CHAPTER 7 STEREOCHEMISTRY SOLUTIONS TO TEXT PROBLEMS 7.1(c) Carbon-2 is a stereogenic center in 1-bromo-2-methylbutane, as it has four different substituents: H, CH3, CH,CH,, and BrCH H BrCh.-c-Chc (d) There are no stereogenic centers in 2-bromo-2-methylbutane CH3-C—CH2CH3 CH 7.2(b) Carbon-2 is a stereogenic center in 1, 1, 2-trimethylcyclobutane CH3 A stereogenic center; the four substituents to which it is directly bonded [H, CH3, CH2, and C(CHa)2l are all different from one another 1, 1, 3-Trimethylcyclobutane however, has no stereogenic centers HC CH 156 Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
CHAPTER 7 STEREOCHEMISTRY SOLUTIONS TO TEXT PROBLEMS 7.1 (c) Carbon-2 is a stereogenic center in 1-bromo-2-methylbutane, as it has four different substituents: H, CH3, CH3CH2, and BrCH2. (d) There are no stereogenic centers in 2-bromo-2-methylbutane. 7.2 (b) Carbon-2 is a stereogenic center in 1,1,2-trimethylcyclobutane. 1,1,3-Trimethylcyclobutane however, has no stereogenic centers. H3C CH3 H3C H Not a stereogenic center; two of its substituents are the same. CH3 CH3 H H3C A stereogenic center; the four substituents to which it is directly bonded [H, CH3, CH2, and C(CH3)2] are all different from one another. CH3 C CH2CH3 CH3 Br BrCH2 C CH2CH3 CH3 H 156 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
STEREOCHEMISTRY 157 7.3 (b) There are two planes of symmetry in(2)-1, 2-dichloroethene, of which one is the plane of the molecule and the second bisects the carbon-carbon bond. There is no center of symmetry. The nolecule is achiral Planes of symmetry (c) There is a plane of symmetry in cis-1, 2-dichlorocyclopropane that bisects the C-1-C-2 bond and passes through C-3. The molecule is achiral Plane of symmetry (d) trans-1, 2-Dichlorocyclopropane has neither a plane of symmetry nor a center of symmetry. Its two mirror images cannot be superposed on each other. The molecule is chiral CI are nonsuperposable 7.4 The equation relating specific rotation [a]to observed rotation a is 100a The concentration c is expressed in grams per 100 mL and the length l of the polarimeter tube in decimeters. Since the problem specifies the concentration as 0.3 g/15 mL and the path length as 10 cm, the specific rotation [a] is 100(-0.78°) 100(0.3g/15mL10cm/10cm/dm 7.5 From the previous problem, the specific rotation of natural cholesterol is [a]=-390. The mixture of natural(-)-cholesterol and synthetic (+)-cholesterol specified in this problem has a specifi rotation[aof-13° Optical purity =%( 33.3%=%(-)-cholesterol -[100-%(-)-cholesterol 133.3%=2 %(-)-cholesterol] 66.7%=%(-)-cholesterol The mixture is two thirds natural(-)-cholesterol and one third synthetic(+)-cholesterol Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
7.3 (b) There are two planes of symmetry in (Z)-1,2-dichloroethene, of which one is the plane of the molecule and the second bisects the carbon–carbon bond. There is no center of symmetry. The molecule is achiral. (c) There is a plane of symmetry in cis-1,2-dichlorocyclopropane that bisects the C-1@C-2 bond and passes through C-3. The molecule is achiral. (d) trans-1,2-Dichlorocyclopropane has neither a plane of symmetry nor a center of symmetry. Its two mirror images cannot be superposed on each other. The molecule is chiral. 7.4 The equation relating specific rotation [] to observed rotation is [] The concentration c is expressed in grams per 100 mL and the length l of the polarimeter tube in decimeters. Since the problem specifies the concentration as 0.3 g/15 mL and the path length as 10 cm, the specific rotation [] is: [] 39° 7.5 From the previous problem, the specific rotation of natural cholesterol is [] 39°. The mixture of natural ()-cholesterol and synthetic ()-cholesterol specified in this problem has a specific rotation [] of 13°. Optical purity %()-cholesterol % ()-cholesterol 33.3% %()-cholesterol [100 % ()-cholesterol] 133.3% 2 [% ()-cholesterol] 66.7% % ()-cholesterol The mixture is two thirds natural ()-cholesterol and one third synthetic ()-cholesterol. 100(0.78°) 100(0.3 g15 mL)(10 cm10 cm/dm) 100 cl Cl H H Cl H Cl and are nonsuperposable mirror images Cl H Cl H Cl H 1 2 3 Plane of symmetry C C Cl Cl H H Planes of symmetry STEREOCHEMISTRY 157 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
