Mohammad Jafarzadeh Faculty of Chemistry, Razi University

Organic Chemistry I

Organic Chemistry, Structure and Function (7th edition) By P. Vollhardt and N. Schore, Elsevier, 2014 1 5-6 Meso Compounds CHAPTER 5 189

Mirror Figure 5-10 The stereochemical relations of the stereoisomers of 2,3-dibromobutane. The lower pair CH3 CH3 consists of identical structures. Br H H Br (Make a model.) H Br H S Br Br R H Br H S S ϭ Enantiomers ϭ RR S R H3C CH3 Br H H Br H3C CH3

CH3 CH3

Diastereomers Diastereomers Diastereomers

CH3 CH3 Br Br S R HH H H H Br Enantiomers? Br H Br Br SR ϭϭRS R S H3C CH3 H Br No! Identical! Br H H3C CH3 (Meso) CH3 CH3

Two identically substituted stereocenters give rise to only three stereoisomers Consider, for example, 2,3-dibromobutane, which can be obtained by the radical bromination of 2-bromobutane. As we did for 2-bromo-3-chlorobutane, we have to consider four structures, resulting from the various permutations in R and S con! gurations (Figure 5-10).

MESO COMPOUNDS H BrH Br2, hv CH *CCH CH H C *C C* CH 3 2 3 ϪHBr 3 3 ϩ 2,3-dibromobutane can be obtained by the radical 5-6 Meso Compounds CHAPTER 5 189 bromination of 2-bromobutane. Br HBr One stereocenter Two stereocenters 2,3-Dibromobutane Mirror Figure 5-10 The stereochemical The ! rst pair of stereoisomers, with R,R and S,S con!relations gurations, of isthe clearly stereoisomers recognizable of as a pair of enantiomers. However, a close look at the second2,3-dibromobutane. pair reveals that (TheS,R )lower and mirror pair consists of identical structures. CH3 image (R,S) are superimposableCH3 and therefore identical. Thus, the S,R diastereomer of Br H H Br (Make a model.) H Br H S Br 2,3-dibromobutaneBr is achiralR Hand notBr optically active,H even though it contains two stereocenters. S S ϭ TheEnantiomers identity of the two structuresϭ can beRR readily con! rmed by using molecular models. S R H3C CH3 Br H A compound thatH containsBr twoH (or,3C as weCH shall3 see, even more than two) stereocenters but is superimposable with its mirror image is a meso compound (mesos, Greek, middle). CH CH 3 A characteristic feature of3 a meso compound is the presence of an internal mirror plane, which divides the molecule such that one half is the mirror image of the other half. For example,Diastereomers in 2,3-dibromobutane, the 2R center is the re" ection of the 3S center. This arrange- Diastereomers ment is best seen in an eclipsed hashed-wedged Diastereomersline structure (Figure 5-11). The presence

CH3 CH3 Identically substituted stereocenters Br Br S R HH H H H Br Enantiomers? Br H Br Br SR ϭϭRS R H3C S H H3C CH3 H3C CH3 H Br No! Identical! Br H Br HRotation3C CH3 R S (Meso) CC C C RS Br CH3 Br CH3 Br 2 H CH3 H H Staggered Mirror Eclipsed Brominating racemic 2-bromobutane at C3 is Two identically substituted stereocentersFigure 5-11 meso -2,3-Dibromobutanegive rise contains an internal mirror plane when rotated into the stereochemically equivalent to to only three stereoisomers eclipsed conformation shown. A molecule with more than one stereocenter is meso and achiral pairing two identical pairs as long as it contains a mirror plane in any readily accessible conformation. Meso compounds (“racemates”) of shoes: only three Consider, for example, 2,3-dibromobutane, whichpossess can identically be obtained substituted by the stereocenters. radical bromination distinct combinations. of 2-bromobutane. As we did for 2-bromo-3-chlorobutane, we have to consider four structures, resulting from the various permutations in R and S con! gurations (Figure 5-10). H BrH

Br2, hv CH *CCH CH H C *C C* CH 3 2 3 ϪHBr 3 3 ϩ Br HBr One stereocenter Two stereocenters 2,3-Dibromobutane

The ! rst pair of stereoisomers, with R,R and S,S con! gurations, is clearly recognizable as a pair of enantiomers. However, a close look at the second pair reveals that (S,R) and mirror image (R,S) are superimposable and therefore identical. Thus, the S,R diastereomer of 2,3-dibromobutane is achiral and not optically active, even though it contains two stereocenters. The identity of the two structures can be readily con! rmed by using molecular models. A compound that contains two (or, as we shall see, even more than two) stereocenters but is superimposable with its mirror image is a meso compound (mesos, Greek, middle). A characteristic feature of a meso compound is the presence of an internal mirror plane, which divides the molecule such that one half is the mirror image of the other half. For example, in 2,3-dibromobutane, the 2R center is the re" ection of the 3S center. This arrangement is best seen in an eclipsed hashed-wedged line structure (Figure 5-11). The presence

