14.7 psi = 530 psi 1 atm
A 250 mL flask contains He at a pressure of 760 torr and a temperature of 25oC. What mass of He is present? First, use the ideal gas equation to calculate the moles of He present. Then, use the molar mass of He to convert to grams of He. P = 760 torr = 1.0 atm, V = 250 mL = 0.25 L, T = 25C + 273.15 = 298 K
PV = nRT
therefore n =
0.010 mol He
6.35
PV RT
=
1.0 atm x 0.25 L 0.0821 L atm/mol K x 298 K
= 0.010 mol He
4.003 g = 0.040 g He 1 mol
a. How many moles of Ar are present in a 2.0 L flask that has a pressure of 1.05 atm at a temperature of 25°C? P = 1.05 atm, V = 2.0 L, T = 25C + 273.15 = 298 K
PV = nRT n =
therefore n =
PV RT
1.05 atm x 2.0 L = 0.086 mol Ar 0.0821 L atm/mol K x 298 K
b. What is the mass of this Ar?
0.086 mol Ar
6.37
39.95 g = 3.4 g Ar 1 mol
A sample of a gas at a temperature of 25.0°C has a pressure of 815 torr and occupies a volume of 9.92 L. a. Use Boyle’s law to calculate the new pressure if the temperature is held constant and the volume is decreased to 5.92 L.
6-8
Boyle’s law: In this problem:
P1V1 = P2V2 P1 = 815 torr V1 = 9.92 L
Solving for P2: P1V1 P2 = = V2
P2 = ? V2 = 5.92 L
815 torr x 9.92 L 5.92 L
= 1.37 x 103 torr
b. Use Gay-Lussac’s law to calculate the new pressure if the volume is held constant and the temperature is increased to 125.0°C.
P1 T1
Gay-Lussac’s law:
P2 T2
=
In this problem: P1 = 815 torr T1 = 25.0°C + 273.15 = 298.2 K Solving for P2: P1 P2 = x T2 T1
P2 = ? T2 = 125.0°C + 273.15 = 398.2 K
815 torr 298.2 K
=
x
398.2 K = 1.09 x 103 torr
c. Use Charles’ law to calculate the new volume if the pressure is held constant and the temperature is increased to 125.0°C. Charles’ law:
V1 T1
=
V2 T2
In this problem: V1 = 9.92 L T1 = 25.0°C + 273.15 = 298.2 K Solving for P2: V1 V2 = x T2 T1
=
V2 = ? T2 = 125.0°C + 273.15 = 398.2 K
9.92 L 298.2 K
x
398.2 K = 13.2 L
d. Use the combined gas law to calculate the new pressure if the temperature is increased to 125.0°C and the volume is decreased to 5.92 L. Combined gas law:
P1 V1 T1
=
P2 V2 T2
6-9
In this problem: P1 = 815 torr V1 = 9.92 L T1 = 25.0°C + 273.15 = 298.2 K Solving for P2: P1V1 P2 = x T1
T2 V2
=
P2 = ? V2 = 5.92 L T2 = 125.0°C + 273.15 = 398.2 K
815 torr x 9.92 L 298.2 K
398.2 K 5.92 L
x
= 1.82 x 103 torr
e. Use the ideal gas law to calculate the number of moles of gas that are present. Ideal gas law: PV = nRT First, we have to convert the pressure from torr to atm: 1 atm 815 torr x = 1.07 atm 760 torr Now, using the ideal gas law to solve for the number of moles, n:
n
6.39
=
PV RT
=
1.07 atm x 9.92 L (0.0821 L atm /mol K) x 298.2 K
= 0.434 mol
The cover of the rock band Led Zeppelin’s first album pictured the Hindenburg, a hydrogen-filled dirigible, going down in flames. At the temperature of 25°C and a pressure of 760 torr, the Hindenburg held 7.062 x 106 cubic feet of hydrogen gas. How many grams of hydrogen did this represent? Use the ideal gas law to calculate the moles of hydrogen gas (H2) using the given information. Then, use the molar mass of H2 to calculate the mass of the hydrogen gas in the dirigible. First, convert the quantities given to units appropriate for the ideal gas law equation. T = 25°C + 273 = 298 K P = 760 torr
x
1 atm 760 torr
= 1.0 atm
12 in V = 7.062 x 106 ft 3 x 1 ft
3
2.54 cm x 1 in
6 - 10
3
x
1 mL 10-3 L x = 2.000 x 108 L 3 1 cm 1 mL
Using the ideal gas law, calculate the moles of H2.
n
=
PV RT
1.0 atm x 2.000 x 108 L (0.0821 L atm /mol K) x 298 K
=
= 8.2 x 106 mol
Calculate the grams of H2 gas using the molar mass of H2. 8.2 x 106 mol
6.41
x
2.0 g 1 mol
= 1.6 x 107 g H2 gas
True or false? Ten liters of helium gas at STP has the same mass as ten liters of neon gas at STP. Explain. False. The same number of moles of each gas is present, but the molar mass of each is different.
6.43
Use concepts discussed in this chapter to explain why an empty plastic milk bottle collapses if the air is pumped out of it. The air around us exerts pressure on everything with which it is in contact, including an empty plastic milk bottle. When open to the atmosphere around it, the pressure of the air inside the bottle exerts a pressure on the inside walls of the empty bottle equal to the pressure exerted by the atmosphere on the outside of the bottle. However, when air inside is removed, that balance is lost. The pressure on the outside of the empty bottle pushes the bottle inward and bottle collapses.
6.45
A mixture of gases contains 0.75 mol of N2, 0.25 mol of O2, and 0.25 mol of He. a. What is the partial pressure of each gas (in atm and in torr) in a 25 L cylinder at 350 K? The partial pressure of a gas can be calculated as though it is the only gas in the container. Assign values to the variables. For calculating the partial pressure of N2: T = 350 K V = 25 L
P= ? n = 0.75 mol
Select the gas law that has just one of each variable. In this case it will be the Ideal Gas Law; PV = n R T. R is the gas constant (0.0821 L atm /mol K). Rearrange the equation so that the required variable is by itself. Replace the
6 - 11
variables in the equation with the values you identified above and solve for the one missing.
PN2 =
nRT 0.75 mol x 0.0821 L atm/ mol K x 350 K = V 25 L
= 0.86 atm
PN2 = 0.86 atm for N2 Convert to torr:
PN2 = 0.86 atm
x
760 torr 1 atm
= 6.5 x 102 torr
PN2 = 6.5 x 102 torr for N2 Now repeat the calculation for O2. Since the temperature and volume are the same, you can use the same set-up from above and simply insert the correct number of moles for O2.
PO2 =
nRT 0.25 mol x 0.0821 L atm/ mol K x 350 K = V 25 L
= 0.29 atm
PO2 = 0.29 atm for O2 Convert to torr:
PO2 = 0.29 atm
x
760 torr 1 atm
= 2.2 x 102 torr
PO2 = 2.2 x 102 torr for O2 Now repeat the calculation for He. Since the temperature and volume are the same, you can use the same set-up from above and simply insert the correct number of moles for He.
PHe =
nRT 0.25 mol x 0.0821 L atm/ mol K x 350 K = V 25 L
= 0.29 atm
PHe = 0.29 atm for He Convert to torr:
PHe = 0.29 atm
PHe = 2.2 x 102 torr for He b. What is the total pressure?
6 - 12
x
760 torr 1 atm
= 2.2 x 102 torr
Dalton’s Law states that the total pressure for all gases in a container is equal to the sum of their partial pressures. Therefore the total pressure is found by adding up the partial pressures calculated above. Ptotal = PN 2 + PO2 + PHe
Ptotal = 0.86 atm + 0.29 atm + 0.29 atm = 1.44 atm 1.44 atm; 1.09 x 103 torr
6.47
For the mixture of gases in Problem 6.45 what are the partial pressures and the total pressure if 0.50 mol of CO2(g) is added? First, since gas pressures are independent of each other, the partial pressures of the first three do not change. Calculate the partial pressure of CO2. Then add this pressure to the partial pressures of the other gases to find the total pressure.
PCO2 =
0.50 mol x 0.0821 L atm/ mol K x 350 K 25 L
= 0.58 atm
PN2 = 0.86 atm N2, 6.5 x 102 torr PO2 = 0.29 atm O2, 2.2 x 102 torr PHe = 0.29 atm He, 2.2 x 102 torr PCO2 = 0.58 atm CO2, 4.4 x 102 torr Total Pressure = 2.02 atm = 1.54 x 103 torr
6.49
Scuba divers sometimes breathe a gas mixture called trimix, which consists of helium, oxygen, and nitrogen gases. If a scuba tank with a volume of 2.5 L, a pressure of 2700 psi, and a temperature of 45°F contains 27% 02, 12% He, and 61% N2,
a. what is the partial pressure of each gas? The partial pressure of each gas can be determined using the percentage of the total pressure given for each gas. First, convert psi to atm for further calculations:
1 atm = 184 atm 14.7 psi The partial pressure of each gas can then be calculated using their respective percentages: 2700 psi
x
6 - 13
PO2
= 0.27 x 184 atm = 50. atm
PHe
= 0.12 x 184 atm = 22 atm
PN2
= 0.61 x 184 atm = 110 atm
b. how many grams of each gas are present? To determine the grams of each gas, calculate the number of moles then convert from moles to grams using the molar mass of each gas. First, convert the temperature from °F to K: 45F - 32 = 7.2C 1.8
7.2C + 273.15 K = 280.4 K
Then, use the ideal gas law to calculate the number of moles given the other know values: PO2 V 50. atm x 2.5 L n O2 = = = 5.4 mol O 2 RT (0.0821 L atm /mol K) x 280.4 K Similar calculations for He and N2 yield: nHe = 2.4 mol He nN2 = 12 mol N2 Convert moles to grams by using their respective molar masses:
32.0 g = 170 g O 2 1 mol 4.0 g grams of He = 2.4 mol x = 9.6 g He 1 mol 28.0 g grams of N 2 = 12 mol x = 340 g N 2 1 mol
grams of O 2 = 5.4 mol x
c. how many molecules of 0 and N and atoms of He are present? 6.02 x 1023 molecules = 3.2 x 1024 molecules O 2 1 mol 6.02 x 1023 atoms atoms of He = 2.4 mol x = 1.4 x 1024 molecules He 1 mol 6.02 x 1023 molecules molecules of N 2 = 12 mol x = 7.2 x 1024 molecules N 2 1 mol molecules of O 2 = 5.4 mol x
6 - 14
6.51
Define the terms solute and solvent. The solute is the component dissolved in the solution. The solvent is the component in the greatest amount in the solution.
6.53
Define the terms pure substance and mixture. A pure substance is made up of just one element or compound. A mixture is a combination of two or more pure substances.
6.55
You add sugar to a pan of cold water and obtain a saturated solution. When you heat the pan on the stove, all of the sugar dissolves. Explain why this happens. The solubility of a solute particle in a given solvent depends on temperature. For gas solutes, the higher the temperature is, the lower their solubility is. For liquid and solid solutes, increasing the temperature increases the solubility. The solubility of sugar, a solid solute, in water is lower at room temperature than at a higher temperature. Thus, when a saturated solution of sugar is heated, the increase in temperature results in higher solubility and all of the sugar dissolves.
6.57
Give an example of a solution in which a. the solute is a solid and the solvent is a liquid. Examples will vary. Some possible answers are: Salt in water, sugar in tea, sugar in coffee. b. the solute is a gas and the solvent is a gas. Examples will vary. Some possible answers are: CO2 in air, O2 in air, He in N2 in a scuba tank.
6.59
At a pressure of 1 atm, in which is O2 gas more soluble, 0oC water 90oC water? 0 °C water. The solubility of a gas is higher when the solution is at a lower temperature.
6.61
a. If 45.0 g of NaCl are added to 100g of water at 20oC, will all of the NaCl dissolve? No. Only 35.9 g of NaCl dissolves per 100 g of water at this temperature. 6 - 15
b. Is the resulting mixture homogeneous or heterogeneous? Heterogeneous because some of the NaCl remains undissolved in the solid form.
6.63
On a hot day, it is often easier to catch fish by casting your line into a deep, cool part of a lake than into a shallow, warm spot. One reason that fish gather in cool water may be related to the levels of oxygen dissolved in the water. Explain. In the cooler part of the lake, more oxygen would remain dissolved because in general, the solubility of a gaseous substance increases with decreasing temperature.
6.65
Predict whether each ionic compound is soluble or insoluble in water. Use the general solubility rules for ions given in the textbook.
6.67
a. (NH4)2SO4
Soluble
b. K2SO4
Soluble
c. CaCO3
Insoluble
d. NaNO3
Soluble
When aqueous copper(II) sulfate is mixed with aqueous sodium sulfide, a precipitate forms. Write a balanced equation for this reaction. Before attempting to balance the equation make sure that you write the correct formulas. Copper(II) sulfate is CuSO4 (equal numbers of Cu2+ and SO42- to give a neutral compound). Sodium sulfide is Na2S (twice as many Na+ as S2- to give a neutral compound.) Now that you have the reactants identified, you need to predict the products. A double replacement reaction can occur when two aqueous solutions of ionic compounds are mixed together. In this case Cu2+ combines with S2- to form CuS and Na+ combines with the SO42- ion to form Na2SO4. Since you are told that a precipitate forms and Table 6.3 shows Na2SO4 to be soluble, the CuS must be the solid formed. Now we can write the equation. CuSO4(aq) + Na2S(aq) →
CuS(s) + Na2SO4(aq)
6 - 16
We must then determine if the equation is balanced. Start by counting each atom. reactant atoms
product atoms
1 Cu atom
1 Cu atom
2 S atoms
2 S atoms
4 O atoms
4 O atoms
2 Na atoms
2 Na atoms
In this case, all of the atoms are balanced on both sides of the equation. The equation is correct as written. CuSO4(aq) +
6.69
Na2S(aq) → CuS(s) + Na2SO4(aq)
Complete and balance each precipitation reaction. All of the reactions involve mixing two aqueous solutions of ionic compounds. In each case a double replacement reaction can occur. Follow the same basic procedure for each of the equations below. Recombine the reactant cations and anions to form new ionic compounds. Refer to Table 6.3 to see if either compound formed is insoluble. If it is insoluble be sure to write (s) with the formula. Use the symbol (aq) for those that are water soluble. Then, write the balanced equation. a. CaCl2(aq) + Li2CO3(aq) → Ca2+ reacts with CO32- to form CaCO3 and Li+ reacts with Cl- to form LiCl. The solubility Table 6.3 indicates that calcium carbonate is insoluble but lithium chloride is soluble. CaCl2(aq) + Li2CO3(aq) → CaCO3(s) + LiCl(aq) Now balance the reaction. (NOTE: do not start balancing until you have written the formulas of all reactants and products.) CaCl2(aq) + Li2CO3(aq) → CaCO3(s) + LiCl(aq) Start by counting each atom. reactant atoms
product atoms
1 Ca atom
1 Ca atom
2 Cl atoms
1 Cl atoms
6 - 17
3 O atoms
3 O atoms
2 Li atoms
1 Li atoms
1 C atom
1 C atom
There are 2 Cl atoms on the left and 1 on the right, so multiply LiCl by 2 (2LiCl) and recount. reactant atoms
product atoms
1 Ca atom
1 Ca atom
2 Cl atoms
2 Cl atoms
3 O atoms
3 O atoms
2 Li atoms
2 Li atoms
1 C atom
1 C atom
This makes the atom count the same on both sides, so the equation is balanced. CaCl2(aq) + Li2CO3(aq) → CaCO3(s) + 2LiCl(aq) b. Pb(NO3)2(aq) +
NaCl(aq) →
Pb2+ reacts with Cl- to form PbCl2 and Na+ reacts with NO3- to form NaNO3. The solubility Table 6.3 indicates that PbCl2 is insoluble and that the other compounds are soluble. Pb(NO3)2(aq) + NaCl(aq) →
PbCl2(s) + NaNO3(aq)
Now balance the reaction. Start by counting each atom. reactant atoms
product atoms
1 Pb atom
1 Pb atom
1 Cl atom
2 Cl atoms
6 O atoms
3 O atoms
1 Na atoms
1 Na atoms
2 N atom
1 N atom
Since there are 2 Cl atoms on the right and only one Cl atom on the left, start by multiplying the NaCl on the left by 2 (2NaCl). Recount the atoms.
6 - 18
reactant atoms
product atoms
1 Pb atom
1 Pb atom
2 Cl atom
2 Cl atoms
6 O atoms
3 O atoms
2 Na atoms
1 Na atoms
2 N atom
1 N atom
The Cl atoms are balanced but there are 2 Na on the left and 1 Na on the right. Multiply NaNO3 by 2 (2NaNO3) and recount. reactant atoms
product atoms
1 Pb atom
1 Pb atom
2 Cl atom
2 Cl atoms
6 O atoms
6 O atoms
2 Na atoms
2 Na atoms
2 N atom
2 N atom
The atoms are now the same on both sides and you can write the balanced equation. Pb(NO3)2(aq) + 2NaCl(aq)
6.71
→
PbCl2(s) + 2NaNO3(aq)
a. Complete and balance the precipitation reaction. Na2CO3 + Pb(NO3)2 → Use the procedure in previous problems to predict the products and to balance the equation. PbCO3 is an insoluble product whereas NaNO3 is a water-soluble product: Na2CO3(aq) + Pb(NO3)2(aq) → 2NaNO3(aq) + PbCO3(s) b. Using Table 6.3, come up with your own precipitation reaction. Show the formation of any insoluble salt, e.g., 2KOH(aq) + MgCl2(aq) → Mg(OH)2(s) + 2KCl(aq)
6 - 19
6.73
Write the ionic equation and the net ionic equation for the reaction in Problem 6.67. The reaction in Problem 6.67 is: CuSO4(aq) + Na2S(aq) →
CuS(s) + Na2SO4(aq)
In an ionic equation electrolytes are represented as individual ions. From the balanced equation above, any ionic compound that is in aqueous form is an electrolyte which can be written as individual ions. Because the product CuS is a solid that does not dissolve in water, this product is written in the original form. Ionic equation: Cu2+(aq) + SO42-(aq) + 2Na+(aq) + S2-(aq) → CuS(s) + SO42-(aq) + 2Na+(aq) The net ionic equation is derived from the ionic equation by removing any ions that appear unchanged between the reactant side and the product side. In this case, both the SO42- and the Na+are unchanged and are considered spectator ions that can be removed from the equation. The reaction that actually takes place is the formation of solid copper(II) sulfide from its constituent ions: Net ionic equation: Cu2+(aq) + S2-(aq) → CuS(s)
6.75
Write the ionic equation and the net ionic equation for each reaction in Problem 6.69.
a. Balanced equation: CaCl2(aq) + Li2CO3(aq) → CaCO3(s) + 2LiCl(aq) Ionic equation: Ca2+(aq) + 2Cl-(aq) + 2Li+(aq) + CO32-(aq) → CaCO3(s) + 2Li+(aq) + 2Cl-(aq) Net ionic equation: Ca2+(aq) + CO32-(aq) → CaCO3(s)
b. Balanced equation: Pb(NO3)2(aq) + 2NaCl(aq)
→
PbCl2(s) + 2NaNO3(aq)
Ionic equation: Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2Cl-(aq) → PbCl2(s) + 2Na+(aq) + 2NO3-(aq)
6 - 20
Net ionic equation: Pb2+(aq) + 2Cl- (aq) → PbCl2(s)
6.77
What happens to the solubility of carbon dioxide gas (CO2) in water in each situation? (Answer as increase, decrease, or no change.) a. The pressure of CO2 over the solution is increased. Increases. The solubility of a gas in a liquid increases as the pressure of the gas above the liquid increases. b. The temperature is increased. Decreases. The solubility of a gas in a liquid decreases as the temperature of the liquid increases.
6.79
Two unopened bottles of carbonated water are at the same temperature. If one is opened at the top of a mountain and the other at sea level, which will produce more bubbles? Explain. The bottle opened at the top of the mountain will produce more bubbles. The solubility of a gas in a liquid will be lower at the top of a mountain where the atmospheric pressure is lower.
6.81
When your body metabolizes amino acids, one of the final end products is urea, a water-soluble compound that is removed from the body in urine. Why is urea soluble in water, when hexanamide, a related compound, is not?
O
O H2N
C Urea urea
NH2
CH3 CH2 CH2 CH2 CH2 C NH2 Hexanamide hexanamide
Urea has more atoms capable of forming hydrogen bonds with water and it does not contain the nonpolar chain of carbon atoms present in hexanamide.
6.83
Explain why diethyl ether is less soluble in water than butyl alcohol. CH3CH2OCH2CH3
CH3CH2CH2CH2OH
Diethyl ether
Butyl alcohol
6 - 21
Molecules of butyl alcohol (CH3CH2CH2CH2OH) are able to form more hydrogen bonds with water.
6.85
Vitamin D is produced in the skin upon exposure to sunlight. Based on its structure, would you expect vitamin D to be hydrophilic or hydrophobic?
HO Vitamin D Hydrophobic First recall what each term means; hydrophilic—soluble in water and hydrophobic—insoluble in water. Note that the vitamin D has an OH group on one end that would make it somewhat like water. However, the large nonpolar part of the molecule is dominant. That makes the compound hydrophobic.
6.87
Explain why adrenaline has a greater solubility in water than dipivefrin (Figure 6.25). Refer to the structures of these two compounds shown in Figure 6.25. The molecules of adrenaline have greater hydrogen bonding capability due to more – OH and –NH groups present in the molecule. In addition, the nonpolar part of adrenaline is smaller than that found in dipivefrin.
6.89
Explain how you would prepare a saturated aqueous solution of baking soda (NaHCO3). A saturated solution is a solution that contains the maximum amount of solute that can be dissolved at a given temperature. To prepare a saturated aqueous solution of NaHCO3, add solid NaHCO3 to water and mix to dissolve all of the NaHCO3. Keep adding NaHCO3 until no more of the solid dissolves.
6 - 22
6.91
Albumin, a protein, is present in normal blood serum at concentrations of 3.5-5.5 g/dL. What is the % (w/v) of albumin in serum that contains 4.0 g of albumin per deciliter? Weight/volume percent is defined as: weight / volume percent =
g solute 100 mL of solution
To calculate the % (w/v) of a solution containing 4.0 g of albumin per deciliter of solution, first convert deciliter to milliliter:
1 dL
1 L 10 dL
1000 mL = 100 mL 1 L
weight / volume percent =
6.93
4.0 g albumin 100 = 4.0 % (w/v) 100 mL of solution
If 100 mL of blood serum contains 5.0 mg of thyroxine, a hormone released by the thyroid gland, thyroxine levels are within the normal range for an adult. Express this concentration of thyroxine in parts per million and parts per billion. Parts per million and parts per billion are concentration units used to describe dilute solutions. The definitions of these two concentration quantities are as follows: parts per million =
parts per billion =
g of solute 106 mL of solution g of solute 109 mL of solution
Use the above equations to calculate the concentrations in parts per million and in parts per billion. First, convert milligrams of thyroxine to grams:
5.0 mg
1g 1000 mg
= 5.0 x 10-3 g thyroxine
parts per million =
5.0 x 10-3 g thyroxine 106 = 5.0 x 101 ppm 100 mL of solution
parts per billion =
5.0 x 10-3 g thyroxine 109 = 5.0 x 104 ppb 100 mL of solution
6 - 23
6.95
Calculate the molarity of each. Molarity is equal to the moles of solute divided by the liters of solution. a. 0.33 mol of NaCl in 2.0 L of solution
molarity =
0.33 mol NaCl = 0.17 M 2.0 L
b. 55.0 g of NaCl in 125 mL of solution First, convert grams of NaCl to moles of NaCl:
55.0 g NaCl
1 mol 58.5 g NaCl
= 0.940 mol NaCl
Convert the volume from mL to L:
125 mL
1L 1000 mL
= 0.125 L
Then, proceed to calculate the molarity: molarity =
6.97
0.940 mol NaCl 0.125 L
= 7.52 M
If 0.30 mg of KCl is present in 1.0 L of aqueous solution, what is the concentration in terms of the following? a. molarity Molarity is a concentration unit defined as the moles of solute per liter of solution. To calculate the molarity of the given solution, first, we need to calculate the moles of KCl in the solution:
0.30 mg KCl
1g 1000 mg
3.0 x 10-4 g KCl
= 3.0 x 10-4 g KCl
1 mol KCl 74.6 g KCl
6 - 24
= 4.0 x 10-6 mol KCl
molarity =
4.0 x 10-6 mol KCl 1.0 L
= 4.0 x 10-6 M
b. weight/volume percent 1.0 L = 1.0 x 103 mL weight / volume percent =
g solute 100 mL of solution
weight / volume percent =
3.0 x 10-4 g KCl 100 = 3.0 x 10-5 % (w / v) 3 1.0 x 10 mL of solution
c. parts per thousand 3.0 x 10-4 g KCl parts per thousand = 1.0 x 103 mL of solution
103 = 3.0 x 10-4 ppt
d. parts per million
parts per million =
3.0 x 10-4 g KCl 1.0 x 103 mL of solution
106 = 3.0 x 10-1 ppm
3.0 x 10-4 g KCl 1.0 x 103 mL of solution
109 = 3.0 x 102 ppb
e. parts per billion
parts per billion =
6.99
For women, normal levels of uric acid in blood serum range from 26 to 60 ppm. If a female patient has 1.2 mg of uric acid in 10.0 mL of blood serum, is she within the normal range? First, calculate the ppm concentration of uric acid in the female’s blood serum:
1.2 mg uric acid
1g 1000 mg
= 1.2 x 10-3 g uric acid
1.2 x 10-3 g uric acid parts per million = 106 = 1.2 x 102 ppm 10.0 mL of solution
6 - 25
The concentration of uric acid in the female’s blood serum is not within the normal range (26 to 60 ppm). It is higher than normal.
6.101 Thyroxine, a thyroid hormone, is present in normal blood serum at 58-167 nmol/L. What is the molar concentration of thyroxine in serum that contains 150 nmol of thyroxine per liter? Since molarity is defined as mol per liter, convert nmol to mol.
150 nmol 1 x 10-9 mol x = 1.5 x 10-7 mol/L L nmol 1.5 x 10-7 mol thyroxine /L or 1.5 x 10-7 M thyroxine
6.103 The serum concentration of cortisol, a hormone, is expected to be in the range 8-20 mg/dL. What is the part per million concentration of cortisol in serum that contains 10 mg of cortisol per deciliter? Parts per million is the weight of solute (in grams) divided by the volume (in mL) of the solution multiplied by 106. Convert mg to g and dL to mL, then multiply by 106.
10 mg 1 x 10-3 g 1 dL x x x 106 = 100 ppm 2 dL mg 1 x 10 mL 100 ppm cortisol 6.105 The normal serum concentration of potassium ion (K+) is 3.5-4.9 mEq/L. Convert this concentration range into mmol/L. For an ion with a 1+ charge, 1 mol = 1 Eq, so 1 mmol = 1 mEq.
3.5 − 4.9 mEq 1 mmol x = 3.5-4.9 mmol/L L mEq
6.107 For adults, a 1.5 ng/dL serum concentration of the hormone thyroxine (776.87 g/mol) is in the normal range. Convert this concentration into a. picograms per milliliter
6 - 26
Convert ng to pg and mL to dL: 1.5 ng 10-9 g 1 pg 1 dL 10−3 L × × × × dL 1 ng 1 mL 10-12 g 10-1 L
=
15 pg mL
b. nanograms per liter Convert dL to L: 1.5 ng 1 dL × dL 10-1 L
=
15 ng L
c. picomoles per liter Convert ng to g, use the molar mass given to convert g to mol, and convert mol to pmol. Then, convert dL to L. 1.5 ng 10-9 g 1 mol 1 pmol 1 dL × × × × -12 dL 1 ng 776.87 g 10 mol 10-1 L
=
19 pmol L
6.109 The normal serum concentration of chloride ion (Cl -) is 95-107 mmol/L. Convert this concentration range into mEq/L. For an ion with a 1- charge 1 mol = 1 Eq, so 1 mmol = 1 mEq. 95 − 107 mmol 1 mEq x = 95 − 107 mEq/L L mmol
6.111 How many milliequivalents of bicarbonate (HCO3-) are present in a 75.0 mL blood serum sample with a concentration of 25 mEq/L of HCO3-? Convert mL to L and multiply by the concentration in mEq/L.
75.0 mL x
1 x 10-3 L 25 mEq x = 1.9 mEq mL 1 L
1.9 mEq HCO3-
6 - 27
6.113 How many moles of sodium ion (Na+) are present in a 10.0 mL blood serum sample with a concentration of 132 mEq/L? Convert mL to L and mEq to mmol (for Na+, 1 mmol = mEq) and then convert mmol to mol.
1 x 10-3 L 132 mEq 1 mmol 1 x 10-3 mol 10.0 mL x x x x = 1.32 x 10-3 mol mL 1 L 1 mEq 1 mmol
1.32 x 10 -3 mol Na+
6.115 If 15.0 mL of 3.0 M HCl are diluted to a final volume of 100.0 mL, what is the new concentration? The dilution equation is Voriginal x Coriginal = Vfinal x Cfinal. To use this equation it is not necessary to have a particular volume or concentration unit, provided that both volume units are the same and that both concentration units are the same. The answer will have the same unit as the value that is given in the problem. Voriginal x Coriginal = Vfinal x Cfinal
Cfinal
=
Voriginal x Coriginal
Cfinal
=
15.0 mL x 3.0 M 100.0 mL
Vfinal =
0.45 M HCl
6.117 A 10.0% (w/v) solution of ethyl alcohol is diluted from 50.0 mL to 200.0 mL. What is the new weight/volume percent? Recall the dilution equation is Voriginal x Coriginal = Vfinal x Cfinal.
Cfinal
=
Voriginal x Coriginal
Cfinal
=
50.0 mL x 10.0% 200.0 mL
Vfinal =
2.50% (w/v)
6 - 28
6.119 How many milliliters of 2.00 M NaOH are needed to prepare 300.0 mL of 1.50 M NaOH? Recall the dilution equation is Voriginal x Coriginal = Vfinal x Cfinal.
Voriginal
=
Vfinal x Cfinal Coriginal
Voriginal
=
300.0 mL x 1.50 M 2.00 M
=
225 mL
6.121 Calculate the final volume required to prepare each solution. a. Starting with 100.0 mL of 1.00 M KBr, prepare 0.500 M KBr. Recall the dilution equation is Voriginal x Coriginal = Vfinal x Cfinal.
Vfinal
=
Voriginal x Coriginal
Vfinal
=
100.0 mL x 1.00 M 0.500 M
Cfinal =
200. mL
b. Starting with 50.0 mL of 0.250 M alanine (an amino acid), prepare 0.110 M alanine. Recall the dilution equation is Voriginal x Coriginal = Vfinal x Cfinal.
Vfinal
=
Voriginal x Coriginal
Vfinal
=
50.0 mL x 0.250 M 0.110 M
Cfinal = 114 mL
6.123 How is a colloid different from a suspension? Suspensions and colloids are special types of mixtures. In a suspension, the mixture is composed of large particles suspended in a liquid. In a colloid, the mixture contains particles that are smaller than particles in a suspension. The smaller size of the particles in a colloid differentiates them from a suspension.
6 - 29
The particles in a colloid are not heavy enough to settle out in the liquid upon standing. 6.125 What do you end up with if you pour dirt into water and stir – a solution, a colloid, or a suspension? When dirt is mixed into water by stirring, a suspension is made because the particles of dirt are heavy enough to separate out upon standing.
6.127 a. Define the term diffusion Diffusion is the movement of substances from areas of higher concentration to areas of lower concentration. b. A process called active transport moves certain ions and compounds across cell membranes from areas of lower concentration to areas of higher concentration. Does active transport involve diffusion? Explain No. In active transport solutes move in a direction opposite to diffusion.
6.129 To make pickles, you soak cucumbers in a concentrated salt solution called brine. Describe how this process is related to osmosis. Osmosis is the net flow of water across a membrane from a solution of lower concentration to a solution of higher concentration. Brine, concentrated salt solution, has a higher concentration of salt than the liquid in cucumber. As a result, there will be a net flow of water from the cucumber into the brine, and a pickle results.
6.131 During severe bleeding, ADH (a hormone released by the hypothalamus) causes vasoconstriction (shrinking of the blood vessels) to take place. What effect does a decrease in blood vessel volume have on blood pressure? A decrease in blood vessel volume causes an increase in blood pressure.
6.133
The chest compressions given during cardiopulmonary resuscitation (CPR) cause the injured person to exhale. Explain why, in terms of Boyle’s law. Chest compressions cause the volume of the lungs to decrease. According to Boyle’s law, if the volume of the lungs and the gas inside decrease, the pressure
6 - 30
of the gas will increase. The gas will then move from the lungs (higher pressure) to the atmosphere that surrounds you (lower pressure).
6.135 Why is the prodrug chloramphenicol palmitate (Figure 6.25) less soluble in water than chloramphenicol? The long hydrocarbon chain in chloramphenicol palmitate makes the molecule less polar and therefore less soluble in water.
6.137 Saliva has some characteristics of a solution. Explain. Saliva is a homogenous mixture of ionic compounds, proteins, and nucleic acids.
6.139 Why is it important that dialyzing solution be isotonic with blood? Dialyzing solution has to be isotonic with blood to ensure that there is no net flow of essential solutes into or out of the blood.
6.141 a. A 1.50 L gas cylinder contains a mixture of nitrogen gas (N2) and hydrogen gas (H2) at a temperature of 35oC and a pressure 895 torr. What is the pressure in atmospheres (atm)? 1 atm = 760 torr or mmHg = 14.7 psi
895 torr x
1 atm = 1.18 atm 760 torr
b. What is the pressure in pounds per square inch (psi)?
895 torr x
14.7 psi = 17.3 psi 760 torr
c. If the partial pressure of N2 is 615 torr, what is the partial pressure of H2? PH2 = 895 torr - 615 torr = 280. torr or 2.80 x 102 torr d. If the temperature is increased to 70.0oC, what is the new partial pressure of each gas and what the new total pressure? New partial pressure of N2: P1 = 615 torr T1 = 35°C + 273.15 = 308 K
P2 = ? T2 = 70.0°C + 273.15 = 343.2 K
6 - 31
New PN2
=
615 torr 308 K
x
343.2 K = 685 torr
New partial pressure of H2: P1 = 2.80 x 102 torr T1 = 35°C + 273.15 = 308 K
New PH2
=
2.80 x 102 torr 308 K
P2 = ? T2 = 70.0°C + 273.15 = 343.2 K
x
343.2 K = 312 torr
New total pressure = 685 torr + 312 torr = 997 torr e. How many moles of N2 and moles of H2 are present in the cylinder? PN2 = 615 torr x
1 atm = 0.809 atm 760 torr
PH2 = 280. torr x
1 atm = 0.368 atm 760 torr
nN2 =
0.809 atm x 1.50 L 0.0821 L atm /mol K x 308 K
= 0.0480 mol N 2
nH2 =
0.368 atm x 1.50 L 0.0821 L atm /mol K x 308 K
= 0.0219 mol H 2
f. How many grams of N2 and grams of H2 are present in the cylinder?
0.0480 mol N 2 x
28.01 g N 2 = 1.34 g N2 1 mol N 2
0.0219 mol H 2 x
2.02 g H 2 = 0.0442 g H 2 1 mol H 2
g. How many molecules of N2 and molecules of H2 are present in the cylinder?
0.0480 mol N 2 x
0.0219 mol H 2
6.02 x 1023 molecules N 2 = 2.89 x 1022 molecules N 2 1 mol N 2
6.02 x 1023 molecules H 2 x = 1.32 x 1022 molecules H 2 1 mol H 2
6 - 32
h. Nitrogen gas and hydrogen gas react according to the equation: N2(g) + 3H2(g) →
2NH3(g)
Assuming that conditions are right for the reaction to take place, which is the limiting reactant, N2 or H2 Calculate the mol of H2 required to react with all of the N2 present:
0.0480 mol N2 x
3 mol H 2 = 0.144 mol H 2 1 mol N2
Because this is more than the amount of H2 available, not all of the N2 will react. H2 is the limiting reactant. i. For the reaction in part h, what is the theoretical yield (in grams) of NH3
0.0219 mol H2 x
2 mol NH3 17.03 g NH3 x = 0.249 g NH3 3 mol H2 1 mol NH3
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 6.2
Which of the following could account for the change to the gas that takes place in Problem 6.1? a. The external pressure was increased. b. The external pressure was decreased. c. The temperature was increased. d. The temperature was decreased.
Answer: a. The external pressure was increased.
SECTION 6.1 GASES AND PRESSURE
6.4
a. Give an example of an element or compound that is a solid at STP. b. Give an example of an element or compound that is a liquid at STP. c. Give an example of an element or compound that is a gas at STP.
6 - 33
Answer: a. Some elemental examples are the metallic elements: iron, copper, aluminum, sodium, etc. Some examples of compounds that are solid at STP are water, sodium chloride, sucrose, etc. Recall that STP means standard temperature (0○C) and pressure (1 atm). b. An elemental example is mercury. Some examples of compounds are hydrogen peroxide, hexane, and ethanol. c. Elemental examples include oxygen, nitrogen, helium, hydrogen, etc. Some examples of compounds include carbon dioxide, methane, ammonia, hydrogen sulfide, etc.
6.6
A pressure of 13.6 psi is how many a. atmospheres? b. torr?
Answer: a. As with any unit conversion, the first step is to know the equivalency between psi and atmospheres (atm). 1 atm = 14.7 psi. Use this as a conversion factor and convert from psi to atm.
13.6 psi
1 atm = 0.925 atm 14.7 psi
b. 760 torr = 14.7 psi. Use this as a conversion factor and convert from psi to torr.
13.6 psi
6.8
760 torr = 703 torr 14.7 psi
A pressure of 597 torr is how many a. atmospheres ? b. psi?
Answer: a. 597 torr
1 atm = 0.786 atm 760 torr
b. 597 torr
14.7 psi = 11.5 psi 760 torr
6 - 34
6.10
In the SI measurement system, the unit of pressure is called the kilopascal (kPa). One atmosphere equals 101.3 kPa. a. Convert 277 kPa into torr. b. Convert 9.22 psi into kilopascals. c. Convert 583 kPa into pounds per square inch.
Answer:
6.12
a. 277 kPa
1 atm 101.3 kPa
b. 9.22 psi
1 atm 14.7 psi
c. 583 kPa
1 atm 101.3 kPa
760 torr = 2080 torr 1 atm
101.3 kPa = 63.5 kPa 1 atm
14.7 psi = 84.6 psi 1 atm
a. Listening to the weather report, you hear that the barometer is falling. What part of the barometer is falling? b. What is happening to the air pressure?
Answer: a. The level of mercury is falling. b. The air pressure is decreasing.
6.14
At an atmospheric pressure of 760 torr and a temperature of 0°C, the mercury in the right arm of the nanometer pictured in Figure 6.4c is 5 mm lower than the mercury in the left arm. What is the gas pressure inside the flask?
Answer: 755 torr. In Figure 6.4 c, the mercury is pushed away from the open-end of the tube. This means that the gas pressure inside the flask is lower than the atmospheric pressure. The difference between the gas pressure inside the flask and the atmospheric pressure can be calculated by subtracting the height difference in mm Hg converted into torr from the atmospheric pressure:
6 - 35
1 torr Gas pressure inside the flask = 760 torr - 5 mmHg = 755 torr 1 mmHg
6.16
Use Table 6.1 to estimate the boiling point of water at a pressure of 2.2 atm.
Answer: Close to 125°C. First, convert 2.2 atm to torr then use this pressure to estimate the boiling point from the table. 2.2 atm
6.18
760 torr 1 atm
= 1,700 torr
If you hang wet sheets outside on a clothesline on a sunny winter day when the temperature is 0°C, they will eventually dry. Explain.
Answer: The water that makes the sheets wet may initially freeze. However, the sheets will eventually dry because the ice will sublime away.
SECTION 6.2
6.20
THE GAS LAWS
At a pressure of 1.5 atm a balloon has a volume of 3.25 L. If the pressure is decreased to 1.0 atm (at constant temperature), what is the new volume of the balloon?
Answer: Boyle’s law relates how the volume and pressure of a system change relative to each other: P1V1 = P2V2. Use Boyle’s law to calculate the new volume (at constant temperature). P1 = initial pressure = 1.5 atm P2 = final pressure = 1.0 atm PV 1 1 = PV 2 2
therefore V2
=
V1 = initial volume = 3.25 L V2 = final volume = ? PV 1 1 P2
=
6 - 36
1.5 atm x 3.25 L = 4.9 L 1.0 atm
Because pressure and volume are inversely proportional (at constant temperature), as the pressure goes down, (1.5 atm to 1.0 atm), the volume increases (3.25 L to 4.9 L).
6.22
At a temperature of 22°C, a balloon has a volume of 8.0 L. If the temperature is increased to 55°C (at constant pressure), what is the new volume of the balloon?
Answer: To determine how the volume changes with the temperature (at constant pressure), use Charles’ law written in equation form as V1/T1 = V2/T2, where T is the Kelvin temperature. T1 = initial temperature = 22C = 22C + 273.15 = 295 K V1 = initial volume = 8.0 L T2 = final temperature = 55C = 55C + 273.15 = 328 K V2 = final volume = ? V1 T1
=
V2 T2
therefore V2
=
T2V1 T1
=
328 K x 8.0 L = 8.9 L 295 K
Because volume and temperature are directly proportional (at constant pressure), we expect the volume to increase (8.0 L to 8.9 L) when the temperature increases (295 K to 328 K).
6.24
At a temperature of 75°C, a gas inside a 5.00 L metal canister has a pressure of 2500 torr. If the temperature is decreased to 45°C (at constant volume), what is the new pressure of the gas?
Answer: In this problem, the two variables that changed (at constant volume) are pressure and temperature. Use Gay-Lussac’s Law: P1/T1 = P2/T2, where T is the Kelvin temperature. T1 = initial temperature = 75C = 75C + 273.15 = 348 K P1 = initial pressure = 2500 torr T2 = final temperature = 45C = 45C + 273.15 = 318 K P2 = final pressure = ? P1 T1
=
P2 T2
therefore P2
=
T2 P1 T1
=
6 - 37
318 K x 2500 torr 348 K
= 2300 torr
Pressure and temperature are directly proportional, therefore when the temperature decreases (348 K to 318 K), the pressure also decreases (2500 torr to 2300 torr).
6.26
A 9.1 L balloon contains 1.25 mol of He(g). At constant pressure and temperature, what is the new volume of the balloon if 0.75 mol of He is added?
Answer: The two properties of gases that changed are the volume and the moles of gas particles. Avogadro’s law relates these two properties through the equation (constant temperature and pressure): V1/n1 = V2/n2, where n = moles of particles. n1 = initial moles = 1.25 mol He V1 = initial volume = 9.1 L n2 = final moles = 1.25 mol He + 0.75 mol He = 2.00 mol He V2 = final volume = ? V1 n1
V2 n2
=
therefore V2
=
n2V1 n1
=
2.00 mol He x 9.1 L = 15 L 1.25 mol He
Because volume and number of moles of particles are directly proportional (at constant pressure and temperature), we expect the volume to increase (9.1 L to 15 L) when the number of moles increases (1.25 mol to 2.00 mol).
6.28
A balloon with a volume of 2.00 L is at a pressure of 1.5 atm and a temperature of 20°C. If the balloon is put into a container and the pressure is increased to 2.0 atm and the temperature is raised to 25°C, what is the new volume of the balloon?
Answer: Because the number of moles of particles inside the balloon remains constant, we only need to relate the pressure, temperature, and volume of the gas using the initial and final conditions. These variables can be related by combining Charles’, Gay-Lussac’s, and Boyle’s Laws to form the combined gas law equation (at constant n, number of moles): PV 1 1 T1
=
P2V2 T2
T1 = initial temperature = 20C = 20C + 273.15 = 293 K P1 = initial pressure = 1.5 atm V1 = initial volume = 2.00 L
6 - 38
T2 = final temperature = 25C = 25C + 273.15 = 298 K P2 = final pressure = 2.00 atm V2 = final volume = ? PV 1 1 T1
6.30
=
PV 2 2 T2
therefore V2
PV T 1 1 2 T1 P2
=
=
1.5 atm x 2.00 L 298 K 293 K 2.00 atm
= 1.5 L
A 1.00 qt metal can contains 0.150 g of He at a temperature of 20°C. What is the pressure (in atm) inside the can? Is this pressure greater than or less than standard atmospheric pressure?
Answer: The properties of a given gas (pressure, P; volume, V; temperature, T; and number of moles, n) can be related by the ideal gas equation: PV = nRT, where R is the universal gas constant and is equal to 0.0821 L atm/mol K. Using the equation, one can calculate the value of any property of a gas given the values for the other three properties. Before plugging in the given values for the known properties, make sure they are converted to the corresponding units in the R constant. 0.946 L = 0.946 L 1 qt n = 0.150 g He 1 mol He = 0.0375 mol He 4.00 g He
V = 1.00 qt
T = 20C = 20C + 273.15 = 290 K R = 0.0821 L atm/mol K P=? PV = nRT P =
nRT V
therefore
0.0375 mol x 0.0821 L atm/mol K x 290 K = 0.94 atm 0.946 L
This pressure is less than standard atmospheric pressure which is 1 atm.
6.32
A 100.0 mL flask contains 400 g of N2 at 0○C. a. What is the pressure in atm? b. What is the pressure in torr? c. What is the pressure in psi?
6 - 39
Answer: a. What is the pressure in atm? Identify the variables. Since grams of N2 is given instead of moles, first convert 400 g to moles. 400 g
1 mol N 2 28.02 g
= 14.3 mol N 2
Convert oC to Kelvin.
K = oC + 273.15 = 0○C + 273 = 273 K
Convert the volume from mL to L.
100 mL
10−3 L = 0.100 L 1 mL
T = 273 K V = 0.100 L
P= ? n = 14.3 mol
Select the gas law that has just one of each variable. In this case it will be the Ideal Gas Law, PV = n R T. R is the gas constant (0.0821 L atm /mol K). Rearrange the equation so that the required variable is by itself. Replace the variables in the equation with the values you identified above and solve for the one missing.
PV = nRT P =
nRT 14.3 mol x 0.0821 L atm/mol K x 273 K therefore = 3210 atm V 0.100 L
b. What is the pressure in torr? Convert the answer from part a to torr. (1 atm =760 torr)
3210 atm
760 torr = 2.44 x 106 torr 1 atm
c. What is the pressure in psi? Convert the answer from part a to psi. (1 atm = 14.7 psi).
3210 atm
6.34
14.7 psi = 4.72 x 104 psi 1 atm
A 750 mL flask contains O2 at a pressure of 0.75 atm and a temperature of 20○C. What mass of O2 is present?
6 - 40
Answer: Solve for the number of moles and then convert from moles of O2 to grams of O2. Convert oC to Kelvin. K = oC + 273.15 = 20○C + 273 = 293 K Select the gas law that has just one of each variable. In this case it will be the Ideal Gas Law; PV = n R T. R is the gas constant (0.0821 L atm /mol K). Since the R constant has the units atm and L in it, before solving for n, it is necessary to convert the volume from mL to L .
750 mL
10−3 L = 0.750 L 1 mL
Assign values to the variables: T = 293 K P = 0.75 atm V = 0.750 L n=?
n =
PV RT
=
0.75 atm x 0.750 L = 2.34 x 10-2 mol 0.0821 L atm/mol K x 293 K
2.34 x 10-2 mol
6.36
32.00 g O2 = 0.749 g O2 1 mol O2
a. How many moles of Ne are present in a 1.0 qt flask that has a pressure of 850 torr at a temperature of 35○C? b. What is the mass of this Ne?
Answer: a. How many moles of Ne are present in a 1.0 qt flask that has a pressure of 850 a temperature of 35○C? First, convert ○C to Kelvin. K = oC + 273.15 = 35○C + 273 = 308 K Convert the volume from qt to L.
1 qt
1L = 0.94 L 1.06 qt
Convert torr to atm.
6 - 41
torr at
1 atm = 1.12 atm 760 torr
850 torr
Assign the values to the variables. T = 308 K P = 1.12 atm V = 0.94 L n=?
n =
PV RT
=
1.12 atm x 0.94 L = 0.042 mol Ne 0.0821 L atm/mol K x 308 K
b. What is the mass of this Ne?
0.042 mol Ne
6.38
20.18 g Ne = 0.85 g Ne 1 mol Ne
A sample of gas at a temperature of 37.0°C has a pressure of 725 torr and occupies a volume of 3.44 L. a. Use Boyle’s law to calculate the new pressure if the temperature is held constant and the volume is increased to 6.17 L. b. Use Gay-Lussac’s law to calculate the new pressure if the volume is held constant and the temperature is decreased to 10.0°C. c. Use Charles’ law to calculate the new volume if the pressure is held constant and the temperature is increased to 150.0°C. d. Use the combined gas law to calculate the new pressure if the temperature is decreased to 25.0°C and the volume is increased to 15.9 L. e. Use the ideal gas law to calculate the number of moles of gas that are present.
Answer: a. Use Boyle’s law to calculate the new pressure if the temperature is held constant and the volume is increased to 6.17 L. Boyle’s law: In this problem: Solving for P2: P1V1 P2 = = V2
P1V1 = P2V2 P1 = 725 torr V1 = 3.44 L
725 torr x 3.44 L 6.17 L
P2 = ? V2 = 6.17 L
= 404 torr
b. Use Gay-Lussac’s law to calculate the new pressure if the volume is held constant and the temperature is decreased to 10.0°C. Gay-Lussac’s law:
P1 T1
=
P2 T2
6 - 42
In this problem: P1 = 725 torr T1 = 37.0°C + 273.15 = 310.2 K Solving for P2: P1 P2 = x T2 T1
P2 = ? T2 = 10.0°C + 273.15 = 283.2 K
725 torr 310.2 K
=
x
283.2 K = 662 torr
c. Use Charles’ law to calculate the new volume if the pressure is held constant and the temperature is increased to 150.0°C. Charles’ law:
V1 T1
V2 T2
=
In this problem: V1 = 3.44 L T1 = 37.0°C + 273.15 = 310.2 K Solving for P2: V1 V2 = x T2 T1
V2 = ? T2 = 150.0°C + 273.15 = 423.2 K
3.44 L 310.2 K
=
x
423.2 K = 4.69 L
d. Use the combined gas law to calculate the new pressure if the temperature is decreased to 25.0°C and the volume is increased to 15.9 L. Combined gas law:
P1 V1 T1
=
In this problem: P1 = 725 torr V1 = 3.44 L T1 = 37.0°C + 273.15 = 310.2 K Solving for P2: P1V1 P2 = x T1
T2 V2
=
P2 V2 T2
P2 = ? V2 = 15.9 L T2 = 25.0°C + 273.15 = 298.2 K
725 torr x 3.44 L 310.2 K
x
298.2 K 15.9 L
= 151 torr
e. Use the ideal gas law to calculate the number of moles of gas that are present. Ideal gas law: PV = nRT First, we have to convert the pressure from torr to atm:
6 - 43
725 torr x
1 atm 760 torr
= 0.954 atm
Now, using the ideal gas law to solve for the number of moles, n: n
6.40
=
PV RT
0.954 atm x 3.44 L (0.0821 L atm /mol K) x 310.2 K
=
=
0.129 mol
If the Hindenburg (Problem 6.39) had been filled with helium instead of hydrogen, assuming the same temperature, pressure, and volume, how many grams of helium would have been present?
Answer: Use the same number of moles and convert this to grams of helium: 8.2 x 106 mol
6.42
x
4.00 g 1 mol
= 3.3 x 107 g He gas
True or false? Two moles of helium gas at STP occupies the same volume as two moles of neon gas at STP. Explain.
Answer: True. The same number of moles of each gas is present so they would occupy the same volume at STP according to the ideal gas law.
6.44 The label on a can of spray paint warns you to keep it away from high temperatures. From the perspective of the ideal gas law, explain why. Answer: If the number of moles and the volume of a gas remain constant, then an increase in temperature results in an increase in pressure (PV = nRT).
SECTION 6.3
6.46
PARTIAL PRESSURE
A mixture of gases contains 0.75 mol of N2, 0.25 mol of O2, and 0.25 mol of He. a. What is the partial pressure of each gas (in atm and in torr) in a 25 L cylinder at 250 K? b. What is the total pressure? 6 - 44
Answer: The partial pressure of each gas can be calculated using the ideal gas law. The total pressure can be calculated by using Dalton’s Law of Partial Pressure which states that the total pressure of a gas is equal to the sum of the partial pressures of all the gases in a mixture. a. Calculate the pressure in atm. To convert to torr, multiply the pressure in atm by the conversion factor 760 torr/1 atm.
PN2
=
nRT V
=
0.75 mol x 0.0821 L atm/mol K x 250 K 25 L
PO2
=
nRT V
=
0.25 mol x 0.0821 L atm/mol K x 250 K = 0.21 atm = 160 torr 25 L
PHe
=
nRT V
=
0.25 mol x 0.0821 L atm/mol K x 250 K = 0.21 atm = 160 torr 25 L
= 0.62 atm
= 470 torr
b. Ptotal = PN2 + PO2 + PHe = 0.62 atm + 0.21 atm + 0.21 atm = 1.04 atm = 790 torr
6.48
For the mixture of gases in Problem 6.46, what are the partial pressures and the total pressure if 0.50 mol of CO2(g) is added?
Answer: The partial pressures of the other gases are not changed by the addition of CO2(g). The partial pressure of CO2(g) is calculated the same way as in the previous problem.
PCO2
=
nRT V
=
0.50 mol x 0.0821 L atm/mol K x 250 K = 0.41 atm 25 L
Ptotal = PN2 + PO2 + PHe + PCO 2 = 1.04 atm + 0.41 atm = 1.45 atm
6.50
If the scuba tank in Problem 6.57 is used until the total pressure drops to 1300 psi, assuming no change in temperature, a. what is the partial pressure of each gas? b. How many grams of each gas are present
6 - 45
c. How many molecules of 02 and N2 and atoms of He are present? Answer: a. Each partial pressure drops proportionately. So, the new partial pressure for each gas is calculated as follows. PO2
=
50. atm
x
1300 psi 2700 psi
=
24 atm
PHe
=
22 atm
x
1300 psi 2700 psi
=
11 atm
PN2
=
110 atm
x
1300 psi 2700 psi
=
53 atm
b. Because the temperature and the volume are the same as in Problem 6.57, the number of moles of each gas also drops proportionate to the drop in pressure. n O2
=
5.4 mol
x
1300 psi 2700 psi
=
2.6 mol
n He
=
2.4 mol
x
1300 psi 2700 psi
=
1.2 mol
n N2
=
12 mol
x
1300 psi 2700 psi
=
5.8 mol
Convert moles to grams by using their respective molar masses: grams of O2 = 2.6 mol x
32.00 g = 83 g O 2 1 mol
grams of He = 1.2 mol x
4.00 g = 4.8 g He 1 mol
grams of N 2 = 5.8 mol x
28.02 g = 160 g N 2 1 mol
c. molecules of O2 = 2.6 mol x
6.02 x 1023 molecules = 1.6 x 1024 molecules O2 1 mol
6 - 46
6.02 x 1023 atoms atoms of He = 1.2 mol x = 7.2 x 1023 atoms He 1 mol molecules of N 2 = 5.8 mol x
SECTION 6.4
6.02 x 1023 molecules = 3.5 x 1024 molecules N 2 1 mol
SOLUTIONS
6.52 A bottle of 190 proof alcohol is 95% ethanol and 5% water. What is the solvent in this solution? Answer: Ethanol is the solvent. When both the solute and the solvent are liquids, the one in largest quantity is considered the solvent.
6.54 Define the terms homogeneous mixture, heterogeneous mixture, solution, and precipitate. Answer: A homogeneous mixture is a mixture whose components are uniformly distributed. A heterogeneous mixture is composed of components that are not evenly distributed. A solution is a homogeneous mixture. A precipitate is a solid substance produced in a solution.
6.56
A pan full of hot salt water, NaCl(aq), is cooled and NaCl(s) precipitates. Explain why this happens.
Answer: The amount of a substance that can dissolve in a solvent is temperature dependent. As the temperature decreases the solubility of the salt goes down and the salt in excess settles out as a precipitate.
6.58
Give an example of a solution in which a. the solute is a gas and the solvent is a liquid b. the solute is a liquid and the solvent is a liquid 6 - 47
Answer: a. Carbonated water. The solute is carbon dioxide gas and the solvent is liquid water. b. Alcoholic beverages, vinegar, etc. In alcoholic beverages, liquid ethanol is dissolved in water, and in vinegar, liquid acetic acid is dissolved in water.
6.60
At which temperature is NaCl more soluble in water, 0°C or 90°C?
Answer: 90°C. The solubility of substances in water generally increases with increasing temperature.
6.62
Explain why an opened bottle of carbonated water keeps its fizz longer if kept in a refrigerator than if kept at room temperature.
Answer: The fizz of carbonated water is due to dissolved carbon dioxide gas being released from solution. The solubility of this gas is higher at lower temperature, i.e. inside the refrigerator, than at higher temperature, i.e. room temperature. Thus, when kept inside the refrigerator, the fizz lasts longer because most of the carbon dioxide gas remains dissolved.
6.64
Once concern about global warming is that increase in water temperature might lead to death of many fish. How might this be related to the solubility of O2 in water?
Answer: An increase in water temperature will cause the solubility of O2 to decrease, depriving fish of the sufficient amount of dissolved O2 necessary to survive.
SECTION 6.5
6.66
PRECIPITATION REACTIONS
Predict whether each ionic compound is soluble or insoluble in water. a. (NH4)3PO4 b. Ba(NO3)2
6 - 48
c. AgCl d. K2CO3 Answer: The solubility of a compound in water can be predicted based on information given in Table 6.3. a. Soluble. Ammonium phosphate is an exception to the insolubility of most phosphate compounds in water. b. Soluble. All nitrate compounds are soluble in water. c. Insoluble. Silver chloride is an exception to the solubility of most chloride compounds in water. d. Soluble. Potassium is an alkali metal and therefore K2CO3 is an exception to the insolubility of most carbonate compounds in water.
6.68
When aqueous barium chloride is mixed with aqueous sodium sulfate, a precipitate forms. Write a balanced equation for this reaction.
Answer: When two aqueous solutions of ionic compounds are mixed, a precipitate may form if there is an insoluble ionic compound formed by combining the soluble ions. When aqueous barium chloride and aqueous sodium sulfate are mixed, the ions that are available to react are Ba2+, Cl-, Na+, and SO42-. Of these four, only Ba2+ and SO42- will combine to form an insoluble compound which will precipitate out of the solution. The balanced equation for this reaction is: BaCl2(aq) + Na2SO4(aq) → BaSO4 (s) + 2 NaCl(aq)
6.70
Complete and balance each precipitation reaction. a. LiCl(aq) + Ag2SO4(aq) → b. CaBr2(aq) + K3PO4(aq) →
Answer: Determine the two possible ionic compounds that may form by double replacement reaction. Use Table 6.3 to predict the insoluble product, if any. a. 2LiCl(aq) + Ag2SO4(aq) → Li2SO4(aq) + 2AgCl(s)
6 - 49
b. 3CaBr2(aq) + 2 K3PO4(aq) → 6KBr(aq) + Ca3(PO4)2(s)
6.72
a. Complete and balance the precipitation reaction. CuCl2 + AgNO3 → b. Using Table 6.3, come up with your own precipitation reaction.
Answer: Determine the two possible ionic compounds that may form by double replacement reaction. Use Table 6.3 to predict the insoluble product, if any. a. CuCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Cu(NO3)2(aq) b. Show the formation of any insoluble salt. For example, 3Mg(NO3)2(aq) + 2Na3PO4(aq) → Mg3(PO4)2(s) + 6NaNO3(aq)
6.74
Write the ionic equation and the net ionic equation for the reaction in Problem 6.68.
Answer: The balanced equation for Problem 6.68 is: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaCl(aq) Balanced Equation The ionic equation is the form of a reaction equation in which electrolytes are written as individual ions. In the balanced equation above, any ionic compound that is in aqueous form is an electrolyte and it can be written as individual ions. Because BaSO4 is a solid that does not dissolve in water, this product is written in the original form. Ba2+(aq) + 2Cl -(aq) + 2Na+(aq) + SO42-(aq) → BaSO4(s) + 2Na+(aq) + 2Cl- (aq) Ionic Equation Each coefficient indicates the actual number of each ion or compound as in the balanced equation. The net ionic equation is derived from the ionic equation by removing any ions that appear unchanged between the reactant side and the product side. In this case, both the Cl -(aq) and the Na+(aq) are unchanged and are considered spectator ions that can be
6 - 50
removed from the equation. The actual reaction that takes place is the formation of solid barium sulfate from its constituent ions: Ba 2+(aq) + SO42+(aq) → BaSO4(s) Net Ionic Equation
6.76
Write the ionic equation and the net ionic equation for each reaction in Problem 6.70.
Answer: a.
Balanced equation: 2LiCl(aq) + Ag2SO4(aq) → Li2SO4(aq) + 2AgCl(s) Ionic equation: 2Li+(aq) + 2Cl-(aq) + 2Ag+(aq) + SO42-(aq) → 2Li+(aq) + SO42- (aq) + 2AgCl(s) Net ionic equation: Cl-(aq) + Ag+(aq) → AgCl(s)
b.
Balanced equation: 3CaBr2(aq) + 2 K3PO4(aq) → 6KBr(aq) + Ca3(PO4)2(s) Ionic equation: 3Ca2+(aq) + 6Br-(aq) + 6K+(aq) + 2PO43-(aq) → 6K+(aq) + 6Br-(aq) + Ca3(PO4)2(s) Net ionic equation: 3Ca2+(aq) + 2PO43-(aq) → Ca3(PO4)2(s)
SECTION 6.6
6.78
SOLUBILITY OF GASES IN WATER
What happens to the solubility of oxygen gas (O2) in water in each situation? (Answer increase, decrease, or no change.) a. The pressure of O2 over the solution is decreased. b. The temperature is decreased.
6 - 51
Answer: Pressure and temperature are two factors that affect the solubility of gases in water. According to Henry’s Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid: the higher the pressure of the gas above, the more soluble it becomes. In terms of temperature, the higher the temperature of the solution the lower the solubility of the gas . a. Decrease. The solubility of O2 gas in water will decrease if its pressure above the solution is decreased. b. Increase. The solubility of O2 gas in water will increase if the temperature is decreased.
6.80
Why does O2 have a greater solubility in blood than in water?
Answer: Hemoglobin in the blood binds O2 molecules increasing its solubility.
SECTION 6.7
6.82
ORGANIC AND BIOCHEMICAL COMPOUNDS
Which of the compounds is more soluble in water? O CH3 CH2 CH2 CH2 CH2 C OH Hexanoic acid
O H C OH
Formic acid
Answer: O H C OH is more soluble in water because it has a smaller hydrocarbon component.
Use the rule “like dissolves like” to predict whether the compound will be soluble in water. The long hydrocarbon part of hexanoic acid makes it less like water.
6.84
Explain why propylene glycol is more soluble in water than isopropyl alcohol.
6 - 52
OH
OH
CH3 CH CH2 OH
CH3 CH CH3
Propylene glycol
2-Propanol
Answer: Propylene glycol is more soluble in water because it is more like water: it has two –OH groups whereas 2-propanol has only one.
6.86
Vitamin A is not water soluble. Why?
OH Vitamin A Answer: Vitamin A is not soluble in water because the part that is like water (-OH) is very small compared to the size of the nonpolar component of the molecule (mostly hydrocarbon). Thus, there is no favorable interaction between Vitamin A molecules and water molecules.
6.88
The artificial sweetener Acesulfame-K is the potassium salt of Acesulfame. Explain why Acesulfame-K has a greater solubility in water than Acesulfame.
O2
O2
Acesulfame-K
Acesulfame-K
Answer: Acesulfame-K has an ionic site. The negative charge on the N can hydrogen bond with the H in water increasing its solubility.
6 - 53
SECTION 6.8
6.90
CONCENTRATION
Explain how you would prepare a saturated aqueous solution of table sugar.
Answer: With stirring, add table sugar to water. Keep adding sugar until no more dissolves.
