Chapter 7 Stereochemistry
Yangpu Suspension Bridge (Shanghai, China) David Richardson
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“One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike -- and yet it is the most precious thing we have.” Albert Einstein
“Whether in the intellectual pursuits of science or in the mystical pursuits of the spirit, the light beckons ahead, and the purpose surging in our nature responds.” Arthur Stanley Eddington, The Nature of the Physical World
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Optical Activity Light possesses the property of a wave, and it vibrates at right angles to the direction in which it is traveling. Light vibrates in an infinite number of planes. Diagram 7.1 illustrates this principle:
Diagram 7.1 ordinary light vibrating in an infinite number of planes
Plane Polarized Light Plane polarized light is light that vibrates in one plane Diagram 7.2
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Diagram 7.2 plane polarized light
Ordinary light can be converted into plane polarized light by passing it through a lens made of polaroid material or through pieces of calcite (a crystalline form of CaCO3) called a Nicol prism. http://en.wikipedia.org/wiki/Nicol_prism An optically active substance is a substance that rotates plane-polarized light. When polarized light vibrating in one plane passes through an optically active substance, the plane-polarized light comes out vibrating in another plane, and a polarimeter detects the degree of rotation of the plane-polarized light. The polarimeter consists of a light source; two lenses (polaroid or Nicol prism); and, between the lenses, a tube to hold the optically active substance. The first lens polarizes the light, then the resulting plane-polarized light passes through the optically active substance, and finally a second lens functions as an analyzer. The observer viewing the plane-polarized light can detect the number of degrees the polarized light rotated- either to the left or right of its original propagation. Diagram 7.3 illustrates this principle.
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Diagram 7.3 plane polarized light rotated after passing through an optically active substance
If a substance is not optically active, then it does not affect plane-polarized light. On the other hand, if a substance is optically active, then it rotates plane-polarized light a specified number of degrees either to the right or to the left. If the optically active substance rotates plane-polarized light to the right, then it is dextrorotatory, and the number of degrees of rotation has a plus, +, sign. This substance rotates planepolarized light clockwise. If the optically active substance rotates plane-polarized light to the left, then it is levorotatory. A minus sign, "-", indicates the number of degrees the substance rotates plane-polarized light counterclockwise.
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One possible identifying mark of an optically active substance is a chiral atom. A chiral atom has four (4) different groups or ligands attached to a central atom. For example, lactic acid has two stereoisomers, compounds I and II. Compounds I and II have four different groups around their central carbon atoms:
These two compounds cannot be superimposed on each other where all the atoms would have the same spatial relationship. They both possess the same physical properties except one rotates plane-polarized light clockwise, i.e., it is dextrorotatory, and the other rotates plane-polarized light counterclockwise, i.e., it is levorotatory. Compound II comes from muscle tissue, and it is dextrorotatory. Since compound II rotates plane-polarized light to the right, (+)-lactic acid is a more descriptive nomenclature for this dextrorotatory lactic acid. Compound I rotates plane-polarized light counterclockwise; therefore, it is (-)-lactic acid. Compounds III, 2-methyl-1-butanol, and IV, 2-methyl-1-butanol, have four (4) different groups about their central carbon atoms. A carbon atom surrounded by four different groups is a chiral carbon atom or a chiral center. Though compounds III and IV have the same chemical and physical properties, they rotate plane-polarized light differently. One rotates plane-polarized light clockwise, and the other rotates plane-polarized light counterclockwise.
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Compound III is levorotatory and is (-)-2-methyl-1-butanol. Compound IV is dextrorotatory, and is (+)-2-methyl-1-butanol Specific Rotation The number of degrees that plane-polarized light rotates depends on the number of molecules of the optically active substance that comes into contact with the planepolarized light passing through the sample tube. Light will encounter twice as many molecules of the optically active substance in a sample tube that is 0.20 m long compared to a sample tube that is 0.10 m long. Consequently, the rotation of plane-polarized light passing through a 0.20 m sample tube containing a specified concentration of an optically active substance would be twice the rotation of plane-polarized light passing through a 0.10 m sample tube containing the same concentration of an optically active substance. Also, light will encounter twice as many molecules in a solution in a 0.20 m sample tube containing 2.00 x 10-2 kg/L compared to 1.00 x 10-2 kg/L. The specific rotation is the number of degrees of rotation of plane-polarized light observed in a 1.00 decimeter (dm) tube when the concentration of the optically active compound is 1.00 g/mL.