Identically substituted stereocenters

H3C H H3C CH3 Br Rotation CC RC C S Br RS Br Br H CH3 H H Staggered Mirror Eclipsed Brominating racemic 2-bromobutane at C3 is Figure 5-11 meso-2,3-Dibromobutane contains an internal mirror plane when rotated into the stereochemically equivalent to eclipsed conformation shown. A molecule with more than one stereocenter is meso and achiral pairing two identical pairs as long as it contains a mirror plane in any readily accessible conformation. Meso compounds (“racemates”) of shoes: only three possess identically substituted stereocenters. distinct combinations. 5-6 Meso Compounds CHAPTER 5 189

CH3 CH3

Diastereomers Diastereomers Diastereomers

CH3 CH3 Br Br S R HH H H H Br Enantiomers? Br H Br Br SR ϭϭRS R S H3C CH3 H Br No! Identical! Br H H3C CH3 (Meso) CH3 CH3

Br2, hv CH *CCH CH H C *C C* CH 3 2 3 ϪHBr 3 3 ϩ Br HBr One stereocenter Two stereocenters 2,3-Dibromobutane

The ! rst pair of stereoisomers, with R,R and S,S con! gurations, is clearly recognizable as a pair of enantiomers. However, a close look at the second pair reveals that (S,R) and mirror image (R,S) are superimposable and therefore identical. Thus, the S,R diastereomer of 2,3-dibromobutane is achiral and not optically active, even though it contains two stereocenters. The identity of the two structures can be readily con! rmed by using molecular models. A compound that contains two (or, as we shall see, even more than two) stereocenters but is superimposable with its mirror image is a meso compound (mesos, Greek, middle). A characteristic feature of a meso compound is the presence of an internal mirror plane, which divides the molecule such that one half is the mirror image of the other half. For A compound that contains twoexample,stereocenters in 2,3-dibromobutane,but is superimposable the 2R center iswith the reits" ectionmirror ofimage the 3S center.is a This arrange- meso compound (mesos, mentGreek, is bestmiddle) seen .inA ancharacteristic eclipsed hashed-wedgedfeature of linea mesostructurecompound (Figure 5-11).is The presence the presence of an internal mirror plane, which divides the molecule such that one half is the

190 mirrorCHAPTERimage 5 of Stereoisomersthe other half. Identically substituted stereocenters

H3C H H3C CH3 of a mirror plane in any energetically accessible conformationBr Rotation of a molecule (Sections 2-8 CC RC C S and 2-9) is suf! cient to make it achiralBr (SectionRS 5-1). As a consequence, 2,3-dibromobutaneBr Br exists in the form of three stereoisomersH only: a pairCH of 3(necessarily chiral) enantiomersH andH an achiral meso diastereomer (see also margin of p. 189). Staggered Mirror Two identically substitutedMeso diastereomersstereocenters can exist in molecules with more than two stereocenters.Eclipsed Examples Brominating racemic give rise to onlyarethree 2,3,4-tribromopentanestereoisomers and 2,3,4,5-tetrabromohexane. 2-bromobutane at C3 is Figure 5-11 meso-2,3-Dibromobutane contains an internal mirror plane when rotated into the stereochemically equivalent to Model Building eclipsedMeso conformationCompounds with shown. Multiple A molecule Stereocenters with more than one stereocenter is meso and achiral pairing two identical pairs as long as it contains a mirror plane in any readily accessible conformation. Meso compounds (“racemates”) of shoes: only three possess identically substituted stereocenters.CH3 distinct combinations. CH3 Br H H Br Br H Br H Br H Br H ´ ´

*C *C H Br ? / C ϭ HBr Mirror plane ϭ & BrH ; ; H Br Br H ≥≥ Br H Br H H Br

CH3 CH3 Mirror plane Mirror 3

Exercise 5-21 Draw all the stereoisomers of 2,4-dibromo-3-chloropentane.

Cyclic compounds may also be meso It is again instructive to compare the stereochemical situation in 2,3-dibromobutane with that in an analogous cyclic molecule: 1,2-dibromocyclobutane. We can see that trans-1,2- dibromocyclobutane exists as two enantiomers (R,R and S,S) and may therefore be optically active. The cis isomer, however, has an internal mirror plane and is meso, achiral, and optically inactive (Figure 5-12). Notice that we have drawn the ring in a planar shape in order to illustrate the mirror Chiral or meso? symmetry, although we know from Chapter 4 that cycloalkanes with four or more carbons in the ring are not " at. Is this justi! able? Generally yes, because such compounds, like their acyclic analogs, possess a variety of conformations that are readily accessible at room Note: The presence of both R and S stereocenters in a temperature (Sections 4-2 through 4-4 and Section 5-1). At least one of these conformations structure does not mean that will contain the necessary mirror plane to render achiral any cis-disubstituted cycloalkane the compound is automatically with identically constituted stereocenters. For simplicity, cyclic compounds may usually be meso. For it to be meso, treated as if they were planar for the purpose of identifying a mirror plane. its stereocenters must have identical sets of substituents, and the molecule as a whole Exercise 5-22 must have an internal mirror Draw each of the following compounds, representing the ring as planar. Which ones are chiral? plane of symmetry. Which are meso? Indicate the location of the mirror plane in each meso compound. (a) cis-1,2- Dichlorocyclopentane; (b) its trans isomer; (c) cis-1,3-dichlorocyclopentane; (d) its trans isomer; (e) cis-1,2-dichlorocyclohexane; (f) its trans isomer; (g) cis-1,3-dichlorocyclohexane; (h) its trans isomer.

