chapter 15 the three-dimensional shape of...

Chapter 15–1

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Chapter 15 The Three-Dimensional Shape of Molecules Solutions to In-Chapter Problems 15.1 Constitutional isomers differ in the way the atoms are connected to one another.

Stereoisomers differ only in the three-dimensional arrangement of atoms.

CH3CH2CHCHCH3 CH3CHCH2CHCH3

CH3 CH3 CH3

CH3

anda. b. C CH

CH3CH3CH2

HC C

H

HCH3CH2

CH3

and

different connection of atomsconstitutional isomers stereoisomers

cis trans

O

OHandc. d.CH3CH CHCH2CH3 CH3CH2CH CHCH3and

identical different connection of atomsconstitutional isomers

15.2 Draw the isomers of trans-2-hexene.

C C

H

CH3 CH2CH2CH3

HCH2 CHCH2CH2CH2CH3a. b. c.C C

H

CH3 H

CH2CH2CH3

trans-2-hexene cis-2-hexenestereoisomer

different connection of atomsconstitutional isomer

different connection of atomsconstitutional isomer

15.3 A molecule (or object) that is not superimposable on its mirror image is said to be chiral.

A molecule (or object) that is superimposable on its mirror image is said to be achiral. a. nail—achiral c. glove—chiral b. screw—chiral d. pencil—achiral

15.4 A molecule that is chiral is not superimposable on its mirror image. A chirality center is a carbon

with four different groups bonded to it. 15.5 To locate the chirality centers, look at each C individually and eliminate those C’s that can’t be

chirality centers. Thus, omit all CH2 and CH3 groups and all multiply bonded C’s. Check all remaining C’s to see if they are bonded to four different groups, as in Example 15.1.

a. b. CH2 CH C CH3

OH

c. CH

BrCH3d.

noneCH3 C CH2CH3

H

Cl(CH3)3CH

H

15.6 Label each chirality center, as in Example 15.1.

a. C

H

CH2CH2NHCH3

O CF3C

CH2CH2N(CH3)2

Br

H

Nb.

Chapter 15–2



15.7 Label each chirality center, as in Example 15.1.

CH3CH2CH2 CH

OHCH3 H2N C

COOHH

CCH3CH2 C

BrCH2CH2

HCCl

CH3HCH3 H

CH2CH3

a. b. c.

15.8 Label the two chirality centers in vitamin K.

CH2CH=CCH2CH2CH2CHCH2CH2CH2CHCH2CH2CH2CH(CH3)2

CH3

O

O CH3 CH3 CH3 15.9 To draw both enantiomers:

[1] Draw one enantiomer by arbitrarily placing the four different groups on the bonds to the chirality center.

[2] Draw a mirror plane and arrange the substituents in the mirror image so that they are a reflection of the groups in the first molecule.

CH3

CCl CH2CH3

H

CH3

CClCH3CH2

HCH3CHCH2CH3

Cl

a. C

Draw the bonds: Add the groups:

mirrorenantiomers

HO

CH

HOCH2

CH3CH2

OH

CH

CH2OHCH2CH3

CH3CH2CHCH2OH

OH

b.C


mirrorenantiomers

15.10 Locate the chirality centers in each compound.

All C's are part of multiple bonds or are bonded to

two or three H's.

CH2CH3Cl

Cl

a. b. c.OH

none

Chapter 15–3



15.11 Label the two chirality centers.

CH3NH Cl

Cl

15.12 Answer each question about propranolol.

OCH2CHCH2NHCH(CH3)2OH

a.

chirality center

H

CCH2

HO(CH3)2CHNHCH2

H

CCH2 OH

CH2NHCH(CH3)2O Ob.

enantiomers

H

CCH2

HO(CH3)2CHNHCH2

H

CCH2 OH

CH2NHCH(CH3)2O Oc.

This enantiomer fits thereceptor.

