STEREOCHEMISTRY 161 7. 13 Examine the structural formula of each compound for equivalently substituted stereogenic centers The only one capable of existing in a meso form is 2, 4-dibromopentane B H-H H-Br H3 2, 4-Dibromopentane Fischer projection of meso-2.4-dibromopentane None of the other compounds has equivalently substituted stereogenic centers. No meso forms are possible for: CH_CHCHCH,CH3 CH,CHCHCH2CH3 CH_ CHCH, CHCH Br Br 4.Brom 7.14 There is a plane of symmetry in the cis stereoisomer of 1, 3-dimethylcyclohexane, and so it is an achiral substance--it is a meso form H3 Plane of symmetry passes through C-2 and C-5 and bisects the ring. The trans stereoisomer is chiral. It is not a meso form 7. 15 A molecule with three stereogenic centers has 2, or 8, stereoisomers. The eight combinations of R Stereogenic center Stereogenic center 123 123 Isomer 1 RRR Isomer 5 SSS Isomer 2 Isomer 6 SSR somer RSR Isomer 7 SRs Isomer 4 SRR Isomer 8 RSS 7.16 2-Hexuloses have three stereogenic centers. They are marked with asterisks in the structural HOCHCCHCHCHCHOH oHOH No meso forms are possible, and so there are a total of 2, or8, stereoisomeric 2-hexuloses Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
7.13 Examine the structural formula of each compound for equivalently substituted stereogenic centers. The only one capable of existing in a meso form is 2,4-dibromopentane. None of the other compounds has equivalently substituted stereogenic centers. No meso forms are possible for: 7.14 There is a plane of symmetry in the cis stereoisomer of 1,3-dimethylcyclohexane, and so it is an achiral substance—it is a meso form. The trans stereoisomer is chiral. It is not a meso form. 7.15 A molecule with three stereogenic centers has 23 , or 8, stereoisomers. The eight combinations of R and S stereogenic centers are: Stereogenic center Stereogenic center 1 2 3 1 2 3 Isomer 1 R R R Isomer 5 SSS Isomer 2 R R S Isomer 6 SSR Isomer 3 R S R Isomer 7 SRS Isomer 4 S R R Isomer 8 RSS 7.16 2-Hexuloses have three stereogenic centers. They are marked with asterisks in the structural formula. No meso forms are possible, and so there are a total of 23 , or 8, stereoisomeric 2-hexuloses. HOCH2CCHCHCHCH2OH OH O OH OH * * * CH3 H3C 1 3 5 4 2 6 Plane of symmetry passes through C-2 and C-5 and bisects the ring. 2,3-Dibromopentane CH3CHCHCH2CH3 Br Br 4-Bromo-2-pentanol CH3CHCH2CHCH3 OH Br 3-Bromo-2-pentanol CH3CHCHCH2CH3 OH Br 2,4-Dibromopentane Equivalently substituted stereogenic centers CH3CHCH2CHCH3 Br * Br * CH3 Br H CH3 H H H Br Fischer projection of meso-2,4-dibromopentane Plane of symmetry STEREOCHEMISTRY 161 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
162 STEREOCHEMISTRY 7.17 Epoxidation of (Z)-2-butene gives the meso(achiral)epoxide Oxygen transfer from the peroxy acid can occur at either face of the double bond, but the product formed is the same because the two mirror-image forms of the epoxide are superposable H-C CH3 CH3 CHA COOH CH COOH H-C H, C meso-2,3-Epoxybutane Epoxidation of (E)r-2-butene gives a racemic mixture of two enantiomeric epoxides 3C CHa COOH H CH, COC HC (2R, 3R)-2, 3-Epoxybutan eS, 3S)-2, 3-Epoxybutane 7.18 The observed product mixture(68% cis-1, 2-dimethylcyclohexane: 32%o trans-1, 2-dimethylcyclo- hexane) contains more of the less stable cis stereoisomer than the trans The relative stabilities of the products therefore play no role in determining the stereoselectivity of this reaction 7.19 The tartaric acids incorporate two equivalently substituted stereogenic centers. (+)-Tartaric acid, as noted in the text. is the 2R 3R stereoisomer. There will be two additional stereoisomers the enan- acid(2S, 3S)and an optically inactive COH COH lane of symmetry HO-H H H COH CO,H S)-Tartaric acid meso Tart optically active) 7.20 No. Pasteur separated an optically inactive racemic mixture into two optically active enantiomers. A meso form is achiral, is identical to its mirror image, and is incapable of being separated into opti- cally active forms Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