Figure 5-12 The trans isomer Br Br Br of 1,2-dibromocyclobutane is H H Br chiral; the cis isomer is a meso R S R compound and optically inactive. i i i Mirror H H H R i i S S i Br Br H

Mirror 1R,2S same as 1S,2R Enantiomers of Meso diastereomer chiral diastereomer trans-1,2-Dibromocyclobutane cis-1,2-Dibromocyclobutane 190 CHAPTER 5 Stereoisomers

of a mirror plane in any energetically accessible conformation of a molecule (Sections 2-8 and 2-9) is suf! cient to make it achiral (Section 5-1). As a consequence, 2,3-dibromobutane exists in the form of three stereoisomers only: a pair of (necessarily chiral) enantiomers and an achiral meso diastereomer (see also margin of p. 189). Meso diastereomers can exist in molecules with more than two stereocenters. Examples are 2,3,4-tribromopentane and 2,3,4,5-tetrabromohexane.

Model Building Meso Compounds with Multiple Stereocenters

CH3 CH3 Br H H Br Br H Br H Br H Br H ´ ´

*C *C H Br ? / C ϭ HBr Mirror plane ϭ & BrH ; ; H Br Br H ≥≥ Br H Br H H Br

CH3 CH3 Mirror plane Mirror

Exercise 5-21 Draw all the stereoisomers of 2,4-dibromo-3-chloropentane.

Cyclic compounds may also be meso It is again instructive to compare the stereochemical situation in 2,3-dibromobutane with that in an analogous cyclic molecule: 1,2-dibromocyclobutane. We can see that trans-1,2- dibromocyclobutane exists as two enantiomers (R,R and S,S) and may therefore be optically active. The cis isomer, however, has an internal mirror plane and is meso, achiral, and optically inactive (Figure 5-12). Notice that we have drawn the ring in a planar shape in order to illustrate the mirror Chiral or meso? symmetry, although we know from Chapter 4 that cycloalkanes with four or more carbons in the ring are not " at. Is this justi! able? Generally yes, because such compounds, like their acyclic analogs, possess a variety of conformations that are readily accessible at room Note: The presence of both R and S stereocenters in a temperature (Sections 4-2 through 4-4 and Section 5-1). At least one of these conformations structure does not mean that will contain the necessary mirror plane to render achiral any cis-disubstituted cycloalkane the compound is automatically with identically constituted stereocenters. For simplicity, cyclic compounds may usually be meso. For it to be meso, treated as if they were planar for the purpose of identifying a mirror plane. its stereocenters must have identical sets of substituents, and the molecule as a wholeCyclic compoundsExercise 5-22may also be meso must have an internal mirror Draw each of the following compounds, representing the ring as planar. Which ones are chiral? plane of symmetry. Trans-1,2- Whichdibromocyclobutane are meso? Indicate theexists locationas of twothe mirrorenantiomers plane in each(R meso,R andcompound.S,S) (a)and cis-1,2-may therefore be opticallyDichlorocyclopentane;active. (b) its trans isomer; (c) cis-1,3-dichlorocyclopentane; (d) its trans isomer; (e) cis-1,2-dichlorocyclohexane; (f) its trans isomer; (g) cis-1,3-dichlorocyclohexane; (h) its trans isomer. The cis isomer has an internal mirror plane and is meso, achiral, and optically inactive.

Figure 5-12 The trans isomer Br Br Br of 1,2-dibromocyclobutane is H H Br chiral; the cis isomer is a meso R S R compound and optically inactive. i i i Mirror H H H R i i S S i Br Br H

Mirror 1R,2S same as 1S,2R Enantiomers of Meso diastereomer chiral diastereomer trans-1,2-Dibromocyclobutane cis-1,2-Dibromocyclobutane

4 STEREOCHEMISTRY IN CHEMICAL REACTIONS

As a model, the conversion of achiral butane into chiral 2-bromobutane will be considered.

The radical mechanism explains why the bromination of butane results in a racemate

The radical center has two equivalent reaction sites (a planar, sp2-hybridized)—the two lobes of the p orbital—that are equally susceptible to attack by bromine in the second step.

The two transition states resulting in the respective enantiomers of 2-bromobutane are mirror images of each other. They are enantiomeric and therefore energetically equivalent. The rates of formation of R and S products are hence equal, and a racemate is formed.

In general, the formation of chiral compounds (e.g., 2-bromobutane) from achiral reactants (e.g., butane and bromine) yields racemates. Or, optically inactive starting materials furnish optically inactive products.