6.92 Potassium iodide (KI) is used to treat iodine deficiencies. What is the %(w/v) of a 75 mL solution containing 2.0 g of KI? Answer: Weight/volume percent is defined as:
6.94
weight / volume percent =
g solute 100% mL of solution
weight / volume percent =
2.0 g KI 100% = 2.7% (w/v) 75 mL of solution
The normal concentration range of vitamin B12 in blood serum is 160-170 ng/L. Express this concentration range in parts per million and parts per billion.
Answer: Parts per million and parts per billion are concentration units used to describe dilute solutions. The definitions of these two concentration quantities are as follows: parts per million =
g of solute 106 mL of solution
parts per billion =
g of solute 109 mL of solution
Use the above equations to calculate the concentrations in parts per million and in parts per billion. First, convert nanograms of vitamin B12 to grams: 160 ng
10-9 g 1 ng
= 1.6 x 10-7 g Vitamin B12
6 - 54
170 ng
10-9 g 1 ng
= 1.7 x 10-7 g Vitamin B12
Using the equivalence 1 L = 1000 mL
parts per million =
1.6 x 10-7 g Vitamin B12 1000 mL of solution
106 = 1.6 x 10-4 ppm
parts per billion =
1.6 x 10-7 g Vitamin B12 1000 mL of solution
109 = 0.16 ppb
Use a similar process for the upper range. The range concentrations in ppm and ppb are: 1.6 x 10-4 – 1.7 x 10-4 ppm and 0.16 – 0.17 ppb.
6.96
Calculate the molarity of each. a. 1.75 mole of NaCl in 15.2 L of solution b. 270 mg of NaCl in 1.00 mL of solution
Answer: The definition of molarity (M) is
molarity =
moles of solute liters of solution
a.
molarity =
1.75 mol NaCl 15.2 L
b. also.
First, convert the mg of NaCl to grams and then to moles. Convert 1.00 mL to L
270 mg NaCl
1.00 mL
molarity =
10-3 L 1 mL
= 0.115 M NaCl
10-3 g 1 mol NaCl 1 mg 58.44 g NaCl
= 4.6 x 10-3 mol NaCl
= 1.00 x 10-3 L
4.6 x 10-3 mol NaCl 1.00 x 10-3 L
= 4.6 M NaCl
6 - 55
6.98 If 15.0 g of CaCl2 are present in 250 mL of aqueous solution, what is the concentration in terms of the following? a. molarity b. weight/volume percent c. parts per thousand d. parts per million e. parts per billion Answer: a.
molarity =
moles of solute liters of solution
15.0 g CaCl2
1 mol CaCl2 110.98 g CaCl2
10-3 L 1 mL
= 0.25 L
250 mL
molarity =
b.
c.
d.
e.
0.135 mol CaCl2 0.25 L
= 0.135 mol CaCl2
= 0.54 M CaCl2
weight / volume percent =
g solute 100% mL of solution
weight / volume percent =
15.0 g CaCl2 100% = 6.0% (w / v) 250 mL of solution
parts per thousand =
g of solute 103 mL of solution
parts per thousand =
15.0 g CaCl2 250 mL of solution
parts per million =
g of solute 106 mL of solution
parts per million =
15.0 g CaCl2 106 250 mL of solution
= 6.0 x 104 ppm
parts per billion =
15.0 g CaCl2 250 mL of solution
= 6.0 x 107 ppb
6 - 56
103
109
= 60. ppt or 6.0 x 101 ppt
6.100 Lead is a toxic metal that can delay mental development in babies. The Environmental Protection Agency’s action level (that level where remedial action must be taken to clean the water) is 15 ppb. Does a 50 mL sample of water that contains 0.35 μg of lead fall above or below this action level? Answer: parts per billion =
g of solute 109 mL of solution
First convert the 0.35 g of lead to grams and then divide that by the 50 mL.
0.35 g Pb
10-6 g 1 g
parts per billion
=
= 3.5 x 10-7 g Pb 3.5 x 10-7 g Pb 50 mL of solution
109
= 7 ppb
7 ppb Pb is below the action level.
6.102 The normal concentration range of lactate in blood serum is 0.6-1.8 mmol/L. What is the molar concentration of lactate in serum that contains 1.2 mmol of lactate per liter? Answer: The molar concentration of a solution is defined as the number of moles of solute in one liter of solution. To find the molar concentration, calculate the number of moles of lactate in 1.2 mmol by converting mmol to mol: 10-3 mol 1.2 mmol 1 mmol
= 0.0012 mol
Therefore, the molar concentration is: molar concentration =
0.0012 mol 1 liter
6 - 57
= 0.0012 M
6.104 The serum concentration of glucose is expected to be in the range 80-100 mg/dL. What is the parts per million concentration of glucose in serum that contains 95 mg of glucose per deciliter? Answer: To calculate parts per million concentration, divide the grams of solute by the milliliters of solution and then multiply by 106. The given concentration of glucose is 95 mg/dL. First, convert 95 mg of glucose to g of glucose.
95 mg glucose
10-3 g 1 mg
= 9.5 x 10-2 g glucose
This amount of solute is contained in 1 dL of the solution. Convert 1 dL to mL: 1 dL
10-1 L 1 mL 1 dL 10-3 L
= 100 mL
Now, calculate parts per million: parts per million =
9.5 x 10-2 g glucose 1.0 x 102 mL of solution
106 = 9.5 x 102 ppm glucose
6.106 The normal serum concentration of phosphate ion (PO43-) is 2.8-4.5 mEq/L. Convert this concentration range into mmol/L. Answer: An equivalent is defined as the number of moles of positive and negative charges released by one mole of solute in a solution. Determine the relationship between the moles of solute and the number of equivalents. For the phosphate ion with a 3- charge, this relationship is given by: 1 mol of PO43- = 3 Eq PO43- or 1 mmol of PO43- = 3 mEq PO43Use this as a conversion factor when calculating the number of millimoles of phosphate ions from the given range of milliequivalents.
2.8 mEq PO43-
1 mmol PO433 mEq PO43-
= 0.93 mmol PO 43-
6 - 58
4.5 mEq PO4
3-
1 mmol PO43 3 mEq PO43-
= 1.5 mmol PO43-
Therefore the range in mmol/L is: 0.93 – 1.5 mmol PO43-/L
6.108 For men below 45 years of age, a serum concentration of prostate specific antigen (PSA) of 2.5 g/L is normal. Convert this concentration into a. nanograms per milliliter b. milligrams per deciliter Answer: a. Convert g to ng and L to mL: 2.5 g 10-6 g 1 ng 10-3 L × × × L 1 g 1 mL 10-9 g
=
2.5 ng mL
b. Convert g to mg and L to dL: 2.5 g 10-6 g 1 mg 10-1 L × × × = L 1 g 1 dL 10-3 g
2.5 x 10-4 mg dL
6.110 The normal serum concentration of magnesium ion (Mg2+) is 1.2-2.5 mmol/L. Convert this concentration range into mEq/L. Answer: For magnesium ion with a 2+ charge, use the following conversion factor: 1 mol Mg2+ = 2 Eq Mg2+ or 1 mmol Mg2+ = 2 mEq Mg2+
1.2 mmol Mg
2+
2 mEq Mg 2+ 1 mmol Mg 2+
= 2.4 mEq Mg 2+
2 mEq Mg 2+ 1 mmol Mg 2+
= 5.0 mEq Mg 2+
2.5 mmol Mg 2+
Therefore, the range in mEq/L is: 2.4 – 5.0 mEq Mg2+/L 6.112 How many equivalents of calcium ion (Ca2+) are present in a 25.0 mL blood serum sample with a Ca2+ concentration of 5.2 mEq/L?
6 - 59
Answer: To calculate the amount of solute in a solution, the concentration can be used as a conversion factor. Calculate the number of mEq of Ca2+ by “converting” 25.0 mL of solution into mEq of Ca2+ using the concentration, 5.2 mEq Ca2+/L as a conversion factor. First, convert 25.0 mL to L: 25.0 mL
10-3 L = 0.025 L 1 mL
5.2 mEq Ca 2+ 0.025 L = 0.13 mEq Ca 2+ 1L
Finally, convert mEq Ca2+ to Eq Ca2+:
0.13 mEq Ca 2+
10-3 Eq = 1.3 x 10-4 Eq Ca 2+ 1 mEq
6.114 How many grams of vitamin B12 are present in a 100.0 mL blood serum sample with a vitamin B12 concentration of 168 ng/L? Answer: The given concentration can be used as a conversion factor between L of solution and grams of vitamin B12. First, do the following conversion: 100.0 mL
0.1000 L
1L = 0.1000 L 1000 mL
168 ng vitamin B12 = 16.8 ng vitamin B12 1L
16.8 ng vitamin B12
10-9 g = 1.68 x 10-8 g vitamin B12 1 ng
6 - 60
SECTION 6.9
DILUTION
6.116 If 3.0 mL of 0.15 M HCl are diluted to a final volume of 250.0 mL, what is the new concentration? Answer: Dilution is a process of reducing the concentration of a solution by adding solvent to it. To calculate the new concentration of a solution given the original concentration, the original volume of the solution, and the final volume of the solution, we can use the following equation: Voriginal x Coriginal = Vfinal x Cfinal In this problem, three of four terms are given: Voriginal = 3.0 mL Coriginal = 0.15 M HCl
Cfinal
=
Voriginal x Coriginal Vfinal
Vfinal = 250.0 mL Cfinal = ?
=
3.0 mL x 0.15 M 250.0 mL
= 1.8 x 10-3 M HCl
6.118 A 5.0% (w/v) solution of ethyl alcohol is diluted from 50.0 mL to 75.0 mL. What is the new weight/volume percent? Answer: Voriginal = 50.0 mL Coriginal = 5.0 % (w/v) ethanol
Cfinal
=
Voriginal x Coriginal Vfinal
Vfinal = 75.0 mL Cfinal = ?
=
50.0 mL x 5.0 % (w/v) 75.0 mL
= 3.3 % (w/v) ethanol
6.120 How many milliliters of 1.50 M NaOH are needed to prepare 225.0 mL of 0.150 M NaOH? Answer: Voriginal = ? Coriginal = 1.50 M NaOH
Vfinal = 225.0 mL Cfinal = 0.150 M NaOH
6 - 61
Vfinal x Cfinal Coriginal
=
Voriginal
225.0 mL x 0.150 M 1.50 M
=
=
22.5 mL
6.122 Calculate the final volume required to prepare each solution. a. Starting with 10 mL of 2.0% (w/v) KI, prepare 0.020% (w/v) KI. b. Starting with 1.0 mL of 60 ppm Cu2+, prepare 18 ppm Cu2+. Answer: a. Voriginal = 10 mL Coriginal = 2.0 % (w/v) KI
Vfinal
=
Voriginal x Coriginal Cfinal
b. Voriginal = 1.0 mL Coriginal = 60 ppm Cu2+
Vfinal
SECTION 6.10
=
Vfinal = ? Cfinal = 0.020 % (w/v) KI
=
= 1000 mL
Vfinal = ? Cfinal = 18 ppm Cu2+
Voriginal x Coriginal Cfinal
10 mL x 2.0 % (w/v) 0.020 % (w/v)
=
1.0 mL x 60 ppm 18 ppm
= 3.3 mL
COLLOIDS AND SUSPENSIONS
6.124 How is a colloid different from a solution? Answer: The particles that make up a colloid are smaller than those present in a suspension. While suspensions settle, colloids do not.
6.126 Give an example of each. a. a solution b. a suspension c. a colloid Answer: a. Answers may vary but examples are sugar in tea, salt in water, and CO2 in soda. b. Answers may vary but examples are muddy water and smoke. c. Answers may vary but examples are soapy water, paint, clay, and mustard.
6 - 62
SECTION 6.11
DIFFUSION AND OSMOSIS
6.128 Define the terms osmosis and osmotic pressure. Answer: Osmosis is the net movement of water across a membrane from the solution of lower concentration to one of higher concentration. Osmotic pressure is the pressure exerted by water as it moves through a semipermeable membrane.
6.130 Soaking in a bath of dissolved Epsom salts is a home remedy that can help to reduce the swelling and discomfort of hemorrhoids. Why is this hypertonic solution of MgSO4 able to reduce the swelling of hemorrhoidal tissue, while pure water is not? Answer: The hypertonic solution of MgSO4 contains a higher concentration of solute than the liquid inside the cells of hemorrhoidal tissue. Through osmosis, there will be a net flow of water from the swollen hemorrhoidal tissue to the hypertonic MgSO4 solution. The swelling of the hemorrhoidal tissue is thereby reduced.
HEALTH LINK
BLOOD PRESSURE
6.132 Ethyl alcohol, the alcohol present in alcoholic beverages, inhibits the release of ADH (see the previous problem). What effect does drinking alcohol have on blood pressure? Answer: ADH causes shrinking of the blood vessels. When ethyl alcohol inhibits the release of the ADH, constriction of the blood vessels is also inhibited. The overall effect is to reduce the pressure exerted on the blood by the vessel walls. Thus, drinking alcohol may result in lowering the blood pressure.
6 - 63
HEALTH LINK
BREATHING
6.134 Sometimes an accident victim will be kept alive by use of a bag valve mask. This mask moves air into the lungs when its bag is squeezed. In terms of the ideal gas law, explain how the bag forces air into an accident victim’s lings. Answer: Squeezing the bag causes its volume to decrease. According to the ideal gas law, when the volume of the gas inside decreases, its pressure increases as long as other properties stay constant. The increase in pressure inside the bag causes air from the bag to move into the lungs where the pressure is lower.
HEALTH LINK
PRODRUGS
6.136 Phenacetin is a prodrug that can be prepared from acetaminophen, a commonly used pain and fever reducer. While available in many parts of the world, phenacetin is not sold in the U.S. due to concerns about its toxicity. What is the theoretical yield of phenacetin (in grams) upon reaction of 325 mg of acetaminophen? Assume that acetaminophen is the limiting reactant.
O NH C CH3 K2CO3
+
+
CH3CH2I
HO Acetaminophen
O NH C CH3 +
KHCO3 +
CH3CH2O Phenacetin
Answer: According to the equation above, the conversion of acetaminophen to phenacetin occurs on a one-to-one mole ratio. Convert 325 mg to g and then convert g of acetaminophen
6 - 64
KI
(C8H9NO2) to mol of acetaminophen. Use the 1:1 mole ratio to calculate the theoretical yield of phenacetin (C10H13NO2) in moles and convert to grams.
325 mg acetaminophen
10-3 g 1 mol acetaminophen 1 mg 151.18 g acetaminophen
2.15 x 10-3 mol acetaminophen
HEALTH LINK
= 2.15 x 10-3 mol acetaminophen
1 mol phenacetin 179.24 g phenacetin = 0.385 g phenacetin 1 mol acetaminophen 1 mol phenacetin
SALIVA
6.138 Saliva has some characteristics of a colloid. Explain. Answer: Saliva is cloudy, indicating the presence of large particles in the mixture that will not settle upon standing.
HEALTH LINK
DIFFUSION AND THE KIDNEYS
6.140 Kidney disease can lead to nephrotic syndrome, which is characterized, in part, by a low blood serum concentration of albumin (a protein). Explain why low serum albumin levels result in edema, swelling caused by the movement of fluid from the blood into tissue. Answer: Albumin is a solute in blood serum. If its concentration is too low, water will diffuse out of the blood serum and into surrounding tissue. The process is known as osmosis. Osmosis causes the movement of water from blood serum (lower solute concentration) to tissues (higher solute concentration).
LEARNING GROUP PROBLEMS
6.142 a. Write the balanced equation for the reaction of lead(II) nitrate with sodium iodide to form sodium nitrate and lead(II) iodide. b. Using Table 6.3, label the reactants and product of this reaction as being aqueous or solid. c. Write the ionic equation for the reaction in part a. d. Write the net ionic equation for this reaction.
6 - 65
e. An aqueous solution is prepared that contains 5.72 g of lead(II) nitrate in 125 mL of solution. What is the weight/volume percent of the solution? f. What is the molarity of the solution in part e? g. If 10.00 mL of the solution in part e is diluted to a final volume of 75.00 mL, what is the weight/volume percent of the new solution? h. An aqueous solution is prepared that contains 9.88 g of sodium iodide in 175 mL of solution. What is the concentration of the solution in parts per million? i. What is the molarity of the solution in part h? j. If 5.00 mL of the solution in part h is diluted to give a final volume of 25.0 mL, what is the molar concentration of the new solution? k. Once the reaction in part a has taken place, what is obtained: a homogeneous mixture or a heterogeneous mixture? Answer: a. Pb(NO3)2 + 2NaI → 2NaNO3 + PbI2 b. Pb(NO3)2(aq) + 2NaI(aq) → 2NaNO3(aq) + PbI2(s) c. Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2I-(aq) → 2Na+(aq) + 2NO3-(aq) + PbI2(s) d. Pb2+(aq) + 2I-(aq) → PbI2(s) e.
weight / volume percent =
g solute 100% mL of solution
weight / volume percent =
5.72 g Pb(NO3 )2 100% = 4.58% (w/v) 125 mL of solution
f. 5.72 g Pb(NO3 ) 2
125 mL
10-3 L 1 mL
molarity =
g.
1 mol Pb(NO3 )2 331.2 g Pb(NO3 ) 2
= 1.73 x 10-2 mol Pb(NO3 ) 2
= 0.125 L
1.73 x 10-2 mol Pb(NO3 )2 0.125 L
= 0.138 M Pb(NO3 )2
Voriginal x Coriginal = Vfinal x Cfinal
10.00 mL x 4.58% (w/v) = 75.00 mL x Cfinal
6 - 66
Cfinal = 0.611% (w/v) h.
i.
parts per million =
g of solute 106 mL of solution
parts per million =
9.88 g of NaI 175 mL of solution
9.88 g NaI
175 mL
10-3 L 1 mL
molarity =
j.
1 mol NaI 149.89 g NaI
106
= 5.65 x 104 ppm
= 6.59 x 10-2 mol NaI
= 0.175 L
6.59 x 10-2 mol NaI 0.175 L
= 0.377 M NaI
Voriginal x Coriginal = Vfinal x Cfinal
5.00 mL x 0.377 M = 25.0 mL x Cfinal Cfinal = 0.0754 M k. A heterogeneous mixture is obtained because of the presence of the solid PbI2.
INTERACTIVE LEARNING PROBLEMS
6.144 How many grams of solute are needed to make 250 mL of 0.100 M K2SO4? Answer: First, convert 250 mL to L then use the molarity concentration as a conversion factor to calculate the number of moles of K2SO4. Convert to grams of K2SO4 to get the final answer. 250 mL
0.250 L
1L = 0.250 L 1000 mL 0.100 mole K 2SO4 = 0.0250 mole K 2SO4 1 L
6 - 67
0.0250 mole K 2SO4
174.3 g K 2SO4 1 mole K 2SO4
6 - 68
= 4.36 g K 2SO4
Chapter 7 Acids, Bases and Equilibrium SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
7.1
a. Write the equilibrium constant expression for the reaction. A
B
[B]
Keq = [A]
b. For this reaction Keq = 2. The drawing below represents an equilibrium mixture of A and B. Explain
As shown in the drawing above, the ratio of the concentration of B to A, [B]/[A], is equal to 2. Because this is the same value as the equilibrium constant, Keq = [B] [A]
= 2 , then this reaction mixture must be at equilibrium.
c. If 3 additional Bs are added, the mixture is no longer in equilibrium. Explain.
For this mixture, the ratio of the concentration of B to A changes to: [B] 5 = = 5 [A] 1
7-1
[B]
The mixture is no longer in equilibrium because [A] is no longer equal to 2. d. To adjust to equilibrium, which will take place for the mixture in part c: a net forward reaction or a net reverse reaction? A net reverse reaction will take place to consume the additional B’s until the ratio of [B]/[A] goes back down to 2. When [B]/[A] = 2, the reaction is at equilibrium. e. Based on your answer to part d, redraw the picture in part c as it will appear once the mixture reaches equilibrium. One B is converted to one A, decreasing the number of B’s to 4 and increasing the number of A’s to 2 bringing the ratio back to its equilibrium value: B
A
A B
B
B
[B] 4 = = 2 [A] 2
7.3
Which of the following observations would indicate that a compound might be an acid? a. Turns litmus pink b. Turns litmus blue c. Has a bitter taste d. Has a sour taste e. Dissolves a metal f. Feels slippery The observations a, d, and e are all properties of an acid. The others (b, c, and f) are properties of a base.
7.5
Name each of the following as an acid and as a binary compound. Covalent binary compounds are named by using the name of the first element and then naming the second element with its end changed to “- ide”. When binary compounds are named as acids, the prefix “hydro-“ is added to the beginning of the second element name and the ending is changed to “-ic acid”.
7-2
a.
7.7
HCl
hydrochloric acid
hydrogen chloride
b. HBr hydrobromic acid hydrogen bromide Give the formula for the conjugate base of each acid. The conjugate base of an acid is the compound or ion that is formed after the acid loses an H+. When an H+ is removed, the resulting particle gains a negative charge. acid
7.9
conjugate base
a. H2CO3
HCO3-
b. HCO3-
CO32-
c. H2SO4
HSO4-
d. H2PO4-
HPO42-
Name the conjugate base of each acid in Problem 7.7 Each of these conjugate bases is a polyatomic ion. conjugate base
7.11
a. HCO3-
hydrogen carbonate ion
b. CO32-
carbonate ion
c. HSO4-
hydrogen sulfate ion
d. HPO42-
hydrogen phosphate ion
Give the formula for the conjugate acid of each base. The conjugate acid of a base is the compound or ion that is formed after the base gains an H+. When an H+ is added, the resulting particle gains a positive charge. base
conjugate acid
a. F-
HF
b. CN-
HCN
c. NH3
NH4+
d. NO2-
HNO2
7-3
7.13
Water is amphoteric (can act as an acid or a base). a. Write a chemical equation that shows water acting as an acid. H2O donates an H+ to NH3: H2O + NH3 ⇋ OH- + NH4+ b. In this reaction, what is the conjugate base of water? OH-. The conjugate base of H2O, OH-, is what is left after water donates an H+ to NH3.
7.15
HCO3- is amphoteric. a. Write the equation for the reaction that takes place between HCO3- and the acid HCl. The acid HCl donates H+ to the base HCO3-. HCO3- (aq) + HCl(aq) → H2CO3(aq) + Cl-(aq) b. Write the equation for the reaction that takes place between HCO3- and the base OH-. The acid HCO3- donates H+ to the base OH-. HCO3- (aq) + OH- (aq) → CO32- (aq) + H2O(l)
7.17
Identify the Bronsted-Lowry acids and bases for the forward and reverse reactions of each. a. F -(aq) + HCl (aq) base acid
HF(aq) + Cl - (aq) acid base
Bronsted-Lowry acids donate H+ and Bronsted-Lowry bases accept H+. In the forward reaction, HCl gives up H+ to F-. In the reverse reaction, HF donates H+ to Cl-.
b. CH3CO2H (aq) + NO3- (aq) acid base
CH3CO2- (aq) + HNO3 (aq) base acid
7-4
Bronsted-Lowry acids donate H+ and Bronsted-Lowry bases accept H+. In the forward reaction, CH3CO2H donates H+ to NO3-. In the reverse reaction, HNO3 donates H+ to CH3CO2-. 7.19
Complete each acid-base reaction. For the forward and reverse reactions, identify each acid and its conjugate base. An acid-base reaction involves the transfer of a H+ ion from the acid to the base. The acid is the ion or molecule that donates the H+ ion and the base is the ion or molecule that accepts the H+ ion. An acid and its conjugate base differ only by one H+ ion. a. NH4+
+
NH4+ + acid
SO4-2 conjugate base
b. CN-
7.21
+
HI
CN+ conjugate base
HI acid
SO4-2 HSO4acid
NH3 + conjugate base
HCN acid
+
Iconjugate base
Which of the following statements are correct at equilibrium? a. The concentration of reactants is always equal to the concentration of products. b. No reactants are converted into products. c. The rate of the forward reaction is equal to the rate of the reverse reaction. Only c. is correct. At equilibrium, the rates of the forward and reverse reactions are equal. Depending on the reaction, the equilibrium concentration of reactants can be greater than or less than product concentrations, therefore a. is not true. Answer b. is not true because at equilibrium the reaction continues to take place in both directions.
7.23
Balance the chemical equation and write the equilibrium constant expression. CH4
+
H2O
H2
For the generalized reaction, aA + bB the equilibrium constant expression is written:
7-5
+
CO
cC
+ dD
K eq =
[C ]c [ D]d [ A]a [ B]b
The balanced equation is: CH4
+
H2O
3H2
+
CO
and the equilibrium constant expression is:
[H 2 ]3[CO] K eq = [CH 4 ][H 2O] 7.25
Balance the chemical equation and write the equilibrium constant expression. SO2
+
O2
SO3
O2
2SO3
The balanced equation is: 2SO2 +
and the equilibrium constant expression is:
K eq =
7.27
[SO3 ]2 [SO 2 ]2 [O 2 ]
Write the equilibrium constant expression for each reaction. In part b, H2O is the solvent. a. C(s) + CO2(g)
2CO(g)
Remember that the concentrations of solids and solvents are not included in the Keq.
K eq =
[CO]2 [CO 2 ]
b. NH3(aq) + H2O(l)
NH4+(aq) + OH -(aq)
Remember that the concentrations of solids and solvents are not included in the Keq.
7-6
K eq =
7.29
[NH 4 ][OH − ] [NH3 ]
Explain why the concentrations of solids do not appear in equilibrium constant expressions. Only concentrations that change should be included in the equilibrium constant. The concentration of a solid is constant at a given temperature.
7.31
Write the reaction equation from which each equilibrium constant expression is derived (assume that no solids or solvents are present). The generalized equilibrium constant expression K eq =
[C ]c [ D]d [ A]a [ B]b
is based on the reaction equation aA + bB cC
a. K eq =
[NOCl]2 [NO]2 [Cl2 ]
2 NO + Cl2 b. K eq = H2
7.33
+ dD
2 NOCl
[HBr]2 [H 2 ][Br2 ]
+ Br2
2 HBr
Keq for the reaction below has a value of 4.2 x10-4. CH3NH2(aq) + H2O(l)
CH3NH3+(aq) + OH-(aq)
a. Write the equilibrium constant expression for the reaction.
7-7
K eq =
[CH3 NH3+ ][OH − ] [CH3 NH 2 ]
b. Which are there more of at equilibrium, products or reactants? Reactants. In the expression for Keq, the reactant concentration is in the denominator. Since the Keq value is very small this tells you that the reactant concentration is much larger than the product concentrations.
7.35
For each reaction, which is greater at equilibrium, the concentration of reactants or products? a. HPO42- + CN -
PO43- + HCN
Keq = 8.6 x 10-4
Reactants. Because Keq is small, reactant concentrations are larger than product concentrations. b. H3PO4 + CN -
H2PO4- + HCN
Keq = 1.5 x 107
Products. Because Keq is big, product concentrations are larger than reactant concentrations.
7.37 The enzyme carbonic anhydrase catalyzes the rapid conversion of CO2 and H2O into H2CO3. CO2 (g) + H2O(l ) H2CO3(aq) a. Write the equilibrium constant expression for this reaction. K eq =
[H 2 CO3 ] [CO 2 ]
b. What effect, if any, does doubling the amount of carbonic anhydrase have on an equilibrium mixture of H2CO3, CO2, and H2O? Explain. No effect. The catalyst increases the rate of the forward and reverse reactions to the same extent, but the concentrations of substances present at equilibrium are not affected.
7.39
When carbon monoxide reacts with hydrogen gas, methanol (CH3OH) is formed. CO(g) + 2H2(g)
CH3OH(g)
7-8
a. For the equilibrium above, what is the effect of increasing [CO]? Of decreasing [H2]? Of increasing [CH3OH]? Increasing [CO] favors the forward reaction (makes more products). Decreasing [H2] slows the rate of the forward reaction (favors the reverse reaction and makes more reactants). Increasing [CH3OH] increases the rate of the reverse reaction (favors the reverse reaction and makes more reactants). b. What would be the effect of continually removing CH3OH from the reaction? The reaction would continually proceed to the right to make CH3OH and would never reach equilibrium.
7.41
The reaction below is exothermic. If the reaction is at equilibrium, what would be the effect of 2 H2(g) + O2(g)
2 H2O(g)
a. decreasing concentration of O2? There would be a net reverse reaction until equilibrium is reestablished. b. increasing the concentration of H2? There would be a net forward reaction until equilibrium is reestablished. c. increasing the pressure? A net forward reaction (toward fewer moles of gas) would take place to reduce the pressure. d. decreasing the volume? Decreasing the volume would increase the pressure. There would be a net forward reaction until equilibrium is reestablished to reduce the pressure. e. decreasing the temperature? The reaction is exothermic 2 H2(g) + O2(g) 2 H2O(g) + heat Decreasing the temperature removes heat. There would be a net forward reaction until equilibrium is reestablished. f. increasing the temperature? Increasing the temperature adds heat. There would be a net reverse reaction until equilibrium is reestablished.
7-9
7.43
Calculate the H3O+ concentration present in water when a. [OH-] = 8.4 x 10-3 M b. [OH-] = 2.9 x 10-9 M c. [OH-] = 5.8 x 10-1 M To calculate the [H3O+], Kw = [H3O+][OH-]
[H3 O+ ] =
Kw [OH − ]
1.0 x 10-14 8.4 x 10-3
= 1.2 x 10-12 M
and
a. [OH-] = 8.4 x 10-3 M
[H3 O + ] = b. [OH-] = 2.9 x 10-9 M [H3O+] =
1.0 x 10-14 -9
2.9 x 10
= 3.4 x 10-6 M
c. [OH-] = 5.8 x 10-1 M
1.0 x 10-14 [H3 O ] = 5.8 x 10-1 +
7.45
= 1.7 x 10-14 M
In Problem 7.43, indicate whether each solution is acidic, basic, or neutral. A solution with [H3O+] > 1.0 x 10-7 M is acidic, with [H3O+] < 1.0 x 10-7 M is basic, and [H3O+] = 1.0 x 10-7 M is neutral.
7.47
a. [H 3O+ ] = 1.2 x 10 -12 M
basic
b. [H 3 O + ] = 3.4 x 10 -6 M
acidic
c. [H 3O+ ] = 1.7 x 10 -14 M
basic
Calculate the OH- concentration present in water when a. [H3O+] = 9.1 x 10-7 M b. [H3O+] = 1.3 x 10-3 M c. [H3O+] = 8.8 x 10-2 M.
7 - 10
[OH − ] =
Kw [H3 O+ ]
a. [H3O+] = 9.1 x 10-7 M
[OH − ] =
1.0 x 10-14 9.1 x 10-7
= 1.1 x 10-8 M
b. [H3O+] = 1.3 x 10-3 M
1.0 x 10-14 [OH ] = 1.3 x 10-3 −
= 7.7 x 10-12 M
c. [H3O+] = 8.8 x 10-2 M.
[OH − ] =
7.49
1.0 x 10-14 8.8 x 10-2
= 1.1 x 10-13 M
In Problem 7.47, indicate whether each solution is acidic, basic, or neutral. A solution with [H3O+] > 1.0 x 10-7 M is acidic, with [H3O+] < 1.0 x 10-7 M is basic, and [H3O+] = 1.0 x 10-7 M is neutral.
7.51
a. [H3O+] = 9.1 x 10-7 M
acidic
b. [H3O+] = 1.3 x 10-3 M
acidic
c. [H3O+] = 8.8 x 10-2 M.
acidic
Calculate the pH of a solution in which a. [H3O+] = 1 x 10-5 M b. [H3O+] = 3.9 x 10-2 M c. [H3O+] = 1 x 10-7 M d. [H3O+] = 7.0 x 10-5 M The pH of a solution can be calculated from the H3O+ concentration using the expression: pH = -log [H3O+] On a standard scientific calculator, enter the value for [H3O+] into the calculator, then press the log button. Multiply the result by -1. The number of digits after the
7 - 11
decimal point in the pH value should be the same as the number of significant digits in the concentration value. a. [H3O+] = 1 x 10-5 M pH
= -log (1 x 10-5)
=
5.0
=
1.41
=
7.0
=
4.15
b. [H3O+] = 3.9 x 10-2 M pH
= - log (3.9 x 10-2)
c. [H3O+] = 1 x 10-7 M pH
= - log (1 x 10-7)
d. [H3O+] = 7.0 x 10-5 M pH
7.53
= -log (7.0 x 10-5)
In Problem 7.51, indicate whether each solution is acidic, basic, or neutral. In a neutral solution pH = 7, in an acidic solution pH < 7, and in a basic solution pH > 7.
7.55
a.
pH = 5.0
acidic
b.
pH = 1.41
acidic
c.
pH = 7.0
neutral
d.
pH = 4.15
acidic
Calculate the pH of a solution in which a. [OH-] = 6.8 x 10-7 M b. [OH-] = 1 x 10-7 M c. [OH -] = 7.0 x 10-5 M d.[OH -] = 1 x 10-1 M First calculate the [H3O+]: Then calculate the pH:
[H3O+ ] =
Kw 1.0 x 10-14 = [OH − ] [OH − ]
pH = -log [H3O+]
7 - 12
a. [OH-] = 6.8 x 10-7 M
1.0 x 10-14 [H3 O ] = 6.8 x 10-7 +
pH
= 1.5 x 10-8 M
= -log (1.5 x 10-8)
= 7.82
b. [OH-] = 1 x 10-7 M
[H3 O + ] =
1.0 x 10-14 1 x 10-7
= 1 x 10-7 M
pH = -log (1 x 10-7)
= 7.0
c. [OH -] = 7.0 x 10-5 M
1.0 x 10-14 7.0 x 10-5
= 1.4 x 10-10 M
= -log (1.4 x 10-10)
= 9.85
[H3O + ] = pH
d. [OH -] = 1 x 10-1 M
[H3O + ] =
1.0 x 10-14 1 x 10-1
= 1 x 10-13 M
pH = -log (1 x 10-13)
7.57
= 13.0
In Problem 7.55, indicate whether each solution is acidic, basic, or neutral. In a neutral solution pH = 7, in an acidic solution pH < 7, and in a basic solution pH > 7.
7.59
a.
pH = 7.82
basic
b.
pH = 7.0
neutral
c.
pH = 9.85
basic
d.
pH = 13.0
basic
What is the concentration of H3O+ in a solution if the pH is a. 7.00 c. 9.37 7 - 13
b. 1.74
d. 10.3
To calculate the concentration of H3O+, [H3O+], from the pH, use the following equation: [H3O+] = 10-pH a. 7.00 [H3O+] = 10-7.00 = 1.0 x 10-7 M b. 1.74 [H3O+] = 10-1.74 = 1.8 x 10-2 M c. 9.37 [H3O+] = 10-9.37 = 4.3 x 10-10 M d. 10.3 [H3O+] = 10-10.3 = 5 x 10-11 M
7.61
What is the concentration of OH- in a solution if the pH is a. 4.12 b. 7.45 c. 2.80 d. 11.61 To calculate the concentration of OH-, [OH-], from the pH, use the following equations: [OH − ] =
[H3O+] = 10-pH a. 4.12
[H3O+] = 10-4.12 = 7.6 x 10-5 M -14
1.0 x 10 [OH - ] = = 1.3 x 10-10 M -5 7.6 x 10 b. 7.45 [H3O+] = 10-7.45 = 3.5 x 10-8 M
7 - 14
Kw [H3 O+ ]
[OH - ] =
1.0 x 10-14 = 2.9 x 10- 7 M -8 3.5 x 10
c. 2.80 [H3O+] = 10-2.80 = 1.6 x 10-3 M -14
1.0 x 10 [OH - ] = = 6.3 x 10-12 M -3 1.6 x 10 d. 11.61 [H3O+] = 10-11.61 = 2.5 x 10-12 M
[OH - ] =
7.63
1.0 x 10-14 = 4.0 x 10-3 M -12 2.5 x 10
Alkali metals react with water to produce alkaline (basic) solutions. Na for example, reacts with water to form NaOH and H2(g). Write a balanced chemical equation for this reaction. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
7.65
In Problem 7.63, what is reduced and what is oxidized? H is reduced because it lost a bond to oxygen. Na is oxidized because it lost electrons, going from neutral Na(s) to Na+.
7.67
A 1.00 mL sample of blood serum has a pH of 7.35. a. What is the concentration of H3O+? [H3O+] = 10-pH [H3O+] = 10-7.35 = 4.5 x 10-8 M b. What is the concentration of OH -? [OH-] = 1.0 x 10-14 / [H3O+] [OH-] = 1.0 x 10-14 / 4.5 x 10-8
7 - 15
[OH-] = 2.2 x 10-7 M c. How many moles of H3O+ are present?
1.00 mL x
4.5 x 10-8 mol 1 x 10-3 L x = 4.5 x 10-11 mol 1L mL
d. How many moles of OH - are present?
1.00 mL x
2.2 x 10-7 mol 1 x 10-3 L x = 2.2 x 10-10 mol 1L mL
e. How many H3O+ ions are present?
4.5 x 10
-11
6.02 x 1023 ions mol x = 2.7 x 1013 H 3O + ions 1 mo l
f. How many OH - ions are present?
2.2 x 10-10 mol x 7.69
6.02 x 1023 ions = 1.3 x 1014 OH - ions 1 mol
Sometimes the basicity of a solution is reported using pOH (pOH = -log[OH-]). What is the a. [OH-] if pOH = 5.20? b. [H3O+] if pOH = 4.33? c. pH if pOH = 8.50? Use the following relationships to calculate the value asked for in each question: [OH-] = 10-pOH
[H3O+] = 10-pH
pOH = -log[OH-]
pH = -log[H3O+]
[OH - ] =
1.0 x 10-14 [H3 O + ]
[H3 O+ ] =
a. [OH-] if pOH = 5.20? [OH-] = 10-5.20 = 6.3 x 10-6 M b. [H3O+] if pOH = 4.33?
7 - 16
1.0 x 10-14 [OH - ]
[OH-] = 10-4.33 = 4.7 x 10-5 M
[H3 O + ] =
1.0 x 10-14 = 2.1 x 10-10 M -5 4.7 x 10
c. pH if pOH = 8.50? [OH-] = 10-8.50 = 3.2 x 10-9 M
[H3 O + ] =
1.0 x 10-14 = 3.1 x 10- 6 M -9 3.2 x 10
pH = -log(3.1 x 10-6) = 5.51
7.71
Write the chemical equation for the reaction of each weak acid with water. Write the corresponding acidity constant expression. Recall that acids donate H+. Remove one H+ from the acid given and write the formula remaining, but lower the charge by 1. The H+ is added to a neutral water molecule to produce a hydronium ion. Follow the rules covered previously to write the equilibrium expression. a. NH4+ NH4+(aq) + H2O(l)
Ka =
H3O+(aq) + NH3(aq)
[H3O+ ][NH3 ] [NH 4 + ]
b. HPO4-2 HPO42-(aq) + H2O(l)
Ka =
7.73
H3O+(aq) + PO43-(aq)
[H3O+ ][PO 43− ] [HPO4 2− ]
a. The Ka of NH4+ equals 5.6 x 10-10. For the reaction in Problem 7.71a, what are there more of at equilibrium, reactants or products? 7 - 17
Reactants. The Ka is small. b. The Ka of HPO42- equals 4.2 x 10-13. For the reaction in Problem 7.71b, what are there more of at equilibrium, reactants or products? Reactants. The Ka is small. c. Which is the stronger acid, NH4+ or HPO42-? NH4+. The Ka of NH4+ is greater than the Ka of HPO42-.
7.75
Calculate the pKa of each acid and indicate which is the stronger acid. pKa is calculated the same way as pH, except that the Ka value is used instead of the [H3O+]. a. HClO, Ka = 3.0 x 10-8 pKa = -log Ka pKa = -log [3.0 x 10-8] = 7.52 On a standard scientific calculator, enter the number 3.0 x 10-8 into the calculator, then press the log button. This gives -7.52. Then, - (-7.52) = 7.52. Report the value to two digits to the right of the decimal point. b. C2O4H -, Ka = 6.4 x 10-5 pKa = -log Ka pKa = -log [6.4 x 10-5] = 4.19 C2O4H- is the stronger acid because acid strength increases as pKa decreases.
7.77
Write the formula of the conjugate base of each acid in Problem 7.75. Which of the two is the stronger base? a. For the acid HClO, ClO- is its conjugate base. b. For the acid HC2O4-, C2O42- is its conjugate base. Because HClO is the weaker acid, its conjugate base (ClO-) is the stronger base.
7 - 18
7.79
0.10 M solutions of each of the following acids are prepared: acetic acid (Ka = 1.8 x 10-5) and hydrofluoric acid (Ka = 6.6 x 10-4). Which acid solution will have the lower pH? Explain. Hydrofluoric acid. Hydrofluoric acid has a Ka of 6.6 x 10-4 while acetic acid has a Ka of 1.8 x 10-5. The higher the Ka value, the stronger the acid is, and the more acidic the solution (lower pH) is.
7.81
a. How many moles of NaOH are present in 31.7 mL of 0.155 M NaOH? Convert the volume of the NaOH to liters then use the molarity of the NaOH to convert to moles of NaOH.
10-3 L 0.155 mol 31.7 mL x x = 4.91 x 10-3 mol NaOH 1 mL 1L
b. How many moles of HC1 are present in a 15.0 mL sample that is neutralized by the 31.7 mL of 0.155 M NaOH? The reaction for the neutralization is: HCl + NaOH → H2O + NaCl. The number of moles of HCl reacted should be equal to the number of moles of NaOH present in the base solution: 4.91 x 10-3 mol NaOH x
1 mol HCl = 4.91 x 10-3 mol HCl 1 mol NaOH
c. What is the molar concentration of the HC1 solution described in part b of this question? The molar concentration is the number of moles of the HCl divided by the volume of the HCl solution given in liters. First, convert the volume of the HCl solution from mL to L and then divide the number of moles from part b by this number:
10-3 L 4.91 x 10-3 mol HCl 15.0 mL x = 0.327 mol/L or 0.327 M mL
7.83
It requires 17.2 mL of 0.100 M KOH to titrate 75.0 mL of an HCl solution of unknown concentration. Calculate the initial HCl concentration. Titration is a technique used to determine the concentration of an unknown solution. In an acid-base titration, the determination of the concentration is 7 - 19
made using the balanced equation for the neutralization reaction between an acid and a base. In this problem, the concentration of HCl can be determined by calculating the number of moles of HCl that react with the given amount of KOH. The balanced equation is given by: KOH(aq) + HCl(aq) → KCl(aq) + H2O(l) First, determine the number of moles of KOH that completely react with all of the HCl in the solution. Since we are working with molarities, convert 17.2 mL to L first.
17.2 mL x
10-3 L = 0.0172 L 1 mL
number of moles of KOH = 0.0172 L
x
0.100 mol KOH 1 L
= 1.72 x 10-3 mol KOH
Use the mole ratio from the balanced equation to determine the number of moles of HCl:
number of moles of HCl = 1.72 x 10-3 mol KOH
x
1 mol HCl 1 mol KOH
= 1.72 x 10-3 mol HCl
Then, calculate the concentration of HCl. The total volume of the original HCl solution is 75.0 mL or 0.0750 L.
1.72 x 10-3 mol HCl concentration of HCl = 0.0750 L
7.85
= 0.0229 M HCl
It requires 25.9 mL of 0.100 M HCl to titrate 15.0 mL of an NaOH solution of unknown concentration. What is the NaOH concentration? In this problem, the concentration of NaOH can be determined by calculating the number of moles of HCl that react with the given volume of NaOH. The balanced equation is given by: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Calculate the number of moles of HCl:
7 - 20
25.9 mL x
10-3 L = 0.0259 L 1 mL
number of moles of HCl = 0.0259 L
0.100 mol HCl 1 L
= 2.59 x 10-3 mol HCl
Use the mole ratio from the balanced equation to determine the number of moles of NaOH:
number of moles of NaOH = 2.59 x 10-3 mol HCl
x
1 mol NaOH 1 mol HCl
= 2.59 x 10-3 mol NaOH
Calculate the concentration of NaOH. The total volume of the original NaOH solution is 15.0 mL or 0.0150 L.
2.59 x 10-3 mol NaOH concentration of NaOH = = 0.173 M NaOH 0.0150 L
7.87
It requires 12.2 mL of 0.100 M NaOH to neutralize 10.0 mL of an unknown H2SO4 solution. Calculate the initial H2SO4 concentration. H2SO4
+
NaOH
Na2SO4
+
H2O
Na2SO4
+
2H2O
First, balance the equation: H2SO4
+
2NaOH
Determine the number of moles of NaOH required to neutralize all of the H2SO4 in the 10.0 mL of solution. Convert mL to L first. 10-3 L 12.2 mL x = 0.0122 L 1 mL moles of NaOH = 0.0122 L x
0.100 mol NaOH = 1.22 x 10-3 mol NaOH 1L
Use the mole ratio from the balanced equation to determine the number of moles of H2SO4 neutralized: 1.22 x 10-3 mol NaOH x
1 mol H2SO4 = 6.10 x 10-4 mol H2SO4 2 mol NaOH 7 - 21
Then, calculate the concentration of H2SO4. The total volume of the original H2SO4 solution is 10.0 mL or 0.0100 L. concentration of H2SO4 =
7.89
6.10 x 10-4 mol H2SO4 = 0.0610 M H2SO4 0.0100 L
It requires 19.2 mL of 0.25 M NaOH to neutralize 15.0 mL of an unknown H3PO4 solution. a. Balance the reaction equation. b. Calculate the initial H3PO4 concentration. a. The balanced equation is H3PO4
+
3NaOH
→
Na3PO4
+
3H2O
b. Follow the steps in the previous problems to calculate the initial H3PO4 concentration. 10-3 L volume of NaOH used = 19.2 mL x = 0.0192 L 1 mL moles of NaOH used = 0.0192 L x
0.25 mol NaOH = 4.8 x 10-3 mol NaOH 1L
moles of H3PO4 neutralized = 4.8 x 10-3 mol NaOH x 1 mol H3PO4 = 1.6 x 10-3 mol H3PO4 3 mol NaOH 1.6 x 10-3 mol H3PO4 concentration of H3PO4 = = 0.11 M H3PO4 0.0150 L
7.91
Ammonium ion (NH4+, pKa = 9.25) reacts with water to produce ammonia (NH3) and hydronium ion (H3O+). NH4+
+
H2O
NH3
+
H3O+
a. At pH = 9.25, which of the following is true? [NH4+] = [NH3], [NH4+] > [NH3], [NH4+] < [NH3]. [NH4+] = [NH3]
7 - 22
When the pH of the solution has the same value as the pKa, the concentration of the acid ([NH4+]) equals the concentration of its conjugate base ([NH3]). b. At pH = 5.25, which of the following is true? [NH4+] = [NH3], [NH4+] > [NH3], [NH4+] < [NH3]. [NH4+] > [NH3] When the pH is lower than the pKa (5.25 < 9.25), the concentration of the acid is greater than that of the conjugate base. c. At pH = 13.25, which of the following is true? [NH4+] = [NH3], [NH4+] > [NH3], [NH4+] < [NH3]. [NH4+] < [NH3] When the pH is greater than the pKa (13.25 > 9.25), the concentration of the acid is less than that of the conjugate base.
7.93
a. The pKa of NH4+ equals 9.25. For the reaction in Problem 7.71a, what is there more of at pH 7.0, NH4+ or NH3? NH4+ The pH is less than the pKa (7.0 < 9.25) therefore, the concentration of the acid ([NH4+]) is greater than that of the conjugate base ([NH3]). b. The pKa of HPO42- equals 7.21. For the reaction in Problem 7.71b, what is there more of at pH 10.0, HPO42- or PO43-? PO43The pH is greater than the pKa (10.0 > 7.21) therefore, the concentration of the acid ([HPO42-]) is less than that of the conjugate base ([PO43-]).
7.95
At low pH, the amino acid aspartic acid has the structure shown below. +
H3 N
CH CO 2H CH2 CO2 H
Aspartic acid a. Redraw aspartic acid, showing the acidic- CO2H and –NH3+ functional groups in their conjugate base form.
7 - 23
H2N CH CO2
CH2 CO2 -
b. In amino acids, -CO2H groups have pKa’s in the range 1.8 – 4.3 and –NH3+ groups have pKa’s in the range 9.1 – 12.5. Draw aspartic acid as it would appear at pH 7. What is the net charge on aspartic acid at this pH? +
H3N
CH CO 2 CH2 CO 2
The net charge is -1.
-
pH 7 is greater than the pKa range for -CO2H groups so the conjugate base form predominates (CO2-). pH 7 is less than the pKa range for –NH3+ groups so the acid form predominates (–NH3+) The net charge is – 1 because +1 -1 -1 = -1.
7.97
An equilibrium mixture of H2CO3 and HCO3- has a pH of 6.5. What happens to the H2CO3 and HCO3- concentrations when the pH of the solution is adjusted to 8.5? First, let’s write the relevant equilibrium reaction: H2CO3(aq) + H2O(l)
HCO3-(aq) + H3O+(aq)
Initially, the pH of the solution is 6.5. Raising the pH to 8.5 means a decrease in [H3O+]. A decrease in [H3O+] will cause a net forward reaction to take place, causing the [HCO3-] to increase and the [H2CO3] to decrease.
7.99
A buffer consisting of a weak acid and its conjugate base is prepared. Over what pH range will the buffer be most effective? pH = pKa ± 1
7.101 If you wish to maintain a pH of 6.0, which is the better buffer, H2CO3 and HCO3- or H2PO4- and HPO42-? H2CO3 and HCO3Buffers are most resistant to pH changes when the pH equals the pKa of the weak acid. The pKa of H2CO3 is 6.36 while the pKa of H2PO4- is 7.21. Because
7 - 24
the pKa of H2CO3 is closer to 6.0, this acid and its conjugate base would make a better buffer for a pH of 6.0. 7.103 A buffer can be prepared using NH4+ and NH3. a. Write an equation for the reaction that takes place when H3O+ is added to this buffer. b. Write an equation for the reaction that takes place when OH- is added to this buffer. A buffer is a solution that resists changes in pH. Buffers commonly contain a weak acid and its conjugate base. The weak acid will neutralize any added base while the conjugate base will neutralize any added acid. a. Write an equation for the reaction that takes place when H3O+ is added to this buffer. NH3 + H3O+ → NH4+ + H2O b. Write an equation for the reaction that takes place when OH- is added to this buffer. NH4+ +
OH-
→
NH3
+
H2O
7.105 Explain how the H2CO3/HCO3- buffer system helps maintain blood serum at a constant pH. This buffer makes use of the equilibrium H2CO3 + H2O
HCO3- + H3O+
In response to an increase in [H3O+] (a drop in pH) the equilibrium shifts to the left. In response to a decrease in [H3O+] (a rise in pH) the equilibrium shifts to the right.
7.107 Explain how respiration helps maintain blood serum at a constant pH. In response to a drop in blood pH (a rise in [H3O+]) the rate of respiration increases. As more CO2 is exhaled, [H2CO3] drops, as does [H3O+]. H2O + CO2 H2CO3 H2CO3 + H2O HCO3- + H3O+
7 - 25
In response to a rise in blood pH (a drop in [H3O+]) the rate of respiration slows. As CO2 accumulates, [H2CO3] increases, as does [H3O+]. H2O + CO2 → H2CO3 H2CO3 + H2O → HCO3- + H3O+ 7.109 Explain how the kidneys help maintain blood serum at a constant pH. The reaction for the H2CO3/HCO3- blood buffer system is: H2CO3 + H2O
HCO3- + H3O+
When the pH of blood serum becomes too low (high H3O+), the kidneys release HCO3- ions that neutralize excess H3O+. When the pH of the blood serum becomes too high, the kidneys do the opposite: HCO3- is removed from the blood causing the concentration of H3O+ to increase, thereby lowering the pH.
7.111 Explain how hyperventilation can lead to respiratory alkalosis. Hyperventilation causes CO2 to be exhaled faster than it can be formed by the cells. The decrease in CO2 causes a decrease in H2CO3. The pH increases resulting in alkalosis.
7.113
In terms of Le Chatelier’s principle, describe how the equilibrium Mb + O2 MbO2 responds to increases in the concentration of a. Mb When [Mb] is increased, the rate of the forward reaction increases and more MbO2 is produced. b. O2 When [O2] is increased, the rate of the forward reaction increases and more MbO2 is produced. c. MbO2 When [MbO2] is increased, the rate of the reverse reaction increases and more Mb and O2 are produced.
7 - 26
7.115 Calculate the logarithm of each number. Assume that each is a measured quantity. a. 1 x 10-9 Using a scientific calculator, enter the number 1 x 10-9, then press the log button. This gives -9. Report the log with one digit to the right of the decimal point. -9.0 b. 1 x 1012 Using a scientific calculator, enter the number 1 x 1012, then press the log button. This gives 12. Report the log with one digit to the right of the decimal point. 12.0 c. 3.4 x 10-2 Using a scientific calculator, enter the number 3.4 x 10-2, then press the log button. This gives -1.468. Report the log with two digits to the right of the decimal point. -1.47 d. 9.7 x 104 Using a scientific calculator, enter the number 9.7 x 104, then press the log button. This gives 4.98677. Report the log with two digits to the right of the decimal point. 4.99
7.117 Calculate the antilogarithm of each number. The numbers given in this problem are not assumed to be measured quantities, so the number of digits reported in the answers is arbitrary. a. 8 Using a scientific calculator, enter the number 8, then press the 2nd button followed by the log button. This gives 1 x 108. 1 x 108
7 - 27
b. –3 Using a scientific calculator, enter the number -3 (remember to put in 3 then press the sign change, NOT the subtraction button) then press the 2nd button followed by the log button. This gives 1 x 10-3. 1 x 10-3 c. 7.9 Using a scientific calculator, enter the number 7.9 into the calculator, then press the 2nd button followed by the log button. This gives 7.943 x 107. 7.9 x 107 d. 15.3 Using a scientific calculator, enter the number 15.3, then press the 2nd button followed by the log button. This gives 1.995 x 1015. 2.0 x 1015
7.119 Someone tells you that the spice turmeric is a natural pH indicator. How might you test this claim, using only substances that you are likely to have available at home? Test glass cleaner (containing basic ammonia) or soap and the pH should indicate a basic solution. Test vinegar and the pH should indicate an acidic solution.
7.121 A buffer may be prepared using acetic acid (pKa = 4.74) and its conjugate base. What is the pH of this buffer when CH3CO2H a. b. c. d.
+
H2O
⇋
CH3CO2-
H3O+
+
the concentration of CH3CO2H is twice that of CH3CO2-. the concentration of CH3CO2H is one-third that of CH3CO2-. [CH3CO2H] = 0.15 M and [CH3CO2-] = 0.62 M. [CH3CO2H] = 0.33 M and [CH3CO2-] = 0.095 M.
Use the Henderson-Hasselbalch equation to calculate the pH for each buffer mixture:
7 - 28
pH = pKa + log
[A-] [HA]
In this buffer, A- is CH3CO2- (the base) and HA is CH3CO2H and pKa = 4.7. pH = 4.74 + log
[CH3CO2-] [CH3CO2H]
a. the concentration of CH3CO2H is twice that of CH3CO2-. If [CH3CO2H] = 2[CH3CO2-], then [CH3CO2-] 1 pH = 4.74 + log = 4.44 - = 4.74 + log 2[CH3CO2 ] 2 b. the concentration of CH3CO2H is one-third that of CH3CO2-. [CH3CO2H] = (1/3) [CH3CO2-] pH = 4.74 + log
[CH3CO2-] = 4.74 + log 3 = 5.22 (1/3)[CH3CO2-]
c. [CH3CO2H] = 0.15 M and [CH3CO2-] = 0.62 M. pH = 4.74 + log
0.62 = 4.74 + log 4.1 = 5.35 0.15
d. [CH3CO2H] = 0.33 M and [CH3CO2-] = 0.095 M. pH = 4.74 + log
0.095 = 4.74 + log 0.29 = 4.20 0.33
7.123 A buffer may be prepared using ammonium ion (Ka = 5.6 x 10-10) and its conjugate base. For this buffer, NH4+
+
H2O
⇋
NH3
+
H3O+
a. What is [NH4+] if [NH3] = 0.15 M and pH = 8.25? b. What is [NH4+] if [NH3] = 0.36 M and pH = 9.30? c. What is [NH3] if [NH4+] = 0.050 M and pH = 8.55? Use the Henderson-Hasselbalch equation. For this type of problem, we are given the concentration of one of the buffer components and the pH of the solution. We will rearrange the Henderson Hasselbalch equation to calculate for the concentration of one of the buffer components as follows: 7 - 29
[A-] pH = pKa + log [HA] pH - pKa = log
[A-] [HA]
Taking the antilog of both sides, we have: [A-]
antilog (pH - pKa ) = [HA] 10 (pH - pKa ) =
[A-] [HA]
a. What is [NH4+] if [NH3] = 0.15 M and pH = 8.25? pKa = -log(5.6 x 10-10) = 9.25 10 (8.25 - 9.25) = 0.10 =
0.15 [NH4+]
0.15 [NH4+]
[NH4+] =
0.15 = 1.5 M 0.10
b. What is [NH4+] if [NH3] = 0.36 M and pH = 9.30? 10 (9.30 - 9.25) = 1.1 =
0.36 [NH4+]
0.36 [NH4+]
[NH4+] =
0.36 = 0.33 M 1.1
7 - 30
c. What is [NH3] if [NH4+] = 0.050 M and pH = 8.55?
10 (8.55 - 9.25) = 0.20 =
[NH3] 0.050
[NH3] 0.050
[NH3] = 0.20 x 0.050 = 0.010 M
7.125 Buffers are effective when the pH is within one unit of the pKa of the weak acid (pH = pKa ± 1). An HCO3-/CO32- buffer has a pH equal to 11.25. At this pH, how much CO32- is present than HCO3-? For HCO3-, the pKa = 10.25. HCO3-
+
H2O
⇋
CO32-
+
H3O+
Using the Henderson-Hasselbalch equation, calculate the ratio [CO32-]/ [HCO3-] using the given pH and pKa values. pH = pKa + log 10
(pH - pKa)
[CO32-] [HCO3-]
[CO32-] = [HCO3-]
[CO32-] (11.25 – 10.25) = 10 - = 10 [HCO3 ] [CO32-] = 10 [HCO3-] The concentration of CO32- is 10 times that of HCO3-.
7.127 a. Write the balanced chemical equation for the reversible acid-base reaction that takes place between nitrous acid (HNO2) and water. HNO2
+
H2O
⇋
7 - 31
NO2-
+
H3O+
b. Write the equilibrium constant expression for this reaction. −
Ka =
[ NO2 ][ H 3O + ] [ HNO2 ]
c. Name and draw the Lewis structure of the conjugate base of nitrous acid. The conjugate base of nitrous acid (HNO2) is the nitrite ion (NO2-). Its Lewis structure is: _
[O N O] d. For nitrous acid, Ka= 4.0 x 10-4. What is the value of pKa? pKa = -log Ka = -log 4.0 x 10-4 = 3.40 e. Which are there more of once the reaction in part a has reached equilibrium, reactants or products? Reactants. The value of the Ka is small. f. For carbonic acid (H2CO3), Ka = 4.4x10-7. Which is the stronger acid, nitrous acid or carbonic acid? HNO2 or nitrous acid. The Ka of nitrous acid (4.0 x 10-4) is greater than the Ka of carbonic acid 4.4 x 10-7.
g. Which is the stronger base, the conjugate base of nitrous acid or the conjugate base of carbonic acid? Conjugate base of carbonic acid. Carbonic acid is a weaker acid than nitrous acid. h. Assuming that the reaction in your answer to part a is at equilibrium, what will happen to restore equilibrium if the pH is lowered by addition of H3O+, a net forward reaction or a net reverse reaction? Net reverse reaction. A net reverse reaction will occur to consume the additional H3O+ and restore the equilibrium ratio of product concentrations to reactant concentrations. i. Over what pH range would a solution containing nitrous acid and its conjugate base be effective as a buffer? pH = pKa ± 1 = 3.40 ± 1 7 - 32
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 7.2
a. Write the equilibrium constant expression for the reaction. C ⇋ D b. For this reaction, Keq =0.25. The mixture of C and D shown below is not in equilibrium. Which is too high, [C] or [D]?
c. To adjust to equilibrium, which will take place: a net forward reaction or a net reverse reaction? d. Based on your answer to part c, redraw the picture in part b as it will appear once the mixture reaches equilibrium. Answer: a. K =
[ D] [C ]
[ D] 2 = = 0.66 . [C ] 3 Because this number is greater than 0.25, the concentration of [D] is too high.
b. The ratio of [D] to [C] is
c. A net reverse reaction will take place to adjust to equilibrium. This will cause the concentration of C to go up and the concentration of D to go down. d. One D is converted back to one C, decreasing the number of D to 1 and increasing the number of C to 4 bringing the ratio back to its equilibrium value: D C C
C
7 - 33
C
[ D] 1 = = 0.25 [C ] 4
SECTION 7.1
7.4
ACIDS AND BASES
Which of the observation in Problem 7.3 would indicate that a compound might be a base?
Answer: b. Turns litmus blue c. Has a bitter taste e. Dissolves a metal (some metals, e.g., aluminum) f. Feels slippery
7.6
Name each of the following as an acid and as a binary compound. a. HF b. HI
Answer: The names of acids that are binary compounds (they contain hydrogen and only one other nonmetal atom) always start with hydro and end with the nonmetal name, with the ending changed to “ic” instead of the usual “ide” ending. When naming the acid as a binary compound, use the regular naming rules for covalent compounds but the prefixes mono, di, tri, etc. are not used. a. Hydrofluoric acid and hydrogen fluoride b. Hydroiodic acid and hydrogen iodide
SECTION 7.2
7.8
BRONSTED-LOWRY ACIDS AND BASES
Give the formula for the conjugate base of each acid. a. HSO4c. HPO42b. H3PO4 d. HNO3
Answer: The conjugate base of an acid is the compound or ion that is formed after the acid loses an H+. When an H+ is removed, the resulting particle gains a negative charge.
7 - 34
7.10
a. SO42-
c. PO43-
b. H2PO4-
d. NO3-
Name the conjugate base of each acid in Problem 7.8
Answer: Consult the Table of Polyatomic Ions in previous chapter for the names of these polyatomic ions.
7.12
a. sulfate ion
c. phosphate ion
b. dihydrogen phosphate ion
d. nitrate ion
Give the formula for the conjugate acid of each base. a. OHc. CH3NH2 b. Cl d. PO32-
Answer: The conjugate acid of a base is the compound or ion that is formed after the base gains an H+. When an H+ is added, the resulting particle gains a positive charge.
7.14
a. H2O
c. CH3NH3+
b. HCl
d. HPO3-
Water is amphoteric (can act as an acid or a base). a. Write a chemical equation that shows water acting as a base. b. In this reaction, what is the conjugate acid of water?
Answer: a. H2O accepts an H+ from HCl: H2O + HCl ⇋ H3O+ + Clb. H3O+. The conjugate acid of H2O is what is formed after water accepts an H+.
7 - 35
7.16
HSO4- is amphoteric. a. Write the equation for the reaction that takes place between HSO4- and the acid HCl. b. Write the equation for the reaction that takes place between HSO4- and the base OH-.
Answer: a. HSO4- + HCl ⇋ H2SO4 + Clb. HSO4- + OH- ⇋ SO42- + H2O
7.18
Identify the Brønsted-Lowry acids and bases for the forward and reverse reactions of each. a. PO43-(aq) + NH4+(aq) ⇋ NH3(aq) + HPO42-(aq) b. HCN(aq) + H2PO4-(aq) ⇋ CN-(aq) + H3PO4(aq)
Answer: Brønsted-Lowry acids release H+ and Brønsted-Lowry bases accept H+. a. PO43- (aq) + NH4+(aq) ⇋ NH3(aq) + HPO42- (aq) base acid base acid b. HCN(aq) + H2PO4-(aq) ⇋ CN-(aq) + H3PO4(aq) acid base base acid
7.20 Complete each acid-base reaction. For the forward and reverse reactions, identify each acid and its conjugate base. a. OH- + HNO3 ⇋ b. HPO42- +
CO32- ⇋
Answer: a.
OH- + conjugate base
HNO3 acid
⇋
H2O acid
+
NO3conjugate base
b.
