Chapter 15–1
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.
CH3
a.
CH3CH2CHCHCH3
CH3CH2
and
b.
CH3CHCH2CHCH3
CH3
CH3
H
and
CH3CH CHCH2CH3
H
C C
H
CH3
H
cis
different connection of atoms
constitutional isomers
c.
and
C C
CH3
CH3CH2
CH3
trans
stereoisomers
d.
CH3CH2CH CHCH3
identical
O
OH
and
different connection of atoms
constitutional isomers
15.2 Draw the isomers of trans-2-hexene.
CH3
CH3
H
C
C
H
C
a.
CH2CH2CH3
trans-2-hexene
H
CH2CH2CH3
C
b. CH2 CHCH2CH2CH2CH3
c.
H
cis-2-hexene
stereoisomer
different connection of atoms
constitutional isomer
different connection of atoms
constitutional 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
b. screw—chiral
c. glove—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.
H
H
a. CH3 C CH2CH3
Cl
b. (CH3)3CH
none
c. CH2 CH C CH3
H
d.
C CH3
Br
OH
15.6 Label each chirality center, as in Example 15.1.
H
a.
Br
C
N
CH2CH2N(CH3)2
H
b.
C
O
CF3
CH2CH2NHCH3
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Chapter 15–2
15.7 Label each chirality center, as in Example 15.1.
H
a.
CH3CH2CH2
COOH
C
CH3
b.
OH
H2N
C
H
CH3
C
H
Br
Cl
c. CH3CH2 C CH2CH2
C
H
CH3
H
CH2CH3
15.8 Label the two chirality centers in vitamin K.
O
CH3
CH2CH=CCH2CH2CH2CHCH2CH2CH2CHCH2CH2CH2CH(CH3)2
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.
Draw the bonds:
Add the groups:
CH3
a. CH3CHCH2CH3
C
C
H
CH2CH3
Cl
Cl
CH3
H
CH3CH2
C
Cl
mirror
enantiomers
Draw the bonds:
b. CH3CH2CHCH2OH
Add the groups:
OH
C
C
OH
H
HO
CH2CH3
CH3CH2
CH2OH
HOCH2
C
H
mirror
enantiomers
15.10 Locate the chirality centers in each compound.
a.
OH
b.
Cl
CH2CH3
none
All C's are part of multiple
bonds or are bonded to
two or three H's.
c.
Cl
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Chapter 15–3
15.11 Label the two chirality centers.
CH3NH
Cl
Cl
15.12 Answer each question about propranolol.
chirality center
OCH2CHCH2NHCH(CH3)2
OH
a.
H
H
C
b.
O
CH2
(CH3)2CHNHCH2
HO
CH2NHCH(CH3)2
OH
C
CH2
O
enantiomers
receptor
receptor
H
C
O
c.
CH2
H
CH2NHCH(CH3)2
OH
(CH3)2CHNHCH2
HO
CH2
O
This enantiomer does
not fit the
receptor.
This enantiomer fits the
receptor.
15.13
C
Convert each molecule to a Fisher projection formula by replacing the chirality center with a
cross.
CH3
a.
H
C
COOH
CH3
Br
H
b. H
Br
Cl
Cl
C
COOH
H
OH
OH
CH2Cl
CH2Cl
15.14 Work backwards to draw a structure with wedges (horizontal bonds) and dashes (vertical bonds).
a. CH3CH2
Br
Cl
CH3CH2
C
Cl
H
H
CH3
CH3
Br
b.
CH3CH2
CH3
NH2
CH3CH2
C
CH3
NH2
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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.
chirality center
a.
Draw the compound.
Then, draw the mirror image.
H
H
CH3CHCH2CH2CH3
OH
CH3
C
CH2CH2CH3
CH3CH2CH2
OH
C
CH3
OH
• Horizontal bonds come
forward.
• Vertical bonds go back.
H
H
CH3
CH2CH2CH3
CH3CH2CH2
OH
Fischer projection of one
enantiomer
chirality center
b.
CH3
OH
Fischer projection of the
second enantiomer
Draw the compound.
Then, draw the mirror image.