5 192 CHAPTER 5 Stereoisomers

Figure 5-13 The creation of ‡ racemic 2-bromobutane from butane by radical bromination at C2. Abstraction of either methylene Bromine attacks Br hydrogen by bromine gives an from the top Br achiral radical. Reaction of Br2 with this radical is equally likely at + either the top or the bottom face, CH3 Br a condition leading to a racemic mixture of products. C Br C H CH2CH3 H CH3 H CH2CH3 (S)-2-Bromobutane

Br −HBr C H CH3 CH2CH3

CH3 Br2 H C CH CH 2 3 Mirror ‡ Achiral radical CH3 H (Two lobes of p orbital CH2CH3 are equivalent) C

CH3 H CH2CH3 Br C

Br + Bromine attacks Br from the bottom Br (R)-2-Bromobutane

Enantiomeric transition states 6 (Equal in energy)

Chlorination of (S)-2-Bromobutane at Either C1 or C4 Remember the use of color to denote group priorities: H H H Cl2, hv Cl2, hv

a Highest — red A at C1 A at C4 A

Br −HBr C H CH3 CH2CH3

CH3 Br2 H C CH CH 2 3 Mirror ‡ Achiral radical CH3 H (Two lobes of p orbital CH2CH3 are equivalent) C

The presence of a stereocenter affects the outcome of the reactionCH3 H CH2CH3 C Halogenation of a chiral and enantiomerically pureBr molecule. In the radical chlorination of the S enantiomer of 2-bromobutane,Brthe chlorine atom has several options for attack: the two terminal methyl groups, the single+ hydrogen at C2, and the two hydrogens on C3. Bromine attacks Br from the bottom Br (R)-2-Bromobutane Chlorination of either terminal methyl group is straightforward. Both of these chlorination products are optically active because the original stereocenter is left intact. Enantiomeric transition states The conversion of the C1 methyl into a chloromethyl(Equal in energy)unit changes the sequence of priorities around C2, its designated configuration changes from S to R. Chlorination of (S)-2-Bromobutane at Either C1 or C4 Remember the use of color to denote group priorities: H H H Cl2, hv Cl2, hv

a Highest — red A at C1 A at C4 A

C C 1 2 1 C 2 3 b Second highest — blue Group C C 4 C ClH C A ϪHCl H C A ϪHCl H C A c Third highest — green priorities 2 ( Br 3 ( Br 3 ( Br have CH3CH2 CH3CH2 ClCH2CH2 d Lowest — black 34 34 21 switched Optically active Optically active Optically active 2R 2S 3S 7 Chlorination of either terminal methyl group is straightforward, proceeding at C1 to give 2-bromo-1-chlorobutane or at C4 to give 3-bromo-1-chlorobutane. In the latter, the original C4 has now become C1, to maintain the lowest possible substituent numbering. Both of these chlorination products are optically active because the original stereocenter is left intact. Note, however, that conversion of the C1 methyl into a chloromethyl unit changes the sequence of priorities around C2. Thus, although the stereocenter itself does not par- ticipate in the reaction, its designated con! guration changes from S to R. What about halogenation at C2, the stereocenter? The product from chlorination at C2 of (S)-2-bromobutane is 2-bromo-2-chlorobutane. Even though the substitution pattern at the stereocenter has changed, the molecule remains chiral. However, an attempt to mea- sure the [a] value for the product would reveal the absence of optical activity: Halogenation 5-7 Stereochemistry in Chemical Reactions CHAPTER 5 193 Chlorination can occur at C2 of (S)-2-bromobutane from either side through enantiomeric transition states of equal energy, producing (S)- and (R)-2-bromo-2-chlorobutane at equal rates and inatequal the stereocenteramounts .leads to a racemic mixture. How can this be explained? For the answer, we must look again at the structure of the radical formed in the course of the reaction Halogenationmechanism.at the stereocenter leads to a racemic mixture (no optical activity). A racemate A racemate forms in this case because hydrogen abstraction from C2 furnishes a planar,2 Note: Radicals lack brains. forms in thisspcase2-hybridized,because achiralhydrogen radical. abstraction from C2 furnishes a planar, sp -hybridized,They do not remember which achiral radical. enantiomer they came from when the hydrogen removed Chlorination of (S)-2-Bromobutane at C2 renders a stereocenter planar and achiral. Thus, in this H Cl H3C Br particular experiment, it Cl H3C Cl2 C C Br C + CH3CH2 C makes no difference whether H C −HCl −Cl H3C 3 CH3CH2 the original starting material CH3CH2 Br CH3CH2 Br Cl was S, R, or any mixture of the Optically active Optically inactive two: The product must always (A racemate) be racemic, because it derives 2S Achiral 50% 2S 50% 2R from an achiral radical.