HPO42-+
CO32-
⇋
PO43-
+
HCO3-
7 - 36
acid
SECTION 7.3
conjugate base
conjugate base
acid
EQUILIBRIUM
7.22 Which of the following statements are correct at equilibrium? a. The forward and reverse reactions stop. b. The concentration of reactants and products does not change. c. The rate of the forward reaction is greater than the rate of the reverse reaction. Answer: Only b. is correct. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. As a result, the concentrations of the reactants and products remain constant because each one is used up as fast as it is produced. Both a. and c. are incorrect. The reactions do not stop. Equilibrium is a dynamic process in which the forward and backward rates are equal. There is no observable change because the concentrations of the reactants and products remain constant.
7.24
Balance the chemical equation and write the equilibrium constant expression. H2 + N2 ⇋ N2H4
Answer: For the general reaction below, the equilibrium constant expression is: aA + bB ⇋ cC
+
K eq =
dD
[C ]c [ D]d [ A]a [ B]b
For the reaction in this problem, the balanced equation and the equilibrium constant expression are 2H2 + N2 ⇋ N2H4 7.26
K eq =
[ N2 H4 ] [ H 2 ]2 [ N 2 ]
Balance the chemical equation and write the equilibrium constant expression. CS2 + H2 ⇋ CH4 + H2S
Answer:
7 - 37
For the general reaction below, the equilibrium constant expression is: aA + bB ⇋ cC
+
dD
K eq
[C ]c [ D]d = [ A]a [ B]b
For the reaction in this problem, the balanced equation and the equilibrium constant expression are CS2 + 4H2 ⇋ CH4 + 2H2S
K eq =
[CH 4 ] [ H 2 S ]2 [CS 2 ] [ H 2 ]4
7.28 Write the equilibrium constant expression for each reaction. In part b, H2O is a solvent. a. PCl5(g) ⇋ PCl3(g) + Cl2(g) b. H2SO4(aq) + 2 H2O(l) ⇋ 2 H3O+(aq) + SO42-(aq) Answer: For the general reaction below, the equilibrium constant expression is: aA + bB ⇋ cC
7.30
+
dD
K eq =
[C ]c [ D]d [ A]a [ B]b
[ PCl3 ][Cl2 ] [ PCl5 ]
a. K eq
=
b. K eq
[ H 3O+ ]2 [ SO4 2- ] = [ H 2 SO4 ]
Explain why the concentrations of solvents do not appear in equilibrium constant expressions.
Answer: Only concentrations that change should be included in the equilibrium constant. Solvent concentrations are constant at a given temperature.
7.32
Write the reaction equation from which each equilibrium constant expression is derived (assume that no solids or solvents are present).
7 - 38
a. K eq
=
[CH 4 ][ H 2 O] [CO][ H 2 ]3
b. K eq
=
[ SO3 ]2 [O2 ][ SO2 ]2
Answer: In an equilibrium constant, the terms in the numerator are the products and the terms in the denominator are the reactants. The exponents correspond to the balancing coefficients of each species in the equation. a. CO(g) + 3 H2(g) ⇋ CH4(g) + H2O(g) b. O2(g) + 2SO2(g) ⇋ 2SO3(g)
7.34
When steam is passed over coal and allowed to come to equilibrium at 800ºC, the equilibrium constant for the reaction has a value of 8.1 x 10-2. C(s) + H2O(g) ⇋ CO(g) + H2(g) a. Explain why the concentration of water appears in the equilibrium constant expression for this reaction, when it is usually omitted from such equations. b. Write the equilibrium constant expression for the reaction. c. Which are there more of at equilibrium, reactants or products?
Answer: a. H2O is not a solvent in this reaction; it is a gas. b.
K eq
=
[CO][H 2 ] [H 2O]
c. Since the Keq is less than 1, the concentration of H2O, a reactant, must be greater than the product of the concentrations of CO and H2. At equilibrium, there are more reactants.
7.36
For each reaction, which is greater at equilibrium, the concentration of the reactants or products? a. HF + HCO3 ⇋ F- + H2CO3 Keq = 1.5 x 103
7 - 39
b. NH4+ + HCO3 -
⇋
NH3 + H2CO3
Keq = 1.5 x 10-3
Answer: a. Products. Because Keq is large, there are more products at equilibrium. b. Reactants. Because Keq is small, there are more reactants at equilibrium.
SECTION 7.4
7.38
LE CHATELIER’S PRINCIPLE
Carbon disulfide (CS2), used in the manufacture of some synthetic fabrics, is prepared by heating sulfur (S2) and charcoal (C). S2(g) + C(s) ⇋ CS2(g) a. For the equilibrium above, what is the effect of increasing [CS2]? Of decreasing [CS2]? Of decreasing [S2]? b. What would be the effect of continually removing CS2 from the reaction?
Answer: Le Chatelier’s principle states that “when a reversible reaction is pushed out of equilibrium, the reaction responds to reestablish equilibrium”. It therefore reverses the effect of the disturbance. Changing the amounts of reactants increases the rate of the forward reaction while changing the amounts of the products increases the rate of the reverse reaction. a. Effect of increasing [CS2]: There is a net reverse reaction causing [CS2] to decrease and [S2] to increase until equilibrium is reestablished. C is a solid reactant and so its concentration is constant and not affected by equilibrium changes. Effect of decreasing [CS2]: There is a net forward reaction causing [S2] to decrease and [CS2] to increase until equilibrium is reestablished. Effect of decreasing [S2]: There is a net reverse reaction causing [CS2] to decrease and [S2] to increase until equilibrium is reestablished. b. Continually removing CS2 from the reaction results in a continuous reaction forward, pushing the conversion of all of the C and S2 reactants to products. The reaction would never reestablish equilibrium.
7.40
Nitrogen gas and hydrogen gas react to form ammonia. 7 - 40
N2(g) + 3H2(g) ⇋ 2NH3(g) a. For the equilibrium above, what is the effect of increasing [NH3]? b. What is the effect of increasing [H2]? c. What is the effect of decreasing [N2]? Answer: Le Chatelier’s principle states that “when a reversible reaction is pushed out of equilibrium, the reaction responds to reestablish equilibrium”. It therefore reverses the effect of the disturbance. Changing the amounts of reactants increases the rate of the forward reaction while changing the amounts of the products increases the rate of the reverse reaction. a. Effect of increasing [NH3]: There is a net reverse reaction to form more N2(g) and H2(g) until equilibrium is reestablished. b. Effect of increasing [H2]: There is a net forward reaction causing more N2(g) to be consumed and more NH3(g) to form until equilibrium is reestablished. c. Effect of decreasing [N2]: There is a net reverse reaction causing more NH3(g) to be consumed to form more N2(g) and H2(g) until equilibrium is reestablished.
7.42
The reaction below is endothermic. If the reaction is at equilibrium, what would be the effect of 3 O2 (g)
2 O3 (g)
a. decreasing concentration of O2? b. decreasing the concentration of O3? c. decreasing the pressure? d. increasing the volume? e. decreasing the temperature? f. increasing the temperature? Answer: a. There would be a net reverse reaction until equilibrium is reestablished. b. There would be a net forward reaction until equilibrium is reestablished. c. A net reverse reaction (toward more moles of gas) would take place until equilibrium is reestablished. d. Increasing the volume would decrease the pressure. There would be a net reverse reaction until equilibrium is reestablished. 7 - 41
e. The reaction is endothermic 3 O2 (g) + heat 2 O3 (g) Decreasing the temperature removes heat. There would be a net reverse reaction until equilibrium is reestablished. f. Increasing the temperature adds heat. There would be a net forward reaction until equilibrium is reestablished.
SECTION 7.5
7.44
IONIZATION OF WATER
Calculate the H3O+ concentration present in water when a. [OH -] = 4.8 x 10-8 M b. [OH -] = 6.6 x 10-2 M c. [OH -] = 1.5 x 10-12 M
Answer: Use the general equation, Kw = 1.0 x 10-14 = [H3O+][OH-] to solve [H3O+]. a.
Kw = 1.0 x 10-14 = [H3O+][OH-] [H3O+] = 1.0 x 10-14 / [OH-] [H3O+] = 1.0 x 10-14 / (4.8 x 10-8 ) [H3O+] = 2.1 x 10-7 M
b.
[H3O+] = 1.0 x 10-14/ [OH-] [H3O+] = 1.0 x 10-14/ (6.6 x 10-2 ) [H3O+] = 1.5 x 10-13 M
c.
[H3O+] = 1.0 x 10-14 / [OH-] [H3O+] = 1.0 x 10-14 / (1.5 x 10-12 ) [H3O+] = 6.7 x 10-3 M
7.46
In Problem 7.44, indicate whether each solution is acidic, basic, or neutral.
7 - 42
Answer: Acidic solutions have [H3O+] greater than 1 x 10-7 M. Basic solutions have [H3O+] less than 1 x 10-7 M. Neutral solutions have [H3O+] equal to 1 x 10-7 M. a. Acidic. [H3O+] = 2.1 x 10-7 M is greater than 1 x 10-7 M. b. Basic. [H3O+] = 1.5 x 10-13 M is less than 1 x 10-7 M. c. Acidic. [H3O+] = 6.7 x 10-3 M is greater than 1 x 10-7 M.
7.48
Calculate the OH - concentration present in water when a. [H3O+] = 6.2 x 10-4 M b. [H3O+] = 8.5 x 10-8 M c. [H3O+] = 1.9 x 10-11 M
Answer: Use the general equation, Kw = 1.0 x 10-14 = [H3O+][OH-] to solve [OH-]. a.
[OH-] = 1.0 x 10-14 / [H3O+] [OH-] = 1.0 x 10-14 / (6.2 x 10-4 M) [OH-] = 1.6 x 10-11 M
b.
[OH-] = 1.0 x 10-14 / [H3O+] [OH-] = 1.0 x 10-14 / (8.5 x 10-8) [OH-] = 1.2 x 10-7 M
c.
[OH-] = 1.0 x 10-14 / [H3O+] [OH-] = 1.0 x 10-14 / (1.9 x 10-11 ) [OH-] = 5.3 x 10-4 M
7.50
In Problem 7.48, indicate whether each solution is acidic, basic, or neutral.
Answer: Acidic solutions have [H3O+] greater than 1 x 10-7 M.
7 - 43
Basic solutions have [H3O+] less than 1 x 10-7 M. Neutral solutions have [H3O+] equal to 1 x 10-7 M. a. Acidic. [H3O+]= 6.2 x 10-4 M is greater than 1 x 10-7 M. b. Basic. [H3O+] = 8.5 x 10-8 M is less than 1 x 10-7 M. c. Basic. [H3O+] = 1.9 x 10-11 M is less than 1 x 10-7 M.
SECTION 7.6
7.52
THE pH SCALE
Calculate the pH of a solution in which a. [H3O+] = 4.9 x 10-3 M b. [H3O+] = 4.9 x 10-4 M c. [H3O+] = 2.2 x 10-13 M d. [H3O+] = 8.5 x 10-7 M
Answer: The pH of a solution can be calculated from the H3O+ concentration using the expression: pH = -log [H3O+] On a standard scientific calculator, enter the value for [H3O+] into the calculator, then press the log button. Multiply the result by -1. The number of digits after the decimal point in the pH value should be the same as the number of significant digits in the concentration value.
7.54
a. pH = -log (4.9 x 10-3)
= 2.31
b. pH = -log (4.9 x 10-4)
= 3.31
c. pH = -log (2.2 x 10-13)
= 12.66
b. pH = -log (8.5 x 10-7)
= 6.07
In Problem 7.52, indicate whether each solution in acidic, basic or neutral.
Answer: Acidic solutions have a pH less than 7. Basic solutions have a pH greater than 7. Neutral solutions have a pH = 7.
7 - 44
a. Acidic. pH = 2.31 is less than 7. b. Acidic. pH = 3.31 is less than 7. c. Basic. pH = 12.66 is greater than 7. d. Acidic. pH = 6.07is less than 7.
7.56 Calculate the pH of a solution in which a. [OH-] = 4.3 x 10-2 M b. [OH-] = 9.7 x 10-13 M c. [OH-] = 5.9 x 10-6 M d. [OH-] = 2.4 x 10-4 M
Answer: First calculate the [H3O+]: Then calculate the pH: a.
[H3O + ] =
pH b.
c.
pH d.
7.58
1.0 x 10-14 9.7 x 10−13
1.0 x 10-14 5.9 x 10−6
= -log (1.7 x 10-9)
[H3O + ] =
pH
1.0 x 10-14 4.3 x 10−2
= -log (1.0 x 10-2)
[H 3O + ] =
Kw 1.0 x 10-14 = [OH − ] [OH − ]
pH = -log [H3O+]
= -log (2.3 x 10-13)
[H 3O + ] =
pH
[H3O+ ] =
1.0 x 10-14 2.4 x 10−4
= -log (4.2 x 10-11)
= 2.3 x 10-13 M
= 12.64 = 1.0 x 10-2 M
= 2.00 = 1.7 x 10-9 M
= 8.77 = 4.2 x 10-11 M
= 10.38
In Problem 7.56, indicate whether each solution is acidic, basic, or neutral. 7 - 45
Answer: Acidic solutions have a pH less than 7. Basic solutions have a pH greater than 7. Neutral solutions have a pH = 7. a. Basic. pH = 12.64 is greater than 7. b. Acidic. pH = 2.00 is less than 7. c. Basic. pH = 8.77 is greater than 7. d. Basic. pH = 10.38 is less than 7.
7.60
What is the concentration of H3O+ in a solution if the pH is a. 5.54 b. 13.8 c. 2.94 d. 9.7
Answer: Use the general equation, [H3O+] = 10-pH to calculate the concentration of H3O+. On a standard scientific calculator, enter the negative of the pH value (e.g., -5.54) then press the 2nd button followed by the log button. The number of significant figures in the concentration is the same as the number of digits after the decimal point in the pH value.
a.
[H3O+] = 10-5.54 [H3O+] = 2.9 x 10-6 M
b.
[H3O+] = 10-13.8 [H3O+] = 2 x 10-14 M
c.
[H3O+] = 10-2.94 [H3O+] = 1.1 x 10-3 M
d.
[H3O+] = 10-9.7
7 - 46
[H3O+] = 2.0 x 10-10 M
7.62
What is the concentration of OH- in a solution if the pH is a. 2.77 b. 0.50 c. 14.19 d. 8.88 To calculate the concentration of OH-, [OH-], from the pH, use the following equations:
[OH − ] =
[H3O+] = 10-pH a. 2.77
[H3O+] = 10-2.77 = 1.7 x 10-3 M
[OH - ] =
1.0 x 10 -14 = 5.9 x 10 -12 M -3 1.7 x 10
b. 0.50 [H3O+] = 10-0.50 = 3.2 x 10-1 M
[OH - ] =
1.0 x 10 -14 = 3.2 x 10 -14 M -1 3.2 x 10
c. 14.19 [H3O+] = 10-14.19 = 6.5 x 10-15 M -14
1.0 x 10 [OH - ] = = 1.5 M 6.5 x 10 -15 d. 8.88 [H3O+] = 10-8.88 = 1.3 x 10-9 M
[OH - ] =
1.0 x 10 -14 = 7.6 x 10 -6 M -9 1.3 x 10
7 - 47
Kw [H3 O+ ]
7.64
Alkaline earth metals react with water to produce alkaline (basic) solution. Ca, for example, reacts with water to form Ca(OH)2 and H2(g). Write the balanced chemical equation for this reaction.
Answer: Ca + 2H2O → Ca(OH)2 + H2
7.66
In Problem 7.64, what is reduced and what is oxidized?
Answer: Ca is oxidized to Ca2+. Hydrogen in water is reduced to elemental H2.
7.68
A 5.00 mL sample of blood serum has a pH of 7.45. a. What is the concentration of H3O+? b. What is the concentration of OH-? c. How many moles of H3O+ are present? d. How many moles of OH- are present? e. How many H3O+ ions are present? f. How many OH- ions are present?
Answer: a.
[H3O+] = 10-pH [H3O+] = 10-7.45 = 3.5 x 10-8 M
b.
[OH-] = 1.0 x 10-14 / [H3O+] [OH-] = 1.0 x 10-14 / 3.5 x 10-8 = 2.9 x 10-7 M
c.
5.00 mL x
d.
10-3 L 2.9 x 10-7 mol 5.00 mL x x = 1.5 x 10-9 mol OH 1 mL 1L
e.
1.8 x 10-10 mol x
f.
6.02 x 1023 ions 1.5 x 10 mol x = 9.0 x 1014 OH - ions 1 mol
10-3 L 3.5 x 10-8 mol x = 1.8 x 10-10 mol H3O + 1 mL 1L
6.02 x 1023 ions = 1.1 x 1014 H 3O + ions 1 mol
-9
7 - 48
7.70
Sometimes the basicity of a solution is reported using pOH (pOH = -log[OH-]). What is the a. [OH-] if pOH = 2.50? b. [H3O+] if pOH = 13.44? c. pH if pOH = 9.84? Use the following relationships to calculate the value asked for in each question: [OH-] = 10-pOH
[H3O+] = 10-pH
pOH = -log[OH-]
pH = -log[H3O+]
[OH - ] =
1.0 x 10-14 [H3 O + ]
[H3 O+ ] =
1.0 x 10-14 [OH - ]
a. [OH-] if pOH = 2.50? [OH-] = 10-2.50 = 3.2 x 10-3 M b. [H3O+] if pOH = 13.44? [OH-] = 10-13.44 = 3.6 x 10-14 M
[H 3 O + ] =
1.0 x 10 -14 = 2.8 x 10 -1 M 3.6 x 10 -14
c. pH if pOH = 9.84? [OH-] = 10-9.84 = 1.4 x 10-10 M
[H 3 O + ] =
1.0 x 10 -14 = 6.9 x 10 -5 M -10 1.4 x 10
pH = -log(6.9 x 10-5) = 4.16
SECTION 7.7
7.72
ACID AND BASE STRENGTH
Write the chemical equation for the reaction of each weak acid with water. Write the corresponding acidity constant expression. a. CH3NH3+ b. HCO3-
7 - 49
Answer: The acidity constant expression for the dissociation of a weak acid is the same as the equilibrium constant expression. a. CH3NH3+(aq) + H2O(l) ⇋ CH3NH2(aq) + H3O+(aq)
Ka
=
[CH3 NH 2 ][H3 O + ] [CH3 NH3+ ]
b. HCO3-(aq) + H2O(l) ⇋ CO32-(aq) + H3O+(aq)
Ka
7.74
=
[CO32− ][H3 O+ ] [HCO3− ]
a. The pKa of CH3NH3+ equals 10.62. For the reaction in Problem 7.72a, what are there more of at equilibrium, reactants or products? b. The pKa of HCO3- equals 10.25. For the reaction in Problem 7.72b, what are there more of at equilibrium, reactants or products? c. Which is the stronger acid, CH3NH3+ or HCO3-?
Answer: Use the Ka value to determine what there is more of at equilibrium, reactants or products. pKa = -log Ka. Therefore, Ka = 10-pKa. To calculate Ka on a standard scientific calculator, enter the negative of the pKa then press the 2nd button followed by the log button. The number of significant figures in the Ka is the same as the number of digits after the decimal point in the pKa value. a. Reactants. Ka = 10-10.62 = 2.4 x 10-11. This is a small Ka value. Therefore, there would be more reactants present at equilibrium. b. Reactants. Ka = 10-10.25 = 5.6 x 10-11. This is a small Ka value. Therefore, there would be more reactants present at equilibrium. c. HCO3- is the stronger acid because its Ka (5.6 x 10-11) is greater than that of CH3NH3+. Acid strength increases as pKa decreases.
7.76
Calculate the pKa of each acid and indicate which is the stronger acid. a. C2O4H2, Ka = 5.9 x 10-2 b. C6O2H8, Ka = 1.7 x 10-5
7 - 50
Answer: The pKa can be calculated from the Ka using the following expression: pKa = -log Ka. Strong acids have low pKa’s and weak acids have high pKa’s. a. C2O4H2
pKa = -log (5.9 x 10-2) = 1.23
b. C6O2H8
pKa = -log (1.7 x 10-5) = 4.77
C2O4H2 is the stronger acid because its pKa is lower.
7.78
Write the formula of the conjugate base of each acid in Problem 7.76. Which of the two is the stronger base?
Answer: a. C2O4H – is the conjugate base of C2O4H2. b. C6O2H7 – is the conjugate base of C6O2H8. The conjugate base of C6O2H8 (C6O2 H7–) is stronger because C6O2H8 is the weaker acid.
7.80
0.50 M solutions of each of the following acids are prepared: HCN (pKa = 9.31) and H2CO3 (pKa = 6.36). a. Which solution will have the lower pH? Explain. b. Which solution will have the lower [OH-]? Explain.
Answer: a. H2CO3 will have the lower pH. H2CO3 is the stronger acid of the two because its pKa is lower. For two solutions of the same concentration, the solution that contains the stronger acid will have a lower pH (higher [H3O+]). b. H2CO3 will have the lower [OH-]. H2CO3 is the stronger acid of the two because its pKa is lower. For two solutions of the same concentration, the solution that contains the stronger acid will the higher [H3O+] and therefore, the lower [OH-].
SECTION 7.8
NEUTRALIZING ACIDS AND BASES
7 - 51
7.82
a. How many moles of NaOH are present in 71.3 mL of 0.551 M NaOH? b. How many moles of HC1 are present in a 25.0 mL sample that is neutralized by the 71.3 mL of 0.551 M NaOH? c. What is the molar concentration of the HC1 solution described in part b of this question?
Answer: a. Convert the volume of the NaOH solution to liters then use its molarity to calculate moles of NaOH. 1 x 10-3 L 0.551 mol 71.3 mL x x = 3.93 x 10-2 mol NaOH mL 1L
b. Use the balanced equation, NaOH + HCl → NaCl + H2O. Note that NaOH has one hydroxide ion and HCl has one hydrogen ion, so they will react 1:1. Convert moles of NaOH to moles of HCl. 3.93 x 10-2 mol NaOH x
1 mol HCl = 3.93 x 10-2 mol HCl 1 mol NaOH
c. The molar concentration is the number of moles of the HCl divided by the total volume of the HCl solution given in liters. First, convert the volume of the HCl solution from mL to L and then divide the number of moles from part b by this number:
molarity of HCl =
7.84
3.93 x 10-2 mol HCl = 1.57 mol/L or 1.57 M 10-3 L 25.0 mL 1 mL
It requires 35.7 mL of 0.250 M KOH to titrate 50.0 mL of an HCl solution of unknown concentration. Calculate the initial HCl concentration.
Answer: Convert the volume of KOH to liters then use the molarity to calculate the moles of KOH. 35.7 mL
10-3 L 1 mL
0.250 mol KOH = 8.93 x 10-3 mol KOH L of solution
Next, convert moles of KOH to moles of HCl. Note that KOH has one hydroxide ion and HCl has one hydrogen ion, so they will react 1:1.
7 - 52
8.93 x 10-3 mol KOH
1 mol HCl = 8.93 x 10-3 mol HCl 1 mol KOH
Finally, convert the volume of HCl to liters. Calculate the molarity by dividing the moles by the volume.
molarity of HCl =
7.86
8.93 x 10-3 mol HCl = 0.179 mol/L or 0.179 M 10-3 L 50.0 mL 1 mL
It requires 45.6 mL of 0.200 M HCl to titrate 25.0 mL of an NaOH solution of unknown concentration. What is the NaOH concentration?
Answer: First, write the balanced equation for the reaction: HCl(aq) + NaOH (aq) → NaCl(aq) + H2O(l) Then, proceed with the calculations based on the given information and the balanced equation above: moles of HCl = 45.6 mL
9.12 x 10-3 mol HCl x
0.200 mol HCl 1L
= 9.12 x 10-3 mol HCl
1 mol NaOH = 9.12 x 10-3 mol NaOH 1 mol HCl
concentration of NaOH =
7.88
10-3 L 1 mL
9.12 x 10-4 mol NaOH = 0.0365 M NaOH 0.0250 L
It requires 23.9 mL of 15.0 mM M HCl to neutralize 20.0 mL of an unknown Ca(OH)2 solution. Calculate the initial Ca(OH)2 concentration. 2HCl + Ca(OH)2 → CaCl2 + 2H2O
Answer: 10-3 L 15.0 mmol HCl 10-3 mol 23.9 mL x x x = 3.59 x 10-4 mol HCl 1 mL 1L 1 mmol
7 - 53
3.59 x 10-4 mol HCl x
1 mol Ca(OH)2 = 1.79 x 10-4 mol Ca(OH)2 2 mol HCl
1.79 x 10-4 mol Ca(OH)2 concentration of Ca(OH)2 = = 0.00896 M Ca(OH)2 0.0200 L 7.90
It requires 21.1 mL of 12.0 mM M H2SO4 to neutralize 15.0 mL of an unknown Ca(OH)2 solution. a. Balance the reaction equation. b. Calculate the initial Ca(OH)2 concentration. H2SO4 + Ca(OH)2 → Ca SO4 + H2O
Answer: a. The balanced equation is H2SO4 + Ca(OH)2 → CaSO4 + 2H2O b.
10-3 L 12.0 mmol H2SO4 10-3 mol 21.1 mL x x x = 2.53 x 10-4 mol H2SO4 1 mL 1L 1 mmol 2.53 x 10-4 mol H2SO4 x
1 mol Ca(OH)2 = 2.53 x 10-4 mol Ca(OH)2 1 mol H2SO4
2.53 x 10-4 mol Ca(OH)2 concentration of Ca(OH)2 = = 0.0169 M Ca(OH)2 0.0150 L SECTION 7.9
7.92
EFFECT OF pH ON ACID AND CONJUGATE BASE CONCENTRATIONS
Lactic acid (CH3CH(OH)CO2H, pKa = 3.85) reacts with water to produce lactate ion (CH3CH(OH)CO2-) and hydronium ion (H3O+). CH3CH(OH)CO2H + H2O ⇋ CH3CH(OH)CO2- + H3O+ a. At pH 3.85, which of the following is true? [lactic acid] = [lactate ion], [lactic acid] > [lactate ion], [lactic acid] < [lactate ion] b. At pH 11.61, which of the following is true? [lactic acid] = [lactate ion], [lactic acid] > [lactate ion], [lactic acid] < [lactate ion] c. At pH 1.66, which of the following is true? [lactic acid] = [lactate ion], [lactic acid] > [lactate ion], [lactic acid] < [lactate ion]
7 - 54
Answer: Use the following general rules to determine the relative concentrations at each pH value. When pH = pKa, [acid] = [conjugate base]. When the pH < pKa, [acid] > [conjugate base]. When the pH > pKa, [acid] < [conjugate base].
7.94
a. At pH 3.85, pH = pKa
[lactic acid] = [lactate ion]
b. At pH 11.61, pH > pKa
[lactic acid] < [lactate ion]
c. At pH 1.66, pH < pKa
[lactic acid] > [lactate ion
a. The Ka of CH3NH3+ equals 2.4 x 10-11. For the reaction in Problem 7.72a, what is there more of at pH 7.0, CH3NH3+ or CH3NH2? b. The Ka of HCO3- equals 5.6 x 10-11. For the reaction in Problem 7.72b, what is there more of at pH 3.0, HCO3- or CO32-?
Answer: First, calculate the pKa from the Ka values (pKa = -log Ka). Then, use the rules below to determine the relative concentrations of the acid and the conjugate base. When pH = pKa, [acid] = [conjugate base]. When the pH < pKa, [acid] > [conjugate base]. When the pH > pKa, [acid] < [conjugate base].
7.96
a.
pKa of CH3NH3+ = -log 2.4 x 10-11 = 10.62 pH < pKa, [CH3NH3+] > [CH3NH2]
b.
pKa of HCO3- = -log 5.6 x 10-11 = 10.25 pH < pKa, [HCO3-] > [CO32-]
At low pH the amino acid lysine has the structure shown below. +
H3N CH CO2H
+
CH2 (CH2)3 NH3 Lysine
a. Redraw lysine, showing the acidic –CO2H and –NH3+ functional groups in their conjugate base form. b. In amino acids, -CO2H group have pKa’s in the range 1.8 – 4.3 and –NH3+ groups have pKa’s in the range 9.1 – 12.5. Draw lysine as it would appear at pH 7. What is the net change on lysine at this pH?
7 - 55
Answer: a. In the conjugate base form, –CO2H and –NH3+ become –CO2- and –NH2, respectively.
H2N CH CO2 CH2 (CH2)3 NH2 Lysine b. For the -CO2H group, the pH > pKa, therefore the conjugate base form predominates (–CO2-) For the –NH3+ group, the pH < pKa, therefore the acid form predominates (– NH3+) +
H3N CH CO2 -
+
CH2 (CH2)3 NH3 Lysine
At pH 7, lysine would have a net charge of 1+.
7.98
An equilibrium mixture of CH3CO2H and CH3CO2- has a pH of 4.8. What happens to the CH3CO2H and CH3CO2- concentrations when the pH of the solution is adjusted to 2.8?
Answer: The concentration of CH3CO2H will increase and the concentration of CH3CO2will decrease. The pH going from 4.8 to 2.8 indicates that H3O+ ions are being added. In accordance to Le Châtelier’s principle, when H3O+ ions are added to the equilibrium, hydrogen ions will produce additional CH3CO2H by reacting with the CH3CO2-.
SECTION 7.10
BUFFERS
7.100 Which weak acid or weak acids from Table 7.4 would be the best choice if you wished to prepare buffers with the following pH values?
7 - 56
a. 7.0
b. 10.0
c. 4.0
Answer: Buffers are prepared by mixing a weak acid with its conjugate base. Buffers are most resistant to pH change when the weak acid and its conjugate base are at equal concentrations when pH = pKa. They are effective when pH = pKa + 1. a. H2CO3 (carbonic acid) and H2PO4- (dihydrogenphosphate ion). In Table 7.4 carbonic acid (pKa = 6.36) and dihydrogen phosphate ion (pKa = 7.21) each has a pKa that is within ± 1 of the pH. b. NH4+ (ammonium ion), HCN (hydrocyanic acid), HCO3- (hydrogencarbonate ion), or CH3NH3+ (methylammonium ion) In Table 7.4 ammonium ion has a pKa = 9.25 , hydrocyanic acid has pKa = 9.31, hydrogen carbonate has a pKa = 10.25, and methylammonium ion has a pKa = 10.62, which puts all of their pKa values within ± 1 of the pH. c. HF (hydrofluoric acid), CH3CH(OH)CO2H (lactic acid), or CH3CO2H (acetic acid) In Table 7.4 acetic acid has a pKa = 4.74 , lactic acid has pKa = 3.85, and hydrofluoric acid has a pKa = 3.18, which puts all of their pKa values within ± 1 of the pH. 7.102 If you wish to maintain a pH of 5.0, which is the better buffer, NH4+ and NH3 or CH3CO2H and CH3CO2-? Answer: CH3CO2H and CH3CO2Buffers are most resistant to pH changes when the pH equals or is within ± 1of the pKa of the weak acid. The pKa of CH3CO2H is 4.74 while the pKa of NH4+ is 9.25.
7.104 A buffer can be prepared using lactic acid [CH3CH(OH)CO2H] and its conjugate base, lactate ion [CH3CH(OH)CO2-]. a. Write an equation for the reaction that takes place when H3O+ is added to this buffer. b. Write an equation for the reaction that takes place when OH - is added to this buffer.
7 - 57
Answer: a.
CH3CH(OH)CO2- + H3O+ → CH3CH(OH)CO2H + H2O
b.
CH3CH(OH)CO2H + OH- → CH3CH(OH)CO2- + H2O
SECTION 7.11
MAINTAINING THE pH OF BLOOD SERUM
7.106 At which pH is the H2CO3/HCO3- buffer system most effective? Answer: Buffers are most effective when the pH is within ± 1 of the pKa of the weak acid. The pKa of H2CO3 is 6.36. So, the H2CO3/HCO3- buffer is most effective between 5.36 and 7.36.
7.108 In metabolic acidosis, the serum Pco2 levels can be lower than normal. Explain this form of compensation by the respiratory system. Answer: Metabolic acidosis is characterized by lower than normal blood serum pH. To correct this problem, the lungs breathe faster and lower Pco2 levels. This causes a net reverse for the reactions below and the blood serum pH rises. H2O + CO2 ← H2CO3 H2CO3 + H2O ← HCO3- + H3O+ 7.110 In respiratory alkalosis, the serum concentration of HCO3- can be lower than normal. Explain this form of compensation by the kidneys. Answer: Respiratory alkalosis occurs when CO2 is exhaled more rapidly than it is produced by cells (too little CO2 results in a drop in [H3O+]). The kidneys respond to a drop in [H3O+] by taking up HCO3-. In accordance with Le Chatelier’s principle, a drop in the concentration of HCO3- will drive the reaction forward leading to an increase in the concentration of H3O+. H2CO3 + H2O → HCO3- + H3O+
7 - 58
7.112 One treatment for respiratory alkalosis is to breathe into a paper bag. Explain how this helps restore the pH of the blood to its normal value. Answer: Any CO2 exhaled in to a paper bag is inhaled. This leads to a rise in the concentrations of H2CO3 and H3O+. H2O + CO2 → H2CO3 H2CO3 + H2O → HCO3- + H3O+
BIOCHEMISTRY LINK
DIVING MAMMALS, OXYGEN, AND MYOGLOBIN
7.114 Studies have shown that the value of the Keq for the reaction in the previous problem is identical for humans and diving animals. Why is the MbO2 in diving animals able to provide more O2 than the MbO2 in humans? Answer: The muscles of diving mammals contain three to ten times more myoglobin than human tissue. This results in more MbO2 being formed and more O2 being stored in the muscle tissues of diving mammals.
MATH SUPPORT
LOGS AND ANTILOGS
7.116 Calculate the logarithm of each number. Assume that each is a measured quantity. a. 1 x 103 b. 1 x 10-6 c. 2.2 x 10-13 d. 8.7 x 109 Answer: When taking the logarithm of a measured quantity, the number of significant digits in the measured quantity determines the number of digits after the decimal place in its logarithm. a. log (1 x 103) = 3.0 b. log (1 x 10-6) = -6.0
7 - 59
c. log (2.2 x 10-13) = -12.66 d. log (8.7 x 109) = 9.94
7.118
Calculate the antilogarithm of each number. a. -1 b. 12 c. 3.5 d. 10.9
Answer: To calculate the antilog of a number, calculate the value of 10 raised to the given number. For measured quantities, the number of significant digits in the answer is equal to the number of digits after the decimal point in the given quantity. In this problem, the numbers were not specified as being measured, so the significant figures reported are arbitrary. a. antilog (-1) = 10-1 = 0.1 b. antilog (12) = 1012 = 1 x 1012 c. antilog (3.5) = 103.5 = 3 x 103 d. antilog (10.9) = 1010.9 = 8 x 1010
BIOCHEMISTRY
PLANTS AS pH INDICATORS
7.120 When you pick a particular flower and immerse it in a pH 7 solution, there is no change in the color. If you place the same flower in a pH 9 solution, the color changes. Why does this happen? Answer: A variation in pH causes a change in the structure of the compound responsible for the color.
BIOCHEMISTRY
THE HENDERSON – HASSELBALCH EQUATION
7.122 A buffer may be prepared using dihydrogen phosphate (pKa = 7.21) and its conjugate base. What is the pH of this buffer when
7 - 60
H2PO4- + e. f. g. h.
H2O
⇋
HPO42-
+
H3O+
the concentration of H2PO4- is one-fifth that of HPO42-. the concentration of H2PO4- is four times greater than that of HPO42-. [H2PO4-] = 0.68 M and [HPO42-] = 0.86 M. [H2PO4-] = 0.55 M and [HPO42-] = 0.15 M.
Answer: Use the Henderson-Hasselbalch equation to calculate the pH for each buffer mixture: [A-] pH = pKa + log [HA] In this buffer, A- is HPO42- (the base) and HA is H2PO4- and pKa = 7.21. pH = 7.21 + log
[HPO42-] [H2PO4-]
a. the concentration of H2PO4- is one-fifth that of HPO42-. If [H2PO4-] = 1/5[HPO42-], then pH = 7.21 + log
[HPO42-] = 7.21 + log 5 = 7.91 1/5[HPO42-]
b. the concentration of H2PO4- is four times greater than that of HPO42-. If [H2PO4-] = 4[HPO42-], then [HPO42-] 1 pH = 7.21 + log = 6.61 2- = 7.21 + log 4[HPO4 ] 4 c. [H2PO4-] = 0.68 M and [HPO42-] = 0.86 M. pH = 7.21 + log
0.86 = 7.21 + log 1.3 = 7.31 0.68
d. [H2PO4-] = 0.55 M and [HPO42-] = 0.15 M. pH = 7.21 + log
0.15 = 7.21 + log 0.27 = 6.65 0.55
7 - 61
7.124 A buffer may be prepared using hydrogen carbonate ion (Ka = 5.6 x 10-11) and its conjugate base. For this buffer, HCO3-
+
H2O
⇋
CO32-
+
H3O+
d. what is [HCO3-] if [CO32-] = 0.053 M and pH = 10.90? e. what is [HCO3-] if [CO32-] = 0.36 M and pH = 9.44? f. what is [CO32-] and [HCO3-] if the combined concentration of CO32- and HCO3- = 0.50M and pH = 10.25? Answer: Use the Henderson-Hasselbalch equation. For this type of problem, we are given the concentration of one of the buffer components and the pH of the solution. We will rearrange the Henderson Hasselbalch equation to calculate for the concentration of one of the buffer components as follows: pH = pKa + log
[A-] [HA]
pH - pKa = log
[A-] [HA]
Taking the antilog of both sides, we have: [A-]
antilog (pH - pKa ) = [HA] 10 (pH - pKa ) =
[A-] [HA]
a. what is [HCO3-] if [CO32-] = 0.053 M and pH = 10.90? pKa = -log(5.6 x 10-11) = 10.25 10 (10.90 - 10.25) = 4.5 =
0.053 [HCO3-]
0.053 [HCO3-]
[HCO3-] =
0.053 = 0.012 M 4.5
b. what is [HCO3-] if [CO32-] = 0.36 M and pH = 9.44? 7 - 62
10 (9.44 - 10.25) = 0.15 =
0.36 [HCO3-]
0.36 [HCO3-]
[HCO3-] =
0.36 = 2.4 M 0.15
c. what is [CO32-] and [HCO3-] if the combined concentration of CO32- and HCO3- = 0.50M and pH = 10.25?
10
(10.25 - 10.25)
[CO32-] = [HCO3-]
[CO32-] 1.0 = [HCO3-] Since [CO32-] = [HCO3-] and [CO32-] + [HCO3-] = 0.50 M, then [HCO3-] = [CO32-] = 0.50 M/2 = 0.25 M 7.126 For the buffer in Problem 7.125, at pH 9.25 how many times less CO32- is present than HCO3-? Answer: [CO32-] pH = pKa + log [HCO3-] 10
(pH - pKa)
[CO32-] = [HCO3-]
[CO32-] (9.25 – 10.25) = 0.10 - = 10 [HCO3 ] [CO32-] = 0.10 [HCO3-] The concentration of CO32- is 1/10 that of HCO3-.
LEARNING GROUP PROBLEMS 7 - 63
7.128 a. Write the balanced equation for the reversible acid-base reaction that takes place between hypochlorous acid (HOCl) and water. b. Write the equilibrium constant expression for this reaction. c. Draw the Lewis structure of the conjugate base of hypochlorous acid. d. For hypochlorous acid, Ka = 3.5 x 10-8. What is the value of pKa? e. Which are there more of once the reaction in part a has reached equilibrium, reactants or products? f. For carbonic acid (H2CO3), Ka = 4.4 x 10-7. Which is the stronger acid, hypochlorous acid or carbonic acid? g. Which is the stronger base, the conjugate base of hypochlorous acid or the conjugate base of carbonic acid? h. Assuming that the reaction in your answer to part a is at equilibrium, what will happen to restore equilibrium if the pH is lowered by addition of H3O+, a net forward reaction or a net reverse reaction? i. Over what pH range would a solution containing hypochlorous acid and its conjugate base be effective as a buffer? Answer: a. HOCl(aq) + H2O(l) ⇋ OCl-(aq) + H3O+(aq) b. K a
[OCl− ][H3O+ ] = [HOCl]
c. O-Cld. pKa = -log 3.5 x 10-8 = 7.46 e. Reactants. The Ka value is very small. f. Carbonic acid (H2CO3) is the stronger acid because its Ka is greater than that of hypochlorous acid. g. The conjugate base of hypochlorous acid is the stronger base because hypochlorous acid is the weaker acid. h.There will be a net reverse reaction. i. 7.46 ± 1 (6.46 to 8.46)
7 - 64
Chapter 8 Organic Reactions 1 – Hydrocarbons, Carboxylic Acids, Amines, and Related Compounds SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
8.1
Identify the term that best describes the two molecules below: constitutional isomers, geometric isomers, different conformations of the same molecule, two unrelated molecules.
Geometric isomers. The two molecules differ only by the configuration around the C=C. The left molecule has two identical groups on the same side of the double bond. The right molecule has two identical groups on opposite sides of the double bond.
8.3
The ester methyl salicylate (wintergreen oil) can be prepared as shown below. Circle the atoms in the reactant molecules that are the source of the water molecule that appears in the product. O
O
C OH
C O CH3 OH
OH
+ 8.5
+
HOCH3
Which has the higher boiling point, CH3CH2CH2CH2CH3 or CH3CH(CH3)CH2CH3? Explain. CH3CH2CH2CH2CH3.
8-1
H2O
These alkanes interact with like molecules through London forces. While both molecules have five carbons, the one on the left has greater surface area and stronger London force interactions.
8.7
Arrange the molecules in order, from the highest boiling point to lowest boiling point: decane, propane, butane. decane > butane > propane Note that these are all “straight chained” molecules so that branching need not be considered. In this, the longer the hydrocarbon chain, the higher the boiling point will be. Remember that boiling points increase as the London force attraction increases. Decane (10 carbon atoms) is the longest and has the strongest London force interactions, followed by butane (4 carbon atoms) and propane (3 carbon atoms).
8.9
Draw a Lewis structure of each alkane. Connect every pair of atoms in the formula with a single line to represent the covalent bond between them. Note how the (CH3) group is written as a side branch. a. CH3CH2C(CH3) 3
H H C H H
H
H C
C
H
H
H C H
C
H
H C H H
8-2
b. CH3CH2CH(CH3)CH(CH3)CH2CH3
H H C H H H H
H H
H C C C C C C H H H
H H H
H C H H
8.11
Draw a skeletal structure for each of the molecules in Problem 8.9. The skeletal structure is drawn by representing all covalent bonds by lines and leaving out all of the carbon and hydrogen atoms.
a.
8.13
b.
Draw a Lewis structure for each molecule.
H H C H H H
H
H C C C C H a.
H H H H
8-3
H H C H
H
H C H
H C H
H H
H
H
H C C C
C
C C H
H H H
H H H C H
b.
H
8.15
8.17
Write a condensed structural formula for each molecule in Problem 8.13.
a.
CH3CH2CH(CH3)2
b.
CH3CH2CH(CH2CH3)CH(CH3)CH(CH3)2
Find and name the parent chain for each molecule, then give the complete IUPAC name for each. a. b. c. CH2CH2CH3
CH3CH2CHCH2CH3 CH3
CH3CHCH2CH3 CH2CH2CH3
CH3CCH2CH3 CH2CH2CH3
First, identify the parent chain by finding the longest chain of carbon. Remember that the longest chain is not necessarily the one that goes straight across. Assign the appropriate name using the numbering prefix and the ending “ane”. Next, identify and name the substituents. Finally, to complete the IUPAC naming process, number the carbons of the parent chain starting from the end nearer the first substituent.
8-4
a. 3-methylpentane. For this molecule, the longest chain has five carbon atoms, making its parent parent chain 5
4
3
2
1
CH3CH2CHCH2CH3 1
2
3 4
5
CH3 methyl chain pentane. The only substituent is a methyl group and the numbering gives the same number (3) for the methyl position whether you count left to right or right to left. b. 3-methylhexane. Note how the parent chain turns and is not straight across. The longest chain parent chain 3
2
1
CH3CHCH2CH3 4 4
methyl
5
5
6 6
CH2CH2CH3 3
2
1
has six carbon atoms making the parent chain hexane. This molecule has only one substituent, the methyl group. In this case numbering top to bottom gives the methyl group a position of 3 but numbering from bottom to top gives a position of 4. The number 3 is the assigned position. c. 4-ethyl-4-methylheptane. Note how the parent chain turns and is not straight across. The longest parent chain 3
methyl
2
1
CH2CH2CH3 5
6
7
4
ethyl
CH3 C CH2CH3 5
6
7
CH2CH2CH3 3
2
1
continuous chain is seven carbon atoms long making the parent chain heptane. This molecule has two substituents, one methyl group and one ethyl group, given
8-5
in alphabetical order in the name. The carbon they are attached to gets the same assigned number from both directions.
8.19
Give the IUPAC name of each molecule. Follow the general IUPAC rules for naming compounds. a.
CH3CH(CH3)CH2CH2CH3
2-methylpentane
b.
CH3CH2CH(CH3)CH2CH3 3-methylpentane CH3
c.
CH3 CH2 CH2 CH2 CH2
hexane CH3 CH CH2 CH3 d.
CH2CH3
3-methylpentane
8.21
Draw and name six of the constitutional isomers with the formula C7H16. Start with the normal straight chain molecule and begin making isomers by shortening the parent chain by one methyl group and placing that methyl group in as many different places as possible. Next, take off two methyl groups and place them on the remaining parent chain to additional structures. Continue the process until you have created six structures. Remember that simply bending or turning the molecule does not make it a different structure. For the molecule C7H16, there are a total of nine constitutional isomers.
heptane
2-methylhexane
8-6
3-methylhexane
2,3-dimethylpentane
2,2-dimethylpentane
3,3-dimethylpentane
2,4-dimethylpentane
2,2,3-trimethylbutane
3-ethylpentane
8.23
Which of the following pairs of molecules are constitutional isomers? Constitutional isomers are molecules that have the same molecular formula but whose atoms are connected differently. If the two molecules have the same
8-7
molecular formula and the atoms are connected the same way, then they are identical molecules. If they have the same molecular formula but different atomic connections, then they are constitutional isomers. The skeletal structures are given to show the atomic connections. a. CH3CH2CH2CH3 and CH3CH(CH3)CH3 constitutional isomers
and
b. CH3CH(CH3)CH2CH3 and CH3CH2CH(CH3)CH3 identical molecules.
and
c. CH3CH(CH2CH3)CH2CH3 and CH3CH2CH(CH3)CH2CH3 identical molecules.
and
8.25
Which pairs of molecules are constitutional isomers? Which are identical?
CH3CHCH3 a.
CH3CH2CH2CH2CH2CH3
and
constitutional isomers
8-8
CH3CH2CH3
CH2CH2CH3
CH2CH2CH2CH3
b.
CH2CH3
CH3CH2CH2
and
identical molecules
CH3 CH3CHCHCH3
CH3
c.
CH3CH CH2CH2CH3
CH3
and
constitutional isomers
8.27
Which are constitutional isomers? a. pentane and 2-methylpentane No. They have different molecular formulas. Pentane is C5H12 and 2methylpentane is C6H14. b. 2-methylpentane and 3-methylpentane Yes. Both molecules have the same formula, C6H14. Different names indicate different atomic connections. c. 2,2-dimethylpropane and pentane Yes. Both molecules have the same formula, C5H12 and different names (different atomic connections). d. 2,2-dimethylpropane and cyclopentane No. The molecules do not have the same molecular formula.
8.29
Which pairs of molecules are constitutional isomers? Which are different conformations of the same molecule? To be constitutional isomers the molecules must have the same molecular formula (which all of the molecules in both a and b do); and they must have different atomic connections. The molecules in part a fail on the second criterion since both
8-9
molecules are butane and have the same atomic connections. In part b, however, the first molecule is butane and the second is 2-methylpropane. This makes the molecules constitutional isomers. Since the molecules in part a are the same molecule with the hydrogen atoms rotated into different special orientations, they are different conformations of the same molecule.
a. Different conformations. The molecules have the same molecular formula and the same atomic connections. They differ only by the confirmation around the second carbon atom from the left.
b. Constitutional isomers. The molecules have the same molecular formula but different atomic connections.
8.31
a. How are constitutional isomers and conformations similar? In order to be constitutional isomers or conformations, molecules must have the same molecular formula. b. How are constitutional isomers and conformations different? Constitutional isomers have different atomic connections. Conformations have the same atomic connections, but different three-dimensional shapes that are interchanged by bond rotation.
8 - 10
8.33
Give the IUPAC name for each molecule.
CH3 a. methylcyclobutane The parent ring is cyclobutane. The methyl group is named as a substituent. No numbering required because there is only one substituent.
H3C
CH3
b. 1,3-dimethylcylohexane The parent ring is cyclohexane. There are two methyl substituent groups at carbon 1 and carbon 3.
CH2CH3 CH3 c. 1-ethyl-2-methylcyclopentane The parent ring is cyclopentane. The substituent groups are named and numbered in alphabetical order: ethyl at carbon 1 and methyl at carbon 2.
H3C
d.
CH3
CH2CH2CH3
1,1-dimethyl-2-propylcyclopropane
8 - 11
The parent ring is cyclopropane. There are 3 substituent groups: 2 methyl and 1 propyl. Name and number in alphabetical order.
8.35
Which molecule(s) in Problem 8.33 can exist as cis and trans isomers? For a cycloalkane to have cis and trans isomers, it must have at least two substituents attached to different ring atoms and the substituents must have different spatial orientations (on the same face of the ring – cis; on opposite sides – trans). a. No. There is only one substituent. b. Yes. There are two methyl substituent groups on different carbon atoms of the ring. c. Yes. There are two substituent groups, ethyl and methyl, on different carbon atoms of the ring. d. No. The same molecule results regardless of whether the propyl group points up or down relative to the two identical methyl substituent groups on the neighboring ring carbon atom.
8.37
Draw and name the three ethylmethylcyclobutane constitutional isomers. To be constitutional isomers, the molecules must have different atomic connections. They also have different IUPAC names.
CH2CH3
CH3
1-ethyl-2-methylcyclobutane
CH2CH3
H3C
1-ethyl-3-methylcyclobutane
CH2CH3 CH3
1-ethyl-1-methylcyclobutane
8 - 12
8.39
a. What is the molecular formula of cyclopropane? C3H6. The expanded structure of cyclopropane is given below. CH2 CH2 CH2
b. Draw a constitutional isomer of cyclopropane. H
H C H
8.41
C
C
H
H
H
Draw a side view of each cycloalkane. (See Figure 8.6 for examples) Trans means that substituents are to be drawn on opposite faces of the ring and cis means they are drawn on the same face.
H
H H
H H
H
CH3
a. trans-1,2-dimethylcyclohexane
H H H
b. trans-1-ethyl-2-methylcyclohexane
8 - 13
CH3
H H H
H CH2CH3
H H
H H H
H
CH3
H
H H
H CH3CH2
c. cis-1,3-diethylcyclopentane
8.43
H
H H CH2CH3 H
Give the complete IUPAC name (including the use of the term cis or trans) for each molecule. First identify the parent by counting the number of carbon atoms in the ring. Next, identify the substituent groups and assign numbering positions. Finally, use the prefix cis if the substituent groups are on the same face of the ring or trans if they are on opposite faces. H
H
CH3CH2CH2 H
H
CH2CH2CH3
H
H
H
H
H
H
a.
cis-1,4-dipropylcyclohexane
H H
CH2CH3 H H H CH3CH2 H b.
H
H
trans-1,3-diethylcyclopentane
8.45
Draw propene, showing the proper three-dimensional shape about each atom. The “ene” ending indicates a hydrocarbon with a double bond. The carbon atoms involved in the double bond are surrounded by a trigonal planar arrangement of atoms. The other carbon atom is surrounded by a tetrahedral arrangement.
8 - 14
8.47
Name each molecule. When naming hydrocarbons with double (alkene) or triple (alkyne) bonds, follow the same basic rules as for the alkanes, except for alkenes use an “ene” ending on the parent chain and for alkynes use an “yne” ending. The longest chain should include the double or triple bond. Number the longest chain from the end closest to the double or triple bond. Indicate the position of the double or triple bond using the lower number. a.
CH2
CHCH2CH3 1-butene
CH3CHCH CCH3
b.
CH3
CH3
2,4-dimethyl-2-pentene CH3C
c.
CCHCH3 CH3
4-methyl-2-pentyne
8.49
Draw each molecule Draw the parent chain first, placing the multiple bond in the position indicated in front of the parent chain name (remember “ene” means double bond and “yne” means triple bond). Next, starting from the same end used to count off the multiple bond position, count to the position number indicated and draw the substituents in their indicated positions.
8 - 15
CH3CHCH3 CH3CH2CH2CH2CHCH
a. 3-isopropyl-1-heptene
CH3C
CH2
CCH3
CH3 CH3
b. 2,3-dimethyl-2-butene
CH3CHCH2CH3 CH3CH2CH2CH2CHC
c. 5-sec-butyl-3-nonyne
8.51
CCH2CH3
The molecule shown is a termite trail marking pheromone. Which double bonds are cis and which are trans?
Recall that cis means that the atoms or groups of atoms being compared (in this case the attached carbon atoms) are on the same side of a line connecting the two double bonded carbon atoms, and trans means they are on opposite sides of that line.
cis
cis
trans
8.53
Name each molecule. Aromatic rings follow the same naming rules as used earlier in the chapter for cycloalkanes. The substituent groups are numbered using the lowest set of digits. Substituent groups are named and numbered in alphabetical order. In the case of only two substituents, the ortho (side by side), meta (one carbon between), para (on opposite side of the ring) designations may be used.
8 - 16
CH2 CH 2 CH 3
H3 C a.
1-methyl-3-propylbenzene CH 3CH CH 3
b. isopropylbenzene
c.
CH 3
CH 3 CH 3
8.55
1,2,3-trimethylbenzene
Draw each molecule. Draw the benzene ring first. Next, draw the substituents at the carbon number location designated in the name.
a. 1,3-dipropylbenzene
b.
p-diethylbenzene
8 - 17
c.
4-isobutyl-1,2-dimethylbenzene
8.57
Draw the organic product of each reaction. If more than one product is possible, draw them all. These reactions are halogenation reactions in which a halogen atom replaces a hydrogen atom on an alkane. The products given here show only one halogen atom substituting for one hydrogen atom on the alkane.
light a. In this halogenation reaction, a chlorine atom replaces a hydrogen atom on the alkane. When only one halogen atom substitutes for one hydrogen atom on the alkane, there is only one possible product. Placing the Cl atom on any of the other C atoms produces the same molecule shown below.
light
light b. In this halogenation reaction, a bromine atom replaces a hydrogen atom on the alkane. There are two possible products.
light
8 - 18
light
c. In this halogenation reaction, a bromine atom replaces a hydrogen atom on the alkane. There are three possible products. light
8.59
Draw the organic product of each reaction. If more than one product is possible, draw them all. These reactions are halogenation reactions in which a halogen atom replaces a hydrogen atom on an alkane. The products given here show only one halogen atom substituting for one hydrogen atom on the alkane.
a.
light
In this halogenation reaction, a chlorine atom replaces a hydrogen atom on the alkane. There is only one possible product.
light
8 - 19
light
b. In this halogenation reaction, a chlorine atom replaces a hydrogen atom on the alkane. The possible products are shown below.
light
c. In this halogenation reaction, a bromine atom replaces a hydrogen atom on the alkane. The possible products are shown below.
8.61
Draw the organic product of each reaction. Pt H2
a.
Pt H2
b.
8 - 20
Pt H2
c.
These reactions are catalytic hydrogenations. The two H atoms will add to the 2 carbon atoms connected by a double bond converting the alkene into an alkane.
Pt H2
a.
Pt H2
b.
Pt H2
c.
8.63
Draw the major organic product of each reaction.
a.
b.
c. Reactions of alkenes with water are hydration reactions. H2O, in the presence of a H+ catalyst adds as a H and an OH to the 2 carbon atoms connected by a double bond producing an alcohol. If the H2O adds to a symmetric alkene, there is only one possible alcohol product. If the H2O adds to an asymmetric alkene, there are
8 - 21
two possible alcohol products. The major and minor products are predicted using Markovnikov’s rule.
a.
b. The alkene in this reaction is an asymmetric alkene and therefore two product alcohols are possible. Markovnikov’s rule states that the major product predicted is the one formed when the H atom adds to the carbon in the C=C that already has more H atoms.
c. The alkene in this reaction is an asymmetric alkene and therefore two product alcohols are possible. Markovnikov’s rule states that the major product predicted is the one formed when the H atom adds to the carbon in the double that already has more H atoms.
8.65
Draw the organic product of each reaction. If more than one product is possible, draw them all.
a.
8 - 22
b.
c.
These are aromatic halogenation reactions, in which a hydrogen atom in an aromatic ring is replaced by a halogen atom.
a. There is only one possible product.
b. There are three possible products.
8 - 23
c. There are three possible products.
8.67
a. Draw the possible products of the substitution reaction.
b.
Draw the major product of the hydration reaction.
8 - 24
a.
Note: Substitutions can also take place on the ethyl substituent but are not shown here.
b.
8.69
a. The phenol molecule below is a tick pheromone (chemical messenger). This is one of the organic products formed when o-chlorophenol undergoes aromatic substitution. Write a chemical equation for this reaction.
The reaction involves o-chlorophenol reacting with Cl2 using a Fe catalyst. The equation is shown below.
HCl
8 - 25
b. Draw all other possible aromatic substitution products of the reaction in part a.
8.71
Name each carboxylic acid. Follow the IUPAC rules for naming carboxylic acids. The parent chain is enclosed in a box and any substituent group is circled. Begin numbering the carbon atoms on the parent chain starting from the C atom of the carboxyl group.
CH3
O
CH3 CH CH2 CH CH2 C OH CH3
a.
3,5-dimethylhexanoic acid
O
b.
HO C CH2 CH3
propanoic acid
O CH3 (CH2)7 CH C OH CH2CH2CH3
c.
8.73
2-propyldecanoic acid
Draw each carboxylic acid. a. octanoic acid Start by drawing CH3 then add (CH2)6 followed by the ending carboxyl group. O CH3
(CH2)6 C OH
8 - 26
b. 3,3-dimethylheptanoic acid First, draw the parent chain with seven (heptane) carbons making the last one the carboxyl group indicated by the “oic acid” ending. Add 2 methyl groups to carbon 3.
CH3
CH2
CH2
CH2
CH3
O
C CH2
C
OH
CH3 c. 3-isopropylhexanoic acid First, draw the parent chain with six (hexane) carbons making the last one the carboxyl group indicated by the “oic acid” ending. Add an isopropyl group to carbon 3.
O CH3 CH2 CH2 CH CH2
C OH
CH3CHCH3 d. 2-bromopentanoic acid First, draw the parent chain with five (pentane) carbons making the last one the carboxyl group indicated by the “oic acid” ending. Add a bromine atom to carbon 2.
O CH3 CH2 CH2 CH C OH Br
8.75
Match each structure to the correct IUPAC name: 3-methylbutanoic acid, 2methylpentanoic acid, 2-methylbutanoic acid. The carboxylic functional group is dominant in the naming so its carbon is carbon 1. In drawing a, there are four carbons and the methyl group is on the second carbon so that the compound is 2-methylbutanoic acid. Drawing b also has four carbons but the methyl group is on carbon 3, making its name 3-methylbutanoic acid. In drawing c, the methyl is on the second carbon but the parent chain has five carbons which makes the name 2-methylpentanoic acid.
8 - 27
a.
parent chain O 2
CH3 CH2 CH C OH methyl
CH3
2-methylbutanoic acid b. parent chain O 3
CH3 CH CH2 C OH methyl
CH3 3-methylbutanoic acid
c.
O 2
methyl
H3C
CH C OH H2C
parent chain
CH2 CH3
2-methylpentanoic acid
8.77
Acetic acid (CH3CO2H) and propyl alcohol (CH3CH2CH2OH) have the same molecular weight, but their boiling points differ by about 20˚C (acetic acid, 118˚C; propyl alcohol, 97.4˚C). Account for this difference. Acetic acid molecules form more hydrogen bonds with one another than do propyl alcohol molecules. Strong hydrogen bonding results in higher boiling points.
8.79
Name each phenol. Follow the IUPAC rules for naming phenol compounds. Begin numbering the C atoms in the ring starting at the C atom to which the –OH group is attached. Count in the direction (clockwise or counterclockwise) that assigns the lower
8 - 28
number to the nearest substituent. For disubstituted rings, the prefixes ortho-, meta-, or para- may be used to indicate the position of a substituent group relative to the –OH group. In some cases, common names are also used. OH
Br
a.
3-bromophenol or m-bromophenol
OH
b.
CH3 CH CH3
4-isopropylphenol
OH OH
c.
8.81
catechol
Draw each phenol. a. 2,4-dimethylphenol Start by drawing the phenol parent (benzene ring with a –OH in the first position). Counting from the –OH carbon add one methyl group to carbon 2 and one methyl group to carbon 4.
OH CH3
CH3
8 - 29
b. 3,5-dimethylphenol Start by drawing the phenol parent (benzene ring with a –OH in the first position). Counting from the –OH carbon add one methyl group to carbon 3 and one methyl group to carbon 5. OH
H3C
CH3
c. 4-propylphenol Start by drawing the phenol parent (benzene ring with a –OH in the first position). Counting from the –OH carbon add a propyl group to carbon 4 (a propyl group is propane attached by its carbon 1).
HO
CH2CH2CH3
d. m-propylphenol Start by drawing the phenol parent (benzene ring with a –OH in the first position). Counting from the –OH carbon add a propyl group to carbon 3 so the two substituents are meta to each other. OH
CH2CH2CH3
8 - 30
8.83
Hydroquinone (Table 8.5) has a higher boiling point than phenol. What might account for this difference? Hydroquinone molecules, with two –OH groups, can form more hydrogen bonds with one another than do phenol molecules.
8.85
Erbstatin has shown antitumor activity in combating skin, breast, and esophageal cancer.
OH H HO
O
C C H
NH C H
Erbstatin a. Is erbstatin a catechol-, a resorcinol-, or a hydroquinone-containing phenol? Erbstatin is a hydroquinone-containing phenol because it contains a phenol group with two –OH groups in para orientation. b. Is the nitrogen atom present in this compound part of an amine functional group? No. The nitrogen atom is not part of an amine functional group because it is bonded to a C=O. It is part of an amide functional group. c. Which stereoisomer (cis or trans) is present? The trans stereoisomer is present.
8.87
Hexylresorcinol (4-hexyl-3-hydroxyphenol) is an antiseptic used in some mouthwashes and throat lozenges. Draw this compound. Start by drawing the phenol parent (benzene ring with a –OH in the first position). Counting from the –OH carbon add another –OH group (called hydroxy when used as a side chain) to carbon 3. Next add a hexyl to carbon 4.
8 - 31
OH
OH CH2(CH2)4CH3 4-hexyl-3-hydroxyphenol
8.89
Draw octanoic acid and its conjugate base. The conjugate base of an acid is the ion or compound left after a H+ ion leaves. O
CH3 CH2 CH2 CH2 CH2 CH2 CH2 C OH octanoic acid
O CH3 CH2 CH2 CH2 CH2 CH2 CH2 C O octanoate ion (conjugate base)
8.91
Draw the conjugate base of each carboxylic acid. a. propanoic acid O CH3 CH2 C O -
b. hexanoic acid O
CH3 CH2 CH2 CH2 CH2 C O -
8 - 32
c. 3-chloropentanoic acid
Cl
O
CH3 CH2 CH CH2 C O -
8.93
Name each of the conjugate bases in your answer to Problem 8.91. O
a.
CH3 CH2 C O -
propanoate ion O
b.
CH3 CH2 CH2 CH2 CH2 C O -
Cl
c.
8.95
hexanoate ion
O
CH3 CH2 CH CH2 C O -
3-chloropentanoate ion
Draw the products of each reaction. In a previous chapter, you saw that acid and base reactions produced a salt and water. The same is true when an organic acid is reacted with a base. The cation of the base forms a salt (ionic compound) with the carboxylate anion formed with the removal of an H+ from the carboxylic acid.
O CH3 CH2 CH C OH a.
