Organic Chemistry Chapter 10 Spectroscopy

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Chapter 10 Spectroscopy

The tunnel (Shanghai, China) David Richardson

“This fascination with computer models is something I understand very well. Richard Feynmann called it a disease. I fear he is right.” Michael Crichton

“It is harder to crack prejudice than an atom.” Albert Einstein

“For me, I am driven by two main philosophies: know more today about the world than I knew yesterday and lessen the suffering of others. You'd be surprised how far that gets you.” Neil deGrasse Tyson

“A process cannot be understood by stopping it. Understanding must move with the flow of the process, must join it and flow with it.” Frank Herbert

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Infrared (IR) Spectroscopy IR spectroscopy deals with that portion of the electromagnetic spectrum defined as the infrared region, and it is small compared to the broad span of the electromagnetic spectrum (Figure 10.1). The wavelengths, Îť, listed in Figure 10.1 are in meters.

Figure 10.1 Electromagnetic spectrum

A more detailed picture of the electromagnetic spectrum may be found at the following website: https://imagine.gsfc.nasa.gov/science/toolbox/emspectrum1.html Table 10.1lists the wavelength, Îť, in meters; the frequency in reciprocal seconds and the energy in joules (J) for regions within the electromagnetic spectrum.

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Region of the Spectrum X-rays Ultraviolet Visible

Wavelength, m 10-9 - 10-11 4x10-7 - 10-9 7x10-7 - 4x10-7

Infrared Microwaves Radio waves Nuclear Magnetic Resonance

10-4 - 7x10-7 0.1-10-4 >0.1 1-5

Frequency, s-1 3x1017 - 3x1019 7.5x1014 - 3x1017 4.3x1014 7.5x1014 3x1012 - 4.3x1014 3x109 - 3x1012 < 3 x 109 3x108 - 5x107

Energy, J 2x10-16 - 2x10-14 5x10-19 - 2x10-16 3x10-19 – 5x10-19 2x10-21-3x10-19 2x10-24 – 2x10-21 < 2x10-24 2x10-25 -3x10-26

Table 10.1 regions within the electromagnetic spectrum

Infrared Spectroscopy The infrared region has a wavelength between 10-4 m to 7x10-7 m, and an energy range between 2 x 10-21J – 3 x 10-19J. The specific region that provides information to chemists about the structural identification of molecules is at wavelengths, λ, 2.5 x 106

m to 1.5 x 10-5 m. Traditionally, this region is in reciprocal centimeters, i.e., 1/λ where

λ is in cm. Therefore, the most useful information for organic chemists is the infrared region between 4000 cm-1 – 666 cm-1. This region is the vibrational infrared region where the frequencies of radiation correspond in energy to the natural vibration frequencies of organic molecules. One may generate an IR spectrum by placing a molecule in an infrared spectrometer and subjecting it to infrared radiation with energies between 2x10-21 J -3x10-19 J. The spectrum consists of percent transmittance versus the wavenumber in reciprocal centimeters or wavelength in micrometers and has the general appearance of Figure 10.2 (the infrared spectrum of an ester). Many of the bands that correspond to transmittance versus wavenumbers are not identified with a stretching or bending vibration; however, those that are will provide important information about the structure of the organic compound.

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Figure 10.2 Sample IR spectrum

An IR spectrum is absorbance versus wavenumbers or wavelengths as well as transmittance versus wavenumbers or wavelengths. The following video on converting per cent transmittance to absorbance: https://www.youtube.com/watch?v=HqqwKvseMLg The relationship between transmittance and absorbance is given by Equation 10.1. Equation 10.1

Where A is the absorbance and T is the transmittance. The percent transmittance is the transmittance multiplied by 100 (%T = 100T). When the frequency of infrared radiation incident on organic molecules equals the

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frequency of the vibrating bonds attached to atoms in the molecule, then the infrared energy absorbed is equal to the energy associated with the vibrating molecule. The radiation absorbed can be measured or the light transmitted can be measured. The light absorbed or transmitted is related to the wavelength or wavenumber corresponding to the light absorbed or transmitted. Every organic compound has its characteristic infrared spectrum; therefore, the infrared spectrum is a fingerprint of the molecule. The position of the infrared absorption is important in identifying certain functional groups. For instance, in Figure 10.2, the carbonyl group in the ester exhibits a characteristic absorption at 1752 cm-1. As indicated earlier, wave numbers are the preferred absorption bands used by organic chemists in identifying functional groups in molecules. Equation 10.2 gives the relationship between the wavenumber and wavelength (in micrometers). Equation 10.2 ! "!"

= #"

! #"

! & !'$ ()

# $

! *)

$

where ! "!"

is the wave number

Equation 10.2 is simply derived from taking the reciprocal of the wavelength in centimeters, i.e.,

1 đ?œ†*)

=

1 1 đ?‘š 100 đ?‘?đ?‘š 'đ?œ†() ( ) . # 1 đ?‘š $ + 1 đ?‘Ľ 10 đ?œ‡đ?‘š

or

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1 đ?œ†*)

1 đ?‘Ľ 10, đ?œ‡đ?‘š = 0 1 1 đ?‘?đ?‘š 'đ?œ†() ( 1

The unit for "

! !"

is cm-1, and is the wavenumber.

Imagine that the atoms of organic molecules behave like balls connected to springs. The bonds connecting the atoms of the organic molecules can vibrate in two ways. They can stretch, and they can bend. In stretching, the distance between the atoms increases, but the atoms remain in the same bond axis. In bending, the molecules deform and the position of the atoms changes relative to the original bond axis. Unlike a ball on a spring, the various molecular vibrations occur at quantized frequencies, i.e., only at specified frequencies that change by an interval of n, where n is a whole number. The various stretching and bending vibrations are fundamental vibrations illustrated in Figure 10.3. Bending Vibrations

Stretching Vibrations

Figure 10.3 Fundamental Absorption Modes for a Tri-Atomic Molecule The “+� and “-“ indicate movement above and behind the plane of the two dimensional image of the tri-

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atomic molecule.

When a ball attached to a spring is allowed to vibrate, the frequency of vibration (oscillation) is a function of the force constant of the spring and the mass of the ball. The frequency of the vibrating ball on the spring relates mathematically to the stretching frequency of vibrating bonds. An equation for this assertion comes from Hooke’s Law (Equation 10.3). Equation 10.3 F = -kx Where F is the restoring force exerted by the atoms; x is the displacement of the end of the spring from its equilibrium position, and k is the force constant of the bond. Equation 10.4 relates the time to complete an oscillation. Equation 10.4

The frequency is given by equation 10.5 Equation 10.5

Therefore, Equation 10.6 represents the frequency of vibration.

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Equation 10.6

and Equation 10.6 can be converted into equation 10.7

Since there are two atoms attached to the bond in the stretching frequency, replace “mâ€? with the reduced mass, Îź

Where đ?œ‡ =

(đ?‘š! )(đ?‘š- ) đ?‘š! + đ?‘š-

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Therefore, the wave numbers of atoms undergoing stretching vibrations in the infrared region of the electromagnetic spectrum is given by equation 10.7. Equation 10.7 1 đ?œ†*)

=

1 đ?‘˜ 2 đ?œ‹ đ?‘? 7 (đ?‘š! )(đ?‘š- ) đ?‘š! + đ?‘š-

Where m1 is the mass of atom 1 and m2 is the mass of atom 2; k is the force constant of the bond holding atoms 1 and 2 together; and c is the velocity of light, 2.998 x 108 m/s or 2.998 x 1010 cm/s. Table 10.2 lists the force constants for some common bonds.

Bond

k, N/m

C-H

500

N-H

650

O-H

760

S-H

400

P-H

310

Si-H

270

C-C

450

C-N

490

C-O

450

C-F

560

C-Cl

340

C-Br

290

C-I

230

C=O

1230

C=C

980

10

N=O

910

S=O

1000

C≥C

1560

C≥N

1750

Table 10.2 k, N/m, for some common bonds Where N(Newton) has units equal to

!" $ %!

The force constant of a bond is directly proportional to the strength of the bond. Table 10.2 indicates that the force constant of a single bond is smaller than the force constant of a double bond, The force constant of a double bond is smaller than the force constant of a triple bond. Covalent bonds of organic and some inorganic molecules absorb certain frequencies of infrared radiation and excite them to higher energy states. The molecular vibrations are quantized and only frequencies of infrared radiation that match the natural frequencies of the bond vibrations are absorbed. The absorbed energy increases the amplitude of the vibrational motion of the bonds and an infrared spectrometer detects and measures the absorption. The spectrophotometer records absorptions occurring at certain frequencies and wavelengths. The stronger the chemical bond, the greater will be the absorption frequency. For example, the C=O bond is stronger than the C-O bond; therefore, the C=O absorption stretch appears at a higher frequency and wavenumber than the C-O absorption stretching frequency. The position of the infrared absorption provides information about the functional group in the organic molecule. Using the data in Table 10.2, one can calculate the wavenumber for the O-H absorption stretching frequency:

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1 𝜆*)

1 𝜆*)

1 𝜆*)

1 𝜆*)

1 𝜆*)

1 𝜆*)

1 𝜆*)

C 𝑁 ⃓ 760 𝑚 ⃓ ⃓ = 𝑚 ⃓ 16.0𝑔 1.00𝑔 (2) ( 3.1416) #2.998 𝑥 10. $ ⃓ # $ # $ -/ 𝑠 ⃓ ⃓ 6.022 𝑥 10 6.022 𝑥 10-/ ⃓ ⃓ 16.0𝑔 1.00𝑔 # $ + # $ -/ 6.022 𝑥 10-/ ⎷ 6.022 𝑥 10 1

C 𝑘𝑔 𝑚 1 ⃓ 760 - #𝑚$ ⃓ 𝑠 ⃓ 𝑠 0!' = 5309 𝑥 10 ⃓ ⃓ 16.0𝑔 1.00𝑔 𝑚⃓ # $ # $ ⃓ -/ ⃓ 6.022 𝑥 10 6.022 𝑥 10-/ ⃓ ⃓ 16.0𝑔 1.00𝑔 # $ + # $ -/ 6.022 𝑥 10-/ ⎷ 6.022 𝑥 10

0!'

𝑘𝑔 𝑚 1 760 - #𝑚$ 𝑠 𝑠 7 1 𝑘𝑔 𝑚 ( 1.56 𝑥 100!' 𝑔) )1000 𝑔.

0!'

𝑘𝑔 760 𝑠 𝑠 7 1 𝑘𝑔 𝑚 ( 1.56 𝑥 100!' 𝑔) )1000 𝑔.

= 5309 𝑥 10

= 5309 𝑥 10

𝑠 = #5309 𝑥 100!' $ (6.98 𝑥 10!, 𝑠 0! ) 𝑚

= 3.7 𝑥 101 𝑚0!

= (3.7 𝑥 101 𝑚0! ) )

1 𝑚 . 100 𝑐𝑚 12

1 đ?œ†*)

= 3700 đ?‘?đ?‘š0!

Therefore, the O-H stretching frequency in wavenumbers is 3700 cm-1 (two significant figures). According to Table 10.3 the calculated value is within range of the observed value. A similar calculation can be done for the C-H bond, and, again, the result is within range of the observed value found in Table 10.3.

1 đ?œ†*)

1 đ?œ†*)

1 đ?œ†*)

1 đ?œ†*)

đ?‘˜đ?‘” đ?‘š 1 500 - #đ?‘š$ đ?‘ đ?‘ = 5309 đ?‘Ľ 100!' 7 1 đ?‘˜đ?‘” đ?‘š ( 1.53 đ?‘Ľ 100!' đ?‘”) )1000 đ?‘”.

đ?‘˜đ?‘” 500 đ?‘ đ?‘ = 5309 đ?‘Ľ 100!' 7 1 đ?‘˜đ?‘” đ?‘š ( 1.53 đ?‘Ľ 100!' đ?‘”) )1000 đ?‘”.

đ?‘ = #5309 đ?‘Ľ 100!' $ (5.71 đ?‘Ľ 10!, đ?‘ 0! ) đ?‘š

= 3.03 đ?‘Ľ 101 đ?‘š0!

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1 đ?œ†*) 1 đ?œ†*)

= (3.03 đ?‘Ľ 101 đ?‘š0! ) )

1 đ?‘š . 100 đ?‘?đ?‘š

= 3030 đ?‘?đ?‘š0!

These calculations are only approximations, because the actual values are determined by the strength of the bonds as well as electrical and steric properties in the vicinity of the bond. The ranges of absorption frequencies for various stretching vibrations are listed in Table 10.3.

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Table 10.3 IR functional group stretching vibrations

http://www.chem.ucla.edu/~webspectra/irtable.html has a more detail table of infrared absorptions. The following is additional information that may be useful in the structure elucidation of organic molecules. The wavenumbers for C-H bending absorption frequencies are between 1475 cm-1 and 1300 cm-1. The wavenumbers for C=C-H bending absorption frequencies are between 1000 cm-1 and 650 cm-1.

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Overtone absorption frequencies can assume values that are ½, ⅓, or ¼ the wavelength ( 2, 3, or 4 times the wave number). Combination bands result from the sum of two or more different wave numbers. Difference bands result from the subtraction of two or more wave numbers. The equation (3n-5) gives the number of vibrations anticipated for a linear molecule where n equals the number of atoms in the linear molecule. For example, carbon dioxide, CO2, would theoretically exhibit four predicted vibration modes since it contains three atoms (one carbon and two oxygen atoms), 3(3)-5= 4. The following is a description of the four possible vibration modes for carbon dioxide.

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The two bending modes, C and D, are degenerate (the same in energy); therefore, only one bending vibration is observed in the infrared (IR). The symmetrical stretching vibration A does not involve a change in dipole moment, i.e., the dipole moment of A is zero; therefore, this stretching vibration is infrared inactive. All infrared vibrations are accompanied by a change in dipole moment. The illustrations above can be repeated for any of the axes, and the results would be the same. No matter how the CO2 molecule moves in space, there are only two vibrations that would be IR active – the asymmetrical stretching vibration (accompanied by a change in dipole moment) and the bending vibration. Figures 10.3 and 10.4 are IR spectra for carbon dioxide. Figure 10.3 represents transmittance versus wave numbers, and Figure 10.4 represents absorbance versus wavenumbers.

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Figure 10.3 Infrared spectrum of carbon dioxide: percent transmittance versus wave numbers, cm-1

http://www.wag.caltech.edu/home/jang/genchem/infrared.htm

The wavelength at approximately 4.3 Îźm (4.3 x 10-6 m) has a wave number 2.33 x 105 m-1 or 2330 cm-1. The predicted value for the unsymmetrical stretch is 2640 cm-1. The wavelength at approximately 14 Îźm (1.4 x 10-5 m) has a wave number 7.14 x 104 m1

or 714 cm-1. The predicted value for the bending vibration is 546 cm-1.

Note that if a vibration leads to a zero dipole moment during vibration, then that vibration is infrared inactive, i.e., in order for a vibration to be observed in the IR spectrum, there must be a net dipole moment during the vibration process.

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Figure 10.4 Infrared spectrum of CO2 : absorbance versus wave numbers

http://science.widener.edu/svb/ftir/ir_co2.html

There are 3n-6 fundamental vibrations for non-linear molecules where n equals the number of atoms in the molecule. Some of the predicted infrared vibrations may not be observed because they are weak or may overlap with other absorptions. Weak bands may also occur that are not predicted due to overtones and combinations of two or more fundamental absorptions. The predicted number of absorptions will not be observed if: a. There is no change in the dipole moment of the molecule during vibration. This was observed for carbon dioxide. b. Absorption occurs outside the region of the spectrum under observation. c. The vibrations result in absorptions so close that they coalesce. d. The absorptions are too weak to be seen.