8 REAL LIFE: MEDICINE 5-4 Chiral Drugs—Racemic or Enantiomerically Pure?

Until the early 1990s, most synthetic chiral medicines were pre- However, in a number of cases, one of the enantiomers pared as racemic mixtures and sold as such. The reasons were of a drug has been found to act as a blocker of the biologi- mainly of a practical nature. Reactions that convert an achiral cal receptor site, thus diminishing the activity of the other into a chiral molecule ordinarily produce racemates (Sec- enantiomer. Worse, one of the enantiomers may have a tion 5-7). In addition, often both enantiomers have comparable completely different, and sometimes toxic, spectrum of physiological activity or one of them (the “wrong” one) is activity. This phenomenon is quite general, since nature’s inactive; therefore resolution was deemed unnecessary. Finally, receptor sites are handed (Real Life 5-5). Several resolution of racemates on a large scale is expensive and adds examples of different bioactive behavior between enantiomers substantially to the cost of drug development. are shown below. O O

H2N S R NH2 OH HO 0≥ ≥ 0 O H2N H H NH2 O Bitter Sweet Asparagine (Amino acid; see Chapter 26)

Cl Cl

≥ H C ≥ H H CH 0 3 0 3 HO OH S O O R In 1806, asparagine was the ﬁ rst amino O Cl Cl O acid to be isolated from nature, from the Active Inactive juice of asparagus. Dichlorprop

(Herbicide)

≥

H ≥ OH HO H 0 H 0 H N CH OH N S 2 HOH2C R

OH HO

Antagonist Bronchodilator Albuterol (see also Chapter Opening) Weed control. 196 CHAPTERThe chlorination 5 Stereoisomersof (S)-2-bromobutane at C3 does not affect the existing chiral center. However, the formation of a second stereocenter gives rise to diastereomers. Chlorination of (S)-2-Bromobutane at C3

1 1 1 CH3 CH3 CH3 2 2 2 H Br Cl , hv H Br H Br 2 ϩ H 3 H ϪHCl H 3 Cl Cl 3 H 4 4 4 CH3 CH3 CH3 2S 2S,3R 2S,3S Optically active Optically active Optically active (Unequal amounts) By abstraction ofTheeither chlorinationone of at theC2 resultshydrogens in a 1:1 mixtureresults ofin enantiomers.a radical Doescenter the reactionat C3 at, twoC3 faces of this radical arealsonot givemirror an equimolarimages mixtureof each of diastereomers?other, because The answerthe isradical no. This retains! nding is thereadilyasymmetry explained on inspection of the two transition states leading to the product (Figure 5-14). of the original Abstractionmolecule ofas eithera result one of theof thehydrogenspresence results inof athe radicalstereocenter center at C3. Inat contrastC2. Thus, with the two sides of the p orbitalthe radicalare formednot equivalentin the chlorination. at C2, however, the two faces of this radical are not mirror images of each other, because the radical retains the asymmetry of the original If the rate of attackmoleculeat asthe a resulttwo offaces the presenceof the of radicalthe stereocenterdiffer, atthe C2. Thus,two diastereomersthe two sides of theshould be different. The twop orbitaltransition are not equivalent.states leading to products are not mirror images of each other Mechanismand are not superimposableWhat are the consequences: They have of thisdifferent nonequivalency?energies If theand raterepresent of attack at thedifferent two facespathways . of the radical differ, as one would predict on steric grounds, then the rates of formation of the 9 two diastereomers should be different, as is indeed found: (2S,3R)-2-Bromo-3-chlorobutane is

Figure 5-14 The chlorination of ‡ (S)-2-bromobutane at C3 produces Br H the two diastereomers of 2-bromo- Chlorine 3-chlorobutane in unequal amounts attacks Cl C C as a result of the chirality at C2. 23 from the Cl H3C H top + H CH3 Br Cl (S)-2-Bromobutane 23 Cl C C H Cl −HCl H3C H CH3 (2S,3S)-2-Bromo- Asymmetric Br 3-chlorobutane C2 carbon C 25% C H CH3 Br H3C H H Cl2 C C Br CH3 ‡ H3C H H C CH3 Chiral radical H3C C (Two lobes of p orbital H are not equivalent) Br H CH3 Cl 23 C C

H3C H Cl + Chlorine attacks Cl Cl from the bottom (2S,3R)-2-Bromo- 3-chlorobutane 75% Diastereomeric transition states (Different in energy) 196 CHAPTER 5 Stereoisomers

Chlorination of (S)-2-Bromobutane at C3

1 1 1 CH3 CH3 CH3 2 2 2 H Br Cl , hv H Br H Br 2 ϩ H 3 H ϪHCl H 3 Cl Cl 3 H 4 4 4 CH3 CH3 CH3 2S 2S,3R 2S,3S Optically active Optically active Optically active (Unequal amounts)

The chlorination at C2 results in a 1:1 mixture of enantiomers. Does the reaction at C3 also give an equimolar mixture of diastereomers? The answer is no. This ! nding is readily explained on inspection of the two transition states leading to the product (Figure 5-14). Abstraction of either one of the hydrogens results in a radical center at C3. In contrast with the radical formed in the chlorination at C2, however, the two faces of this radical are not mirror images of each other, because the radical retains the asymmetry of the original molecule as a result of the presence of the stereocenter at C2. Thus, the two sides of the p orbital are not equivalent. Mechanism What are the consequences of this nonequivalency? If the rate of attack at the two faces of the radical differ, as one would predict on steric grounds, then the rates of formation of the two diastereomers should be different, as is indeed found: (2S,3R)-2-Bromo-3-chlorobutane is

H3C H Cl + Chlorine attacks Cl Cl from the bottom (2S,3R)-2-Bromo- 3-chlorobutane 75%

Diastereomeric transition states 10 (Different in energy) 198 CHAPTER 5 Stereoisomers

Exercise 5-26 Try It Yourself Write the structures of the products of monobromination of (S)-2-bromopentane at each carbon atom. Name the products and specify whether they are chiral or achiral, whether they will be formed in equal or unequal amounts, and which will be in optically active form.

