CH 3 CHCH CH
4. Isomerization
5.5 Synthesis and Optical Activity
1. Optically inactive reactants with achiral catalysts or solvents yield optically inactive products. With a chiral catalyst, for instance, an enzyme, any chiral product will be optically active.
2. A second chiral center generated in a chiral compound may not have an equal chance for R and S configurations; a 50 : 50 mixture of diastereomers is not usually obtained.
3. Replacement of a group or atom on a chiral center can occur with retention or inversion of configura-tion or with a mixture of the two (complete or partial racemizaconfigura-tion), depending on the mechanism of the reacconfigura-tion.
Problem 5.20 (a) What two products are obtained when C3of (R)-2-chlorobutane is chlorinated? (b) Are these diastereomers formed in equal amounts? (c) In terms of mechanism, account for the fact that
is obtained when (R)-ClCH2-C*H(CH3)CH2CH3is chlorinated.
In the products, C2retains the R configuration, since none of its bonds were broken and there was no change in priority. The configuration at C3, the newly created chiral center, can be either R or S. As a result, two diastereomers are formed, the optically active RR enantiomer and the optically inactive RS meso compound.
(b) No. The numbers of molecules with S and R configurations at C3are not equal. This is so because the pres-ence of the C2-stereocenter causes an unequal likelihood of attack at the faces of C3. Faces which give rise to diastereomers when attacked by a fourth ligand are diastereotopic faces.
(c) Removal of the H from the chiral C leaves the achiral free radical ClCH2C.
(CH3)CH2CH3. Like the radical in Problem 5.21, it reacts with Cl2to give a racemic form.
The C3of 2-chlorobutane, of the general type RCH2R′, which becomes chiral when one of its H’s is replaced by another ligand, is said to be prochiral.
Problem 5.21 Answer True or False to each of the following statements and explain your choice. (a) There are two broad classes of stereoisomers. (b) Achiral molecules cannot possess chiral centers. (c) A reaction catalyzed by an enzyme always gives an optically active product. (d) Racemization of an enantiomer must re-sult in the breaking of at least one bond to the chiral center. (e) An attempted resolution can distinguish a race-mate from a meso compound.
(a) True. The two classes are enantiomers and diastereomers.
(b) False. Meso compounds are achiral, yet they possess chiral centers.
(c) False. The product could be achiral.
(d) True. Only by breaking a bond could the configuration be changed.
(e) True. A racemate can be resolved, but a meso compound cannot be, because it does not consist of enantiomers.
Problem 5.22 In Fig. 4.4, give the stereochemical relationship between (a) the two gauche conformers, (b) the anti and either gauche conformer.
(a) They are stereomers, because they have the same structural formulas but different spatial arrangements.
However, since they readily interconvert by rotation about a σ bond, they are not typical, isolatable, con-figurational stereomers; rather, they are conformational stereomers. The two gauche forms are non-super-imposable mirror images; they are conformational enantiomers.
(b) They are conformational diastereomers, because they are stereomers but not mirror images.
Configurational stereomers differ from conformational stereomers in that they are interconverted only by breaking and making chemical bonds. The energy needed for such changes is of the order of 200–600 kJ/mol, which is large enough to permit their isolation, and is much larger than the energy required for interconversion of conformers.
SUPPLEMENTARY PROBLEMS
Problem 5.23 (a) What is the necessary and sufficient condition for the existence of enantiomers? (b) What is the necessary and sufficient condition for measurement of optical activity? (c) Are all substances with chiral atoms optically active and resolvable? (d) Are enantiomers possible in molecules that do not have chiral carbon atoms?
(e) Can a prochiral carbon ever be primary or tertiary? ( f ) Can conformational enantiomers ever be resolved?
(a) Chirality in molecules having nonsuperimposable mirror images. (b) An excess of one enantiomer and a specific rotation large enough to be measured. (c) No. Racemic forms are not optically active but are resolv-able. Meso compounds are inactive and not resolvresolv-able. (d ) Yes. The presence of a chiral atom is a sufficient but not necessary condition for enantiomerism. For example, properly disubstituted allenes have no plane or center of symmetry and are chiral molecules even though they have no chiral C’s:
(e) No. Replacing one H of σ 1° CH3group by an X group would leave an achiral ⎯ CH2X group. In order for a 3° CH group to be chiral when the H is replaced by X, it would already have to be chiral when bonded to the H.
( f ) Yes. There are molecules which have a large enthalpy of activation for rotating about a σ bond because of severe steric hindrance. Examples are properly substituted biphenyls, for example, 2,2 ′-dibromo-6,6′-dinitro-biphenyl (Fig. 5.5). The four bulky substituents prevent the two flat rings from being in the same plane, a require-ment for free rotation.
Problem 5.24 Select the chiral atoms in each of the following compounds:
Figure 5.5
(a) There are eight chiral C’s: three C’s attached to a CH3group, four C’s attached to lone H’s, and one C attached to OH. (b) Since N is bonded to four different groups, it is a chiral center, as is the C bonded to the OH.
(c) There are no chiral atoms in this molecule. The two sides of the ring ⎯ CH2CH2⎯ joining the C’s bonded to Cl’s are the same. Hence, neither of the C’s bonded to a Cl is chiral.
Problem 5.25 Draw examples of (a) a meso alkane having the molecular formula C8H18and (b) the simplest alkane with a chiral quaternary C. Name each compound.