7.17 Epoxidation of (Z)-2-butene gives the meso (achiral) epoxide. Oxygen transfer from the peroxy acid can occur at either face of the double bond, but the product formed is the same because the two mirror-image forms of the epoxide are superposable. Epoxidation of (E)-2-butene gives a racemic mixture of two enantiomeric epoxides. 7.18 The observed product mixture (68% cis-1,2-dimethylcyclohexane: 32% trans-1,2-dimethylcyclohexane) contains more of the less stable cis stereoisomer than the trans. The relative stabilities of the products therefore play no role in determining the stereoselectivity of this reaction. 7.19 The tartaric acids incorporate two equivalently substituted stereogenic centers. ()-Tartaric acid, as noted in the text, is the 2R,3R stereoisomer. There will be two additional stereoisomers, the enantiomeric ()-tartaric acid (2S,3S) and an optically inactive meso form. 7.20 No. Pasteur separated an optically inactive racemic mixture into two optically active enantiomers. A meso form is achiral, is identical to its mirror image, and is incapable of being separated into optically active forms. (2S,3S)-Tartaric acid (optically active) (mp 170°C, []D 12°) meso-Tartaric acid (optically inactive) (mp 140°C) CO2H CO2H HO H OH H H H CO2H CO2H OH OH Plane of symmetry O CH3 CH3 H H S S (2S,3S)-2,3-Epoxybutane O H3C H3C H H R R (2R,3R)-2,3-Epoxybutane CH3COOH O CH3COOH O H CH3 H E H3C C C O CH3 CH3 H H S R meso-2,3-Epoxybutane O H3C H3C H H R S meso-2,3-Epoxybutane H3C CH3 H Z H C CH3COOH C O CH3COOH O 162 STEREOCHEMISTRY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
STEREOCHEMISTRY 163 7.21 The more soluble salt must have the opposite configuration at the stereogenic center of 1-phenylethylamine, that is, the S configuration. The malic acid used in the resolution is a single enantiomer, S. In this particular case the more soluble salt is therefore(S)-1-phenylethylammonium (S)-malate 7.22 In an earlier exercise(Problem 4.23)the structures of all the isomeric CsH,O alcohols were pre sented. Those that lack a stereogenic center and thus are achiral are CH3 CH,CH,CH,CH,OH CH CHCH,CH,OH (CH3)CCH,OH 3. MethyI-I-butanol 2.2-Dimethyl-1-propano CH CH3CH_CHCH,CH3 CH, CH, COH 3-Pentanol 2. MethyI-2-butanol The chiral isomers are characterized by carbons that bear four different groups. These are CH,CHCH, CH,CH3 CH3 CHCH(CH3)2 CHCH,CHCH,OH 7.23 The isomers of trichlorocyclopropane are aCI YCI H Enantiomeric forms of 1, 1, 2-trichlorocycloprd 7.24 (a) Carbon-2 is a stereogenic center in 3-chloro-1, 2-propanediol Carbon-2 has two equivalent substituents in 2-chloro-1, 3-propanediol, and is not a stereogenic center CICH, CHCH,OH HOCH CHCHOH 3-Chloro-1, 2-propanediol 2-Chloro-1, 3-propanediol (b) The primary bromide is achiral; the secondary bromide contains a stereogenic center and is CH CH=CHCH, Br CH_CHCH=CH2 B Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
7.21 The more soluble salt must have the opposite configuration at the stereogenic center of 1-phenylethylamine, that is, the S configuration. The malic acid used in the resolution is a single enantiomer, S. In this particular case the more soluble salt is therefore (S)-1-phenylethylammonium (S)-malate. 7.22 In an earlier exercise (Problem 4.23) the structures of all the isomeric C5H12O alcohols were presented. Those that lack a stereogenic center and thus are achiral are The chiral isomers are characterized by carbons that bear four different groups. These are: 7.23 The isomers of trichlorocyclopropane are 7.24 (a) Carbon-2 is a stereogenic center in 3-chloro-1,2-propanediol. Carbon-2 has two equivalent substituents in 2-chloro-1,3-propanediol, and is not a stereogenic center. (b) The primary bromide is achiral; the secondary bromide contains a stereogenic center and is chiral. CH3CH CHCH2Br Achiral CH3CHCH CH2 * Br Chiral ClCH2CHCH2OH * OH 3-Chloro-1,2-propanediol Chiral HOCH2CHCH2OH Cl 2-Chloro-1,3-propanediol Achiral cis-1,2,3-Trichlorocyclopropane (achiral—contains a plane of symmetry) Cl H H Cl Cl trans-1,2,3-Trichlorocyclopropane (achiral—contains a plane of symmetry) Cl H H Cl Cl Cl Cl H Cl Cl H Cl Enantiomeric forms of 1,1,2-trichlorocyclopropane (both chiral) Cl 2-Pentanol 3-Methyl-2-butanol CH3CHCH2CH2CH3 OH * 2-Methyl-1-butanol CH3CH2CHCH2OH CH3 * CH3CHCH(CH3)2 OH * 3-Pentanol 2-Methyl-2-butanol CH3CH2COH CH3 CH3 CH3CH2CHCH2CH3 OH 1-Pentanol 3-Methyl-1-butanol 2,2-Dimethyl-1-propanol CH3CH2CH2CH2CH2OH (CH3) CH3CHCH2CH2OH 3CCH2OH CH3 STEREOCHEMISTRY 163 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