H
H
CH3CH2CHCH2Cl
Cl
ClCH2
C
CH2CH3
CH3CH2
Cl
C
CH2Cl
Cl
• Horizontal bonds come
forward.
• Vertical bonds go back.
H
ClCH2
H
CH2CH3
CH3CH2
Cl
Fischer projection of one
enantiomer
15.16
CH2Cl
Cl
Fischer projection of the
second enantiomer
A chiral compound is optically active and an achiral compound is optically inactive.
COOH
a.
CH3(CH2)6CH3
achiral
optically inactive
b.
C
CH3
H
OH
chiral
optically active
CH3
c.
C
CH3
H
OH
achiral
optically inactive
d.
achiral
optically inactive
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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.
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.
COO–
a.
C
H
H3N
CH3
(R)-alanine
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
CH3
CH3
CH3
H C Cl
H C Cl
Cl C H
Cl C H
H C Br
Br C H
H C Br
Br C H
CH2
CH2
CH2
CH2
CH3
CH3
CH3
CH3
W
X
Y
c. Z and W are diastereomers of Y.
d. X and Y are diastereomers of Z.
Z
b. X and Y are enantiomers.
a. W and Z are enantiomers.
15.21 Convert each structure to a Fischer projection by replacing each chirality center with a cross.
CH3
CH3
H
C
Cl
H
H
C
Br
Br
CH3
CH3
C
Cl
Cl
C
H
Cl
C
H
C
H
H
C
Br
Br
C
H
CH2
CH2
CH2
CH2
CH3
CH3
CH3
CH3
CH3
CH3
CH3
CH3
H
Cl
H
Cl
Cl
H
Cl
H
H
Br
Br
H
H
Br
Br
H
CH2CH3
W
CH2CH3
X
CH2CH3
Y
CH2CH3
Z
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Chapter 15–6
15.22 Identify the chirality center and draw the enantiomers.
H
CH3
CH3
H
C CH2
CH2 C
CH3
CH3
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
b. shoe—chiral
15.24
c. baseball glove—chiral
d. soccer ball—achiral
Use the definitions in Answer 15.3 to label each object as chiral or achiral.
a. boot—chiral
b. index card—achiral
c. scissors—chiral
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.
CF3
a.
C
Br
Cl
H
CF3
Cl
H
C
b.
Cl
C
Br
Cl
mirror image
molecule
nonsuperimposable
chiral
15.28
H
H
H
H
C
Cl
Cl
mirror image
molecule
superimposable
achiral
Draw the mirror image for each compound.
CH2OH
a. CH3CH2
O
CH2CH3
CH3CH2
O
CH2CH3
molecule
mirror image
superimposable
achiral
b.
C
HOCH2
OH
H
HOCH2
HO
H
molecule
C
CH2OH
mirror image
superimposable
achiral
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Chapter 15–7
15.29
A and B are mirror images of each other—enantiomers. A and C are superimposable and
identical.
rotate
C
B
A
cannot rotate to align
enantiomer
identical
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
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.
CH3 Cl
a.
b.
CH3CH2CHCl
CH3CHCHCl2
Cl
CH3CHCH3
four different
groups bonded to C
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.
CH3
C
O
CH2CH3
b.
CH3
C
CH3CHO
c.
CHCH2CH3
CH3
d.
CH3CHCHO
Cl
CH3
four different
groups bonded to C
no chirality center
four different
groups bonded to C
two CH3's bonded
to middle C
no chirality center
Label each chirality center.
a.
b.
OH
OH
CH3
c.
CH3
d.
none
none
15.34
CH3CCH2CHCH3
H
two CH3's bonded
to middle C
no chirality center
four different
groups bonded to C
O
15.33
d.
Cl
Cl
two Cl's bonded to end C
no chirality center
15.32
c.
Label each chirality center.
CH3
a.
CH3
none
b.
CHCH3
OH
c.
O
d.
O
none
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Chapter 15–8
15.35
Draw a compound to fit each description.
OH
CH3
a.
b.
CH3CH2 C CH2CH2CH3
CH3CH2 C CH2CH2CH3
H
H
C7H16
one chirality center
15.36
C6H14O
one chirality center
Draw a compound to fit each description.