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[Îą ]20D C o
=
Îą observed length of the tube in decimeters x concentration in g/mL
Where D is the D line of Na with a wavelength of 5.893 x 10-7 m The specific rotation is as much of a physical constant as the melting point. http://en.wikipedia.org/wiki/Specific_rotation Enantiomers While doing some work on salts of tartaric acid, Louis Pasteur discovered that optically inactive sodium ammonium tartrate (a mixture of compounds V and VI) exists as a mixture of two different kinds of crystals that are mirror images of each other. Pasteur physically separated the mixture. When he had accomplished this meticulous task, he discovered that the mixture of compounds V and VI was optically inactive; however, after carefully separating compound V from compound VI and dissolving them in water, each compound exhibited optical activity. One rotated plane-polarized light clockwise and the other rotated plane-polarized light counterclockwise.
Pasteur proposed the existence of a new type of isomer whose structures differ only in being mirror images of one another, and whose properties differ in the direction each
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rotates plane-polarized light. These compounds are enantiomers. Compounds I and II are enantiomers, compounds III and IV are enantiomers, and compounds V and VI are also enantiomers. Enantiomers are non-superimposable compounds that are mirror images of each other. It is impossible to superpose one isomer on the other without breaking bonds. Compounds 1-IV possess one chiral center, and compound V and VI possess two chiral centers. Molecules possessing at least one chiral center are capable of existing as enantiomers. Chiral centers are also described as asymmetrical carbon atoms. Molecules that can exist as enantiomers are chiral molecules, and molecules that cannot exist as enantiomers are achiral molecules. However, a chiral center is not a criterion for the molecule existing as enantiomers, because the primary property of optical activity is the ability of the isomers to be non-superimposable or dissymmetric. Molecules with carbon atoms containing three different groups have mirror images, but their mirror images are superimposable; therefore, they are identical since they can be superimposed atom for atom on one another. Many common objects possess an enantiomeric relationship. For example, shoes, a helix, gloves, human hands, and right and left-handed screws. Properties of Enantiomers Enantiomers have identical physical properties such as boiling point, melting point, spectrum, refractive index, solubility, and specific gravity. One cannot separate enantiomers by conventional techniques such as distillation, chromatography, or crystallization since enantiomers have identical physical properties. One would need special separation techniques to separate enantiomers into their mirror image components. Enantiomers also have the same chemical properties; therefore, they would react identically with reagents that are not themselves enantiomeric. The reactivity of enantiomers differs with enantiomeric reagents. These reaction differences can separate enantiomers into their mirror image components.
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An equal mixture of the levorotatory isomer and the dextrorotatory isomer will not rotate plane-polarized light. Each enantiomer would cancel the rotation of the other. When the enantiomers exist in equal proportion, the mixture is a racemic modification. The dextrorotatory compound has a “+” sign and the levorotatory compound has a “-“ sign; therefore, the racemic modification has a “±” or dl designation. If there is no designation before the name of the chiral compound, the compound is a racemic modification. Practically all the naturally occurring chiral molecules exist as only one of the enantiomers. Nature rarely produces racemic modifications. The opposite occurs in the laboratory where the syntheses of molecules exhibiting chirality generally produce a racemic modification. The physiological properties of enantiomers can differ. There are multiple examples where one enantiomer has a pronounced physiological activity while its mirror image is devoid of such activity. The difference in physiological activity can be attributed to a difference in reactivity of enantiomeric pairs toward biochemical compounds such as enantiomeric enzymes and proteins. Also, the difference in physiological activity can be attributed to the difference in the products formed when enantiomers react with other enantiomers. Later we will see that this factor is the basis for separating enantiomers. Compounds VII and VIII, have chiral carbon atoms indicated by asterisks; therefore, they exist as enantiomers.
Both enantiomers, found in nature, have identical physical properties. Compounds VII
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and VIII are carvones. Compound VIII, β-carvone, isolated from spearmint, has the odor of spearmint. Compound VII, α-carvone, isolated from caraway seeds, has the odor of caraway seeds. One anticipates these distinct odors since the odor receptors in the nose are chiral and can distinguish between the two enantiomers. http://www.leffingwell.com/chirality/carvotanacetone.htm gives another perspective of compounds VII and VIII. Chirality is not an exclusive property of carbon compounds. Certain nitrogen, phosphorus and silicon compounds also possess chirality. For example, the
phosphorus compounds IX and X are enantiomers. The silicon containing compounds XI and XII are enantiomers.