This enantiomer doesnot fit thereceptor.

receptor receptor

15.13 Convert each molecule to a Fisher projection formula by replacing the chirality center with a

cross.

CH Br

CH3

Cl

CH OH

COOH

CH2Cl

a. b.HCH3

BrCl

H OHCOOH

CH2Cl 15.14 Work backwards to draw a structure with wedges (horizontal bonds) and dashes (vertical bonds).

C BrCH3CH2

Cl

C CH3CH3CH2

H

NH2

CH3

CH3CH2 Br

CH3

Cl

a. b. CH3CH2 CH3

H

NH2

Chapter 15–4



15.15 Convert each molecule to a Fisher projection formula as in Example 15.2. • Draw the tetrahedron of one enantiomer with the horizontal bonds on wedges and the vertical

bonds on dashed lines. Arrange the four groups on the chirality center arbitrarily in the first enantiomer.

• Draw the second enantiomer by arranging the substituents in the mirror image so they are a reflection of the groups in the first molecule.

• Replace the chirality center with a cross to draw the Fischer projections.

CH3CHCH2CH2CH3

OH

a.

HCH2CH2CH3CH3

OH

HCH3CH3CH2CH2

OH

• Horizontal bonds come forward.• Vertical bonds go back.

chirality center

CCH3 CH2CH2CH3

H

OHCCH3CH2CH2 CH3

H

OH

Draw the compound. Then, draw the mirror image.

Fischer projection of one enantiomer

Fischer projection of the second enantiomer

CH3CH2CHCH2Clb.

Cl

H

ClCH2CH3ClCH2

H

ClCH2ClCH3CH2

• Horizontal bonds come forward.• Vertical bonds go back.

chirality center

CClCH2 CH2CH3

H

Cl

CCH3CH2 CH2Cl

H

Cl

Draw the compound. Then, draw the mirror image.

Fischer projection of one enantiomer

Fischer projection of the second enantiomer

15.16 A chiral compound is optically active and an achiral compound is optically inactive.

CH3

CCH3 OH

H

COOH

CCH3 OH

HCH3(CH2)6CH3a. b. c. d.

chiraloptically active

achiraloptically inactive



Chapter 15–5



15.17 Because two enantiomers rotate the plane of polarized light to an equal extent but in opposite directions, the rotation of one enantiomer is cancelled by the rotation of the other enantiomer and the sample is optically inactive.

15.18 Answer the questions about cholesterol as in Sample Problem 15.3.

a. The specific rotation of cholesterol is a negative value, so cholesterol is levorotatory. b. The specific rotation of the enantiomer has the same value, but is opposite in sign—that is,

+32. 15.19 Answer each question about alanine, as in Sample Problem 15.3.

(R)-alanine

COO–

CCH3H3N

H

a.

b. Two enantiomers have the same melting point, so the melting point of (R)-alanine is 297 oC.

c. The specific rotation of (R)-alanine is –8.5.

15.20 Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other.

Look for two compounds that when placed side-by-side have the groups on both chirality centers drawn as a reflection of each other. A given compound has only one possible enantiomer. Diastereomers are stereoisomers that are not mirror images of each other.

CH3

CH ClCCH2

H Br

CH3

CH ClCCH2

Br H

CH3

CCl HCCH2

Br H

CH3

CCl HCCH2

H Br

CH3 CH3 CH3CH3

W X Y Z

a. W and Z are enantiomers.

b. X and Y are enantiomers.

c. Z and W are diastereomers of Y.

d. X and Y are diastereomers of Z.