OH OH
Br
a.
b.
CH3CH2 C CH3
CH3 C C CH2CH3
H H
H
C4H9Br
one chirality center
C5H12O2
two chirality centers
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.
15.39
Locate the chirality centers in each compound.
NH2
a.
b.
N
N
CO2CH3
HO2CCH2CHCNHCHCH2
O
CH3
15.40
Locate the chirality centers in each compound.
O
CH3
CHCO2H
a.
N
b.
CH
CO2CH3
15.41
Locate the chirality center and draw the enantiomers.
Draw the bonds:
Add the groups:
CH3
H
CH2CHNCH3
CH3
C
C
CH2
CH3
H
NHCH3
H C
CH2
CH3NH
mirror
enantiomers
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Chapter 15–9
15.42
Locate the chirality center and draw the enantiomers.
Draw the bonds:
HO
H
HO
CH2CH2NCHCH2CH2
C
OH
CH3
Add the groups:
HO
CH3
HO
CH2CH2NH
C
OH
CH3
H
CH2CH2
OH
H C
NHCH2CH2
CH2CH2
HO
OH
mirror
enantiomers
15.43
Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other.
NH2
CH3
a.
H
C
CH3
CH3
and
OH
CH3
C
b.
H
OH
The central C is bonded to two CH3 groups.
There is no chirality center.
identical molecules
Br
c.
C
CH3
C
CH3
CO2H
H
NH2
and
HO2C
H
C
CH3
The central C is bonded to four different groups.
The compounds are nonsuperimposable
mirror images.
enantiomers
Br
Cl
H
and
Cl
H
C
CH3
The central C is bonded to four different groups.
The compounds are nonsuperimposable
mirror images.
enantiomers
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Chapter 15–10
15.44
Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other.
COOH
a.
C
Br
H
CH3CH2
OCH3
COOH
and
c.
C
Br
H
The central C is bonded to four different groups.
The compounds are nonsuperimposable
mirror images.
enantiomers
COOH
b. HOH2C
C
C
CHO
CH3
CH3O
CH2CH3
OCH3
and
OHC
H3C
C
OCH3
The central C is bonded to two OCH3 groups.
There is no chirality center.
identical molecules
COOH
and
OH
OH
CH3
C
CH2OH
CH3
The central C is bonded to four different groups.
The compounds are nonsuperimposable
mirror images.
enantiomers
15.45
Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other.
Diastereomers are stereoisomers that are not mirror images of each other.
Br
a. CH3CH2CH2OCH3
and
CH3CH2OCH2CH3
different connectivity
constitutional isomers
CH3
b.
c. CH3CH2CH2
and
mirror images
enantiomers
HO
Br
CH3
and
CH3
Br
C
CH2CH2CH3
Br
There is no chirality center.
identical molecules
CH3
OH
C
CH3
CH3
d.
Cl
H
H
Cl
CH2CH3
and
Cl
H
Cl
H
CH2CH3
two chirality centers
not mirror images
diastereomers
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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.
a.
OH
OH
C
C
CH3CH2
CO2H and HO2C
H
H
O
c.
CH2CH3
CH3CH2OCH2CH3 and
C4H10O
mirror images
enantiomers
b. CH3CH2CH2CHOH
and
H3C
C
CH2CH3
C4H8O
not isomers of each other
CH3CH2OCH(CH3)2
d.
and
CH3
H
OH
HO
H
different connectivity
constitutional isomers
H
OH
HO
H
CH2CH3
CH2CH3
two chirality centers
mirror images
enantiomers
15.47
Answer each question with a compound of molecular formula C5H10O2.
H
a.
CH3CH2CCOOH
b.
c.
no chirality center
same molecular formula
constitutional isomer
CH3
chirality center
15.48
CH3CH2CH2CH2COOH
OH
O
CH3CH2CHCH2CH
no COOH
same molecular formula
constitutional isomer
Answer each question with a compound of molecular formula C5H12O.
H
a.
CH3CH2COCH3
b.
no chirality center
same molecular formula
constitutional isomer
CH3
chirality center
15.49
CH3CH2CH2OCH2CH3
c.