Compounds IX and X are nonsuperimposable mirror images, and compounds XI and XII are nonsuperimposable mirror images. For further clarification of the nonsuperimposability of these molecules, try building models of their mirror images and attempt to overlap them atom for atom. Nitrogen compounds interconvert to their enantiomers by an inexplicable process identified as the “umbrella effect;” therefore, canceling the nonsuperimposability of the enantiomers (the criterion for optical activity).
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Absolute Configuration The (-) or (+) before the name of the enantiomer only specifies the direction of rotation of plane-polarized light. It has nothing to do with the configuration of the enantiomer (i.e., it does not indicate the arrangement of the atoms in the molecule). How, then, do we specify one enantiomer as opposed to its mirror image? One can distinguish the left hand and the right hand from one another by simply referring to the right hand as the right hand and the left hand as the left hand. Enantiomers are not distinguishable in such a simple manner. One distinguishes them by designating their chiral centers as R (Rectus, Latin for right) and S (Sinister, Latin for left). An R designation means that when the atom of the fourth priority hides behind the chiral center the remaining three attached atoms assume a clockwise movement from the first priority atom to the second priority atom and, finally, to the third priority atom. An S designation means that when the atom of the fourth priority hides behind the chiral center the remaining three attached atoms assume a counterclockwise movement from the first priority atom to the second priority atom and, finally, to the third priority atom. The rules for specifying absolute configuration follow the following Cahn-Ingold-Prelog priority rules: (1) Assign priority to the four atoms attached to the chiral carbon atom based on their atomic numbers. The atom with the highest atomic number attached to the chiral center has the first priority. For example, let’s consider an enantiomer of compound XIII, 1-bromoethano.
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l
Br, atomic number 35, would have the first priority O, atomic number 8, would have the second priority C, atomic number 6, would have the third priority H, atomic number 1, would have the fourth priority (2) The molecule is rotated so that the fourth priority atom hides behind the chiral center. The remaining three atoms are identified as positioned from the first priority to the second priority and, finally, to the third priority atom in a clockwise arrangement or a counterclockwise arrangement. For example, the chiral center in compound XIII has an “R� designation as a consequence of following the Cahn-Ingold-Prelog priority rules.
On the other hand, if the enantiomer had the atoms attached as in compound XIV
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And the molecule is rotated so that the fourth priority atom is placed behind the chiral center. The remaining three atoms would have the following arrangement:
Therefore, the chiral center in compound XIV would have an “S” designation as a consequence of following the Cahn-Ingold-Prelog priority rules. (3) The “R” configuration is specified when the atoms are arranged from first priority to third priority in a clockwise manner. (4) The “S” configuration is specified when the atoms are arranged from first priority to third priority in a counterclockwise manner. By observing these rules, both the configuration and the direction of rotation of plane polarized light of the enantiomer can be specified. Remember that the rotation of plane polarized light can only be determined by using a polarimeter. Also, the R and S configurations do not determine if a compound is levorotatory or dextrorotatory. A compound can be S and dextrorotatory, or R and levorotatory, or S and levorotatory, or R and dextrorotatory. The enantiomers of 2-chlorobutane, compounds XV and XVI, have the following
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structures:
Measurements with a polarimeter showed that compound XV, with the S designation at the chiral center is dextrorotatory; and that compound XVI, with the R designation at the chiral center, is levorotatory. Therefore, the IUPAC name for compound XV is S(+)-2-chlorobutane and the IUPAC name for compound XVI is R-(-)-2-chlorobutane. If two or more of the atoms attached to the chiral center are identical as in compound XVII, 2-bromobutane, then the following rules apply. Br H3C
C
CH2CH3
H XVII
If two or more atoms attached to the chiral center are identical, then go to the next atoms attached to the identical atoms. This process continues until one finds an atom of higher priority. Also, a triple bond attached to a chiral center would have priority over a double bond because in establishing priorities a double bond would be (a), and a triple bond would be (b).