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Non-Fundamental Absorptions As previously indicated, the three types of non-fundamental vibrations in IR spectra are combination, overtone, and difference bands. Generally, these bands have low intensities. The absorption bands of different functional groups appear at different frequencies. Consequently, the positions of the absorptions (and their intensities) are the basis for interpreting infrared spectra. For example, a strong transmittance band at 1887 cm-1 would suggest the presence of a carbonyl group in the molecule. A strong transmittance band in the region between 3750 cm-1 – 3000 cm-1 would be attributed to the OH stretching vibration. Depending upon other characteristic transmittance bands, the compound could be an alcohol, acid, or phenol. Frequently, infrared spectroscopy obtains structural features of molecules; however, comparing infrared spectra with that of known samples is a viable method for identifying organic compounds. Definitive identification of organic compounds (without comparison to known IR spectra) occurs with supporting spectrometric methods, e.g., Ultraviolet (UV) spectroscopy, nuclear magnetic resonance (NMR) spectrometry, and mass spectrometry (MS). As mentioned previously, the two most important regions of an infrared spectrum are 4000 cm-1 – 1300 cm-1 and 909 cm-1 – 650 cm-1. Table 10.3 lists some important transmittance or absorbance bands located between 4000 cm-1 – 1300 cm-1 and between 909 cm-1 – 650 cm-1. The following are some characteristic transmittance bands for organic molecules in the infrared region of the electromagnetic spectrum. Hydrocarbons, CnH(2n+2) Relatively simple bending and stretching vibrations showing only C-H and C-C transmittances/absorbances characterize saturated straight-the chain hydrocarbons. Figure 10.5, the spectrum of octane, is an example of a straight chain hydrocarbon

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spectrum.

Figure 10.5 Infrared Spectrum of octane

http://orgchem.colorado.edu/Spectroscopy/Spectroscopy.html

Figure 10.5 shows the C-H stretching (asymmetrical and symmetrical) of the CH3 group at 2971 cm-1; the –CH2- bending (asymmetrical and symmetrical at 1470 cm-1; the C-H bending at 1383 cm-1; and the –CH2- rocking at 728 cm-1. Branched Hydrocarbons The following observations characterize the IR of branched hydrocarbons: (1) The isopropyl group has a strong doublet of about equal intensity between 1388 cm-1 – 1380 cm-1 as a consequence of bending vibrations. (2) The t-butyl group has two bending vibrations at 1395 cm-1 – 1385 cm-1 and near 1370 cm-1, and the latter transmittance band is more intense (Figures 10.6). (3) If a compound has an internal carbon with two methyl groups attached, the IR spectrum has a doublet in the same region as the isopropyl and tertiary butyl bending vibrations at 1395 cm-1 – 1380 cm-1.

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Figure 10.6 Infrared spectrum of 2,2,4-trimethylpentane the spectra was taken from

http://www.google.com/search?hl=en&lr=&rlz=1R2GGLL_en&q=%22infrared+spectrum+of+2%2C2%2 C4-trimethylpentane%22&btnG=Search&aq=f&oq=&aqi

Cyclic Hydrocarbons, CnH2n Cyclic hydrocarbons, unless there is ring strain, have infrared spectra similar to straight-chain hydrocarbons. Cyclic hydrocarbons have a slight shift in the -CH2 – bending vibration to higher wavelength (lower wavenumber). For example, the -CH2 – bending in n-octane (Figure 10.5) occurs at 1470 cm-1, but the CH2 – bending in cyclohexane (Figure 10.7) occurs at 1461 cm-1.

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Figure 10.7 Infrared Spectrum of cyclohexane

Alkenes, CnH2n Alkenes possessing a terminal double bond, C=C-H, exhibit a C-H transmittance or absorbance band at 3058 cm-1 unless other transmittance bands obscure the transmittance at 3058 cm-1. The transmittance band at 3058 cm-1 is due to the terminal alkene C-H stretch. Figure 10.8 illustrates this feature. Also, there is a weak to moderate transmittance band between 1667 cm-1 – 1634 cm-1 due to C=C stretching vibration. The C-H alkene bending vibration exhibits two transmittance bands between 1000 cm-1 – 650 cm-1. These bands are more intense than the other alkene transmittance bands.

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Figure 10.8 Infrared spectrum of 1-pentene

The C=C stretching vibrations for trans alkenes are generally weaker than the C=C vibrations for the cis alkenes.

Figure 10.9 Infrared Spectrum of cyclohexene taken from

http://neon.otago.ac.nz/chemlect/chem111/module4/overlays/lecture4/lecture.pdf

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If the alkene is di-substituted, symmetrical, and internally located, then the C=C stretching vibration will be very weak or absent. Akynes, CnH(2n-2) The spectra of alkynes exhibit C≡C stretching vibration between 2260 cm-1 – 2100 cm-1 (Figure 10.10). This observation does not occur if the triple bond is symmetrical. The infrared spectrum of 1-hexyne shows a transmittance band for the C-H stretching vibration of the terminal triple bond at 3324 cm-1. The C-H stretching vibration between 3333 cm-1 – 3267 cm-1 is generally strong for terminal alkynes.

Figure 10.10 Infrared spectrum of 1-heptyne taken from http://orgchem.colorado.edu/hndbksupport/irtutor/alkynesir.html

Remember that the C≡C stretching vibration at 2126 cm-1 will not occur with a symmetrical triple bond. Amines, RNH2, R2NH and R3N The transmittance bands of primary and secondary amines are due to N-H stretching

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vibrations. The IR spectra for dilute solutions of primary amines exhibit asymmetrical and symmetrical transmittance bands at approximately 3496 cm-1 and 3390 cm-1. Figure 10.11 is an example of the infrared spectrum of primary amines.

Figure 10.11 Infrared spectrum of n-butylamine spectra taken from

http://books.google.com/books?id=ega5c11VHvkC&pg=PA900&lpg=PA900&dq=%22infrared+spectrum +of+nbutylamine%22&source=bl&ots=tM4rWQhGup&sig=lyLYg13ewFXj3fbc6YaXYmPWCHM&hl=en&ei=Hmp 3SpW9LYfatgPIqOjlBA&sa=X&oi=book_result&ct=result&resnum=3#v=onepage&q=%22infrared%20spe ctrum%20of%20n-butylamine%22&f=true

The IR spectrum of n-butylamine exhibits the characteristic –NH2 stretching vibration at 3496 cm-1 and 3390 cm-1. The N-H stretching vibration appears as a single transmittance band at around 3367 cm-1 – 3322 cm-1. Ethers, ROR Aliphatic ethers exhibit strong transmittance bands in the region between 1150 cm-1 – 1085 cm-1.

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The infrared spectrum of dibutyl ether shows a broad and strong transmittance band at approximately 1136 cm-1. Aryl alkyl ethers show symmetrical and asymmetrical stretching vibration between 1275 cm-1 – 1275 cm-1 and 1075 cm-1 – 1020 cm-1 respectively. Amides

The characteristic transmittance bands of amides result from the N-H and C=O stretching vibrations. The C=O stretching vibration of amides depends on the physical state of the amide and the extent of hydrogen bonding. The C=O stretching vibrations usually occur at shorter wavenumbers than the C=O stretching vibration of ketones and aldehydes. Table 10.4 lists some common wavenumbers for various structural classes of amides.

amide primary secondary tertiary

solid state, wave number, cm-1 1650 1640 1680 – 1630

solution, wave number, cm-1 1690 1680-1700 1647-1615

Table 10.4

The wavenumbers for the stretching vibrations of amides are impacted by other factors including the physical state of the amide and the electrical effects and the nature of the solvent dissolving the amide.

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The nature of the N-H stretching vibration of amides also depends upon structural features of the amide, i.e., whether it is a primary, secondary, or tertiary amide. Primary amides in dilute solution show two symmetrical and asymmetrical N-H stretching vibrations near 3520 cm-1 and 3400 cm-1. In the solid state, these transmittance bands near 3350 cm-1 and 3180 cm-1 are due to the ability of these systems to molecularly associate (hydrogen bond).

Amides can also molecularly associate linearly.

The N-H stretching vibration for secondary amides in dilute solution occurs near 3500 cm-1 – 3400 cm-1. In the solid state, the N-H stretching vibration appears as multiple transmittance bands near 3330 cm-1 – 3060 cm-1. Figure 10.13, the infrared spectrum of butyramide, and Figure 10.14 is the infrared spectrum of N-methylformamide.

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Figure 10.13 Infrared spectrum of butyramide

Figure 10.14 Infrared spectrum of N-methylformamide

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Observe the multiple bands in Figure 10.14 around 3330 cm-1 – 3400 cm-1 for the N-H stretching vibration of N-methylformide.

Organic Acids and Acid Halides

Organic acids show characteristic absorption for the O-H and C=O stretching vibration. Organic acids in the solid and liquid states and in solution (unless the solution is very dilute) exist as dimers. Hydrogen bonding causes the dimerization of organic acids.

The O-H stretching vibration is influenced by hydrogen bonding and appears as a transmittance broad and an intense band in the region between 3300 cm-1 – 2500 cm-1. The carbonyl stretching frequencies for acids are usually more intense than for ketones and aldehydes. The C=O stretching vibration occurs between 1720 cm-1 – 1695 cm-1 for aliphatic acids and 1710 cm-1 – 1680 cm-1 for aromatic acids in which the carbonyl group of the carboxylic acid is attached to the aromatic ring. Figure 10.15 is the infrared spectrum of propanoic acid, an aliphatic acid. Observe the

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sharp absorption at around 1700 cm-1. Also, observe the broad band due to 3500 cm-1 – 2500 cm-1.

Figure 10.15 Infrared spectrum of propanoic acid

Figure 10.16 is the infrared spectrum of benzoic acid, an aromatic acid. Observe the carbonyl stretching vibration at 1687 cm-1.

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Figure 10.16 Infrared spectrum of benzoic acid

The characteristic transmittance peaks observed in the infrared spectrum of stearic acid are listed in Table 10.5.

Transmittance broad intense OH stretch vibration carbonyl, C=O, stretch vibration O-H out of phase bending vibration

Wave number 2841 cm-1 1692 cm-1 931 cm-1

Table 10.5 primary transmittance signals for stearic acid

Acid Halides Acid halides exhibit C=O stretching vibration at 1815 cm-1 – 1785 cm-1 in contrast to the carbonyl stretching vibration of non-conjugated aliphatic acids that occur between 1720 cm-1 – 1695 cm-1.

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The predominant transmittance for butyryl chloride is at about 1785 cm-1. The inductive effect of the chloride strengthens the C=O bond; therefore, the stronger bond strength of the carbonyl bond will lead to a higher wavenumber and lower wavelength.

butyryl chloride

Esters

Two strong transmittance bands are characteristics of esters. These transmittance bands occur at higher wavenumbers than ketones as a consequence of inductive effects. The C=O stretching vibration occurs at 1750 cm-1 – 1735 cm-1 for aliphatic esters and C-O stretching vibration at 1250 cm-1 and 1051 cm-1. For aromatic esters, the transmittance bands are in the region between 1730 cm-1 – 1715 cm-1. The C-O transmittance band is generally broad and strong. Depending on the nature of the compound, one may observe only one C-O transmittance. Table 10.6 lists the characteristic transmittance peaks observed in the infrared spectrum of ethyl acetate.

Transmittance carbonyl, C=O, stretching vibration C-O stretching vibration

Wave number 1751 cm-1 1250 cm-1 and 1051 cm-1 (two asymmetric coupled vibrations)

Table 10.6 primary transmittance signals for ethyl acetate

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Nitriles R-C≡N Nitriles exhibit a moderate and sharp transmittance between 2260 cm-1 – 2240 cm-1. The transmittance band for aromatic nitriles is between 2200 cm-1 – 2222 cm-1. The transmittance of the C≡N stretching vibration is more intense than the C≡C stretching vibration transmittance band and the C=C stretching vibration band. Table 10.7 compares the wavenumbers for the stretching vibrations for these three systems.

Transmittance bands C≡N C≡C C=C

Wave number 2260 cm-1 – 2222 cm-1 2260 cm-1 – 2100 cm-1 1667 cm-1 – 1640 cm-1

Table 10.7 primary transmittance bands for C≡N; C≡C and C=C

Aldehydes and Ketones

Ketones and aldehydes exhibit characteristic transmittance bands due to C=O stretching vibrations between 1870 cm-1 – 1540 cm-1. Transmittance bands between these wavenumbers are high intensity and are seldom obscured by other transmittance bands. The C-H stretching vibration for aldehydes exhibits a transmittance band between

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2830 cm-1 – 2695 cm-1. However, the aldehydic C-H stretching vibration can be obscured by C-H stretching vibrations of aromatic and aliphatic compounds. The strong C=O stretching vibration for aldehydes and ketones are in the infrared spectra of acetone and propanal. Acetone exhibits a strong transmittance band at 1724 cm-1 due to C=O stretching vibration. Also, propanal exhibits a strong transmittance band at 1724 cm-1 due to C=O stretching vibration. Aldehydes exhibit two weak aldehydic C-H transmittance bands at about 3390 cm-1 and 2941 cm-1

Alcohols, ROH The O-H stretching vibration is very characteristic of alcohols. The position of the transmittance band depends upon hydrogen bonding of the OH group. An O-H group that is not involved in hydrogen bonding exhibits a strong transmittance band between 3650 cm-1 – 3584 cm-1. Hydrogen bonding causes the O-H stretching vibration to shift to smaller wavenumbers that appear between 3550 cm-1 – 3200 cm-1. This is due to the decrease in the force constant caused by hydrogen bonding. Hydrogen bonding also causes a broadening of the transmittance band. The transmittance spectrum of 1-butanol exhibits an O-H stretching vibration between 3000 cm-1 – 3200 cm-1. For neat solutions (pure) of primary, secondary, and tertiary alcohols the position of the C-O stretching vibration increases in wavenumbers from primary alcohol to tertiary alcohol. Table 10.8 demonstrates this phenomenon.

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Transmittance bands C-O broad stretching vibration for CH3CH2CH2CH2OH, 1-butanol, a primary alcohol C-O broad stretching vibration for CH3CH2CH(CH3)OH, 2-butanol, a secondary alcohol C-O broad stretching vibration for (CH3)3COH, 2-methyl-2-propanol, a tertiary alcohol

Wavenumber 1080 cm-1 – 1020 cm-1

1160 cm-1 – 1060 cm-1

1250 cm-1 – 1120 cm-1

Table 10.8 C-O broad stretching vibrations for primary, secondary and tertiary alcohols

Observe the shift toward higher wave numbers of the C-O stretching vibration as we move from primary to secondary to tertiary alcohols.

Anhydrides

Anhydrides exhibit two strong transmittance bands due to asymmetric and symmetric C=O stretching vibrations at about 1786 cm-1 and 1724 cm-1. Maleic anhydride, compound I, is an unsaturated five-member cyclic anhydride with a higher wavenumber than noncyclic anhydrides.

36

I Compound I exhibits a transmittance band for the C=O stretching vibration at about 1865 cm-1 and 1782 cm-1. Table 10.9 is a tabulation of the wavenumbers for saturated noncyclic anhydrides, conjugated noncyclic anhydrides, and conjugated cyclic anhydrides.