CH3CH2CHCH2OH
CH3
no ether
same molecular formula
constitutional isomer
Draw the constitutional isomers and then label the chirality centers.
chirality center
CHO
CH3
C
CHO
CH3
CH3
15.50
CH3
C
CH2CH3
CH3CH2CH2CH2CHO
CH3CHCH2CHO
CH3
H
Draw the enantiomers in three dimensions.
Cl
C
H3C
Cl
H
CH2CH3
H C
H3CH2C
CH3
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Chapter 15–12
15.51
Convert the given ball-and-stick model to a Fischer projection by replacing the chirality center
with a cross.
chirality center
CHO
Br
H
CH3
15.52
Convert the given ball-and-stick model to a Fischer projection by replacing the chirality center
with a cross.
chirality center
H
Cl
CH3
C C Br
H H H
15.53
H
Convert each molecule to a Fisher projection formula by replacing the chirality center with a
cross.
OCH3
CHO
a.
H
C
CHO
OH
H
b.
OH
CH3
CH3
CH3
CH3
15.54
Cl
Br
C
C
OCH3
H
CH3
H
OH
CH3
OH
CH2CH3
CH2CH3
Convert each molecule to a Fisher projection formula by replacing the chirality center with a
cross.
OCH2CH3
a. H3C
H
C
COCH2CH3
OCH2CH3
H
H3C
N(CH3)2
N(CH3)2
b.
COCH2CH3
H
C
Br
H
H
C OH
H
OH
CH3
CH3
15.55
Br
Work backwards to draw three-dimensional representations from the Fischer projections.
a.
H
F
CH(CH3)2
H
C
COOH
COOH
CH3
CH3
F
CH(CH3)2
b.
H
CH3
NH2
H
CH2CH3
H
C
NH2
CH3
C
H
CH2CH3
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Chapter 15–13
15.56
Work backwards to draw three-dimensional representations from the Fischer projections.
CH2CH3
CH2CH3
a. HO
CH3
HO
C
CH3
CH3
CH3
H
b.
H
Cl
H
C
Cl
Cl
H
C
Cl
CH2Cl
15.57
CH2Cl
Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other.
CH3
a.
Cl
CH3
H
H
and
CO2H
c.
Cl
CH3
H
CH3
CO2H
NH2
and
COOH
CH3
mirror images
enantiomers
C(CH3)3
COOH
OH
and
HO
H
CH3
d.
(CH3)3C
C(CH3)3
and
OH
HO
H
H
mirror images
enantiomers
15.58
H
CH2OH
The central C is bonded to two CH3 groups.
There is no chirality center.
identical molecules
b.
H2N
CH2OH
C(CH3)3
H
The central C is bonded to two C(CH3)3 groups.
There is no chirality center.
identical molecules
Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other.
COOH
a.
H
COOH
H
and
H
CH2OH
H
NH2
c.
NH2
H3C
OH
Cl
CH3CH2
H3C
Cl
HO
CH2OCH3
CH2CH3
Cl
The central C is bonded to two chlorines.
There is no chirality center.
identical molecules
CH3
H
mirror images
enantiomers
Cl
and
CH3
and
H
The central C is bonded to two hydrogens.
There is no chirality center.
identical molecules
b.
CH2OH
d.
CH2CH3
CH3
CH2OCH3
and
H3C
Cl
CH2CH3
Cl
mirror images
enantiomers
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
C
CH3CH2
H
OH
chiral and optically active
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Chapter 15–14
15.60
O
O
a.
b.
H
chirality center
C5H10O
optically inactive
15.61
H
C5H10O
optically active
A chiral compound is optically active and an achiral compound is optically inactive.
CH3
a.
C
Br
CH2CH3
H
chiral
optically active
15.62
b.
c.
OH
achiral
optically inactive
CH3CH2CH2CH2OH
achiral
optically inactive
A chiral compound is optically active and an achiral compound is optically inactive.
a. (CH3)2CHCH2CHO
b.
O
c.
O
H3C
achiral
optically inactive
achiral
optically inactive
chiral
optically active
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 planepolarized 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 planepolarized light, so the boiling points of taxol and its enatiomer are identical.
c. Two enantiomers have identical physical properties except for their interaction with planepolarized 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.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
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.