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Therefore, the priority around the chiral center of compound XVI would be indicated in the following manner:
Therefore, the ethyl (CH3CH2) group would have priority over the methyl (CH33) group. In the case of the ethyl group, the next atom attached to the carbon atom (attached to the chiral center) is another carbon atom. In the case of the methyl group, the next atom attached to the carbon (attached to the chiral center) is a hydrogen atom. In compound XVII, the priority is Br > CH3CH3 > CH3 3> H, and the absolute configurations of the enantiomers of compound XVII would be:
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In considering the enantiomers of compound XVIII, 3-methylhexane, the propyl group would have the first priority, the ethyl group the second priority, and the methyl group the third priority. CH3CH2CH2
CH2CH2CH3
H H H3C
CH3CH2
CH2CH3
CH3
S-3-methylhexane
R-3-methylhexane XVIII
Compounds Containing Two or More Chiral Centers The maximum number of stereoisomers possible for compounds containing two or more chiral centers is 2n, where n is the number of chiral centers. If two or more of the chiral centers are identical, i.e., contain the same substituents, then the number of stereoisomers decreases. Compound XIX, 2,3-dichlorbutane, has two identical chiral centers.
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CH3
H
H
C
C
Cl
Cl
CH3
XIX
The enantiomers of compound XIX can be written in the following manner:
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A and B are different, and this difference can be determined by rotating A and B about the C2 – C3 sigma bond until the methyl groups are eclipsed.
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Rotation about the C2-C3 sigma bond in A results in eclipsing the methyl groups, the hydrogen atoms and the chlorine atoms. A and B are isomers, and the configurations A and B have mirror images. Let’s see if their mirror images are superimposable. The mirror image of A, A’, is
The mirror image of B, B’, is
A and A’ are superimposable. Simply rotate A’ 180o about the y axis as illustrated in Diagram 6.4 and A will be generated.
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Therefore, A and A’ are identical. In addition, the eclipsed conformation of A has a plane of symmetry and the staggered conformation of A has a point of symmetry.
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Stereoisomer A does not have a nonsuperimposable mirror image and is, therefore, optically inactive; therefore, compound A is a meso compound. Carbon atoms 2 and 3 of stereoisomer A have opposite configurations; consequently, there is an internal cancellation of rotation of plane-polarized light; therefore, the meso compound is a compound with an internal racemic modification. Stereoisomer B has a nonsuperimposable mirror image and can, therefore, exist as an enantiomeric pair. The enantiomers of B will rotate plane-polarized light an equal number of degrees to the right or the left. The enantiomers of B have identical physical properties and chemical properties except in reactions with other enantiomers. The meso compound A has physical and chemical properties different from those of the enantiomers of B. This is the case since A and B are not mirror images. The meso compound is the RS stereoisomer where there is an internal cancellation making the compound optically inactive. The enantiomers of B are the RR and SS stereoisomers, and one stereoisomer rotates plane-polarized light to the right and the other rotates plane-polarized light to the left. Diastereoisomers In review, there are three stereoisomers of compound XIX.
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CH3
H
H
C
C
Cl
Cl
CH3
XIX
These stereoisomers are the enantiomers B and B’, the RR and SS compounds, and the meso, RS, compound A. A and B and A and B’ are diastereoisomers. B and B’ are enantiomers.
A is a stereoisomer, but not a mirror image of B or B’. A is a diastereoisomer, a stereoisomer that is not a mirror image, of B or B’. Another example of diastereoisomers is the isomers of compound XX, 1,2dichlorocyclobutane.
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XX
Cyclobutane appears to be a rigid square geometric structure. In reality, this is not the case, because the cyclobutane ring is slightly puckered. Compounds C and C’ are non-superimposable mirror images; therefore, they are enantiomers. Compound D does not have a non-superimposable mirror image; therefore, it does not have an enantiomer. However, D has an isomeric relationship with C and C’. D is a stereoisomer that is not a mirror image of C and C’; therefore, D is a diastereoisomer of C and C’. In summary, C and C’ are enantiomers; C and D are diastereoisomers, and C’ and D are diastereoisomers.
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Compounds Containing Two Non-Identical Chiral Centers Recall that the maximum number of stereoisomers possible for a compound containing n number of chiral centers is 2n. For a compound with two non-identical chiral centers, the maximum number of stereoisomers is 22 = 4. Figure 7.1 illustrates the configuration for the four (4) expected stereoisomers and their relationship to each other for 2-chloro-3-bromobutane:
Figure 7.1 the enantiomers and diastereoisomers of 2-chloro-3-bromobutane
Figure 7.2 is the staggered arrangement of E, E’, F, and F’, the isomers of 2-chloro-3bromobutane.
