Chapter 13–1
Chapter 13 Unsaturated Hydrocarbons
Solutions to In-Chapter Problems
13.1
To draw a complete structure for each condensed structure, first draw in the multiple bonds. Then
draw in all the other C’s and H’s, as in Example 13.1.
H
H H
a.
CH2 CHCH2OH
=
H
CH(CH2)2CH3 =
b. (CH3)2C
H C C C O H
H C
H H H H
C C C C C H
H
H
H H H
H C H
H
H
H
c. (CH3)2CHC
CCH2C(CH3)3 =
H
H
H C
C C C C
H
H
H C H
H C H
C
H C H
H
To determine whether each molecular formula corresponds to a saturated hydrocarbon, alkene, or
alkyne, recall that the formula for a saturated hydrocarbon is CnH2n + 2, the formula for an alkene
is CnH2n, and the formula for an alkyne is CnH2n – 2.
a. C3H6 = CnH2n = alkene
b. C5H12 = CnH2n + 2 = saturated hydrocarbon
13.3
C H
H
H
13.2
H
c. C8H14 = CnH2n – 2 = alkyne
d. C6H12 = CnH2n = alkene
Use the general formulas [saturated hydrocarbon (CnH2n + 2), alkene (CnH2n), and alkyne
(CnH2n – 2)] to determine the molecular formula for each compound.
a. alkene = CnH2n, 4 × 2 = 8, C4H8
b. saturated hydrocarbon = CnH2n + 2, (6 × 2) + 2 = 14, C6H14
c. alkyne = CnH2n – 2, (7 × 2) – 2 = 12, C7H12
d. alkene = CnH2n, 5 × 2 = 10, C5H10
13.4
Give the IUPAC name for each compound using the following steps, as in Example 13.2:
[1] Find the longest chain containing both carbon atoms of the multiple bond.
[2] Number the chain to give the double bond the lower number.
[3] Name and number the substituents and write the complete name.
1
a.
CH2 CHCHCH2CH3
2 3
CH3
5 C's in the longest chain
pentene
4
5
CH2 CHCHCH2CH3
Answer: 3-methyl-1-pentene
CH3
double bond at C1
methyl at C3
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Chapter 13–2
b.
(CH3CH2)2C
CHCH2CH2CH3
ethyl at C3
CH2CH3
CH3CH2C CHCH2CH2CH3
3
2
CH2CH3
CH3CH2C CHCH2CH2CH3
1
5
4
7 C's in the longest chain
heptene
double bond at C3
7
c.
CH3CH2CH CHCH=CHCH3
4
3
2
2
5
CH2CH3
Answer: 3-ethylcyclopentene
CH2CH3
4
5 C's in the ring
cyclopentene
ethyl at C3
Give the IUPAC name for each compound using the following steps, as in Example 13.2:
[1] Find the longest chain containing both carbon atoms of the multiple bond.
[2] Number the chain to give the multiple bond the lower number.
[3] Name and number the substituents and write the complete name.
CH3CH2CH2CH2CH2C CCH(CH3)2
CH3
CH3CH2CH2CH2CH2C CCHCH3
CH3
CH3CH2CH2CH2CH2C CCHCH3
5
9 C's in the longest chain
nonyne
CH2CH3
CH3CH2 C C CH2 C CH3
CH3
8 C's in the longest chain
octyne
13.6
Answer: 2,4-heptadiene
3
d.
b.
1
double bonds at C2 and C4
1
a.
5
6
CH3CH2CH CHCH=CHCH3
7 C's in the longest chain
heptadiene
13.5
Answer: 3-ethyl-3-heptene
7
6
4
32
Answer: 2-methyl-3-nonyne
1
triple bond at C3
6
1
2
3
4
5
7
8
CH2CH3
CH3CH2 C C CH2 C CH3
Answer: 6,6-dimethyl-3-octyne
CH3
triple bond at C3
2 methyl groups at C6
To draw the structure corresponding to each name, follow the steps in Example 13.3.
• Identify the parent name to find the longest carbon chain or ring, and then use the suffix to
determine the functional group; the suffix -ene = alkene and -yne = alkyne.
• Number the carbon chain or ring and place the functional group at the indicated carbon. Add
the substituents and enough hydrogens to give each carbon four bonds.
1 2
a. 4-methyl-1-hexene
6 carbon chain
double bond at C1
3
4
5
6
C C C C C C
CH3
CH2 CHCH2CHCH2CH3
CH3
methyl at C4
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Chapter 13–3
1 2
b. 5-ethyl-2-methyl-2-heptene
3
4
5
6
7
CH3C CHCH2CHCH2CH3
C C C C C C C
CH3
7 carbon chain
double bond at C2
methyl at C2
1
c. 2,5-dimethyl-3-hexyne
CH3
CH2CH3
ethyl at C5
2
3
4
5
6
CH3
C C C C C C
6 carbon chain
triple bond at C3
CH2CH3
CH3
CH3CHC
CH3
methyl at C2
CCHCH3
CH3
methyl at C5
CH2CH2CH3
1 CH2CH2CH3
4C C
!d. 1-propylcyclobutene
3
4 carbon ring
double bond at C1
C C
propyl at C1
2
1
e. 1,3-cyclohexadiene
6 carbon ring
2 double bonds (C1 and C3)
3
5
4
10
f. 4-ethyl-1-decyne
2
6
9
8
7
6
5
4
2
1
CH3CH2CH2CH2CH2CH2CHCH2C
CH2CH3
10 carbon chain
triple bond at C1
13.7
3
C C C C C C C C C C
CH
CH2CH3
ethyl at C4
To draw the structures of the cis and trans isomers, follow the steps in Example 13.4.
• Use the parent name to draw the carbon skeleton and place the double bond at the correct
carbon.