CHO
CHO
CHO
H3C
H
H
CH3
H3C
H
H
HO
H
H
OH
HO
H
HO
Cl
Cl
Cl
enantiomers
15.67
CHO
CH3
H
Cl
diastereomers
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.
CHO
a.
CHO
H
C
OH
HO
C
H
H
C
OH
H
C
OH
CH2OH
H
C
OH
HO
HO
C
H
HO
CH2OH
A
B
CHO
CHO
enantiomers
C
H
C
H
CH2OH
CH2OH
C
D
enantiomers
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.
CH2OH
b.
C O
H
C
OH
CH2OH
15.68
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.
CH2OH
a.
CH2OH
C O
H C OH
C O
C O
HO C H
HO C H
H C OH
H C OH
CH2OH
CH2OH
E
CH2OH
HO C H
F
CH2OH
C O
H C OH
HO C H
CH2OH
CH2OH
G
H
enantiomers
enantiomers
The figure above shows the enantiomers. All other relationships are diastereomers.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
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.
CHO
b.
H C OH
H C OH
H C OH
CH2OH
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.
CH3
C
H
CH3
CH3
cis
H
C
C
C
H
H
trans
not mirror images
diastereomers
CH3
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.
H
a.
c.
CH3CHCO2H
chirality center
H
CH3
COOH
OH
OH
HOOC
CH3
OH
Fischer projections
H
b. CH3
H
C
COOH
OH
15.72
HOOC
C
CH3
OH
enantiomers
Locate the chirality centers in vitamin E and vitamin C.
CH3
chirality centers
HO
(CH2)3CH(CH2)3CH(CH2)3CH(CH3)2
a.
H3C
O
CH3
CH3
CH3
CH3
chirality center
OH
b.
HOCH2CH
chirality centers
HO
O
O
OH
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 15–17
15.73
Answer each question about hydroxydihydrocitronellal.
OH
a.
CH3
CH3CCH2CH2CH2CH
CH3
chirality center
CH3
HO
b.
c.
CH3CCH2CH2CH2
C
H
HO
CH3
CH3CCH2CH2CH2
CH3
CH2CHO
CH3
CH3
15.74
CH2CHO
enantiomers
CH2CHO
H
OHCCH2
C
CH2CH2CH2CCH3
H
CH3
CH3
OH
OHCCH2
Fischer projections
OH
CH2CH2CH2CCH3
CH3
H
Answer the questions for L-Leucine.
chirality center
COOH
COOH
COOH
a.
H2N
b.
H
CH2CH(CH3)2
COOH
c.
H2N
H
CH2CH(CH3)2
C
(CH3)2CHCH2
H
NH2
and
H
H2N
C
CH2CH(CH3)2
COOH
H
NH2
CH2CH(CH3)2
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 15–18
15.75
Answer each question about Darvon.
chirality center
CH2
CH2
a.
CH3CH2CO2
C
CH3
C
chirality center
c.
CH3CH2CO2
C
H
C
H
CH3
CH2N(CH3)2
CH2N(CH3)2
diastereomer
CH2
b.
H
C
O2CCH2CH3
C
CH3
CH2
d.
CH2N(CH3)2
CH3
C
CH3CH2CO2
C
H
CH2N(CH3)2
enantiomer
constitutional isomer
(Novrad is Darvon
spelled backwards.)
e.
CH2
CH3CH2CO2
CH3
H
CH2N(CH3)2
Fischer projection
15.76
Answer the questions about Plavix.
chirality center
a.
S
b.
N C
H
S
C
S
N
Cl
H
and
H
C
N
S
Cl
Cl
CO2CH3
c.
CO2CH3
CO2CH3
CO2CH3
CO2CH3
N
N
H
S
H
Cl
Cl
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 15–19
15.77
Each chirality center in sucrose is labeled with a star.
OH
HOCH2
O
HO
*
* *
**
HO
15.78
HOCH2
O
*
* **
O
OH
OH
*
= chirality center
CH2OH
Each chirality center in nandrolone is labeled with a star.
CH3 OH
**
* *
**
*
= chirality center
O
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.