+ KOH
CH3 O +
CH3 CH2 CH C O K CH3
8 - 33
+ H2O
O CH2CH2C OH + NaOH
CH3
b. O +
CH3
8.97
CH2 CH2C O Na
+ H2O
Dinoseb, an herbicide and insecticide, is sold as a water-soluble ammonium salt. Draw this ammonium salt, which is produced by reacting Dinoseb with the base ammonia.
CH3
OH
CH3CH2CH
NO2
NO2 Dinoseb The anion of the salt has the original acid structure with the hydrogen removed from the –OH and a negative charge placed on the oxygen. The cation of the salt is drawn beside the O-. CH3
O NH4+
CH3CH2CH
NO2
NO2
8 - 34
8.99
In 1875 sodium salicylate, the sodium salt of salicylic acid, was introduced as an analgesic (painkiller).
O C OH OH
Salicylic acid a. Which functional group of salicylic acid is the most acidic, the phenol group or the carboxyl group? The carboxylic acid group is more acidic than the phenol group. b. Draw the ionic compound sodium salicylate. (Hint: Only one H+ is removed from salicylic acid.)
O C O- Na
+
OH
8.101 p-Ethylphenoxide ion is the conjugate base of p-ethylphenol. Draw both compounds.
CH3 CH2
OH
CH3 CH2
O-
p-ethylphenoxide
p-ethylphenol
8.103 a. Draw 4-chloropentanoic acid. First, draw pentanoic acid. Starting from carbon 1 (-COOH), count to the 4th carbon and attach a chlorine atom.
8 - 35
O CH3CH CH2CH2 C OH Cl
b. Draw and name the conjugate base of 4-chloropentanoic acid. The conjugate base of an acid is has a H+ ion removed and replaced by a negative charge: O CH3CH CH2 CH2 C O Cl 4-chloropentanoate ion
c. Which predominates in water at pH 7, 4-chloropentanoic acid or its conjugate base Its conjugate base. Since most carboxylic acids have pKa values around 5, carboxylate ions predominate at pH 7. 8.105 a. Draw malic acid (Table 8.6) and circle the least acidic –OH hydrogen atom.
malic acid
b. Draw sodium malate, the salt formed when malic acid and NaOH react in a 1:1 ratio. One possible sodium salt of malic acid is
sodium malate
Losing H+ from the carboxyl group on the left instead would also be an acceptable answer.
8 - 36
c. Draw salicylic acid (Table 8.6) and circle the least acidic –OH hydrogen atom.
salicylic acid d. Draw sodium salicylate, the salt formed when salicylic acid and NaOH react in a 1:1 ratio.
sodium salicylate
e. Which is more acidic, malic acid or salicylic acid? (Refer to Table 8.6). Salicylic acid is more acidic. Its pKa (2.97) is smaller than the pKa of malic acid (3.40).
8.107 Draw the products of each reaction. The reaction of a carboxylic acid and alcohol in the presence of an acid catalyst produces an ester and a water molecule.
O a.
H+
C OH + HOCH2CH2CH3
O C O CH2CH2CH3 + H2O
8 - 37
O b.
H+
CH3CH2C OH + HOCH2
O CH3 CH2 C O CH2
+ H2O
8.109 Name the organic products in Problem 8.107 Hint: -CH2C6H5 is a benzyl group. a. propyl benzoate
b. benzyl propanoate
8.111 Draw and name the ester formed when p-ethylbenzoic acid reacts with isopropyl alcohol, CH3CH(OH)CH3, in the presence of H+.
O CH3CH2
C OH
CH3
+ HO CH
H+
O CH3CH2
C O CH CH3
CH3 p-ethylbenzoic acid
isopropyl alcohol
isopropyl p-ethylbenzoate
8.113 Draw the products obtained from each reaction. These are hydrolysis reactions in which an ester is broken apart by water to form a carboxylic acid plus an alcohol.
8 - 38
CH3
H+
a.
H+
b.
8.115 Draw the products obtained when a. cyclopentyl acetate reacts with H2O in the presence of H+.
O CH3C O
+
H+
H2 O
O CH3C OH
+ HO
b. cyclopentyl acetate reacts with NaOH(aq). In the presence of NaOH, a H+ ion is removed from the carboxylic acid product and a caboxylate salt is formed.
(aq)
8 - 39
8.117 Draw the products of each reaction.
a.
(aq)
b.
(aq)
c.
(aq)
8.119 Draw a skeletal structure for each of the products formed when the ester below, one of the primary compounds in beeswax, is reacted with NaOH(aq). O CH3(CH2)14
C O CH2(CH2)28CH3
8 - 40
The carboxylate ion and alcohol products are: O O
+
Na
+ HO
8.121 The local anesthetic procaine can be hydrolyzed under acidic or basic conditions. Draw the products of each reaction. a.
b.
NaOH
8 - 41
8.123 Draw each compound. Review the general IUPAC rules for naming amines and alkylammonium ions to derive and draw the structure for each compound or ion below: a. 1-pentanamine
CH3 CH2 CH2 CH2 CH2 NH2 b. N-isopropyl-1-pentanamine
CH3 CH3 CH2 CH2 CH2 CH2 NH CH CH3 c. N-ethyl-N-methyl-2-hexanamine
CH3 CH2 CH2 CH2 CH CH3 CH3 N
CH2 CH3
d. dimethyldipropylammonium ion
CH3 +
CH3 N
CH2 CH2 CH3
CH2 CH2 CH3
8.125 Draw each amine. a. ethylamine
CH3 CH2 NH2 b. isopropylamine CH3 CH CH3 NH2
8 - 42
c. butylmethylamine
CH3 CH2 CH2 CH2 NH CH3 d. diethylpropylamine CH3 CH2 N
CH2 CH3
CH2 CH2 CH3
8.127 Give the IUPAC name of each amine. The IUPAC naming of amines uses the name of the corresponding alkane parent chain with the “e” replaced with “amine”. If a substituent is attached to the nitrogen of the amine, it is preceded by N–. The amine position on the parent chain is given by a number preceding the parent chain.
a. CH3NH2 methanamine
b.
CH3CH CH3 NH2
2-propanamine CH3CH CH3 c.
HN CH2CH2CH2CH3
N-isopropyl-1-butanamine d.
CH3CH2CH2N CH2CH3 CH3
N-ethyl-N-methyl-1-propanamine
8.129 Give the common name of each amine in Problem 8.127. The common name of amines puts the name of all substituents attached to the N in front of “amine”.
8 - 43
a. methylamine b. isopropylamine c. butylisopropylamine d. ethylmethylpropylamine 8.131 Identify each amine in Problem 8.127 as being 1°, 2°, or 3°. A 1° amine has only one carbon atom bonded to the amine nitrogen atom. A 2° amine has two carbon atoms each one directly bonded to the amine nitrogen atom. A 3° amine has three carbon atoms each one directly bonded to the amine nitrogen atom.
a. CH3NH2 1° amine b.
CH3CH CH3 NH2
1° amine CH3CH CH3 c.
HN CH2CH2CH2CH3
2° amine d.
CH3CH2CH2N CH2CH3 CH3
3° amine 8.133 Identify each compound in Problem 8.123 as being 1o, 2o, 3o, or 4o. A 1° amine has only one carbon atom bonded to the amine nitrogen atom. A 2° amine has two carbon atoms each one directly bonded to the amine nitrogen atom. A 3° amine has three carbon atoms each one directly bonded to the amine nitrogen atom. A 4° has four carbon atoms each directly bonded to the amine nitrogen atom.
8 - 44
a.
CH3 CH2 CH2 CH2 CH2 NH2
1° amine
CH3 CH3 CH2 CH2 CH2 CH2 NH CH CH3
b.
2° amine
CH3 CH2 CH2 CH2 CH CH3 CH3 N
c.
CH2 CH3
3° amine
CH3 +
CH3 N d.
CH2 CH2 CH3
CH2 CH2 CH3
4° amine
8.135 Identify each compound as a pyridine, a pyrimidine, or a purine (see Figure 8.26a). A pyridine has only nitrogen atom replacing a carbon in a benzene ring, a pyrimidine has two, and a purine has two in a benzene ring fused to a fivemembered ring containing two additional nitrogen atoms. See text problem for the structures. CH3
N (CH3)2CH
S N
O
OCH2CH3
P
OCH2CH3 a.
Diazinon (an insecticide)
pyrimidine
8 - 45
O N
b.
NH C CH2
Difenpiramide (an anti-inflammatory drug)
pyridine
8.137 Account for the fact that ethylamine is more water soluble than hexylamine. Although both compounds contain a polar amine group, ethylamine contains a smaller hydrocarbon group than hexylamine, making ethylamine more polar and thus more soluble in water.
8.139 Draw and name the conjugate acid of a. ammonia H +
H N H
ammonium ion
H
b. propylamine
H +
CH3 CH2 CH2 N H
propylammonium ion
H c. methylethylamine
H +
CH3 N CH2CH3
ethylmethylammonium ion
H
8 - 46
d. triethylamine
H +
CH3 CH2 N CH2CH3
triethylammonium ion
CH2CH3
8.141 a. Write the equilibrium equation for the acid-base reaction that takes place between propylamine and water. Propylamine is basic and so water behaves as the Bronsted-Lowry acid: CH3CH2CH2NH2
+
H2O
⇋
CH3CH2CH2NH3+
+
OH-
b. The conjugate acid of propylamine has a pKa of 10.6. Which predominates at pH 7, propylamine or its conjugate acid? Its conjugate acid. The pH of the solution (7) is less than the pKa of the conjugate acid of propylamine (10.6).
8.143 a. Write a balanced equation for the reaction of propylamine with HCl. CH3CH2CH2NH2 +
HCl
⇋
CH3CH2CH2NH3+ Cl-
b. Name the salt that forms. CH3CH2CH2NH3+ Clpropylammonium chloride
8.145 Anabasine hydrochloride, the salt of anabasine (an amine present in tobacco), is sold as an insecticide. Draw this salt. (Hint: The nitrogen atom in the pyridine ring is not basic.)
anabasine
8 - 47
Cl-
anabasine hydrochloride
8.147 Phenylephrine is the active ingredient in many non-prescription decongestants. Draw the product of each acid-base reaction.
NaOH
phenylephrine
a.
HCl
phenylephrine
b.
8.149 Draw each amide. a. hexanamide
8 - 48
This is an amide that has 6 carbon atoms in the carboxylic acid residue. O CH3CH2CH2CH2CH2
C NH2
b. N-propylacetamide There is a propyl group bonded to the N atom and there are 2 carbon atoms (“acet”) in the carboxylic acid residue. O CH3 C NH CH2CH2CH3
c. N-butyl-N-methylbenzamide
There is a butyl group and a methyl group bonded to the N atom the carboxylic acid residue is derived from benzoic acid.
O C N CH2CH2CH2CH3 CH3
8 - 49
8.151 Name each amide. Use the general IUPAC rules for naming amides. Begin counting from the carbonyl carbon (C=O) to indicate the location of each substituent group on the hydrocarbon chain. Substituent groups bonded to the N atom are identified by placing N- before the name of the substituent group. In some cases, a common name is used. O C NH2
a.
benzamide
O CH3 C N CH2CH3 CH3
b.
N-ethyl-N-methylacetamide
O CH3 CH2 CH C NH2 CH3
c.
2-methylbutanamide
8.153 Draw the amide that will be produced by each reaction. When a carboxylic acid reacts with an amine, a bond is formed between the carbonyl group and the nitrogen of the amine group forming an amide. O CH3
C OH
+
NH3
heat
a. O C NH2
CH3
The amide product is O
b.
CH3 CH2 CH2 CH2 C OH
+
8 - 50
heat NH2CH3
O The amide product is
CH3 CH2 CH2 CH2 C NH CH3
8.155 Draw the carboxylic acid and amine from which lidocaine (Problem 8.148) can be prepared.
8.157 Name the amide product of each reaction in Problem 8.153. O CH3
C NH2
p-methylbenzamide
a.
O b.
CH3 CH2 CH2 CH2 C NH CH3
N-methylpentanamide
8.159 Draw the products formed when each amide in Problem 8.149 is reacted with H2O in the presence of H+. The reaction of an amide with water in the presence of H+ is a hydrolysis reaction. The amide breaks down into a carboxylic acid and a protonated amine.
8 - 51
a. O
H+ +
CH3CH2CH2CH2CH2 C NH2
H2O O CH3CH2CH2CH2CH2C OH
+
+
NH4
b. O
H+
CH3 C NH CH2CH2CH3
H2O
+
H
O CH3 C OH
+
+ H N CH2CH2CH3 H
c. O
H+
C N CH2CH2CH2CH3
+ H2O
CH3
H
O C OH
+
+
H N CH2CH2CH2CH3 CH3
8.161 Draw the products formed when each amide in Problem 8.151 is heated in the presence of H2O and H+. The reaction of an amide with water in the presence of H+ is a hydrolysis reaction. The amide breaks down into a carboxylic acid and a protonated amine.
O C NH2
H+ +
H2O O C OH
a.
8 - 52
+
+ NH4
O
H+ H2O
CH3 C N CH2CH3 + CH3
H
O
+
CH3 C
OH + H N CH2 CH3 CH3
b.
O
H+
CH3 CH2 CH C NH2 +
H2O
CH3
O CH3 CH2 CH C OH
+
+ NH4
CH3
c.
8.163 Draw the products obtained when erbstatin (Problem 8.85) is hydrolyzed under acidic conditions. Hydrolysis of erbstatin under acidic conditions results in the breakdown of the amide bond producing a protonated amine and a carboxylic acid:
OH
O H
HO
+
HO C H
C C H
+
NH3
8.165 Lidocaine (Problem 8.148) can be hydrolyzed under acidic or basic conditions. Draw the products of each reaction.
8 - 53
a.
lidocaine
b.
lidocaine
8.167 The boiling point of propanamide is 213○C and that of methyl acetate is 57.5○C. Account for this difference in boiling points for these two molecules with very similar molecular weights.
O
O
CH3CH2C NH2
CH3C O CH3
Propanamide
Methyl acetate
8 - 54
Boiling points are determined by the strength of noncovalent forces holding the liquid molecules together; the stronger the noncovalent forces, the higher the boiling point for similar size molecules. Propanamide molecules have stronger noncovalent forces because of their ability to hydrogen bond with each other through their –NH groups. Methyl acetate molecules interact only through the weaker dipole-dipole and London forces.
8.169 Olvanil, a compound structurally related to capsaicin (Figure 8.19), has been studied as a potential analgesic. Locate the phenol, amide, and alkene functional groups in the molecule.
amide O CH2NH C CH2CH2CH2CH2CH2CH2CH2
H C
alkene C H CH2(CH2)6CH3
CH3O phenol HO
olvanil
8.171 a. What is exfoliation and why is it done? Exfoliation is the process of removing the outermost layer of the skin. According to claims made by cosmetic companies, exfoliation “will unblock pores, remove, wrinkles and age spots, repair sun damage, and improve the overall feel of the skin.” b. How does mechanical exfoliation differ from chemical exfoliation? During mechanical exfoliation, the skin is rubbed with abrasive materials to remove the outermost layer. In chemical exfoliation, acids are applied to the skin to peel the outermost layer. c. How is exfoliation done by a dermatologist similar to exfoliation done at home using skin care products? Exfoliation done at home and done by dermatologists both make use of acids to chemically remove the outermost layer of the skin. d. How does exfoliation done by a dermatologist differ from exfoliation done at home using skin care products?
8 - 55
Exfoliation done at home makes use of less concentrated acids than those used by dermatologists. 8.173 a. Classify each of the compounds in Figure 8.27 as being 1o, 2o, or 3o amine. epinephrine (2°)
amphetamine (1°)
phenylephrine (2°)
pseudoephedrine (2°)
methamphetamine (2°)
b. Which functional groups are present in epinephrine, but not in methamphetamine? Alcohol and phenol groups c. Draw and name the product of the reaction that takes place between pseudoephedrine and HCl.
pseudoephedrine hydrochloride
8.175 a. What is the function of quorum sensing? Quorum sensing is a switch from individual to group activity among bacteria. b. What role do autoinducers play in quorum sensing? Autoinducers are signaling molecules. At certain concentrations, autoinducers cause metabolic changes producing compounds that benefit the group initiating quorum sensing. c. How does biofilm formation benefit bacteria? The biofilm keeps nutrients close at hand and has channels that expedite water uptake and removal of waste products.
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8.177 Draw the products from the hydrolysis of Lufenuron under acidic conditions.
NH4+
8.179 a. Draw methylcyclobutane. C H3
b. Draw a constitutional isomer of methylcyclobutane that has a ring and is a cis geometric isomer.
C H3 C H 3
c. Draw a constitutional isomer of methylcyclobutane that has a ring, but has no geometric isomers.
CH2 CH3
8 - 57
d. Draw a constitutional isomer of methylcyclobutane that has no ring and is a cis geometric isomer. H
H C CH3
C CH2CH3
e. Draw a constitutional isomer of methylcyclobutane that has no ring and has no geometric isomers. CH2=CH-CH2-CH2-CH3 f. Name the major noncovalent force that attracts any one of the molecules in this problem to a similar molecule. London force
8.181 In addition to capsaicin (Figure 8.19), peppers contain smaller amounts of homocapsaicin and 6,7-dihydrocapsaiciin.
a. Which best describes the relationship between capsaicin and homocapsaicin? They are: constitutional isomers, different conformations of the same molecule, stereoisomers of one another, identical molecules, entirely different molecules. Entirely different molecules. The two molecules do not have the same molecular formula. Homocapsaicin has 2 fewer hydrogen atoms. b. Which stereoisomer is present in homocapsaicin, cis or trans? 8 - 58
Trans c. Which reaction might you use to convert capsaicin into 6,7-dihydrocapsaicin? Hydrogenation (across the double bond) d. Draw the products formed when homocapsaicin is hydrolyzed under acidic conditions. O CH3 O
+
CH2 NH3
+
HO C CH2 CH2 CH2 CH2 CH2
H
C C HO
H
CH(CH3)2
e. Draw the phenoxide salt that forms when 6,7-dihydrocapsaicin is reacted with NaOH.
-
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 8.2
Identify the term that best describes the two molecules below: constitutional isomers, geometric isomers, different conformations of the same molecule, two unrelated molecules.
Answer: Constitutional isomers.
8 - 59
The two molecules differ only by the atomic connections of the two substituent groups to the ring. The left molecule has two substituent groups on adjacent carbon. In the right molecule, the two substituent groups are two carbon atoms apart.
8.4
The ester acetylsalicylic (aspirin) can be hydrolyzed as shown below. Circle the atoms in the product molecules that were originally in the reactant water molecule.
Answer:
O C OH
O
O C CH3
+
+
H H2O
O
Acetylsalicylic acid
C OH O
OH
+
HO C CH3
Salicylic acid
SECTION 8.1 8.6
ALKANES
Which has the higher boiling point? Explain.
or Answer:
The larger a molecule is, the higher its boiling point will be. Comparing
(5-carbon chain) and
8 - 60
(8-carbon chain), the latter is expected to have the higher boiling point.
8.8
Arrange the molecules in order, from highest boiling point to lowest boiling point: pentane, octane, methane.
Answer: From highest boiling point to lowest boiling point: octane > pentane > methane Boiling point decreases as London force attraction decreases. London force attraction decreases as the size of the molecule decreases.
8.10
Draw a Lewis structure of each alkane. a. CH3CH(CH2CH2CH3)CH2CH3 b. CH3C(CH2CH3) 2CH(CH3)CH3
Answer: Connect every pair of atoms in the formula with a single line to represent the covalent bond between them.
H H H H a.
H C C C C H H H H H C H H C H H C H H
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H H C H H C H H b.
H
H C H H
H C C
C
C H
H H H C H
H
H C H H
8.12
Draw a skeletal structure for each of the molecules in Problem 8.10.
Answer:
b.
a.
8.14
Draw a Lewis structure for each molecule.
a.
b.
Answer: H H C a.
H
H H
H C
C C
H H
H H
H C H H C
H C H H C H
H H H C
C
H
H
H
b.
H C
H
C H
H H C H H C
C
C
H H C H H H C H H H
H
H
8 - 62
C H H
8.16
Write a condensed structural formula for each molecule in Problem 8.14.
Answer: a. CH3CH(CH3)CH2C(CH3)2CH(CH3)CH3 b. CH3C (CH3)2CH(CH3)CH(CH3)CH3
8.18
Find and name the parent chain for each molecule, then give the complete IUPAC name for each.
a.
CH3CH2CH2 CH3 CH2CH3
b.
CH3CH CHCH2CH2CH2CH3 CH3CH2CH2 CH3
c. CH3CH2CH2CH2 CHCH2CH CH3 CH2CH3
Answer: First, identify the parent chain by finding the longest chain of carbon. Remember that the longest chain is not necessarily the one that goes straight across. Assign the appropriate name using the numbering prefix and the ending “ane”. Next, identify and name the substituents. Finally, to complete the IUPAC naming process, number the carbons of the parent chain starting from the end nearer the first substituent. a. butane The longest chain contains 4 carbons so the parent name is butane. There are no alkyl group substituents. b. 5-ethyl-4-methylnonane The parent name is nonane because the longest chain contains 9 carbon atoms. There are 2 substituent groups: a methyl group on the 4th carbon and an ethyl group on the 5th carbon. c. 4-ethyl-2-methyloctane
8 - 63
The longest chain is an 8-carbon chain so the parent name is octane. There are two alkyl substituent groups: a methyl group at the 2nd carbon and an ethyl group at the 4th carbon.
8.20
Give the IUPAC name of each molecule. a. CH3CH2C(CH3)2CH2CH2CH3 b. CH3CH(CH3)CH(CH3)CH2CH2CH3
c.
d. Answer: a. 3,3-dimethylhexane b. 2,3-dimethylhexane c. 2-methylheptane d. 4-ethyl-3-methylheptane
SECTION 8.2
8.22
CONSTITUTIONAL ISOMERS
Draw and name six of the constitutional isomers with the formula C8H18.
Answer: Constitutional isomers are molecules that have the same formula but atomic connections. Below are some examples of the constitutional isomers with the formula C8H18. CH3CH2CH2CH2CH2CH2CH2CH3
octane
CH3 CH3 C CH2CH2CH2CH3
2,2 - dimethylhexane
CH3
8 - 64
CH3 2,2,3,3 - tetramethylbutane
CH3 C CH3 CH3 C CH3 CH3 CH3 CH3CH CHCH2CH2CH3
2,3 - dimethylhexane
CH3
CH3 CH3CHCH2CHCH2CH3
2,4 - dimethylhexane
CH3 CH3 CH3CHCH2CH2CHCH3
2,5 - dimethylhexane
CH3 CH3 CH3CH2CHCH2CH2CH2CH3
3-methylheptane
CH2CH3 CH3CH2CHCH2CH2CH3
3-ethylhexane
CH3CH2CH3 3-isopropylpentane
CH3CH2CH CH2CH3
CH3 2-methylheptane
CH3CHCH2CH2CH2CH2CH3 CH3
4-methylheptane
CH3CH2CH2CHCH2CH2CH3
8.24
Which of the following pairs of molecules are constitutional isomers? a. CH3CH2CH2CH3 and CH3(CH2) 2CH3
8 - 65
b. CH3C(CH3) 2CH2CH3 and CH3CH2 CH2 CH2 CH2CH3 c. CH3CH(CH3) CH(CH3)CH3 and CH3CH2 CH2CH2CH3 Answer: Constitutional isomers are molecules that have the same molecular formula but whose atoms are connected differently. If the two molecules have the same molecular formula and the atoms are connected the same way, then they are identical molecules. If they have the same molecular formula but different atomic connections, then they are constitutional isomers. The skeletal structures are given to show the atomic connections. a. Not constitutional isomers. These are the same molecules with same molecular formula and the same atomic connections. and
b. Constitutional isomers. These two molecules have the same molecular formula (C6H14) but different atomic connections. and
c. Not constitutional. These molecules do not have the same molecular formulas.
and
8.26
Which pairs of molecules are constitutional isomers? Which are identical?
CH3 a.
CH3CH2CH2CH3
CH3CH2CH2
CH3 b. CH3CHCH3
CH3CH2CH2 CH3
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CH3 c.
CH3CHCHCH3
CH3CHCH CH3 CH3
H3C CH3
Answer: Each of these pairs has the same molecular formula. Check the bonding arrangements (atomic connections) to see if the molecules are constitutional isomers or are identical molecules. If the bonding arrangements are the same, then they are identical molecules. If they have different atomic connections, then they are constitutional isomers. a. identical molecules b. constitutional isomers c. identical molecules
8.28
Which are constitutional isomers? a. hexane and 3-methylpentane b. hexane and 2,3-dimethylpentane c. hexane and 2,3-dimethylbutane d. hexane and cyclohexane
Answer: Constitutional isomers have the same molecular formulas but different atomic connections. a. Constitutional isomers. The two molecules have the same molecular formulas (C6H14) but different atomic connections. b. Not constitutional isomers. The two molecules have different molecular formulas (C6H14 and C7H16) c. Constitutional isomers. The two molecules have the same molecular formulas (C6H14) but different atomic connections. d. Not constitutional isomers. The two molecules have different molecular formulas (C6H14 and C6H12)
SECTION 8.3
8.30
CONFORMATIONS
Which pairs of molecules are constitutional isomers? Which are different conformations of the same molecule?
8 - 67
a.
b.
Answer: Constitutional isomers have the same molecular formulas but different atomic connections. Different conformations of the same molecule arise from free rotation around single bonds. a. Constitutional isomers. The two molecules have the same molecular formulas (C6H14) but different atomic connections. b. Identical molecules. The two molecules are exactly the same except we are viewing them from different directions.
8.32
True or false? a. Constitutional isomers have the same IUPAC name. b. The different conformations of an alkane have the same IUPAC name.
Answer: a. False. Constitutional isomers have different IUPAC names because they are different compounds with different atomic connections. b. True. Different conformations of an alkane differ only in their three-dimensional shapes by free rotation around a single bond.
SECTION 8.4
8.34
CYCLOALKANES
Give the IUPAC name for each molecule.
H3C
CH3
a.
8 - 68
CH3 b.
C H CH3 CH2CH2CH3
c. CH2CH2CH2CH3
CH3 H3C C CH3 d.
Answer: a. b. c. d.
8.36
1,1-dimethylcyclopentane isopropylcyclopentane 1-butyl-3-propylcyclohexane t-butylcyclopropane
Which molecule(s) in Problem 8.34 can exist as cis and trans isomers?
Answer: Cis/trans isomers of cycloalkanes exist when two substituents can be oriented either on the same side relative to the face of the ring (cis) or on opposite sides (trans). Among the choices given in Problem 8.34, only choice c. has two substituents that can have two different geometric orientations:
CH2CH2CH3 CH2CH2CH3 CH2CH2CH3 CH2CH2CH2CH3 cis-1-butyl-3-propylcyclohexane
trans-1-butyl-3-propylcyclohexane
8.38
Draw and name three ethyldimethylcyclohexane constitutional isomers.
8 - 69
Answer: Below are some examples of the constitutional isomers of the compound.
1-ethyl-2,3-dimethyl-cyclohexane
2-ethyl-1,3-dimethyl-cyclohexane
1-ethyl-2,4-dimethyl-cyclohexane
4-ethyl-1,2-dimethyl-cyclohexane
1-ethyl-3,5-dimethyl-cyclohexane
2-ethyl-1,4-dimethyl-cyclohexane
8.40
a. What is the molecular formula of methylcyclopropane? b. Draw three constitutional isomers of methylcyclopropane.
Answer: a. C4H8. The expanded structure of methylcyclopropane is given below.
8 - 70
CH3 CH H 2C
CH 2
b. Draw three constitutional isomers of methylcyclopropane. H
H
C
C
C
H
H
H
H C H
8.42
H
CH3 CH CH CH3
CH 3 H 2C
C
CH3
Draw a side view of each cycloalkane. (See Figure 8.6 for examples.) a. cis-1,2-dimethylcyclopentane b. trans-1,2-dimethylcyclopentane c. cis-1-ethyl-2-isopropylcyclohexane Answer:
a. CH3 CH3
b.
CH3 CH3
c. CH3CH2 CH CH3 CH3
8.44
Give the complete IUPAC name (including the use of the term cis or trans) for each molecule.
8 - 71
H H
a. H3C
H H H H H
H
b. H
H
CH3
H
H
H H
H H
CH3CH2CH2
CH2CH2CH3 H
H
Answer: a. cis-1,3-dimethylcyclopentane b. trans-1,3-dipropylcyclohexane
SECTION 8.5
8.46
ALKENES, ALKYNES, AND AROMATIC COMPOUNDS
Draw propyne showing the proper three-dimensional shape about each atom.
Answer:
linear H C
H
C
C
H
H tetrahedral
8.48
Name each molecule. CH3 a. CH3CH C CH3
8 - 72
CH3 b.
HC
C C CH3 CH3
c.
CH3CHCH CH CH3 CH3
Answer: a. 2-methyl-2-butene b. 3,3-dimethyl-1-butyne c. 4-methyl-2-pentene
8.50
Draw each molecule. a. 2,3-dimethyl-3-hexene b. 3,4-dimethyl-3-hexene c. 4,5-dimethyl-1-hexyne
Answer: CH3 a. CH CH C CH 3
CH2 CH3
CH3
CH3 b.
CH3 CH2 C C CH2 CH3 CH3
c.
8.52
Label the indicated carbon-carbon double bonds as cis or trans.
Answer:
8 - 73
trans H3C
CH3
cis
CH3
H3C
CH3
O
H
11-cis-Retinal (a derivative of vitamin A)
8.54
Name each molecule. CH 3
H3C b.
a .
CH3 CH2CH2CH3
CH 3
CH3 CH2
c .
CH3 CH3 CH2
CH3 CH3
Answer: a. 1-methyl-3-propylbenzene b. 1,2,4-trimethylbenzene c. 4-t-butyl-1,2-diethylbenzene
8.56
Draw each molecule. a. 1,2-dipropylbenzene b. m-diisopropylbenzene c. 1,2,4-trimethylbenzene
8 - 74
Answer: CH 2CH 2CH3 CH 2CH2CH3
a .
CH3CHCH 3
b. CH 3CHCH 3
CH3 CH3 c.
CH3
SECTION 8.6
8.58
REACTIONS OF HYDROCARBONS
Draw the organic product of each reaction. If more than one product is possible, draw them all.
d. e. f.
2,2-dimethylpropane
2,3-dimethylbutane 2,2-dimethylbutane
light
light light
Answer: These reactions are halogenation reactions in which a halogen atom replaces a hydrogen atom on an alkane. The products given here show only one halogen atom substituting for one hydrogen atom on the alkane.
8 - 75
a.
2,2-dimethylpropane
light
The structure of 2,2-dimethylpropane is shown below. In this halogenation reaction, a bromine atom replaces a hydrogen atom on the alkane. When only one halogen atom substitutes for one hydrogen atom on the alkane, there is only one possible product. Placing the Br atom on any of the other C atoms produces the same molecule shown below.
light
b.
2,3-dimethylbutane
light
In this halogenation reaction, a chlorine atom replaces a hydrogen atom on the alkane. There are two possible products.
light
c.
2,2-dimethylbutane
light
In this halogenation reaction, a chlorine atom replaces a hydrogen atom on the alkane. There are three possible products.
8 - 76
light
8.60
Draw the organic product of each reaction. If more than one product is possible, draw them all (ignore stereoisomers).
a. cyclobutane light
b. methylcyclobutane light
c. 1,3-dimethylcyclobutane light Answer: These reactions are halogenation reactions in which a halogen atom replaces a hydrogen atom on an alkane. The products given here show only one halogen atom substituting for one hydrogen atom on the alkane.
a. cyclobutane light In this halogenation reaction, a bromine atom replaces a hydrogen atom on the alkane. There is only one possible product.
8 - 77
light
b. methylcyclobutane light The possible products are shown below.
light
c. 1,3-dimethylcyclobutane light The possible products are shown below.
light
8 - 78
8.62
Draw the organic product of each reaction. a. 1-butyne
2 H2
b. 2-methyl-2-butene
Pt
Pt
H2
c. 1,5-dimethylcyclopentene
Pt
H2
Answer: These reactions are catalytic hydrogenations. H atoms will add to the 2 carbon atoms connected by a double or a triple bond converting an alkyne or an alkene into an alkane. a.
Pt 2H2 b. Pt
H2
Pt H2
c.
8.64
Draw the major organic product of each reaction. 2-methyl-1-hexene
a. 2-methyl-2-butene
b. 1,5-dimethylcyclopentene
c. Answer: Reactions of alkenes with water are hydration reactions. H2O, in the presence of a H+ catalyst adds as a H and an OH to the 2 carbon atoms connected by a double 8 - 79
bond producing an alcohol. If the H2O adds to a symmetric alkene, there is only one possible alcohol product. If the H2O adds to an asymmetric alkene, there are two possible alcohol products. The major and minor products are predicted using Markovnikov’s rule.
a. The alkene in this reaction is an asymmetric alkene and therefore two product alcohols are possible. Markovnikov’s rule states that the major product predicted is the one formed when the H atom adds to the carbon in the C=C that already has more H atoms.
b.
c. The alkene in this reaction is an asymmetric alkene and therefore two product alcohols are possible. Markovnikov’s rule states that the major product predicted is the one formed when the H atom adds to the carbon in the double that already has more H atoms.
8.66
Draw the organic product of each reaction. If more than one product is possible, draw them all.
8 - 80
a. b. c.
p-chlorotoluene o-xylene 1,3,5-trimethylbenzene
Answer: These are aromatic halogenation reactions, in which a hydrogen atom in an aromatic ring is replaced by a halogen atom. a. There are two possible products.
b. There are two possible products.
c. There is only one possible product.
8.68
Draw the organic product of each reaction. If more than one product is possible, draw them all.
8 - 81
light
a.
Fe
b. Answer:
light
a.
b.
Fe
8.70
Eugenol gives cloves their odor and flavor.
Eugenol
a. Draw the organic products that can form when eugenol reacts with Br2 and Fe.
8 - 82
b. Draw the organic product that can form when eugenol reacts with Br2 and light. c. Draw the major product formed when eugenol reacts with H2O and H+. Answer: a. Fe
Eugenol
8 - 83
b. light
Eugenol
c.
Eugenol
SECTION 8.7
8.72
CARBOXYLIC ACIDS
Name each carboxylic acid. O CH3 CH2 CH2 CH CH2 C OH
a.
CH3 CH2 CH2
8 - 84
O HO C CH CH2 CH CH3 CH3
b.
CH2 CH3
O CH3 (CH2)5 CH C OH CH3 CH CH3
c. Answer:
Follow the IUPAC rules for naming carboxylic acids. The parent chain is enclosed in a box and any substituent group is circled. Begin numbering the carbon atoms on the parent chain starting from the C atom of the carboxyl group. O CH3 CH2 CH2 CH CH2 C OH CH3 CH2 CH2
a.
3-propylhexanoic acid
O HO C CH CH2 CH CH3 CH3
CH2 CH3
b.
2,4-dimethylhexanoic acid O CH3 (CH2)5 CH C OH
c.
8.74
CH3 CH CH3
2-isopropyloctanoic acid
Draw each carboxylic acid. a. 4-methylheptanoic acid b. 2-propylhexanoic acid c. 3-chlorobutanoic acid
Answer: CH3 a.
.
O
CH3 CH2 CH2 CH CH2 CH2 C OH 4-methylheptanoic acid
8 - 85
O b.
CH3 CH2 CH2 CH2 CH C OH CH2CH2CH3 2-propylhexanoic acid
Cl c.
O
CH3 CH CH2 C OH 3-chlorobutanoic acid
8.76
Match each structure to the correct IUPAC name: o-bromobenzoic acid, mbromobenzoic acid, p-bromobenzoic acid.
a.
b.
c.
Answer:
8.78
a.
b.
c.
p-bromobenzoic acid
o-bromobenzoic acid
m-bromobenzoic acid
Butanoic acid is more soluble in water than hexanoic acid. Account for this difference.
8 - 86
Answer: Both acids have carboxyl groups that are polar. Butanoic acid has a shorter hydrocarbon chain than hexanoic acid. Therefore, it is more polar and more soluble in water.
SECTION 8.8
8.80
PHENOLS
Name each phenol.
a.
b.
c.
Answer: a. 2,4-dichlorophenol
8.82
b. 2-ethylphenol
Draw each phenol. a. p-t-butylphenol b. 2,6-dichlorophenol c. 4-isopropylphenol d. o-isopropylphenol
Answer:
a.
8 - 87
c. hydroquinone
b.
c.
d.
8.84
If you come into contact with urushiol (Figure 8.18b), the phenol partially responsible for the itching and burning caused by poison ivy, you cannot wash it off with water. Explain.
Answer: The urushiol molecule has a long hydrocarbon side chain that makes it insoluble in water.
8.86
Some varieties of rice plants are allelopathic, which means that they release chemicals that block the growth of other plants that might compete for nutrients. The compound below is one of the allelopathic chemicals produced by rice. (Arabinose is a simple sugar.)
8 - 88
a. Is the compound a catechol-, a resorcinol-, or a hydroquinone-containing phenol? b. Which steroisomer (cis or trans) is present? Answer: a. Resorcinol-containing phenol. b. cis
8.88
Thymol is an active ingredient in some mouthwashes. Name thymol as a phenol.
Thymol Answer: 2-isopropyl-5-methylphenol
SECTION 8.9
8.90
CARBOXYLIC ACIDS AND PHENOLS AS WEAK ORGANIC ACIDS
Draw m-chlorobenzoic acid and its conjugate base.
Answer: In order to draw m-chlorobenzoic acid, draw benzoic acid which has a benzene ring with the carboxyl group on carbon 1. Next, add a chloro substituent (chlorine atom) on carbon 3. To make the conjugate base, remove the hydrogen from the carboxyl group and add a negative charge.
8 - 89
O
O C OH
8.92
C O
Cl
Cl
m-chlorobenzoic acid
conjugate base
Draw the conjugate base of each carboxylic acid. a. 3-methylhexanoic acid b. formic acid c. 3,5-dibromobenzoic acid
Answer: a. Draw a six carbon (hexane) chain with the carbon 1 being the carboxyl group. Add a methyl group to carbon 3. Finally, remove hydrogen from the –OH and place a negative charge on the oxygen.
O
CH3
CH3CH2CH2CHCH2C O Conjugate base b.Draw one carbon (methane) as the carboxyl group. Remove hydrogen from the –OH and place a negative charge on the oxygen. O HC
O
-
Conjugate base c. Start by drawing the carboxylic acid parent chain (benzene ring with a –COOH in the first position). Counting from the carbon containing the –COOH group add a bromo substituent on carbon 3 and another bromo substituent on carbon 5. Remove hydrogen from the –OH and place a negative charge on the oxygen.
8 - 90
O C O
Br
Br Conjugate base
8.94
Name each of the conjugate bases in you answer to Problem 8.92.
Answer: To name the conjugate base of a carboxylic acid, the “ic” ending is replaced with “ate” followed by “ion” instead of “acid”. a. 3-methylhexanoate ion b. formate ion c. 3,5-dibromobenzoate ion
8.96
Draw the products of each reaction.
O a.
H C OH
+
KOH
NaOH
b.
Answer: These are neutralization reactions that lead to the formation of carboxylate salts and water.
O
a.
H C O- K
+
8 - 91
-
b.
8.98
Due to its high toxicity, the pesticide Acifluorfen is approved for use on just a short list of crops. This pesticide is sold as its sodium salt. Draw this salt, which is produced by reacting Acifluorfen with sodium hydroxide. O F3 C
O Cl
C
OH
NO 2
Acifluorfen
Answer: The carboxylic acid group reacts with the sodium hydroxide to form the sodium salt.
O O
CF3 Cl
+
C O Na NO2
Acifluorfen
8.100 Gentisic acid (Figure 8.18b) can be used as an anti-rheumatism drug. a. Which functional group of gentisic acid is the most acidic, one of the phenol groups or the carboxyl group? b. Draw the ionic compound potassium gentisate. (Hint: only one H+ is removed from gentisic acid.) Answer: a. The carboxyl group is most acidic. Carboxylic acids are more acidic than phenols. b. The “ate” ending requires that you draw the organic acid with the hydrogen removed from the carboxyl group and a negative charge placed on the oxygen. The cation of the salt is drawn beside the O-.
8 - 92
HO O - + C O K OH
8.102 2,4-Dichlorophenoxide is the conjugate base of 2,4-dichlorophenol. Draw both compounds. Answer: To draw 2,4-dichlorophenol, start by drawing the phenol parent (benzene ring with a – OH in the first position). Counting from the first carbon add a chloro substituent on carbon 2 and another chloro substituent on carbon 4. To draw 2,4-dichlorophenoxide, remove hydrogen from the –OH and place a negative charge on the oxygen.
O-
OH
Cl
Cl
Cl
Cl 2,4-dichlorophenol
2,4-dichlorophenoxide ion
8.104 a. Draw o-chlorophenol. b. Draw and name the conjugate base of o-chlorophenol. c. Which predominates in water at pH 7, o-chlorophenol or its conjugate base? Answer: a.
o-chlorophenol b. The conjugate base of o-chlorophenol is one where an H+ ion is removed from the –OH group.
8 - 93
-
o-chlorophenoxide ion c. The acidic form would predominate at pH 7 because its pKa is approximately 10.
8.106 a. For trichloroacetic acid (Table 8.6), pKa = 0.70 and for acetic acid, pKa = 4.74. Which is the stronger acid? b. What is the value of Ka for each of these acids? c. Draw sodium trichloroacetate, the salt formed when trichloroacetic acid and NaOH react in a 1:1 ratio. Answer: a. b.
Trichloroacetic acid is the stronger acid. It has the smaller pKa value. For trichloroacetic acid:
Ka = 10-0.70 = 0.20
For acetic acid:
Ka = 10-4.74 = 1.8 x 10-5
c.
SECTION 8.10
PREPARING ESTERS
8.108 Draw the products of each reaction. O a. CH3 CH CH2 C OH
+
H+ HOCH3
CH3
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H+ b.
Answer: These reactions lead to the formation of esters:
O a.
CH3 CH CH2C O CH3 CH3
b.
8.110 Name the organic products in Problem 8.108. Answer:
O a.
CH3 CH CH2C O CH3
methyl 3-methylbutanoate
CH3
b.
propyl 3-bromobenzoate
8.112 Draw and name the ester formed when benzoic acid reacts with pentyl alcohol, HOCH2CH2CH2CH2CH3, in the presence of H+.
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Answer: O
H+
C OH
+ HOCH2CH2CH2CH2CH3
benzoic acid
pentyl alcohol
O C O CH2CH2CH2CH2CH3 pentyl benzoate
8.114 Draw the products obtained from each reaction. O
a.
CH3 CH CH2 CH2 C O CH3 + NaOH(aq) CH3
O + NaOH(aq)
C O b. Answer:
These are hydrolysis reactions carried out in a basic solution. The products are a carboxylate salt and an alcohol. O +
CH3 CH CH2 CH2 C O Na
a.
+
CH3OH
CH3
O +
C O Na
+
HO
b.
8.116 Draw the products obtained when a. methyl 3,3-dimethyl hexanoate reacts with H2O in the presence of H+. b. methyl 3,3-dimethyl hexanoate reacts with NaOH(aq) .
8 - 96
Answer:
CH3
O
a. CH3 CH CH C CH C O CH + 2 2 2 3
H2O
H+
CH3 CH3
O
CH3 CH2 CH2 C CH2 C OH
+
HOCH3
+
HOCH3
CH3
O
CH3
b. CH3 CH2 CH2 C CH2 C O CH3 +
H2O
OH-
CH3 CH3
O
CH3 CH2 CH2 C CH2 C O CH3
8.118 Draw the product of each reaction. O CH3 CH CH2 C O CH2 CH CH3 + NaOH(aq)
a.
CH3
CH3
O CH3 C
O
CH 2
+ KOH(aq)
b. O
c.
CH3 O C CH2 (CH2)4 CH3
+ NaOH(aq)
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Answer: These are hydrolysis reactions carried out in a basic solution. The products are a carboxylate salt and an alcohol. O CH3 CH CH2 C O
+
+
Na
HO CH2 CH CH3 CH3
CH3
a.
O CH 3 C
O– K +
+
HO
CH 2
b. O +
c.
Na O C CH2 (CH2)4 CH3
+
CH3OH
8.120 Draw a skeletal structure for each of the products formed when the ester below, one of the primary compounds in carnauba wax, is reacted with NaOH(aq). O CH3(CH2)24
C O CH2(CH2)28CH3
Answer: A hydrolysis reaction takes place when the ester is reacted with the base, NaOH(aq). O CH3(CH2)24
C O CH2(CH2)28CH3
+
NaOH(aq) O
CH3(CH2)24
+
C O Na
+
8.122 Aspirin can be hydrolyzed under acidic or basic conditions. Draw the products of each reaction.
8 - 98
HO CH2(CH2)28CH3
a.
NaOH(aq)
b.
Answer:
a.
b.
NaOH(aq)
8 - 99
SECTION 8.11
AMINES
8.124 Draw each compound. a. 2-hexanamine b. N-methyl-3-pentanamine c. N,N-dimethyl-1-butanamine d. tetramethyammonium ion Answer: a. Draw the parent chain hexane, remove one hydrogen from carbon 2 and add the amino group, –NH2.
NH2 CH3CH2CH2CH2CHCH3 b. Draw the parent chain pentane. Remove a hydrogen atom from carbon 3 and add the amino group. Next, remove one hydrogen atom from the amino group and replace it with a methyl group (this is what N-methyl requires).
NHCH3 CH3CH2CHCH2CH3 c. Draw the parent chain butane. Remove a hydrogen atom from carbon 1 and add the amine group. Next, remove both hydrogen atoms from the amino group and replace them with two methyl groups (this is what N,N-dimethyl indicates).
CH3 CH3CH2CH2CH2NCH3 d. Draw the ammonium ion, NH4+. Replace each of the four hydrogen atoms with a methyl group.
CH3
CH3 + N CH3 CH3
8 - 100
8.126 Draw each amine. a. t-butylamine b. propylamine c. isobutylmethylamine d.triisopropylamine Answer: a. Draw a t-butyl group and attach an amino group –NH2 to the central C atom. CH3 CH3 C NH2 CH3
b. Draw a propyl group and attach an amino group –NH2 to first C atom. CH3 CH2 CH2 NH2
c. Attach an isobutyl group and a methyl group to –NH. CH3 CH CH2 NH CH3
CH3
d. Attach three isopropyl groups to N. CH3 CH N CH3
CH CH3
CH3 CH CH3 CH3
8.128 Give the IUPAC name of each amine.
a.
CH3 NH CH3
b.
CH3 CH NH CH3 CH3
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c.
CH3 CH3 CH2 N CH CH3 CH3
d. CH3 CH CH2 NH2 CH3
Answer: First, identify the longest chain to which the amine N atom is bonded. Change the ending of the alkane name of the longest chain from “e” to “amine”. Substituent groups on the parent chain are numbered starting from the C closest to the N atom. Substituent groups bonded directly to the N atom of the amine are preceded by N-. a. N-methylmethanamine b. N-methyl-2-propanamine c. N-ethyl-N-methyl-2-propanamine d. 2-methyl-1-propanamine
8.130 Give the common name of each amine in the Problem 8.128. Answer: Name all of alkyl groups bonded to the N atom using their alkyl names and add “amine”. a. Dimethylamine b. Isopropylmethylamine c. Ethylisopropylmethylamine d. isobutylamine
8.132 Identify each amine in Problem 8.128 as being 1°, 2°, or 3°.
8 - 102
Answer: 1° amines have one C atom bonded to the N. 2° amines have 2 C atoms bonded to the N atom. 3° amines have 3 C atoms bonded to the N atom.
a.
CH3 NH CH3
b.
CH3 CH NH CH3 CH3
c.
2° amine
2° amine
CH3 CH3 CH2 N CH CH3 CH3
3° amine
d. CH3 CH CH2 NH2 CH3
1° amine
8.134 Identify each compound in Problem 8.124 as being 1°, 2°, 3° or 4°. Answer: 1° amines have one C atom bonded to the N. 2° amines have 2 C atoms bonded to the N atom. 3° amines have 3 C atoms bonded to the N atom.
NH2 a.
CH3CH2CH2CH2CHCH3
1° amine
NHCH3 b.
CH3CH2CHCH2CH3
2° amine
CH3 c.
CH3CH2CH2CH2NCH3
3° amine
8 - 103
CH3
CH3 + N CH3 CH3
d.
4° amine
8.136 Identify each compound as a pyridine, a pyrimidine, or a purine (see Figure 8.26a).
NH2 N
N
H3C
N
HO CH2
O
H
N
N
HO
H
HO
Vitamin B6
H OH
CH2OH CH2OH
a.
F
b.
H
Fludarabine (an anti-cancer drug)
Answer: a. pyridine
b. purine
8.138 Account for the fact that propylamine is more water soluble than trimethylamine. Answer: Primary amines, like CH3CH2CH2NH2, have two N–H hydrogen atoms and one nitrogen atom available for forming hydrogen bonds with water. Tertiary amines, like (CH3)3N, have only one nitrogen atom capable of forming hydrogen bonds.
SECTION 8.12
AMINES AS WEAK ORGANIC BASES
8.140 Draw and name the conjugate acid of
8 - 104
a. butylamine b. dipropylamine c. trimethylamine d. t-butylamine Answer: a. Combine NH2 and a butyl group. Next, add one hydrogen ion to the amino group to make NH3. The hydrogen ion bound to the nitrogen carried a positive charge so place a “+” to the right of the –NH3. To name the ion, change the amine part of the name to ammonium ion. CH3CH2CH2CH2NH3+
butylammonium ion b. Combine the NH and two propyl groups. Next, add one hydrogen ion to the NH to make it NH2. Place a “+” to the right of the –NH2. To name the ion, change the amine part of the name to ammonium ion.
+ CH3CH2CH2NH2CH2CH2CH3 dipropylammonium ion c. Combine N and three methyl groups. Next, add one hydrogen ion to the N to make it NH. Place a “+” to the right of the –NH. To name the ion, change the amine part of the name to ammonium ion.
H + CH3NCH3 CH3 trimethylammonium ion d. Combine NH2 and a t-butyl group. Next add one hydrogen ion to the NH2 to make it NH3. Place a “+” to the right of the –NH2. To name the ion, change the amine part of the name to ammonium ion.
CH3 + CH3CNH3 CH3 t-butylammonium ion
8 - 105
8.142 a. Write the equilibrium equation for the acid-base reaction that takes place between aniline and water.
Aniline
b. Being attached to an aromatic ring causes the –NH2 group of aniline to be less basic than nonaromatic amines, such as propylamine. Because aniline is a weak base, its conjugate acid is relatively strong (pKa = 4.6). Which predominates at pH 7, aniline or its conjugate acid? Answer: a. Aniline is a basic compound:
Aniline b. Aniline predominates at pH 7 because this pH is greater than its pKa.
8.144 a. Write a balanced chemical equation for the reaction of aniline (Problem 8.142) with HCl. b. Name the salt that forms. Answer: a. and b.
Aniline
Anilinium chloride
8.146 In the early 1980s a street drug sold as synthetic heroin appeared in southern California. This drug contained an impurity called MPTP, which caused
8 - 106
irreversible symptoms of Parkinson’s disease (including immobility, slurred speech, and tremors). The MPTP existed as a hydrochloride salt. Draw this salt. N
CH3
MPTP Answer: The nitrogen atom of MPTP is basic.
+ CH3 N Cl H
8.148 Nitrogen atoms attached to an aromatic ring or belonging to an amide group are weaker bases than the nitrogen atoms in other amines. Draw the product of the acid-base reaction between the local anesthetic lidocaine and HCl.
Answer:
SECTION 8.13
AMIDES
8.150 Draw each amide. a. N-isopropylacetamide
8 - 107
b. N,N-dimethyl-3,4-dimethylhexanamide c. N-ethyl-N-methylbutyramide Answer: a. The name indicates an isopropyl group on the N atom. “acet” refers to an acyl group bonded to the N. O CH3 C NH CH CH3 CH3
b. There are 2 methyl groups bonded to the N atom, there are 6 carbon atoms in the carboxylic acid residue, and 2 methyl groups are bonded to the carboxylic acid chain at C-3 and C-4, counting from the C=O group. CH3
O
CH3 CH2 CH CH CH2 C N CH3 CH3
CH3
c. There is an ethyl group and a methyl group bonded to the N atom. The main chain has 4 C atoms including the carbonyl C. O CH3 CH2 CH2 C N CH2CH3 CH3
8.152 Name each amide. O a.
CH3 CH2 C N CH3 CH3
O
CH3 b.
CH3 CH CH C NH CH2 CH3 CH3
8 - 108
c.
Answer: Use the general IUPAC rules for naming amides. Begin counting from the carbonyl carbon (C=O) to indicate the location of each substituent group on the hydrocarbon chain. Substituent groups bonded to the N atom are identified by placing N- before the name of the substituent group. In some cases, a common name is used. a. N,N-dimethylpropanamide b. N-ethyl-2,3-dimethylbutanamide c. N-methyl-4-chlorobenzamide
8.154 Draw the amide that will be produced by each reaction. O
CH3CH2CH2C OH
a.
+
NH2CH3
+
CH3NHCH3
heat
O CH3CH2CH2C OH
b.
heat
Answer: Amides are produced by heating a carboxylic acid in the presence of an amine or ammonia. Remove a hydrogen from the nitrogen and the –OH group from the carboxylic acid. Attach the nitrogen group to the carbonyl group where the –OH group was originally.
a.
O CH3CH2CH2C OH
+
NH2CH3
O CH3CH2CH2C NHCH3
8 - 109
heat
b.
O CH3
O CH3CH2CH2C OH
+
CH3NHCH3
heat
CH3CH2CH2C NCH3
8.156 Draw the carboxylic acid and amine from which the pain reliever acetaminophen can be prepared.
Answer:
8.158 Name the amide product of each reaction in Problem 8.154. Answer: a. N-methylbutananamide. There is one methyl group attached to the N atom and there are 4 carbon atoms in the carboxylic acid residue. b. N, N-dimethylbutanamide. There are 2 methyl groups bonded to the N atom and there are 4 carbon atoms in the carboxylic acid residue.
8.160 Draw the products formed when each amide in Problem 8.150 is reacted with H2O in the presence of H+. Answer: The reaction of an amide with water in the presence of H+ results in the amide breaking apart to form a carboxylic acid and an amine. a. N-isopropylacetamide
8 - 110
O
O
H+
CH3 C NH CH CH3
CH3 C OH
H2O
+
+
+
H3N CH CH3 CH3
CH3
b. N,N-dimethyl-3,4-dimethylhexanamide CH3
O
H+ +
CH3 CH2 CH CH CH2 C N CH3 CH3
H2O
CH3 CH3
O
CH3 CH2 CH CH CH2 C OH
+ H2N CH3
+
CH3
CH3
c. N-ethyl-N-methylbutyramide
O
H+
CH3 CH2 CH2 C N CH2CH3
+ H2O
CH3 O CH3 CH2 CH2 C
OH
+ + H2N CH2CH3 CH3
8.162 Draw the products formed when each amide in Problem 8.152 is heated in the presence of H2O and H+. Answer: The reaction of an amide with water in the presence of H+ results in the amide breaking apart to form a carboxylic acid and an amine.
O a.
+
H
CH3 CH2 C N CH3
+
H2O
O CH3 CH2 C OH +
+
H2N CH3 CH3
CH3
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O
CH3
+
b. CH3 CH CH C NH CH2 CH3
H +
H2O
CH3 O
CH3
+
CH3 CH CH C OH + NH3 CH2 CH3 CH3
c.
8.164 Draw the products obtained when capsaicin (Figure 8.19) is hydrolyzed under acidic condition.
O CH3 O
CH2 NH C CH2 CH2 CH2 CH2
H
C C H
HO Capsaicin Answer:
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CH(CH3)2
O CH2 NH C CH2 CH2 CH2 CH2
CH3 O
H
C C
+
H2O
CH(CH3)2
H
HO
+
H
Capsaicin O CH3 O
+
CH2 NH3
+
HO C CH2 CH2 CH2 CH2
H
C C H
HO
8.166 Acetaminophen (Problem 8.156) can be hydrolyzed under acidic or basic conditions. Draw the products of each reaction.
acetaminophen a. acetaminophen
b. Answer:
a.
b.
8.168 The boiling point of propanamide is 213°C and that of pentanamide is 232°C. Account for this difference in boiling points. O O
CH3 CH2 C NH2
CH3 CH2 CH2 CH2 C NH2
Propanamide
Pentanamide
8 - 113
CH(CH3)2
Answer: The difference in boiling points can be accounted for by the size and the strength of the intermolecular attractive forces between the molecules of each compound. Both propanamide and pentanamide have amide groups that are polar. Pentanamide has a longer hydrocarbon chain and thus greater capability for London dispersion interactions between its molecules.
HEALTH LINK
A CHILI PEPPER PAINKILLER
8.170 a. Draw the products formed when Olvanil (Problem 8.169) is hydrolyzed under acidic conditions. b. Draw the phenoxide salt that forms when Olvanil is reacted with NaOH. Answer: a. The products for the hydrolysis in acid are shown below.
Olvanil
b. The phenoxide salt product for the hydrolysis in NaOH is shown below.
-
8 - 114
HEALTH LINK
ALPHA HYDROXY ACIDS
8.172 a. Of phenol and resorcinol (see the footnote of Table 8.6), which is the weaker acid? b. Write the equation for the acid-base reaction that takes place between this acid and water. c. Which is there more of at equilibrium, the acid in your answer to part a or its conjugate base? Answer: a. Phenol is the weaker acid. Phenol has a pKa of 9.96 and resorcinol has a pKa of 9.81. Strong acids have low pKa’s and weak acids have high pKa’s.
-
b. c. There will be more acid at equilibrium because the Ka is small Ka = 10-pKa = 10- 9.96 = 1.1 x 10-10
HEALTH LINK
ADRENALINE AND RELATED COMPOUNDS
8.174 a. What is an over-the-counter drug? b. Why are pseudoephedrine-containing drugs no longer sold over the counter? Answer: a. An over-the-counter drug is a nonprescription drug. b. Pseudoephedrine-containing drugs can be used as a precursor for the illegal synthesis of methamphetamine.
HEALTH LINK
BIOFILMS
8.176 Homoserine (Figure 8.32) forms a lactone, but serine, a related amino acid, does not. Explain why.
8 - 115
O NH2 CH C OH CH2 OH Serine Answer: Because 4-membered rings are quite strained, serine would be unstable as a lactone.
BIOCHEMISTRY LINK
A CURE FOR FLEAS
8.178 Precor, used in flea collars and animal sprays, is a hormone that prevents flea pupae (the stage of life between larval and adult forms) from developing.
O
CH3
O O
Precor a. Precor has low water solubility. Explain why. b. Draw the products obtained when Precor is hydrolyzed under basic conditions. c. Draw the products obtained when Precor is hydrolyzed under acidic conditions. Answer: a. The contribution of the polar ether and ester groups, which can form hydrogen bonds with water, is outweighed by the nonpolar nature of the rest of the molecule. b. Under basic conditions the ester bond is broken producing a carboxylate ion and an alcohol.
O
CH3
O O
+
HO
c. Under acidic conditions the ester bond is broken producing a carboxylic acid and an alcohol.
8 - 116
CH3
O
O OH
+
HO
LEARNING GROUP PROBLEMS
8.180 a. Draw a side view for the cis isomer of the molecule. OH
CH
3
b. Draw a side view for the trans isomer of the molecule. c. Draw a constitutional isomer of the molecule in part a that has a ring and is a cis geometric isomer. d. Draw a constitutional isomer of the molecule in part a that has a ring, but has no geometric isomers. e. name the major noncovalent force that attracts one of the molecules in part a to a similar molecule. f. Draw a constitutional isomer of the molecule in part a that contains an alkene group and has no geometric isomers. g. Draw a constitutional isomer of the molecule in part a that contains an alkene group and is a trans isomer. Answer:
a.
CH3
OH
OH
b.
c.
CH3
CH3
CH2OH
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CH2OH
d. e.
hydrogen bonding
f.
CH2=CHCH2CH2CH2-OH CH2 CH2 OH
H
g.
C CH3
C H
8.182 a. Which acid in Table 8.6 is the strongest? b. Name the conjugate base of this acid. c. Write the equation for the reaction that takes place between this acid and water. d. Which is there more of at equilibrium, the acid in your answer to part a or its conjugate base? e. Draw and name the salt that forms when the acid from part a reacts with KOH. f. Draw and name the products that form when the acid from part a reacts with HOCH2CH(CH3)2 and H+. g. Write a balanced equation for the reaction that takes place between the organic product from part f and NaOH(aq). h. Draw and name the products that form when the acid from part a is heated in the presence of (CH3CH2)2NH. i. Write a balanced equation for the reaction that takes place when the organic product for part h is heated in the presence of H2O and H+. Answer: a. Trichloroacetic acid (pKa = 0.70) is the strongest acid. The strength of the acid increases as the pKa decreases. b. Trichloroacetate is the conjugate base of trichloroacetic acid. c. An H+ ion is donated to a water molecule:
Cl O
Cl O Cl
C C OH
+
H2O
Cl
C C O Cl
Cl
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+
+
H3O
d. There is more acid than the conjugate base at equilibrium because the Ka is small. Ka = 10-pKa = 10-0.70 = 0.20. e. The salt formed is potassium trichloroacetate. Cl O Cl
+
C C O K Cl
f. The ester formed is sec-butyl trichloroacetate. Cl O Cl
C C O CH2CH(CH3)2
+
H2O
Cl
g. The products are a carboxylate salt and an alcohol.
Cl O Cl
C C O CH2CH(CH3)2
+
NaOH
Cl Cl O Cl
+
C C O Na Cl
h. An amide is formed. Cl O Cl
C C N(CH2CH3)2 Cl
i. The products are a carboxylic acid and an ammonium ion.
8 - 119
+
HO CH2CH(CH3)2
Cl O Cl
+
C C N(CH2CH3)2
+ H2O
Cl
H
heat
Cl O Cl
C C OH Cl
8 - 120
+
+ H2N
(CH2CH3)2
Chapter 9 Organic Reactions 2 - Alcohols, Ethers, Aldehydes, and Ketones SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
9.1
a. Add the missing hydrogen atoms to the ring on each molecule.
heat
menthol
H
alkene products
CH3 H
H H H CH3 CH CH3
H
H
H
CH3 H
H
H
H H
H H
OH
H CH3 CH
H H H
H
CH3 H
H H
H H CH3 CH CH3
CH3
menthol
alkene products
b. When heated in the presence of H+, menthol reacts to form two alkenes. In the reaction above, which alkene is the major product? When menthol is heated in the presence of H+, a dehydration reaction occurs. When there are two possible products, the major product is that formed by the removal of –OH from one carbon atom and removal of –H from the neighboring C atom that carries fewer H atoms. 9-1
H
CH3 H
H H
H
H H H
H CH3 CH
H
dehydration
CH3 H
H
H H
H H H CH3 CH CH3
OH
CH3 menthol
major alkene product The H atom is removed from this C atom because it contains fewer H atoms than the other C atom adjacent to the -OH group.
c. Circle the atoms in menthol that are the source of the water molecule that appears as one of the reaction products.
H
CH3 H
H
H H H
H H CH3 CH
H
OH
CH3
menthol 9.3
Identify each alcohol as being 1°, 2°, or 3°. Alcohols are classified according to the number of carbon atoms bonded to the carbon atom to which the hydroxyl group is attached (C-OH). If there is only C atom bonded to the C-OH, the alcohol is 1○. If there are two C atoms each one directly bonded to the C-OH, the alcohol is 2○. If there are three C atoms each one directly bonded to the C-OH, the alcohol is 3○.
OH a.
CH3CH2CH CH2CH3
2°
9-2
OH b. CH3C CH2CH3 CH2CH3 3°
CH3 c.
HOCH2 CH CH2CH3
1°
9.5
Give the IUPAC name of each alcohol in Problem 9.3. a. The parent chain has five carbon atoms which makes it a pentane. To name the parent chain, drop the “e” and add “ol”. The hydroxyl group is on carbon 3 so place a 3- in front of pentanol. 3-pentanol b. The parent chain has five carbon atoms which makes it a pentane. To name the parent chain drop the “e” and add “ol”. The hydroxyl group is on carbon 3 so place a 3- in front of pentanol. There is also a methyl substituent on carbon three so assign it the number 3 as well. 3-methyl-3-pentanol c. The parent chain has four carbon atoms which makes it a butane. To name the parent chain drop the “e” and add “ol”. The hydroxyl group is on carbon 1 so place a 1- in front of butanol. There is also a methyl substituent on carbon 2 so assign it the number 2. 2-methyl-1-butanol
9.7
Draw each alcohol molecule. a. 2-hexanol Draw a six carbon (hexane) parent chain. Add –OH to carbon 2.