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Figure 7.2 staggered conformations of the stereoisomers of 2-chloro-3-bromebutane Figure 7.2 shows that E and E’ are enantiomers; F and F’ are enantiomers; E’ and F are diastereoisomers; E and F are diastereoisomers; E and F’ are diastereoisomers. There are two sets of enantiomers and three sets of diastereoisomers. The number of stereoisomers possible for a compound containing four non-identical chiral centers would be 16 (24) different stereoisomers. For example, compound XXI, would have sixteen different stereoisomers.
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H C
O
H
C
OH
H
C
OH
H
C
OH
H
C
OH
H
C
OH
H XXI
Try drawing the sixteen stereoisomers of compound XXI.
Symmetry, a Better Method for Determining Chirality A chiral center conveys chirality to molecules; however, often it is not clear that a molecule is chiral. This is especially true for some cyclic compounds. A molecule can be chiral, but yet does not possess a chiral center. Three elements that should be looked for to ascertain if a molecule is nonchiral or achiral. Molecules are achiral if they possess a plane of symmetry; a point of symmetry; or a rotating axis of symmetry. Plane of Symmetry If a molecule has a plane of symmetry, it is achiral, and the molecule is nondissymmetric. A plane of symmetry is an imagined mirror that bisects the molecule in a way that half of the bisected molecule is the mirror reflection of the other half. For
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example, as indicated in Figure 7.3, a plane of symmetry can be drawn through compound XXII, dichloromethanol; compound XXIII, 1,2-dichloro-1,2-dibromoethane; compound XXIV, 1,2-dichlorocyclopropane; compound XXV, 1,2-dichlorocycbutane; compound XXVI, 1,2-dichloro-1,2-dibromoethene; and compound XXVII, trichloromethane. H Br C
OH
Cl C
C
H
Cl
Cl
Cl
Br H XXIII
XXII
H
H H
H
Cl Cl
XXIV
H
H
Cl
Cl
XXV
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Cl
Br Cl
C
Cl C
H
C Br
Cl
Cl XXVII
XXVI
Compounds XXI – XXVI are achiral; therefore, they are superimposable on their mirror images and optically inactive. Compound XXVIII, 3-bromo-1-chloro-1-phenylallene does not have a plane of symmetry; consequently, the molecule is chiral. This molecule is chiral even though it
H
H C
C
Br
C
C Br
Cl
C
C Cl
XXVIII
does not have a chiral center. Often it is necessary to construct models of molecules before attempting to establish chirality. Nevertheless, with experience and practice, one can use diverse techniques to determine the chirality of molecules. The important factor is that if one finds a plane, point, or rotating axis of symmetry, then the molecule is achiral, if not, then the molecule is chiral. Let’s use the plane of symmetry model to determine if compound XXIX, chloroallene; compound XXX, 1,1-dibromoallene; and compound XXXI, 1-chloro-1,2-butadiene are chiral or achiral. 29
Cl
H C
C
Cl
H
C
H
C
C H
C H
H
XXIX
Br
H C
C
Br
H
C
Br
C H
C
C H
Br
XXX
Cl
H C
H
C
Cl
C
H C
H
CH3
C
C CH3
XXXI
Compound XXIX has a plane of symmetry; therefore, it has a superimposable mirror image.
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Cl C
C
C C C
C H
H
Cl
H
H
H
H
These two compounds overlap atom for atom and p orbitals with p orbitals, and there is a plane of symmetry in the molecule. Consequently, the molecule is achiral.
H
Cl C
C
H
C H
Compound XXX has a plane of symmetry; therefore, its mirror image is superimposable.
Br
H C
C
Br
Br
H C C C
C H
H
Br
These two compounds overlap atom for atom and p orbitals with p orbitals, and, analogous to compound XXIX, there is a plane of symmetry in the molecule. Consequently, the molecule is achiral. Compound XXXI does not have a plane of symmetry; therefore, its mirror image is non superimposable.