Transmittance bands C=O stretching vibration for saturated noncyclic anhydrides C=O stretching vibration for conjugated noncyclic anhydrides C=O stretching vibration for conjugated cyclic anhydrides

Wave number 1818 cm-1 and 1750 cm-1 1175 cm-1 – 1720 cm-1 1865 cm-1 – 1782 cm-1

Table 10.9 C=O stretching vibrations for cyclic and noncyclic anhydrides

Aromatic Hydrocarbons The C-H stretching vibrations of aromatic hydrocarbons occur in the region between 3100 cm-1 – 3000 cm-1. The C-H in-plane bending vibration is more predominant than the C-H aromatic stretching vibration. The bending vibrations are strong transmittance bands between 1300 cm-1 – 1000 cm-1. The C-C stretching vibrations for aromatic carbons are in the region between 1600 cm1

– 1585 cm-1 and 1500 cm-1 – 1400 cm-1.

The nature of ring substitution in mononuclear aromatic compounds can be

37

ascertained in the infrared spectra of neat solutions. This is accomplished by studying the weak overtones and combination transmittance bands in the region between1627 cm-1 – 2000 cm-1. The following is a sketch of the pattern for the weak overtone and combination for substituted aromatic compounds: 1. Sketch of the transmittance bands between1627 cm-1 – 2000 cm-1 for monosubstituted aromatic compounds.

2. Sketch of the transmittance bands for the weak overtone and combination bands between1627 cm-1 – 2000 cm-1 for di-substituted aromatic compounds (ortho, meta, and para patterns).

38

3. Sketches of the weak overtone and combination transmittance bands between1627 cm-1 – 2000 cm-1 for tri-substituted aromatic compounds.

39

4. Sketches of the weak overtone and combination transmittance bands between1627 cm-1 – 2000 cm-1 for tetra-substituted aromatic compounds.

40

5. Sketch of the weak overtone and combination transmittance bands between1627 cm-1 – 2000 cm-1 for penta-substituted aromatic compounds. 41

6. Sketch of the weak overtone and combination transmittance bands between1627 cm-1 – 2000 cm-1 for hexa-substituted aromatic compounds.

For example, Figure 10.17 is the infrared spectrum of toluene, compound I.

I 42

Figure 10.17 http://orgchem.colorado.edu/Spectroscopy/Spectroscopy.html

Note the following about Figure 10.17: a. The overtone transmittance bands and its pattern between 2000 cm-1 – 1667 cm-1 which is the characteristic pattern for monosubstituted aromatic compounds. b. C=C aromatic stretching vibration at 1614 cm-1, 1506 cm-1 and 1465 cm-1 c. The characteristic strong aromatic C=C out-of-plane bending vibration at 738cm -1 and 694 cm-1

43

Ultraviolet-Visible Spectroscopy UV-Visible Absorption spectroscopy is effective in identifying molecules with a high degree of conjugation. The Beer-Lambert Law (absorbance is directly proportional to the concentration of the absorbing species and the path length of the light) is represented by Equation 10.8. Equation 10.8

A = -log

I =ÎľcL Io

This equation can be used to determine the concentrations of the absorbing species; where Îľ is the molar absorptivity or extinction coefficient in

Where L is the path length through the sample in cm; c is the concentration of the absorbing species in moles/L. Many organic compounds exhibit color, e.g., chlorophyll is green; 2,4dinirophenylhydrozones of aldehydes and ketones exhibits colors between. yellow and red depending on the extent of conjugation in the molecules.

44

These compounds are colored because they exhibit wavelengths from electronic excitations in the visible region of the electromagnetic spectrum. There are four kinds of electronic excitations in ethylene. Ethylene contains two types of bonding molecular orbitals, the Ď€ molecular orbital and the Ďƒ molecular orbital. The following figure illustrates the four types of electronic transitions possible in ethylene (ethene).

45

Ďƒâ†’Ďƒ* with an energy difference, ∆đ??¸ = đ??¸2∗ − đ??¸2 , that is greater than 1.2 x 10-18 J and a wavelength that is less than 165 nm. The diagram indicates that the Ďƒâ†’Ďƒ* transition has the largest value for ΔE and the smallest Îť , ∆đ??¸ =

3* "

The π→π* transition has an energy smaller than 1.2 x 10-18 J and a wavelength about 190 nm which is greater than 165 nm. The π→π* transitions are susceptible to signals in ultraviolet-visible spectra only when conjugation results in decreasing the energy difference between the π→π* transition; thereby, increasing the value for Îť. Functional groups or chromophores absorb light resulting in electrons being excited to antibonding molecular orbitals in a unique pattern. For example, electrons in the sigma frame exhibit Ďƒâ†’Ďƒ* transitions, electrons in nonbonding molecular orbitals exhibit n→π* transitions, electrons in unsaturated systems can exhibit π→π* transitions; and electrons in nonbonding molecular orbitals adjacent to the Ď€ frame can exhibit n→π* transitions as well as π→π* transitions. Cyclopentene exhibits Ďƒâ†’Ďƒ* transitions; Ďƒâ†’Ď€* transitions; Ď€â†’Ďƒ* transitions; and π→π* transitions. The π→π* transitions has the smallest value for ΔE but the largest value for 46

λ. In addition to σ→σ* transitions; σ→π* transitions; π→σ* transitions; and π→π* transitions, aldehydes and ketones exhibit n→σ* transitions and n→π* transitions. Considering these transitions, the n→π* transition is the longest wavelength, i.e.., between 200nm-400 nm. The π→π* transitions are 10 to 100 times more intense than the n→π* transitions. Both transitions are useful in identifying unknowns. Absorptions below 200 nm (the vacuum ultraviolet) are of little use in interpreting UV spectroscopy. Absorptions in this region are σ→σ* and σ→π*. As indicated previously, conjugated dienes exhibit longer wavelengths than isolated systems.

The conjugated diene system exhibits delocalization energy due to resonance; therefore, the energy difference between the highest occupied bonding molecular orbitals and the lowest unoccupied molecular orbital decreases, and the wavelength increases.

The more conjugated the system, the smaller the energy difference

between the highest occupied molecular orbital and the lowest unoccupied molecular orbital. The π* antibonding molecular orbital, the excited state, is more polar than the π

47

bonding molecular orbital, the ground state; consequently, polar solvents would solvate the excited state to a greater extent than the ground state. This would result in a decrease in the value of ΔE and an increase in λ toward the visible (a bathochromic shift or redshift) region of the electromagnetic spectrum. This effect is demonstrated in the following diagram:

ΔE > ΔE’ and λ < λ’ For molecules exhibiting n→π* transitions, the ground state molecularly associates with hydrogen bonding solvents better than the excited state. This is a blueshift or a hypsochromic shift. Following is an illustration of the hypsochromic or blueshift:

48

ΔE < ΔE’ and λ > λ’ Solvents that molecularly associate would interact with the ground state better than the excited state resulting in a blueshift or hypsochromic shift in the ultraviolet-visible spectrum. The conjugated diene system exhibits delocalization energy due to resonance; therefore, the energy difference between the highest occupied bonding molecular orbitals and the lowest unoccupied molecular orbital decreases and the wavelength increases.

The more conjugated the system, the smaller the energy difference

between the highest occupied molecular orbital and the lowest unoccupied molecular orbital. The π* antibonding molecular orbital, the excited state, is more polar than the π bonding molecular orbital, the ground state; consequently, polar solvents would solvate the excited state to a greater extent than the ground state. This would result in a decrease in the value of ΔE and an increase in λ toward the visible (a bathochromic shift or red shift) region of the electromagnetic spectrum. This effect is demonstrated in the following diagram:

49

Woodward Correlation Rules Dr. Robert Woodward of Harvard developed a correlation table (Table 10.10) which can provide approximate λmax for conjugated systems that have certain functional groups associated with the conjugation. For example, the wavelength, λmax, for π→π* transition for 1,3-butadiene, the parent conjugated system, is 217 nm, and if this system has any of the functional groups in table 10.10 attached to diene systems., then the Woodward correlation Table would add the designated nanometers in Table 10.10 to 217 nm.

50

Functionality Associated with the Conjugated System

Additions to the Parent System, nm

Exocyclic* Double Bond (Conjugated Double Bond)

5

Extension of double bond

30

Alkyl, R, Groups

5

Homoannularity** (double bonds in the same ring)

36

-Cl

5

-Br

5

-NR3

60

-SR

30

-OR

6

-OOCCH3

0

Table 10.10 Woodward Correlation Table for approximating λmax of conjugated systems *double bond directly attached, but external to a ring, e.g.,

**Example of homoannularity

The Woodward rules can be used to approximate the λmax for the following terpene.

51

Parent

217 nm

Extension of the double bond system

30 nm

Homoannularity

36 nm

Four alkyl groups at 5nm each

20 nm

Approximate λmax

303 nm

The Woodward rules can be used to approximate the λmax for the following steroid.

52

Parent

217 nm

Extension of the double bond system

30 nm

Homoannularity

36 nm

Four alkyl groups at 5 nm each

20 nm

Exocyclic double bond

5 nm

Approximate λmax

308 nm

The Woodward rules can be used to approximate the λmax for β-carotene.

Parent

217 nm

9 x Extension of the double bond system

9 x 30 nm= 270 nm

ten alkyl groups at 5nm each

50 nm

Approximate λmax

537 nm

The literature values for the λmax of β-carotene is 450 nm and 478 nm.

53

Table 10.11 gives the additional nanometers added to 215 nm for n→ π* transition of the parent carbonyl system.

Functionality Associated with the Conjugated Carbonyl System

Additions to the Parent System

Extension to the Conjugated Carbonyl

30 nm

R α β γ and higher -OH α β γ and higher

10 nm 12 nm 18 nm 35 nm 30 nm 50 nm

CH3COOα, β, and γ -OCH3 α β γ δ -Cl α β -Br α β Exocyclic double bond Homoannularity

6 nm 35 nm 30 nm 17 nm 31 nm 15 nm 12 nm 25 nm 30 nm 5 nm 39 nm

Table 10.11 gives approximate nanometers that should be added to 215 nm for n→ π* transitions in carbonyl systems.

Following are applications of the table for n→ π* transitions for two carbonyl systems.

54

Parent

215 nm

Extension of the double bond system

30 nm

Exocyclic double bond

5 nm

γ- alkyl substituent

18 nm

δ - alkyl substituent

18 nm

Approximate λmax

286 nm

Predicted λmax for

Parent

215 nm

Extension of the double bond system

30 nm

Homoannularity

39 nm

γ- alkyl substituent

18 nm

2 x δ - alkyl substituent

2 x 18 nm

Approximate λmax

338 nm

55

Aromatic compounds exhibit π→π* transitions between 229 nm – 330 nm depending on the type of functional group attached to the aromatic system (Table 10.12).

Compound

Wavelength, nm

Benzene, C6H6

255

Phenol, C6H5OH

210

Nitrobenzene, C6H5NO2

252 and 280

Biphenyl, , C6H5 - C6H5,

246 and 330

Styrene, C6H5CH=CH2

244 and 280

Table 10.12 Approximate π→π* transitions for aromatic systems

56

Magnetic Resonance Spectroscopy Infrared spectra result from bond vibrations, UV-visible spectra result from loosely attached electrons transitioning to higher energy levels, and nuclear magnetic resonance (NMR) spectra result from spinning nuclei in different chemical environments. NMR spectrometry gives more information about molecular structure than infrared or ultraviolet-visible spectrometry. NMR spectrometry in conjunction with infrared spectroscopy and mass spectrometry can definitively identity many organic compounds. The nuclei of certain atoms are constantly spinning (analogous to a spinning top or gyroscope) and since atomic nuclei have positive charges, the spinning results in a nuclear angular momentum “I”, the spin angular momentum. Quantization of the spin angular momentum is analogous to vibrational frequencies. The value of “I” depends on the number of protons and neutrons in atoms. If the number of protons and neutrons is even, then the spin angular momentum of the nucleus is zero, i.e., the nucleus does not spin. If the number of protons is odd and the number of neutrons is even or vice versa, the spin angular momentum is ½ or greater. Nuclei having “I” equal to ½ are ideally suited for NMR study while nuclei with an I value of zero or a value greater than ½ are not suited for magnetic resonance studies. Examples of nuclei with a value of ½ are 1H; 19F; 13C; and

31

P. Molecules containing

these elements are suitable for Nuclear Magnetic Resonance studies; however, 1H is the most studied nucleus since it is found in large isotopic abundance and is common to all organic molecules. Nuclear magnetic resonance studies on the 13C nuclei exploded over the last several years because of extraordinary improvements in computer techniques and increased sensitivity in magnetic resonance instrumentation.

13

C NMR studies result in the

determination of the carbon skeleton of organic compounds. A nucleus with a spin angular momentum “I,” according to quantum theory, can have 2(I)+1 orientations when placed in an applied magnetic field. Thus the nucleus of 1H can have two orientations, 2(½)+1; it can align itself with the applied magnetic field 57

(parallel to the field) or against the field (anti-parallel to the field). This situation is analogous to the needle of a compass which aligns itself with the earth’s magnetic field. Alignment with the field represents a lower energy state, -μHHo, and alignment against the field represents a higher energy level, +μHHo. The magnitude of the magnetic moment of the proton, μH, as well as the strength of the applied magnetic field, Ho, determine the difference in energy, ∆E, between these two orientations. ∆E = E2 – E1 = μHHo – (-μHHo) = 2 μHHo Figure 10.18 represents this relationship.

Figure 10.18 representation of the change in energy for the nuclear energy levels generated by an external magnetic field 58

According to Figure 10.18, đ??ť'4 is greater than đ??ť5 ; therefore, ∆đ??¸ 4 > ∆đ??¸. As the applied magnetic field increases, the energy difference, ∆đ??¸ 4 , between orientations increases. In addition to alignment with or against the applied field, the nucleus will also rotate about the field. This nuclear phenomenon is called precession and again is analogous to a spinning gyroscope. Figure 10.19 is a representation of the motion of a nucleus in a magnetic field.

Figure 10.19 the motion of a nucleus in a magnetic field

The principle of NMR spectrometry requires that the precessing proton changes orientation from a lower to higher energy state. One can accomplish that by holding the applied magnetic field, Ho, constant and applying electromagnetic radiation (of the radio frequency range) in such a way that its magnetic component H1 is rotating at right angles to the applied field and in the same direction as the precessing proton. Increasing the frequency of the electromagnetic radiation results in increasing the rotation of H1. When the rate of rotation of H1 is equal to the angular velocity of the precessing proton, energy is absorbed and the proton flips to the higher energy orientation (Figure 10.20).

59

Figure 10.20 illustration of proton transition from a low energy level to high energy orientation

At this point, the proton is in resonance, and the absorption of energy appears as a peak in a spectrum. Holding H1 constant and varying Ho. will achieve the same results. This procedure is easier to accomplish and is what NMR instruments do. Protons located in different environments in the molecule will absorb different amounts of energy generating a spectrum of the molecule. Figures 10.21 and 10.22 respectively display a rough schematic of an NMR spectrometer and an example of an NMR spectrum.

Figure 10.21 schematic diagram of an NMR spectrometer

60

Figure 10.2 is an example spectrum of the NMR of p-xylene.