Recalling the rule that states that “optically inactive starting materials furnish optically inactive products,” we expect racemates for all products. Thus, attack at C1, C2, or C4 will result in racemic 2-bromo-1-chlorobutane, 2-bromo-2-chlorobutane, and 3-bromo- 1-chlorobutane, respectively. Importantly, attack at C3 will still give two compounds, even though racemic, namely, the 2S,3S/2R,3R (25%) and 2S,3R/2R,3S diastereomers (75%) of 2-bromo-3-chlorobutane. What are the conventions of writing chemical equations when racemates are involved? Attack at C1Unless, C2 ,specior! callyC4 indicatedwill result by thein R/Sracemic notation, the2- bromosign of -optical1-chlorobutane, rotation, or some2- bromosur- -2- chlorobutane,roundingand 3- bromo text, it - is1 - assumedchlorobutane, that all ingredientsrespectively in a. reactionImportantly, are racemic.attack To at avoidC3 thewill still give two compounds,clutter of writingnamely, both enantiomersthe 2S,3 Sin/ 2suchR,3 cases,R (25 only% )oneand is shown,2S,3 Rthe/2 equimolarR,3S diastereomers presence of the other being tacitly assumed. The chlorination of racemic 2-bromobutane at C3 is then (75%) of 2-bromowritten-3 -aschlorobutane follows: . Br H Br Br H Br H

/∑ Cl2, hv /∑ /∑

∑ ∑ / or / ϩ Cl H H Cl 75% 25%

Stereoselectivity is the preference for one stereoisomer Exercise 5-27 A reaction that leadsWrite theto productsthe predominant of the monochlorination(or exclusive) of bromocyclohexaneformation at C2.of (oneCaution:of Isseveral the startingpossible stereoisomeric productsmaterial chiral?)is stereoselective.

The chlorination of (S)-2-bromobutane at C3 is stereoselective, as a result of the chirality of the radical intermediateStereoselectivity. The corresponding is the preferencechlorination for oneat C 2stereoisomeris not stereoselective: The intermediate isA achiralreaction thatand leadsa racemate to the predominantis formed (or. exclusive) formation of one of several possible stereoisomeric products is stereoselective. For example, the chlorination of (S)-2-bromobutane at C3 is stereoselective, as a result of the chirality of the radical intermediate. The corre- 11 sponding chlorination at C2, however, is not stereoselective: The intermediate is achiral and a racemate is formed. How much stereoselectivity is possible? The answer depends very much on substrate, reagents, the particular reaction in question, and conditions. In the laboratory, chemists use enantiomerically pure reagents or catalysts to convert achiral compounds into one enantiomer of product (enantioselectivity; see Real Life 5-4). In nature, enzymes perform this job (see Real Life 5-5). In all cases, it is the handedness of that reagent, catalyst, or enzyme Dopamine plays a major role that is responsible for introducing the stereocenter compatible with their own chirality. An in reward-driven learning. The example from nature is the enzyme-catalyzed oxidation of dopamine to (2)-norepinephrine, pleasure you will experience discussed in detail in Problem 65 at the end of the chapter. The chiral reaction environment when you receive your created by the enzyme gives rise to 100% stereoselectivity in favor of the enantiomer shown. graduation diploma is due The situation is very similar to shaping " exible achiral objects with your hands. For exam- to an increase of the level of ple, clasping a piece of modeling clay with your left hand furnishes a shape that is the dopamine in your brain. mirror image of that made with your right hand. HHO %´ HO CH2CH2NH2 Dopamine HO C ␤-monooxygenase, O2 CH2NH2

HO HO Dopamine (؊)-Norepinephrine 198 CHAPTER 5 Stereoisomers

Recalling the rule that states that “optically inactive starting materials furnish optically inactive products,” we expect racemates for all products. Thus, attack at C1, C2, or C4 will result in racemic 2-bromo-1-chlorobutane, 2-bromo-2-chlorobutane, and 3-bromo- 1-chlorobutane, respectively. Importantly, attack at C3 will still give two compounds, even though racemic, namely, the 2S,3S/2R,3R (25%) and 2S,3R/2R,3S diastereomers (75%) of 2-bromo-3-chlorobutane. What are the conventions of writing chemical equations when racemates are involved? Unless speci! cally indicated by the R/S notation, the sign of optical rotation, or some sur- rounding text, it is assumed that all ingredients in a reaction are racemic. To avoid the clutter of writing both enantiomers in such cases, only one is shown, the equimolar presence of the other being tacitly assumed. The chlorination of racemic 2-bromobutane at C3 is then written as follows: Br H Br Br H Br H

/∑ Cl2, hv /∑ /∑

∑ ∑ / or / ϩ Cl H H Cl 75% 25%

Exercise 5-27 Write the products of the monochlorination of bromocyclohexane at C2. (Caution: Is the starting material chiral?)