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Cl
H C
C
Cl
H C C C
C CH3
H
H
H3C
These two compounds cannot overlap atom for atom and p atomic orbitals with p atomic orbitals, and there is no plane of symmetry in the molecule. Consequently, the molecule is chiral and can exist as enantiomers. Compounds XXIX and XXX are achiral, and compound XXXI is chiral. Compounds XXIX and XXX are achiral, because they have planes of symmetry; however, compound XXXI is chiral, because there is no plane of symmetry, point or symmetry or rotating axis of symmetry. Resolution of Racemic Modifications Two compounds of different physical properties form when a racemic modification reacts with an optically active resolving agent. These two compounds have diastereomeric relationships to one another; therefore, they can be separated using techniques like crystallization, chromatography, etc. The pure enantiomers are isolatable (usually by hydrolysis) once the diastereoisomers are separated. Following is a general schema for resolving enantiomers: (+) E + (-) E racemic modification
+
2 (+) R
(+) E (+)-R + (-) E (+) R
resolution agent
diastereoisomers (different physical properties)
(+) E (+)-R
(+) E
+
(+) R
pure enantiomer
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(-) E (+)-R
(-) E
+
(+) R
pure enantiomer
The choice of a resolution agent depends on the structure of the racemic modification. The resolving agent must react with the racemic modification, producing diastereoisomers that can be hydrolyzed (or undergo some other kind of separation reaction) to give the pure enantiomers and the original resolving agent. After separating and hydrolyzing the diastereoisomers, there is generally little or no problem in separating the resolving agent from the enantiomer. If the racemic mixture is an amine, usually the resolving agent is an optically active acid. The reaction of the acid and the amine results in two diastereomeric amine salts. The salts can then be separated, and the pure enantiomers regenerated by hydrolysis. The following equations illustrate the resolution of racemic 2-aminobutane into optically pure levorotatory and dextrorotatory enantiomers.
(2S)-2-aminobutane
(2R)-2-aminobutane
The Racemic Modification
R (+) Lactic Acid The Resolving Agent
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S
R
R
R
SR and RR represent a racemic mixture (a racemic modification) of the diastereoisomeric salts. Diastereoisomers can be physically separate, and the enantiomers isolated or resolved. When the diastereoisomers are physically separated, they can be treated with sodium hydroxide to regenerate the separated pure enantiomers.
The S enantiomer
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The R enantiomer
If the enantiomers were acids, then the racemic acids could be resolved with an optically active base. The series of reactions would be similar to the previous discussion except the resolving agent would be an acid. Loss of optical Activity Whenever a bond to a chiral atom breaks, the optical activity of the compound is either partially or completely destroyed. For example, the free radical chlorination of 1chloro-3-methylpentane will result in the formation of a compound that will be a racemic modification. This observation can be rationalized by examining the mechanistic pathway for the conversion of 1-chloro-3-methylpentane to 1,3-dichloro2-methylpentane. 1.
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2. Step two is a propagation step that leads to two possible isomers, i.e., attack of a chlorine radical on the right of the flat radical and attack of a chlorine radical on the left of the flat radical.
(3R)-1,3-dichloro-2-methylpentane
(3S)-1,3-dichloro-2-methylpentane
Other dichloroisomers are possible in this free radical chlorination reaction in addition to (3R)-1,3-dichloro-2-methylpentane and (3S)-1,3-dichloro-2-methylpentane. However, this discussion focuses on disrupting the chiral center in 1-chloro-3methylpentane with the generation of 1,3-dichloro-2-methylpentane. When 1,3-dichloro-2-methylpentane forms, the chlorine radical abstracts the tertiary hydrogen atom on the chiral carbon. The abstraction of the tertiary hydrogen results in the formation of a flat tertiary alkyl radical and destroys the chiral center. Then,
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chlorine attacks the flat tertiary alkyl radical in a propagation step. The chlorine can attack either side of the flat radical to generate a racemic modification.
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Problems
1. A measured mass of an optically active compound is dissolved in an aprotic solvent making a total volume of 50.0 mL. When this solution is placed in a 10.0 cm polarimeter tube, the observed rotation is +2.79o. If the specific rotation is +93o g-1 dm-1 mL, calculate the mass of the optically active sample in the polarimeter tube. 2. Specify the absolute configuration of the chiral centers, if any, in the following compounds. Indicate whether the compound is optically active. a. H H3C
C
CH2CH3
Br
b. CH3
H C CH3
C
C H
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c. CH3
H C
C
C
H
CH3
d.
H CH3
e.
O
f. H Br H Br
39
g.
.. N CH2CH3
CH3
H
h.