Chemical Shift The environment of the proton determines the extent the nucleus would experience the effect of the applied magnetic field. Protons in different environments experience different magnetic field effects. For instance, a proton on an OH group would experience a different applied field effect than the protons of a methyl, CH3, group. The differences are due to the differences in electron densities surrounding the nuclei of protons. For instance, the proton bonded to an oxygen atom has a lower electron density than the protons bonded to the carbon atom of a methyl group. This difference is due to the electron-withdrawing capability of the oxygen atom bonded to the proton of the OH group. Protons exhibiting this effect are deshielded. On the other hand, the methyl protons are shielded since they experience little or no electron-withdrawing effect; consequently, the higher the electron density around the proton, the greater the shielding effect. As the shielding increases, the applied field necessary for resonance to occur increases. Therefore, the radio frequency radiation necessary for resonance to occur by any proton (shielded or deshielded depends on the spin angular momentum, ÎźH, the shielding constant, Ďƒ and the strength of the applied field, Ho, and can be expressed by Equation 10.8. 61

Equation 10.8

where h is Planck’s constant Equation 10.8 can be rearranged to give equation 10.9 where Ho is the applied field strength necessary to execute resonance for any proton. Equation 10.9

Equation 10.9 indicates that an increase in the shielding constant would increase the applied field necessary to cause resonance. It is difficult to measure absolute values for Ho ; therefore, the applied field, Happlied field, necessary to produce resonance for a particular proton is measured against a reference. The difference between the field, Ho, of the actual proton and the field, Hreference, of the standard divided by the applied field, Happlied field, is the chemical shift (Equation 10.10). Equation 10.10 chemical shift = δ and

The most popular reference is tetramethylsilane (TMS), (CH3)4Si. The chemical shift of TMS has a value of zero (assigned), and the resonances of other protons are measured relative to the TMS standard since the protons of TMS are highly shielded. TMS protons appear at a higher applied field strength than most organic products because of the electron density about them. Protons of other organic compounds have a field effect that is lower than TMS; therefore, they would have chemical shifts higher than 0, 62

the chemical shift for TMS. Since the frequency necessary for resonance to take place in the sample and the references are large values with slight differences, it is more convenient to express chemical shifts in parts per million; consequently, the equation for the chemical shift, δ can be written as: đ?œˆ5 − đ?œ?678 đ?›ż = ) . 10+ đ?œˆ5 (đ?‘“đ?‘&#x;đ?‘’đ?‘žđ?‘˘đ?‘’đ?‘›đ?‘?đ?‘Ś đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘ đ?‘?đ?‘’đ?‘?đ?‘Ąđ?‘&#x;đ?‘œđ?‘šđ?‘’đ?‘Ąđ?‘’

The NMR spectrum of p-xylene exhibits the chemical shifts for the different types of protons relevant to the standard, TMS, at 0 ppm or 0 δ. The single line at about 2.25 ppm (δ) is the signal from the six protons of the two methyl groups in the para position. These protons are more shielded than the single line for the four aromatic protons at 7.0 ppm (δ). Table 10.12 list the chemical shifts for some protons in various chemical environments. These values are important in deciding the nature of the environment of the proton and lead to the identification of the compound. For example, let’s consider the protons in diethyl ether, compound I, and 2-bromoethanol, compound II.

The six protons labeled “a� of compound I are equivalent, i.e., they are in the same environment; consequently, they will experience the same effect of the applied field. The four protons labeled “b� are in the same environment but are in a different environment than the “a� protons. The “b� protons are more deshielded than the “a�

63

protons since they are adjacent to an electron-withdrawing group, the oxygen atom. Therefore, the “b” protons will experience a higher δ value than the “a” protons. In compound II, the two a’ protons are equivalent; therefore, they will have a chemical shift value that is different from the b’ protons. Oxygen is more electronegative than Br; consequently, the two b’ hydrogen atoms will be more deshielded than the two a’ protons, and the NMR spectrum of compound II will show two different resonance signals. Table 10.13 is a tabulation of approximate additional δ values that would add to the standard δ values in the proton magnetic spectra of methyl hydrogen atoms, CH3; methylene hydrogen atoms, CH2; and methine hydrogen atoms, CH, in proximity to a specified functional group. The standard proton magnetic resonance position of the hydrogen atoms of a methyl group is 0.90 δ. The standard proton magnetic resonance position of the hydrogen atoms of a methylene group is 1.20 δ. The standard proton magnetic resonance position of the hydrogen atoms of a methine group is 1.55 δ. The α hydrogen atom (Figure 10.23) is the hydrogen atom adjacent to a specified functional group. The β hydrogen atom (Figure 10.23) is the hydrogen atom that is one carbon atom removed from the specified functional group.

Figure 10.23

Table 10.14 list chemical shift values for methyl protons with tetramethylsilane (TMS) as a standard

64

Designated Protons

Chemical Shift range, hertz (cps)

Chemical Shift range, ppm (δ): hertz/60.

R-CH2-R

78

1.5

R3CH

120

2.0

R2C=CH2

2.0 x 102

5.0

R2C=CHR

2.7 x 102 – 3.9 x 102

4.5 – 6.5

3.9 x 102 – 5.1 x 102

6.5 - 8.5

RC≡C-H

1.08 x 102 – 1.86 x 102

1.8 – 3.1

R2C=CRCH3

90 - 1.56 x 102 -

1.5 – 2.6

R-CH2Cl

2.22 x 102

3.7

R-CH2Br

1.38 x 102

2.3

R-O-CH3

3.48 x 102

5.8

RCHO

5.4 x 102 – 6.0 x 102

9-10

RO-H

30-3.00 x 102

0.5 – 5

3.6 x 102 – 4.8 x 102

6.0-8.0

6.0 x 102 – 7.8 x 102

10 – 13

RCOOH

Table 10.12 chemical shift values for methylene protons with tetramethylsilane (TMS) as a standard

65

Proximity Functional Group -Cl -Br -I

-OH -OR

-N, amines -SH

-NO2

CH3 -

α

Β

2.3 0 1.8 0 1.3 0 1.4 5 1.2 5 1.2 0 0.9 0 2.4 5 2.4 5 2.9 5 2.9 0 1.2 5 1.2 0 1.1 0 3.5 0 0.9 0

0.6 0 0.8 0 1.1 0 0.3 5 0.2 5 0.2 5 0.0 5 0.4 0 0.3 0 0.4 0 0.4 0 0.2 0 0.4 0 0.4 5 0.6 5 0.1 5

-CH2 -

α

β

2.3 0 2.1 5 1.9 5 1.4 5 1.1 0 1.0 0 0.7 5 2.3 0 2.3 0 3.0 0 2.9 5 1.4 0 1.3 0 1.1 0 3.1 5 0.8 0

0.5 5 0.6 0 0.6 0 0.5 5 0.3 5 0.3 0 0.1 0 0.2 0 0.1 5 0.4 5 0.4 5 0.1 5 0.3 0 0.4 0 0.8 5 0.0 5

CH-

α

β

2.5 5 2.2 0 2.7 0 1.3 5 0.9 5 0.9 5 1.2 5 2.3 0 2.1 0 3.3 0 3.4 5 1.3 5 1.3 0 1.0 5 3.0 5 0.3 5

0.1 5 0.2 5 0.3 5 0.0 0 0.0 0 0.0 0 0.0 0 0.0 0 0.0 0 0.0 0 0.0 0 0.0 0 0.0 0 0.0 0 0.0 0 0.0 0

Table 10.13 predicts the rough proximity locations of the chemical shifts (ppm) of methyl, methylene, and methine protons attached to primary (alpha) and secondary (beta) electron withdrawing groups.

The results of adding the α and β values to the standard values give only close approximations to the experimental values.

66

Methyl Protons

Chemical Shift range, hertz (cps)

Chemical Shift range, ppm (δ): hertz/60.

R-CH3

54

0.90

C=C-C-CH3

57

0.95

0-C-CH3

78

1.3

C=C-CH3

110

1.8

1.41 x 102

2.35

CH3COR

1.29 x 102

2.15

1.6 x 102

2.6

R-O-CH3

2.01 x 102

3.35

C=C-O-CH3

2.3 x 102

3.8

2.31 x 102

3.85

R-COOCH3

2.3 x 102

3.8

N-CH3

1.29 x 102

2.15

+N- CH3

2.0 x 102

3.3

1.8 x 102

3.0

S-CH3

1.3 x 102

2.1

O2N-C-CH3

96

1.6

1.1 x 102

1.8

1.4 x 102

2.3

RCO-N-CH3

Table 10.14 Chemical shift values for methyl protons with tetramethylsilane (TMS) as a standard

There are four non-equivalent protons in 1,2-dichloropropane, compound III.

67

Protons “a” and “b” are in different environments; therefore, they will give rise to separate 1H NMR signals. At first glance, it would appear that protons a and b are equivalent; however, a careful examination of the molecule (including making a model) in the format shown in III would indicate that protons “a” and “b” are in different environments. There are four different kinds of protons in t-butyl isobutyl ether, compound IV.

The different protons are: nine protons for “a”; two protons for “b”; six protons for “c”; and one proton for “d”; therefore, the ratio of different types of protons would be 9:2:6:1. How many 1H NMR signals would cyclobutane exhibit? How many 1H NMR signals would ethanol, CH3CH2OH, exhibit? Proton magnetic resonance spectroscopy cannot distinguish between mirror images. Compounds IV and V, enantiomers of 2-chlorobutane, would not give separate 1

H NMR signals.

68

The geometric isomers cis-2-bromo-2-butene, compound VI, and trans-2-bromo-2butene, compound VII, have the vinylic proton in different environments. Consequently, the chemical shifts would be slightly different.

Spin-spin Interaction High resolution 1H NMR spectra of organic compounds exhibit peak patterns due to spin interaction of neighboring nuclei. Spin-spin interactions or peak splitting provides information about protons on adjacent carbon atoms. Spitting patterns are helpful in structure elucidation. Figure 10.24 is the low resolution 1H NMR spectrum of ethyl alcohol. Figure 10.25 is the high resolution 1H NMR spectrum of ethyl alcohol.

69

Figure 10.24 the low resolution 1H NMR spectrum of CH3CH2OH

Figure 10.25 the high resolution 1H NMR spectrum of CH3CH2OH found at http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/nmr1.htm

Spin-spin interaction results from the tendency of bonding electrons to pair its spin with the spin of the nearest proton nucleus. The spin transfer effect will, in turn, affect the second bonding electron and this effect will be felt by the nearest adjacent protons. One proton has influence on the spin orientation of an adjacent proton through the bonding electrons. The terminology used "is that the protons couple each other". The coupling causes the splitting of protons on adjacent carbon atoms. For example, the following system would exhibit two types of signals: 70

The magnetic field the “a” protons experience is slightly increased or decreased by the spin of the tertiary proton “b.” The field experienced by proton “a” increases if the tertiary proton is aligned with the applied field or decreased if the tertiary proton is aligned against the applied field. Therefore, for ½ of the molecules, absorption by the “a” protons is shifted slightly downfield and for the remaining ½ of the molecules, the absorption is slightly upfield. The singlet for protons “a” would be split into a doublet of equal intensity. On the other hand, the field experienced by proton “b” can be increased, decreased, or unaffected by the spins of protons “a.” Protons “a” can align with the applied field, align against the field, or have no impact on the field in two different ways as demonstrated in the following manner.

Proton “b” would appear as a triplet with peak intensities of 1:2:1 as indicated in Figure 10.26.

Figure 10.26 spin-spin interaction of protons in 1,1-dichloro-2-bromoethane The number of peaks found in the multiplet, resulting from proton coupling, depends on the number of ways the spin of the interacting nuclei can arrange to give different magnetic moments. 71

For example, the spin-spin interaction of the methyl and methylene protons in CH3CH2X would give rise to the following spectrum.

Replacing one of the methyl hydrogen atoms with a deuterium atom gives XCH2CH2D, and the spin-spin interaction would give rise to the following spectrum:

Replacing two of the methyl hydrogen atoms with deuterium atoms gives XCH2CHD2, and the spin-spin interaction would give rise to the following spectrum.

Replacing three of the methyl hydrogen atoms with deuterium atoms gives XCH2CD3 , and the spin-spin interaction would give rise to the following spectrum.

72

The coefficients of the binomial expansion (a+b)n where n is equal to the number of protons on adjacent carbon atoms are equal to the peak ratios of the multiplet.

For example, protons on a carbon atom adjacent to carbon atoms containing six protons would exhibit a septet (seven lines) with intensities calculated by expanding the binomial equation:

Therefore, the ratio for the septet would be 1:6:15:20:15:6:1. The proton magnetic resonance spectrum of ethanol, CH3CH2OH, exhibits a triplet for the methyl hydrogen atoms because of the spin-spin interactions of the hydrogen atoms on the carbon atom adjacent to the methyl group, the methylene group. The hydrogen atoms on the methyl group split the signal of the hydrogen atoms on the methylene group, -CH2-, into a quartet. The hydrogen atom of the OH group exchanges rapidly with the hydrogen atoms of other ethanol molecules; therefore, it does not remain in position long enough to see splitting unless conditions exist to eliminate the exchange process. Figures 10.27 and 10.28 represent the spin-spin interactions of the methylene and methyl protons. 73

Figure 10.27 possible spin orientations for the methylene, –CH2-, protons

Figure 10.28 possible spin orientations for the methyl, -CH3, protons

As indicated in Figure 10.28, the methyl protons can have eight possible spin combinations. Two sets of three combinations are equivalent; therefore, the adjacent protons of the methylene group would experience a quartet with a peak ratio of 1:3:3:1. The 1H NMR spectrum of 1,1-dichloro-2,2-dibromethane, compound VIII, would exhibit two doublets as showed in Figure 10.29.

74

Figure 10.29 1H NMR spectrum of compound VIII

The Ha and Hb protons are in different environments; therefore, each proton would be split into a doublet as a consequence of the number of possible spin orientations for each proton as indicated in Figure 10.30.

Figure 10.30 the spin orientations of Ha and Hb protons in compound VIII

Figure 10.30 shows that each proton has two spin combinations, +½ (↑) and -½(↓). This means that the protons will split each other into doublets of equal intensity. The Hb proton will be located farther downfield from the Ha proton because chlorine has a greater electronegativity value than bromine; therefore, the Hb is more deshielded and would experience a lower field effect. This information is useful for predicting the 1H NMR spectrum for 2-chloropropane

75

compound IX.

The six protons of the two methyl groups are equivalent. These six protons will be split into a doublet by the proton on the neighboring methine group. The proton on the methine group will be split into a septet with peak intensities 1:6:15:20:15:6:1. Figure 10.31 illustrates the 1H NMR spectrum of compound IX.

Figure 10.31 1H NMR spectrum of 2-chloroproane

The septet signal for the methine proton would occur at a lower field since it is adjacent to an electronegative chlorine atom. The methine proton’s septet signal and the methyl protons’ doublet would exhibit a peak ratio of 1:6 with the intensity of the septet, as calculated from the binomial theorem, (a+b)6, 1:6;15:20:15:6:1, and the intensity of the doublet would be 1:1. Figure 10.32 reflects these details.

76

Figure 10.32 1H NMR spectrum of 2-chloropropane showing the spin-spin integration and integration to obtain the ratio of 1:6

The splitting patterns of protons are due to the spin-spin interaction of protons on adjacent carbon atoms. The observed splitting pattern for a particular absorption is always one more than the number of adjacent protons. If the adjacent protons are equivalent, then no splitting pattern will occur, e.g., the 1H NMR spectra for 1,2dibromoethane, compound X, and dioxane, compound XI, will exhibit singlets (no splitting, because all the hydrogen atoms are equivalent) as shown in Figures 10.33 and 10.34 respectively.