9-3
OH CH3CH2CH2CH2CHCH3 b. 3-methyl-1-pentanol Draw a five carbon (pentane) parent chain. Add –OH to carbon 1. Add –CH3 to carbon 3.
CH3 CH3CH2CHCH2CH2OH c.
4-isopropylcyclohexanol
Draw a six-membered ring (cyclohexane). Add an –OH. The –OH is carbon 1. Count to carbon 4 and add an isopropyl group.
OH
CH3CHCH3
9.9
Identify each alcohol in Problem 9.7 as being 1°, 2°, or 3°.
OH a.
CH3CH2CH2CH2CHCH3
2°
CH3 b.
CH3CH2CHCH2CH2OH
1°
9-4
OH
c.
9.11
9.13
2°
To which organic family does each molecule belong?
a.
sulfides
b.
ethers
c.
alcohols
Name each molecule in Problem 9.11
a. CH3CH2SCH2CH2CH3
ethyl propyl sulfide
b. CH3CH2OCH2 CH2CH3
ethyl propyl ether
c.
9.15
CH3CHCH3
2-pentanol
Are the compounds in parts a and b of Problem 9.11 identical molecules, different conformations of the same molecule, cis/trans isomers, constitutional isomers, or entirely different molecules? Entirely different molecules. CH3CH2SCH2CH2CH3 is a sulfide and CH3CH2OCH2CH2CH3 is an ether.
9-5
9.17
a. What is the shape around the S atom for the molecule in Problem 9.11a? bent b. What is the shape around the O atom for the molecule in Problem 9.11b? bent c. One of the molecules in parts a and b of Problem 9.11 is polar and the other is not. Which is which? a. is not polar, b. is polar
9.19
Give the common name of each ether.
a.
s-butyl propyl ether
b. CH3 CH2 CH CH3 O CH3 s-butyl methyl ether
c.
cyclohexyl methyl ether
9.21
Draw each molecule. a.dimethyl ether
CH3 O
CH3
b. dicyclopropyl ether
9-6
O c. butyl ethyl ether
CH3CH2CH2CH2 O CH2CH3
9.23
The molecule CH3CH2CH2SH is partly responsible for the strong odor that is associated with onions. Give its IUPAC and common name. IUPAC name: Common name:
9.25
1-propanethiol propyl mercaptan
a. Allyl mercaptan is the major compound that can be detected on the breath immediately after eating garlic. Draw this compound (allyl = -CH2CH=CH2).
H2C CH CH2
SH
b. Give the IUPAC name for allyl mercaptan. 2-propen-1-thiol
9.27 a. Which of the following is a liquid at STP: CH3CH2OH or CH3OCH3?(see Table 9.1) CH3CH2OH is a liquid at room temperature. b. Which of the following is a liquid at STP: CH3CH2OH or CH3OCH3? CH3CH2OH has an –OH group that can hydrogen bond with other molecules of CH3CH2OH, leading to strong intermolecular attractive forces. CH3OCH3 does not have hydrogen bonding capability, has a low boiling point, and is a gas at STP (see Table 9.1).
9.29
Explain why CH3CH2OCH2CH3 is less soluble in water than its constitutional isomer CH3CH2CH2CH2OH. CH3CH2CH2CH2OH has a greater capability to hydrogen bond with water. This enhances its solubility in the solvent.
9-7
9.31
Ethanol (molecular weight = 46.0 amu) has a boiling point of 75.5°C and propane (molecular weight = 44.0 amu) has a boiling point of -42°C. Account for the difference in boiling point. Ethanol is composed of polar molecules that are capable of hydrogen-bonding to each other. Propane is composed of nonpolar molecules held together only by weaker London forces. The stronger intermolecular forces between ethanol molecules cause its boiling point to be much higher than propane’s.
9.33 1-Propanol has a boiling point of 97.4°C and 1-butanol has a boiling point of 117.3°C. Account for the difference in boiling point. Both have polar molecules capable of hydrogen-bonding to one another. But butanol has stronger London forces due to its longer hydrocarbon chain.
9.35
Methanethiol (molecular weight = 48.0 amu) has a boiling point of 6°C and ethanol (molecular weight = 46.0 amu) has a boiling point of 78.5°C. Account for the difference in boiling point. Although both ethanol and methanethiol contain polar molecules, ethanol molecules are capable of forming the strong hydrogen-bonding intermolecular force and thus ethanol has the higher boiling point.
9.37
Isoimpinellin, a naturally occurring ether, is found at very low levels in celery. Tests have shown that at high concentrations this compound can be a carcinogen (cancer-causing agent). This is not cause to avoid eating celery, however, because many foods naturally contain toxic substances at very low levels. O CH3
O
O
O
O CH3 Isoimpinellin
9-8
a. How many ether groups does isoimpinellin contain? Isoimpinellin has 3 ether groups: ether
O CH3
O
O
ether ether
O
O CH3
b. What other functional groups are present in this molecule? alkene, aromatic, ester
9.39
Draw the products of each nucleophilic substitution reaction.
a.
- OH
b.
- OCH2CH2CH3
+
CH3 Br CH3 O CH2 CH2 CH3 + Br
9-9
c.
- SCH3 + CH3 CH2 CH CH3 Cl
CH3 CH2 CH CH3
+ Cl
S CH3
d.
- SH + CH3 CH2 CH2 CH2 Br CH3 CH2 CH2 CH2 SH
9.41
+ Br
Name each of the products in Problem 9.39. a. 3-methyl-1-butanol b. methyl propyl ether c. s-butyl methyl sulfide d. 1-butanethiol or butyl mercaptan
9.43
9.45
Draw the missing alkyl bromide for each reaction.
a.
- OCH3
+
CH3 CH2 Br
CH3 CH2 O CH3
b.
- SH
+
CH3 CH2 CH2 Br
CH3 CH2 CH2 SH
c.
- SCH2CH3
+
CH3 CH2 Br
CH3 CH2 S CH2 CH3
Write a chemical reaction that shows how each compound can be produced from an alkyl bromide.
9 - 10
a. CH3CH2CH2OH
- OH
+
CH3 CH2 CH2 OH
CH3 CH2 CH2 Br
b. CH3OCH2CH2CH3
- OCH2CH2CH3
+
CH3O CH2 CH2 CH3
CH3 Br or
- OCH3
9.47
CH3CH2CH2 Br
+
CH3O CH2 CH2 CH3
The molecule CH3Cl is named methyl chloride. Name each of the following molecules: These are all non-IUPAC names:
9.49
a. CH3CH2Br
ethyl bromide
b. CH3CHFCH3
isopropyl fluoride
c. CH3CH2CH2CH2Br
butyl bromide
d. CH3CH(CH3)CH2Cl
isobutyl chloride
Complete the series of reactions by adding the necessary reactants or conditions to the reaction arrows. a. b. The two reactions involved in each of these are alkyl halogenation and nucleophilic substitution. -OCH 3
light a.
alkyl halogenation
nucleophilic substitution 9 - 11
-SCH 3
light b.
9.51
alkyl halogenation
nucleophilic substitution
Draw the organic product (if any) expected from each reaction. Alcohols undergo oxidation to aldehydes or carboxylic acids or ketones when they react with K2Cr2O7. When thiols are oxidized by I2, two molecules of the thiol react to form a disulfide bond.
OH a.
+
K2Cr2O7
O
b.
HO CH2 CH2 CH2 CH2 CH3 + K2Cr2O7 O HO C CH2 CH2 CH2 CH3
HO c.
CH3 +
K2Cr2O7
no reaction (3o alcohol)
9 - 12
d.
2 CH3 SH
I2
+
CH3 S
e.
2 CH3 CH CH3
+
S CH3
I2
SH
CH3 CH CH3 S S CH3 CH CH3
9.53
Describe the difference in the products obtained when 1-butanol and 1butanethiol are oxidized. When 1-butanol is oxidized, a carboxylic acid is produced. When 1-butanethiol is oxidized, a disulfide is produced.
9.55
When oxidized, which alcohol produces the ketone shown below?
Secondary alcohols, when oxidized, produce a ketone. The secondary alcohol below, when oxidized, will produce the ketone above.
9 - 13
9.57
a. Lactate, which builds up in muscle cells during exercise, is sent to the liver where an enzyme catalyzes the oxidation of this 2° alcohol to pyruvate. Draw pyruvate.
-
enzyme
?
NAD+
Pyruvate
Lactate
When oxidation takes place, the 2° alcohol group is converted to a ketone:
O O
CH3 C C O pyruvate b. Lactate and pyruvate are the conjugate bases of lactic acid and pyruvic acid, respectively. Draw these carboxylic acids.
OH O
O O
CH3 CH C OH
CH3 C C OH
lactic acid
pyruvic acid
c. At pH 7, why does each of these acids appear in its conjugate base form? pH = 7 is greater than the pKa of a typical carboxylic acid.
9.59
Draw the major product of each reaction. In the presence of heat and H+, the –OH group of an alcohol is removed, along with a neighboring hydrogen, to form a double bond. The hydrogen atom removed is from the neighboring carbon atom that carries the fewest H atoms.
CH3
a.
CH3
H+
CH3CH CH CH CH3
heat
CH3
CH3CHC CH3
OH
9 - 14
CHCH3
NADH + H+
b.
CH3
H+
CH3C CH2CH3
heat
CH3 CH3C
OH
9.61
CHCH3
Draw each molecule named below and draw the major organic product expected when each is reacted with H+ and heat. In the presence of heat and H+, the –OH group of an alcohol is removed, along with a neighboring hydrogen, to form an alkene. The hydrogen atom removed is from the neighboring carbon atom that carries the fewest H atoms. a. 2,3-dimethyl-2-butanol
yields
b. 2,3-dimethyl-3-hexanol
yields
c. 2-methylcyclopentanol
OH CH3 yields
9.63
CH3
a. When a particular alcohol is heated in the presence of H+, 2-methyl-2 butene is the major product. Draw and name this alcohol.
9 - 15
In the presence of heat and H+, the –OH group of an alcohol is removed, along with a neighboring hydrogen, to form an alkene. The hydrogen atom removed is from the neighboring carbon atom that carries the fewest H atoms.
OH H+, heat CH3 C CH CH3
CH3 C CH2 CH3
CH3
CH3
2-methyl-2-butene (major product)
2-methyl-2-butanol
b. When a different alcohol is heated in the presence of H+, 2-methyl-2 butene is the major product. Draw and name this alcohol.
OH CH3 CH CH CH3
H+, heat CH3 C CH CH3
CH3
CH3
3-methyl-2-butanol
9.65
2-methyl-2-butene (major product)
Complete the series of reactions by adding the necessary reactants or conditions to the reaction arrows.
a.
b.
The reaction conditions and reactants are shown below for each set of reactions.
9 - 16
a.
hydration
oxidation
b.
heat nucleophilic substitution
9.67
dehydration
Menthol flavor is due to the alcohol named menthol (below). Complete each reaction of menthol by adding the necessary reactants or conditions to the reaction arrows.
a.
menthol
b.
menthol
The reaction conditions and reactants are shown below.
9 - 17
heat
a.
menthol This is a dehydration reaction where an alcohol is converted to an alkene.
b.
menthol This is an oxidation reaction of a secondary alcohol to a ketone.
9.69
Name each of the following molecules.
CH3 a.
O
CH3 CH CH C H CH3
2,3-dimethylbutanal
O b.
CH3 CH2 C CH CH2CH3 CH2CH3
4-ethyl-3-hexanone
CH3 CH2 O c.
CH3 C
C H
CH3 CH2
9 - 18
2-ethyl-2-methylbutanal
O CH3 d.
CH3 C C CH3 CH3
3,3-dimethyl-2-butanone
9.71
Draw each molecule. a. pentanal
O CH3 CH2 CH2 CH2 C H b. 3-bromohexanal
Br
O
CH3 CH2 CH2 CH CH2 C H c. dipropyl ketone
O CH3 CH2 CH2 C CH2 CH2 CH3 d. 2,5-dibromocyclohexanone
O Br Br
9.73
2-Hexanone has a boiling point of 150°C and 2-pentanone has a boiling point of 102°C. a. Draw each molecule.
9 - 19
O
O
CH3 C CH2 CH2 CH2 CH3
CH3 C CH2 CH2 CH3
2-hexanone
2-pentanone
b. Account for the difference in boiling point. 2-hexanone has a higher boiling point than 2-pentanone because it is a larger molecule and has stronger London dispersion forces between its molecules.
9.75
Propanone (molecular weight = 58.0 amu) has a boiling point of 56°C and 1propanol (molecular weight = 60.0 amu) has a boiling point of 97.4°C. Account for the difference in boiling point. 1-propanol molecules are able to hydrogen-bond whereas propanone molecules are not.
9.77
Draw the product (if any) of each reaction.
O
O
K2Cr2O7
CH3 CH2 CH CH2 C H
CH3 CH2 CH CH2 C OH CH3
CH3
a.
CH3
K2Cr2O7
HO CH2 CH2 CH2 C CH3
O
CH3
HO C CH2 CH2 C CH3 CH3
CH3
b.
OH
O
O
K2Cr2O7
CH3 C CH C OH
CH3 CH CH C H c.
9.79
O CH3
CH3
Draw each molecule. Draw the product (if any) obtained when each is reacted with K2Cr2O7. a. propanal
9 - 20
When an aldehyde is oxidized by K2Cr2O7 the product is a carboxylic acid.
yields b. butanone Ketones are not oxidized by K2Cr2O7.
yields
no reaction
c. 1-butanol First, the alcohol is oxidized by K2Cr2O7 to an aldehyde. Then the aldehyde is oxidized to a carboxylic acid.
yields
d.3-chloro-3-methylpentanal When an aldehyde is oxidized by K2Cr2O7 the product is a carboxylic acid.
yields
e. 3-hydroxybutanal When an aldehyde is oxidized by K2Cr2O7 the product is a carboxylic acid. This molecule also has a secondary alcohol group which is oxidized to a ketone.
yields
9 - 21
9.81
Draw the product of each reaction. When aldehydes and ketones are reacted with H2 in the presence of a Pt catalyst, the aldehydes are converted to a primary alcohol and the ketones are converted to secondary alcohols.
O a.
+
CH3CH2C CH CH3
Pt
H2
CH2CH3
OH CH3CH2CHCHCH 3 CH2CH3
b.
O
CH3CH2
Pt
+
H2
OH
CH3CH2
CH3
CH3 O c.
Pt
+
CH3C CH2C H
CH3CCH2CH2OH
H2
CH3
9.83
CH3
One step in the synthesis of fatty acids by the body involves the enzyme catalyzed reduction of a ketone group present in a carboxylic acid derivative. In this compound, “ACP” is acyl carrier protein, which will be discussed in Chapter 14. Draw the reduction product. O
O
CH3 CH2 CH2 C CH2 C ACP
+
NADPH
9 - 22
+
+
H
enzyme ?
+
NADP+
Reduction converts the ketone group into an alcohol group:
9.85
The flavor of vanilla is largely due to the molecule named vanillin.
vanillin a. Which organic families are present in vanillin?
aldehyde
phenol ether
vanillin
b. Draw a product of each reaction
9 - 23
vanillin This is an oxidation reaction. Potassium dichromate is an oxidizing agent that oxidizes the aldehyde group to a carboxylic acid.
vanillin This is an oxidation reaction. Cu2+ is found in Benedicts' reagent which is basic. The aldehyde functional group is oxidized to a carboxylate ion.
H2
vanillin This is reduction reaction. H2 , in the presence of the Pt catalyst, reduces the aldehyde functional group to an alcohol.
9 - 24
vanillin This is an acid-base reaction. The phenol group is acidic and can react with the base NaOH to form the sodium salt of vanillin and water.
vanillin This is an aromatic substitution reaction. A hydrogen atom on the ring is replaced by a bromine atom using an Fe catalyst. Only one possible product is given above.
9.87
Draw the product of each reaction. In the presence of H+ as a catalyst, alcohols can undergo an addition reaction with the carbonyl group of an aldehydes or ketone to form hemiacetals. When an aldehyde or a ketone reacts with 2 molecules of the alcohol, an acetal is formed.
H+ a.
9 - 25
O H+ b.
+ 2
CH3 CH2 CH2 CH2 OH
CH3CH2CH2CH2O
OCH2CH2CH2CH3
H+ c.
9.89
Draw the missing reactant for each reaction.
H+ a.
Since the product is a hemiacetal at position 3 of a 5 carbon chain, the original reactant had to be 3-pentanone.
9 - 26
H+ 2
b.
Since the product is an acetal formed by addition of two alcohol molecules, the reactant contained C=O in a 5-carbon ring.
CH3 H+ c.
+
CH3CH O CH3CH
2 CH3CH OH
CH3CH O
CH3
CH3
Since the product is an acetal formed by addition of two alcohol molecules, the reactant was an aldehyde. O
CH3C H
9.91
Draw the missing reactant for each reaction. OH
O H+ CH3 C
CH2 CH3 + CH3 CH OH CH3
CH3 C
CH2 CH3
O
CH CH3 CH3
a.
9 - 27
O CH3 CH2 C
H+ CH2 CH3 + CH3 CH2 CH2 OH OH CH3 CH2 C CH2 CH3 O CH2 CH2 CH3
b.
H+
2 c.
9.93
a. How are cyclic hemiacetals similar to lactones (Figure 8.32)? Cyclic hemiacetals and lactones each have a ring that contains a C-O-C linkage. b. How are cyclic hemiacetals different from lactones? In cyclic hemiacetals, one of the carbon atoms in the C-O-C linkage is also attached to an –OH group. In lactones, this carbon atom is double-bonded to an oxygen atom (C=O).
9.95
Draw the product of each reaction. The structure of vanillin is given in Problem 9.85. a.
vanillin This is a hemiacetal formation reaction. The alcohol adds to the C=O forming a hemiacetal.
9 - 28
b.
2
vanillin This is an acetal formation reaction. The two alcohol molecules add to the C=O forming an acetal.
9.97
Draw the product of each reaction. The structure of menthol is given in Problem 9.67. a.
This is a hemiacetal formation reaction. The alcohol group in methanol adds to the C=O group in the ketone forming a hemiacetal.
9 - 29
b.
2
This is an acetal formation reaction. The 2 menthol molecules add to the C=O group in the ketone forming an acetal.
9.99
The toxicity of methanol is mainly due to the aldehyde produced when it is oxidized in the liver. Draw methanol, then draw and name the aldehyde formed on its oxidation. methanol
methanal or formaldehyde
O CH3OH
H C
H
9.101 O2 is reduced when it is converted into H2O2. Explain. Hydrogen atoms are added to O2 to form H2O2. 9.103 In the breakdown of superoxide, which product results from the reduction of O2and which results from the oxidation of O2-? Explain. 2O2- + 2H+ → H2O2 + O2 The product that results from reduction of O2- is H2O2. The change is classified as reduction because it involves a gain of H. The product that results from oxidation of O2- is O2. This change is classified as oxidation because it involves a loss of electrons.
9.105 The anti-depressant fluoxetine (Prozac), which has been detected in biosolids, is typically sold as the hydrochloride salt. Draw this salt.
9 - 30
Cl-
Hydrochloride salt of fluoxetine
9. 107 The anti-depressant Venlefaxine has been detected in ground water at concentrations of 50 g/L. a. At this concentration, how many grams of this drug are present in 3.0 L of water? First, calculate the g of the drug in 3.0 L of water using the given concentration then convert this to g. 3.0 L x
50 g 1x 10-6 g x = 1.5 x 10-4 g or 2 x 10-4 g with 1 sig. fig. 1 L 1 g
b. Convert 50 g /L into parts per million.
Parts per million =
g of solute x 106 mL of solution
50 g 1x 10-6 g 1L x x x 106 = 5 x 10-2 ppm 1 L 1 g 1000 mL c. Convert 50 g /L into parts per billion.
Parts per billion =
g of solute x 109 mL of solution
50 g 1x 10-6 g 1 L x x x 109 = 5 x 101 ppb 1 L 1 g 1000 mL
9.109 a. An alkyl halide is converted into the alcohol below. Write the chemical equation for this reaction.
9 - 31
CH2 OH CH3
CH3
CH3 Alcohol A
CH2 Cl CH3
CH2 OH CH3
+
CH3
CH3
OH–
CH3
b. Draw the reaction product Alkene B.
H+ Alcohol A
Alkene B
heat
CH2 CH3
CH3
CH3 Alkene B c. Draw the reaction product, Alcohol C. Alkene B
CH3 CH3
+
H+
H2O
Alcohol C
OH CH3
CH3 Alcohol C d. Draw the reaction product, Alkene D. H+ Alcohol C heat
9 - 32
Alkene D
CH3
+
Cl–
CH3 CH3
CH3
CH3 Alkene D e. Give the IUPAC name of Alkene D. 1,2,3,3-tetramethylcyclohexene f. Draw the reaction product.
Alcohol A
H+
g. Draw the reaction product.
2 Alcohol A
H+
9 - 33
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 9.2
a. Add the missing hydrogen atoms to the ring on each molecule.
CH3
CH3
CH3
+
H
or
H2O
+
OH CH3 CH CH3
CH3 CH
CH3 CH CH3
CH3
b. When reacted with water in the presence of H+, two alcohol products are formed. In the reaction above, which alcohol is the major product? c. Circle the atoms in the major product that were provided by the water molecule.
Answer: H3C H H
H
H +
H
H
H
H2O
+
H H CH3 CH CH3 HH
CH3
H3C H
H H
H H H CH3 CH CH3
H
H OH
or
H
H
H
H
H H CH3 CH
H H OH
CH3
a. b. The major product is the alcohol below. According to Markovnikov’s rule, the major product is that formed when the H atom bonds to the C atom in the double that already has the higher number of H atoms.
9 - 34
OH
CH3
OH
CH3 CH CH3
CH3 H H CH3 CH CH3
c.
SECTION 9.1
9.4
OH
ALCOHOLS, ETHERS, AND RELATED COMPOUNDS
Identify each alcohol as being 1°, 2°, or 3°. CH3 CHCH2 OH a.
b.
CH2CH2CH3
CH3CHCH3 CH3CH2 CHOH
H3C
OH
c.
Answer: The carbon to which the hydroxyl group is attached is bonded to only one other carbon atom in a 1° alcohol, to two carbon atoms in a 2° alcohol, and to three carbon atoms in a 3° alcohol. a. 1° alcohol
9 - 35
b. 2° alcohol c. 3° alcohol
9.6
Give the IUPAC name of each alcohol in Problem 9.4.
Answer: In naming alcohols, identify the longest chain that contains the –OH functional group. The parent name is derived from the alkane name of the longest chain with the ending changed to “-ol”. Counting from the C atom closest to the –OH group, assign a number to the C atom to which the –OH group is bonded. Name and number any substituent groups to complete the name. a. 2-methyl-1-pentanol The longest chain containing the –OH group has 5 C atoms. The –OH group is bonded to the first C atom and a methyl group is bonded to second C atom. b. 2-methyl-3-pentanol The longest chain containing the –OH group has 5 C atoms and the –OH group is bonded to the third C atom. On C number 2, a methyl group is bonded. c. 1-methyl-1-cyclopentanol The parent chain is a cyclic alkane that has 5 C atoms. Both the –OH group and a methyl group are bonded to the same C atom designated number one.
9.8
Draw each alcohol molecule. a. 2,3-dimethyl-3-hexanol b. 4,4-dimethyl-2-hexanol c. cis-3-methylcyclohexanol
Answer:
CH3 CH3 a.
CH3 CH C CH2 CH2 CH3 OH
OH b.
CH3
CH3 CH CH2 C CH2 CH3 CH3
9 - 36
c. OH
9.10
CH3
Identify each alcohol in Problem 9.8 as being 1°, 2°, or 3°.
Answer: The carbon to which the hydroxyl group is attached is bonded to only one other carbon atom in a 1° alcohol, to two carbon atoms in a 2° alcohol, and to three carbon atoms in a 3° alcohol. a. 3° alcohol b. 2° alcohol c. 2° alcohol
9.12
To which organic family does each molecule belong? a.
CH3SSCH3
O CH3 b.
CH3 CH CH3
c.
CH3CH2SH
Answer: a. disulfide b. ether c. thiol
9.14
Name each molecule in Problem 9.12.
Answer: a. dimethyl disulfide
9 - 37
b. isopropyl methyl ether c. ethanethiol
9.16
Are the compounds in parts b and c of Problem 9.11 identical molecules, different conformations of the same molecule, cis/trans isomers, constitutional isomers, or entirely different molecules?
Answer: Constitutional isomers. The two molecules below have the same molecular formula (C5H12O) but different connectivities of atoms.
CH3CH2CH2
CH3
CH3CH2OCH2CH2CH3
9.18
CH OH
a. Of the molecules in parts b and c of Problem 9.11, one has a boiling point of 117.3°C and the other has a boiling point of 39.6°C. Which is which? b. Account for this difference in boiling point.
Answer: a. The alcohol has the higher boiling point, 117.3 °C. CH3 CH2 CH2 CH OH CH3
b. Strong hydrogen bonding between molecules of the alcohol increases the boiling point. The ether molecules have weaker intermolecular forces.
9.20
Give the common name of each ether. CH3 a.
CH3 O C CH3 CH3
b.
CH3 CH O CH2CH2CH2CH3 CH3
c.
CH3CH2CH2CH2CH2OCH2CH2CH2CH2CH3
9 - 38
Answer: The common naming system gives the names of the alkyl group attached to the O atom followed by “ether”. a. t-butyl methyl ether There is a t-butyl group on one side and a methyl on the other. b. butyl isopropyl ether There is a butyl group on one side and an isopropyl group on the other. c. dipentyl ether There is a pentyl group on each side. Instead of using pentyl pentyl, the name dipentyl is used.
9.22
Draw each molecule. a. dipropyl ether b. cyclopentyl ethyl ether c. isobutyl isopropyl ether
Answer: a.
CH3 CH2 CH2 O CH2 CH2 CH3
O CH2 CH3 b.
c.
CH3 CH CH2 O CH CH3 CH3
9.24
CH3
Give the IUPAC and common name of the molecule. SH CH3 C CH3 CH3
Answer: IUPAC name: 2-methyl-2-propanethiol
9 - 39
In the IUPAC naming system, compounds with a –SH group have the ending “thiol”. After naming the parent chain of the corresponding alkane, the “e” is not dropped prior to adding thiol to the name. The longest chain is three carbons long making it a propane and the –SH is on carbon 2 making the parent chain 2propanethiol. There is also a –CH3, methyl, on carbon 2. Common name: t-butyl mercaptan In the common naming system, compounds with a –SH group are given the family name mercaptan. All common names with this group end in mercaptan and the organic group is named as an alkyl group. Since the carbon that is attached to the –SH has three other carbon groups attached it is a tertiary carbon and the group has four carbons, it is t-butyl.
9.26
a. Diallyl disulfide is the compound second highest in concentration in the breath immediately after eating garlic. Draw this compound (allyl = CH2CH=CH2). b. Allyl propyl disulfide is one of the compounds responsible for the odor of onions. Draw this compound.
Answer: a.
CH2 CH CH2 S S CH2 CH CH2
b.
CH2 CH CH2 S S CH2 CH2 CH3
9.28 Account for the fact that dipropyl ether has a higher boiling point than diethyl ether. Answer: Dipropyl ether has a longer hydrocarbon chain than diethyl ether and, therefore, stronger London force interactions.
9.30
Account for the fact that dipropyl ether has a lower solubility in water than diethyl ether.
Answer: The longer hydrocarbon chains in dipropyl ether make it more hydrophobic (nonpolar, water insoluble).
9 - 40
9.32
Ethanol (molecular weight = 46.0 amu) is miscible in water, while propane (molecular weight = 44.0 amu) is insoluble. Account for the difference in water solubility.
Answer: Ethanol is polar and has the ability to form hydrogen bonds with water whereas propane is polar cannot form hydrogen bonds with water.
9.34
1-Propanol is miscible in water and 1-butanol has a water solubility of 8.0g/100 mL. Account for this difference.
Answer: Both can form hydrogen bonds with water molecules, but 1-butanol contains more carbon atoms and is more nonpolar.
9.36
Methanethiol (molecular weight = 48.0 amu) has a water solubility of 2.3 g/100 mL and ethanol (molecular weight = 46.0 amu) is miscible in water. Account for the difference in water solubility.
Answer: Ethanol molecules can form hydrogen bonds with water molecules. Methanethiol cannot.
9.38
Thioctic acid is a growth factor for many bacteria. Which functional groups does this molecule contain? O
S S
CH2 CH2 CH2 CH2 C OH Thioctic acid
Answer: Disulfide and carboxylic acid.
SECTION 9.2
9.40
PREPARATION OF ESTERS
Draw the products of each nucleophilic substitution reaction. 9 - 41
Answer: A nucleophilic substitution replaces a leaving group such as chlorine, bromine, or iodine with an electron rich group.
a.
b.
c. d.
9.42
-
Name each of the organic products in Problem 9.40.
Answer: a. 2-propanol or isopropyl alcohol The parent chain has three carbons (propane), so drop the “e” and add “ol”. The – OH is on the carbon 2 so place 2- in front of propanol. b. methyl pentyl ether The O between two carbon groups indicates this compound is an ether. One group is a methyl and the other is a pentyl. c. methanethiol or methyl mercaptan The SH means this is a thiol. The only group attached is methyl. d. dipropyl sulfide The S has two organic groups attached to it making the molecule a sulfide. Both groups attached to S are propyl, so the term dipropyl is used with the family name sulfide.
9 - 42
9.44
Draw the missing alkyl bromide for each reaction. a.
+
O–
S–
b.
O
S
+ SH
c.
H S–
+
Answer: In these reactions a given nucleophile reacts with an alkyl bromide to make a product molecule. Examine the product molecule and identify the alkyl group which must be added to the nucleophile to produce it. That alkyl group is supplied by the alkyl bromide. Br
a. b.
CH3Br Br
c.
9.46
Write a chemical reaction equation that shows how each compound can be produced from an alkyl bromide. a. CH3CH2SCH2CH2CH2CH3 b. CH3CH(CH3)CH2CH2SH
Answer: a. CH3CH2Br + -SCH2CH2CH2CH3 → CH3CH2SCH2CH2CH2CH3 + Br b. CH3CH(CH3)CH2CH2Br + -SH → CH3CH(CH3)CH2CH2SH + Br -
9.48
Methyl chloride has the formula CH3Cl. Draw each of the following molecules: a. propyl chloride b. s-butyl bromide c. isopropyl chloride d. t-butyl bromide
9 - 43
Answer: a. CH3CH2CH2Cl
or
CH3CHBrCH2CH3
CH3 CH CH2 CH3 Br
b.
or c.
or d.
9.50
Complete the series of reactions by adding the necessary reactants or conditions to the reaction arrows. a.
b.
9 - 44
Answer: a.
light
b.
light
SECTION 9.3
9.52
REACTIONS
Draw the organic product (if any) expected from each reaction. K2Cr2O7
a. K2Cr2O7
b.
K2Cr2O7
c.
d.
I2
2
SH e.
2
+
I2
9 - 45
Answer: When an alcohol is oxidized by reaction with K2Cr2O7, the CH-OH is converted to C=O. Primary alcohols become carboxylic acids, secondary alcohols become ketones and tertiary alcohols do not react.
O a.
CH3 CH2 CH2 C OH O
b.
CH3 CH2 C CH3 O
c. H3C
When a thiol is reacted with I2, the hydrogen will be removed from the –SH on two molecules and the molecules bond together to make a disulfide. d. CH3CH2CH2SSCH2CH2CH3
e.
9.54
S S
Describe the difference in the products obtained when 1-butanol and 2-butanol are oxidized.
Answer: 1-butanol is a primary alcohol; its oxidation first produces an aldehyde then a carboxylic acid: oxidation
oxidation
9 - 46
2-butanol is a secondary alcohol; its oxidation produces a ketone only:
oxidation
9.56
When oxidized, which thiol produces the disulfide shown below?
Answer:
9.58
a. One step in the breakdown of fatty acids by the body involves the enzyme catalyzed oxidation of a 2° alcohol group present in a carboxylic acid derivative. In this compound, “CoA” is coenzyme A, whose structure will be introduced in Chapter 14. Draw the oxidation product.
OH
O
CH3 CH2 CH2 CH CH2 C CoA
+
enzyme
NAD+
? + NADH + H+ b. Name the carboxylic acid that provides the carboxylic acid residue present in the reactant. c. To which carbon atom in the reactant is the –OH group attached, or ? Answer: O
a.
O
CH3 CH2 CH2 C CH2 C CoA
b. hexanoic acid c. carbon atom
9 - 47
9.60
Draw the major product of each reaction.
H+
a.
heat
b.
H+ heat
Answer: These are dehydration reactions that produce alkenes from the alcohols after removal of water.
a.
b.
9.62
Draw each molecule named below and draw the major organic product expected when each is reacted with H+ and heat. a. 2,3-dimethyl-3-pentanol b. 2-hexanol c. 1-methylcyclopentanol
Answer: When an alcohol reacts with H+ and heat, a dehydration reaction occurs in which a water molecule is removed, leading to the formation of a C=C. The major product is the one formed when an H atom is removed from the neighboring carbon atom that carries the fewest H atoms. Water is also a product but is not shown in the following equations. 9 - 48
CH3 CH3 a.
H+, heat
CH3CH3 CH3 C C CH2 CH3
CH3 CH C CH2 CH3 OH
b.
H+, heat CH3 CH2 CH2 CH2 CH CH3
CH3 CH2 CH2 CH CH CH3
OH H3C
CH3
OH H+, heat
c.
9.64
a. When a particular alcohol is heated in the presence of H+, 1methylcyclopentene is the major product. Draw and name this alcohol. b. When a different alcohol is heated in the presence of H+, 1methylcyclopentene is the major product. Draw and name this alcohol.
Answer:
H3C
OH 1-methylcyclopentanol
a. CH3 OH b.
9.66
2-methylcyclopentanol
Complete the series of reactions by adding the necessary reactants or conditions to the reaction arrows.
a.
9 - 49
b.
Answer: a.
b.
heat
9.68
Citronellol is a component of rose oil. Give the expected product of each reaction of citronellol.
a.
heat
b.
9 - 50
c. H2
d. Answer:
a.
b.
major product
c.
d. SECTION 9.4
9.70
ALDEHYDES AND KETONES
Name each of the following molecules.
O a.
CH3C(CH3)2CH2 C H O
b.
CH3C CH2CH(CH3)2
O c. CH3CH(CH3) C CH(CH3)2
9 - 51
O d. CH3CH2CH(CH2CH2CH3) C H
Answer: a. 3,3-dimethylbutanal Since the C=O is a terminal C atom, this is an aldehyde. In the IUPAC system, the parent chain for the corresponding alkane is named and the “e” is dropped and replaced by “al”. Because the C=O is always carbon 1 for aldehydes, a number is not placed in front of the name of the parent chain. The parent chain contains 4 C atoms (butane) and there are two methyl groups located on carbon 3. b. 4-methyl-2-pentanone Since the C=O is not a terminal C atom, this is a ketone. In the IUPAC system, the parent chain for the corresponding alkane is named and the “e” is dropped and replaced by “one”. A number is placed in front of the name of the parent chain to indicate the location of the keto group (Number carbon atoms from the end that gives the C=O the smallest possible number.). The parent chain has 5 carbon atoms (pentane) and there is one methyl group located on carbon 4. c. 2,4-dimethyl-3-pentanone This is a ketone. The parent chain has 5 C atoms (pentane) and there are two methyl groups located on carbon 2 and carbon 4. d. 2-ethylpentanal This is an aldehyde. The parent chain has 5 C atoms (pentane) and there is one ethyl group located on carbon 2.
9.72
Draw each molecule. a. octanal b. 2,5-dimethylcyclopentanone c. 2-octanone d. 2-isopropylpentanal
Answer: a. There must be eight carbon atoms (octane). Draw a CH3 followed by (CH2)6 and then end the molecule with the aldehyde group (indicated by the “al” ending on the parent chain name).
9 - 52
O CH3(CH2)6
C H
b. Draw cyclopentane. Add = O to one carbon atom (this becomes position 1). Connect CH3 groups to carbon atoms 2 and 5. O H3C
CH3
c. There must be eight carbon atoms (octane). Draw a CH3 followed by (CH2)5 and then add the keto group (indicated by the “one” ending) followed by a CH3. This puts the C=O on the second carbon, accounting for the 2- in the name. O
CH3(CH2)5 C CH3 d. Draw the parent chain with five carbons (pentane) with C=O on the end carbon (carbon 1). Add an isopropyl group to carbon 2. O CH3CH2CH2 CH C H CH3 CH CH3
9.74
Acetone has a boiling point of 56○C and isopropyl alcohol has a boiling point of 82oC. a. Draw each molecule b. Account for the difference in boiling point.
Answer: a. The structural formulas for acetone and isopropyl alcohol are:. O
OH
CH3 C CH3
CH3 CH CH3
acetone
isopropyl alcohol
b. Alcohol molecules can interact through relatively strong hydrogen bonds, while ketone molecules interact through relatively weak dipole-dipole forces. The stronger the noncovalent interactions are, the higher the boiling point is for the substance. 9 - 53
9.76
Propanone (molecular weight = 58.0 amu) has a boiling point of 56˚C and butane (molecular weight = 58.0 amu) has a boiling point of 0˚C. Account for the difference in boiling point.
Answer: Propanone molecules are primarily attracted to one another by dipole-dipole interactions. Butane molecules are attracted to one another by relatively weaker London forces.
SECTION 9.5
9.78
OXIDATION OF ALDEHYDES
Draw the product of each reaction.
O a.
CH3 CH2 CH CH2 C H
Cu2+
CH3 CH3
Cu2+
b. HO CH2 CH2 CH2 C CH3 CH3
OH c.
O
CH3 CH CH C H
Cu2+
CH3
Answer: a. This is an oxidation reaction. The oxidizing reagent Cu2+ (Benedict’s reagent) converts the aldehyde to a carboxylic acid. O
CH3 CH2 CH CH2 C OH CH3
b. The oxidizing reagent Cu2+ will not oxidize an alcohol. No Reaction Product 9 - 54
c. This is an oxidation reaction. The oxidizing reagent Cu2+ converts the aldehyde to a carboxylic acid and the alcohol remains unchanged. OH
O
CH3 CH CH C OH CH3
9.80
Draw each molecule. Draw the product (if any) obtained when each is reacted with Benedict’s reagent. a. propanal b. butanone c. 1-butanol d. 3-chloro-3-methylpentanal e. 3-hydroxybutanal
Answer: Benedict’s reagent is an oxidizing agent that will oxidize aldehydes to carboxylic acids, but will not oxidize alcohols or ketones. O a.
CH3 CH2 C OH
CH3 CH2 C H
O b.
Benedict's reagent
no reaction
CH3 CH2 C CH3
OH c.
Benedict's reagent
no reaction
CH3 CH2 CH2 CH2 Cl
d.
O
Benedict's reagent
O
Cl
Benedict's reagent
CH3 CH2 C CH2 C
CH3 CH2 C CH2 C H
CH3
CH3
OH e.
O
O
Benedict's reagent
OH
O
CH3 CH CH2 C OH
CH3 CH CH2 C H
9 - 55
OH
SECTION 9.6
REDUCTION OF ALDEHYDES AND KETONES
9.82 Draw the product of each reaction. O a.
CH3 C CH2 CH CH3
H2
+
Pt
CH3
O +
b. CH3 CH2 CH2 CH2 C H
H2
Pt
O c.
H3C
CH3
+
H2
Pt
Answer: When reacted with H2 under a Pt catalyst, aldehydes and ketones are reduced to alcohols. OH a.
CH3 CH CH2 CH CH3 CH3
OH b. CH3 CH2 CH2 CH2 CH2 OH H3C
CH3
c.
9.84
The first step in the biochemical breakdown of progesterone is a reaction involving 20α- hydroxysteroid dehydrogenase (20 α-HSDH), an enzyme which catalyzes the reduction of a ketone group at position 20 in steroids. a. Draw the product that forms when progesterone is reduced by the action of 20α-HSDH and NADH.
9 - 56
O 21
C CH3
20 18 12
CH3
17
11 19 1
13
CH3
16
9
2
14
+
H+
15
8
10
+ NADH
3 5
O
7
4
OH
6
20-HSDH
Progesterone
?
+
NAD+
b. Draw the product that forms when 1 mol progesterone is reacted with 3 mol of H2 in the presence of Pt. Answer: a. When a ketone group is reduced, an alcohol is produced. The enzyme 20α-HSDH only reduces the C=O at carbon 20. OH 21
CH CH3
20 18 12
CH3
17
11 19 1
13
CH3
16
9
2
14
15
8
10
3 5
O
7
4
OH
6
b. In this reaction, a sufficient quantity of H2 was provided to reduce both C=O groups and to add hydrogen in place of the double bond between carbons 4 and 5. OH 21 20 18 12
CH3
11 19 1
17
13
CH3
16
9
2
CH CH3
14 8
10
15
3
HO
9.86
5 4
7 6
OH
The flavor of cinnamon is largely due to the molecule named cinnamaldehyde.
9 - 57
cinnamaldehyde a. Which organic families are present in cinnamaldehyde? b. Draw a product of each reaction.
cinnamaldehyde
cinnamaldehyde 2 H2 cinnamaldehyde
cinnamaldehyde Answer: aromatic
aldehyde
alkene
a.
cinnamaldehyde
cinnamaldehyde b.
This is an oxidation reaction. Potassium dichromate is an oxidizing agent that oxidizes the aldehyde group to a carboxylic acid.
9 - 58
cinnamaldehyde This is an oxidation reaction. Cu2+ is found in Benedicts' reagent which is basic. The aldehyde functional group is oxidized to a carboxylate ion.
2 H2
cinnamaldehyde This is reduction reaction. 2 mol of H2, in the presence of the Pt catalyst, are enough to both reduce the aldehyde functional group to an alcohol and add H atoms to the double bond.
cinnamaldehyde This is an aromatic substitution reaction. A hydrogen atom on the ring is replaced by a chlorine atom using an Fe catalyst. Only one possible product is given above.
SECTION 9.7 KETONES 9.88
REACTIONS OF ALCOHOLS WITH ALDEHDYES AND
Draw the product of each reaction.
O a.
H+
CH3 CH2 C CH2 CH3 + CH3 OH
9 - 59
H+ b.
2
H+
c.
Answer: a. When a ketone reacts with an alcohol in the presence of an acid catalyst, the alcohol adds across the C=O bond. The H of CH3OH adds to the carbonyl O and CH3O adds to the carbonyl C. OH CH3 CH2 C CH2 CH3 O CH3
b. When an aldehyde is reacted with two molecules of alcohol in the presence of an acid catalyst, an acetal is formed. In this case the double bonded O is replaced by two CH3CH2O- from the alcohol molecules.
c. Refer to the comments from part a.
9.90
Draw the missing reactant for each reaction.
9 - 60
OH CH3 C CH3
H+ a.
+
O b. CH3 CH2 C H
CH3 CH2 CH2 O
CH3CH2CH2OH
OH
H+
+
CH3CH2
CH OCH2CH2CH3
CH3O
H+ +
c.
O CH3
2 CH3OH
Answer:
O a.
CH3 C CH3
b.
CH3CH2CH2OH
O c.
9.92
Draw the missing reactant for each reaction.
CH3 a.
O
CH3 CH CH CH2 CH2 C H
H+ +
CH3 CH3
OH
CH3 CH CH CH2 CH2 C H CH3
9 - 61
O CH3
O b.
H+
CH3 C CH2 CH CH3
+
CH3
CH3 O CH3 C CH2 CH CH3 CH3 O
O c.
CH3 CH2 CH CH2 C H
CH3
H+ +
CH3 CH3 CH2 O CH3 CH2 CH CH2 C H CH3
OH
Answer: a. CH3OH b. 2 CH3OH c. CH3CH2OH
9.94
What product is formed in the presence of H+ when an alcohol molecule is reacted with each of the following? a. an aldehyde b. a carboxylic acid
Answer: a. A hemiacetal is formed when an aldehyde is reacted with an alcohol molecule in the presence of H+. b. An ester is formed when a carboxylic acid is reacted with an alcohol molecule in the presence of H+.
9.96
Draw the product of each reaction. The structure of cinnamaldehyde is given in Problem 9.86.
9 - 62
Cinnamaldehyde a.
Cinnamaldehyde b. Answer:
a. cinnamaldehyde
2
b. cinnamaldehyde
9.98
Draw the product of each reaction. The structure of citronellol is given in Problem 9.68.
citronellol
a.
2 citronellol
b.
9 - 63
Answer:
a.
b. HEALTH LINK
ALDEHYDE DEHYDROGENASE
9.100 Describe the physiological action of disulfiram. Answer: When ethanol is first oxidized in the body, acetaldehyde is formed. Disulfiram slows down the further oxidation of acetaldehyde to acetic acid. The acetaldehyde concentration increases in the blood producing disagreeable side effects of nausea and vomiting.
HEALTH LINK
PROTECTIVE ENZYMES
O2 is reduced when it is converted into O2-. Explain.
9.102 Answer:
O2 gains an electron when it is converted to O2-.
9.104 In the breakdown of hydrogen peroxide, which product results from the oxidation of H2O2 and which product results from the reduction of H2O2? Explain. 2 H2O2 → 2 H2O + O2 9 - 64
Answer: H2O is the product of the reduction of H2O2 because an oxygen atom is lost. O2 is formed as the oxidation production because H atoms are lost.
HEALTH LINK
DRUGS IN THE ENVIRONMENT
9.106 The hormone estrone has been detected in river, stream, and well water. Draw the product formed when the ketone group in estrone is reduced by H2 and Pt.
Estrone Answer:
H2 Pt Estrone
9.108 Paraxanthine, a product of caffeine metabolism in humans, has been detected in ground water at concentrations of 150 µg/L. a. At this concentration, how many grams of paraxanthine are present in 25.0 mL of water? b. Convert 150 µg/L into parts per million. c. Convert 150 µg/L into parts per billion.
9 - 65
Answer: a. Convert 25.0 mL to L, use the concentration to calculate the µg of paraxanthine, and then, convert µg to g of paraxanthine.
25.0 mL x
1 x 10-3 L 150 g 1 x 10-6 g x x = 3.8 x 10-6 g 1 mL 1L 1 g
b.
Parts per million =
g of solute x 106 mL of solution
150 g 10-6 g 10-3L x x x 106 = 0.15 ppm 1L 1 g 1 mL c.
Parts per billion =
g of solute x 109 mL of solution
150 g 1x 10-6 g 10-3 L x x x 109 = 150 ppb 1L 1 g 1 mL
LEARNING GROUP PROBLEMS
9.110 a. Draw butyl bromide. b. Draw the reaction product, Alcohol E. Butyl bromide + OH- → Alcohol E c. Give the IUPAC and common name of Alcohol E. d. Draw the reaction product, Alkene F. H+ Alkene F Alcohol E heat e. Name alkene F. f. Draw the reaction product, Alcohol G. Alkene F
+
H2O
H+ Alcohol G
g. Draw the reaction products, carboxylic acid H and ketone I.
9 - 66
K2Cr2O7
Alcohol E
Carboxylic acid H
K2Cr2O7
Ketone I Alcohol G h. Draw the product formed when Ketone I is reacted with H2/Pt. i. Draw the reaction product, Thiol J.
Thiol J
Butyl bromide + SHj. Draw the reaction product. 2 Thiol J
+
I2
Answer: a.
CH3 CH2 CH2 CH2 Br
CH3 CH2 CH2 CH2 OH b.
Alcohol E
c. 1- butanol is the IUPAC name. Butyl alcohol is the common name.
CH3 CH2 CH CH2 Alkene F
d. e. 1-butene
OH CH3 CH2 CH CH3 f. Alcohol G, the major product
O
O
g.
CH3 CH2 CH2 C OH
CH3 CH2 C CH3
Carboxylic acid H
Ketone I
OH h.
CH3 CH2 CH CH3
9 - 67
CH3 CH2 CH2 CH2 SH
i. j.
Thiol J
CH3 CH2 CH2 CH2 S S CH2 CH2 CH2 CH3
9 - 68
Chapter 10 Carbohydrates SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
10.1
a. Which pairs of glasses are mirror images?
A and D are mirror images. B and C are also mirror images:
b. Which, if any, pairs of glasses can be considered to be enantiomers? B and C can be considered enantiomers: they are mirror images of each other (see above) but they are not superimposable (see below).
c. Which, if any, pairs of glasses can be considered to be diastereomers? Diastereomers are stereoisomers that are not mirror images of each other. The following pairs can be considered diastereomers: A and B, A and C, B and D, and C and D.
10.3
How do oligosaccharides differ from polysaccharides?
10 - 1
Oligosaccharides contain from 2 to10 monosaccharide residues while polysaccharides contain more than 10.
10.5
Classify each monosaccharide in terms of functional group and number of carbon atoms.
O HO CH2 CH C H OH
a.
Aldotriose. There is an aldehyde functional group (“aldo”) and there are 3 carbon atoms (“triose”). O HO CH2 CH C CH2 OH OH
b.
Ketotetrose. There is a ketone function group (“keto”) and 4 carbon atoms (“tetrose”).
O
OH
HO CH2 C CH CH CH CH2 OH OH
c.
OH
Ketohexose. There is a ketone functional group and 6 carbon atoms (“hexose”).
O H d.
OH
C CH CH CH2 OH OH
Aldotetrose. There is an aldehyde functional group and 4 carbon atoms.
10.7
Draw an example of each type of monosaccharide a. an aldoheptose The name heptose indicates a seven carbon sugar. The “aldo” means it is an aldehyde. Therefore, draw a seven carbon chain with a –OH on every carbon except carbon 1 as this is where the C=O goes. 10 - 2
OH OH OH OH OH OH O H
C
C
C
C
C
C
H
H
H
H
H
H
C
H
an aldoheptose b. a ketononose The name nonose indicates a nine carbon sugar. The “keto” means it is a ketone. Therefore, draw a nine carbon chain with a –OH on every carbon except for that carbon where the C=O goes.
H
OH OH OH OH OH OH OH O
OH
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
C
H
H
a ketononose
10.9
Using wedge and dashed line notation, draw the enantiomer of each molecule.
a.
b.
10 - 3
An enantiomer is a non-superimposable mirror-image of the molecule. a.
b.
10.11 Label the chiral carbon atom in each molecule. A chiral carbon atom must have four different atoms or groups of atoms attached to it.
*
a.
b.
CH3CH2CH CH2CH2CH3 Br
*
O
CH2CH C OH NH2
10.13 Draw both enantiomers of each molecule in Problem 10.11, using wedge and dashed line notation to show the three-dimensional shape about each chiral carbon atom.
10 - 4
Enantiomers are molecules that are mirror images of each other and are not superimposable. a. Note that in each structure below the propyl is out of the plane of the paper and the hydrogen points into the plane. The structures are mirror images and they are not superimposable.
b. Note that in each structure the amino group is out of the plane of the paper and the hydrogen points into the plane. The structures are mirror images and they are not superimposable.
10.15 How many chiral carbon atoms does each molecule have and how many total stereoisomers are possible? A chiral carbon atom has four different atoms or groups of atoms bonded to it. The chiral carbon atoms are marked by an asterisk (*) in each of the structures. The number of possible stereoisomers can be calculated using the formula 2n, where n is the number of chiral carbon atoms.
10 - 5
Br
a.
* CH3 CH CH2 CH CH3 * CH2CH3
2n = 22 = 4 possible stereoisomers
OH
b.
CH3 CH CH2 CH CH3 * CH3
OH HO c.
2n = 21 = 2 possible stereoisomers
O
* CH2 CH CH C * OH
CH2 OH 2n = 22 = 4 possible stereoisomers
10.17 (+)-Propoxyphene is an analgesic (a painkiller).
(+)-Propoxyphene
10 - 6
a. Label the chiral carbon atom(s) in this molecule, using an asterisk.
*
*
b. How many other stereoisomers of propoxyphene exist? Explain. Four stereoisomers exist for propoxyphene. The number of stereoisomers is equal to 2n where n is the number of chiral carbons in the molecule. For propoxyphene with 2 chiral carbons, the total number of stereoisomers that exist is equal to 22 or 4. c. The enantiomer of (+)-propoxyphene is an antitussive (a cough suppressant). Draw this enantiomer.
(+)-Propoxyphene
Enantiomer
10.19 Pick the statement that best describes the relationship between each pair of molecules: constitutional isomers, stereoisomers, different conformations of same molecule, or identical molecules.
10 - 7
a. Stereoisomers. These two molecules have the same molecular formula and the same atomic connections but different three-dimensional orientation of the groups that can be interchanged only by breaking bonds.
b. Different conformations of the same molecules/ identical molecules. Flipping one of them horizontally by a 180° (or rotating the molecule) will produce a molecule identical to the other.
CH3
CH3 C c.
H
C
C H
H
CH3 C
H
CH3
Stereoisomers. These two molecules have the same molecular formula and the same atomic connections but different three-dimensional orientation of the groups that can be interchanged only by breaking bonds.
d. Constitutional isomers. These 2 molecules have the same molecular formulas but the atomic connections are different.
10 - 8
10.21 a. Using Fischer projections, draw the two enantiomers of the monosaccharide b. Label the molecules in your answer to part a as being D or L.
a. and b. In drawing a Fischer projection for a monosaccharide, the carbon atoms run vertically with the aldehyde or ketone group at or near the top. In D sugars, the –OH bonded to the chiral carbon atom farthest from the C=O is pointed to the right. In L sugars, this same –OH group is pointed to the left.
O
O C
C
H
H
OH
H
HO
H
CH2OH
CH2OH
D
L
c. Classify the monosaccharide in part a in terms of functional group and number of carbon atoms. Aldotriose. There is an aldehyde group and 3 carbon atoms.
10.23 a. How many chiral carbon atoms does D-lyxose contain?
O C H HO
H
HO
H
H
OH CH2OH
D-Lyxose
10 - 9
3 chiral carbon atoms, as marked by asterisks below.
O C H HO * H HO * H * OH
H
CH2OH D-Lyxose b. How many total stereoisomers are possible for this aldopentose? 8 possible stereoisomers: 23 = 8. c. Draw and name the enantiomer of D-lyxose. The enantiomer of D-lyxose is obtained by drawing the mirror image of D-lyxose.
O C H H
OH
H
OH
HO
H CH2OH
L-Lyxose d. D-Ribose is a diastereomer of D-lyxose. Draw two additional D-diastereomers of this monosaccharide. Diastereomers are stereoisomers that are not mirror images. Move one or more of the –OH groups in directions unlike those found in D-lyxose. Be sure that you haven’t drawn L-lyxose (part c) and that you have drawn D sugars.
10 - 10
O
O
C H
C H
H
OH
HO
H
HO
H
H
OH
H
OH
H
OH CH2OH
CH2OH D-diastereomers D-diastereomers
10.25 Which statement(s) best describe the pair of molecules.
a. b. c. d. e.
They are stereoisomers. They are enantiomers. They are diastereomers. They are epimers. They are anomers.
The answers are a, c, and d. They are diastereomers, stereoisomers that are not mirror images. They are also epimers, diastereomers that differ at only one chiral carbon atom (3rd carbon from the top).
10 - 11
10.27 a. Draw all of the D-2-ketohexose stereoisomers.
b. In the answer to part a, identify two pairs of epimers. Epimers are diastereomers that differ only at one chiral carbon atom. For each of these pairs, the two molecules differ only at the first chiral carbon closest to the C=O.
a pair of epimers
a pair of epimers
c. In the answer to part a, identify two pairs of diastereomers that are not epimers. For each of these pairs of diastereomers, the two molecules differ at two chiral carbon atoms. For each pair, these two chiral carbon atoms are the two closest to the C=O.
10 - 12
a pair of diastereomers that are not epimers
a pair of diastereomers that are not epimers
10.29 Pick the statement that describes the relationship between each pair of molecules: enantiomers, diastereomers, constitutional isomers, molecules that are not stereoisomers or constitutional isomers. a. D-ribose and L-ribose Enantiomers. These two molecules are mirror-mages of each other as indicated by the D- and L- difference in orientation. b. D-glucose and D-galactose Diastereomers. These two are stereoisomers that are not mirror-images of each other. c. L-glucose and D-galactose Diastereomers. These two are stereoisomers that are not mirror-images of each other. d. D-glucose and D-2-deoxyribose Molecules that are not stereoisomers or constitutional isomers. These two molecules do not have the same molecular formula.
10.31 Draw each of the following glucose derivatives. a. D-2-deoxyglucose The –OH group at carbon number 2 is replaced (“2-deoxy”) by a H atom.
10 - 13
O C H H
H
HO
H
H
OH
H
OH CH2 OH
b. D-glucosamine An –OH group is replaced by an amine group. O
C H H
NH2
HO
H
H
OH
H
OH CH2 OH
c. D-glucuronic acid The terminal –CH2OH group is replaced by a –COOH group. O
C H H
OH
HO
H
H
OH
H
OH C OH O
d. sorbitol The aldehyde group (-CHO) at carbon 1 is replaced by a –CH2OH group.
10 - 14
CH2 OH H
OH
HO
H
H
OH
H
OH CH2 OH
10.33 For the molecules in Problem 10.31, a. To which class of monosaccharide derivative does each belong? b. How many chiral carbon atoms does each have? c. How many stereoisomers are possible for each? 10.31 a D-2-deoxyglucose
10.31 b D-glucosamine O
10.31 c D-glucuronic acid O
10.31 d sorbitol
C H H * NH2 HO * H
C H H * OH HO * H
CH2OH H * OH HO * H
* *
H
* OH
H
* OH
H
* OH
*
H
* OH
H
* OH
H
* OH
CH2OH
C OH
CH2OH
alcohol sugar
24 = 16
a.
deoxy sugar
amino sugar
b.
3
4
O carboxylic acid sugar 4
c.
23 = 8
24 = 16
24 = 16
4
10.35 When D-gluconic acid is produced from D-glucose, which functional group is oxidized? The aldehyde on carbon 1.
10.37 a. Draw all possible aldotetroses. b. Which are D and which are L sugars? The name tetrose indicates a four carbon sugar. The “aldo” means it is an aldehyde. Therefore, draw a four carbon chain with a –OH on every carbon
10 - 15
except carbon 1, as this is where the C=O goes. To make all possible aldotetroses, arrange the –OH groups of carbons 2 and 3 in as many different ways as possible. (Note: Two chiral carbons tell you there are 22 = 4 different structures possible). D structures have the –OH group on the right at carbon 3, while L structures have the –OH group on the left.
O
O
O
O
C H
C H
C H
C H
HO
H
H
OH
H
OH
HO
H
H
OH
H
OH
HO
H
HO
H
CH2OH
CH2OH
CH2OH
CH2OH
D
D
L
L
10.39 Draw each galactose derivative (see Figure 10.9). a. D-2-deoxygalactose Draw the galactose as shown in Figure 10.9 and then remove the –OH group from carbon 2 and replace it with hydrogen.
O C H H
H
HO
H
HO
H
H
OH CH2OH
b. galactitol Draw galactose as shown in Figure 10.9. Convert the aldehyde group into an alcohol.
10 - 16
CH2OH H
OH
HO
H
HO
H
H
OH CH2OH
c. D-galacturonic acid Draw galactose as shown in Figure 10.9. Convert the alcohol group on carbon 6 to a carboxylic acid.
10.41 Draw the product obtained when D-talose.
O C H HO
H
HO
H
HO
H
H
OH CH2OH
D-Talose
10 - 17
a. reacts with H2 and Pt. When a monosaccharide is reduced, the carbon-oxygen double bond of the aldehyde or ketone group is reduced to an alcohol:
CH2OH HO
H
HO
H
HO
H
H
OH CH2OH
reduced form of D-talose b. reacts with Benedict’s reagent Oxidizing carbon 1 of D-talose produces a carboxylic acid group at carbon 1:
O C OH HO
H
HO
H
HO
H
H
OH CH2OH
oxidized form of D-talose at carbon 1
10.43 Define the term reducing sugar. A reducing sugar is a carbohydrate that gives a positive Benedict’s test (in the process of being oxidized, the sugar reduces Cu2+ ion present in the reagent).
10.45 a. Draw the D-aldohexose that gives the following alcohol sugar, when treated with H2 and Pt.
10 - 18
CH2OH H
OH
HO
H
HO
H
H
OH CH2OH
H2 and Pt will reduce an aldehyde to an alcohol. Start with the structure shown in the problem. Convert the alcohol into an aldehyde.
O C H H
OH
HO
H
HO
H
H
OH CH2 OH
b. Is the aldohexose a reducing sugar? Yes. The aldehyde group can be oxidized by Cu2+. c. Is the aldohexose a deoxy sugar? No. Deoxy means that an –OH group has been replaced by hydrogen. This is not the case for the D-aldohexose. d. Is the aldohexose an amino sugar? No. Amino sugar means that an –OH has been replaced by –NH2. This is not the case for the D-aldohexose. e. Draw the L- aldohexose that also gives the alcohol sugar above, when treated with H2 and Pt.
10 - 19
O C H HO
H
H
OH
H
OH
HO
H CH2 OH
10.47
Draw the alcohol sugar, aldonic acid, and uronic acid that can be formed from each monosaccharide. a. D-2-deoxyribose In an aldonic acid, the aldehyde group has been oxidized. In uronic acids, the primary alcohol group at the end of the molecule opposite the aldehyde group has been oxidized.
D-2-deoxyribose
the alcohol sugar of D-2-deoxyribose
the aldonic acid of D-2-deoxyribose
the uronic acid of D-2-deoxyribose
the aldonic acid of L-2-deoxyribose
the uronic acid of L-2-deoxyribose
b. L-2-deoxyribose
L-2-deoxyribose
the alcohol sugar of L-2-deoxyribose
10 - 20
c. The C3 epimer of D-2-deoxyribose
C3 epimer of D-2-deoxyribose
10.49
the alcohol sugar of C3 epimer of D-2-deoxyribose
the aldonic acid of the C3 epimer of D-2-deoxyribose
the uronic acid of the C3 epimer of D-2-deoxyribose
How does an anomer differ from a β anomer? As pyranoses and furanoses are typically drawn , in an anomer, the hemiacetal -OH points down. In a anomer, the hemiacetal -OH points down.
10.51 Draw each molecule. a. -D-glucopyranose
b. -D-ribofuranose
10 - 21
c. -D-galactopyranose
CH2OH HO H
O
OH
H OH
H
H
OH
H
d. - D-arabinopyranose (see Practice Problem 10.6)
H H HO
O H H
HO
OH
H
H OH
10.53 Each molecule in Problem 10.51 has how many chiral carbon atoms and is one of how many total stereoisomers?
Molecule
Number of Chiral Carbons
Total Number of Stereoisomers
β-D-glucopyranose
5
32
-D-ribofuranose
4
16
β-D-galactopyranose
5
32
-D-arabinopyranose
4
16
10.55 D-lyxose (see Problem 10.23) is a diastereomer of D-ribose. Draw α-Dlyxofuranose. A lyxofuranose molecule is a lyxose molecule that has formed a five-membered cyclic hemiacetal. The five-membered ring is drawn with the oxygen at the back
10 - 22
and the hemiacetal carbon on the right side. The α-notation indicates that the –OH on the hemiacetal carbon is pointing down.
HOCH2
H
O OH
OH OH
H H
H
10.57 Draw β-D-mannopyranose (see Figure 10.11).
HOCH2 H
O H OH
OH
OH H
HO H
H
10.59 a. Draw the pyranose anomer of D-2-deoxyglucose.
O C H H
H
HO
H
H
OH
H
OH
H
CH2 OH O H OH
H
H
H
OH
H OH
CH2 OH D-2-deoxyglucose
pyranose anomer
b. Draw the β pyranose anomer of D-glucuronic acid.
10 - 23
O
O
C H H
OH
HO
H
H
OH
H
OH
H
C OH O H OH
H
H
OH
OH
OH H
C OH O D-glucuronic
pyranose anomer
10.61 Define the term mutarotation. Mutarotation is the process of converting back and forth from an α anomer to the open form to the β anomer.
