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