77

Figure 10.33 representation of the 1H NMR spectrum of compound X

Figure 10.34 representation of the 1H NMR spectrum of compound XI

The J coupling constant The coupling constant, J, cited in cps, is the distance between the peaks of the

78

multiplex. Figure 10.35 illustrates the J coupling constant exhibited by 1,1-dichloro-2bromoehane.

1, 1-dichloro-2-bromoethane

Figure 10.35 the J coupling constants for the splitting patterns in 1,1-dichloro-2- bromoethane

The J coupling constant is a measure of the effectiveness of spin coupling, and it is independent of the applied field. Diamagnetic Anisotropy Electron density surrounding the proton determines the chemical shift. The inductive effect of the bonded groups is a major factor in influencing the chemical shift.

79

However, this type of shielding or deshielding effect is not sufficient to explain the low field resonance of the aldehydic proton (Figure 10.36), and the aromatic protons (Figure 10.36), or the high field resonance of acetylenic protons (Figure 10.36). Diamagnetic Anisotropy is the primary factor that influences the unexpected signals for aldehydic protons, aromatic protons, or acetylenic protons.

Figure 10.34 aldehydic proton; aromatic proton, and acetylenic proton

Diamagnetic Anisotropic is the tendency of the π electrons to circulate when placed in an applied magnetic field. The electron circulation produces a small magnetic field that opposes or augments the applied field. Opposition to the applied magnetic field results in a shielding effect which causes the protons to resonate at a high field. If the anisotropic effect augments the applied field, the protons will experience deshielding and will resonate at low field. The magnetic anisotropic effect is common with multiple bonds such as in aromatic compounds, acetylenic compounds, and carbonyl compounds. Figure 10.37 illustrates the Diamagnetic Anisotropic Effect produced by the circulating electrons in benzene

80

Figure 10.37 anisotropic effect generated by the circulating π electrons in the benzene ring

The spinning nuclei of the aromatic ring generate a magnetic field that augments the external magnetic field; therefore, these protons exhibit a downfield shift in its 1H NMR spectrum. A close examination of Figure 10.37 suggests that protons in the region above the aromatic ring should be shielded and, therefore, should exhibit an up field shift. This is the case in compound XI where protons b, c, d, and e would exhibit proton resonance as indicated in Table 10.15

81

Proton a b c d e

Chemical Shift, δ(ppm) 3.81 2.50 1.30 0.90 0.30

Table 10.15 the chemical shifts a proton a, b. c, d and e in compound XI

Figure 10.38, a representation of the spectrum of compound XI, shows that protons d and e reside directly above the aromatic ring and are, therefore, shielded by the magnetic field generated by the circulating π electrons, and exhibit a high field shift ( an upfield chemical shift or a low δ value).

Figure 10.38 theoretical representation of the 1H NMR spectrum of compound XI

Diamagnetic anisotropy can be used to explain the low field resonance of aldehydic protons. Figure 10.39 illustrates the anisotropy for aldehydic protons.

82

Figure 10.39 Diamagnetic Anisotropy in Acetylenic Systems

Analogous to the aromatic protons in the plane of the ring, the aldehydic protons are situated so that the field generated by circulating Ď€ electrons in the carbonyl group generates a field that augments the applied field; therefore, causing the aldehydic proton to be deshielded and exhibit a downfield δ value around 9 ppm, a much lower field than would have been predicted without the consideration of the Diamagnetic Anisotropic Effect. 13

C NMR Spectrometry

Wide-band heteronuclear decoupled 13C spectra indicate the different types of carbon atoms in a compound. The coupled 13C is complex and gives insight into the number of hydrogen atoms attached to the carbon atoms. However, coupled 13C spectra exhibit large J values. Noise generators generate noise or heteronuclear decoupled spectra. The following is an example of a 13C NMR decoupled spectrum of diethyl phthalate.

83

Diethyl phthalate

The spectrum exhibits six lines.

The carbon atoms in dethyl phthalate are labeled in the following manner.

84

The labeled carbon atoms would exhibit the following C-13 downfield signals relevant to TMS as a reference.

13

C NMR like 1H NMR gives useful structural information; however, 13C NMR provides direct information about the carbon skeleton of a molecule. C-13 spectra give information about the number and type of carbons and the chemical shifts relevant to the chemical environment of the carbon. The intensity of the signals is weaker than the proton signals; therefore, an increase in the signal-to-noise ratio is required - this is done by pulsed Fourier Transform (FT) – NMR. Fourier Transformation is the mathematical manipulation of the data to plot spectra.

Tetramethylsilane TMS) is the reference compound that relates 13C NMR chemical shifts. 85

p-Ethoxy ethoxybenzene exhibits the following decoupled 13C NMR spectrum.

p-ethoxy ethoxybenzene

The spectrum exhibits four signals in reference to TMS. This indicates that the molecule has four carbon atoms in different chemical environments. The labeled carbon atoms give rise to the different signals in the carbon-13 spectrum.

The C-13 decoupled signals would correspond to the following spectrum.

86

13

C NMR spectra are not necessarily concerned with electrical integration and spinspin interaction. The Fourier-Transformed pulse technique distorts the signal intensities for carbons without hydrogen atoms; therefore, the integration is misleading. Fundamentally, the different signals in the 13C NMR spectra give rise to the different types of carbon atoms in the molecule, i.e., if the carbon-13 spectrum exhibits five signals, then the molecule has carbon atoms in five different chemical environments. Table 10.18 lists the chemical shifts for several carbon atoms relative to tetramethylsilane.

87

Type of Carbon

Approximate Chemical Shift, PPM, relevant to TMS R-CH3 8-35 R2CH2 15-50 R3CH 20-60 R4C 30-40 C-O 50-80 C-N 40-60 C-Cl 35-80 C-Br 25-65 C-I 0-40 RCONR2 165-175 RCO2R 165-175 RCO2H 175-185 RCOH 190-200 RCOR 205-220 ≥CR 65-85 alkene carbon 100-150 Aromatic C6Hx 110-170

Table 10.18 C-13 carbon Shifts Relative to TMS

Other types of Magnetic Resonance Spectrometry In addition to the traditional proton magnetic spectroscopy encountered in an introductory organic chemistry course, there are other types of magnetic resonance spectrometric research tools that chemist have reported in the literature including the following: (a) Distortionless Enhancement of NMR Signals by Polarization Transfer (DEPT), a one Dimensional NMR technique, that enhances the intensities of 13C signals and provides information about the number of attached protons to the carbon atoms (b) Insensitive Nuclei Enhanced by Polarization Transfer (INEPT), a one dimensional NMR technique, that gives similar results as DEPT but less detail. (c) Nuclear Overhauser Effect Spectroscopy (NOE), a one Dimensional NMR technique, that involves irradiation of the sample at specific frequencies before a signal is acquired, and enhancing the intensities of nearby nuclei. (d) Insensitive Nuclei Enhanced by Polarization Transfer (INEPT), a one dimensional

88

NMR technique. This technique is similar to DEPT, but with less detail. (e) Nuclear Overhauser Effect Spectroscopy (NOE), a one Dimensional NMR Technique, where the sample is irradiated at specific frequencies before the signal is acquired, and enhancing the intensities of nearby nuclei. (f) Nuclear Overhauser Effect Spectroscopy (NOESY), a two-dimensional version of the NOE experiment that yields a display of all atoms that are close in space. The diagonal and the projection on each axis gives rise to one dimensional spectra. (g) Correlated Spectroscopy (COSY), a two-dimensional NMR technique which displays all protons that are coupled. The diagonal and the projection on each axis are the one-dimensional spectra. The off-diagonal peaks indicate the presence of coupling between pairs of protons. The following is an example of a COSY spectrum of ethylbenzene copied from http://chem.ch.huji.ac.il/nmr/techniques/2d/cosy/cosy.html

(h) Spin Echo Spectroscopy (SECSY), a two dimensional NMR technique, yields the same information as COSY, but with a different format. (i) J-Resolved 2D NMR, a two Dimensional NMR technique, has a normal spectrum on one axis and coupling on the other axis. (j) Heteronuclear Shift Correlation (HETCOR), a two dimensional NMR technique, correlates the 13C spectrum on one axis and 1H spectrum of the other axis. 89

(k) Incredible Natural Abundance Double Quantum Transfer Experiment (INADEQUATE), a two dimensional NMR technique, that directly obtains carboncarbon connectivity, and, ultimately, the carbon skeleton of a molecule. (l)

Inadequate Sensitivity Improvement by Proton Indirect Detection (INSIPID), a two Dimensional NMR technique, is a reverse detection INADEQUATE experiment that greatly reduces the acquisition time.

DEPT 13C NMR Let’s examine one of these other types of NMR spectra, the DEPT 13C NMR spectrum, in more detail. DEPT uses a second radio frequency transmitter to irritate the sample. The frequency excites the protons affecting the appearance of 13C spectra by transferring polarization to 13C nuclei. The resulting spectra provide information about the number of protons attached to carbon atoms. DEPT spectra are “edited” spectra since the peak intensities and phases are modified relative to normal 13C spectra. DEPT spectra unambiguously identify the CH3, CH2, and CH carbon peaks. The technique uses three separate spectra: (1) At 45o, DEPT-45 (2) At 90o, DEPT-90 (3) At 135o, DEPT-135. DEPT does not detect the quaternary carbons or solvents. The following is a representation of the 13C NMR spectrum of 4-hydroxy-3-methyl-2butanone (compound I).

O CH3 d

C a

c CH

b CH2OH

CH3 e

90

I c

b

e

d

a CDCl3

240

220

200

180

160

140

120

100

80

40

60

0

20

The three DEPT spectra of 4-hydroxy-3-methyl-2-butanone are: (a) DEPT-45 c

b

240

220

200

180

160

140

120

100

80

60

e

d

40

20

0

DEPT-45 spectra exhibit positive phases (above the horizontal line) for methyl, CH3, carbon atoms; methylene, CH2, carbon atoms; and methane, CH, carbon atoms. The carbon atoms without hydrogen atoms (examples quaternary carbon atoms and carbonyl carbon atoms) and carbon atoms from solvents will not appear in DEPT-45 spectra. Consequently, the carbonyl carbon, carbon atom “a,” and the carbon atom for the solvent deuterochloroform disappear. Carbon atoms “b”, “c”, “d”, and “e” are in positive phase (above the horizontal line)

91

(b) DEPT-90 c

240

220

200

180

160

140

120

100

80

40

60

d

b

0

20

e

DEPT -90 provides spectra in which carbon atoms with three hydrogen atoms and two hydrogen atoms attached are in negative phase (below the horizontal line) and carbon atoms with one hydrogen atom attached are in positive phase (above the horizontal line). Consequently, the “c” carbon atom of 4-hydroxy-3-methyl-2-butanone has one hydrogen atom attached and, therefore, is in a positive phase (above the horizontal line). Carbon atoms “d”, three hydrogen atoms attached; “e”, three hydrogen atoms attached; and “b”, two hydrogen atoms attached, are in negative phase (below the horizontal line). (c) DEPT-135 c

220

200

180

160

140

120

100

80

60

e

d

40

20

0

b

DEPT-135 provides spectra in which carbon atoms with two hydrogen atoms attached are in negative phase and carbon atoms with one hydrogen atom attached and three hydrogen atoms attached are in positive phase. Consequently, the “b” carbon atom of 4-hydroxy-3-methyl-2-butanone has two 92

hydrogen atoms attached and is in a negative phase. Carbon atoms “c”, one hydrogen atom attached; “d”, three hydrogen atoms attached; and “e”, three hydrogen atoms attached, are in a positive phase. DEPT spectra assist spectroscopists to identify the types of carbon atoms in molecules.

93

Mass Spectrometry Mass spectrometry is an analytical instrumental method that determines the structural features of organic molecules using their fragmentation patterns. The following equation defines the fragmentation patterns:

Volatilizing a liquid sample in vacuo and passing it through an ionization chamber generates the fragmentation patterns that define a mass spectrum. The edge of a special insertion probe serves as a way to inject the sample into the mass spectrometer. Bombarding the sample with 70 electron volts (eV) of energy removes one electron from the bonding or nonbonding molecular orbital, therefore, ionizing the molecule. The resulting ion is called the parent ion or molecular ion. The molecular ion may fragment into ions of smaller

) 9

ratios.

A positive potential repels the positively charged fragment from the ionization chamber, and an accelerator plate (2000 V) accelerates the positive ion through a magnetic field. The fragments deflect with a certain radius of curvature depending on the

) 9

ratio

according to Equation 10.11. Equation 10.11 Centrifugal force = Centripetal force

94

Where H is the strength of the magnetic field e is the charge on the ion v is the velocity of the ion m is the mass of the ion r is the radius of curvature of the ion in the magnetic field Diagram 10.1 is the representation of a working mass spectrometer (http://www.chemguide.co.uk/analysis/masspec/howitworks.html)

Diagram 10.1 Mass Spectrometer

Figure 10.40 is the output data for the mass spectrum of a molecule with a mass equal to 132 amu.

95

Figure 10.40 Representation of the Mass Spectrum of a molecule

The mass spectrum represented by Figure 10.40 shows a molecular ion peak at m/e ratio equal to 132 (this means the molecular mass is 132 g/mol) and fragmentation patterns at e/m at 103, 77, 63, 51 and 39. Equations 10.12 explains the data in the spectrum. The mathematics begins by recognizing that potential energy = kinetic energy. Equation 10.12

eV = the potential energy of the ion accelerated by a potential voltage, V, and ! đ?‘šđ?‘Ł - is the kinetic energy. -

V is the voltage; e is the charge on the particle; m is the mass of the particle; and v is the velocity of the particle.

The radius of curvature of an accelerating particle in a magnetic field is given by Equation 10.13.

96

Equation 10.13

where H is the strength of the magnetic field; e is the charge on the particle; m is the mass of the particle; r is the radius of curvature of the particle; and v is the velocity of the particle. Manipulating Equation 10.13 leads to Equation 14

Equation 10.14

H2r 2 m = 2V e From the rearrangement of

: &9 &;& )&

= đ?‘Ł - to give đ??ť- đ?‘&#x; - =

)& < & 9&

!

and using Equation 10.12, đ?‘’đ?‘‰ = - đ?‘šđ?‘Ł 2 đ?‘’đ?‘‰ = đ?‘šđ?‘Ł - đ??ť- đ?‘&#x; - =

�(2 ��) �97

đ??ť- đ?‘&#x; - = : &;& - =

�(2 �) �

)

= 9

Equation 10.14

Equation 10.14 is a mathematical expression for calculating the mass-to-charge (m/e) ratio of a moving particle of a specified charge in a magnetic field, H, with a voltage V with a designated radius of curvature. Following is the derivation of Equation 10.14 from Equations 10.12 and the manipulated form of Equation 10.13. Therefore, every particle accelerated in a magnetic field would have a

) 9

ratio equal to

the square of the strength of the magnetic field times the square of its radius of curvature in the magnetic field divided by twice the voltage within the magnetic field. Every fragmentation particle would have exposure to the same voltage and the same magnetic strength; therefore, the difference in their charge-to-mass ratios would be the radius of curvature within the magnetic field. The larger the particle, the larger the radius of curvature and larger fragments would give rise to greater mass-to-charge values. Most of the ions that pass through the collector slip have a plus one charge (+1). The mass spectrometer generally functions by holding r and H constant and vary V, the )

accelerating potential. There is an indirect relationship between V and 9 . The result is that the mass of a high

) 9

ratio registers first.