158 STEREOCHEMISTRY 7.6 Draw the molecular model so that it is in the same format as the drawings of (+)and(-)-2-butanol HO H Redraw H3C CH, CH3 Reorient the molecule so that it can be compared with the drawings of(+)and(-)-2-butanol HO H which becomes C-OH H, C CH-CH H,C The molecular model when redrawn matches the texts drawing of (+)-2-butanol .7(b) The solution to this problem is exactly analogous to the sample solution given in the text to part(a) -CHF CH CH Order of precedence: CHF>CH3 CH,>CH3>H The lowest ranked substituent(H) at the stereogenic center points away from us in the drawing. The three higher ranked substituents trace a clockwise path from CH,F to CH, CH3 HC、C CH,CH3 The absolute configuration is R; the compound is(R)-(+)-l-fluoro-2-methylbutane (c) The highest ranked substituent at the stereogenic center of l-bromo-2-methylbutane is CH,B and the lowest ranked substituent is H. Of the remaining two, ethyl outranks methyl Order of precedence: CH, Br > CH, CH3>CH3>H The lowest ranking substituent(H) is directed toward you in the drawing, and therefore the molecule needs to be reoriented so that H points in the opposite direction -CHBr t brch CHCH CH, CH3 turn180° (+)-1-Bromo-2-methylbutane Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
158 STEREOCHEMISTRY 7.6 Draw the molecular model so that it is in the same format as the drawings of () and ()-2-butanol in the text. Reorient the molecule so that it can be compared with the drawings of () and ()-2-butanol. The molecular model when redrawn matches the text’s drawing of ()-2-butanol. 7.7 (b) The solution to this problem is exactly analogous to the sample solution given in the text to part (a). Order of precedence: CH2F CH3CH2 CH3 H The lowest ranked substituent (H) at the stereogenic center points away from us in the drawing. The three higher ranked substituents trace a clockwise path from CH2F to CH2CH3 to CH3. The absolute configuration is R; the compound is (R)-()-1-fluoro-2-methylbutane. (c) The highest ranked substituent at the stereogenic center of 1-bromo-2-methylbutane is CH2Br, and the lowest ranked substituent is H. Of the remaining two, ethyl outranks methyl. Order of precedence: CH2Br CH2CH3 CH3 H The lowest ranking substituent (H) is directed toward you in the drawing, and therefore the molecule needs to be reoriented so that H points in the opposite direction. H CH3 BrCH2 CH2CH3 ()-1-Bromo-2-methylbutane turn 180 C CH3 H CH2Br CH3CH2 C C H3C CH2F CH2CH3 C H H3C CH2F CH3CH2 ()-1-Fluoro-2-methylbutane Reorient which becomes H3C CH2CH3 C HO H H OH H3C CH3CH2 C H H H3C CH2CH3 C HO H O C C C H H H H H H H C H Redraw as Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
STEREOCHEMISTRY 159 The three highest ranking substituents trace a counterclockwise path when the lowest ranked lbstituent is held away from The absolute configuration is S, and thus the compound is(S)-(+)-1-bromo-2-methylbutane (d) The highest ranked substituent at the stereogenic center of 3-buten-2-ol is the hydroxyl group ind the lowest ranked substituent is H. Of the remaining two, vinyl outranks methy Order of precedence: HO> CH,CH>CH3>H The lowest ranking substituent(H) is directed away from you in the drawing. We see that the order of decreasing precedence appears in a counterclockwise manner. (+)-3-Buten-2-o1 The absolute configuration is S, and the compound is(S)-(+)-3-buten-2-o1 7.8 (b) The stereogenic center is the carbon that bears the methyl group Its substituents are -CF2CH2>-CH, CF2>CH3> H When the lowest ranked substituent points away from you, the remaining three must appe ty ding order of precedence in a counterclockwise fashion in the S difluoro-2-methylcyclopropane is therefore 7.9(b) The Fischer projection of(R)-(+)-1-fuoro-2-methylbutane is analogous to that of the alcohol in part(a). The only difference in the two is that fluorine has replaced hydroxyl as a sub stituent at C-1 H,F ChF is the same as which becomes the Fischer projection H--CH CH,CH CHCH CHCH Although other Fischer projections may be drawn by rotating the perspective view in other di rections, the one shown is preferred because it has the longest chain of carbon atoms oriented on the vertical axis with the lowest numbered carbon at the top (c) As in the previous parts of this problem, orient the structural formula of (S)-(+)-l-bromo- 2-methylbutane so the segment BrCH,C-CH,CH3 is aligned vertically with the lowest numbered carbon at the top. C-CH, Br is the same as H3C which becomes the Fischer projection HC-H CHa CH CH, CH3 CHCH Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