15.21 Convert each structure to a Fischer projection by replacing each chirality center with a cross.

CH3CH ClCCH2

H Br

CH3CH ClCCH2

Br H

CH3CCl HCCH2

Br H

CH3CCl HCCH2

H Br

CH3 CH3 CH3CH3

CH3

CH2CH3

ClHBrH

CH3

CH2CH3

ClHHBr

CH3

CH2CH3

HClBrH

CH3

CH2CH3

HClHBr

W X Y Z

Chapter 15–6



15.22 Identify the chirality center and draw the enantiomers.

CH3CH3H H

C CCH3 CH3

CH2 CH2

Solutions to End-of-Chapter Problems 15.23 Use the definitions in Answer 15.3 to label each object as chiral or achiral.

a. chalk—achiral c. baseball glove—chiral b. shoe—chiral d. soccer ball—achiral

15.24 Use the definitions in Answer 15.3 to label each object as chiral or achiral.

a. boot—chiral c. scissors—chiral b. index card—achiral d. drinking glass—achiral

15.25 All objects and molecules have a mirror image. Butane does not have an enantiomer since the

molecule and its mirror image are superimposable, so they are identical. 15.26 The human body is chiral because it is not superimposable on its mirror image. 15.27 Draw the mirror image for each compound.

CCl H

H

Cl

CF3

CBr H

Cla. b.

CF3

CBrH

Cl

nonsuperimposablechiral

C ClH

H

Cl

superimposableachiral

molecule mirror image molecule mirror image

15.28 Draw the mirror image for each compound.

HOCH2

CCH2OHH

HOa. b.

CH2OH

CHOCH2 H

OH


molecule mirror image

OCH3CH2 CH2CH3

OCH3CH2 CH2CH3


molecule mirror image

Chapter 15–7



15.29 A and B are mirror images of each other—enantiomers. A and C are superimposable and identical.

A B C

cannot rotate to alignenantiomer

identical

rotate

15.30 D and E are superimposable and identical. D and F are mirror images of each other—

enantiomers. 15.31 To locate the chirality centers, look at each C individually and eliminate those C’s that can’t be


two Cl's bonded to end Cno chirality center

CH3CH2CHCl

Cl

CH3CHCHCl2

Cl

CH3CHCH3

Cl

CH3CCH2CHCH3

CH3

H

Cla. b. c. d.

four different groups bonded to C


two CH3's bonded to middle C

no chirality center 15.32 To locate the chirality centers, look at each C individually and eliminate those C’s that can’t be


no chirality center

CH3CHO

CH3

CH3CHCHOCl

a. b. c. d.



two CH3's bonded to middle C

no chirality center

CO

CH2CH3CH3CO

CHCH2CH3CH3CH3

15.33 Label each chirality center.

a. b. c. d.OH CH3 CH3OH

none none 15.34 Label each chirality center.

a. b. c. d.CHCH3

none

CH3OH

O

none

O

CH3

Chapter 15–8



15.35 Draw a compound to fit each description.

CH3CH2 CCH3

HCH2CH2CH3a. b. CH3CH2 C

OHCH2CH2CH3

HC7H16

one chirality centerC6H14O

one chirality center 15.36 Draw a compound to fit each description.

CH3CH2 CBr

HCH3a. b. CH3 C

OHC

HC4H9Br

one chirality centerC5H12O2

two chirality centers

OH

HCH2CH3

15.37 A chirality center must have a carbon bonded to four groups and a carbonyl carbon has only three

groups. Therefore, a carbonyl carbon can never be a chirality center. 15.38 A chirality center must have a carbon bonded to four groups and a carbon atom that is part of a

triple bond has only two groups. Therefore, a carbon atom that is part of a triple bond can never be a chirality center.


N NCH3

HO2CCH2CHCNHCHCH2

NH2

O

CO2CH3a. b.


Na. b.

OCHCO2HCH3

CH

CO2CH3 15.41 Locate the chirality center and draw the enantiomers.