The current of the ions entering the collection slit is recorded as a peak on a chart. The height of the peak is directly proportional to this current. Compounds containing C, H, O, and S have molecular ion peaks that are even, and

98

odd

) 9

ratio fragmentation patterns unless rearrangement occurs or there is a loss of a

neutral fragment. If rearrangement occurs or if there is a loss of a neutral fragment from a radical-carbocation, the molecular ion, then the m/e ratio will be even. Compounds containing C, H, and N usually have odd ) 9

) 9

molecular ion peaks, and even

fragmentation patterns unless rearrangement occurs or there is a loss of a neutral

fragment. If rearrangement occurs or if there is a loss of a neutral fragment from a radical-carbocation, the molecular ion peak will be odd. The base peak in the mass spectrum is the peak with the highest m/e ratio. The parent peak or molecular ion peak is generally the peak of appreciable intensity at the highest m/e ratio. Isotope peaks also exist as the molecular ion peak “plus 1” peak or the molecular ion “plus 2” peak. Depending on the number of ions present, isotope peaks may exist at mass-to-charge ratios higher than the parent peak “plus 2” mass-tocharge ratio. The mass-to-charge ratios for other peaks represent major fragmentation patterns that are characteristic of the molecular structure of the molecule. The following are some examples of the major fragmentation patterns of molecules that are characteristic of the molecular structures of molecules. These fragmentation patterns occur in a constant magnetic field with slight variations in electrical potentials. Primary Alcohols

A m/e of 31 is the base peak for primary alcohols. The molecular ion peak depends on )

the size of the primary alcohol but would follow a # $ pattern of 32, 46, 60, 74, 88, 102, 9

99

116, 130, etc. )

Also, there would be a variety of other # 9 $ peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions. Secondary Alcohols The mass-to-charge ratio for secondary alcohols depends on the alkyl groups attached. First, the bulkier alkyl group would be cleaved from the molecular ion leaving )

a fragment that would exhibit # 9 $ equal to the mass of the smaller alkyl group + carbon atom + hydrogen atom + hydroxyl group. For example, 2-pentanol would lose )

a propyl radical from the molecular ion to give rise to a base peak at # 9 = 45$. This base peak would consist of a methyl group + hydrogen atom + carbon atom + oxygen atom + hydrogen atom = 15 + 1 +12 + 16 + 1 = 45.

100

The mass-to-charge ratio for the base peak of secondary alcohols would depend on the alkyl groups attached to the carbon atom that has the -OH group attached. Tertiary Alcohols The mass-to-charge ratio for tertiary alcohols depends on the alkyl groups attached. First, the bulkier alkyl group would be cleaved from the molecular ion leaving a )

fragment that would exhibit a # 9 $ ratio equal to the mass of the smaller alkyl groups + carbon atom + hydroxyl group. For example, 2-methyl-2-pentanol would lose a propyl )

radical from the molecular ion to give rise to a base peak at # 9 = 59$. This base peak would consist of two methyl groups + carbon atom + oxygen atom + hydrogen atom = 15 + 15 +12 + 16 + 1 = 59.

101

Primary Amines

)

A # 9 = 30$ would be the base peak for primary amines. The molecular ion peak would 102

)

depend on the size of the primary amine, but would follow a # 9 $ pattern of 31, 45, 59, 73, 87, 101, 115, 129, etc. Also, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions. Secondary Amines The mass-to-charge ratio for secondary amines depends on the alkyl groups attached. First, the bulkier alkyl group minus a methylene group would be cleaved from the )

molecular ion leaving a fragment that would exhibit # 9 $ equal to the mass of the smaller alkyl group + hydrogen atom + nitrogen atom + carbon atom + two hydrogen atoms. For example, N-methylpropylamine would lose an ethyl radical from the )

molecular ion to give rise to a base peak at # 9 = 44$ This base peak would consist of a methyl group + hydrogen atom + nitrogen atom + carbon atom + two hydrogen atoms = 15 + 1 +14 + 12 + 2 = 44.

103

Tertiary Amines The mass-to-charge ratio for tertiary amines depends on the alkyl groups attached. First, the bulkier alkyl group minus a methylene group would be cleaved from the molecular ion leaving a fragment that would exhibit a

) 9

ratio equal to the mass of the

smaller alkyl groups + nitrogen atom + carbon atom + two hydrogen atoms. For example, N, N-dimethylpropylamine would lose an ethyl radical from the molecular ion )

to give rise to a base peak at # 9 = 58$ . This base peak would consist of two methyl groups + nitrogen atom + carbon atom + two hydrogen atoms = 15 + 15 +14 + 12 + 2 = 58.

104

Straight Chain Alkanes

)

Straight chain alkanes would exhibit a base peak at # 9 = 43$, and the molecular ion peak would depend on the size of the alkane but would change by 14, the mass of a methylene, CH2, group. )

Also, there would be a variety of other # 9 $ peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

105

Branched Alkanes

The base peak would be formed at the highest branch as illustrated with isopentane; therefore, the base peak would be at a mass to-charge ratio of 43 for iso-alkanes, and the molecular ion peak would depend on the size of the iso-alkane, but would follow a pattern of 14, the mass of a methylene group.

106

Again, the base peak would be formed at the highest branch; therefore, t-butyl carbocation exhibiting a mass -to-charge ratio of 57 would be the base peak. The molecular ion peak would depend on the size of the 2,2-dimethylalkane, but would follow a pattern of 14 (the mass of a methylene group), 28, 42, etc. the mass of a methylene group for straight chain alkyl groups. )

Also, there would be a variety of other # 9 $ peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

107

Alkenes

)

The base peak for propene would be at a # 9 $ ratio of 27. )

In addition, there would be a variety of other # 9 $ peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

108

)

The base peak for terminal alkenes would be at # 9 = 41$ because the allyl carbocation (stabilized by resonance) is more stable than a vinyl carbocation. Also, there would be a variety of other peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

Cyclohexene loses a neutral fragment (ethene) as illustrated by the fragmentation patterns above where the ethene is lost from the molecular ion, a radical-carbocation; )

therefore, the base would have an even # 9 $ ratio. This type of fragmentation pattern was mentioned previously. )

Also, there would be a variety of other # 9 $ peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

109

Carbonyl Compounds

)

)

Methyl ketones give rise to a base peak at # 9 = 43$. The peak at # 9 $ equal to 43 is stabilized by resonance , and is referred to as the methyl acylium ion. The molecular ion peak of methyl ketones depends on the size of the methyl ketone.

)

)

Phenyl ketones give rise to a base peak at # 9 = 105$. The peak at # 9 $ equal to 105 is stabilized by resonance , and is referred to as the phenyl acylium ion (oxophenylmethyllium ion). The molecular ion peak of phenyl ketones depends on the size of the phenyl ketone.

110

)

In addition, there would be a variety of other # 9 $ peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions. Primary Alkyl Bromides

A m/e of 93 would be the base peak for primary alkyl bromides. The molecular ion peak would depend on the size of the primary alkyl bromide, but would follow a pattern of 108, 122, 136, 150, 164, 178, 192, 206, etc. Primary alcohol bromides would also exhibit a characteristic molecular ion + 2 peak that would be about 97% as intense as the molecular ion peak. )

In addition, there would be a variety of other # 9 $ peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

111

Esters

The base peak of esters depends on the group attached to the carbonyl group. If the

112

)

group is a methyl group, then the base peak would be at # 9 = 43$. If the group is a )

propyl group, then base peak would be at # 9 = 71$. Esters with alkyl groups that have two or more carbon atoms attached to the oxygen of the Ester can undergo the McLafferty rearrangement.

As illustrated in the above fragmentation pattern, the McLafferty rearrangement results in rearrangement of a hydrogen atom followed by the loss of a neutral fragment. Consequently, the base peak has an even mass to charge ratio. The mass to charge ratio of the base peak depends on the size of the group attached to the carbonyl group. For example, once the molecular ion forms, if a methyl group is attached to the carbonyl group, and an ethyl group is attached to the oxygen atom, then a hydrogen atom on the methyl group of the ethyl group rearranges to the oxygen of the carbonyl group. Movement of electrons would occur as illustrated in the McLafferty fragmentation pattern resulting in the loss of ethene (a neutral fragment), and the formation of an even

) 9

fragment at

) 9

= 60.

Other neutral fragments can be lost, but the rearrangement would involve the transfer of a hydrogen atom. For example, if a propyl group is attached to the oxygen atom, 113

the neutral fragment lost would be propene. If an ethyl group is attached to the )

carbonyl group, then the base peak would be at # 9 = 74$. If a propyl group is )

attached to the carbonyl group, then the base peak would be at # 9 = 88$. If a butyl )

group is attached to the carbonyl group, then the base peak would be at # 9 = 102$. The McLafferty rearrangement is not unique to esters. Compounds with the following molecular arrangements can exhibit the McLafferty rearrangement.

Where Z and Q are electronegative atoms with electrons in nonbonding molecular orbitals. In addition, there would be a variety of other

) 9

peaks resulting from the fragmentation

of the molecular ion or fragmentations from other ions. Arenes Aromatic nuclei with alkyl side chains, arenes, form a base peak at a mass to charge )

ratio equal to 91. The # 9 $at 91 is due to the formation of benzyl carbocation that rearranges to a tropyllium ion. The benzyl carbocation is formed by cleavage at the 114

carbon atom attached to the aromatic nucleus. The mass-to-charge ratio of the molecular ion or parent peak depends on the structure of the R group.

The benzyl carbocation then rearranges to the tropyllium ion by the following pathway:

115

Incidentally, the tropyllium ion obeys Hückel’s rule for aromaticity. Isotope Peaks Other isotope peaks appear in mass spectra. For example, sulfur has a molecular ion “plus 2” isotope peak that is approximately 5% of the molecular ion peak, and chlorine has a molecular ion “plus 2” isotope peak that is approximately 33% of the molecular ion peak.

116

The statement throughout this discussion “Also, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions� can be explained by evaluating the mass spectra of a couple of compounds. For example, the evaluation of the spectrum of benzophenone would lead to the following fragmentation patterns.

benzophenone

Mass Spectrum of Benzophenone

117

The molecular ion peak The base peak

118

The m/e =77 is the result of the heterolytic cleavage of the bond between the phenyl group and the carbonyl group. This is an example of “other m/e peaks resulting from the fragmentation of the molecular ion.” The m/e of 51 is the result of the loss of a neutral fragment, acetylene, from the phenyl )

carbocation. The # 9 $ ratio is odd because the neutral fragment is the result of the )

fragmentation of a carbocation by homolytic cleavage. This an example of “other # 9 $ peaks resulting from the fragmentations of other ions.”

119

The peak at mass-to-ratio 152 is more difficult to explain. It is a small peak; therefore, the mechanism for its formation must be more involved than the mechanism for the formation of the other fragmentation patterns. A mass/charge ratio of 152 involves rearrangement and the loss of a neutral fragment. The process described is a stepwise solution that attempts to show the various homolytic cleavages required to arrive at the desired fragment. Consequently, the proposed fragmentation pattern concerted, and the following represents a rationale for the formation of the fragment.

120

Following is a representation of resonance stabilization of the fragment at m/e at 152.

121

The evaluation of the spectrum of n-butylamine leads to the following fragmentation patterns.

n-butylamine The mass spectrum of n-butylamine

122

)

The # 9 = 43$ is the result of the heterolytic cleavage of the bond between the propyl )

group and the carbon attached to the amino group. This is an example of “other # 9 $ peaks resulting from the fragmentation of the molecular ion.�

123

Problems 1. Convert 3.96 Îźm to cm-1 2. Give a brief explanation about the theory of infrared spectroscopy.

3. Calculate the approximate stretching vibration, in wavenumbers, for the C F bond if the force constant, k, is approximately 760 N/m. 4. How man fundamental vibrations are predicted for the non-linear water molecule? How many fundamental vibrations are predicted for the linear carbon dioxide molecule? 5. Lists four reasons why all predicted fundamental vibrations of a bond may not appear in the infrared spectrum. 6. Draw the expected fundamental vibrations for CO2 and label the degenerate 124

modes. 7. Using the data in Table 10.2, calculate the wave numbers for the C≡N and C≡C stretching vibrations. 8. Predict which of the following compounds would give little or no C≡C stretching transmittance bands: (a) H-C≡C-CH3

(b) H-C≡C-CH2CH3

(c) CH3C≡CCH3

(d) CH3CH2 CH2C≡CCH3

9. The carbon-oxygen double bond appears at a higher wavenumber (higher frequency of infrared radiation) than the carbon-oxygen single bond. Give an explanation for this observation.

10. A certain compound gives a carbonyl absorption at approximately 1775 cm-1. Which of the following compounds would give rise to a transmittance band at 1775 cm-1? Give an explanation for your selection. O CH3

C a

O

O

O

CH3 b

c

d

11. When a carbonyl group is conjugated with a C-C double bond, the conjugation lowers the observed wave number by approximately 20 cm-1 as compared to a nonconjugated carbonyl. For example, compound A has a strong transmittance band at 1715 cm-1 and compound B has a strong transmittance band at 1685 cm-1.

125

O

CH3 CH

CH2

C

O

CH3 C

CH3

CH

C

CH3

CH3

CH3

B

A

Based on the transmittance band for cyclohexanone at 1715 cm-1 and 1724 cm1 for cyclopentanone, predict the absorption for

O

O and

12. How could you use infrared spectroscopy to distinguish between the following compound pairs? Indicate the characteristic absorptions expected for each structure. (a) O

O CH3CH2CH2

C

OH

C

CH3CH2CH2

H

(b) O

O CH3CH2CH2

C

CH3

CH3CH2CH2

C

OCH3

126

(c)

O

O

(d)

CH3CH2C

N

CH3

C

C

CH3

(e)

O

O

(f) CH3 CH3

C

OH

CH3CH2CH2CH2OH

CH3

(g)

CH3CH2CH2CH2CH3

CH3CH CH3

CH2

CHCH3 CH3

127

(h) CH3CH2C

CH3

CH3CH2CH2 OCH3

O

(i)

CH3CH2CH2 OH

CH2

C

OH

O

(j)

CH3 H3C

O

CH

CH3

CH

C OH

C O

CH3

13.

CH

H3C

O H3C

O

C

Sketch the 1H NMR spectrum for a.

b. H C H

H

H

H C

C C

CH3

H

C O

CH3

O

128

14. The aroma of coffee is, in part, related to compound A, C5H8O2. Compound A can be represented by the following theoretical H1 NMR spectrum. C5H8O2, exhibits a strong IR transmittance at 1750 cm-1. Suggest a structure for compound A.

1

H NMR of Compound A

129

15.

2’-deoxyuridine, compound B, has the following H1 NMR spectrum:

H 1'

H 2'

CH2OH 5'

H

N

4 3 5

H

4'

HO 3'

H H

H

O

6

2 1

N

H

O Compound B

Assign as many signals as you can to the appropriate protons in 2’-deoxuridine.