H C
O
H
C
OH
H
C
OH
CH2OH i.
CH2CH3 H
C
Br
H
C
Br
CH2CH3
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j, CH2CH3 C H
CH3
CH2CH2CH3
3. Show how differences in the solubility of diastereomeric salts may be used to resolve optically active alcohols, ROH. Hint: O
O C O
C
OR
C
OH
+ ROH
C O phthalic anhydride
O an acid-ester
4. In the following reaction, determine if the chiral center in the reactants has been disrupted in synthesizing the products:
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(a)
H
H +
CH3
C
+
COOH
H
ROH
CH3
OH
C
COOR
+ H2O
OH
(b) H CH3CH2
H
C
CN
+ 2 H2O
+
HCl
C
CH3CH2
COOH
+
NH4Cl
CH3
CH3
(c) H
H
C
CH3CH2
+ NaOH
Br
CH3CH2
C
OH
+
NaBr
CH3
CH3
(d) H H AlCl3 +
Cl
C
CH2CH3
C
CH2CH3 +
HCl
CH3
CH3
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5. The specific rotation of an optically active dextrorotatory compound is +56o. Calculate the percentage composition of a mixture of enantiomers of this compound whose rotation is -22o. 6. Which of the following structures are superimposable on their enantiomers? (a)
CH3 CH3
Br
H
Br H
(b) CH3 H
H
CH3
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(c)
CH3 H
C
OH
H
C
OH
CH3
(d)
H
H3C
+ O ..
(e)
CH3 CH3CH2
+
-
Cl N
CH2CO2CH3
H
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(f)
H
H
Cl
Cl
g. H Br Br H
7. Calculate the number of stereoisomers possible for the following compound:
H
H H3CCH2 HO
H S
OH
CH3
HO H
H
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8. Using chemical equations, suggest a pathway for resolving the following enantiomers:
CH3 C
CH3CH2
N
N H3C
CH3
H
CH3
H
H
C
CH2CH3
H
9. Determine which compounds are nonsuperimposable (dissymmetric) on their mirror images: a.
Cl C
Cl C
C H
H
b. CH3
H3C
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c.
COOH H
C
OH
H
C
OH
COOH
d.
Cl
H C
H
C
C H
e.
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10. A compound with the molecular formula of C7H16 was found to be optically active. When C7H16 is treated with Br2 and light, the resulting monobrominated mixture is optically inactive, but resolvable. (a) Suggest structures for C7H16 and the resulting monobrominated product. (b) Write a mechanism for the conversion of the optically active C7H16 to the optically inactive monobrominated mixture.
11. Anisatin, C15H20O8, is a poisonous compound found in the Shikimi plant. Anisatin has eight chiral centers. Following is the molecular structure of anisatin. O HO O
HO
H OH O
HO O
A three-dimensional model for anisatin can be represented by Figure 1.
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Figure 1
Rotating the three-dimensional model of anisatin down and to the left would generate the following structure for anisatin.
O
OH
O HO HO
H O
O OH
A three-dimensional model for this structure may be represented by Figure 2.
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Figure 2
Identify the eight chiral carbon atoms in anisatin and determine the R/S notation (the absolute configuration) at each chiral center.
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Solutions to Problems Chapter 7 1. A measured mass of an optically active compound is dissolved in an aprotic solvent making a total volume of 50.0 mL. When this solution is placed in a 10.0 cm polarimeter tube, the observed rotation is +2.79o. If the specific rotation is +93o g-1 dm-1 mL , calculate the mass of the optically active sample in the polarimeter tube.
+93o g −1 dm −1 mL =
+2.79 o 1m 10 dm xg 10.0 cm x x x 100 cm 1m 50.0 mL
+93o g −1 dm −1 mL x 1.0 dm x
xg = +2.79 o 50.0 mL
x = 1.5 g 2. Specify the absolute configuration of the chiral centers, if any, in the following compounds. Indicate whether the compound is optically active.
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a. H H3C
C
CH2CH3
Br
S Optically active Chiral center The molecule is chiral b. CH3
H C
C
C H
CH3
No plane of symmetry The molecule is chiral Optically active c.
CH3
H C H
C
C CH3
Plane of symmetry The molecule is achiral Optically inactive
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d.
H CH3 R Optically active Chiral center The molecule is chiral e.
O Plane of symmetry The molecule is achiral optically inactive
f.