CH2CHNCH3

CH3

H CH3

CCH2 NHCH3

H

CH3

CCH2CH3NH

HC


mirrorenantiomers

Chapter 15–9



15.42 Locate the chirality center and draw the enantiomers.

CH2CH2NCHCH2CH2

CH3

CCH2CH2

H

CH3

CCH2CH2

H

C

Draw the bonds:

Add the groups:

mirrorenantiomers

HO

HOH

CH3

OH

OH

HO

HO CH2CH2NH NHCH2CH2

OH

OHHO


a. CH CH3

CH3

OHCCH3 H

CH3

OH

and

NH2

CCH3 H

CO2H

NH2

CCH3H

HO2Cb. and

The central C is bonded to two CH3 groups.There is no chirality center.

identical molecules

The central C is bonded to four different groups.The compounds are nonsuperimposable

mirror images.enantiomers

Br

CCH3 H

Cl

Br

CCH3H

Clandc.



Chapter 15–10




a.

CHOH2C OH

COOH

CH3

COH CH2OH

COOH

CH3

and

COOH

CCH3CH2 H

Br

COOH

CCH2CH3H

Br

OCH3

CCH3O CH3

CHO

OCH3

COCH3H3C

OHC

b.

and andc.





The central C is bonded to two OCH3 groups.There is no chirality center.

identical molecules


Diastereomers are stereoisomers that are not mirror images of each other.

There is no chirality center.identical molecules

mirror imagesenantiomers

CH3CH2CH2OCH3 CH3CH2OCH2CH3anda.

b.

CH3

OH

CH3

HOand

c.

d. ClCH3

H

H Cl

CH2CH3

Cl

CH3

H

Cl H

CH2CH3

and

CCH3CH2CH2 CH3

Br

BrCCH3 CH2CH2CH3

Br

Br

anddifferent connectivity

constitutional isomers

two chirality centersnot mirror imagesdiastereomers

Chapter 15–11



15.46 Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other. Diastereomers are stereoisomers that are not mirror images of each other.

two chirality centersmirror imagesenantiomers

a.

b.

c.

d.H OH

H OHCH2CH3

different connectivityconstitutional isomers

OH

CCH3CH2 H

CO2H

OH

CCH2CH3H

HO2Cand


CH3CH2OCH2CH3 and CO

CH2CH3H3C

C4H10O C4H8O

not isomers of each other

CH3CH2CH2CHOHCH3

and CH3CH2OCH(CH3)2

HO H

HO HCH2CH3

and

15.47 Answer each question with a compound of molecular formula C5H10O2.

CH3CH2CCOOHH

CH3

CH3CH2CH2CH2COOHCH3CH2CHCH2CH

OH Oa. b. c.

chirality center

no chirality centersame molecular formulaconstitutional isomer

no COOHsame molecular formulaconstitutional isomer

15.48 Answer each question with a compound of molecular formula C5H12O.

CH3CH2COCH3

H

CH3

CH3CH2CH2OCH2CH3a. b. c.

chirality center

no chirality centersame molecular formulaconstitutional isomer no ether

same molecular formulaconstitutional isomer

CH3CH2CHCH2OH

CH3

15.49 Draw the constitutional isomers and then label the chirality centers.

C CH3CH3

CHO

CH3

C CH2CH3CH3

CHO

H

CH3CH2CH2CH2CHO

chirality center

CH3CHCH2CHOCH3

15.50 Draw the enantiomers in three dimensions.

Cl

CH3C CH2CH3

Cl

CCH3H3CH2C

H H

Chapter 15–12



15.51 Convert the given ball-and-stick model to a Fischer projection by replacing the chirality center with a cross.

BrCHO

HCH3

chirality center

15.52 Convert the given ball-and-stick model to a Fischer projection by replacing the chirality center

with a cross.

H

CH3

ClBr

chirality center

CH

HC

H

Cl

BrH


cross.

CHOOH

CH3

H

OCH3HCH3

CH2CH3

OHCH3CH OH

CHO

CH3

CCH3 H

OCH3

Ca. b.

CH3 OHCH2CH3


cross.

OCH2CH3H

N(CH3)2H3C

COCH2CH3BrH

CH3

OHH

CH3C H

OCH2CH3

N(CH3)2

CH Br

COCH2CH3

C

a.b.