130

16. Determine structures for the products from the following pathway:

(CH3)3CCH=CHC(CH3)3 B2H6

C10H23B (infrared spectrum at 2500 cm-1) A (1) 30% H2O2 (2) H2O

C10H22O B Following is the H1 NMR of B:

1

H NMR of Compound B

131

17. Suggest structures for molecules exhibiting the following spectra. Give rationales for your answers by explaining the chemical shifts, spin-spin interactions and, if provided, the electrical integration for each spectrum. (a) Determine the structural formula for C7H7ClO; the compound exhibits the following spectra:

1

H NMR for C7H7ClO

13

C 132

NMR for C7H7ClO

(b) Determine the structural formula for C9H9ClO; the compound exhibits the following spectra:

1

13

H NMR for C9H9ClO

C NMR for C9H9ClO

133

(c) Determine the structural formula for C12H14O4 , the compound exhibits the following spectra:

1

13

H NMR for C12H14O4

C NMR for C12H14O4

(d) Determine the structural formula for C5H9O4N , the compound exhibits the following spectra:

134

1

H NMR for C5H9O4N

13

C NMR for C5H9O4N

(e) Determine the structural formula for C6H14O, the compound exhibits the following spectra:

135

1

H NMR for C6H14O

13

C NMR for C6H14O

(f) Determine the structural formula for C7H12O4, the compound exhibits the following proton NMR and C-13 spectra:

136

H1 NMR for C7H12O4

13

C NMR for C7H12O4

137

(g) Determine the structural formula for C4H10O, the compound exhibits the following spectra:

1

H NMR for C4H10O

13

C NMR for C4H10O

138

(h) Determine the structural formula for C11H9NO4, (challenging)the compound exhibits the following spectra:

4 3

1

δ 8.2

δ 5.7

1 TMS

δ 2.5

δ1.7

δ0

Characteristic infrared absorptions at about 2222 cm-1 3333 cm-1 1751 cm-1.

139

18. Determine the structure of C7H8S if the compound exhibits the following spectra

data:

1

13

H NMR for C7H8S

C NMR for C7H8S

140

partial mass spectrum of C7H8S

19. Determine the structure of C9H10 if the compound exhibits the following proton magnetic resonance and carbon-13 magnetic resonance spectra.

141

20. Determine the structure of C12H14O4 if the compound exhibits the following the following proton magnetic spectrum:

142

21. Suggest a reasonable structure for C8H10 from its 13C NMR spectrum, its 13C NMR DEPT-45 spectrum, its 13C NMR DEPT-90 spectrum, and its 13C NMR DEPT-135 spectrum.

The 13C NMR spectrum for C8H10

13

C NMR DEPT-45 spectrum for C8H10

143

13

C NMR DEPT-90 spectrum for C8H10

13

C NMR DEPT-135 spectrum for C8H10

144

Solutions to Problems Chapter 10

1. Convert 3.96 μm to cm-1

3.96 x 10 −6 x

ν cm −1 =

1m 100 cm x = 3.96 x 10 −4 cm = λ 6 1 x 10 m 1m

1 1 = = 2.53 x 10 3 cm −1 −4 λ 3.96 x 10 cm

2. Give a brief explanation about the theory of infrared spectroscopy. An infrared spectrum occurs when the frequency of infrared radiation incident on organic molecules equals the frequency of the vibrating bonds attached to atoms in the molecule. The infrared energy absorbed is equal to the energy associated with the vibrating molecule. The radiation absorbed can be measured, or the light transmitted can be measured. The light absorbed or transmitted is related to the wavelength or wave number corresponding to the light absorbed or transmitted.

3. Calculate the approximate stretching vibration, in wavenumbers, for the C-F bond if the force constant, k, is approximately 760 N/m.

145

1 s = ν m −1 = 5.309 x 10 −10 λm m

1 s = ν m −1 = 5.309 x 10 −10 λm m

1 s = ν m −1 = 5.309 x 10 −10 λm m

760 kg-m/ms 2 (12.0 g/6.022 x10 23 ) x (19.0 g/6.022 x 10 23 ) 12.0 g 19.0 g + 23 6.022 x 10 6.022 x 10 23 760 kg-m/ms 2 (1.99 x 10 −23 g) x (3.155 x 10 −23 g) 1.99 x 10 −23 g + 3.155 x 10 −23 g

760 kg/s 2 6.278 x 10 −46 g 2 5.145 x 10 −23 g

1 760 kg/s 2 −10 s = ν m −1 = 5.309 x 10 λm m 1.22 x 10 −23 g 1 s 760 kg/s 2 = ν m −1 = 5.309 x 10 −10 λm m 1.22 x 10 −23 g x 1 kg 1000 g

1 1 = ν m −1 = 5.309 x 10 −10 m −1 x 2.50 x 1014 = 1.33 x 10 5 m −1 λm s

1 1m = ν cm −1 = 1.33 x 10 5 m −1 x = 1.33 x 10 3 cm −1 λcm 100 cm

146

4. How many fundamental vibrations are predicted for the non-linear water molecule? How many fundamental vibrations are predicted for the linear carbon dioxide molecule? The theoretical number of fundamental vibrations for water would be determined by 3n -6 3(3) - 6 = 3 The theoretical number of fundamental vibrations for the linear molecule carbon dioxide, CO2, would be determined by 3n - 5 3(3) - 5 = 4 5. Lists four reasons why all predicted fundamental vibrations of a

bond may not

appear in the infrared spectrum. There is no change in the dipole moment of the molecule during vibration. Absorption occurs outside the region of the spectrum under observation. The vibrations result in absorptions so close that they coalesce. The absorptions are too weak to be seen. 6. Draw the expected fundamental vibrations for CO2 and label the degenerate modes. The number of vibrations expected for carbon dioxide, a linear molecule, would be 3 (3)-5 = 4. The four possible vibration modes are:

147

The two bending modes, C and D, are degenerate (the same in energy); therefore, only one bending vibration is observed in the infrared (IR). The symmetrical stretching vibration A does not involve a change in dipole moment, i.e., the dipole moment of A is zero; therefore, this stretching vibration is infrared inactive. These illustrations can be repeated for any of the axes, and the results would be the same. So no matter how the molecule moves in space, there are only two vibrations that would be IR active – the asymmetrical stretching vibration and the bending vibration. Therefore, the IR spectrum of carbon dioxide would be:

148

7. Using the data in Table 10.2, calculate the wave numbers for the C≡N and C≡C stretching vibrations. The calculated wavenumber for the carbon-nitrogen triple bond is: 1 s = ν m −1 = 5.309 x 10 −10 λm m

1750 kg-m/ms 2 (12.0 g/6.022 x 10 23 )(14.0 g/6.022 x 10 23 ) 12.0 g 14.0 g + 23 6.022 x 10 6.022 x 10 23

1 1750 kg-m/ms 2 −10 s = ν m −1 = 5.309 x 10 λm m 1.07 x 10 −23 g 1 s 1750 kg-m/ms 2 = ν m −1 = 5.309 x 10 −10 λm m 1.07 x 10 −23 g x 1 kg 1000 g

1 s 1 = ν m −1 = 5.309 x 10 −10 x 4.045 x 1014 = 2.15 x 10 5 m −1 λm m s 149

1 1m = ν cm −1 = 2.15 x 10 5 m −1 x = 2.15 x 10 3 cm −1 λcm 100 cm The calculated wavenumber for the carbon-carbon triple bond is: 1 s = ν m −1 = 5.309 x 10 −10 λm m

1560 kg-m/ms 2 (12.0 g/6.022 x 10 23 )(12.0 g/6.022 x 10 23 ) 12.0 g 12.0 g + 23 6.022 x 10 6.022 x 10 23

1 1560 kg-m/ms 2 −10 s = ν m −1 = 5.309 x 10 λm m 9.95 x 10 −24 g 1 s 1560 kg-m/ms 2 = ν m −1 = 5.309 x 10 −10 λm m 9.95 x 10 −24 g x 1 kg 1000 g

1 s 1 = ν m −1 = 5.309 x 10 −10 x 3.96 x 1014 = 2.10 x 10 5 m −1 λm m s

1 1m = ν cm −1 = 2.10 x 10 5 m −1 x = 2.10 x 10 3 cm −1 λcm 100 cm

8. Predict which of the following compounds would give little or no C≡C stretching transmittance bands: (a)

H-C≡C-CH3

(c) CH3C≡CCH3

(b) H-C≡C-CH2CH3 (d) CH3CH2 CH2C≡CCH3

150

The infrared stretching vibration for the triple bond in “c” would not be observed because there is no change in the dipole moment of the molecule during vibration. 9. The carbon-oxygen double bond appears at a higher wave number (higher frequency of infrared radiation) than the carbon-oxygen single bond. Give an explanation for this observation. The carbon-oxygen double bond has a stronger force constant than the carbonoxygen single bond, and the force constant is directly proportional to the wave number. 10. A certain compound gives a carbonyl absorption at approximately 1775 cm-1. Which of the following compounds would give rise to a transmittance band at 1775 cm-1? Give an explanation for your selection.

O CH3

C a

O

O

O

CH3 b

c

d

The infrared spectrum of “d” (cyclobutanone) would exhibit a carbonyl absorption at 1775 cm-1 . Cyclobutanone has angle strains of about 19.5o, and this strain increases the numerical value of the force constant, k. The force constant is directly proportional to the wave number; consequently, the wave number increases from an expected value of about 1724 cm-1 to 1775 cm-1.

151

11. When a carbonyl group is conjugated with a C-C double bond, the conjugation lowers the observed wave number by approximately 20 cm-1 as compared to a nonconjugated carbonyl. For example, compounds A has a strong transmittance band at 1715 cm-1 and compound B has a strong transmittance band at 1685 cm-1.

O

CH3 CH

CH2

O

CH3

C

C

CH3

CH

C

CH3

CH3

CH3

B

A

Based on the transmittance band for cyclohexanone at 1715 cm-1 and 1724 cm-1 for cyclopentanone, predict the absorption for

O

O and

The transmittance band for 2-cyclohexenone would be 1695 cm-1, and the transmittance band for 2-cyclopentenone would be 1704 cm-1.

152

12. How could you use infrared spectroscopy to distinguish between the following compound pairs? Indicate the characteristic absorptions expected for each structure.

(a)

O

O CH3CH2CH2

C

OH

C

CH3CH2CH2

H

The carboxylic acid would exhibit a broad 0H stretching vibration between 3500 cm1

-2500 cm-1 and a carbonyl stretching vibration between 1720 cm-1 - 1696 cm-1.

The aldehyde would exhibit a C-H stretching vibration of the hydrogen attached to the carbonyl group between 2830 cm-1 - 2695 cm-1, and a carbonyl stretching vibration at about 1724 cm-1.

(b)

O

O CH3CH2CH2

C

CH3

CH3CH2CH2

C

OCH3

The ketone would exhibit a stretching vibration between at approximately 1724 cm1

.

The ester would exhibit a carbonyl stretching vibration between 1750 cm-1 - 1735 cm-1, and carbon-oxygen transmittance bands at 1250 cm-1 and 1051 cm-1.

153

(c)

O

O

2-Cyclohexenone would exhibit a carbonyl stretching vibration at 1695 cm-1, and cyclohexanone would exhibit a carbonyl stretching vibration at 1715 cm-1. (d)

CH3CH2C

N

CH3

C

C

CH3

The nitrile would exhibit a sharp transmittance band between 2260 cm-1 - 2222 cm-1. The Alkyne would not exhibit a transmittance band between 2260 cm-1

- 2100 cm-1 ,

because the molecule is symmetrical; therefore, there is no change in the dipole moment of the molecule during vibration. (e)

O

O

Cyclobutanone (transmittance band at approximately 1780 cm-1) is more strained than cyclohexanone (transmittance band at approximately 1715 cm-1); therefore, the carbonyl transmittance band (approximately 1780 cm-1) for cyclobutanone would appear at a higher wavenumber.

154

(f)

CH3 CH3

C

OH

CH3CH2CH2CH2OH

CH3 The t-butyl group is characterized by two bending vibrations at 1395 cm-1 – 1385 cm-1 and a more intense transmittance band near 1370 cm-1. These transmittance bands are not exhibited 1-butanol. (g)

CH3CH2CH2CH2CH3

CH3CH

CH2

CH3

CHCH3 CH3

The isopropyl group is characterized by a strong doublet between 1388 cm-1 – 1380 cm-1 as a consequence of bending vibrations. This doublet is not exhibited by npentane.

(h) CH3CH2C

CH3

CH3CH2CH2 OCH3

O

2-Butanone would exhibit a strong transmittance band at about 1724 cm-1. This is not the case for ethyl methyl ether; however, the ether would exhibit a strong transmittance band between 1150 cm-1 – 1085 cm-1.

155

(i)

CH3CH2CH2 OH

CH2

C

OH

O The carboxylic acid, cyclohexylacetic acid, would exhibit a broad transmittance band between 3300 cm-1 – 2500 cm-1 and an intense transmittance band for the carbonyl group between 1720 cm-1 – 1695 cm-1. 1-Propanol would exhibit a transmittance band between 3200 cm-1 – 3000 cm-1. (j)

CH3 H3C

CH

O

CH CH3

O

C O

H3C

CH3 H3C

CH

C OH

C O

2-Methylpropanoic acid would exhibit a broad transmittance band between 3300 cm1

– 2500 cm-1 and an intense transmittance band for the carbonyl group between 1720

cm-1 – 1695 cm-1. -1

1786 cm

The anhydride would exhibit two strong transmittance bands at

and 1724 cm-1

156

13. Sketch the H1 NMR spectrum for

a.

b. H C H

H

H

H C

C C

CH3

H

C O

CH3

O

157

a.

J is the spacing coupling; therefore, Jac has a larger value than Jab ; Jac has a larger value than Jbc; and Jab has a larger value than Jbc.

158

b.

The hydrogen atoms on the methyl group would appear as a doublet at about 1.2 ppm. The hydrogen atoms on methine group would appear as a septet (a multiplex) farther down field (at approximately 2.8 ppm) than the hydrogen atoms on the methylene group. The hydrogen atoms on the methylene group would appear as a doublet at about 2.3 ppm.

159

14. The aroma of coffee is, in part, related to compound A, C5H8O2. Compound A can be represented by the following theoretical H1 NMR spectrum. C5H8O2, exhibits a strong IR transmittance at 1750 cm-1. Suggest a structure for compound A.

H1 NMR of Compound A

160

15. 2’-deoxyuridine, compound B, has the following H1 NMR spectrum:

H 1'

H 2'

CH2OH 5'

H

N

4 3 5

H

4'

HO 3'

H H

H

O

6

2 1

N

H

O Compound B

NMR spectrum of compound B

161

Assign as many signals as you can to the appropriate protons in 2’-deoxyuridine. The hydrogen atoms that may be distinguishable are: (1) The two hydrogen atoms on the carbon-carbon double bond labeled 4 and 5. The hydrogen atom on carbon 4 is a distorted doublet that appears at about 6.2 ppm, and the hydrogen atom on carbon 5 is a distorted doublet that appears at about 5.6 ppm. (2) The two hydrogen atoms on the carbon 2 that appear as a singlet at about 5.2 ppm. (3) The hydrogen atom on carbon 4’ that appears as a distorted triplet close to 4.0 ppm, and the two hydrogen atoms on carbon 5’ that appear as a distorted triplet around 3.9 ppm. The remaining hydrogen atoms are difficult to identify from the spectrum because they exhibit overlapping signals in proximity fields.