The three highest ranking substituents trace a counterclockwise path when the lowest ranked substituent is held away from you. The absolute configuration is S, and thus the compound is (S)-()-1-bromo-2-methylbutane. (d) The highest ranked substituent at the stereogenic center of 3-buten-2-ol is the hydroxyl group, and the lowest ranked substituent is H. Of the remaining two, vinyl outranks methyl. Order of precedence: HO CH2?CH CH3 H The lowest ranking substituent (H) is directed away from you in the drawing. We see that the order of decreasing precedence appears in a counterclockwise manner. The absolute configuration is S, and the compound is (S)-()-3-buten-2-ol. 7.8 (b) The stereogenic center is the carbon that bears the methyl group. Its substituents are: When the lowest ranked substituent points away from you, the remaining three must appear in descending order of precedence in a counterclockwise fashion in the S enantiomer. (S)-1, 1- difluoro-2-methylcyclopropane is therefore 7.9 (b) The Fischer projection of (R)-()-1-fluoro-2-methylbutane is analogous to that of the alcohol in part (a). The only difference in the two is that fluorine has replaced hydroxyl as a substituent at C-1. Although other Fischer projections may be drawn by rotating the perspective view in other directions, the one shown is preferred because it has the longest chain of carbon atoms oriented on the vertical axis with the lowest numbered carbon at the top. (c) As in the previous parts of this problem, orient the structural formula of (S)-()-1-bromo- 2-methylbutane so the segment BrCH2@C@CH2CH3 is aligned vertically with the lowest numbered carbon at the top. C CH3 CH3CH2 CH2Br H H3C C CH2CH3 CH2Br H H3C CH2CH3 CH2Br is the same as which becomes the Fischer projection H C H3C CH3CH2 CH2F H C CH3 CH2CH3 CH2F H CH3 CH2CH3 CH2F is the same as which becomes the Fischer projection H H CH3 F F CF2CH2 CH2CF2 CH3 H Highest rank Lowest rank ()-3-Buten-2-ol CH2 CH2 HO C H H3C CH2 C H3C CH OH C BrCH2 CH3 CH2CH3 STEREOCHEMISTRY 159 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
160 STEREOCHEMISTRY (d) Here we need to view the molecule from behind the page in order to write the Fischer projec tion of (S)-(+)-3-buten-2-ol CH H CH3 -CH=CH, is the same as which becomes the Fischer I H-OH HO CH=CH, CH=CH, 7.10 In order of decreasing rank, the substituents attached to the stereogenic center in lactic ad -OH, -CO H, -CH3, and-H. The Fischer projection given for(+)-lactic acid (a)corres to the three-dimensional representation(b), which can be reoriented as in(c). When(c)is from the side opposite the lowest ranked substituent(H), the order of decreasing precedence is anti- clockwise, as shown in(d).(+)-Lactic acid has the S configuration. COH CO,H HO CO2H HOrC-H c-H HOCOH CH CH CH 7.11 The erythro stereoisomers are characterized by Fischer projections in which analogous substituents in this case oh and nh are on the same side when the carbon chain is vertical. there are two erythro stereoisomers that are enantiomers of each other OH NH2 H,N-H Analogous substituents are on opposite sides in the threo isomer: CH H-OH Ho-t-H H Threo Threo 7.12 There are four stereoisomeric forms of 3-amino-3-butanol (2R, 3R)and its enantiomer(2S, 3S) In the text we are told that the(2R, 3R)stereoisomer is a liquid. Its enantiomer(2S, 3S)has the same physical properties and so must also be a liquid. The text notes that the(2R, 3S) stereoisomer is a solid(mp 49C). Its enantiomer(2S, 3R)must therefore be the other stereoisomer that is a crystalline Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(d) Here we need to view the molecule from behind the page in order to write the Fischer projection of (S)-()-3-buten-2-ol. 7.10 In order of decreasing rank, the substituents attached to the stereogenic center in lactic acid are @OH, @CO2H, @CH3, and @H. The Fischer projection given for ()-lactic acid (a) corresponds to the three-dimensional representation (b), which can be reoriented as in (c). When (c) is viewed from the side opposite the lowest ranked substituent (H), the order of decreasing precedence is anticlockwise, as shown in (d). ()-Lactic acid has the S configuration. 7.11 The erythro stereoisomers are characterized by Fischer projections in which analogous substituents, in this case OH and NH2, are on the same side when the carbon chain is vertical. There are two erythro stereoisomers that are enantiomers of each other: Analogous substituents are on opposite sides in the threo isomer: 7.12 There are four stereoisomeric forms of 3-amino-3-butanol: (2R,3R) and its enantiomer (2S,3S) (2R,3S) and its enantiomer (2S,3R) In the text we are told that the (2R,3R) stereoisomer is a liquid. Its enantiomer (2S,3S) has the same physical properties and so must also be a liquid. The text notes that the (2R,3S) stereoisomer is a solid (mp 49°C). Its enantiomer (2S,3R) must therefore be the other stereoisomer that is a crystalline solid. CH3 OH H CH3 H H2N Threo CH3 H NH2 CH3 HO H Threo CH3 OH NH2 CH3 H H Erythro CH3 H H CH3 HO H2N Erythro H3C HO CO2H C H (c) CH3 HO CO2H H (b) CH3 HO CO2H H (a) CH3 HO2C OH (d) C C CH3 HO CH CH2 CH CH2 CH CH2 H C OH CH3 H OH CH3 is the same as which becomes the Fischer projection H 160 STEREOCHEMISTRY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
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