164 STEREOCHEMISTRY (c) Both stereoisomers have two equivalently substituted stereogenic centers, and so we must be alert for the possibility of a meso stereoisomer. The structure at the left is chiral. The one at the right has a plane of symmetry and is the achiral meso stereoisomer CH ymmetry H,N-H H-1NH2 H-+NH H-1NH2 Chiral Meso: achiral (d) The first structure is achiral; it has a plane of symmetry Plane of symmetry passes through C-1, C-4, and C-7 The second structure cannot be superposed on its mirror image; it is chiral Cl Mirror image Reoriented mirror image 7. 25 There are four stereoisomers of 2, 3-pentanediol, represented by the Fischer projections shown. All CH H-OH H HO→H H-OH Ho-t-H HO H-OH CH,CH3 CH,CH3 CH, CH3 CH,CH3 There are three stereoisomers of 2, 4-pentanediol. The meso form is achiral; both threo forms are H3 H3 H-OH Ho-H H-OH HO→H meso-2. 4-Pentanediol Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(c) Both stereoisomers have two equivalently substituted stereogenic centers, and so we must be alert for the possibility of a meso stereoisomer. The structure at the left is chiral. The one at the right has a plane of symmetry and is the achiral meso stereoisomer. (d) The first structure is achiral; it has a plane of symmetry. The second structure cannot be superposed on its mirror image; it is chiral. 7.25 There are four stereoisomers of 2,3-pentanediol, represented by the Fischer projections shown. All are chiral. There are three stereoisomers of 2,4-pentanediol. The meso form is achiral; both threo forms are chiral. meso-2,4-Pentanediol CH3 CH3 H H H OH OH H CH3 CH3 H H HO OH H H CH3 CH3 H H HO OH H H Enantiomeric threo isomers Enantiomeric erythro isomers CH2CH3 CH3 H H OH OH CH2CH3 CH3 HO HO H H Enantiomeric threo isomers CH2CH3 CH3 HO H OH H CH2CH3 CH3 HO H H OH Cl Cl Cl Reference structure Mirror image Reoriented mirror image Cl 1 4 7 Plane of symmetry passes through C-1, C-4, and C-7. H2N NH2 CH3 CH3 H H Chiral Plane of symmetry NH2 NH2 CH3 CH3 H H Meso: achiral 164 STEREOCHEMISTRY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
STEREOCHEMISTRY 165 7.26 Among the atoms attached to the stereogenic center, the order of decreasing precedence is Br CI>F>H. When the molecule is viewed with the hydrogen pointing away from us, the order Br-Cl-F appears clockwise in the R enantiomer, anticlockwise in the S enantiomer. -Cl F.nCHI-CI 727(a )-2-Octanol has the R configuration at C-2. The order of substituent precedence is HO> CH, CH2>CH3>H The molecule is oriented so that the lowest ranking substituent is directed away from you and the order of decreasing precedence is clockwise HO H CH3 CH,CH, (b) In order of decreasing sequence rule precedence, the four substituents at the stereogenic cen- m L-glutamate are NH,>CO>Ch>H CO CO H3N he same as henOCh CH, CH, CO, Na+ CHaCH,CO2 Na When the molecule is oriented so that the lowest ranking substituent(hydrogen) is directed away from you, the other three substituents are arranged as CH CH CO Nat The order of decreasing rank is counterclockwise: the absolute configuration is s. 7.28 (a) Among the isotopes of hydrogen, T has the highest mass number (3), D next(2), and H low est(1). Thus, the order of rank at the stereogenic center in the reactant is CH3>T>D>H The order of rank in the product is HO>CH3>T>D Orient with lowest ranked substituent H, C Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
7.26 Among the atoms attached to the stereogenic center, the order of decreasing precedence is Br Cl F H. When the molecule is viewed with the hydrogen pointing away from us, the order Br → Cl → F appears clockwise in the R enantiomer, anticlockwise in the S enantiomer. 7.27 (a) ()-2-Octanol has the R configuration at C-2. The order of substituent precedence is HO CH2CH2 CH3 H The molecule is oriented so that the lowest ranking substituent is directed away from you and the order of decreasing precedence is clockwise. (b) In order of decreasing sequence rule precedence, the four substituents at the stereogenic center of monosodium L-glutamate are N H3 CO2 CH2 H When the molecule is oriented so that the lowest ranking substituent (hydrogen) is directed away from you, the other three substituents are arranged as shown. The order of decreasing rank is counterclockwise; the absolute configuration is S. 7.28 (a) Among the isotopes of hydrogen, T has the highest mass number (3), D next (2), and H lowest (1). Thus, the order of rank at the stereogenic center in the reactant is CH3 T D H. The order of rank in the product is HO CH3 T D. Orient with lowest ranked substituent away from you. CH3 D T H3C T OH C D H CH3 T C D HO T CH3 biological oxidation CH2CH2CO2 NH3 O2C Na H3N H CH2CH2CO2 CO2 H3N H CH2CH2CO2 CO2 is the same as C Na Na OH CH3 CH2CH2 HO H Cl H Cl F Br H F C Cl Br R-() S-() R-() S-() C Br H F Br Cl C F H STEREOCHEMISTRY 165 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website