10.63 Cellobiose is a reducing sugar. Write a reaction equation that shows why. When cellobiose ring opens, it contains an aldehyde that can be oxidized by Cu2+.
10 - 24
CH2OH O H OH H
H
H HO
CH2OH O H OH H H
H
HO
H
10.65
OH
H Benedict’s Solution
CH2OH OH H OH H
H
H
H
O
OH
CH2OH O H OH H
OH
O C H
O H
OH
H
OH
a. Gentiobiose consists of two D-glucopyranose residues joined by a β-(1→6) glycosidic bond.Draw this disaccharide. Draw two -D-glucopyranose molecules. From one of them, remove the hydrogen from the –OH on the hemiacetal carbon (carbon 1). From the other remove the –OH from carbon 6. Draw a bond connecting the two molecules.
H HO
CH2OH O H OH H
O CH2
H H
H
OH HO
O H OH
H
H
OH
OH H
b. Is gentiobiose a reducing sugar? Explain. Yes. When the hemiacetal (at the right side of the molecule) undergoes mutarotation, the resulting aldehyde group can be oxidized by Cu2+. 10 - 25
10.67 Vanillin β-D-glucoside gives vanilla extract its flavor. Draw the new glycoside that would form if the monosaccharide residue in this glycoside was changed to the pyranose form of the C-3 epimer of glucose.
Vanillin -D-Glucoside In the open form structure of the C3 epimer of glucose, the –OH on the third carbon is pointing to the right. In the cyclic form, that same –OH group would be pointing down. The pyranose form has the O of the hemiacetal –OH group pointing down.
10.69 A glycoside formed by combining -D-glucopyranose and 1-octanol can be used as a detergent. Draw this glycoside.
10 - 26
10.71 A patent application has been made for the use of the glycoside below to prevent or treat hair loss. Draw the five molecules formed when this glycoside is hydrolyzed. The three product monosaccharides should be drawn in their open form.
Under hydrolysis, glycosidic bonds an ester bonds are broken. The five molecules produced are given below. For the monosaccharides, both the cyclic and open forms are given for reference.
10 - 27
or
or
or
10.73 Draw a disaccharide consisting of two D-glucose residues that is not a reducing sugar. The disaccharide can be made by linking two D-glucose molecules with an α, β(1 1) glycosidic bond. The resulting structure contains no hemiacetal group. Therefore, it is unable to mutarotate, and it is not a reducing sugar.
10 - 28
H HO
H HO
CH2OH O H OH H
H
H OH CH2OH O H OH H H
O
H
OH
10.75 a. Is trehalose, a disaccharide found in a wide range of living things, a reducing sugar? CH2OH O H H H
HO
OH
H
H
OH O
CH2OH H HO
O H OH
H
H
OH
H
Trehalose No. The molecule contains no hemiacetal groups. b. Which describes the glycosidic bond in this disaccharide: -(1↔ 1), -(1→ 2), -(1→ 3), -(1→ 4), -(1→ 5), or -(1→ 6)? α,α-(1↔ 1) The glycosidic bond involves the hemiacetal carbon (carbon 1) of each ring, each of which is the anomer.
10 - 29
10.77
Draw the product obtained when lactose is reacted with Benedict’s reagent. CH2OH O OH H H CH2OH OH H O OH H O H H OH OH H H H
H
OH lactose ( anomer)
The carbonyl group of the open form of the monosaccharide residue has been oxidized into a carboxyl group.
Benedict's Reagent
CH2OH H CH2OH OH H
O H OH
H
H
OH
O
OH H OH
H
H
OH
O C OH
H
lactose (open-ring)
10.79 True or False? a. Lactose and cellobiose are stereoisomers. True. Lactose and cellobiose have the same molecular formula and atomic connections but have different three-dimensional shapes that are interchanged only by breaking bonds. b. Lactose and cellobiose are enantiomers. False. They are not enantiomers because they are not mirror images of each other.
10 - 30
c. Lactose and cellobiose are diastereomers True. They are diastereomers because they are stereoisomers that are not enantiomers.
10.81 What is lactose intolerance and what are some of the options for dealing with it? Lactose intolerance is a deficiency in β-galactosidase, the enzyme that catalyzes the hydrolysis of the β-(1→4) glycosidic bond in lactose. Persons with this disorder can deal with it by avoiding dairy products, by using dairy products from which lactose has been removed, and by taking tablets containing β-galactosidase.
10.83 Human milk contains lactose at a concentration of 7% (w/v). a. How many grams of lactose are present in 15 mL of human milk? Use the given concentration above to calculate the grams of lactose remembering that
% w/v =
g solute x 100% mL solution
In 15 mL of human milk,
7% w / v =
7 g solute x 100% 100 mL solution
Therefore, a 7% (w/v) solution contains 7 g lactose/100 mL milk.
15 mL milk x
7 g lactose = 1 g lactose 100 mL milk
b. What is the molar concentration of lactose in human milk? First, convert 100 mL to 0.100 L then use the molar mass of lactose (342 g/mol) to calculate its molar concentration in milk.
1 mol lactose 7 g lactose x 0.100 L = 0.2 M 342 g lactose c. What is the parts per million concentration of lactose in human milk? Recall the formula in for calculating parts per million:
10 - 31
Parts per million =
g of solute x 106 mL of solution
This formula gives
7 g lactose x 106 = 7 x 104 ppm 100 mL milk 10.85 a. How is the structure of amylose similar to that of amylopectin? Amylose and amylopectin each contain glucose residues connected by −(1→4) glycosidic bonds. b. How is the structure of amylase different from that of amylopectin? Amylose contains only −(1→4) glycosidic bonds while amylopectin contains −(1→4) glycosidic bonds and − (1→6) glycosidic bonds at the branching points.
10.87 How is the structure of glycogen similar to that of amylopectin? Glycogen and amylopectin each consist of glucose residues joined by α-(1→4) and α-(1→6) glycosidic bonds.
10.89 Why can humans use starch as a food source, but not cellulose? Humans have the enzymes necessary for hydrolyzing and digesting starch (amylase, maltase, and debranching enzymes) but not the enzymes required for hydrolyzing the type of glycosidic bonds found in cellulose.
10.91 Oral bacteria have an enzyme that converts sucrose into the polysaccharide dextran. Dextran is one component of dental plaque, which holds bacteria that attack tooth enamel. Dextran is a polysaccharide built from -D-glucopyranoses joined by alternating -(1→3) and -(1→6) glycosidic bonds. Draw a portion of dextran.
10 - 32
-(1
3)
-(1
6)
-(1
3)
-(1
10 - 33
6)
10.93 A particular lichen produces the polysaccharide below.
1
CH2 H HO
O
O
H OH
H
CH2 H
O H
H
OH HO A
2 O
H OH
H
H
OH
H H HO
CH2 O H OH
H
H
OH
O H
B
C a. Name the monosaccharide residues A-C and specify whether each is a furanose or a pyranose. A, B, and C are D-glucose residues (pyranose form). b. Label [β(1 → 4), etc.] the glycosidic bonds 1 and 2. Both glycosidic bonds 1 and 2 are -(1→6) c. Is the molecule a homopolysaccharide or is it a heteropolysaccharide? Homopolysaccharide. The monosaccharide residues are the same.
10.95 What is invert sugar and why is it used in foods? Invert sugar is the sweetening mixture produced when sucrose is hydrolyzed into glucose and fructose. It is used in foods because it is sweeter than sucrose and it is cost-effective.
10 - 34
10.97 The structure of sucralose is closest to which of the four disaccharides discussed in Section 10.6? Sucralose is a derivative of sucrose, a naturally occurring disaccharide.
10.99 a. Name each monosaccharide residue in rebaudioside A (Figure 10.22) and specify whether each is a furanose or a pyranose. All monosaccharide residues are D-glucose residues (pyranose). See figure below for the structure of rebaudioside A. b. Identify [β-(1→4), etc.] each glycosidic bond in rebaudioside A. -(1→3) and -(1→2) c. Is this compound a reducing sugar? No, it is not a reducing sugar. The molecule contains no hemiacetal groups. -(1→3)
-(1→2)
10 - 35
10.101 a. Draw a Fischer projection of D-2-deoxyribose.
b. Draw a diastereomer of D-2-deoxyribose that is a D-sugar.
c. Draw L-2 deoxyribose.
d. Draw a diastereomer of D-2-deoxyribose that is an L-sugar, but is not L-2-deoxyribose.
10 - 36
e. Which, if any of the molecules in parts a-d are pairs of enantiomers? a. and c., b. and d. f. Is D-2-deoxyribose a reducing sugar? Why or why not? Yes. D-2-Deoxyribose has an aldehyde group that can be oxidized by Cu2+. g. Draw -D-2-deoxyribofuranose.
h. Draw β-D-2- deoxyribofuranose.
i. Are the molecules in parts g and h enantiomers, diastereomers, or neither? diastereomers j. Is -D-2-deoxyribofuranose a reducing sugar? Why or Why not? Yes. When the hemiacetal group undergoes mutarotation, the resulting aldehyde group can be oxidized by Cu2+.
10 - 37
k. Draw a disaccharide that consists of two β-D-2- deoxyribofuranose residues joined by a β(1→5) glycosidic bond.
l. Is the disaccharide a reducing sugar? Why or why not? Yes. When the hemiacetal group (at the right side of the disaccharide molecule) undergoes mutarotation, the resulting aldehyde group can be oxidized to Cu2+.
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 10.2
The aldehyde O atom and the –OH group on carbon 4 of the Fischer projection of D- glucose (A) are shown in color. Which atoms in the wedge and dotted line (B) and side view (C) of D-glucose and which atoms in α-D-glucopyranose (D) should appear in these same colors?
Answer:
10 - 38
SECTION 10.1
10.4
A
B
C
D
MONOSACCHARIDES
How are oligosaccharides similar to polysaccharides?
Answer: Both oligosaccharides and polysaccharides are synthesized from monosaccharides, the building blocks of carbohydrates.
10.6
Classify each monosaccharide in terms of functional group and number of carbon atoms. O
a.
HO CH2 C CH2 OH OH
OH
O
b. HO CH2 CH CH CH CH C H OH
OH
10 - 39
O
OH
HO CH2 C CH CH CH2 OH
c.
OH
O HO CH2 C CH CH2 OH
d.
OH Answer: a. Ketotriose. There is a ketone functional group and 3 C atoms. b. Aldohexose. There is an aldehyde functional group and 6 C atoms. c. Ketopentose. There is a ketone functional group and 5 C atoms. d. Ketotetrose. There is a ketone functional group and 4 C atoms.
10.8
Draw an example of each type of monosaccharide. a. an aldononose b. a ketooctose
Answer: a. An aldononose is a monosaccharide that contains an aldehyde group and a total of 9 carbon atoms in the chain. b.A ketooctose is a monosaccharide that contains a ketone group and a total of 8 carbon atoms in the chain. OH OH OH OH OH OH OH OH O H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
C
H
H
OH OH OH OH OH OH O
OH
C
C
C
C
C
C
C
H
H
H
H
H
H
a ketooctose
an aldononose
Other examples may include variations of the structures above with other orientations of the hydroxyl groups.
SECTION 10.2
STEREOISOMERS
10 - 40
C
H
H
10.10
Using wedge and dashed line notation, draw the enantiomer of each molecule.
Answer: The given molecules are on the left side and the enantiomers are on the right side.
a.
b.
10.12 Label the chiral carbon atom in each molecule.
CH3 a.
CH3 CH CH CH2 Br CH3
b.
Answer: Chiral carbons are bonded to four different groups. Each chiral carbon is labeled by an asterisk (*) in the following molecules.
10 - 41
CH3 CH3 CH *CH CH2 Br
a.
CH3
*
b.
10.14 Draw both enantiomers of each molecule in Problem 10.12, using wedge and dashed line notation to show the three-dimensional shape about each chiral carbon. Answer:
a.
b.
10.16
How many chiral carbon atoms does each molecule have and how many total stereoisomers are possible? O
a.
HO
C CH CH3 NH2
10 - 42
Cl CH3 CH2 CH CH2 CH3
b.
CH3 c.
CH3 CH CH CH2 CH2 NH2 OH
Answer: a. 1 chiral carbon atom and 2 (21) possible stereoisomers.
O C *CH CH3
HO
NH2 b. No chiral carbon atoms present. c. 2 chiral carbon atoms and 4 (22) possible stereoisomers.
CH3 CH3 *CH *CH CH2 CH2 NH2 OH
10.18 Draw two diastereomers of the molecule shown in the previous problem. Answer: Refer to the drawing in Problem 10.17. A diastereomer is a stereoisomer that is not an enantiomer. Draw one molecule that switches the position of two groups on one chiral carbon atom (here the –CH3 and the –CH2N(CH3)2 on the right side). Then draw its enantiomer.
O CH3CH2C O
CH3 C
CH2
C
CH2
H
CH2N(CH3)2 C
CH2N(CH3)2
O C and
CH3CH2
10 - 43
O
C
H CH3
10.20 Pick the statement that best describes the relationship between each pair of molecules: constitutional isomers, stereoisomers, different conformations of the same molecule, identical molecules.
CH3
CH3 a.
C
C
C
H
H
H
CH3 CH3
C H
b.
c.
d.
Answer: a. Constitutional isomers. b. Identical molecules. c. Stereoisomers. d. Different conformations of the same molecule.
10.22 .
a. Using Fisher projections, draw the two enantiomers of the monosaccharide
10 - 44
O HO CH2 CH C CH2 OH OH b. Label the molecules in your answer to part a as being D or L. c. Classify the monosaccharide in part a in terms of functional group and number of carbon atoms. Answer:
CH2OH
CH2OH
C O
C O
H
a. and b.
OH
HO
H
CH2OH
CH2OH
D
L
c. ketotetrose
10.24 a. How many chiral carbon atoms does D-gulose contain? O
C H H
OH
H
OH
HO
H
H
OH CH2OH
D-gulose b. How many total stereoisomers are possible for this aldohexose? c. Draw and name the enantiomer of D-gulose. d. Draw two D-diastereomers of this monosaccharide. Answer: a. D-gulose contains 4 chiral carbons. b. 24 = 16 possible stereoisomers
10 - 45
O
O C H
c.
H
OH
HO
H
H
OH
HO
H
HO
H
H
OH
H
OH
HO
H
CH2OH
d.
H C
HOCH2
D-gulose
L-gulose
O
O
C H
C H
HO
H
HO
H
H
OH
H
OH
HO
H
H
OH
H
OH
H
OH
CH2OH
CH2OH
2 examples of D-diastereomers of D-gulose. 10.26 Which statement(s) best describe the pair of molecules.
a. b. c. d. e.
They are stereoisomers. They are enantiomers. They are diastereomers. They are epimers. They are anomers.
10 - 46
Answer: The answers are a and c. They are diastereomers, stereoisomers that are not mirror images.
10.28 a. Draw the eight D-aldohexose stereoisomers. b. In the answer to part a, identify four pairs of epimers. c. In the answer to part a, identify four pairs of diastereomers that are not epimers. Answer:
a. b. Epimers are diastereomers that differ only at one chiral carbon atom. For each of these pairs, the two molecules differ only at the first chiral carbon closest to the C=O.
10 - 47
c. Diastereomers are stereoisomers that are not mirror images of each other. If not epimers, each pair must differ at more than 1 carbon atom.
10 - 48
SECTION 10.3 IMPORTANT MONOSACCHARIDES AND MONOSACCHARIDE DERIVATIVES
10.30 Pick the statement that best describes the relationship between each pair of molecules: enantiomers, diastereomers, constitutional isomers, molecules that are not stereoisomers or constitutional isomers. a. D-ribose and D-2-deoxyribose b. D-glucose and D-fructose c. D-fructose and D-ribose d. D-fructose and L-fructose Answer: a. molecules that are neither stereoisomer nor constitutional isomers b. constitutional isomers c. molecules that are neither stereoisomer nor constitutional isomers d. enantiomers
10 - 49
10.32 Draw each of the following. a. ribitol b. N-acetyl-D-glucosamine c. D-gluconic acid d. L-rhamnose Answer:
O C H CH2OH
H
H
OH
HO
H
H
OH
H
OH
H
OH
H
OH
CH2OH a.
O
NH C CH3
CH2OH b. N-acetyl-D-glucosamine
ribitol O
O
C OH
C H
H
OH
H
OH
HO
H
H
OH
H
OH
HO
H
H
OH
HO
H
CH2OH c. D-gluconic acid
CH3 d. L-rhamnose
10.34 For the molecules in Problem 10.32, a. to which class of monosaccharide does each belong? b. how many chiral carbon atoms does each have? c. how many stereoisomers are possible for each? Answer:
10 - 50
O C H
OH
H
HO
H
H
OH
H
OH
H
OH
HO
H
H
OH
H
OH
HO
H
OH
HO
H
OH
H
OH CH2OH
NH C CH3
CH2OH
CH2OH
D-gluconic acid carboxylic acid sugar 4
N-acetyl-D-glucosamine amino sugar
3
c.
C H H
H
b.
C OH OH
H
a.
O
H
CH2OH
ribitol alcohol sugar 3
O
O
2 =8
4 4
4
2 = 16
2 = 16
CH3 L-rhamnose deoxy sugar 4 4
2 = 16
10.36 When D-glucuronic acid is produced from D-glucose, which functional group is oxidized? Answer: The alcohol on carbon 6:
O
O
C H
C H H
H
OH
HO
H
H
OH
H
OH
OH
HO
H
H
OH
H
OH C OH
CH2OH
O
D-glucose
D-glucuronic acid
10.38 a. Draw all possible 2-ketopentoses. b. Which are D and which are L sugars? Answer: a. Ketopentoses contain 5 carbons in the chain and a ketone group at carbon 2. A ketopentose has 2 chiral carbons and 4 possible stereoisomers:
10 - 51
CH2OH O C
CH2OH O C
CH2OH O C
CH2OH O C
HO
H
H
OH
H
OH
HO
H
HO
H
H
OH
HO
H
H
OH
CH2OH L-sugar
CH2OH
CH2OH
CH2OH
D-sugar
L-sugar
D-sugar
b. In D sugars, the –OH attached to the chiral carbon farthest from the C=O group points to the right. In L sugars, it points to the left.
10.40 Draw each monosaccharide derivative of D-lyxose (see Problem 10.23). a. D-2-deoxylyxose b. lyxitol c. D-lyxonic acid Answer: a. The prefix “deoxy” refers to the replacement of an –OH group by an H. The replacement occurs at carbon 2 in the case of D-2-deoxylyxose:
O C H H
H
HO
H
H
OH CH2OH
D-2-Deoxylyxose b. Lyxitol is the alcohol sugar derived from lyxose by reducing the aldehyde group to an alcohol group:
10 - 52
CH2OH HO
H
HO
H
H
OH CH2OH
Lyxitol c. D-Lyxonic acid is derived from lyxose by oxidizing the aldehyde group of lyxose to a carboxylic acid group:
O C OH HO
H
HO
H
H
OH CH2OH
D-Lyxonic acid
SECTION 10.4
REACTIONS OF MONOSACCHARIDES
10.42 Draw the product obtained when D-lyxose (Problem 10.23) a. reacts with H2 and Pt b. reacts with Benedict’s reagent. Answer: a. The aldehyde group is reduced to an alcohol group.
CH2OH HO
H
HO
H
H
OH CH2OH
b. The aldehyde functional group is oxidized to a carboxylic acid.
10 - 53
O C OH HO
H
HO
H
H
OH CH2OH
10.40 D-Fructose does not contain an aldehyde group, yet it is a reducing sugar. Explain why. Answer: Fructose is a ketohexose. Ketone groups in ketoses can rearrange to form aldehyde groups under the basic condition of oxidizing agents such as Benedict’s reagent. After being converted into an aldose, it is a reducing sugar.
10.46 a. Draw the D-2-ketopentose that gives the following alcohol sugar, when treated with H2 and Pt.
CH2OH H
OH
H
OH
H
OH CH2OH
b. Is the ketopentose a reducing sugar? c. Is the ketopentose a deoxy sugar? d. Is the keto pentose an amino sugar? e. Draw the L-2-ketopentose that also gives the alcohol sugar above, when treated with H2 and Pt. Answer: a. The alcohol sugar shown above is derived from the following D-2ketopentose:
10 - 54
CH2OH C O H
OH
H
OH CH2OH
b. Yes, the D-2-ketopentose shown above is a reducing sugar. c. No, the D-2-ketopentose shown above is not a deoxy sugar. d. No, the D-2-ketopentose shown above is not an amino sugar. e. The L-2-ketopentose sugar is the enantiomer of the D-2-ketopentose sugar shown above:
CH2OH O C HO
H
HO
H CH2OH
10.48 Draw the alcohol sugar, aldonic acid, and uronic acid that can be formed from each monosaccharide. a. D-glucosamine b. 4-deoxy-L-galactose c. 3-deoxy-L-mannose Answer: In an aldonic acid, the aldehyde group has been oxidized. In uronic acid, the primary alcohol group at the end of the molecule opposite the aldehyde group has been oxidized.
10 - 55
D-glucosamine a.
4-deoxy-L-galactose
b.
3-deoxy-L-mannose c.
the aldonic acid of D-glucosamine
the aldonic acid of 4-deoxy-L-galactose
the aldonic acid of 3-deoxy-L-mannose
10 - 56
the uronic acid of D-2-deoxyribose
the uronic acid of 4-deoxy-L-galactose
the uronic acid of 3-deoxy-L-mannose
SECTION 10.5
MONOSACCHARIDES IN THEIR CYCLIC FORM
10.50 How does a pyranose differ from a furanose? Answer: A pyranose contains a 6-membered cyclic hemiacetal ring and a furanose contains a 5-membered cyclic hemiacetal ring.
10.52 Draw each molecule. a. α-D-glucopyranose b. β-D-ribofuranose c. α-D-galactopyranose d. β-D-arabinopyranose (see Practice Problem 10.6) Answer: a. A glucopyranose molecule is a glucose molecule that has formed a sixmembered cyclic hemiacetal. The six-membered ring is drawn with the oxygen at the back and the hemiacetal carbon on the right side. The α-notation indicates that the –OH on the hemiacetal carbon is pointing down.
H
CH2OH O H OH H
H OH
HO H
OH
b. A ribofuranose molecule is a ribose molecule that has formed a five-membered cyclic hemiacetal. The five-membered ring is drawn with the oxygen at the back and the hemiacetal carbon on the right side. The β-notation indicates that the –OH on the hemiacetal carbon is pointing up.
HOCH2
OH
O H
H H
H OH
OH
c. A galactopyranose molecule is a galactose molecule that has formed a sixmembered cyclic hemiacetal. The six-membered ring is drawn with the oxygen at
10 - 57
the back and the hemiacetal carbon on the right side. The α-notation indicates that the –OH on the hemiacetal carbon is pointing down.
CH2OH OH O H OH H H H
H OH
OH
d. An arabinopyranose molecule is an arabinose molecule that has formed a sixmembered cyclic hemiacetal. The six-membered ring is drawn with the oxygen at the back and the hemiacetal carbon on the right side. The β-notation indicates that the –OH on the hemiacetal carbon is pointing up.
H
H O H H
OH
OH
H
OH
OH H
10.54 Each molecule in Problem 10.52 has how many chiral carbon atoms and is one of how many total stereoisomers? Answer: Molecule -D-glucopyranose β-D-ribofuranose -D-galactopyranose β-D-arabinopyranose
Number of Chiral Carbons 5 4 5 4
Total Number of Stereoisomers 32 16 32 16
10.56 D-Gulose (see Problem 10.24) is a diastereomer of D-glucose. Draw -Dgulopyranose. Answer:
10 - 58
CH2OH O
HO
H
H
H
H
OH
OH
OH H
-D-gulopyranose
10.58 Draw –D-allopyranose.
O C H H
OH
H
OH
H
OH
H
OH CH2OH
D-Allose Answer: CH2OH H HO
O H H
H
OH
OH
H OH
-D-Allopyranose
10.60 a. Draw the α pyranose anomer of N-Acetyl-D-glucosamine. b. Draw the β pyranose anomer of D-galactosamine (the C-4 epimer of Dglucosamine).
10 - 59
Answer: a.
CH2OH
O C H H
O
O
H
NHCCH3
HO
H
H
OH
H
OH
H
H OH
HO
H
OH O NH C CH3
H
CH2OH -pyranose anomer
N-Acetyl-D-glucosamine b.
O CH2OH
C H H
NH2
HO
H
HO
H
H
OH
HO H
O H OH
H
H
NH2
OH H
CH2OH -pyranose anomer
D-Galactosamine
10.62 Explain why –D-glucopyranose is a reducing sugar. Answer: –D-glucopyranose is a reducing sugar because it is capable of mutarotation (ring-opening) at carbon 1. The mutarotation exposes an aldehyde group that can undergo oxidation.
SECTION 10.6
OLIGOSACCHARIDES
10.64 Lactose is a reducing sugar. Write a reaction equation that shows why.
10 - 60
Answer: Lactose is a reducing sugar because during mutarotation an aldehyde group is exposed and can undergo oxidation:
CH2OH H CH2OH O
HO
H
H
OH
H
H
OH
O
O
OH
H OH
H
H
OH
H
H
Lactose ( anomer)
CH2OH H CH2OH O
HO
H
H
OH
H
H
OH
O
OH H OH
H
H
OH
O aldehyde group
C H
H
Lactose (open-ring) 10.66 a. N-Acetyllactosamine consists of a -D-galactopyranose residue joined to an N-acetyl-D-glucosamine residue by a −(1→4) glycosidic bond. Draw this disaccharide. b. Is N-acetyllactosamine a reducing sugar? Explain. Answer: a. N-acetyllactosamine is shown below:
10 - 61
CH2OH O
H
H
CH2OH O
HO
OH
O
H
H
H
OH
H
H
OH
OH
H
O H NH C CH3
H N-Acetyllactosamine
b. N-acetyllactosamine is a reducing sugar because it is capable of mutarotation (ring-opening) at carbon 1 of the N-acetyl-D-glucosamine residue. The mutarotation exposes an aldehyde group.
10.68 Daunorubicin is a glycoside prodrug that shows promise for the treatment of certain forms of cancer. A monosaccharide is released when this compound is hydrolyzed. Draw the monosaccharide in its -pyranose form and in its open form.
O
O
OH
C CH3 OH
CH3O
O
OH
H H HO
O O
CH3 H
H
NH2
H
H
Daunorubicin
10 - 62
Answer:
10.70 The color of some flowers comes from glycosides called anthocyanidins. The glycoside formed by combining -D-galactopyranose with the –OH group at carbon 3 of the anthocyanidin called petunidin has a purple color. Draw this glycoside.
8 7
2
6 5
3 4
Petunidin
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Answer:
8 7
2
6 3 5
4
10.72 The glycoside below, ginsenoside Rb 1, is present in ginseng. The biochemical effects of this and other compounds present in ginseng are being studied. Draw the molecules formed when this glycoside is hydrolyzed. Any monosaccharides formed should be drawn in their open form.
10 - 64
Answer:
3
or
3
or
10 - 65
10.74 Draw a disaccharide consisting of two D-galactose residues that is a reducing sugar. Answer:
CH2OH HO H
CH2OH O
H OH
H
H
OH
O H H
O H OH
H
H
OH
OH H
10.76 a. Is raffinose (Figure 10.19) a reducing sugar? b. Identify the type of glycosidic bonds [-(1→4), -(1→4), etc.] in this trisaccharide. Answer: a. Raffinose is not a reducing sugar because it is not capable of ring-opening or mutarotation. All three of the anomeric carbons of raffinose are locked in glycosidic bonds. b. The types of glycosidic bonds are shown in the following structure.
10 - 66
CH2OH HO H
O
H
H
O −(1
OH
H
H
OH
6)
CH2 O
H
H
H OH
H
H
OH
HO CH2
O
HO
H
H
HO
OH
H
10 - 67
O -(1
CH2OH
2)
10.78 Draw the product obtained when cellobiose is reacted with Benedict’s reagent. Answer:
CH2OH O
H CH2OH O
H H OH
OH
H
OH
H
H
OH
O
OH
H
H
OH
H
H
cellobiose ( anomer)
Benedict's Reagent The carbonyl group of the open form of the monosaccharide residue has been oxidized into a carboxyl group.
CH2OH H CH2OH H OH
O H OH
H
H
OH
O
OH H OH
H
H
OH
O C OH
H
cellobiose (open-ring)
10.80 True or false? a. Maltose and sucrose are enantiomers. b. Maltose and sucrose are diastereomers. c. Maltose and sucrose are constitutional isomers.
10 - 68
Answer: a. False. They are not enantiomers because they are not mirror images of each other. b. False. Maltose and sucrose are not diastereomers because they are not stereoisomers of each other. Their atoms are bonded differently: maltose consists of 2 pyranose rings while sucrose consists of a pyranose ring and a furanose ring. c. True. Maltose and sucrose have the same molecular formula (C12H22O11) but different atomic connections.
10.82 When the enzyme lactase is used to hydrolyze lactose present in milk, lactose-free milk is obtained. Explain why lactose-free milk is sweeter than regular milk (see Table 10.2). Answer: Lactose hydrolyzes into galactose and glucose which have higher relative sweetness than lactose according to Table 10.2. 10.84 Cow’s milk contains lactose at a concentration of 5% (w/v). a. How many grams of lactose are present in 150 µL of cow’s milk? b. What is the molar concentration of lactose in cow’s milk? c. What is the parts per million concentration of lactose in cow’s milk? Answer: a. Convert mL to mL and. To calculate the grams of lactose, use the following equation. weight / volume percent =
5% w / v =
g solute 100 mL solution
5 g solute 100 mL solution
In 150 µL of cow’s milk,
150 L
1 x 10-6 L 1 L
1 mL milk 5 g lactose = 8 x 10-3 g lactose -3 1 x 10 L 100 mL milk
10 - 69
b. Note that 100 mL of milk contains 5 g of lactose. Convert 100 mL to 0.100 L then use the molar mass of lactose (342.0 g/mol) to calculate its molar concentration in milk.
1 mol lactose 5 g lactose x 0.100 L = 0.1 M . 342.0 g lactose c. Parts per million =
SECTION 10.7
g of solute 5g 106 = 106 = 5 x 104 ppm mL of solution 100 mL
POLYSACCHARIDES
10.86 a. How is the structure of cellulose similar to that of amylose? b. How is the structure of cellulose different from that of amylose? Answer: a. Both cellulose and amylose are homopolysaccharides consisting of a chain of glucopyranose residues joined by 1→4 glycosidic linkages. b. Cellulose has β-(1→4) glycosidic bonds and amylose has α-(1→4) and α(1→6) glycosidic bonds.
10.88 How is the structure of glycogen different from that of amylopectin? Answer: The glycogen polysaccharide chains are more branched than those of amylopectin.
10.90 Why can horses use cellulose as a food source? Answer: Horses have the ability to use cellulose as a food source because they are able to digest cellulose. The digestive tract of horses contains symbiotic bacteria that can catalyze the breakdown of cellulose.
10.92 Inulins, polysaccharides that contain mostly fructose residues, are one component of dietary fiber. A typical inulin molecule is a chain of -(2→1)
10 - 70
linked -D-fructofuranoses attached to carbon 1 of the fructose residue in a sucrose molecule. Draw a portion of an inulin molecule. Answer:
-(2
1)
10.94 Arabinoxylan is a polysaccharide that is present in the bran of many grasses. A portion of its structure is shown. This polysaccharide contains the residues of two different monosaccharides, D-xylose (Sample Problem 10.7) and Larabinose (Practice Problem 10.6). In the following structure, which residues are D-xylose and which are L-arabinose?
10 - 71
Answer: The residues that are circled (pyranose) are derived from L-arabinose while the residues that are boxed (furanose) are derived from D-xylose.
HEALTH LINK
RELATIVE SWEETNESS
10.96 How is high fructose corn syrup made? Answer: Corn syrup (D-glucose) is treated with glucose isomerase. This produces a fructose-glucose mixture known as high fructose corn syrup. 10.98 What does the “K” in Acesulfame-K stand for? Why is Acesulfame-K sold in this form? Answer: The K stands for potassium. In this form, Acesulfame has a greater solubility in water.
10 - 72
HEALTH LINK
STEVIA
10.100 a. Name four different organic functional groups that are present in rebaudioside A (Figure 10.22). b. Draw the products obtained upon complete hydrolysis of the compound. c. Even though rebaudioside A contains glucose residues, it is a no-calorie sweetener. Why might this be the case? Answer: a. Alcohol, acetal, alkene, ester b. The products are 4 D-glucose rings and a multi-ring molecule.
OH CH2OH O
H
H
4 HO
OH
H
H
OH
CH3
OH H
+ O C HO
CH3
c. Rebaudioside A does not broke down into glucose molecules that can be further metabolized.
LEARNING GROUP PROBLEMS
10.102 a. Classify D-tagatose in terms of functional group and number of carbon atoms.
CH2OH C O HO
H
HO
H
H
OH CH2OH
D-tagatose
10 - 73
CH2
b. Draw the product obtained when D-tagatose reacts with H2 and Pt. c. D-tagatose is a reducing sugar. Explain how this can be the case. d. Draw the monosaccharide derivative(s) formed when D-tagatose reacts with Benedict’s reagent. e. Draw and name the enantiomer of D-tagatose. f. Draw a diastereomer of D-tagatose that is a D-sugar, but is not D-fructose. g. Draw -D-tagatofuranose. h. Draw -D-tagatofuranose. i. Is -D-tagatofuranose a reducing sugar? Why or why not? j. Draw a disaccharide that consists of two D-tagatofuranose residues joined by an (2↔2) glycosidic bond. k. Is the disaccharide a reducing sugar? Why or why not? Answer: a. Ketohexose b. The ketone group is reduced to an alcohol group.
CH2OH HC OH HO
H
HO
H
H
OH CH2OH
c. Ketone groups in ketoses can rearrange to form aldehyde groups under the basic condition of oxidizing agents such as Benedict’s reagent. After being converted into an aldose, it becomes a reducing sugar. d. The aldehyde group formed from the rearrangement of the ketone functional group is oxidized to a carboxylic acid group.
O C OH HC OH HO
H
HO
H
H
OH CH2OH
10 - 74
CH2OH
CH2OH
C O
O C
HO
H
H
OH
HO
H
H
OH
H
OH
HO
H
CH2OH
CH2OH
e. D-tagatose
L-tagatose
CH2OH C O
f.
H
OH
HO
H
H
OH CH2OH
HOCH2 H
g.
O OH
HO
H
H
OH
-D-tagatofuranose
HOCH2 H
h.
CH2OH
O
OH
OH
HO
H
H
CH2OH
-D-tagatofuranose
i. -D-tagatofuranose is a reducing sugar because it is capable of mutarotation (ring-opening) and transformation to an aldehyde group that can undergo oxidation.
10 - 75
HOCH2 H
O
CH2OH
OH
HO
H
H
O
CH2OH j.
CH2OH
O OH
HO
H
H
H
k. No. There is no hemiacetal group that can undergo mutarotation or ringopening.
10 - 76
Chapter 11 Lipids and Membranes SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
11.1
In Chapter 8 we saw that, in the presence of H+, a carboxylic acid and an alcohol will react to form an ester plus water. Triglycerides are formed in an enzymecatalyzed version of this same reaction. a. For the reaction below, circle the reactant atoms that combine to form the water molecules.
b. Draw the triglyceride product by combining the remaining (not circled) parts of the reactant molecules. O O CH2 O O CH O O CH2
11.3
Why is the melting point of lauric acid (Table 11.1) lower than that of myristic acid?
11 - 1
Lauric acid has a shorter hydrocarbon tail than myristic acid, so London force interactions between lauric acid molecules are weaker.
11.5
Draw skeletal structures of the following (see Table 11.1). a. lauric acid.
b. linolenic acid.
11.7
Sodium palmitate, CH3(CH2)14CO2Na, has a higher melting point than palmitic acid, CH3(CH2)14CO2H. Why? The ionic bonds that hold the ions in sodium palmitate to one another are stronger than the noncovalent interactions that hold one palmitic acid molecule to another.
11.9
In terms of structure, what distinguishes fatty acids from the carboxylic acids that were discussed in Chapter 8? Fatty acids are carboxylic acids that typically have between 12 and 20 carbon atoms.
11.11 Fatty acids are carboxylic acids that typically contain 12-20 carbon atoms. Some biological molecules contain carboxylic acid residues longer than 20 carbon atoms. Would these fatty acids with more than 20 carbon atoms in length be considered lipids? Explain. Lipids are defined as biochemical compounds that are water insoluble. Fatty acids longer than 20 carbons follow this definition and therefore are considered lipids.
11 - 2
11.13 Draw skeletal structures for the products formed when the beeswax ester (Table 11.2) is saponified (hydrolyzed). Saponification is the hydrolysis of the ester bonds found in waxes.
O OH
and
HO
11.15 List the biological functions of waxes. Some of the biological functions of waxes are to form a protective layer for retaining water or keeping out water of an organism and to act as a form of energy storage.
11.17 The fragrance of spermaceti, a wax produced by whales, once made it important to the perfume industry. One of the main constituents of spermaceti is cetyl palmitate, which is formed from palmitic acid and cetyl alcohol, CH3(CH2)14CH2OH. Draw the condensed structure of cetyl palmitate. First find the formula for palmitic acid, CH3(CH2)14CO2H in Table 11.1. Next remember that the reaction between a carboxylic acid and an alcohol forms an ester. Remove the –OH from the acid and the –H from the alcohol and join the two ends together.
O CH3(CH2)14C OCH2(CH2)14CH3
11.19 Some of the ester molecules present in beeswax have an –OH group in the ω-3 position of the alcohol residue. Draw a possible structure for one of these esters. Refer to Health Link: Omega-3 Fatty Acids.
11 - 3
O CH3(CH2)14
C O CH2(CH2)26 CH CH2CH3
or CH3(CH2)14CO2CH2(CH2)26CH(OH)CH2CH3
OH -3 carbon
-3 carbon
11.21 a. Some moths produce tetradecyl acetate and use it as a sex pheromone (attractant). Draw this ester. (Hint: The prefix “tetradec” means “14”.) The tetradecyl group comes from the alcohol component and the acetate group comes from the carboxylic acid component:
or
b. How does the structure of this ester differ from that of a typical wax ester? The fatty acid residue in a typical wax ester has 14 to 36 carbon atoms and the alcohol residue has 16 to 30 carbon atoms. In tetradecyl acetate, the carboxylic acid residue (the “acetate” component) only has 2 carbon atoms and the alcohol residue (the “tetradecyl” component) only has 14 carbon atoms. c. Which would you expect to have the higher melting point, tetradecyl acetate or beeswax ester (Table 11.2)? The beeswax ester is expected to have the higher melting point because its molecules are larger and attracted by stronger London forces.
11.23 List four biological functions of triglycerides. Energy source, thermal insulation, padding, and buoyancy.
11.25 Write a balanced reaction equation for the complete hydrogenation of palmitoleic acid (Table 11.1).
11 - 4
Hydrogenation is the process of adding H atoms to carbon atoms of a double bond. H2 → CH3(CH2)5CH2CH2(CH2)7CO2H Pt
CH3(CH2)5CH═CH(CH2)7CO2H
11.27 Draw a triglyceride made from glycerol, myristic acid, palmitic acid, and oleic acid. Would you expect this triglyceride to be a liquid or a solid at room temperature? Explain. Triglycerides are composed of three fatty acid residues and a glycerol residue. Look up the formulas for each of the fatty acids and attach them to the glycerol molecule in the same way that esters were drawn in the previous problems.
O O
CH3(CH2)12C
CH2
O O CH
CH3(CH2)14C O CH3(CH2)7CH
O
CH(CH2)7C
CH2
This triglyceride contains more saturated than unsaturated fatty acid residues, so it should be a solid at room temperature.
11.29 a. Draw the products formed when the triglyceride is saponified.
O O
CH3(CH2)14C
CH2
O CH3CH2(CH=CHCH2)3(CH2)6C O CH3(CH2)7CH=CH(CH2)7C
O CH
O
CH2
Saponification is another term for ester hydrolysis in the presence of OH-. To show the products, take the molecule apart the same as any other ester. The products will be the three fatty acid anions and glycerol.
11 - 5
O CH3(CH2)14C O
HO CH2 O
CH3CH2(CH=CHCH2)3(CH2)6C O
HO CH
O CH3(CH2)7CH=CH(CH2)7C O
HO CH2
O b. For each product from part a., state whether it is hydrophobic, hydrophilic, or CH (CH ) C O 3 2 14 amphipathic. O CH3(CH2)14C O amphipathic
HO CH2
O
CHHO (CH=CHCH2)3(CH2)6C O 3CH2CH 2 amphipathic O CHHO 3(CHCH 2)7CH=CH(CH2)7C O HO CH2HO CH2
O O O )6CO-O CH3CH2(CH=CHCH )3(CH 2C 2 CH (CH ) 3 O 2 14 CH3(CH2)14C O O - O O (CH ) C CH3(CH2)7CH=CH CH 2 7 (CH ) C C O CH3CH2(CH=CHCH O 2)6HO CH3CH2(CH=CHCH HO 2CH HO CH 2 62)3(CH 2)3
O O O 2)7C O HO CH2HO CH2 CH3(CH2CH )7CH=CH 2)7C (CH 3(CH2)(CH 7CH=CH amphipathic
hydrophilic
11.31 Vegetable oils tend to become rancid more quickly than animal fats. Why? Vegetable oils have more carbon-carbon double bonds than animal fats. It is the oxidation of these bonds that causes oils to become rancid.
11.33 What does it mean to partially hydrogenate vegetable oil? Partial hydrogenation of vegetable oil involves the addition of hydrogen atoms to convert some of the double bonds in unsaturated fatty acid residues into single bonds.
11.35 Which two phospholipids are most prevalent in plants and animals? Phosphatidylcholine (lecithin) and phosphatidylethanolamine (cephalin) are the two phospholipids most prevalent in plants and animals.
11 - 6
HO CH
HO CH2
11.37 a. To which class of glycerophospholipids does the compound belong?
O CH3(CH2)16 C O CH2 O CH3(CH2)5CH=CH(CH2)7 C O CH O
O CH2 O P
O-
OCH2 CH C O +
NH3
It belongs to the phosphatidylserine family of glycerophospholipids. b. How do other members of this class of phospholipids differ from one another? The other members of this class have different fatty acids bonded to the glycerol. c. Draw the products obtained when the compound is saponified into its component parts. See below. d. For each product from part c, state whether it is hydrophobic, hydrophilic, or amphipathic
O CH3(CH2)16
C
O- +
CH3(CH2)5CH=CH(CH2)7
O
CH2 OH
C
O - + CH OH CH2 OH
amphipathic
amphipathic
hydrophilic O
O +
O
-
P
O - + HO
O
CH2 CH C +
NH3
hydrophilic
11 - 7
hydrophilic
O-
11.39 How do phosphatidylethanolamines differ from phosphatidylcholines? The two families differ by the alcohol group attached to the phosphate group in a glycerophospholipid. In phosphatidylethanolamines, the alcohol is derived from ethanolamine: HOCH2CH2NH3+. In phosphatidylcholine, it is derived from choline:
CH3 HO
+
CH2 CH2 N
CH3
CH3
11.41 Draw a sphingomyelin that contains oleic acid.
CH3(CH2)11CH2CH=CH CH OH O CH3 (CH2)6 CH2 C H
CH2 (CH2)6 C NH CH C O
H CH2 O
P
+
O CH2 CH2 NH3
O 11.43 In certain membranes sphingomyelin serves as a replacement for phosphatidylcholine. What structural similarities between these classes of phospholipids make this a good substitution? Sphingomyelin and phosphatidylcholine have similar structures. They both contain a positively charged ammonium residue:
CH3 +
CH2 CH2 N
O CH3
CH3
P
CH3 O
O-
CH3 sphingomyelin
phosphatidylcholine
11 - 8
+
CH2 CH2 N
CH3
11.45 A particular cerebroside contains a glucopyransose residue attached to a sphingosine residue by a β glycosidic bond (see Section 10.6). Based on the sphingolipid shown in Figure 11.18, draw this cerebroside.
CH3(CH2)11CH2CH=CH CH OH O CH3(CH2)5
CH CH (CH2)7 C NH CH CH2OH
H HO
O H OH
H
H
OH
O
CH2
H
11.47 The sphingomyelins present in cow brain include residues of saturated fatty acids having between 16 and 32 carbon atoms and/or residues of unsaturated fatty acids with 18 or 22 carbon atoms. Draw one of these sphingomyelins.
11 - 9
CH3(CH2)11CH2CH=CH CH OH O CH3(CH2)5CH=CH(CH2)11
C NH
CH CH3
O CH2 O
P
OCH2
+
CH2 N
O-
CH3
CH3
or
fatty acid residue
CH3(CH2)11CH2CH=CH CH OH O CH3(CH2)14
C NH CH CH3
O CH2 O
P O -
OCH2
+
CH2 N
CH3
CH3
11.49 a. Phospholipase A2, an enzyme present in snake venom, catalyzes the hydrolysis (removal of) the middle fatty acid residue in triglycerides. Draw a diglyceride that might be the product of such a hydrolysis reaction. Given the following triglyceride, remove the middle fatty acid residue by hydrolysis, leaving behind an –OH group attached to the middle carbon atom of the glycerol residue.
11 - 10
hydrolysis of middle fatty acid residue
a triglyceride
middle fatty acid residue removed product diglyceride
11.51 How is cholesterol used in the human body? The body uses cholesterol primarily as a precursor for the synthesis of other steroids.
11.53 Label the hydrophobic and hydrophilic parts of taurocholate (Figure 11.21). Hydrophilic parts will have many –OH, –NH, or –O bonds while hydrophobic will be mostly long chains of hydrocarbons.
11 - 11
Hydrophobic Hydrophilic HO CH 3
Hydrophobic
CH3
CHCH2CH2CNH
H3C
CH2 CH2 SO-
OH
HO
O
Hydrophilic
3
Hydrophilic
11.55 Which molecule is the starting point for the synthesis of sex hormones and bile salts? Cholesterol.
11.57 a. What are the biological functions of HDL and LDL? HDL transports cholesterol and phospholipids from cells back to the liver. LDL transports cholesterol and phospholipids from liver to cells. b. In the news, you often hear HDL referred to as “good cholesterol” and LDL as “bad cholesterol”. In terms of the structure HDL and LDL, is use of the term “cholesterol” totally correct? Explain. Partially. Cholesterol is the major component of LDH. HDL contains more protein and phospholipids than cholesterol. c. What makes HDL “good” and LDL “bad”? Having high LDH and low HDL levels is a warning sign for atherosclerosis and an increased risk of stroke and heart disease.
11.59 Cortisol, cortisone, and other anti-inflammatory steroids block the action of an enzyme that catalyzes the hydrolysis of unsaturated fatty acids, including arachidonic acid, from membrane lipids. How does this result in reduced inflammation? Hydrolysis of a particular phospholipid to release arachidonic acid is the first step in the production of prostaglandins. When cortisol, cortisone, and other antiinflammatory steroids inhibit the enzyme that catalyzes this hydrolysis, the formation of prostaglandins is blocked.
11 - 12
11.61 Dexamethasone, an alternative to COX-2 inhibitors, blocks the production of COX-2. What are the benefits of taking a drug that prevents formation of COX-2, but has no effect on COX-1? Drugs that prevent formation of COX-2 but have no effect on COX-1 avoid the side effects associated with the inhibition of COX-1 enzymes namely ulcers and kidney damage.
11.63 How are facilitated diffusion and active transport different? Facilitated diffusion moves substances from areas of higher concentration to areas of lower concentration, and does not require the input of energy. Active transport moves substances in the opposite direction and requires the input of energy.
11.65 How are facilitated diffusion and diffusion different? In facilitated diffusion, diffusion across a membrane takes place with the assistance of proteins.
11.67 To function properly, membranes must be flexible or fluid. In light of this fact, propose an explanation of why the cell membranes in the feet and legs of a reindeer contain a higher percentage of unsaturated fatty acids than do the cell membranes in the interior of its body. Reindeer are typically found in cold, snow-covered regions. Since the melting point of fatty acids goes down as the degree of unsaturation goes up, having more unsaturated fatty acids in the feet and legs reduces the likelihood of the fats solidifying. This helps keep the membranes more flexible.
11.69 Oleic acid (Figure 11.1) is not an omega-3 fatty acid. Which type of omega fatty acid is it? Omega-9 fatty acid. The first C=C bond from the free end of the hydrocarbon chain occurs at carbon-9.
11.71 What are the health benefits of consuming omega-3 fatty acids? The known health benefits of consuming omega-3 fatty acids include: increased HDL; lowered LDL and triglycerides; decreased blood pressure; lowered risk for heart disease; and less rheumatoid arthritis inflammation.
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11.73 When unsaturated fatty acids were discussed in Section 11.1, it was said that their carbon-carbon double bonds are usually cis. Name one food source that is a natural source of unsaturated fatty acids with trans double bonds. Beef and dairy products contain natural trans fatty acids.
11.75 Draw a trans fat that might form if the following triglyceride is subjected to partial hydrogenation O
CH3(CH2)12C O CH2 O CH3(CH2)4CH2 H CH3(CH2)4CH2
CH2(CH2)6
C=C
H
H
O
CH2(CH2)6
C=C
C O CH
C O CH2
H
Hydrogenation is the chemical reaction that involves breaking the double bond between two carbon atoms and adding a hydrogen atom to each carbon. Since the question specifies partial hydrogenation, only one of the double bonds is hydrogenated. The other bond undergoes cis to trans conversion. Two options are possible for the triglyceride in the problem. One of them is shown below. O CH3(CH2)12C O CH2 O H CH3(CH2)4CH2
C=C
CH2(CH2)6
C O CH
H H H
O
CH3(CH2)4CH2C CCH2(CH2)6 C O CH2 H H
11 - 14
And the other is shown here.
O CH3(CH2)12C O CH2 O
H H
CH3(CH2)4CH2C CCH2(CH2)6 C O CH H H O H CH3(CH2)4CH2
C=C
CH2(CH2)6
C O CH2
H
11.77 Draw the products formed if olestra (Figure 11.15) is saponified. During saponification, all ester bonds are hydrolyzed into carboxylate ions and alcohols.
8
-
11.79 a. Draw nandrolone and lauric acid, the hydrolysis products obtained when nandrolone laurate (Figure 11.22) is saponified. The structures of nandrolone and the anion of lauric acid are shown below. In saponification, the fatty acid is produced in a basic solution which results in the H+ being removed from the fatty acid leaving behind the carboxylate anion. 11 - 15
and
-
anion of lauric acid
nandrolone
b. How different is the structure of nandrolone from that of testosterone? The structures of nandrolone and testosterone are shown side-by-side below. The only difference between the two molecules is that the testosterone molecule has an extra methyl group (-CH3) indicated by the arrow below.
testosterone
nandrolone
11.81 a. To which class of lipids does each molecule belong?
fatty acid
steroid
b. The molecule on the left in part a is estradiol. What is the function of this lipid?
11 - 16
Estradiol is a female sex hormone. Female sex hormones regulate menstruation, breast development, and other female traits. c. Draw the product formed when the molecule on the right in part a is completely hydrogenated (reacted with 3H2 and Pt). O HO
d. Which has a higher melting point, the molecule on the right in part a or the molecule in your answer to part C? The molecule in part c is expected to have a higher melting point because it is a saturated or a completely hydrogenated fatty acid. e. In the body, some estradiol is combined with fatty acids to produce esters. It is believed that these esters serve as a way to store estradiol for later use. Using the two molecules in part a, draw one of these esters. Estradiol’s alcohol group is involved in ester formation, not its phenol group. O CH3 O
HO
f. Which would you expect is more hydrophilic, estradiol or the estradiolcontaining ester in your answer to part e? Estradiol. Estradiol has a smaller nonpolar or hydrophobic component. The long hydrocarbon chain in the estradiol-containing ester in part e. makes it more hydrophobic. g. Estradiol fatty acid esters are carried through the blood in lipoproteins and are stored in body fat. Why are these esters soluble in fat? Estradiol fatty acid esters have a significant hydrophobic (nonpolar) component.
11 - 17
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 11.2
In Chapter 8 we saw that in the presence of H+, an ester can be hydrolyzed to give a carboxylic acid plus an alcohol. a. Draw the fatty acid and alcohol obtained when the wax ester below is hydrolyzed. b. Circle the atoms in the product molecules that were supplied by water. O CH3(CH2)14
H+
C OCH2(CH2)28CH3
+
H2O
Answer:
O a. and b.
SECTION 11.1
11.4
CH3(CH2)14
C OH
+
HO CH2(CH2)28CH3
FATTY ACIDS
Why is the melting point of palmitoleic acid (Table 11.1) lower than that of oleic acid?
Answer: Stronger London force interactions exist between molecules that have longer hydrocarbon chains. Oleic acid has a longer hydrocarbon chain (18 C) than palmitoleic acid (16 C). The London force interactions are stronger between molecules of oleic acid and thus, its melting point is higher. Both molecules have one carbon-carbon double bond.
11.6
Draw skeletal structures of the following (See Table 11.1). a. palmitic acid b. palmitoleic acid
11 - 18
Answer:
a. palmitic acid
b. palmitoleic acid
11.8
a. Which term best describes palmitic acid (see Problem 11.7): hydrophobic, hydrophilic, or amphipathic b. Which term best describes sodium palmitate (see Problem 11.7): hydrophobic, hydrophilic, or amphipathic?
Answer: a. Hydrophobic. Palmitic acid is hydrophobic because it is not soluble in water. The molecules of palmitic acid are largely nonpolar. b. Amphipathic. Sodium palmitate is amphipathic because it has both a hydrophobic component (large hydrocarbon chain) and a hydrophilic component (ionic end).
11.10 In terms of structure, what distinguishes unsaturated fatty acids from the alkenes discussed in Chapter 8? Answer: Unsaturated fatty acids are carboxylic acids that typically have 12 to 20 carbon atoms. At least one of these C-C bonds is a double bond, typically in the cis configuration. Alkenes are a general group of molecules with as few as two carbon atoms that contain at least one double bond, which may be in the trans or cis configuration.
11.12 Fatty acids are carboxylic acids that typically contain 12-20 carbon atoms.
11 - 19
Some biological molecules contain carboxylic acid residues shorter than 12 carbon atoms. What would determine whether or not carboxylic acids are lipids? Answer: Fatty acids that have fewer than 12 carbon atoms can still be considered lipids as long as they are insoluble in water.
SECTION 11.2
WAXES
11.14 Draw skeletal structures for the products formed when the carnauba wax ester (Table 11.2) is saponified (hydrolyzed). Answer: The structural formula for carnauba wax is given in Table 11.2. Draw the molecule. Note that the CO2 group is in the middle of the formula, which tells you this is an ester. Hydrolysis breaks the ester bond. ↓ O O
Saponification, the hydrolysis of an ester in the presence of OH – forms a carboxylate ion.
O O
and an alcohol:
HO
11.16 What property of waxes allows them to fulfill their primary biological function? Answer: Waxes are hydrophobic which allows them to carry out their primary protective function of keeping water in or out of an organism.
11 - 20
-
11.18 a. Draw skeletal structures for the fatty acid and alcohol from which the ester in insect wax (Table 11.2) is made. b. Draw the aldehyde that could be reduced to form the alcohol in your answer to part a. c. Draw the primary alcohol that could be oxidized to form the carboxylic acid in your answer to part a. Answer: a. O OH
and HO 48
O
b.
CH3 (CH2)48 C H
OH
c.
CH3 (CH2)24 CH2
11.20 The virulence (disease causing ability) of the bacterium that causes tuberculosis is partly due to the presence of a particular class of wax ester in its outer cell membrane. This ester contains two fatty acid residues and one alcohol residue. Draw the skeletal structure of one of these esters, formed from the carboxylic acid and alcohol molecules shown.
O CH3 (CH2)18 (CH2CH(CH3))4 C OH CH3(CH2)20CH(OH)CH2CH(OH)(CH2)4CH(CH3)CH(OCH3)CH3 Answer: To form the ester, connect two fatty acid residues to the alcohol using ester bonds. The alcohol has two –OH sites.
11 - 21
2 fatty acid residues
C O C O O
O
alcohol residue
11.22 a. Webworms produce 9-cis-tetradecenyl acetate and use it as a sex pheromone (attractant). Draw this ester. (Hint: The prefix “tetradec” means “14”) b. Which would you expect to have the higher melting point, 9-cis-tetradecenyl acetate or tetradecyl acetate (see Problem 21)? Answer:
a. b. Tetradecyl acetate is expected to have a higher melting point because the alcohol residue is saturated. Saturated chains have larger surface area of contact between molecules and thus stronger London forces.
SECTION 11.3
11.24
TRIGLYCERIDES
True or false? a. Saturated fats contain only saturated fatty acids. b. Unsaturated fats contain only unsaturated fatty acids. c. Triglycerides are carboxylic acids.
11 - 22
O CH3
Answer: Both a. and b. are false. Saturated fats contain more saturated than unsaturated fatty acid residues. Unsaturated fats contain more unsaturated than saturated fatty acid residues. c. False. Triglycerides are esters formed from glycerol and fatty acids.
11.26 Write a balanced reaction equation for the complete hydrogenation of linolenic acid (Table 11.1). Answer: Hydrogenation involves the addition of hydrogen atoms across a C-C double bond. This reaction usually occurs in the presence of a catalyst like Pt.
3
Pt
11.28 Draw a triglyceride made from glycerol, linoleic acid, and two linolenic acid molecules. Would you expect this triglyceride to be a liquid or a solid at room temperature? Explain. Answer: A triglyceride is a lipid that consists of 3 hydrocarbon chains attached by ester bonds to a glycerol molecule. O
CH3(CH2)4(CH=CHCH2)2(CH2)6 C
O CH2
linoleic residue
O CH3CH2(CH=CHCH2)3(CH2)6 C
O CH
linolenic residue
O CH3CH2(CH=CHCH2)3(CH2)6 C
O CH2
linolenic residue
11 - 23
This triglyceride can be expected to be a liquid. Its many cis double bonds cause the hydrocarbon tails to be bent and not to pack closely together. This decreases the strength of intermolecular forces and lowers the melting point of the compound.
11.30 a. Draw the products formed when the triglyceride is saponified.
O CH3(CH2)5CH=CH(CH2)7 C
O CH2
O CH3(CH2)12 C
O CH
O CH3(CH2)7CH=CH(CH2)7 C
O CH2
b. For each product from part a, state whether it is hydrophobic, hydrophilic, or amphipathic. Answer: a. Saponification requires the breaking of the ester bond to form the carboxylate ion and alcohol group. O
O
CH3(CH2)5CH=CH(CH2)7 C O -
+
CH3(CH2)12 C O -
O
+
CH3(CH2)7CH=CH(CH2)7 C O -
CH2 OH
+
CH OH CH2 OH
b. For each product from part a, state whether it is hydrophobic, hydrophilic, or amphipathic.
11 - 24
O
O
CH3(CH2)5CH=CH(CH2)7 C O -
+
CH3(CH2)12 C O -
O
+
CH3(CH2)7CH=CH(CH2)7 C O -
CH2 OH
+
CH OH CH2 OH
11.32 List two different ways that you can slow the spoilage of vegetable oil. Answer: Refrigerating vegetable oil in a tightly capped bottle can slow the spoilage of vegetable oil. Partial hydrogenation removes some of the oxidizable double bonds that lead to rapid spoilage. Antioxidants can also be mixed in with the oil to reduce breakdown by oxidation.
11.34 Why do food companies partially hydrogenate vegetable oil? Answer: Partial hydrogenation converts vegetable oil into a semisolid product that can be used for cooking.
SECTION 11.4
PHOSPHOLIPIDS AND GLYCOLIPIDS
11.36 True or false? a. All phospholipids contain a phosphate residue. b. All phospholipids contain two fatty acid residues. c. All phospholipids contain a glycerol residue. Answer: a. True b. False, only glycerophospholipids do. c. False, in sphingolipids, sphingosine replaces glycerol and one fatty acid.
11 - 25
11.38 a. To which class of phospholipids does the compound belong?
CH3(CH2)11CH2CH=CH CH OH O CH3(CH2)4(CH=CHCH2)2 (CH2)6 C NH CH O CH2O P
CH3 OCH2
O
+
CH2 N
CH3
CH3
b. How do other members of this class of phospholipids differ from one another? c. Draw the products obtained when the molecule is saponified into its component parts. d. For each product from part c, state whether it is hydrophobic, hydrophilic, or amphipathic. Answer: a. Sphingolipid. The class sphingolipids is recognized by the incorporation of a sphingosine residue into the structure of the phospholipid. b. The other members of this class have may have different alcohol and fatty acid components bonded to the phosphate group. c. and d.
CH3(CH2)11CH2CH=CH CH OH +
NH3 CH CH2OH amphipathic O
O CH3(CH2)4(CH=CHCH2)2 (CH2)6 C O
O P
CH3 OCH2
O
amphipathic
CH3
CH3 hydrophilic
11 - 26
+
CH2 N
11.40 Lecithin (phosphatidylcholine in Figure 11.17) is an emulsifying agent, but triglycerides are not. Account for the difference. Answer: Lecithin is amphipathic and triglycerides are hydrophobic. Being amphipathic, lecithins can interact with both nonpolar compounds (oil) and polar ones (water) to keep an emulsion from separating into its two components.
11.42 Draw a sphingomyelin that contains palmitoleic acid. Answer:
CH3(CH2)11CH2CH=CH CH OH O CH3(CH2)5CH=CH (CH2)7 C NH CH O CH2O P O
CH3 +
OCH2 CH2 N
CH3
CH3
11.44 Which is better at forming hydrogen bonds, phosphatidylethanolamine or phosphatidylcholine? Explain. Answer: Phosphatidylethanolamine. Both compounds contain O atoms capable of forming hydrogen bonds. The –CH2CH2NH3+ part of phosphatidylethanolamine contains three N-H bonds able to form hydrogen bonds, while the –CH2CH2N(CH3)3+ part of phosphatidylcholine does not.
11.46 A particular cerebroside contains a galactopyranose residue attached to a sphingosine residue by a β glycosidic bond (see Section 10.6). Based on the sphingolipid shown in Figure 11.18, draw this cerebroside.
11 - 27
Answer:
CH3(CH2)11CH2CH=CH
CH OH
O CH3(CH2)5
CH CH (CH2)7 C NH CH CH2OH
HO H
O H OH
H
H
OH
O
CH2
H
11.48 The membrane of an animal cell nucleus contains phosphatidylcholine in which a significant number of the fatty acid residues are saturated. After reading Section 11.7, would you expect this to increase or decrease the fluidity of the membrane structure? Answer: Decrease. These molecules would decrease the fluidity of the membrane because they would decrease the proportion of unsaturated fatty acids, which give membranes greater fluidity.
11.50 a. Often, the fatty acid released by the action of phospholipase A2 (see Problem 11.49) is arachidonic acid. Indirectly, the release of this fatty acid is linked to inflammation that arises at the site of a snake bite. Explain. b. The diglyceride (lysophospholipid) formed by the phospholipase A2-catalyzed breakdown of a triglyceride acts as a detergent that can disrupt cell membranes. Which more closely resembles membrane phospholipids, triglycerides or diglycerides? Answer: a. Arachidonic acid is a precursor for the synthesis of different eicosanoids including prostaglandins which cause inflammation. b. Membrane phospholipids have two fatty acid residues and therefore resemble diglycerides more closely.
11 - 28
SECTION 11.5
STEROIDS
11.52 A person whose diet contains no cholesterol can still have high cholesterol levels. Explain. Answer: Cholesterol is manufactured by the liver.
11.54 Label the hydrophobic and hydrophilic parts of glycocholate (Figure 11.21). Answer: The hydrophilic parts are the polar sites and the hydrophobic parts are the nonpolar (hydrocarbon) portions of the molecule. hydrophobic
hydrophilic
hydrophilic CH3 OH
hydrophobic
CH3
O
CH CH2 CH2 C
NH CH2
CH3
C O
O
hydrophilic HO
OH
hydrophilic
hydrophilic
11.56 Bile salts aid in the digestion of triglycerides by acting as emulsifiers. Explain. Answer: Emulsifiers are substances that allow the formation of emulsions. In an emulsion, two insoluble liquids form a colloidal dispersion through stabilization of the polar and nonpolar sites of the two liquids. Bile salts act as emulsifying agents in the digestion of triglycerides. They have hydrophilic sites that have favorable interaction with the aqueous environment of the digestive system and hydrophobic sites that interact with the water insoluble triglycerides.