Stereoselectivity is the preference for one stereoisomer Asymmetric synthesisA reaction that leads to the predominant (or exclusive) formation of one of several possible stereoisomeric products is stereoselective. For example, the chlorination of (S)-2-bromobutane at C3 is stereoselective, as a result of the chirality of the radical intermediate. The corre- In the laboratory,spondingchemists chlorinationuse enantiomericallyat C2, however, is not stereoselective:pure reagents The intermediateor catalysts is achiralto andconvert achiral compounds into aone racemateenantiomer is formed. of product (enantioselectivity). In nature, enzymes perform How much stereoselectivity is possible? The answer depends very much on substrate, this job. reagents, the particular reaction in question, and conditions. In the laboratory, chemists use enantiomerically pure reagents or catalysts to convert achiral compounds into one enantiomer of product (enantioselectivity; see Real Life 5-4). In nature, enzymes perform this job In all cases, reagent,(see Realcatalyst, Life 5-5). or In allenzyme cases, it isthat the handednessis responsible of that reagent,for introducing catalyst, or enzymethe stereocenter Dopamine compatibleplays a major role with theirthat is ownresponsiblechirality for introducing. the stereocenter compatible with their own chirality. An in reward-driven learning. The example from nature is the enzyme-catalyzed oxidation of dopamine to (2)-norepinephrine, pleasure you will experience discussed in detail in Problem 65 at the end of the chapter. The chiral reaction environment when you receiveAn yourexample fromcreated natureby the enzymeis givesthe rise toenzyme 100% stereoselectivity-catalyzed in favoroxidation of the enantiomerof shown.dopamine to (-)- graduationnorepinephrine diploma is due . The situationchiral is reactionvery similar toenvironment shaping " exible achiralcreated objectsby withthe yourenzyme hands. Forgives exam- rise to 100% to an increasestereoselectivity of the level of ple,in favor claspingof a the piece enantiomer of modeling clayshown with your. left hand furnishes a shape that is the dopamine in your brain. mirror image of that made with your right hand. HHO %´ HO CH2CH2NH2 Dopamine HO C ␤-monooxygenase, O2 CH2NH2

HO HO Dopamine (؊)-Norepinephrine

12 RESOLUTION: SEPARATION OF ENANTIOMERS

The generation of a chiral structure from an achiral starting material furnishes a racemic mixture. How can pure enantiomers of a chiral compound be obtained?

One possible approach is to start with the racemate and separate one enantiomer from the other. This process is called the resolution of enantiomers.

Some enantiomers, such as those of tartaric acid, crystallize into mirror-image shapes, which can be manually separated. However, this process is time consuming, not economical for anything but minute-scale separations, and applicable only in rare cases.

A better strategy for resolution is based on the different physical properties of diastereomers. Find a reaction that converts a racemate into a mixture of diastereomers.

The R forms of the original enantiomer mixture can be separable from the corresponding S forms by fractional crystallization, distillation, or chromatography of the diastereomers.

13 How can such a process be developed?

The trick is to add an enantiomerically pure reagent that will attach itself to the components of the racemic mixture.

For example, we can imagine reaction of a racemate, XR,S (in which XR and XS are the two enantiomers), with an optically pure compound YS. The reaction produces two optically active diastereomers, XRYS and XSYS, separable by standard techniques.

Now the bond between X and Y in each of the separated and purified diastereomers is broken, liberating XR and XS in their enantiomerically pure states.

In addition, the optically active agent YS may be recovered and reused in further resolutions.

14 200 CHAPTER 5 Stereoisomers

Figure 5-15 Flowchart for the Racemic mixture (enantiomers): Components X ϩ X separation (resolution) of two R S have the same physical properties enantiomers. The procedure is based on conversion into separable diastereomers by means of reaction Optically pure reagent YS with an optically pure reagent. Diastereomers: Components have X Y ϩ X Y R S S S the different physical properties

Separate diastereomers

XRYS XSYS

Cleave Cleave

XR ϩϩYS XS YS

Separate Separate and recover YS and recover YS

XR Pure XS Pure

OH@ COOH & NH2 H C A ϩ ( OH CH3CHCqCH HOOC H Racemic (R,S)-3-butyn-2-amine (؉)-Tartaric acid

H2O, several days O O

ϩ ϩ

r Ϫ r Ϫ

OH COO 3N HH @ COO 3N HH & OH &

C H * C ϩ H C C * R S ( OH H3C ( OH H3C HOOC H HOOC H Dextrorotatory (؉)-tartrate salt of R-amine Levorotatory (؉)-tartrate salt of S-amine 22°C 22°C [␣]D ϭϩ24.4 [␣]D ϭϪ24.1 Crystallizes from solution Stays in solution

K2CO3, H2O K2CO3, H2O

H2N H H NH2 * C * C R S H3C H3C 47% 51% R)-(؉)-3-Butyn-2-amine (S)-(؊)-3-Butyn-2-amine) 22°C 20°C [␣]D ϭϩ53.2(±1) [␣]D ϭϪ52.7(±1) b.p. 82°–84°C b.p. 82°–84°C

Figure 5-16 Resolution of 3-butyn-2-amine with (1)-2,3-dihydroxybutanedioic [(1)-tartaric] acid. It is purely accidental that the [a] values for the two diastereomeric tartrate salts are similar in magnitude and of opposite sign. An example is (+)-2,3-dihydroxybutanedioic acid [(1)-(R,R)-tartaric acid].