H OHCH3

15.55 Work backwards to draw three-dimensional representations from the Fischer projections.

CH F

CH3

CH(CH3)2

CH NH2

COOH

CCH3 HCH2CH3

a.CH3

H FCH(CH3)2

COOHH NH2

CH3 HCH2CH3

b.

Chapter 15–13



15.56 Work backwards to draw three-dimensional representations from the Fischer projections.

CHO CH3

CH2CH3CH Cl

CH3

CH ClCH2Cl

a.

CH2CH3HO CH3

CH3H ClH Cl

CH2Cl

b.


b.

a. c.CH3

Cl HCH3

CH3

H ClCH3

CO2HH NH2

CH2OH

CO2HH2N H

CH2OH

C(CH3)3

(CH3)3C OHH

C(CH3)3HO C(CH3)3

H

COOHCH3 OH

H

COOHHO CH3

H

d.

and

and

and

and

The central C is bonded to two CH3 groups.There is no chirality center.

identical molecules


The central C is bonded to two C(CH3)3 groups.There is no chirality center.

identical molecules



b.

a. c.COOH

H HNH2

COOHH H

NH2

CH2OHH3C OH

H

CH2OHHO CH3

H

CH2OCH3

CH2CH3 CH3

Cl

CH2OCH3

H3C CH2CH3

Cl

ClCH3CH2 CH3

Cl

ClH3C CH2CH3

Cld.

and

and

and

and

The central C is bonded to two hydrogens.There is no chirality center.

identical molecules

The central C is bonded to two chlorines.There is no chirality center.

identical molecules



15.59

a. An optically active compound rotates the plane of polarized light. An optically inactive compound does not rotate the plane of polarized light.

b. One possibility: CH3CH2CH2CH2OH is achiral and optically inactive. c. One possibility:

CH3

CCH3CH2 OH

H

chiral and optically active

Chapter 15–14



15.60

H

O

C5H10Ooptically inactive

a. H

O

C5H10Ooptically active

b.chirality center


OH c.CH3

CBr H

CH2CH3a. b. CH3CH2CH2CH2OH





a. (CH3)2CHCH2CHO


b. O


c. O

H3C


15.63

a. Lactose is dextrorotatory. b. The specific rotation of two enantiomers is the same numerical value, but opposite in sign.

Therefore, the specific rotation of the enantiomer of lactose is –52. c. Two enantiomers have identical physical properties except for their interaction with plane-

polarized light, and therefore, 22 g of the enantiomer of lactose dissolves in 100 mL of water. 15.64

a. The specific rotation of two enantiomers is the same numerical value, but opposite in sign. Therefore, the specific rotation of the enantiomer of taxol is +49.

b. Two enantiomers have identical physical properties except for their interaction with plane-polarized light, so the boiling points of taxol and its enatiomer are identical.

c. Two enantiomers have identical physical properties except for their interaction with plane-polarized light, so the solubility properties of taxol and its enatiomer are identical.

15.65 Enantiomers are stereoisomers that are nonsuperimposable mirror images of one another, whereas

diastereomers are stereoisomers that are not mirror images of one another.

Chapter 15–15



15.66 The two major types of stereoisomers are enantiomers and diasteromers. Enantiomers are stereoisomers that are nonsuperimposable mirror images of one another, whereas diastereomers are stereoisomers that are not mirror images of one another.

CHOH3C H

CHOH CH3

enantiomers

HO HCl

H OHCl

CHOH3C H

CHOH CH3

HO HCl

HO HCl

diastereomers 15.67 Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other.


CH OHC

CH2OH

CHOH OH

CH OHC

CH2OH

CHOHO H

C HHOC

CH2OH

CHOHHO

C HHOC

CH2OH

CHOOHH

A B C Denantiomersenantiomers

a.

The figure above shows the enantiomers. All other relationships are diastereomers.

[1] A and B, diastereomers; [2] A and C, diastereomers; [3] A and D, enantiomers; [4] B and C, enantiomers; [5] B and D, diastereomers; [6] C and D, diastereomers.

b.