11 - 29
11.58 A cholesterol screening finds that a person’s blood serum has 35 mg/dL HDL and 150 mg/dL LDL. a. Are these values within the normal range? b. How many grams of HDL are present in 1.0 mL of the serum? c. How many micrograms of LDL are present in 2.5 x 10-4 L of the serum? d. What is the LDL/HDL ratio? Is this value considered to be healthy? e. Can total cholesterol be calculated from the HDL and LDL concentrations? Explain. Answer: a. No, HDL is too low (<40 mg/dL) and LDL is too high (>100 mg/dL). b. The steps involve converting 1.0 mL to dL, calculating the mg of HDL using 35 mg/dL HDL as a conversion factor, and converting mg HDL to g HDL.
1.0 mL x
10-3 L 1 dL x x 1 mL 10-1 L
35 mg 10-3 g x = 3.5 x 10-4 g HDL 1 dL 1 mg
c. The steps involve converting 2.5 x 10-4L to dL, calculating the mg of LDL using 150 mg/dL LDL as a conversion factor, and converting mg LDL to g HDL.
2.5 x 10-4 L x
d.
1 dL x 10-1 L
150 mg 10-3 g 1 g x x 1 dL 1 mg 10-6 g
= 3.8 x 102 g LDL
LDL 150 mg/dL = = 4.3 HDL 35 mg/dL
No, a healthy person’s cholesterol ratio should be less than 2.5. e. No, because HDL and LDL are not the only lipoproteins that transport cholesterol. We would also need the concentration for VLDL to calculate the total cholesterol. SECTION 11.6
EICOSANOIDS
11.60 Low doses of aspirin can reduce the risk of heart attack and stroke, but can also increase the risk of gastrointestinal bleeding. Aspirin’s effect on the synthesis of which type of eicosanoid leads to the greater risk of bleeding?
11 - 30
Answer: Most likely thromboxane A2 since it is involved in blood clotting.
11.62 To which class of lipid does dexamethasone (Problem 11.61) belong?
O
C CH2OH
CH3
HO
OH
H3C
CH3 F
O Dexamethasone Answer: Steroids. Dexamethasone has the basic fused multi-ring structure typical that defines steroids as a class of lipids.
SECTION 11.7
MEMBRANES
11.64 How are facilitated diffusion and active transport similar? Answer: In both kinds of movement across a cell membrane, proteins play a role in assisting the passage of compounds and ions through a membrane.
11.66 How are facilitated diffusion and diffusion similar? Answer: Both processes involve the movement of solute from an area of high concentration to an area of low concentration.
11.68 The lipids present in the cell membranes of fish, especially fish that live in cold water, contain more unsaturated fatty acid residues than do the lipids found in the membranes of land animals. Explain why.
11 - 31
Answer: Lipids that contain a higher percentage of unsaturated fatty acids have a lower freezing point which allows the cell membrane to maintain its fluidity even in very cold temperatures.
HEALTH LINK
OMEGA – 3 FATTY ACIDS
11.70 a. In the name “eicosapentaenoic acid”, the prefix eicos indicates that there are twenty carbon atoms in the molecule (Figure 11.4). To what does the pentaen part of the name refer? b. In the name “docosahexaenoic acid”, what does the prefix docos indicate? To what does the hexaen part of the name refer? Answer: a. There are 5 C=C bonds in the molecule. b. Docos indicates that there are 22 C atoms in the molecule and hexaen indicates that there are 6 C=C bonds.
11.72 Why is it recommended that some of the fish that are a good source of omega-3 fatty acids not be eaten? Answer: They may contain high levels of mercury.
HEALTH LINK
TRANS FATS
11.74 Which is likely to have the most trans fats, vegetable oil, margarine, or butter? Explain. Answer: Margarine is the food that is most likely to have trans fats because margarine is made from the partial hydrogenation of vegetable oil. In this process, cis bonds can be converted into trans bonds.
11 - 32
11.76 Draw a trans fat that might form if the following triglyceride is subjected to partial hydrogenation. O CH3 (CH2)6CH2 CH2(CH2)6 C O CH2 C C H
H
O
CH3(CH2)16 C O CH O CH3(CH2)3CH2 C
C
H
CH2
C
CH2 (CH2)6 C O CH2
C
H
H H
Answer: Partial hydrogenation can cause one or more of the double bonds to reform into a trans orientation. O CH3(CH2)6CH2
CH2 (CH2)6 C O CH2
C
C
H
H
O CH3(CH2)16 C O CH
trans bonds O CH3(CH2)3CH2
HH C
C CH2
H
HEALTH LINK
C
C
CH2 (CH2)6 C O CH2 H
OLESTRA
11.78 Olestra passes through the digestive system untouched, taking some dietary vitamin A, vitamin E, and other fat-soluble (water-insoluble) vitamins with it. Which noncovalent interaction allows these vitamins to be associated with olestra? Answer: London forces
11 - 33
HEALTH LINK
ANABOLIC STEROIDS
11.80 a. Testosterone replacement therapy often involves the use of testosterone esters. Draw testosterone acetate (formed by reacting the alcohol group of testosterone with acetic acid) and testosterone state. b. Once administered, testosterone esters move to fatty tissue and, over time, released into the bloodstream and hydrolyzed to yield testosterone. Which of the testosterone esters in part a do you think is released more quickly from fatty tissue? Explain. Answer:
a.
testosterone acetate
testosterone stearate
b. Testosterone acetate is more likely to be released more quickly. Because it is a smaller, nonpolar molecule, it is less soluble in fatty tissue than the bigger, more hydrophobic testosterone stearate.
11 - 34
LEARNING GROUP PROBLEMS
11.82 a. It has been found that certain fatty acid esters of plant sterols (steroids produced by plants) are effective in lowering cholesterol levels in the body. For the sterol ester below, label the fatty acid residue and the sterol residue.
CH3 CH3 O O
b. Draw the fatty acid salt and sterol obtained when the compound in part a is hydrolyzed using NaOH(aq). c. Draw the fatty acid from your answer to part b, as it would appear at pH 1, at pH 7, and at pH 12. d. Of the compounds in your answer to part b, which (if any) are hydrophobic and which (if any) are amphipathic? e. Draw a product that might be formed when the fatty acid whose residue appears in this problem is partially hydrogenated (reacted with H2 and Pt). f. Draw the organic product formed when the sterol in your answer to part b is reacted with K2Cr2O7. Answer: a.
sterol residue fatty acid residue
11 - 35
b. O +
+
+
+
O Na
CH3 CH3
HO
c. O OH pH = 1 O O pH = 7 and 12
d. O O Na amphipathic CH3 CH3
HO
hydrophobic
e. 1 or more of the double bonds become saturated when the fatty acid is reacted with H2 and Pt (partial hydrogenation).
11 - 36
O OH
f. Draw the organic product formed when the sterol in your answer to part b is reacted with K2Cr2O7.
CH3 oxidized
CH3
O
11 - 37
Chapter 12 Peptides, Proteins, and Enzymes SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
12.1
Three amino acids are combined to produce a tripeptide.
a. How many peptide bonds does the tripeptide have? Two, as enclosed below.
b. Which amino acid residue is at the N-terminus?
c. Which amino acid residue is at the C-terminus?
12 - 1
d. How many different tripeptides could be produced using one of each amino acid? 6 e. How many different tripeptides could be produced using one, two, or three of each amino acid? 27
12.3
a. For lactic acid (2-hydroxypropanoic acid), pKa =3.86. Draw lactic acid as it would appear at pH 7.
When the pH is greater than the pKa, the basic form (lactate ion) predominates.
b. For diethylammonium ion, pKa = 11.1. Draw diethylammonium ion as it would appear at pH 7.
When the pH is less than the pKa, the acidic form (diethylammonium ion) predominates.
12.5
Draw methionine as it would appear at each of the following pHs. a. pH 1
12 - 2
At pH 1, the carboxyl group appears as –CO2H and the amino group as NH3+. pH 1 is less than the typical values for the pKa’s of the carboxyl groups and the amine conjugate acid groups. When pH < pKa, the acidic form predominates.
O + H3NCHC OH CH2CH2SCH3 b. pH 7 At a pH 7 the carboxyl group appears as –CO2- and the amino group as NH3+.
O + H3NCHC O CH2CH2SCH3 pH 7 is greater than the typical values for the pKa’s of the carboxyl groups; the basic form predominates. pH 7 is less than the typical pKa’s for amine conjugate acid groups; the acidic form predominates. c. pH 14 At a pH 14 the carboxyl group appears as –CO2- and the amino group as -NH2.
O H2NCHC O
-
CH2CH2SCH3 pH 14 is greater than the typical values for the pKa’s of the carboxyl groups and the amine conjugate acid groups. When pH > pKa, the basic form predominates.
12.7
a. What is the net charge on arginine at pH 1? 2+ At pH 1, the carboxyl group appears as –CO2H and the amino (-NH2) group as NH3+. The side chain is basic and is positively charged.
12 - 3
+
NH3 H2 N
+
O
NH NH2
OH
b. What is the net charge on arginine at pH 7? 1+ At pH 7, all the carboxyl groups are negatively charged and the amine groups are positively charged. +
NH3 +
O
NH
H2N
NH2
O
c. What is the net charge on arginine at pH 14? 1At pH 14, the carboxyl group appears as –CO2- and the amino group as –NH2. The side chain has lost a H+ and has no charge.
NH2 HN
NH NH2
12.9
O O
-
Using Fischer projections draw each amino acid as it would appear at pH 7. A Fischer projection is drawn with the chiral carbon atom at the intersection of a vertical and a horizontal line. For amino acids, the carboxyl group points up, the R group points down and the amino group points either left (L amino acid) or right (D amino acid). Consult the text for the structure of each amino acid at pH 7.
12 - 4
a. L-isoleucine
O C
O
+ H3N
-
H CHCH3 CH2 CH3
b. D-aspartic acid
O O
C
-
+ NH3
H CH2
O
C O
c. L-tyrosine
O C O+ H3N
H CH2
OH
12 - 5
d. D-phenylalanine
O C O NH3+
H CH2
12.11 Using a Fischer projection, draw each amino acid from Problem 12.9 as it appears at pH 1. At a pH of 1, carboxyl groups appear as –CO2H and amino groups as –NH3+. a. L-isoleucine O C + H3 N
OH H
CHCH3 CH2 CH3
12 - 6
b. D-aspartic acid
O OH
C
+ NH3
H CH2 C
OH
O c. L-tyrosine
O C OH + H3N
H CH2
OH d. D-phenylalanine
O C OH NH3+
H CH2
12.13 Two of the amino acids in Table 12.1 have two chiral carbon atoms. Which ones?
12 - 7
The two amino acids in Table 12.1 that have two carbon atoms with four different atoms or groups of atoms are threonine and isoleucine.
*
*
*
*
threonine isoleucine 12.15 Monosodium glutamate (MSG), used to enhance the flavor of certain foods, has the structure shown below. Draw this molecule as it would appear at a pH of 1.
At a pH of 1, carboxyl groups appear as –CO2H and the amino group as –NH3+.
12.17 Other amino acids than the twenty common ones shown in Table 12.1 appear in proteins. One of these, selenocysteine (Sec), is identical to cysteine except that the sulfur atom in Cys is replaced by a selenium atom. a. The side chain in Sec is as acidic as the side chain in Asp. Draw Sec as it would appear at pH 7. 12 - 8
O +
H3 N CH C O CH2Se b. To which classification of amino acids (nonpolar, polar-acidic, polar-basic, polar-neutral) does Sec belong? Polar-acidic, just like Asp.
12.19 Which specific class of bonds holds one amino acid residue to the next in the primary structure of a protein? Are these bonds covalent or noncovalent? Peptide bonds, also known as amide bonds. Covalent.
12.21 Draw the products obtained when Asp-Phe-OMe (Section 12.4) is hydrolyzed under acidic conditions.
O
O +
H3N
+
H3N
CH C OH CH2
CH C OH CH2
+
+
HOCH3
C OH O
12.23 a. Is Ala-Phe-Thr-Ser an oligopeptide or a polypeptide? Oligopeptide. The compound has four amino acid residues. Oligopeptides are defined to have between 2 and 10 amino acid residues. b. How many peptide bonds does the molecule contain? Three. A peptide bond is found between each of the joined residues c. Which is the N-terminal amino acid? Alanine. The N-terminal residue is written first, therefore alanine is the Nterminal amino acid.
12 - 9
12.25 Circle the backbone for the pentapeptide shown in Figure 12.4a. The backbone includes all of the peptide bonds.
12.27 Name each of the amino acid residues in the oligopeptide shown in Figure 12.4a.
glycine
glycine valine phenylalanine
tyrosine 12.29 Draw Gly-Phe-Lys-Lys as it would appear at a. pH 1
12 - 10
O +NH
3
CH2
O
C
NH
CH
O
C
NH
CH2
CH
O
C
NH CH
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
NH+3
NH3+
C
OH
b. pH 7 O +NH
3
CH2
O
C
NH
CH
O
C
NH
CH2
CH
O
C
NH CH
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
NH+3
NH3+
O–
C
c. pH 14 O NH2
CH2
C
O NH
CH
C
O NH
CH2
12.31 What is the net charge on Asp-Lys at each pH? a. pH 1
total charge = 2+
12 - 11
CH C
O NH CH
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
NH2
NH2
C
O-
O
O
NH3 CH C
NH CH C
CH2
CH2
C O
CH2
OH
CH2
+
OH
CH2
+
NH3 b. pH 7
total charge = 0
O
O
NH3 CH C
NH CH C
CH2
CH2
C O
CH2
+
O-
O-
CH2 CH2
+
NH3 c. pH 14
total charge = 2-
O
O
NH2 CH C
NH CH C
CH2
CH2
C O
CH2
O-
O-
CH2 CH2 NH2
12.33 What is the net charge on the oligopeptide in Figure 12.4a at a. pH 1 b. pH 14 The oligopeptide in Figure 12.4a has the following structure. Only the forms of the N-terminal amino group and the C-terminal carboxyl group are affected by pH.
12 - 12
-
a. pH 1 1+ At pH 1, the carboxyl group at the C-terminal is in its acidic form (-CO2H) and the amino group at the N-terminal is also in its acidic form (-NH3+). Therefore, there is a net charge of 1+. b. pH 14 1At pH 14, the carboxyl group at the C-terminal is in its basic form (-CO2-) and the amino group at the N-terminal is also in its basic form (-NH2). Therefore, there is net charge of 1-.
12.35 List the primary structure of all possible tripeptides containing one residue each of aspartic acid, phenylalanine, and valine. Asp-Phe-Val Phe-Val-Asp
12.37
Asp-Val-Phe Val-Asp-Phe
Phe-Asp-Val Val-Phe-Asp
In an aqueous environment will the following peptide fragment more likely be buried inside a globular protein or located on its surface? Explain. O
O
O
O
NH CH C NH CH C
NH CH C
NH CH C
CH3CHCH3
CH2
CH CH3
CH2
CH2
CH2
CH3
S CH3
Inside. This arrangement allows the nonpolar side chains to interact through London forces and does not disrupt hydrogen bonding between water molecules.
12 - 13
12.39 List the chemical bonds or forces that are primarily responsible for maintaining a. the primary structure of a protein Peptide bonds b. the secondary structure of a protein Hydrogen bonds c. the tertiary structure of a protein Hydrogen bonds, ionic bonds (or salt bridges), hydrophobic effects, and disulfide bonds d. the quaternary structure of a protein Hydrogen bonds, hydrophobic effects, disulfide bonds, and ionic bonds (or salt bridges),
12.41 Which amino acids have side chains that can participate in ionic bond formation at pH 7. Aspartic acid, glutamic acid, lysine, and arginine. An ionic bond is formed when a positively charged side chain is attracted to a negatively charged side chain. The only amino acids that have side chains that will have a charge at a pH of 7 are: arginine (=NH2+), aspartic acid (-CO2-), glutamic acid (CO2-), and lysine (-NH3+). Table 12.1 also lists histidine as having a charged side chain at pH 7 (=NH+-). However, at pH 7, this side chain has mostly lost its H+ and is therefore not charged.
12.43 What distinguishes a protein that has quaternary structure from one that does not? A protein that has a quaternary structure consists of more than one polypeptide chain whereas a protein that has no quaternary structure consists of only a single polypeptide chain.
12.45 Draw structures that show a. hydrogen bonding between His and Glu side chains.
12 - 14
O H3N CH C O +
CH2
+
O
H3N
- - - - - - - - O C CH2 CH2 CH N H C O
+
N
-O
H
hydrogen bonding
b. hydrogen bonding between Thr and Ser side chains. O
H +N CH C O 3
O H3+N
CH2 H
O ----------
CH
C O-
CH
CH3
H O
hydrogen bonding
12.47 Which is more likely to lead to a change in the biological activity of a protein, the replacement of a leucine residue with a valine residue or the replacement of the same leucine residue with a lysine? Explain. Both leucine and valine are nonpolar residues while lysine is a polar-basic residue. Replacing leucine with a lysine residue would more likely lead to a change in the three-dimensional structure of the protein and its biological activity. This is because the two residues differ in the types of noncovalent interactions in which they participate.
12.49 What term is used to describe an enzyme whose tertiary structure has been unfolded? Denatured.
12.51 Which types of bonds or interactions are disrupted during the denaturation of a protein? The bonds and interactions that are disrupted when a protein is denatured are those that contribute to the secondary, tertiary, and quaternary structures of the
12 - 15
protein. These include: hydrogen bonds, ionic bonds, disulfide bonds, and hydrophobic interactions.
12.53 Is it possible for an enzyme to show both absolute specificity and relative specificity? No. Absolute specificity means an enzyme can modify only one specific substrate whereas relative specificity allows an enzyme to accept a wider range of substrates that belong to a particular group. The range of substrates cannot be both narrow and wide.
12.55 The enzyme glucokinase catalyzes the transfer of a phosphate group from ATP to carbon atom #6 of D-glucose. D-glucose is the only substrate for the enzyme.
hexokinase
D-glucose + ATP
D-glucose-6-phosphate + ADP
Which of the following terms describe the specificity of glucokinase: absolute specificity, relative specificity, stereospecificity? Both absolute specificity and stereospecificity apply because only D-glucose is the substrate for the enzyme.
12.57 The enzyme trypsin, found in the small intestine, operates best at pH 8. Draw a graph for trypsin, similar to that found in Figure 12.21a.
Reaction Rate
At pH 8, the reaction rate would be highest.
1 2 3
4
5
6
7
8
9
10
pH
12 - 16
11
12
12.59 A thermophilic bacterium found in a hot spring at Yellowstone Park produces an enzyme with a temperature optimum of 98°C. For this enzyme, draw a graph similar to that found in Figure 12.21b. Assume that the enzyme denatures at 100°C.
Reaction Rate
The reaction rate peaks at the optimum temperature of 98 °C. At 100 °C, the enzyme denatures and loses its activity, thus a sharp drop-off of the reaction rate.
60
70
80
90
100 110
120
130
Temperature (○C)
12.61 The coenzyme NADH is a cofactor but is not a prosthetic group. Explain. NADH is a cofactor required by an enzyme for its catalytic action. It is not a prosthetic group because it is not bound to the enzyme protein and does not contribute to its tertiary structure.
12.63 An inhibitor is added to a Michaelis-Menten enzyme. How could you tell whether the inhibitor is reversible or irreversible? When an irreversible inhibitor acts on an enzyme, its effect on the catalytic activity of the enzyme is permanent. The activity of the enzyme cannot be restored. When a reversible inhibitor binds to an enzyme, it interferes with the catalytic activity only on a temporary basis. When the inhibitor dissociates from the enzyme, the catalytic activity of the enzyme is restored.
12.65 Explain why most competitive inhibitors are structurally similar to a substrate for the enzyme they inhibit, while most noncompetitive inhibitors are not. Competitive inhibitors bind at the substrate binding site. They can accomplish this easily if they are structurally similar to the substrate. Noncompetitive inhibitors do not bind at the substrate binding site.
12 - 17
12.67 a. To which reaction type (synthesis, decomposition, single replacement, double replacement) does the reaction in Figure 12.24 belong? Double replacement. The acetyl group in aspirin switches places with the H atom in the –OH group of the serine side chain in COX. b. Due to its action on COX enzymes, aspirin is considered to be a “suicide inhibitor”. Explain. In the process of transferring an acetyl group to COX, the aspirin itself is chemically altered and is no longer able to function as an inhibitor. c. If transfer of an acetyl group from aspirin to the side chain of a serine residue in a COX enzyme caused a change in the tertiary structure of the enzyme, which interaction was most likely disrupted: a hydrogen bond, an ionic bond, or a disulfide bond? Hydrogen bond. The serine side chain lost its ability to hydrogen bond when the –OH group was replaced by the acetyl group.
12.69 Celecoxib (Celebrex) is an inhibitor of COX-2 (Section 11.6). Given that arachidonic acid (Figure 11.23) is a substrate for COX-2, is celecoxib more likely to be a competitive inhibitor or a noncompetitive inhibitor?
CH3
N N
CF3
NH2 SO2 Celecoxib Noncompetitive inhibitor. Because celecoxib and the arachidonic acid substrate do not have similar structures, their binding sites are probably different. Therefore, celecoxib is more likely a noncompetitive inhibitor.
12 - 18
12.71 Describe the effect of positive and negative effectors on allosteric enzymes. A positive effector enhances substrate binding and increases the rate of the enzyme-catalyzed reaction. A negative effector reduces substrate binding and slows the reaction. 12.73 a. Why are some enzymes produced initially as zymogens? Some enzymes are harmful to internal organs and may catalyze unwanted reactions. To control the activity of these enzymes, they are initially produced in their inactive form called zymogens. The zymogens are converted to their active, catalytic form when their specific substrates are present. b. A person is found to produce trypsinogen molecules that have an abnormal primary structure. The trypsin produced from this zymogen is normal. How is this possible? The abnormal primary structure is on the part of the polypeptide chain removed during activation of the trypsinogen.
12.75 Describe the quaternary structure of hemoglobin. A hemoglobin molecule is a tetramer consisting of four separate polypeptide chains (two identical α chains and two identical β chains). Each α chain consists of 141 amino-acid residues coiled into seven α-helical regions and each β chain consists of 146 amino acid residues coiled to form eight α-helical regions.
12.77 List one unusual amino acid residue found in collagen. Is this residue present in a newly assembled collagen polypeptide? Explain. Hydroxyproline (Hyp). No, Hyp is formed by the enzyme-catalyzed oxidation of a hydroxyl group of a proline residue, but this occurs after the protein chain has been formed.
12.79 Describe the quaternary structure of a typical antibody. A typical antibody is a tetrameric protein, consisting of four subunits. Two subunits are light chains that consist of 215 amino acids each. The other two subunits are heavy chains consisting of about 400 amino acids each. The subunits are held together by disulfide bridges.
12 - 19
12.81 What do names such as H1N1 and H5N1 indicate about the structure of an influenza virus? The H in the name stands for hemagglutinin, a molecule that extends from the protein surface of a flu virus. The N stands for neuraminidase, an enzyme found on the surface of a flu virus. The number following each of these letters in the names indicate one of 16 types of hemagglutinin and one of 9 versions of neuraminidase.
12.83 Describe the role of neuraminidase in the release of newly formed influenza viruses from the cell that they have infected. The neuraminidase catalyzes the reaction that breaks the bond attaching the newly formed virus to the cell, releasing the virus from the infected cell.
12.85 Tamiflu is actually a prodrug, a drug that is administered in an inactive form and later activated (see Chapter 6 Health Link “Prodrugs”). In the case of Tamiflu, activation takes place in the liver, where the ester group of this drug is hydrolyzed. Draw the activated form of Tamiflu, as it would exist at pH 7. (Hint: Amide nitrogen atoms are not basic.)
Tamiflu
After hydrolysis of the ester bond, the structures of the products are shown below as they would exist at pH 7.
12 - 20
12.87
True or false? If liver cells are the only cells in the human body that produce enzyme x, an increase in serum levels of enzyme x following a virus infection probably indicates that the infection has resulted in some liver damage. Explain. True. When cells die their contents are released into the bloodstream. If enzyme x is detected in the bloodstream in elevated concentrations it probably indicates liver cells have been damaged.
12.89 a. Give the complete name of each amino acid: Cys, Pro, Phe, and Val. cysteine, proline, phenylalanine, valine b. Draw each of the amino acids in part a as it would appear at pH 7.
c. Draw each of the amino acids in part a as it would appear at pH 1.
d. Draw each of the amino acids in part a as it would appear at pH 14.
12 - 21
e. The pentapeptide Cys-Pro-Phe-Val-Cys is known to inhibit the growth of Hantavirus. Draw this pentapeptide as it would appear at pH 7.
f. Which amino acid residue is at the N-terminus? At the C-terminus? cysteine is at the N-terminus and cysteine is also at the C-terminus g. In its active form, the pentapeptide is bent into a horseshoe shape and held in that position by a disulfide bond. Draw this structure.
h. Does the pentapeptide have a primary structure? Explain. Yes. The sequence of amino acids (Cys-Pro-Phe-Val-Cys) gives the primary structure.
12 - 22
i. If your answer to part h was yes, what holds the primary structure together? Peptide bonds. j. Does the pentapeptide have a secondary structure? Explain. No. There is no H-bonding that takes place between amide N-H and C=O groups along the polypeptide backbone. k. If your answer to part j was yes, what holds the secondary structure together? No secondary structure. l. Does the pentapeptide have a tertiary structure? Explain. Yes. The overall three-dimensional shape of the pentapeptide is horseshoe shape. m. If your answer to part l was yes, what holds the secondary structure together? Disulfide bond. n. Does the pentapeptide have a quaternary structure? Explain. No. There is only one polypeptide chain. o. If your answer to part n was yes, what holds the quaternary structure together? No quaternary structure.
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 12.2
a. The tripeptide in Problem 12.1 is incorporated into a longer peptide. Show the hydrogen bonds that could form if two identical peptide strands are arranged side-by-side.
12 - 23
b. Would the hydrogen bonding arrangement in part a more likely be part of an -helix or of a -sheet? c. Is the arrangement of peptide strands parallel or is it antiparallel? Answer: a. Hydrogen bonding could form between the H bonded to an N atom and the O of the carbonyl group on the other segment.
b. This is more likely to be a part of a beta sheet because hydrogen bonding takes place between two different segments of a peptide backbone. c. The strands are parallel because they run in the same direction: N-terminus at the left end and C-terminus at the right end.
SECTION 12.1
12.4
AMINO ACIDS
a. For 4-ethylphenol, Ka = 1 x 10-10. Draw this compound as it would appear at pH 7. b. For triethylammonium ion, Ka = 1 x 10-11. Draw this compound as it would appear at pH 7.
Answer: a. pH 7 is less than the pKa (=10) so the acidic form exists.
12 - 24
b. pH 7 is less than the pKa (=11) so the acidic form exists. (CH3CH2)3NH+
12.6
Draw proline as it would appear at each of the following pH’s. a. pH 1 b. pH 7
c. pH 14
Answer: The structure of an amino acid is affected by the pH of the solution in which it is dissolved because it has both acidic and basic sites. The following are the structures of the amino acid proline at the given pH’s: a. pH 1: This pH is less than the typical pKa for both –CO2H and –NH3+ (5 and 11, respectively) so the acidic forms predominate. H +
H N
H O C C OH CH2
CH2 CH2
b. pH 7: This pH is less than the typical pKa for –NH3+ but higher than the typical pKa for –CO2H so the carboxyl group is in its conjugate base form.
H +
H N
H O
C C OCH2
CH2 CH2
c. pH 14: This is higher than the typical pKa for both –CO2H and –NH3+ so the conjugate base forms predominate.
H O H N
C C O-
CH2
CH2 CH2
12 - 25
12.8
a. What is the net charge on valine at pH 1? b. at pH 7? c. at pH14?
Answer: a. pH 1: This pH is less than the typical pKa for both –CO2H and –NH3+ (5 and 11, respectively) so the acidic forms predominate.
O +
H3N CH C OH CH CH3 CH3 pH 1
At pH 1, valine has a net charge of 1+.
b. pH 7: This pH is less than the typical pKa for –NH3+ but higher than the typical pKa for –CO2H so the carboxyl group is in its conjugate base form.
O +
H3N CH C O CH CH3 CH3 pH 7
At pH 7, valine has a net charge of 0.
c. pH 14: This is higher than the typical pKa for both –CO2H and –NH3+ so the conjugate base forms predominate.
O
NH2 CH C O CH CH3 CH3 pH 14
At pH 14, valine has a net charge of 1–.
12.10 Using Fischer projections, draw each amino acid as it would appear at pH 7. a. D-methionine c. L-lysine b. L-leucine d. D-asparagine
12 - 26
Answer: a. D-methionine
O C O+
H
NH3
CH2CH2S CH3 b. L-leucine
O C O+
NH3
H
CH2CH CH3 CH3 c. L-lysine
O
C O-
+
NH3
H +
CH2CH2CH2CH2NH3 d. D-asparagine
12 - 27
O
C O+
H
NH3 O CH2C NH2
12.12 Using a Fischer projection, draw each amino acid from Problem 12.10 as it appears at pH 1. Answer: a. D-methionine
O C OH +
H
NH3 CH2CH2SCH3
b. L-leucine O C OH +
NH3
H
CH2CHCH3 CH3
c. L-lysine
12 - 28
O C OH +
NH3
H +
CH2CH2CH2CH2NH3 d. D-asparagine
O C OH +
H
NH3 O CH2C NH2
12.14 One of the amino acids in Table 12.1 has no chiral carbon atoms. Which one? Answer: A chiral carbon must have four different groups bonded to it. Among the 20 amino acids, only glycine has a central carbon that does not have four different groups around it: O C O+
H
NH3 H
12.16 One metabolic fate of glutamic acid is its conversion into -aminobutyric acid. Draw -aminobutyric acid. [Hint: A gamma () carbon atom is attached to a beta () carbon atom.]
12 - 29
Answer:
O H2N CH2CH2CH2C OH
12.18 Pyrrolysine is a naturally occurring amino acid used by some bacteria. O
NH3CHC O+
(CH2)4 O NH2C
N
CH3
Pyrrolysine a. Hydrolysis of the amide bond in pyrrolysine yields which amino acid? b. Assuming that the side chain in pyrrolysine is as basic as the side chain in arginine, draw pyrrolysine as it would appear at pH 7. Recall that amide nitrogen atoms do not act as bases. c. To which classification of amino acids (nonpolar, polar-acidic, polar-basic, polar-neutral) does pyrrolysine belong? Answer: a. Hydrolysis of the amide bond yields the amino acid lysine: O NH3CHC O +
O NH3CHC O -
(CH2)4
hydrolysis
O NH C
O- O C
+
N
(CH2)4 +
NH3
CH3
Pyrrolysine
Lysine
b. At pH 7, all the acidic and basic sites are charged. 12 - 30
+ CH3
N
O +
NH3CHC O
-
(CH2)4 O NH C
+
NH
CH3
Pyrrolysine at pH 7 c. Assuming that the side chain is basic, pyrrolysine would be classified as polarbasic.
SECTION 12.2
THE PEPTIDE BOND
12.20 a. To which functional group from organic chemistry do peptide bonds belong? b. When a peptide bond is hydrolyzed, which two functional groups are produced? Answer: a. amide functional group b. carboxylic acid and amine functional groups
12.22 Draw the products obtained when Phe-Asp-OMe (Section 12.4) is hydrolyzed under acidic conditions.
12 - 31
Answer:
O
O
+
H3N CH C NH CH C O CH3 CH2
CH2
+
C O -
H +
2 H2O
O Phe-Asp-OMe O
O
+
H3N CH C OH
+
+
H3N CH C OH
CH2
+
CH3OH
CH2 C OH O
12.24 a. How many peptide bonds does each of the oligopeptides in Figure 12.4 contain? b. Which amino acid residue is at the N-terminus of each of these oligopeptides? c. Which amino acid residue is at the C-terminus of each? Answer: a. The oligopeptide in Figure 12.4 a has 4 peptide bonds. The oligopeptide in Figure 12.4 b has 8 peptide bonds. b.
N-terminus of Figure 12.4 a oligopeptide: N-terminus of Figure 12.4 b oligopeptide:
tyrosine cysteine
b.
C-terminus of Figure 12.4 a oligopeptide: C-terminus of Figure 12.4 b oligopeptide:
valine glycine
12.26 Circle the backbone for the nonapeptide shown in Figure 12.4b.
12 - 32
Answer: The backbone includes all the peptide bonds. O
O
O O O H3N CH C NH CH C NH CH C NH CH C NH CH C +
CH2
CH2
S
O O O O CH NH C N CH C NH CH C NH CH2 C NH2
CH3 CH
CH2
CH2
CH2
CH2
CH2
C NH2
S
CH3
C NH2 O
O
CH2
CH2 CH2
CH2 CH CH3 CH3
OH
12.28 Name each of the amino acid residues in the oligopeptide shown in Figure 12.4b. Answer: Cysteine, Tyrosine, Isoleucine, Glutamine, Asparagine, Cysteine, Proline, Leucine, Glycine.
SECTION 12.3
PEPTIDES, PROTEINS, AND pH
12.30 Draw Asp-Ser-Lys-Val as it would appear at a. pH 1 b. pH 7 b. pH 7 Answer: a. pH 1 Start with drawings provided with Table 12.1. Asp is at the N-terminus. Draw the individual residues connecting each with a peptide bond. At a pH of 1, carboxyl groups appear as –CO2H and amino groups appear as –NH3+.
12 - 33
+ H3N
H
O
C
C
NH
CH2 O
H
O
C
C
NH
H
O
C
C
H NH
C
O C OH
CH2
CH2
CHCH3
OH
CH2
CH3
C OH
CH2 CH2 + NH3
b. pH 7 Start with drawings provided with Table 12.1. Asp is at the N-terminus. Draw the individual residues connecting each with a peptide bond. At a pH of 7, carboxyl groups appear as –CO2- and amino groups as –NH3+.
+ H3N
H
O
C
C
CH2 O
H NH
C
O C
NH
H
O
C
C
H NH
C
O C O
CH2
CH2
CHCH3
OH
CH2
CH3
C O
-
CH2 CH2 + NH3
b. pH 7 Start with drawings provided with Table 12.1. Asp is at the N-terminus. Draw the individual residues connecting each with a peptide bond. At a pH 14 carboxyl groups appear as –CO2- and amino group as –NH2.
12 - 34
H H2N
C
C
CH2 O
O NH
H
O
C
C
NH
H
O
C
C
H NH
C
O C O
CH2
CH2
CHCH3
OH
CH2
CH3
C O
-
CH2 CH2 NH2
Match the side chains of the amino acid residues with drawings in Table 12.1. Start at the N-terminus (one with an unreacted amino group) and list the individual residues. Cysteine, Tyrosine, Isoleucine, Glutamine, Asparagine, Cysteine, Proline, Leucine, Glycine.
12.32 What is the net charge on Phe-Asp at each pH? a. pH 1 b. pH 7 c. pH 14 Answer: Draw the structure for Phe-Asp using the drawings shown in Table 12.1. a. pH 1 At a pH of 1, carboxyl groups appear as –CO2H and amino groups as –NH3+. Phe-Asp has net 1+ charge at this pH.
12 - 35
O O + H3N CH C NH CH C OH CH2
CH2 C OH O
b. pH 7 At a pH of 7, carboxyl groups appear as –CO2- and amino groups as –NH3+. Phe-Asp has a net 1- charge at this pH. O O + H3N CH C NH CH C O CH2
CH2 C O
-
O
c. pH 14 At a pH 14 the carboxyl groups appear as –CO2- and amino group as NH2. Phe-Asp has a net 2- charge at this pH. O
O
H2N CH C NH CH C O CH2
CH2 C O
-
O
12.34 What is the net charge on the oligopeptide in Figure 12.4b at a. pH 1? b. pH 14?
12 - 36
Answer: a. Net charge is +1. At a pH of 1, carboxyl groups appear as –CO2H and amino groups as –NH3+. O O O O O + H3N CH C NH CH C NH CH C NH CH C NH CH C CH2
CH2
CH2
CH2
CH2
C NH2
S
C NH2 O
O
CH3 CH
CH2
CH2 CH3
S
O O O O NH CH C N CH C NH CH C NH CH2 C NH2 CH2
CH2
CH2 CH CH3
CH2
CH3
OH
b. Net charge is 0. At a pH 14 the carboxyl groups appear as –CO2- and amino group as NH2. O
O
O O O H2N CH C NH CH C NH CH C NH CH C NH CH C CH2
CH2
O O O O NH CH C N CH C NH CH C NH CH2 C NH2
CH3 CH
CH2
CH2
CH2
CH2
CH2
C NH2
S
CH3
C NH2 O
O
S
CH2
CH2 CH2
CH2 CH CH3 CH3
OH
SECTION 12.4
PROTEIN STRUCTURE
12.36 How many tetrapeptides can be produced that contain one residue each of serine, methionine, arginine, and tyrosine? Answer: Twenty-four tetrapeptides. Starting with a different residue each time, make as many different arrangements as possible. Count the different possible arrangements. Ser-Met-Arg-Tyr Ser-Met-Tyr-Arg Ser-Arg-Tyr-Met Ser-Arg-Met-Tyr Ser-Tyr-Met-Arg Ser-Tyr-Arg-Met
Met-Ser-Arg-Tyr Met-Ser-Tyr-Arg Met-Arg-Ser-Tyr Met-Arg-Tyr-Ser Met-Tyr-Ser-Arg Met-Tyr-Arg-Ser
12 - 37
Arg-Tyr-Met-Ser Arg-Tyr-Ser Met Arg-Met-Tyr-Ser Arg-Met-Ser-Tyr Arg-Ser-Tyr-Met Arg-Ser-Met-Tyr
Tyr-Met-Arg-Ser Tyr-Met-Ser-Arg Tyr-Ser-Arg-Met Tyr-Ser-Met-Arg Tyr-Arg-Ser-Met Tyr-Arg-Met-Ser
12.38 In an aqueous environment will the following peptide fragment more likely be buried inside a globular protein or located on its surface? Explain.
O
O
O
O
NH CH C NH CH C NH CH C NH CH C CH2OH
CH2
C O-
CH3 CH OH
O
CH2 C NH2 O
Answer: In an aqueous environment, the peptide fragment would most likely be found on the surface of a globular protein. The R groups of the fragment are all polar groups and they will interact well with water.
12.40 Describe the two types of secondary structure. Answer: The two types of secondary structure are the -helix and the -pleated sheet. The general shape of an -helix resembles that of a coiled spring. This shape is stabilized by hydrogen bonding between an –NH groups and–C=O groups four residues farther along the chain. A -pleated sheet structure is formed when segments of an amino acid chain align adjacent to each other to form a sheet-like structure. This structure is stabilized by hydrogen bonding between –NH groups and –C=O groups from adjacent chains.
12.42 Which nonpolar amino acid has a side chain that can participate in hydrogen bonding? Answer: Tryptophan is a nonpolar amino acid that can participate in hydrogen bonding because it has an –NH in its R group:
12 - 38
O
H3N CH C O +
CH2
N H
12.44 What distinguishes a simple protein from a conjugated protein? Answer: A conjugated protein requires a prosthetic group for its biological function whereas a simple protein does not. A prosthetic group is a nonpeptide component joined to the polypeptide chain of a protein.
12.46 Draw structures that show a. an ionic bond between Asp and Arg side chains b. hydrophobic interactions between Trp and Val side chains. Answer: a. Draw the side chains of Asp and Arg. The ionic bond is the attraction between the –Oof Asp and the =NH2+ group of Arg. Ionic bond
O CH2C O- + NH2 H2N C NHCH2CH2CH2 b. Draw the side chains of Trp and Val. The hydrophobic interaction is the attraction between the two nonpolar side chains. hydrophobic interaction CH2 CH CH3 CH3
12 - 39
12.48 A particular mutation (permanent change to DNA) results in the production of the insulin A chain (Figure 12.13) in which one of the Cys residues involved in a disulfide bridge is replaced by a Tyr residue. This mutation, which produces insulin with one less disulfide bond than normal, leads to diabetes and pancreatic disease. Insulin has three disulfide bonds. Which is most likely, that the missing disulfide bond affects only the tertiary structure of the B chain, that it affects only the quaternary structure of insulin, or that it affects both? Explain. Answer: The missing disulfide bond in the insulin A chain most likely affects only the tertiary structure of the A chain. The quaternary structure is held together by the two other disulfide bonds that connect the A chain to the B chain. Assuming that that the tertiary structure of A is not significantly changed, we can assume that the disulfide bonds to chain B will maintain the integrity of the quaternary structure.
SECTION 12.5
DENATURATION
12.50 Which is/are true for a denatured globular protein? a. is biologically inactive b. contains no peptide bonds c. has an abnormal primary structure Answer: a. is biologically inactive
12.52 List some of the ways to denature a protein. Answer: a. change the temperature b. change the pH c. add detergent or soap d. agitate the protein
12 - 40
SECTION 12.6
ENZYMES
12.54 Is it possible for an enzyme to show both relative specificity and stereospecificity? Answer: Yes. Relative specificity is the ability to react with a range of substrates, such as many different carbohydrates. Stereospecificity is the ability to react with or form a particular stereoisomer, such as the D isomers of those carbohydrates. Therefore an enzyme can show both relative specificity and stereospecificity.
12.56 The enzyme aconitase catalyzes the conversion of citrate into isocitrate. Several related molecules are also substrates for the enzyme.
CO2 -
CO2 CH2
HO C CO2 -
aconitase
CH2
H C CO2 -
H C H
HO C H
CO2 -
CO2 -
Citrate Isocitrate a. Label any chiral carbon atoms in citrate and isocitrate. b. How many stereoisomers are possible for citrate and for isocitrate? Answer: a. The chiral carbon atoms are labeled with an asterisk. Citrate does not have any chiral carbon atoms.
CO2 -
CO2 CH2
HO C CO2 -
aconitase
CH2
H C* CO2 -
H C H
HO C * H
CO2 -
CO2 -
Citrate
Isocitrate
b. 4 stereoisomers possible for isocitrate because there are two chiral carbon atoms. Citrate does not have any chiral carbon atoms so no possible stereoisomers exist.
12 - 41
12.58 Why does a change in pH usually produce a change in the ability of an enzyme to act as a catalyst? Answer: Enzymes are proteins whose function and activity are determined by their specific three-dimensional shape. When the pH changes, the enzyme may lose its native conformation and lose its ability to function as a catalyst.
12.60 Why does a change in temperature produce a change in the ability of an enzyme to act as a catalyst? Answer: Increasing the temperature increases the motion of the enzyme which may disrupt the hydrogen bonds and other noncovalent interactions that maintain the native conformation of the protein. 12.62 The ion Zn2+ is a cofactor for some enzymes, but is not a coenzyme. Explain. Answer: Zn2+ is an inorganic ion which makes it a cofactor. Coenzymes are organic compounds.
SECTION 12.7
CONTROL OF ENZYME-CATALYZED REACTIONS
12.64 An inhibitor is added to a Michaelis-Menten enzyme. How could you tell whether the inhibitor being competitive or noncompetitive? Answer: A competitive inhibitor alters KM but not the Vmax. A noncompetitive inhibitor alters Vmax but not the KM.
12.66 Some bacterial proteins contain D-amino acid residues, which are produced from L-amino acids once the polypeptide chain has been produced. One such case involves the conversion of L-alanine into D-alanine by the enzyme alanine racemase.
12 - 42
a. Using Fischer projections, write a reaction equation for the transformation of L-alanine into D-alanine. b. To what does the term “racemase” apply? (Hint: See Section 10.2) c. Scientists are investigating the possible use of L-fluoroalanine as an antibiotic. By inhibiting alanine racemase, this compound would slow bacterial growth. Is L-fluoroalanine more likely to be a competitive inhibitor or a noncompetitive inhibitor of alanine racemase? Explain.
O C O+
H3N
H
CH2F L-fluoroalanine d. To which classification of amino acids (nonpolar, polar-acidic, polar-basic, polar-neutral) does alanine belong? e. To which classification of amino acids (nonpolar, polar-acidic, polar-basic, polar-neutral) does fluoroalanine belong? Answer: a.
O
O C O
+
H3 N
C O
racemase H
H
+
NH3
CH3
CH3
L-alanine
D-alanine
b. A racemic mixture is a 50:50 mixture of enantiomers. As L-alanine is converted into its enantiomer D-alanine, the alanine mixture moves toward becoming racemic. c. It is likely a competitive inhibitor because it resembles the substrate L-alanine. d. Alanine is a nonpolar amino acid because its side chain is a hydrocarbon. e. Because C-F bonds are polar, fluoroalanine would be a polar-neutral amino acid.
12 - 43
12.68 By inhibiting COX enzymes, low doses of aspirin can reduce the risk of heart attack and stroke. However, if aspirin and acetaminophen are taken at the same time, some of the beneficial effects of aspirin are lost. Explain why acetaminophen (a competitive inhibitor of COX enzymes) has this effect. Answer: While acetaminophen is attached to the active site of a COX enzyme, aspirin is unable to irreversibly inhibit the enzyme. By the time that acetaminophen (a reversible competitive inhibitor) leaves the active site, the aspirin may no longer be present.
12.70 Methanol (CH3OH) is poisonous because, once ingested, the enzyme alcohol dehydrogenase catalyzes its oxidation into formaldehyde (CH2O). This aldehyde damages tissues and often causes blindness. Ethanol (CH3CH2OH) given intravenously is a common treatment for methanol poisoning. Alcohol dehydrogenase becomes so “busy” oxidizing ethanol that methanol gets metabolized in a way that does not produce formaldehyde. a. Which of these two alcohols binds more effectively to alcohol dehydrogenase: (see Section 12.7). b. Which of these two alcohols, once bound to the enzyme, is converted to product more quickly? c. In terms of its ability to act as a substrate for alcohol dehydrogenase, would 1octanol be as effective as ethanol when it comes to treating methanol poisoning? d. Besides its ability to act as a substrate, what additional information might you want about 1-octanol before using it to treat methanol poisoning? Answer: a. The enzyme binds ethanol 10 times more effectively than methanol. b. Ethanol is converted 20 times faster than methanol. c. Yes. 1-octanol binds more effectively than ethanol. d. Is 1-octanol poisonous or is it converted into poisonous compounds? You would want to make sure that the 1-octanol product does not interfere with other biochemical pathways or has some other toxicity.
12.72 Explain how addition or removal of a phosphate group can be used to control the activity of an enzyme.
12 - 44
Answer: Phosphorylation or dephosphorylation alters the structure of an enzyme in a way that activates or deactivates it (or changes its ability to act as a catalyst).
12.74 What is feedback inhibition? Answer: Feedback inhibition is the mechanism by which a reaction or a series of catalytic reactions are inhibited by their own products. This inhibition prevents the overformation of catalytic products.
BIOCHEMISTRY LINK
HEMOGLOBIN, A GLOBULAR PROTEIN AND COLLAGEN, A FIBROUS PROTEINS
12.76 The binding of oxygen to hemoglobin is cooperative. Explain this term. Answer: Hemoglobin consists of four polypeptide subunits. In its cooperative binding, the binding of oxygen to the active site on one subunit affects the oxygen binding at other subunits.
12.78 Give a molecular explanation for the fact that steak from old cattle tends to be tougher than steak from young cattle. Answer: Steak is composed of structural protein called collagen. When it is first formed, collagen consists of protein fibers that are held by hydrophobic interactions and hydrogen bonds. As collagen grows old, covalent cross-linking forms between these protein fibers which results in a structurally stronger binding between the individual fibers. The steak from an old steer is tougher for that reason.
HEALTH LINK
12.80
IMMUNOTHERAPY
Rituximab is an antibody for the CD20 antigen. a. Rituximab is a monoclonal antibody. What makes it monoclonal? b. What is an antigen?
12 - 45
c. How is Rituximab involved in the destruction of B-cells? Are only cancerous B-cells destroyed? Answer: a.
Rituximab is a monoclonal antibody because it is cloned from one type of immune cell. b. An antigen is a protein attached to invading cells or substances. The attachment of an antibody to an antigen marks the foreign cell or substance for destruction by immune cells. c. Rituximab attaches to the CD20 antigen on B-cells, marking it for destruction. Both normal and cancerous cells are destroyed in the process. HEALTH LINK
TAMIFLU AND RELENZA AS ENZYME INHIBITORS
12.82 What structural feature of respiratory cells makes them susceptible to influenza viruses? Answer: Respiratory cells have receptors that attach to the hemagglutinin found on the surface of flu viruses. This allows the virus to penetrate the outer cell membrane and infect the cell.
12.84 Vaccines stimulate the production of antibodies, specialized proteins that the immune system uses to identify and neutralize foreign objects. Influenza vaccines help the immune system react to neuraminidase and hemagglutinin. Why do you suppose that people are given a different flu vaccine every year? Answer: Antibodies must recognize the specialized proteins on viruses to mark them for destruction. Viruses may undergo mutations or genetic changes that may lead to modification of the neuraminidase and hemagglutinin make-up of the surface that render them unrecognizable to specific vaccine antibodies.
12.86 A fast-acting anti-influenza prodrug named cs-8958 is undergoing testing (see the Chapter 6 Health Link: Prodrugs). When the ester group of this prodrug is hydrolyzed, octanoic acid and a Relenza derivative are formed. Draw the Relenza derivative as it would appear at pH 7. (Hints: Amide nitrogen atoms are not basic. In the group attached to the bottom of the six-membered ring, only the double-bonded N is basic.)
12 - 46
Answer:
HEALTH LINK
PROTEINS IN MEDICINE
12.88 a. Explain how asparaginase treatment kills certain leukemia cells, while having no impact on normal cells. b. Would orally administered asparaginase be as effective as intravenously injected asparaginase in reducing serum asparagines levels? Explain. Answer: a. Asparaginase is an enzyme that catalyzes the hydrolysis of the amino acid asparagine to produce aspartic acid and ammonia. Asparagine is an essential nutrient for protein synthesis. While normal cells are able to produce all of the asparagine necessary for protein synthesis, some leukemia cells depend on asparagine supplied by the blood. Asparaginase treatment in the blood depletes the asparagine supply to leukemia cells. Because the asparaginase is unable to penetrate cell membranes, normal cells are not affected by the presence of this enzyme in the blood.
12 - 47
b. No. The asparaginase, like other proteins, would be broken down during digestion and would not enter the blood stream.
LEARNING GROUP PROBLEMS
12.90 a. Give the complete name of each amino acid: Gly, Tyr, Asp, Pro, Glu, Thr, and Trp. b. Draw each of the amino acids in part a as it would appear at pH 7. c. Draw each of the amino acids in part a as it would appear at pH 1. d. Draw each of the amino acids in part a as it would appear at pH 14. e. The synthetic peptide name Chignolin has the structure Gly-Tyr-Asp-Pro-GluThr-Gly-Thr-Trp-Gly. Draw this decapeptide as it would appear at pH 7. f. Which amino acid residue is at the N-terminus? At the C-terminus? g. Chignolin is bent into a horseshoe shape and held in the position by a very short region of β-sheet that involves hydrogen bonding between the Tyr-Asp residues and the Thr-Trp residues. Draw this structure showing the hydrogen bonds. h. Is the region of β-sheet parallel or is it antiparallel? i. Does the decapeptide have primary structure? Explain. j. Does the decapeptide have secondary structure? Explain. k. Does the decapeptide have tertiary structure? Explain. l. Does the decapeptide have quaternary structure? Explain. Answer: a. glycine (Gly) proline (Pro) glutamic acid (Glu)
tyrosine (Tyr)
aspartic acid (Asp)
threonine (Thr)
tryptophan (Trp)
b. At a pH of 7, carboxyl groups appear as –CO2- and amino groups as –NH3+.
12 - 48
O +
H3N CH C O CH2
O +
H3N CH2 C O
OH
Gly
Tyr O
+
H3N CH C O CH2 C O O Asp
O +
O
H3N CH C O
H2N
CH C O
CH2
CH2
CH2
+
CH2 C O
CH2
O
Pro
Glu O
+
H3N CH C O CH CH3 OH Thr
O +
H3N CH C O CH2
NH Trp
12 - 49
c. At a pH of 1, carboxyl groups appear as –CO2H and amino groups as –NH3+.
O +
H3N CH C OH CH2
O +
H3N CH2 C OH
OH
Gly
Tyr
O +
H3N CH C OH CH2 C OH O Asp
O +
O
H3N CH C OH
H2N
CH C OH
CH2
CH2
CH2
+
CH2 C OH
CH2
O
Pro
Glu
O +
H3N CH C OH CH CH3 OH Thr
12 - 50
O +
H3N CH C OH CH2
NH Trp d. At a pH 14 the carboxyl groups appear as –CO2- and amino group as NH2.
O H2N CH C O CH2
O H2N CH2 C O
OH
Gly
Tyr
O H2N CH C O CH2 C O O Asp
12 - 51
O O
H2N CH C O
HN
CH C O
CH2
CH2
CH2
CH2 C O
CH2
O
Pro
Glu O
H2N CH C O CH CH3 OH Thr
O H2N CH C O CH2
NH Trp e. O
O
O
H3N CH2 C
NH CH C
NH CH C
N
CH2
CH2
+
CH2
C O O
O
O
O
O
O
O
CH C
NH CH C
NH CH C
NH CH2 C
NH CH C
NH CH C
CH2 CH2
CH2
CH CH3
CH CH3
CH2
OH
OH
O NH CH2 C O
CH2
C O
NH
O OH Gly
Tyr
Asp
Pro
Glu
Thr
Gly
Thr
f. Glycine (Gly) is at the N-terminus. Glycine (Gly) is also at the C-terminus. g. The region of hydrogen bonding is labeled in the following structure.
12 - 52
Trp
Gly
O NH O C CH2 CH2
CH
C
O NH
CH
C
CH2
CH CH3
NH
CH2
OH
CH2
H-bonding
C O
CH
C O
N
O
CH2 O C
NH O
HO CH CH
CH CH2 C O
CH3 C O
NH
NH O C CH CH2
OH NH
CH2 CH C O
NH NH O C
H-bonding CH2
CH2
+
C O
NH3
O
h. Antiparallel. The two segments run in opposite directions. i. Yes. The primary structure is the sequence of amino acids. j. Yes, in the form of a -sheet held together by hydrogen bonding interactions, k. No. No other structure exists except for a very short -sheet region. l. No. No other chain exists.
12 - 53
Chapter 13 Nucleic Acids SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
13.l
The picture shows RNA polymerase attached to a strand of DNA
a. Which reaction is catalyzed by RNA polymerase? Addition of nucleotide residues to a growing RNA strand. b. What is the name of this process? Transcription c. Which building blocks are used to form the final product? Ribonucleotides d. What is the name of the DNA strand that gets “read” by RNA polymerase? Template strand e. In which direction does RNA polymerase move along the DNA? 3’ to 5’ on the template strand f. To which end of the growing product chain are building blocks attached? 3’ end
13 - 1
13.3
a. Which monosaccharide is used to make DNA? 2-Deoxyribose b. Draw this monosaccharide in its -furanose form. The structure of -2-deoxyribofuranose is given below:
-2-deoxyribofuranose
13.5
Name the four bases that are present in DNA. Cytosine, guanine, thymine, and adenine
13.7
Which of the bases present in nucleotides are purines? Adenine and guanine are purines.
13.9
The structure of phosphate is pH dependent. a. Draw the four forms in which phosphate appears. The four forms of phosphate are: O HO
P
O OH
OH phosphoric acid
O O-
O
OH
O-
- O P OO-
dihydrogen phospate ion
hydrogen phosphate ion
phosphate ion
HO
P
HO
P
O-
b. Which form(s) appear at physiological pH? At physiological pH, phosphate exists in the forms of the dihydrogen phosphate ion and the hydrogen phosphate ion
13 - 2
13.11 Draw the phosphate monoester in Figure 13.1 as it would appear at low pH.
O CH3 O
P
OH
OH
13.13 a. How many different phosphate monoesters can β-D-ribofuranose form? Four phosphate esters can form because it has four -OH groups. b. When reacted with NH2CH3, how many different N-glycosides can β-D-ribofuranose form? One β-N-glycoside can form when an amine reacts at the hemiacetal carbon atom of the β-D-ribofuranose.
13.15 a. What structural feature makes a base found in nucleic acids a purine? A purine is a base with the structure:
N
N N
N H
b. Of the bases adenine, thymine, guanine, cytosine, and uracil, which are purines? Adenine and guanine are purines.
13.17 Draw the complete structure of guanosine 5'-monophosphate (see Figure 13.6). The guanosine molecule is shown in Figure 13.5 and the nucleoside 5' monophosphate structure is shown in Figure 13.6. Draw the structure for guanosine and then remove the –OH from carbon 5 of the ribose ring and replace it with phosphate.
13 - 3
O N
O-
-O P O
O
CH2 H
NH
N
O H
H
OH
OH
N
NH2
H
13.19 How many phosphoester bonds and how many phosphoanhydride bonds are present in the following? a. a nucleotide One phosphoester bond and no phosphoanhydride bond. A phosphoester bond connects a phosphate and a nucleoside. There is one phosphate connected to the nucleoside and therefore one phosphoester bond. There are no phosphoanhydride bonds (a chemical bond formed between two phosphate groups) since there is only one phosphate (see Figure 13.6). b. a nucleoside diphosphate One phosphoester bond and one phosphoanhydride bond. There is one phosphate connected to the nucleoside and therefore one phosphoester bond. There are two phosphates and therefore one phosphoanhydride bond (see Figure 13.7a).
13.21 a. Draw 3'-dCTP. 3’-dCTP is an abbreviation for 3’-deoxycytidine triphosphate. Draw deoxycytidine . Remove the –OH on carbon 3 and add three phosphate groups.
13 - 4
b. Draw 3'-CTP. 3’-CTP is an abbreviation for 3’cytidine triphosphate. Draw cytidine . Remove the –OH on carbon 3 and add three phosphate groups.
13.23 a. Name the types of bonds broken when ATP is completely hydrolyzed. Phosphoanhydride, phosphoester, and N-glycoside (see Figure 13.7a). b. What products are obtained when ATP is completely hydrolyzed? 3 phosphates, D-ribose, and adenine.
13 - 5
13.25 a. Name the types of bonds broken when the molecule in Problem 13.21a is completely hydrolyzed. Phosphoanhydride, phosphoester, and N-glycosidic bonds are broken when the 3’-dCTP molecule is completely hydrolyzed. b. What products are obtained when this molecule is completely hydrolyzed? Three phosphate ions, a -D-2-deoxyribofuranose molecule, and a cytosine molecule are obtained when the 3’-dCTP molecule is completely hydrolyzed.
13.27 In the sugar phosphate backbone of DNA a. which sugar residue is present? 2-Deoxyribose b. which type of bonds connect the sugar and phosphate residues? Phosphoester bonds c. where are bases attached? They are attached at C-1'.
13.29 In terms of DNA and RNA structure, what do the terms 3'-terminus and 5'terminus mean? At 3’- terminus of a DNA or RNA strand, the 3’ hydroxyl group has no attached nucleotide residue. At the 5’-terminus, the 5’ phosphate group has no attached nucleotide residue.
13.31 Is each of the strands for the DNA in Figure 13.10 an oligonucleotide or polynucleotide? Oligonucleotide. Each DNA strand in Figure 13.10 consists of 10 nucleotide residues which makes each one an oligonucleotide.
13.33 RNA consists of ribonucleotide residues connected to one another by what type of bond? Phosphodiester bonds. 13 - 6
13.35 The term primary structure refers to the sequence of what in DNA? The term primary structure refers to the sequence of nucleotide residues in DNA.
13.37 a. Draw the complete structure of the DNA dinucleotide dGC. The dinucleotide dGC consists of guanosine and cytidine groups bonded by a phosphoester bond between the 3' carbon of the guanosine and the 5' carbon of cytidine. See figure shown. b. Label the 3' and 5' ends of the molecule.
O 5' end
N
O-
- O P O CH2 O
H
NH
N
O H
H
O
H
N H
guanine
NH2 NH2 N
- O P O CH2 O
N
O
H
H
H
3' end
OH
H
cytosine O
H
The 5' end is the free phosphate group at carbon 5 of the guanosine residue and the 3' end is the free OH group at the 3' carbon of the cytidine residue.
13.39 a. What does the term base pairing mean? The term base pairing refers to the hydrogen bonding that occurs between two complementary bases attached to the two sugar-phosphate backbones of a double-stranded DNA molecule. b. In DNA, which bases are complementary to one another? In DNA, cytosine (C) is complementary to guanine (G) and thymine (T) is complementary to adenine (A).
13 - 7
13.41 Using the P-S-P backbone representation of DNA shown in Figure 13.8b, draw the dinucleotide that is complementary to dGC and label the 3’ and 5’ ends. dGC has a guanosine residue at the 5’ end cytidine residue at the 3’ end. For an antiparallel pairing, a dinucleotide complementary to this would also have a cytidine residue at the 3’ end and a guanosine residue at the 5’ end:
complementary dinucleotide
13.43 Sometimes, not all of the bases in a region of double stranded DNA or RNA are paired with their complementary base. This often has little effect on the secondary and tertiary structure. Find an example of this base pairing mismatch in the tRNA molecule shown in Figure 13.18.
base pairing mismatch (G-U) in tRNA
13 - 8
13.45 a. What are histones? Histones are proteins that interact with the phosphate groups of the DNA backbone. b. What are nucleosomes? Nucleosomes are groups of histones wrapped by DNA. c. What is chromatin? A chromatin is a coiled string of nucleosomes.
13.47 When DNA is denatured, which of the following is disrupted? a. primary structure b. secondary structure c. tertiary structure b. and c. only. When DNA is denatured, only the secondary (double-stranded helix held together by hydrogen bonds) and tertiary (overall three-dimensional shape including secondary structure) structures are disrupted. The primary structure is not affected by denaturing.
13.49 What molecule is made during the following processes? a. DNA replication New DNA molecules b. transcription RNA molecules c. translation Protein molecules d. reverse transcription DNA molecules
13.51 Explain the term semiconservative replication. 13 - 9
In semiconservative replication, a double-stranded DNA is synthesized from a daughter DNA strand and a parent DNA strand. The inclusion of a parent DNA as one of the two strands ensures the correct passing on of information contained in the parent DNA.
13.53 What is a replication fork? A replication fork, found at each end of the origin, is where replication takes place.
13.55 a. In which direction along an existing DNA strand does DNA polymerase move? Polymerases move along an existing DNA strand in the 3' to 5' direction. b. In which direction does DNA polymerase synthesize a new DNA strand? Polymerases synthesize new DNA in the 5' to 3' direction. c. If an origin opens and a DNA polymerase attaches to each of the exposed single strands of DNA, do the polymerases move in the same direction or in opposite directions as they make new DNA? Opposite directions.
13.57 Name the three types of RNA and describe their function. Transfer RNAs carry the correct amino acid to the site of protein synthesis. Messenger RNAs carry the information that specifies which protein should be made. Ribosomal RNAs combine with proteins to form ribosomes, the multi-subunit complexes in which protein synthesis takes place.
13.59
a. In which direction does RNA polymerase move along a strand of DNA? From 3’ to 5’ on the DNA template strand b. In which direction is RNA synthesized? 5’ to 3’ 13 - 10
13.61 Enzymes called aminoacyl tRNA synthetases catalyze the combination of amino acids with the proper tRNAs. The first step in this process is the reaction of an amino acid with ATP to form an aminoacyl-AMP. An aminoacyl-AMP consists of an amino acid residue attached through its carboxyl group to the 5’ phosphate group of AMP (Figure 13.6) by an anhydride bond. Draw the aminoacyl-AMP that involves the amino acid alanine.
NH2 N O
+ NH3 CH C O CH3
N
O P
-
O CH2
O
O
N
H
HH
OH
OH
H
N
13.63 The sequence dGGCAT appears in a template strand of DNA. a. Which base is at the 3'-terminus of this primary structure? Thymine. Unless otherwise noted, a sequence of bases is listed from the 5’-terminus to the 3’-terminus. Therefore, the base at the 3’-terminus is thymine (T). b. What is the sequence in the DNA strand that is complementary to this sequence? (Label the 3' and 5' ends.) 3'-terminus: dCCGTA : 5'-terminus A and T are complementary bases, as are G and C. The complementary strand will run from 3’ to 5’. c. What is the sequence in the RNA strand that is synthesized from dGGCAT? (Label the 3' and 5' ends.) 3' terminus: CCGUA : 5' terminus In the new RNA strand, U pairs with A in the DNA template and A pairs with T in the DNA template. G and C are complementary bases. The new RNA strand will run from 3’ to 5’.