H Br H Br
optically active No plane of symmetry Two chiral centers
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The molecule is chiral g.
.. N CH3
CH2CH3 H
Nitrogen molecules are not chiral because of the umbrella effect; therefore, the molecule is optically inactive.
h.
H C
O
H
C
OH
H
C
OH
CH2OH
Optically active No plane of symmetry Two chiral centers The molecule is chiral
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i.
CH2CH3 H
C
Br
H
C
Br
CH2CH3
The molecule is optically inactive , because of internal compensation. The molecule possesses a plane of symmetry. The molecule has two chiral centers. The molecule is achiral. J,
CH2CH3 C H
CH3
CH2CH2CH3
optically active
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3. Show how differences in the solubility of diastereomeric salts may be used to resolve optically active alcohols, ROH. Hint:
O
O C O C O phthalic anhydride
C
OR
C
OH
+ ROH
O an acid-ester
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4. In the following reaction, determine if the chiral center in the reactants has been disrupted in synthesizing the products: (a) H
H +
CH3
C
COOH
+
H
ROH
CH3
C
OH
+ H2O
COOR
OH
There is no disruption to the chiral center.
(b) H CH3CH2
H
C
CN
+ 2 H2O
+
HCl
C
CH3CH2
+
COOH
NH4Cl
CH3
CH3
There is no disruption to the chiral center
(c) H CH3CH2
C
H Br
+ NaOH
CH3
CH3CH2
C
OH
+
NaBr
CH3
There is a disruption to the chiral center
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(d) H H AlCl3 +
Cl
C
CH2CH3
C
CH2CH3 +
HCl
CH3
CH3
There is disruption to the chiral center 5. The specific rotation of an optically active dextrorotatory compound is +56o. Calculate the percentage composition of a mixture of enantiomers of this compound whose rotation is -22o. Let x = the fraction of the dextrorotatory compound Then, 1-x = the fraction of levorotatory compound x (+56o) + (1-x)( -56o) = -22o 56 x - 56 + 56x = -22 112x = 56 – 22 112x = 34 x = 34/112 x = 0.30 the fraction of the mixture that is dextrorotatory 1-x = 0.70 the fraction of the mixture that is levorotatory
6. Ascertain whether the following structures are superimposable on their enantiomers: (a)
CH3 Br
H
CH3
Br
H This molecule does not superpose on its mirror image. 59
(b) CH3 H
H
CH3
This molecule does not superpose on its mirror image. (c)
CH3 H
C
OH
H
C
OH
CH3 This molecule superposes on its mirror image.
(d)
H
H3C
+ O ..
In the static model, the molecule does not superpose on its mirror image, but the other lone pair can be protonated; therefore, the molecule is optically inactive.
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(e)
CH3 CH3CH2
+
-
Cl N
CH2CO2CH3
H This molecule does not superpose on its mirror image. (f)
H
H
Cl
Cl
This molecule superposes on its mirror image.
(g) H Br Br H
This molecule superposes on its mirror image.
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7. Calculate the number of stereoisomers possible for the following compound:
H
H H
H3CCH2 HO
S
OH
CH3
HO H
H
2n = 25 = 32 isomers
8. Using chemical equations, suggest a pathway for resolving the following enantiomers:
CH3 H CH3CH2
C
N
N CH3
H
R
CH3
H
H3C
C
CH2CH3
H
S
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9. Determine which compounds are nonsuperimposable (dissymmetric) on their mirror images: a.
Cl C
Cl C
C H
H
This molecule is nonsuperposable on its mirror image. b. CH3
H3C
This molecule is nonsuperposable on its mirror image.
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c.
COOH H
C
OH
H
C
OH
COOH This molecule is superposable on its mirror image. d.
Cl
H C
H
C
C H
This molecule is superposable on its mirror image. e.
This molecule is nonsuperposable on its mirror image. 10. A compound with the molecular formula of C7H16 was found to be optically active. When C7H16 is treated with Br2 and light, the resulting monobrominated mixture is optically inactive, but resolvable. Suggest structures for C7H16 and the resulting monobrominated product.
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(b) Write a mechanism for the conversion of the optically active C7H16 to the optically inactive monobrominated mixture. 1.
2.
3.
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Step 3 may be envisioned in the following manner:
11. Following are the identified chiral centers in anisatin, C15H20O8, and the absolute configuration (R/S designation) at each chiral center.
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