A popular reaction employed in the resolution of enantiomers is salt formation between acids and bases. For example, (+)-tartaric acid functions as an effective resolving agent of racemic amines.

The racemate is first treated with (+)-tartaric acid to form two diastereomeric tartrate salts.

The salt incorporating the R-amine crystallizes on standing and can be filtered away from the solution, which contains the more soluble salt of the S-amine.

Treatment of the (+) salt with aqueous base liberates the free amine, (+)-(R)-3-butyn-2- amine.

16 200 CHAPTER 5 Stereoisomers

Separate diastereomers

XRYS XSYS

Cleave Cleave

XR ϩϩYS XS YS

Separate Separate and recover YS and recover YS

XR Pure XS Pure

OH@ COOH & NH2 H C A ϩ ( OH CH3CHCqCH HOOC H Racemic (R,S)-3-butyn-2-amine (؉)-Tartaric acid

H2O, several days O O

ϩ ϩ

r Ϫ r Ϫ

OH COO 3N HH @ COO 3N HH & OH &

K2CO3, H2O K2CO3, H2O

H2N H H NH2 * C * C R S H3C H3C 47% 51% R)-(؉)-3-Butyn-2-amine (S)-(؊)-3-Butyn-2-amine) 22°C 20°C [␣]D ϭϩ53.2(±1) [␣]D ϭϪ52.7(±1) b.p. 82°–84°C b.p. 82°–84°C

Figure 5-16 Resolution of 3-butyn-2-amine with (1)-2,3-dihydroxybutanedioic [(1)-tartaric] 17 acid. It is purely accidental that the [a] values for the two diastereomeric tartrate salts are similar in magnitude and of opposite sign. A very convenient way of separating enantiomers without the necessity of isolating diastereomers is by so-called chiral chromatography.

The optically active auxiliary [such as (+)-tartaric acid or any other suitable cheap optically active compound] is immobilized on a solid support (such as silica gel, SiO2, or aluminum oxide, Al2O3).

This material is then used to fill a column, and a solution of the racemate is allowed to pass through it.

The individual enantiomers will reversibly bind to the chiral support to different extents (because this interaction is diastereomeric) and therefore be held on the column for different lengths of time (retention time).

Therefore, one enantiomer will elute from the column before the other, enabling separation.

18 The Big Picture CHAPTER 5 201

Figure 5-17 Resolution of a racemate on a chiral column. The sample is applied to the top of a column ! lled with an enantiopure Eluting chiral support. One enantiomer solvent (green) interacts more strongly with the support than the other Two enantiomers (red) and is relatively slow to pass begining to separate through the column. Therefore, the red enantiomer is eluted before its green mirror image. Commercial columns often use the glucose polymer cellulose (Section 24-12) as the chiral stationary phase.

Racemate applied

Solid chiral support

Test tube

Red enantiomer eluted before green one. 19 Time this interaction is diastereomeric) and therefore be held on the column for different lengths of time (retention time). Therefore, one enantiomer will elute from the column before the other, enabling separation.

THE BIG PICTURE The end of this chapter marks a milestone in your learning of organic chemistry. From here on, we will add very few fundamentally new ideas to our understanding of molecular structure. Rather, we will build on the basic principles outlined so far. You may ! nd it useful to refer back to Chapters 1–5 as often as needed as a reminder of the several complementary viewpoints from which we have examined molecular structure: 1. Molecules consist of nuclei surrounded by electrons. The rules that govern their interactions are Coulomb’s law, Lewis structures, and orbitals (Chapter 1). 2. The hydrocarbon skeletons of organic molecules, as exempli! ed by the alkanes, can be modeled as chains of linked carbon atoms that rotate, " ip, have spatial requirements, and can be distorted away from ideal bond angles (Chapters 2 and 4). 3. The diversity of structures in organic chemistry is due to the ability of carbon atoms to assemble chains in linear, branched, cyclic, or polycyclic forms, all of which can bear multiple substituents and functional groups. Because of carbon’s tetrahedral bonding, organic molecules assume a variety of three-dimensional shapes, with variations that have important consequences in their physical properties and reactivity. 4. When tetrahedral carbon bears four different substituents, it constitutes a stereocenter, giving rise to enantiomers—image and nonsuperimposable mirror image. Two or more such stereocenters generate new stereoisomers, differing only in the spatial arrangement of their component groups.