13.65 a. What are codons?
13 - 11
A codon is a series of three bases carried by mRNA that specifies a particular amino acid in the genetic code. b. Which type of RNA has codons? Messenger RNA.
13.67 Which amino acid is specified by each codon (listed 5' to 3')? See Table 13.1. Find the beginning code letter on the left column. Locate the last code letter in the far right column. This establishes the row. Find the middle code letter at the top of the table and follow that column down to the row in line with your other two letters and locate the amino acid. a. CCU Pro b. AGU Ser c. GUU Val d. GAA Glu
13.69 Which codon(s) specify each amino acid? Find the amino acid on Table 13.1 in the middle of the chart. The column in which it is located establishes the middle code letter. The left and right code letters are found on the same row in the left and right columns, respectively. Every row that contains the amino acid gives a different codon for it. a. Phe UUU and UUC b. Lys AAA and AAG 13 - 12
c. Asp GAU and GAC
13.71 Which anticodons are complementary to the codons in Problem 13.67? a. b. c. d.
13.73
GGA (listed 3' to 5') UCA (listed 3' to 5') CAA (listed 3' to 5') CUU (listed 3' to 5')
Name the components that combine with one another during initiation of translation. An mRNA and a tRNA combine with ribosome subunits.
13.75 A tripeptide has the primary structure Ser-Lys-Asp. a. Assuming that no post-translational modification took place, what is a sequence of bases in the mRNA that would code for this tripeptide? (Ignore start and stop codons and label the 3' and 5' ends.) Ser has the following codons: UCU, UCC, UCA, UCG, AGU, AGC Lys has the following codons: AAA, AAG Asp has the following codons: GAU, GAC A total of 24 sequences of bases in mRNA would code for Ser-Lys-Asp. They are shown below in the left column. b. Assuming that no post-transcriptional modification took place, what is the sequence of bases in the template DNA strand used to make the mRNA? (Label the 3' and 5' ends.) The corresponding sequence of template DNA strand is shown on the right column below. Any one of them would satisfy the requirement of writing a sequence. (a) Possible sequences of mRNA bases (5' end) Ser – Lys – Asp UCUAAAGAU UCCAAAGAU UCAAAAGAU UCGAAAGAU AGUAAAGAU
(3' end)
(b) Possible sequences of DNA bases (3' end)
(5' end) AGATTTCTA AGGTTTCTA AGTTTTCTA AGCTTTCTA UCATTTCTA
13 - 13
(a) Possible sequences of mRNA bases (5' end) Ser – Lys – Asp AGCAAAGAU
(3' end)
(b) Possible sequences of DNA bases (3' end)
(5' end) UCGTTTCTA
UCUAAGGAU UCCAAGGAU UCAAAGGAU UCGAAGGAU AGUAAGGAU AGCAAGGAU
AGATTCCTA AGGTTCCTA AGTTTCCTA AGCTTCCTA TCATTCCTA TCGTTCCTA
UCUAAAGAC UCCAAAGAC UCAAAAGAC UCGAAAGAC AGUAAAGAC AGCAAAGAC
AGUTTTCTG AGGTTTCTG AGTTTTCTG AGCTTTCTG TCATTTCTG TCGTTTCTG
UCUAAGGAC UCCAAGGAC UCAAAGGAC UCGAAGGAC AGUAAGGAC AGCAAGGAC
AGATTCCTG AGGTTCCTG AGTTTCCTG AGCTTCCTG TCATTCCTG TCGTTCCTG
13.77 What is an operon? An operon is a group of genes whose expression is controlled by one promoter site. The transcription of this group of genes is initiated at the promoter site.
13.79 Explain how allolactose is able to influence transcription of the lacZ, lacY, and lacA genes. Allolactose is a modified form of lactose. Its formation is catalyzed by a small amount of -galactosidase in the cells when lactose is present. Allolactose binds to the repressor protein that controls the transcription of the lacZ, lacY, and lacA genes. Upon binding, allolactose modifies the shape of the repressor protein so that it is no longer able to attach to and repress the operator sites. Once the repressor protein is released from the operator sites, RNA polymerase can bind and begin the transcription of the three lac genes.
13.81
Define the term mutation. 13 - 14
A mutation is any permanent modification of the primary structure of DNA.
13.83
Which of the levels of protein structure (primary, secondary, tertiary, quaternary) can be affected by a mutation? A mutation directly affects primary structure. A change in primary structure can affect secondary, tertiary, and quaternary structure.
13.85 Describe the general procedure used to make recombinant DNA and include the role of restriction enzymes, plasmids, and ligases. Recombinant DNA is formed by combining DNA segments from two different sources. A common type of recombinant DNA synthesis is the insertion of human DNA segments into bacterial plasmids. In this technique, the DNA segment of interest is cleaved from one source using a restriction enzyme specific to the isolation of the segment of interest. The same restriction enzyme can be used to cleave the bacterial plasmid at the appropriate site. Using ligases, these two segments attach to each other to form the recombinant DNA. 13.87 What is “sense” RNA and “antisense” RNA? A sense RNA is an RNA that would normally be translated to produce a particular protein. An antisense RNA is a complementary RNA that is transcribed from a new gene inserted into a piece of DNA. The antisense RNA base-pairs with the sense RNA locking it into a double-stranded RNA and preventing translation of the sense RNA and production of the protein.
13.89 What are short tandem repeats and how are they used in DNA fingerprinting? Short tandem repeats are relatively small stretches of DNA that contain short repeating sequences of bases. Restriction enzymes are used to cut the DNA into small segments and the number of base repeats of the STRs is determined.
13.91 Beginning with one double strand of DNA, how many double-stranded DNAs will be present after 15 cycles of PCR? 32,768 double-stranded DNA’s Each PCR cycle doubles the number of double-stranded DNA’s. Therefore, after 15 cycles the number will be 215 = 32,768.
13 - 15
13.93 What is a snRNP and what function does it have? snRNPs are small nuclear ribonucleoproteins that consist of RNA and proteins. They function in a form of post-transcriptional modification called RNA splicing.
13.95 Describe the role of each in RNA interference. a. double-stranded RNA RNA strands fold back on themselves to produce double-stranded RNA. This is the first step in RNA interference. The double-stranded DNA is the source of small interfering RNA (siRNA). b. nuclease Nuclease cuts double-stranded RNA into 21-23 base pair fragments called small interfering RNA. c. small interfering RNA Small interfering RNA’s, the 21-23 base pair fragments created by nuclease, are the source of guide strands. d. guide strand A guide strand, one strand of each small interfering RNA, is complementary to the mRNA targeted for destruction. e. RISC RISC (RNA-Induced Silencing Complex) binds a guide strand to itself. When complementary mRNA binds to the guide strand, RISC cuts the mRNA, preventing it from undergoing translation and silencing the gene that produced the mRNA.
13 - 16
13.97 Why did scientists studying a disease of cats insert a jellyfish gene into cat eggs? The jellyfish gene inserted into unfertilized cat eggs codes for the production of a protein that glows green when exposed to blue light. The gene acted as a marker that allowed the scientists to detect that gene transfer has occurred. 13.99 a. Draw and name the 5’ deoxyribonucleotide formed by combining phosphate, β-2-deoxyribofuranose, and adenine.
NH2 phosphoester bond
N
O-
- O P O CH O 2 O
N
H
HH
OH
H
H
N N N-glycosidic bond
deoxyadenosine 5'-monophosphate b. For this nucleotide, name the bond that joins the phosphate and deoxyribose residues and the bond that joins the deoxyribose and adenine residues. See figure in part a. c. Draw the dinucleotide formed by combining two of the nucleotides from part a. Label the 5’ and 3’ end.
13 - 17
NH2 5' end
N
O-
-O P O CH O 2 O
N
H
HH
O
H
H
N N NH2 N
- O P O CH O 2 O
N
H
HH
OH
H
H
N N
3' end d. For this dinucleotide, name the bond that joins the two nucleotide residues. Phosphodiester bond. e. Suppose that a trinucleotide is formed by adding a third nucleotide from part a to the dinucleotide from part c, and that this trinucleotide is incorporated into a DNA template strand that codes for mRNA. What codon will be transcribed from this trinucleotide residue? UUU (listed 5’-3’) f. Which amino acid is specified by the codon in part e? Phenylalanine g. What is anticodon of the tRNA that carries the amino acid in part f? AAA (listed 3’-5’) h. If a mutation changes base residue at the 5’ end of the trinucleotide in part e from adenine to guanine, which new codon will be transcribed? UUC (listed 5’-3’) i. Which amino acid is specified by the codon in part h? Phenylalanine 13 - 18
j. What is the anticodon of the tRNA that carries the amino acid in part i? (listed 3’-5’) AAG
ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS
13.2
The picture below shows a ribosome attached to a strand of mRNA.
a. Which reaction is catalyzed by the ribosome? b. What is the name of this process? c. Which building blocks are used to form the final product? d. In which direction does the ribosome move along the mRNA? e. To which end of the growing product chain are building blocks attached? Answer: a. Peptide bond formation. b. The formation of the peptide bond occurs during the elongation stage of the translation (protein synthesis) process. c. Amino acids are bonded together to form the growing polypeptide chain. d. From the 5’ to 3’ along the mRNA. e. Each amino acid is attached to the end opposite the first amino acid (fMet)
SECTION 13.1
13.4
NUCLEIC ACID BUILDING BLOCKS
a. Which monosaccharide is used to make RNA? b. Draw this monosaccharide in its β-furanose form.
13 - 19
Answer: a. Ribose. The R in RNA is for “ribo” indicating that the monosaccharide is ribose. b. Draw the five-membered ribose ring structure with the hemiacetal –OH group pointing upward.
HOH2C
OH
O H
H H
H OH
13.6
OH
Name the four bases that are present in RNA.
Answer: Adenine, guanine, cytosine, and uracil.
13.8
Which of the bases present in nucleotides are pyrimidines?
Answer: Cytosine, thymine, and uracil.
13.10 Figure 13.1 shows the phosphorus atom in a phosphate ion as having five covalent bonds. Draw a phosphate ion in which the phosphorus atom has an octet of valence electrons. Answer:
O O
P
O
O
13.12 Draw the phosphate diester in Figure 13.1 as it would appear at low pH.
13 - 20
Answer:
O CH3 O
P
OH
CH3 O
13.14 a. How many different phosphate monoesters can β-D-2-deoxyribofuranose form? b. When reacted with NH2CH3, how many different N-glycosides can β-D-2deoxyribofuranose form? Answer: a. Three. Since there are 3 –OH groups, 3 esters can form. b. One. Only one N-glycoside can form because there is only one hemiacetal group that can react with an amine group.
13.16 a. What structural feature makes a base found in nucleic acids a pyrimidine? b. Of the bases adenine, thymine, guanine, cytosine, and uracil, which are pyrimidines? Answer: a. A pyrimidine is a base with the structure:
N N
b. Thymine, cytosine, and uracil are pyrimidines.
13.18 Draw the complete structure of thymidine 5'-monophosphate (see Figure 13.6).
13 - 21
Answer:
-
SECTION 13.2
NUCLEOSIDE DI- AND TRIPHOSPHATES, CYCLIC NUCLEOTIDES
13.20 How many phosphoester bonds and how many phosphoanhydride bonds are present in the following? a. a nucleoside triphosphate b. a cyclic nucleotide Answer: a.
2 phosphoanhydride bonds 1 phosphoester bond O-
O-
O-
-O P O P O P O CH2 O
O
O
H
base
O H
H
OH
OH
b. 2 phosphoester bonds and no phosphoanhydride bond
13 - 22
H
CH2 O
2 phosphoester bonds
-O
base
O
H
P
H
H
O
OH
H
O
13.22 a. Draw 5'-dGTP b. Draw 5'-GTP Answer: a. The structure of 5'-dGTP, deoxyguanosine 5'-triphosphate, is shown below:
O O-
O-
N
O-
-O P O P O P O CH2 O
O
O
N
O H
H
OH
H
H
NH N
NH2
H
b. The structure of 5'-GTP, guanosine 5'-triphosphate, is shown below:
O O-
O-
N
O-
-O P O P O P O CH O
O
O
H
N
O
2
NH
H
H
OH
OH
N
NH2
H
13.24 a. Name the types of bonds broken when cAMP is completely hydrolyzed. b. What products are obtained when cAMP is completely hydrolyzed?
13 - 23
Answer: a. Phosphoester and N-glycoside. b. 1 phosphate, D-ribose, and adenine.
13.26 a. Name the types of bonds broken when the molecule in Problem 13.22a is completely hydrolyzed. b. What products are obtained when this molecule is completely hydrolyzed? Answer: a. Phosphoanhydride, phosphoester, and N-glycosidic bonds are broken when the 5'-dGTP molecule is completely hydrolyzed. b. Three phosphate ions, a -D-2-deoxyribofuranose molecule, and a guanine molecule are obtained when the 5’-dGTP molecule is completely hydrolyzed
SECTION 13.3
POLYNUCLEOTIDES
13.28 In the sugar-phosphate backbone of RNA a. which sugar residue is present? b. which type of bonds connect the sugar and phosphate residues? c. where are bases attached? Answer: a. Ribose b. Phosphoester bonds c. Carbon 1 of each ribose residue 13.30 How does the 5’ end of DNA or RNA typically differ from the 3’ end? Answer: For the RNA structure, the 3' terminus refers to the end of the molecule that has a free –OH group at carbon 3 of the ribose residue and the 5' terminus refers to the end of the molecule that has a phosphate residue at carbon 5 of the ribose residue.
13 - 24
13.32 Is the tRNA in Figure 13.18 an oligonucleotide or a polynucleotide? Answer: Polynucleotide. The t-RNA contains more than 8-10 nucleotide residues.
13.34 RNA consists of a sugar-phosphate backbone to which the bases A, G, C, and U are attached by what type of bond? Answer: N-glycosidic bond.
SECTION 13.4
DNA STRUCTURE
13.36 Describe the secondary and tertiary structure of DNA. Answer: The secondary structure of DNA consists of antiparallel double strands twisted into a helix. The tertiary structure of DNA involves supercoiling of the double helix.
13.38 a. Draw the complete structure of the DNA trinucleotide dTAT. b. Label the 3' and 5' ends of the molecule. Answer: a. and b. See the following structure.
13 - 25
O 5' end
H3C
NH
O O P O CH2 O
H
O
N
H
H
O
H
O H
NH2 N
O P O CH2 O
thymine
O
N
H
H
O
H
H
N N
H
O
H H3C
O P O CH2 O
adenine
NH N
O H
H
OH
H
thymine
O H
3' end
13.40 Double-stranded DNA is antiparallel. Explain this term. Answer: Antiparallel means that when the two strands of the double-stranded DNA are placed side-by-side, one strand will have a sequence running from the 5' to the 3' end and the other strand will have a sequence going from the 3' to the 5' end.
13.42 Using the P-S-P backbone representation of DNA shown in Figure 13.8b, draw the trinucleotide that is complementary to dTAT and label the 3’ and 5’ ends. Answer:
5'
3'
13.44 A palindrome is a word or phrase that reads the same forward and backward. “Wow” and “never odd or even” are examples. Double stranded DNA sometimes contains regions of palindrome, where the sequence of one strand is the reverse of 13 - 26
its complementary strand. In the DNA palindrome shown below, each strand has the primary structure dAGCT.
Find the regions of palindromic double-stranded DNA in Figures 13.10 and 13.22. Answer: In Figure 13.10, the palindromic region has the following complementary pairing of bases. 5’ – CATGGCCATG – 3’ 3’ – GTACCGGTAC – 5’
In Figure 13.22, the palindromic region has the following complementary pairing of bases. 5’ – GAATTC – 3’ 3’ – CTTAAG – 5’
13.46 a. Ionic bonds play a role in the interaction between DNA and histones. Amino acid side chains in histones must carry what net charge? b. Are the histone amino acid residues referred to in part a polar-acidic or are they polar-basic? Answer: a. A net charge of 1+. The amino acid side chains in histones must be positively charged because they interact with the negatively charged phosphate groups of the DNA backbone. b. The amino acid residues in histones are polar-basic. Basic sites have a net positive charge when they have accepted a H+.
SECTION 13.5
DENATURATION
13.48 What does it mean to renature DNA? 13 - 27
Answer: To renature DNA means to allow the DNA to return to its original double stranded formation.
SECTION 13.6
NUCLEIC ACIDS AND INFORMATION FLOW
13.50 The primary structure of what molecule is read during the following processes? a. DNA replication b. transcription c. translation d. reverse transcription Answer: a. DNA. b. DNA. c. mRNA. d. viral RNA.
SECTION 13.7
DNA REPLICATION
13.52 a. What is an origin? b. During DNA replication in plants and animals does just one origin form? Answer: a. A site on DNA where the double strand has been pulled apart to expose single strands. DNA replication takes place at an origin. b. No, during replication in plants and animals, many origins are opened at one time along a double-stranded DNA helix.
13.54 Which nucleoside triphosphates are substrates for DNA polymerase?
13 - 28
Answer: DNA polymerase is the enzyme that catalyzes the synthesis of DNA from nucleotides. The substrates for this enzyme are the nucleotides that form the DNA polymer: dATP, dGTP, dCTP, and dTTP.
13.56 In addition to adding nucleotides to a growing DNA strand, what is another important function of DNA polymerases? Answer: Another function of DNA polymerases is to proofread the newly synthesized DNA strand and correct any wrong addition by removing the wrong nucleotide and inserting the correct one.
SECTION 13.8
TRANSCRIPTION AND RNA
13.58 Which nucleoside triphosphates are substrates for RNA polymerase? Answer: The substrates for RNA polymerase are ATP, GTP, CTP, and UTP.
13.60 a. What types of post-transcriptional modifications do RNAs undergo? b. How are rRNAs and zymogens similar? Answer: a. Transfer RNA is shortened and some of the major bases are converted into minor bases. Messenger RNA may be trimmed at either end, the 3’ and 5’ ends may be altered, and sections of the middle may be removed. Ribosomal RNA is cut into smaller pieces. b. rRNAs and zymogens are molecules that are both inactive until a modification takes place. Zymogens are enzyme precursors that are activated by the breaking of one or more specific peptide bonds in a protein. rRNAs are activated when ribosomal RNA is clipped into smaller pieces during post-transcriptional modification.
13.62 The second of the reactions catalyzed by aminoacyl tRNA synthetases (Problem 13.61) is reaction of an aminoacyl-AMP with a tRNA molecule to form an ester. 13 - 29
The ester consists of a carboxylic acid residue (supplied by the amino acid) and an alcohol residue (supplied by an –OH group attached to the ribose on the 3'terminus of the tRNA). a. Class 1 aminoacyl tRNA synthetases attach the amino acid to the 2' –OH group of the terminal nucleotide residue. Draw the ester formed when alanine is attached to the 2' –OH group at the 3'-terminus of the tRNA below.
remainder of tRNA O Adenine O P O CH2 O -O H HH H OH OH 3'-Terminus b. Class 2 aminoacyl tRNA synthetases attach the amino acid to the 3' –OH group of the terminal nucleotide residue. Draw the ester formed when serine is attached to the 3' –OH group at the 3'-terminus of the tRNA above. Answer:
a.
remainder of tRNA O Adenine O P O CH2 O -O H HH H OH O C O CH CH3 +
NH3
13 - 30
b.
remainder of tRNA O Adenine O P O CH2 O -O H HH H O OH C O CH CH2OH +
NH3
13.64 The sequence dATCCCC appears in a template strand of DNA. a. Which base is at the 3'-terminus of this primary structure? b. What is the sequence in the DNA strand that is complementary to this sequence? (Label the 3' and 5' ends.) c. What is the sequence in the RNA strand that is synthesized from dATCCCC? (Label the 3' and 5' ends.) Answer: a. The primary structure of DNA is typically listed from the 5' to the 3' end. C is typically at the 3' end. b. 3'- dTAGGGG -5' c. 3'- dUAGGGG -5'
13.66 a. What are anticodons? b. Which type of RNA has anticodons? Answer: a. An anticodon consists of a triplet of bases that is complementary to a codon. b. Transfer RNA (Trna) has anticodons.
13.68 Which amino acid is specified by each codon (listed 5' to 3')? a. UUU b. CCC c. AAA 13 - 31
d. GGG Answer: Refer to Table 13.1 to determine which amino acid is specified by each codon. a. Phe b.Pro c. Lys d.Gly
13.70 Which codon(s) specify each amino acid? a. Asn b. Tyr c. Gly Answer: Read from 5' to 3': a. AAU and AAC b.UAU and UAC c. GGU , GGC, GGA, and GGG
13.72 Which anticodons are complementary to the codons in Problem 13.68? Answer: a. UUU
AAA (listed 3' to 5')
b. CCC
GGG (listed 3' to 5')
c. AAA
UUU (listed 3' to 5')
d. GGG
CCC (listed 3' to 5')
13 - 32
SECTION 13.9
TRANSLATION
13.74 Describe what happens during elongation. Answer: During elongation, amino acids are joined by peptide bonds to the growing polypeptide chain. 13.76 a. For the tripeptide in Problem 13.75, again assuming that no post-translational modification took place, what is a different sequence of bases in the mRNA that would code for this tripeptide? (Ignore start and stop codons and label the 3' and 5' ends.) b. Assuming that no post-transcriptional modification took place, what is the sequence of bases in the template DNA strand used to make the mRNA? (Label the 3’ and 5’ ends.) Answer: a. The tripeptide Ser-Lys-Asp can be represented by several different code sequences. Ser has the following codons: UCC, UCG, UCA, UCU, AGU, AGC; Lys has the following codons: AAA, AAG; and Asp has the following codons: GAC, GAU. Recall that a tripeptide has many different possible combinations of coding anyone of which could be used to answer this question. The 24 possible combinations for Ser-Lys-Asp are shown in the table after part b. Example: (listed 5’ to 3’) UCU-AAG-GAC
13 - 33
b. The DNA sequences are listed in corresponding order to the answers for part a. a. Possible sequence of mRNA bases (5' end)
b. Possible sequence of DNA bases
(3' end)
(3' end)
(5' end)
UCUAAAGAU UCCAAAGAU UCAAAAGAU UCGAAAGAU AGUAAAGAU AGCAAAGAU
dAGATTTCTA dAGGTTTCTA dAGTTTTCTA dAGCTTTCTA dUCATTTCTA dUCGTTTCTA
UCUAAGGAU UCCAAGGAU UCAAAGGAU UCGAAGGAU AGUAAGGAU AGCAAGGAU
dAGATTCCTA dAGGTTCCTA dAGTTTCCTA dAGCTTCCTA dTCATTCCTA dTCGTTCCTA
UCUAAAGAC UCCAAAGAC UCAAAAGAC UCGAAAGAC AGUAAAGAC AGCAAAGAC
dAGUTTTCTG dAGGTTTCTG dAGTTTTCTG dAGCTTTCTG dTCATTTCTG dTCGTTTCTG
UCUAAGGAC UCCAAGGAC UCAAAGGAC UCGAAGGAC AGUAAGGAC AGCAAGGAC
dAGATTCCTG dAGGTTCCTG dAGTTTCCTG dAGCTTCCTG dTCATTCCTG dTCGTTCCTG
SECTION 13.10
CONTROL OF GENE EXPRESSION
13.78 How does the presence of lactose act as an “on” switch for the lac operon? Answer: Lactose is converted to allolactose by β-galactosidase. Allolactose binds to the repressor protein and changes its shape in such a way that it can no longer attach to the operator sites. This allows the RNA polymerase to bind to the promoter site turning “on” transcription. . 13 - 34
13.80 a. Draw lactose and allolactose. b. What products are obtained when allolactose is hydrolyzed? Answer: a. The structures for lactose and allolactose are shown in the following drawings.
HO CH2 O
H HO CH2 O
HO
H
H
OH
H
H
OH
OH
H
O
OH
H
H
OH
H
H lactose
HOCH2 HO H
O
O H OH
H
CH2
H
O
H H
OH HO
H OH
H
H
OH
allolactose b. One galactose and one glucose.
SECTION 13.11
MUTATION
13.82 a. Are all mutations harmful to an individual? Explain. b. Are all mutations passed on to offspring? Explain.
13 - 35
OH H
Answer: a. No. Mutations that take place outside of genes will have little or no effect on gene expression. Because in many cases more than one codon specifies the same amino acid residue, some mutations have no effect on the protein specified by a gene. b. No. Only mutations that occur in the egg or sperm cells can be passed on to offspring. 13.84 Could a person’s health be negatively impacted by a mutation that produces DNA polymerase with a lower than normal effectiveness at proofreading and repair? Explain. Answer: Possibly. The error rate during replication may increase if the DNA polymerase is less effective than normal but if these errors occur outside of genes, there is little or no effect on gene expression that may have a negative impact on health. In addition, if the DNA with lower than normal effectiveness at proofreading and repair does not catch an error, there are other DNA repair enzymes that are likely to catch the mistakes.
SECTION 13.12
RECOMBINANT DNA
13.86 The building blocks used to construct human DNA are identical to those used to construct goat DNA. Why, then, are humans and goats different? Answer: DNA determines the identity of a living thing by specifying which enzymes and other proteins are synthesized. Human DNA and goat DNA are constructed from the same building blocks but contain different genes that control what enzymes and other proteins are made.
13.88 Although each involves double-stranded RNA, RNA interference is different than the use of antisense mRNA (see Health Link RNA Interference). Explain. Answer: In RNA interference, the mRNA transcribed from a particular gene is destroyed, which prevents the gene product (protein) from being produced. In the use of antisense mRNA, a gene is inserted into DNA. It produces the antisense strand 13 - 36
that base-pairs with the sense RNA locking it into a double-stranded RNA and preventing its translation into protein.
SECTION 13.13
DNA FINGERPRINTING
13.90 If a person’s DNA is “fingerprinted” and the STR information is saved in a database, should he/she be concerned that information regarding genetic diseases that he/she might carry will be released to insurance companies? Explain. Answer: A DNA fingerprint only contains information about short tandem repeats (STRs) which are found in the noncoding portion of the DNA. STR analysis does not contain information about genes and does not convey information that relates to genetic diseases.
13.92 Which primer would be appropriate if you wanted to initiate copying of the single strand of DNA that has the primary structure dCCTAGGCGAATCCG? Recall that the primary structure of nucleotides is listed 5' to 3'. a. dCCTAG b. dGGATC c. dCGGAT d. dGGCGC e. dGCCTA Answer: b. dGGATC. This primer is complementary to the first five base pairs in the primary structure.
HEALTH LINK
13.94
LUPUS
How might an immune response against snRNPs be linked to disease?
Answer: An immune response against snRNPs causes the formation of antibodies that may lead to loss of or inactivation of the splicing function of snRNPs. Loss of the splicing function may affect post-transcriptional modification producing defective proteins that may cause disease. 13 - 37
HEALTH LINK
RNA INTERFERENCE
13.96 Using RNA interference to treat disease requires synthesizing the appropriate small interfering RNA. a. True or False? Knowing the primary structure of the mRNA targeted for destruction would allow the appropriate small interfering RNA to be produced. b. True or False? Knowing the primary structure of the gene that codes for the mRNA targeted for destruction would allow the appropriate small interfering RNA to be produced. c. True or False? Knowing the primary structure of the native protein coded for by the mRNA targeted for destruction would allow the appropriate small interfering RNA to be produced. Answer: a. True. One strand of small interfering RNA will be complementary to the mRNA. b. False. mRNA undergoes post-transcriptional modifications, so the mRNA may not exactly reflect the primary structure of the gene that codes for it. c. False. Proteins undergo post-translational modifications, so the native protein may not exactly reflect the primary structure of the mRNA from which it is produced.
HEALTH LINK
GLOWING CATS
13.98 In addition to FIV research, what other uses have scientists found for the GFP gene? Answer: The GFP gene has also been used as a marker in stem cell research in pigs. It also has been used to detect arsenic and TNT through insertion of the gene in bacteria causing them to glow when exposed to these substances.
13 - 38
LEARNING GROUP PROBLEMS
13.100 a. A portion of the RNA in a retrovirus is shown below. Draw the new double stranded DNA that will be produced if the viral RNA is inserted into a cell and reverse transcription takes place.
b. Suppose that the strand of new DNA that is complementary to the RNA in part a codes for two amino acids. If this DNA strand is transcribed into mRNA, which two codons will be produced? c. Which amino acids are specified by the codon in part b? d. What anticodons will be present in the tRNAs that carry the amino acids in part c? e. An N-glycosidic bond connects which residues in RNA? f. A 3’, 5’-phosphodiester bond connects which two residues in RNA? g. If the RNA in part a is completely hydrolyzed, which three classes of compounds will be produced? Answer: a. The two strands of the new double stranded DNA will have the following sequence of bases. 3’ – GAATCG – 5’ 5’ – CTTAGC – 3’ b. New DNA strand complementary to the RNA in part a mRNA complementary to this new strand Two codons produced
3’ – GAATCG – 5’ 5’ – CUUAGC – 3’ 5’ – CUU – 3’
5’ – AGC – 3’
c. Two codons produced
5’ – CUU – 3’
5’ – AGC – 3’
Amino acids specified
Leu
Ser
Two codons produced
5’ – CUU – 3’
5’ – AGC – 3’
Anticodons
3’ – GAA – 5’
3’ – UCG – 5’
d.
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e. Base and monosaccharide residues. f. Two monosaccharide residues. g. Organic base, monosaccharide, and phosphate ions.
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Chapter 14 Metabolism SOLUTIONS TO ODD-NUMBERED END OF CHAPTER PROBLEMS (SOLUTIONS TO EVEN-NUMBERED PROBLEMS FOLLOW BELOW)
14.1
In the accompanying drawing, indicate where each of the following takes place. a. a 6-carbon molecule is converted into two 3-carbon compounds. b. a 3-carbon compound loses one of its carbon atoms at the same time that it is activated. c. a 4-carbon compound is converted into 6-carbon compound, which is then broken down into the same 4-carbon compound.
a. b.
c.
14.3
Define the term metabolic pathway. A metabolic pathway is a group of biochemical reactions.
14.5
In the first step of glycolysis, the following two reactions are coupled Glucose + Pi → glucose 6-phophate + H2O
ΔG = +3.3 kcal
ATP + H2O → ADP + Pi
ΔG = -7.3 kcal
a. Is the reaction Glucose + Pi → glucose 6-phosphate + H2O spontaneous?
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No. A positive ΔG (ΔG = +3.3 kcal) indicates the reaction is nonspontaneous. b. Write the net reaction equation and calculate ΔG for the coupled reaction. The overall reaction is the sum of both reactions. Write ALL reactants on the left and ALL products on the right. Cancel out the reactants and products that are the same on both sides. Write the remaining reactants and products as the overall reaction. Add the ΔG values, +3.3 kcal + (-7.3 kcal) = -4.0 kcal. glucose + H2O + ATP + Pi → glucose 6-phosphate + ADP + Pi + H2O glucose + ATP →
glucose 6-phosphate + ADP
G = -4.0 kcal
c. Is the first step in glycolysis spontaneous? Yes. A negative ΔG (ΔG = -4.0 kcal) indicates the first step in glycolysis is spontaneous.
14.7
Suppose that an uncatalyzed reaction is spontaneous because G has a value of -10 kcal/mol. An enzyme that catalyzes the reaction is identified. What effect will the enzyme have on the rate of the reaction? A catalyst lowers the activation energy. A lower activation energy increases the rate of reaction. The catalyst does not affect the G value, however.
14.9
What is metabolism? Metabolism is the sum of all reactions that take place in a living thing.
14.11 What provides the energy used to produce ATP from ADP and Pi, anabolism or catabolism? Catabolism
14.13 Define the term reduction and explain how it applies to the difference in the structures of NAD+ and NADH. Reduction is the gain of electrons. In organic and biochemical molecules, a gain of hydrogen and /or a loss of oxygen is indication that reduction has taken place. NADH, the reduced form of the coenzyme has one more hydrogen atom than NAD+, the oxidized form of the coenzyme.
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14.15 Describe the role of the oxidized and reduced forms of NAD+ and FAD in metabolism. In the oxidation reactions that take place in catabolism, the oxidized coenzymes NAD+ and FAD are converted to their reduced forms, NADH, and FADH2 by accepting hydrogen atoms. The energy of these reduced forms can be used to produce ATP and the hydrogen atoms can be used during anabolism.
14.17 The abbreviation Pi is used to represent phosphate. a. Draw phosphate ion. The phosphate ion has the following structure:
O
-O P O Ob. Depending on the pH of a solution, phosphate ion can exist in three other forms. Draw and name them. The three other forms of phosphate ion vary according to the number of hydrogen atoms attached to the oxygen: O HO
P
O-
O HO
P
O-
O HO
P
OH
O-
OH
OH
hydrogen phosphate ion
dihydrogen phosphate ion
phosphoric acid
14.19 Write a reaction equation for the hydrolysis of GTP to produce Pi. GTP + H2O → GDP + Pi
14.21 What products are formed when each of the following undergoes digestion and is broken down to its building blocks? a. a polysaccharide
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Monosaccharides are formed when a polysaccharide is broken down during digestion. b. a triglyceride Fatty acids, glycerol, and monoacylglycerides are produced when triglycerides undergo digestion. c. a protein Amino acids are produced when a protein is broken down to its building blocks.
14.23 Name the functional group hydrolyzed when each undergoes digestion. a. a polysaccharide Glycosidic bonds are hydrolyzed. b. a triglyceride Ester functional groups are hydrolyzed. c. a protein Amide functional groups are hydrolyzed.
14.25 a. How do the structures of amylose and amylopectin, the two homopolysaccharides that make up starch, differ? Amylopectin contains α-(1 → 6) glycosidic bonds. Amylose does not. b. How are the structures of amylose and amylopectin, the two homopolysaccharides that make up starch, similar? Both amylose and amylopectin contain glucose units connected by (1→4) glycosidic bonds.
14.27 Name three common disaccharides and name the monosaccharide(s) formed when they are hydrolyzed in the small intestine. Upon hydrolysis, the following disaccharides will form monosaccharides as shown:
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Sucrose → glucose + fructose Maltose → 2 glucose Lactose → glucose + galactose
14.29 What is the role of bile acids in digestion? Bile acids are emulsifying agents in the digestion of triglycerides in the small intestine.
14.31 Studies have shown that, in newborns, pancreatic lipases are poor catalysts. Lingual lipase, whose secretion is stimulated by feeding, is active, however. Explain how the digestion of triglycerides by newborns differs from than that of adults. In newborns, the hydrolysis of triglycerides begins in the stomach using lingual lipase, whereas in adults the triglyceride digestion occurs mostly in the small intestine using pancreatic lipases.
14.33 What is the role of glycolysis? Glycolysis converts a glucose molecule into 2 pyruvate ions with accompanying production of 2 ATP and 2 NADH. b. What is the net change in ATP and NADH from the passage of one glucose molecule through glycolysis? 2 ATP molecules and 2 NADH molecules are produced from one glucose molecule that undergoes glycolysis.
14.35 a. What is the relationship between glucose 6-phosphate and fructose 6phosphate (step 2 of glycolysis)? Are they constitutional isomers, cis/trans isomers, stereoisomers, different conformations of the same compound, or identical compounds? Constitutional isomers. Glucose 6-phosphate and fructose 6-phosphate have the same chemical formulas but different atomic connections. b. Are these two compounds reducing sugars? Explain. Yes, the two compounds are reducing sugars. Glucose 6-phosphate opens up into an aldose form and fructose 6-phosphate opens up into a ketose form which is transformed into an aldose in the basic conditions of the Benedict’s reagent
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c. Draw each as it appears in its open form.
O
CH2OH
C H H
OH
HO
H
H
OH
O H
OH
C O HO
H
H
OH
O H
OH
- O P O CH2 O-
- O P O CH2 O-
open form of fructose 6-phosphate
open form of glucose 6-phosphate
14.37 a. In humans, pyruvate is converted into what compound under anaerobic conditions? Under anaerobic conditions, pyruvate is reduced to lactate in humans. b. What purpose does this reaction have? Reduction of pyruvate to lactate results in the oxidation of NADH under anaerobic conditions to produce the NAD+ required to continue glycolysis. The lactate is sent to the liver and used to make glucose.
14.39 What is the net change in ATP and NADH from the passage of one glucose molecule through glycolysis, followed by the conversion of pyruvates into lactates? 2 ATP molecules and 2 NADH molecules are produced when one glucose molecule goes through glycolysis and produces 2 pyruvate molecules. The conversion of the 2 pyruvate ions to 2 lactate ions uses up the 2 NADH formed in glycolysis. The net change is a gain of 2 ATP and no change in NADH.
14.41 Draw pyruvic acid, the acidic form of pyruvate. Which predominates at physiological pH, the acidic form or the conjugate base form of this acid?
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pyruvic acid Physiological pH (7.4) is a slightly basic condition. Under this pH, the conjugate base form of pyruvic acid will predominate:
pyruvate ion 14.43 a. Is the conversion of acetaldehyde to ethanol an oxidation or a reduction? Reduction. The conversion of acetaldehyde to ethanol is shown below. This reaction is a reduction reaction because the carbonyl group is reduced to a single bond to oxygen with an increase in bonds to hydrogen atoms.
OH
O
CH3 CH2
CH3 C H
ethanol
acetaldehyde
b. When this reaction takes place during alcoholic fermentation, what is the oxidizing agent? acetaldehyde c. What is the reducing agent? NADH
14.45 a. In what part of a cell does glycolysis take place? In the cytoplasm of the cell. b. In what part of a cell does the conversion of pyruvate into acetyl-CoA take place? In the mitochondria of the cell.
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14.47 Glycolysis can be described as a process in which energy is invested “up front” in exchange for a greater return of energy later on. Which steps in glycolysis involve investment of energy? Steps 1 and 3 involve “investment” or input of energy through coupling of the reaction with ATP hydrolysis.
14.49 Monosaccharides other than glucose are converted into compounds that are intermediates in glycolysis. Why is this more efficient than having a different catabolic pathway for the conversion of each monosaccharide into pyruvate? Each different pathway would require a different set of enzymes.
14.51 a. Define the term gluconeogenesis. Gluconeogenesis is a metabolic pathway that synthesizes glucose from noncarbohydrate sources such as amino acids, lactate, and glycerol. b. What is the role of this pathway? One role of this pathway is to recycle lactate into glucose. This pathway is the main source of glucose during fasting or starvation.
14.53 Which three steps in glycolysis cannot be directly reversed during gluconeogenesis? Steps 1, 3, and 10.
14.55 From which intermediate shared by glycolysis and gluconeogenesis is glycogen produced? Glucose 6-phosphate
14.57 a. Describe the role that insulin plays in glycogenesis and glycogenolysis. Insulin activates glycogen synthetase (speeds up glycogenesis) and deactivates glycogen phosphorylase (slows down glycogenolysis). b. Describe the role that glucagon plays in glycogenesis and glycogenolysis.
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Glucagon deactivates glycogen synthetase (slows down glycogenesis) and activates glycogen phosphorylase (speeds up glycogenolysis).
14.59 In the first step of the citric acid cycle, citrate (6 carbon atoms) is formed. The final product of the cycle is oxaloacetate (4 carbon atoms). Where do the missing two carbon atoms end up? The 2 carbon atoms are released as CO2 molecules in steps 3 and 4.
14.61 a. Which step is the control point of the citric acid cycle? Step 3 of the citric acid cycle, the conversion of isocitrate to -ketoglutarate, acts as the control point for this cycle. b. Name the enzyme that catalyzes the reaction that takes place at this step and list its positive and negative effectors. The enzyme that catalyzes this reaction is isocitrate dehydrogenase. The negative effectors for this allosteric enzyme are ATP and NADH and the positive effectors are ADP and NAD+.
14.63 a. Which two compounds donate their electrons to electron transport chain? NADH and FADH2 b. Which molecule is the final electron acceptor of the electron transport chain? The O2 molecule.
14.65 a. The electron transport chain and oxidative phosphorylation typically generate how many ATP from 1 NADH? 2.5 moles of ATP’s are produced for every 1 mole of NADH2that goes through the electron transport chain and oxidative phosphorylation. b. The electron transport chain and oxidative phosphorylation typically generate how many ATP from 1 FADH2? 1.5 moles of ATP’s are produced for every 1 mole of FADH2 that goes through the electron transport chain and oxidative phosphorylation.
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14.67 Glycerol can be converted into a compound used in glycolysis and gluconeogenesis. Explain. Glycerol is produced by the hydrolysis of triglycerides. It can be converted into glycerol-3-phosphate which is transformed into dihydroxyacetone phosphate, an intermediate in both glycolysis and gluconeogenesis.
14.69
In one pass through the β oxidation spiral a fatty acyl-CoA is shortened by two carbon atoms. What other products are formed? One acetyl-CoA, one FADH2, and one NADH
14.71 Calculate the net number of ATPs produced when one 14-carbon fatty acid is activated, undergoes β oxidation, and the products pass through the citric acid cycle and the electron transport chain. Activation in the first step requires the energy equivalent of 2 ATP’s (see explanation in Section 14.9). Each step in the spiral produces one FADH2 = 1.5 ATP and one NADH = 2.5 ATP. This represents ATP production with each pass through the cycle. The number of passes = (number of carbons – 2)/2 so for a 14carbon fat there are (14-2)/2 = 6 passes with an acetyl-CoA produced on the final pass. FADH2 6 x 1.5 = 9 ATP
NADH 6 x 2.5 = 15 ATP
Total ATP = 24 ATP Then, as the acetyl-CoA enters the citric cycle, each pass generates 1 GTP (Same as 1 ATP), 3 NADH and 1 FADH2. For the citric cycle: Acetyl-CoA 7 x 1 = 7 GTP
NADH (7 x 3) x 2.5 = 52.5 ATP
FADH2 7 x 1.5 = 10.5 ATP
Total ATP = 70 from citric cycle. Total ATP for reaction = Total of oxidation spiral + Total of citric cycle Activation = 24 ATP + 70 ATP – 2 ATP = 92 ATP Total ATPs produced = 92
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14.73 a. Name the three ketone bodies. Acetoacetate, 3-hydroxybutyrate, and acetone b. Are they all ketones? No. 3-Hydroxybutyrate contains no ketone group. c. Why is acidosis (low blood pH) associated with the overproduction of ketone bodies? Two of the ketone bodies are carboxylic acids.
14.75 Where is the NADPH used in fatty acid biosynthesis produced? NADPH is produced in the pentose phosphate pathway which occurs in the cytoplasm.
14.77 Where in a cell does fatty acid biosyntheses take place? Cytoplasm
14.79 During transamination, which molecule is a common amino group acceptor? -Keto acids are amino group acceptors during transamination. A common example is -ketoglutarate. 14.81 Oxidative deamination of glutamate produces which compound besides NH4+? α-Ketoglutarate
14.83 Which compound is used for nitrogen excretion in the following? a. humans urea in humans b. birds uric acid in birds
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c. fish NH4+ in fish
14.85 How does the way that you generate some of your additional body heat when you are cold differ from the way that an infant does so? Adults primarily generate heat by shivering. The heat is produced by the metabolism resulting from muscle action. Newborn infants rely on the oxidation of triglycerides in brown fat cells to produce body heat.
14.87 a. During digestion, a maltose molecule is hydrolyzed to produce two glucose molecules. Does this reaction take place in the mouth, the stomach, or the small intestine? Small intestine. b. Which enzyme catalyzes the reaction in part a? Maltase. c. The two glucoses from part a enter glycolysis and are broken down into pyruvates. Under anaerobic conditions the pyruvates are converted into lactates. Starting from the two glucose molecules and ending at lactates, what is the net change in ATP and NADH? 4 ATP and 0 NADH d. Under aerobic condition, the pyruvates from part c are converted in acetylCoAs, which enter the citric acid cycle. Beginning with two glucose molecules, producing acetyl-CoA’s, all of which enter the citric acid cycle, what is the net change in ATP (or its equivalent), NADH, and FADH2? 4 ATP, 4 GTP, 16-20 NADH, and 4-8 FADH2 e. Starting with your answer to part d, and having all of the NADH and FADH2 enter the electron transport chain and oxidative phosphorylation, what will be the net production of ATP, starting with two glucose molecules? 60-64 ATP
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ANSWERS TO EVEN-NUMBERED END OF CHAPTER PROBLEMS 14.2
In the drawing in Problem 14.1, indicate where each of the following takes place. a. reduction of NAD+ b. reduction of FAD c. oxidation of NADH d. oxidation of FADH2 e. a net production of ATP
Answer:
a. and e.
a.
a., b., and e.
c., d., and e.
SECTION 14.1
14.4
METABOLIC PATHWAYS, ENERGY, AND COUPLED REACTIONS
After reviewing this chapter, name a metabolic pathway that a. is linear c. is a spiral b. is circular
Answer: a. Glycolysis
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b. citric acid cycle c. fatty acid catabolism
14.6
In the last step of gluconeogenesis, glucose 6-phosphate is converted into glucose. Which of the reaction equations below is spontaneous? (Refer to Problem 14.5) a. Glucose 6-phosphate + ADP → Glucose + ATP b. Glucose 6-phospate + H2O → Glucose + Pi
Answer: a. Not spontaneous. The G is positive for this reaction Glucose 6-phosphate + ADP → Glucose + ATP because this is the reverse of the coupled reaction (see the answer for 14.5 b.) Glucose + ATP →
Glucose 6-phosphate + ADP
which has a negative G value (-4.0 kcal). b. Spontaneous. The G is negative for this reaction Glucose 6-phospate + H2O → Glucose + Pi because this is the reverse of the reaction given in Problem 14.5 Glucose + Pi → glucose 6-phophate + H2O which has a positve ΔG value (+3.3 kcal).
14.8
For the reaction described in Problem 14.7, what effect will the enzyme have on the value of ΔG for the reaction?
Answer: No effect. An enzyme increases the rate of a reaction, but it has no effect on G.
SECTION 14.2
OVERVIEW OF METABOLISM
14.10 How does catabolism differ from anabolism?
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Answer: In catabolism, substances are broken down into smaller compounds, usually resulting in the release of energy. In anabolism, larger compounds are synthesized from smaller compounds, usually requiring the input of energy.
14.12 What process does the energy provided by hydrolyzing ATP to ADP and Pi drive, anabolism or catabolism? Answer: Anabolism is the synthesis of larger compounds from smaller ones. It typically requires the input of energy, such as from the hydrolysis of ATP.
14.14 Define the term oxidation and explain how it applies to the difference in the structures of FAD and FADH2. Answer: In general, oxidation is the loss of electrons from any chemical species. It is also defined as a chemical reaction in which an atom gains a bond to oxygen or loses a bond to hydrogen. FADH2 loses hydrogen when it is converted to FAD. Therefore, it is oxidized.
14.16 Describe the role of coenzyme A in metabolism. Answer: Coenzyme A is often used to activate (raise the energy of) a compound.
14.18 The abbreviation PPi is used to represent pyrophosphate ion. a. Draw pyrophosphate. b. Draw pyrophosphate as it would appear at a very acidic pH. Answer: a. Draw two phosphate ions connected by a phosphoanhydride bond.
O
O
O P O P O O
O
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b. A very acidic pH results in H+ attaching to all of the –O- sites.
O
O
HO P O P OH OH
OH
14.20 Write a reaction equation for the hydrolysis of UTP to produce PPi. Answer: UTP + H2O → UMP + PPi
SECTION 14.3
DIGESTION
14.22 Where (in the mouth, the stomach, and/or the small intestine) does digestion of each take place? a. polysaccharides b. triglycerides c. proteins Answer: a. Mouth and small intestine. b. Stomach and small intestine. c. Stomach and small intestine.
14.24 Name an enzyme that catalyzes a reaction involved in the digestion of a. a polysaccharide b. a triglyceride c. a protein Answer: a. amylase b. lipase c. pepsin
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14.26 Only a small fraction of the starch in food is digested before it enters the small intestine. Why? Answer: Salivary amylase, which initiates the digestion of carbohydrates in the mouth, is denatured when it reaches the stomach. Enzymes responsible for the majority of carbohydrate digestion are present in the small intestine.
14.28 What two types of glycosidic bonds must be hydrolyzed to break amylopectin down into individual glucose molecules? Does the same enzyme hydrolyze both types? Answer: The α-(1-6) glycosidic bonds and α-(1-4) glycosidic bonds must be broken. No. The α-(1-4) glycosidic bonds are hydrolyzed by the amylase and the α-(1-6) glycosidic bonds are hydrolyzed by a debranching enzyme.
14.30 To which class of lipids do bile acids belong? Answer: They are steroids.
14.32 a. The saliva of dogs contains no amylase. Why might this be expected to be the case for carnivores? b. Dogs do produce pancreatic amylase, which is released into the intestine. Based on this fact, some claim that dogs are not strict carnivores. Explain Answer: a. Carnivores consume mostly meat, which is a protein. b. The presence of amylase indicates the ability to digest polysaccharides (e.g., starch).
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SECTION 14.4
GLYCOLYSIS
14.34 a. Which step in glycolysis is the major control point? b. What compounds act as positive effectors of the enzyme that catalyzes this reaction? c. What compounds are negative effectors? Answer: a. Step 3, the addition of phosphate to fructose 6-phosphate. b. Positive effectors are ADP and AMP. c. Negative effectors are ATP and citrate.
14.36 a. Which class of reaction is involved in the conversion of 2-phosphoglycerate into phosphoenolpyruvate (step 9 of glycolysis): synthesis, decomposition, single replacement, or double replacement? b. Draw each of these compounds as it would appear at low pH. Answer: a. Decomposition. Water is eliminated. b.
O
O
HO P O
HO P O
OH HOCH2
OH
O CH C OH
2-phosphoglycerate
O
H2C C C OH phosphoenolpyruvate
14.38 a. What products are formed when pyruvate undergoes alcoholic fermentation? b. What purpose does this reaction have? Answer: a. Ethanol and CO2. b. NADH is recycled back into NAD+, which allows glycolysis to continue.
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14.40 What is the net change in ATP and NADH from the passage of one glucose molecule through glycolysis, followed by the conversion of pyruvates into ethanol and CO2 molecules? Answer: One glucose molecule is converted into two pyruvate ions, with the production of 2 ATP and 2 NADH. Conversion of the two pyruvate ions into ethanol and CO2 requires 2 NADH. Net change in ATP = 2
Net change in NADH = 0
14.42 Draw lactic acid, the acidic form of lactate. Which predominates at physiological pH, the acidic form or the conjugate base form of this acid? Answer:
OH O CH3 CH C OH lactic acid The conjugate base form will predominate at physiological pH.
14.44 a. Is the conversion of pyruvate to lactate an oxidation or a reduction? b. When this reaction takes place in a cell, what is the oxidizing agent? c. What is the reducing agent? Answer: a. Reduction. The conversion of pyruvate to lactate involves the conversion of a ketone functional group to an alcohol by addition of hydrogen atoms. b. Pyruvate. The oxidizing agent in a reaction is the substance that is reduced. c. NADH. The NADH is losing hydrogen and an electron. In doing so, it reduces the pyruvate and it is therefore the reducing agent.
14.46 Is the conversion of pyruvate into acetyl-CoA an oxidation or a reduction?
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Answer: Oxidation. The oxidation of an organic compound involves the loss of hydrogen or the gain of oxygen. The conversion of pyruvate to acetyl-CoA results in the pyruvate losing hydrogen and an electron to NAD+. Therefore the pyruvate is oxidized.
14.48 Not all of the free energy (G) released during glycolysis ends up in ATP. What happens to the remainder of the energy? Answer: The remainder of the energy is released as heat. Only 33.2% of the total energy produced is used to form ATP.
14.50 a. Which steps in glycolysis involve a coupled reaction? b. Are the reactions coupled to make the net reaction spontaneous or to make it nonspontaneous? c In the coupled reactions, is G greater than zero or is it less than zero? Answer: a. Steps 1, 3, 6, 7 and 10 involve coupled reactions. b. Reactions are coupled to make the net reaction spontaneous. c. G is always less than zero for spontaneous reactions.
SECTION 14.5
GLUCONEOGENESIS
14.52 True or False? It requires more ATP and NADH to make one glucose from two pyruvates than is obtained when one glucose undergoes glycolysis to form two pyruvates. Answer: True. The conversion of one glucose into two pyruvates, the net change is +2 NADH and +2ATP. In the conversion of two pyruvates into one glucose, the net change is -2 NADH and -6 ATP.
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14.54 Which step in gluconeogenesis involves a compound also used in the citric acid cycle? Answer: Step 10 in gluconeogenesis uses oxaloacetate which also participates in the citric acid cycle.
SECTION 14.6
GLYCOGEN METABOLISM
14.56 During glycogenolysis, the splitting of a glucose residue off of glycogen is spontaneous. During glycogenesis, the addition of a glucose residue to glycogen is also spontaneous. How is it possible for both processes to be spontaneous? Answer: The splitting and addition reactions are not the reverse of one another. In glycogenolysis, the glucose residue is split off from glycogen in the form of glucose 1-phosphate. In glycogenesis, the glucose is added to glycogen in the form of glucose-UDP.
14.58 a. Describe the role that glycogen phosphorylase plays in glycogen metabolism. b. Glycogen phosphorylase is controlled by covalent modification. Explain this term. c. Glycogen phosphorylase is an allosteric enzyme controlled by positive and negative effectors. What are positive and negative effectors? Answer: a. Glycogen phosphorylase is the enzyme that catalyzes the breakdown of glycogen. In its active form the enzyme has phosphate groups attached to it. These groups are used to split glucose from glycogen in the form of glucose 1phosphate. b. Covalent modification is the enzyme-catalyzed breaking and formation of covalent bonds within an enzyme. When there is a metabolic demand for the breakdown of glycogen, a kinase produces the highly active form of glycogen phosphorylase by attaching a phosphate group to the inactive form. When the breakdown of glycogen needs to be slowed, a phosphorylase switches the enzyme off by converting the active form into the inactive form. c. Glycogen phosphorylase is an allosteric enzyme because its activity is also regulated by allosteric effectors. The binding of a small molecule or ion
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changes the three-dimensional shape of the glycogen phosphorylase subunits and either enhances the binding ability of substrates (positive effector) or reduces its binding ability (negative effector).
SECTION 14.7
CITRIC ACID CYCLE
14.60 What is the net change in GTP, NADH, and FADH2 from the passage of two acetyl-CoA through the citric acid cycle? Answer: The passage of two acetyl-CoA through the citric cycle creates 2 GTP, 6 NADH, and 2 FADH2.
14.62 a. Some of the acetyl-CoA used in the citric acid cycle is produced from pyruvate. List the reactants and products of this reaction. b. Name an enzyme complex that catalyzes this reaction and list its negative effectors. Answer: a. The reactants are pyruvate and Coenzyme A. The products are acetyl-CoA and CO2. In the process, an NAD+ is reduced to NADH. b. Pyruvate dehydrogenase is the enzyme complex that catalyzes this reaction. Its negative effectors are ATP, NADH, and acetyl-CoA.
SECTION 14.8
ELECTRON TRANSPORT CHAIN AND OXIDATIVE PHOSPHORYLATION
14.64 H+ moves from the intermembrane space, through ATP synthase, and into the mitochondrial matrix. Which term best describes this process: facilitated diffusion or active transport? Answer: Facilitated diffusion best describes this process of ions or molecules moving through a membrane assisted by a protein.
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14.66 Account for the 30-32 ATPs generated when one glucose molecule is catabolized by glycolysis and the citric acid cycle. Answer: In glycolysis, one glucose molecule yields 2 pyruvates, 2 ATP and 2 NADH. Depending on the system used to shuttle the electrons from NADH into mitochondria, these 2 NADH become 2 NADH or 2 FADH2. Converting the 2 pyruvates into 2 acetyl-CoA’s produces 2 NADH. Two passes through the citric acid cycle starting with acetyl-CoA produces 2 GTP (equivalent to 2 ATP), 2 FADH2, and 6 NADH. In the electron transport chain, each FADH2 generates 1.5 ATP and each NADH generates 2.5 ATP. Glycolysis (glucose → 2 pyruvates) 2 pyruvates → 2 acetyl CoAs 2 passes 2 acetyl CoA in citric acid cycle
2NADH x 2.5 ATP/NADH = or, 2 FADH2 x 1.5 ATP/ FADH2 = 2NADH x 2.5 ATP/NADH = 2 GTP = 2 FADH2 x 1.5 ATP/ FADH2 = 6NADH x 2.5 ATP/NADH =
TOTAL
SECTION 14.9
2 ATP 5 ATP 5 ATP 2 ATP 3 ATP 15 ATP 32 ATP
2 ATP 3 ATP 5 ATP 2 ATP 3 ATP 15 ATP 30 ATP
LIPID METABOLISM
14.68 Describe how fatty acids are activated and then moved from the cytoplasm into the mitochondrion. Answer: Fatty acids are activated by the attachment of a coenzyme A (CoA) residue to produce fatty acyl-CoA. The CoA residue is then replaced by a compound called carnitine and the fatty acyl- carnitine is moved into the mitochondria by active transport. 14.70 How does oxidation get its name?
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Answer: -oxidation refers to the oxidation of the carbon atom of the fatty acid into a carbonyl group in the catabolic pathway.
14.72 Calculate the net number of ATPs produced when one 10-carbon fatty acid is activated, enters the mitochondrion, and undergoes complete oxidation to produce acetyl-CoA and reduced coenzymes. Answer: Use Figure 14.18 as an example to track the passage of a 10-carbon fatty acid through the steps of fatty acid catabolism: -oxidation spiral: A 10-carbon fatty acid makes 4 passes through the -oxidation spiral. One pass will produce 1 FADH2, 1 NADH, and 1 acetyl-CoA. The total number of FADH2, NADH, and acetyl-CoA produced after 4 passes are:
4 passes x (1 FADH2/pass) 4 passes x (1 NADH/pass) 4 passes x (1 acetyl-CoA/pass) 1 remaining acetyl-CoA
= 4 FADH2 = 4 NADH = 4 acetyl-CoA = 1 acetyl-CoA
Citric acid cycle: When 1 acetyl-CoA enters the citric acid cycle, 1 GTP, 3 NADH, and 1 FADH2 are formed. 5 passes x (1 acetyl-CoA/pass) x (1 GTP/ acetyl-CoA) = 5 GTP (5 ATP) 5 passes x (1 acetyl-CoA/pass) x (3 NADH/ acetyl-CoA) = 15 NADH 5 passes x (1 acetyl-CoA/pass) x ( 1 FADH2 / acetyl-CoA) = 5 FADH2 Electron transport chain: In the electron transport chain, the conversion of 1 NADH to NAD+ produces 2.5 ATP and the oxidation of 1 FADH2 produces 1.5 ATP: (4 NADH + 15 NADH) (4 FADH2 + 5 FADH2)
x x
(2.5 ATP/1 NADH) (1.5 ATP/1 FADH2)
= 47.5 ATP = 13.5 ATP
Activation of first step in the fatty acid catabolism: This step uses 2 ATP = -2 ATP Total number of ATP produced = 5 ATP + 47.5 ATP + 13.5 ATP - 2 ATP = 64 ATP 14.74 One effect of the “low carb” diet craze that swept the United States in 2004 was that a greater number of people had very bad breath, described as “sickeningly
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sweet”. From the standpoint of metabolism, what molecule produced by a diet that is low in carbohydrates and high in fat and protein might cause bad breath? Answer: In a diet low in carbohydrates but high in fat and protein, glucose metabolism is substantially slowed down and different pathways are activated. One is fatty acid oxidation which results in increased production of acetyl-CoA. The other is gluconeogenesis which involves the conversion of oxaloacetate to phosphoenolpyruvate. The consumption of oxaloacetate decreases the amount available to combine with acetyl-CoA in the citric acid cycle. As a result, when these two oxaloacetate-competing pathways are activated, acetyl-CoA may accumulate. In both cases discussed, acetyl-CoA production leads to overproduction of ketone bodies that cause a “sickeningly sweet” odor in breath.
14.76 a. Draw linoleic and linolenic acid. b. Which carbon-carbon double bonds in these molecules cannot be formed by human desaturase enzymes? Answer: O CH3(CH2)4CH=CHCH2CH=CH(CH2)7
C OH
linoleic acid
a.
O CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7
C OH
linolenic acid
b. The double bonds beyond carbon 10.
14.78 Where in a cell does fatty acid oxidation take place? Answer: Fatty acid oxidation occurs in the mitochondria of the cell.
SECTION 14.10
AMINO ACID METABOLISM
14.80 a. Draw the α-keto acid obtained when cysteine undergoes transamination. b. Draw the -keto acid obtained when phenylalanine undergoes transamination.
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Answer: a. Draw cysteine and replace the –NH3+ and hydrogen on the carbon with a =O. O O HS CH2 C C OH
b. Draw the -keto acid obtained when phenylalanine undergoes transamination.
O O
CH2 C C O -
14.82 When aspartate undergoes deamination, the reaction product can take part in the citric acid cycle to produce ATP or in gluconeogenesis to produce glucose. Explain. Answer: When aspartate undergoes oxidative deamination, the amine group is removed and the carbon to which it is attached is oxidized to a C=O. The resulting oxaloacetate ion can take part in either the citric acid cycle to produce ATP or in gluconeogenesis to produce glucose.
O
O
- O C CH2 CH C O -
deamination
O O
O
-O C CH C C O 2
+
NH3 aspartate
oxaloacetate
14.84 What is the source of NH4+ produced during amino acid catabolism? Answer: Oxidative deamination. The release of NH4+ from amino acids commonly requires two reactions. The first is transamination, wherein an amino group from an amino acid is transferred to an α-keto acid. The second is oxidative deamination, wherein the amino group is replaced by a carbonyl group. The amino group is released as NH4+.
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HEALTH LINK
BROWN FAT
14.86 Hibernating animals produce brown fat which helps them to survive during the winter months. Explain how brown fat makes better use of the animal’s triglyceride supply during hibernation than other fat does. Answer: The presence of the uncoupler thermogenin allows energy released during catabolism to be used in the production of heat, rather than production of ATP.
LEARNING GROUP PROBLEMS 14.88 a. In a transamination reaction, the amino acid aspartic acid reacts with ketoglutarate to produce oxaloacetate plus glutamate. Draw the structures of these reactants and products. b. The glutamate from part a undergoes oxidative deamination to reform ketoglutarate . Write the reaction equation for this process. c. In mammals, what is the fate of the NH4+ produced by the reaction in part b? d. The oxaloacetate produced by the reaction in part a reacts with acetyl-CoA to start one pass through the citric acid cycle. At the end of one cycle, when oxaloacetate has been formed again, how many NADH, FADH2, and ATP equivalents will have been produced? e. Two oxaloacetates enter the gluconeogenesis pathway to produce one glucose 6-phosphate. What is the net change in ATP (or its equivalent) and NADH? f. The glucose 6-phosphate produced in part e enters glycogenesis and is added to a glycogen molecule. Which two enzymes are involved in this process and how do the hormones insulin and glucagon affect the activity of one of these enzymes? Answer: a. In a transamination reaction, an amino group is transferred from an amino acid to -ketoglutarate.
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O
O
O
O
C CH2 CH C O +
+
O
O O
transamination
C CH2 CH2 C C O
NH3
-ketoglutarate
aspartate O O
O
O O
+
C CH2 C C O
O
O
C CH2 CH2 CH C O +
NH3 oxaloacetate
glutamate
b. The glutamate from part a undergoes oxidative deamination to reform ketoglutarate. Write the reaction equation for this process.
O
O O
C CH2 CH2 CH C O +
NH3
O deamination
O O
O
oxidative
C CH2 CH2 C C O
+
-ketoglutarate
glutamate c. In mammals, NH4+ is converted into urea. d. 3 NADH, 1 FADH2, and 1 ATP equivalent (1 GTP)
e. -4 ATP (actually -2ATP and -2 GTP) and -2 NADH f. Glycogen synthase and UDP-glucose phosphorylase. Insulin activates glycogen synthase when blood glucose levels rise. Glucagon switches off glycogenesis by covalent modification of the enzyme when blood glucose levels drop.
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NH4+