(b) The chiral C in this alkane must be attached to the four simplest alkyl groups. These are CH3⎯, CH3CH2⎯, CH3CH2CH2⎯, and (CH3)2CH⎯, and the compound is
Problem 5.26 Relative configurations of chiral atoms are sometimes established by using reactions in which there is no change in configuration because no bonds to the chiral atom are broken. Which of the following re-actions can be used to establish relative configurations?
(a)
(b) and (e). The others involve breaking bonds to the chiral C.
Problem 5.27 Account for the disappearance of optical activity observed when (R)-2-butanol is allowed to stand in aqueous H2SO4and when (S)-2-iodooctane is treated with aqueous KI solution.
Optically active compounds become inactive if they lose their chirality because the chiral center no longer has four different groups, or if they undergo racemization. In the two reactions cited, C remains chiral, and it must be concluded that in both reactions, racemization occurs.
Problem 5.28 For the following compounds, draw projection formulas for all stereoisomers and point out their R,S specifications, optical activity (where present), and meso compounds: (a) 1,2,3,4-tetrahydroxybutane, (b) 1-chloro–2,3-dibromobutane, (c) 2,4-diiodopentane, (d) 2,3,4-tribromohexane, (e) 2,3,4-tribromopentane.
(a) HOCH2C*HOHC*HOHCH2OH, with two similar chiral C’s, has one meso form and two optically active enantiomers.
(b) ClCH2C*HBrC*HBrCH3has two different chiral C’s. There are four (22) optically active enantiomers.
The two sets of diastereomers are differentiated by the prefix erythro for the set in which at least two identical or similar substituents on chiral C’s are eclipsed. The other set is called threo.
(c) CH3C*
HICH2C*
HICH3has two similar chiral C’s, C2and C4, separated by a CH2group. There are two enantiomers comprising a () pair and one meso compound.
(d) With three different chiral C’s in CH3C*HBrC*HBrC*HBrCH2CH3, there are eight (23) enantiomers and four racemic forms.
(e) C1H3CH*
2
BrC3HBrC*
4
HBrC5H3has two similar chiral atoms (C2and C4). There are two enantiomers in which the configurations of C2and C4are the same, RR or SS. When C2and C4have different configurations, one R and one S, C3becomes a stereocenter and there are two meso forms.
Problem 5.29 The specific rotation of (R)-()-2-bromooctane is 36°. What is the percentage composition of a mixture of enantiomers of 2-bromooctane whose rotation is 18°?
Let x mole fraction of R, 1 x mole fraction of S.
x(−36°)+( – )(1 x 36°)=18° or x=14 The mixture has 25% R and 75% S; it is 50% racemic and 50% S.
Problem 5.30 Predict the yield of stereoisomeric products, and the optical activity of the mixture of products, formed from chlorination of a racemic mixture of 2-chlorobutane to give 2,3-dichlorobutane.
The (S)2-chlorobutane comprising 50% of the racemic mixture gives 35.5% of the meso (SR) product and 14.5% of the RR enantiomer. The R enantiomer gives 35.5% meso and 14.5% RR products. The total yield of meso product is 71%, and the combination of 14.5% RR and 14.5% SS gives 29% racemic product. The total reaction mixture is optically inactive. This result confirms the generalization that optically inactive starting materials, reagents, and solvents always lead to optically inactive products.
Problem 5.31 For the following reactions, give the number of stereoisomers that are isolated, their R,S con-figurations, and their optical activities. Use Fischer projections.
(a) This meso alcohol is oxidized at either terminal CH2OH to give an optically inactive racemic form. The chiral C next to the oxidized C undergoes a change in priority order; CH2OH (3) goes to COOH (2). Therefore, if this C is R in the reactant, it becomes S in the product; if S, it goes to R.
(b) Replacement of Cl by the isopentyl group does not change the priorities of the groups on the chiral C. There is one optically active product, whose two chiral C’s have R configurations.
(c) This reduction generates a second chiral center.
RR and SS enantiomers are formed in equal amounts to give a racemic form. The meso and racemic forms are in unequal amounts.
(d) Reduction of the double bond makes C3chiral. Reduction occurs on either face of the planar π bond to form molecules with R and molecules with S configurations at C3. These are in unequal amounts because of the adjacent chiral C that has an S configuration. Since both chiral atoms in the product are structurally identi-cal, the products are a meso structure (RS) and an optically active diastereomer (SS).
Problem 5.32 Designate the following compounds as erythro or threo structures.
(a) Erythro (see Problem 5.28).
(b) Erythro; it is best to examine eclipsed conformations. If either of the chiral C’s is rotated 120° to an eclipsed conformation for the two Br’s, the H’s are also eclipsed.
(c) Threo; a 60° rotation of one of the chiral C’s eclipses the H’s but not the OH’s.
Problem 5.33 Glyceraldehyde can be converted to lactic acid by the two routes shown below. These results reveal an ambiguity in the assignment of relative D,L configuration. Explain.
In neither route is there a change in the bonds to the chiral C. Apparently, both lactic acids should have the D configuration, since the original glyceraldehyde was D. However, since the CH3and COOH groups are inter-changed, the two lactic acids must be enantiomers. Indeed, one is () and the other is (). This shows that, for an unambiguous assignment of D or L, it is necessary to specify the reactions in the chemical change. Because of such ambiguity, R,S is used. The () lactic acid is S, the () enantiomer is R.
Problem 5.34 Deduce the structural formula for an optically active alkene, C6H12, which reacts with H2to form an optically inactive alkane, C6H14.
The alkene has a group attached to the chiral C, which must react with H2to give a group identical to one already attached, resulting in loss of chirality.
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