CH OHC

CH2OH

CH2OHO



E F G Henantiomers

enantiomers

a.CH2OHC OC OHHC OHHCH2OH

CH2OHC OC HHOC OHHCH2OH

CH2OHC OC HHOC HHOCH2OH

CH2OHC OC OHHC HHOCH2OH

The figure above shows the enantiomers. All other relationships are diastereomers.

Chapter 15–16



[1] E and F, diastereomers; [2] E and G, enantiomers; [3] E and H, diastereomers; [4] F and G, diastereomers; [5] F and H, enantiomers; [6] G and H, diastereomers.

b.CHOCC OHHC OHHCH2OH

H OH

15.69 Enantiomers are stereoisomers that are mirror images. cis-2-Butene and trans-2-butene are

diastereomers because they are stereoisomers that are not mirror images of one another.

C CH

CH3 CH3

HC C

H

CH3 H

CH3cis trans

not mirror imagesdiastereomers

15.70 cis-2-Butene and trans-2-butene are achiral because four different groups are needed for a chiral

center. Since two of the carbons in each compound share a double bond, the molecules are achiral.

15.71 Answer each question about lactic acid.

HCOOH

OHCH3

CCH3 COOH

H

OH

CH3CHCO2HOH

C CH3HOOC

H

OH

HHOOC

OHCH3a.

b.

c.chirality center

enantiomers

Fischer projections

15.72 Locate the chirality centers in vitamin E and vitamin C.

a.O

CH3HO

H3CCH3

(CH2)3CH(CH2)3CH(CH2)3CH(CH3)2

CH3CH3 CH3

chirality centers

chirality center

O O

OHHO

HOCH2CHOH

chirality centersb.

Chapter 15–17



15.73 Answer each question about hydroxydihydrocitronellal.

chirality center

a. CH3CCH2CH2CH2CH CH2CHOCH3OH

CH3

enantiomers

CCH3CCH2CH2CH2 CH2CHO

CH3

H

HO

CH3

COHCCH2 CH2CH2CH2CCH3

CH3

H

OH

CH3

b.

Fischer projections

c.CH3

CH2CHO

H

CH3CCH2CH2CH2HO

CH3

OHCCH2 CH2CH2CH2CCH3

CH3

H

OH

CH3

15.74 Answer the questions for L-Leucine.

COOHH2N H

CH2CH(CH3)2

chirality center

a.

COOH

C(CH3)2CHCH2 NH2

H

COOH

CCH2CH(CH3)2H2N

Handb.

COOHH

CH2CH(CH3)2

H2Nc.COOH

NH2

CH2CH(CH3)2

H

Chapter 15–18



15.75 Answer each question about Darvon.

chirality center

enantiomer

Fischer projection

C O2CCH2CH3

CH2

CH CH3

CH2N(CH3)2

b.

a. CCH3CH2CO2

CH2

CCH3 HCH2N(CH3)2

CCH3CH2CO2

CH2

CH CH3

CH2N(CH3)2

c.

CCH3

CH2

CCH3CH2CO2 HCH2N(CH3)2

d.

CH2

CH2N(CH3)2

CH3CH2CO2

HCH3

e.

chirality center

diastereomer

constitutional isomer(Novrad is Darvon

spelled backwards.)

15.76 Answer the questions about Plavix.

chirality center

a.

CO2CH3

CN H

CO2CH3

CNHandb.

CO2CH3

HNc.

CO2CH3

NH

NS C

CO2CH3

H Cl Cl

S S

Cl

S

Cl

S

Cl

Chapter 15–19



15.77 Each chirality center in sucrose is labeled with a star.

OO

O

HOCH2

HOCH2OH

OH

HO

OH

OHHOCH2***

* ** ** *

* = chirality center

15.78 Each chirality center in nandrolone is labeled with a star.

* = chirality center

O

CH3OH

****

**

chapter 15 the three-dimensional shape of...

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