162

16. Determine structures for the products of the following pathway:

(CH3)3CCH=CHC(CH3)3 B2H6

C10H23B (infrared spectrum at 2500 cm-1) A (1) 30% H2O2 (2) H2O

C10H22O B Following is the H1 NMR of B:

163

1

H NMR of Compound B The structural formula for C10H23B would be:

The structural formula for C10H22O would be:

164

17. Suggest structures for molecules exhibiting the following spectra. Give rationales for your answers by explaining the chemical shifts, spin-spin interactions and, if provided, the electrical integration for each spectrum. Determine the structural formula for C7H7ClO; the compound exhibits the following spectra:

1

H NMR for C7H7ClO

Electrical Integration: 2:2:3 Chemical Shifts: Para-substituted pattern between 6.9 ppm- 7.3 ppm for a disubstituted aromatic compound in which the groups attached to the aromatic nucleus are different. Methyl protons at 3.2 ppm suggests that the carbon atom of the methyl group is attached to an electronegative atom. Spin-spin Interaction: No spin-spin interaction for the methyl protons; therefore, the protons of the methyl group appear as a singlet.

165

The spin-spin interaction of the aromatic protons are complicated, but signal of paradisubstituted aromatic protons resemble what appears to be a quartet, but aromatic protons attached to the carbon atoms of para-disubstituted aromatic compounds resemble what appears to be a quartet, but, in reality, the spin-spin interaction process is somewhat more complicated.

13

C NMR for C7H7ClO

The 13C NMR spectrum suggests that there are five types of carbons in C7H7ClO. (1) An aliphatic carbon atom attached to an oxygen atom at about 63 ppm (2) Four (4) aromatic carbon atoms between 110 ppm-170 ppm. Oxygen is more electronegative than chlorine. The aromatic carbon atom at 159 ppm is attached to an oxygen atom; the carbon atom at 130 ppm is attached to a chlorine atom; the carbon atom at about 123 ppm is attached to a carbon atom adjacent to the carbon atom attached to the oxygen atom; and the carbon atom at about 118 ppm is attached to the carbon atom adjacent to the carbon atom attached to the chlorine atom. 166

The data from the spectra suggest that the structure is p-chloroanisole.

Determine the structural formula for C9H9ClO; the compound exhibits the following spectra:

1

H NMR for C9H9ClO

Electrical Integration: 2:2:2:3 Chemical Shifts: Para-substituted pattern between 7.5 ppm- 8.0 ppm for a disubstituted aromatic compound in which the groups attached to the aromatic nucleus are different.

167

The spectrum exhibits methyl protons at 1.0 ppm and methylene protons at about 2.6 ppm. This suggests that the methylene carbon atom is attached to an electronegative group. Spin-spin Interaction: The protons on the carbon atom of the methyl group are split into a triplet. The triplet suggests that the methyl group is attached to a carbon atom with two protons attached (the methylene group). The protons on the carbon atom of the methylene group are split into a quartet. The quartet suggest that the methylene group is attached to a carbon atom with three protons (the methyl group). The methyl group and the methylene group form an ethyl group, CH3CH2 -

The adjacent protons of the methylene group would experience a quartet with a peak ratio of 1:3:3:1.

168

The adjacent protons of the methyl group would experience a triplet with a peak ratio of 1:2:1. The spin-spin interaction of the aromatic protons are complicated, but aromatic protons attached to the carbon atoms of para-disubstituted aromatic compounds resemble what appears to be a quartet, but, in reality, the spin-spin interaction process is somewhat more complicated.

13

C NMR for C9H9ClO

A careful analysis of the 13C NMR spectrum suggests that there are seven types of carbons in C9H9ClO. Two aliphatic carbon atoms between 13 ppm-38 ppm. One carbonyl carbon atom at 220 ppm Four (4) aromatic carbon atoms between 110 ppm-170 ppm. The carbon atom of the methyl group is at 13 ppm. The carbon atom of the methylene group is at 38 ppm. Chorine is more electronegative than the carbonyl group.

169

The aromatic carbon atom at 138 ppm is attached to a chlorine atom; the carbon atom at 126 ppm is attached to a carbonyl group; the carbon atom at 125 ppm is adjacent to the carbon atom attached to the carbonyl group; and the carbon atom at about 124 ppm is attached to the carbon atom adjacent to the carbon atom attached to the carbonyl group. The data from the spectra suggest that the structure is p-chlorophenyl ethyl ketone (2[p-chlorophenyl]-1-propanone).

(a) Determine the structural formula for C12H14O4 , the compound exhibits the following spectra:

170

1

H NMR for C12H14O4

Electrical Integration: 4:2:3 Chemical Shifts: Para-substituted appears apparent at about 8.2 ppm for a disubstituted aromatic compound in which the groups attached to the aromatic nucleus are identical. The spectrum exhibits methyl protons at 1.2 ppm and methylene protons at about 3.9 ppm. The methylene carbon atom is attached to a strong electronegative group. Spin-spin Interaction: The protons on the carbon atom of the methyl group are split into a triplet. The triplet suggests that the methyl group is attached to a carbon atom with two protons (a methylene group). The protons on the carbon atom of the methylene group are split into a quartet. The quartet suggests that the methylene group is attached to a carbon atom with three protons (a methyl group). The methyl group and the methylene group form an ethyl group, CH3CH2 -

13

C NMR for C12H14O4

171

Analysis of the 13C NMR spectrum suggests that there are five types of carbons in C12H14O4. (1) Two types of aliphatic carbon atoms - one at 17 ppm and the other at 61 ppm. (2) Carbonyl carbon atom at about 175 ppm (3) Two (2) types of aromatic carbon atoms -one at 125 ppm and the other at 125 ppm. The carbon atom of the methyl group is at 13 ppm. The carbon atom of the methylene group is at 38 ppm. The shift at 38 ppm suggests that one of the aliphatic groups, the methylene group, is attached to an oxygen atom. The aromatic carbon atom at 135 ppm is attached to a carbonyl group; the carbon atom at 125 ppm is adjacent to the carbon atom attached to the carbonyl group. The data from the spectra suggest that the structure is diethyl terephthalate (diethyl benzene-1,4-dicarboxylate).

(d) Determine the structural formula for C5H9O4N , the compound exhibits the following spectra:

172

1

H NMR for C5H9O4N Electrical Integration: 2:1:3:3

Chemical Shifts: There are four (4) types of protons exhibited by the spectrum. The signals include: (1) Methyl protons at about 1.2 ppm (2) Another methyl group at about 1.4 ppm (3) Methine proton at about 2.6 ppm. The methine proton is attached to an electronegative atom (probably nitrogen) (4) Methylene protons at about 4.0 ppm. Spin-spin Interaction: The methyl protons at 1.2 ppm are split into a doublet. The doublet suggests that the methyl group is attached to a carbon atom with one proton attached (a methine group). The methine proton at 2.6 ppm is split into a quartet. This suggest that the methine proton is on a carbon atom attached to a carbon atom with three protons (a methyl group).

173

The methyl protons at 1.4 ppm are split into a triplet. The triplet suggests that the methyl group is attached to a carbon atom with two protons attached (a methylene group). The methylene proton at 4.0 ppm is split into a quartet. This suggests that the methylene proton is on a carbon atom attached to a carbon atom with three protons (a methyl group). The methyl group and the methylene group form an ethyl group, CH3CH2 The methyl group and the methine group form

Where the X and the Y represent atoms with protons attached.

13

C NMR for C5H9O4N

Analysis of the 13C NMR spectrum suggests that there are five types of carbons in C5H9O4N.

174

(1) Four aliphatic carbon atoms - one at about 17 ppm; one at about 18 ppm; one at about 38 ppm; and one at about 61 ppm. (2) Carbonyl carbon atom at about 183 ppm The shift at 38 ppm suggests that one of the aliphatic groups, the methine group, is attached to a nitrogen atom. The shift at 61 ppm suggests that one of the aliphatic groups, the methylene group, is attached to an oxygen atom. The data from the spectra suggest that the structure is ethyl 2-nitropropanoate.

(e) Determine the structural formula for C6H14O, the compound exhibits the following spectra:

175

1

H NMR for C6H14O

Electrical Integration: 1:6 Chemical Shifts: There are two (2) types of protons exhibited by the spectrum. The signals include: Methyl protons at about 1.2 ppm Methine proton at about 3.7 ppm. The methine proton is attached to an electronegative atom (oxygen) Spin-spin Interaction: The methyl protons at 1.2 ppm are split into a doublet. The doublet suggests that the methyl group is attached to a carbon atom with one proton attached (a methine group). The methine protons at 3.7 ppm are split into a multiplet (a septet). The septet suggests that the methine group is attached to carbon atoms with six protons (two methyl groups).

176

13

C NMR for C6H14O

Analysis of the 13C NMR spectrum suggests that there are two types of carbons in C6H14O. (1) One type of carbon atom at about 21 ppm (2) Another type of carbon atom at about 75 ppm The shift at 75 ppm suggests that the carbon atom of the methine group is attached to an oxygen atom. The data from the spectra suggest that the structure is diisopropyl ether.

(f) Determine the structural formula for C7H12O4, the compound exhibits the following NMR and C-13 spectra:

177

H1 NMR for C7H12O4 Chemical Shifts: There are three (3) types of protons exhibited by the spectrum. The signals include: (1) One at about 1.5 ppm (2) Another at about 3.4 ppm. (3) The final one at 4.0 ppm The chemical shifts at 3.4 ppm and 4.0 ppm suggest that the protons reside on carbon atoms that are attached to electronegative atoms. Spin-spin Interaction: The protons at 1.5 ppm are split into a triplet. The triplet suggests that the protons are attached to a carbon atom that is attached to a carbon atom that has two protons attached (a methylene group). The protons at 3.4 ppm form a singlet. Therefore, the protons exhibiting no spin-spin interaction (a singlet suggests that they are attached to a carbon atom that is attached to atoms that do not have protons attached).

178

where X and Y do not have protons attached The protons at 4.0 ppm form a quartet. The quartet suggests that the protons are attached to a carbon atom that is attached to a carbon atom that has three protons attached (a methyl group).

13

C NMR for C7H12O4

Analysis of the 13C NMR spectrum suggests that there are four types of carbons in C7H12O4. (1) Three aliphatic carbons at 17 ppm; 39 ppm; and 61 ppm respectively (2) A carbonyl carbon at about 179 ppm The shift at 61 ppm suggests that the carbon atom is attached to an oxygen atom, and the shift at 39 ppm suggests that the carbon atom is attached to a carbonyl group.

179

The data from the spectra suggest that the structure is diethyl malonate.

(g) Determine the structural formula for C4H10O, the compound exhibits the following spectra:

1

H NMR for C4H10O

Electrical Integration: 2:1:1:6 Chemical Shifts: There are four (4) types of protons exhibited by the spectrum. The signals include: (1) Six methyl protons at about 1.0 ppm

180

(2) A methine proton at about 1.8 ppm. The methine proton is attached to a carbon atom that is attached to an electronegative atom (oxygen) (3) Two methylene protons at 3.5 ppm that are attached to a carbon atom attached to an oxygen atom. (4) A proton attached to an oxygen atom at 2.9 ppm. Spin-spin Interaction: The methyl protons at 1.0 ppm are split into a doublet. The doublet suggests that the methyl group is attached to a carbon atom with one proton attached (a methine group). The methine protons at 1.8 ppm are split into a multiplet. The multiplex suggests that the methine group is attached to carbon atoms with several protons (in this case about 8 protons (two methyl groups and a methylene group). The methylene group at 3.5 ppm are spit into a doublet. The shift at 2.9 ppm is due to the proton directly attached to an oxygen atom. The protons of alcohol (OH) generally undergo rapid exchange with neighboring molecules; therefore, the OH shift is normally observed as a singlet.

13

C NMR for C4H10O

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Analysis of the 13C NMR spectrum suggests that there are three types of carbons in C4H10O. The three aliphatic carbons exhibit chemical shifts at approximately 19 ppm; 32 ppm; and 72 ppm respectively The shift at 72 ppm suggests that the carbon atom is attached to an oxygen atom, and the shift at 39 ppm suggests that the carbon atom is attached to a carbon atom that is attached to an oxygen atom. The data from the spectra suggest that the structure is isobutyl alcohol (2-methyl-1propanol).

(a) Determine the structural formula for C11H9NO4, the compound exhibits the following spectra (a challenging problem):

4 3

1

δ 8.2

δ 5.7

1 TMS

δ 2.5

δ1.7

δ0

proton magnetic spectrum Electrical Integration: 4:1:1:3 182

Chemical Shifts: There are four (4) types of protons exhibited by the spectrum. The signals include: (1) methyl protons at 1.7 ppm (2) A methine proton at about 5.7 ppm. The methine proton is attached to a carbon atom that is attached to an electronegative atom (oxygen) (3) proton on an oxygen atom giving a signal at 2.5 ppm (4) Four aromatic protons at 8.2 ppm. There is no spitting of the aromatic hydrogens because the electronegative groups attached in a para arrangement are similar, but not exactly the same. Such a phenomenon is frequently observed in magnetic resonance spectroscopy. Spin-spin Interaction: The methyl protons at 1.7 ppm are split into a doublet. The doublet suggests that the methyl group is attached to a carbon atom with one proton attached (a methine group). The methine protons at 5.7 ppm are split into a quartet, and each signal of the quartet is spit a doublet indicating that the splitting occurs with different J-coupling constants. The resulting signal suggests that the methine group is attached to a carbon atom with three protons attached and another atom with one proton attached.

This other atom indicated as X in the partial structure must be oxygen. Therefore, the partial structure would be:

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Characteristic infrared absorptions at about: 2222 cm-1 3333 cm-1 1751 cm-1 The transmittance band at 2222 cm-1 is the carbon-nitrogen triple bond stretching vibration of a nitrile. The transmittance band 3333 cm-1 is the O-H stretching vibration. The transmittance band 1751 cm-1 is the a partial carbonyl stretching vibration of an anhydride. These data suggest that the structure is 2-hydroxypropanoate p-cyanobenzenoate anhydride.

18. Determine the structure of C7H8S if the compound exhibits the following spectra data:

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1

H NMR for C7H8S

13

C NMR for C7H8S

185

partial mass spectrum of C7H8S

186

19. Determine the structure of C9H10 if the compound exhibits the following proton magnetic resonance and carbon-13 magnetic resonance spectra.

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20. Determine the structure of C12H14O4 if the compound exhibits the following the following proton magnetic spectrum:

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21. Suggest a reasonable structure for C8H10 from its 13C NMR spectrum, its 13C NMR DEPT-45 spectrum, its 13C NMR DEPT-90 spectrum, and its 13C NMR DEPT-135 spectrum. The 13C NMR spectrum for C8H10

The spectra suggest that there are five types of hydrogen atoms in C8H10 . These carbon atoms are methyl carbon atoms that appear between 8-35 ppm relevant to TMS, and four kinds of aromatic carbon atoms that appear between 110 - 170 ppm relevant to TMS. 13

C NMR DEPT-45 spectrum for C8H10

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The quaternary signals vanish in 13C NMR DEPT-45 spectra; therefore, a signal disappears in the DEPT-45 spectrum for C8H10 indicating that the structure contains a quaternary carbon atom. 13

C NMR DEPT-90 spectrum for C8H10

13

C NMR DEPT-90 spectra exhibit negative phases for methyl and methylene groups

and a positive phase for methine groups. The 13C NMR DEPT-90 of C8H10 exhibit a negative phase at 20 ppm suggesting that C8H10 contains a methyl group. 13

C NMR DEPT-135 spectrum for C8H10

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13

C NMR DEPT-135 spectra exhibit negative phases for methylene groups and positive

phases for methine and methyl groups. The 13C NMR DEPT-135 spectrum for C8H10 does not exhibit any negative phases. This suggests that C8H10 does not contain methylene groups. Therefore, the structure consistent with the C-13 spectra data is:

m-Xylene (1,3-dimethylbenzene)

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