• Use the definitions of cis and trans to draw the isomers. When the two alkyl groups are on the
same side of the double bond, the compound is called the cis isomer. When they are on
opposite sides, it is called the trans isomer.
H
a. cis-2-octene
8 carbon chain
CH3CH CHCH2CH2CH2CH2CH3
1
2
3 4
5
6
7
8
H
C
CH3
C
CH2CH2CH2CH2CH3
cis isomer
both alkyl groups on the same side
H
b. trans-3-heptene
7 carbon chain
CH3CH2CH CHCH2CH2CH3
1
2
3
4
5
6
7
CH2CH2CH3
C
CH3CH2
C
H
trans isomer
alkyl groups on opposite sides
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Chapter 13–4
CH3
H
c. trans-4-methyl-2-pentene
C
CH3CH CHCH(CH3)CH3
1
5 carbon chain
2
3
4
CHCH3
C
CH3
5
H
trans isomer
alkyl groups on opposite sides
13.8
Whenever the two groups on each end of a C=C are different from each other, two isomers are
possible.
CH3
a. CH3CH2CH CHCH3
c.
Each C has one H and one alkyl group.
Cis and trans isomers are possible.
CH3 C CHCH2CH2CH3
two CH3's
cannot have cis or trans isomers
b. CH2 CHCH2CH2CH3
two H's
cannot have cis or trans isomers
13.9
When the two alkyl groups are on the same side of the double bond, the compound is called the
cis isomer. When they are on opposite sides, it is called the trans isomer.
cis
H
H
H
C
C
C
C
HOCH2(CH2)7CH2
CH2CH2CH3
H
trans
13.10
a.
Stereoisomers differ only in the three-dimensional arrangement of atoms.
Constitutional isomers differ in the way the atoms are bonded to each other.
CH3CH CHCH2CH3
and
C bonded to one H, one CH3
CH2 CHCH2CH2CH3
C bonded to two H's
different connectivity
constitutional isomers
b.
CH3CH2
C C
H
CH3CH2
CH3
C C
and
H
H
C C
CH3
H
CH3
H
CH3CH2
c.
CH3CH2
and
H
C bonded to one CH2CH3
and one CH3
CH3
C C
H
H
C bonded to one H
and one CH2CH3
different connectivity
constitutional isomers
cis isomer
trans isomer
same connectivity
different 3-D arrangement
stereoisomers
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Chapter 13–5
13.11
Double bonds in naturally occurring fatty acids are cis.
H
H
CH3CH2CH2CH2CH2
C C
CH2
H
H H
H H
H
C C
C C
C C
CH2
O
CH2CH2CH2COH
CH2
arachidonic acid
13.12
13.13
Fats are solids at room temperature because of their higher melting point. They are formed from
fatty acids with few double bonds. Oils are liquids at room temperature because of their lower
melting points. They are also formed from fatty acids, but have more double bonds.
CH3(CH2)14COOH
CH3(CH2)5CH=CH(CH2)7COOH
palmitic acid
no double bonds
higher melting point
63 °C
palmitoleic acid
one double bond
lower melting point
1 °C
The functional groups in tamoxifen are labeled.
ether
amine
OCH2CH2N(CH3)2
aromatic ring
aromatic ring
C C
CH3CH2
aromatic ring
alkene
13.14
The functional groups in RU 486 and levonorgestrel are labeled.
amine
aromatic ring
hydroxyl
hydroxyl
alkyne
(CH3)2N
CH3 OH
C
alkene
C
alkyne
CH3CH2 OH
C CH
CH3
ketone
ketone
O
RU 486
O
alkene
alkene
levonorgestrel
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Chapter 13–6
13.15
To draw the product of a hydrogenation reaction, use the following steps, as in Example 13.5:
• Locate the C=C and mentally break one bond in the double bond.
• Mentally break the H–H bond of the reagent.
• Add one H atom to each C of the C=C, thereby forming two new C–H single bonds.
a.
CH2CH(CH3)2
CH3
b.
C
Pd
H2
C
CH3
Pd
H
CH3
CH3CH2CH2CH2CH2CH3
CH3CHCH2CH2CH(CH3)2
CH3
CH3
H2
c.
13.16
H2
CH3CH2CH CHCH2CH3
Pd
To draw the product of each halogenation reaction, add a halogen to both carbons of the double
bond.
Cl Cl
a.
CH3CH2CH CH2
+
Cl2
CH3CH2C C H
H H
Br
CH3
CH3
b.
+
Br2
CH3
13.17
CH3
Br
In hydrohalogenation reactions, the elements of H and Br (or H and Cl) must be added to the
double bond. When the alkene is unsymmetrical, the H atom of HX bonds to the carbon that has
more H’s to begin with.
H H
a.
CH3CH CHCH3
+
HBr
CH3C CCH3
Br H
This C does not have any H's.
Add Br here.
Br
CH3
CH3
+
b.
HBr
This C has more H's.
Add H here.
H
H
This C does not have any H's.
Add Cl here.
CH3 H
c.
(CH3)2C
CHCH3
+
HCl
CH3 C
Cl
CCH3
H
This C has more H's.
Add H here.
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Chapter 13–7
d.
+
H
Cl
HCl
H
H
13.18
In hydration reactions, the elements of H and OH are added to the double bond. In unsymmetrical
alkenes, the H atom bonds to the less substituted carbon.
a.
HO H
H OH
CH3CH CHCH3
CH3C CCH3
H2SO4
H H
This C has one H so the OH bonds here.
b.
CH3CH2CH CH2
HO H
H OH
CH3CH2C C H
H2SO4
H H
This C has 2 H's so the H bonds here.
CH3
c.
CH3
13.19
OH
CH3
H OH
H2SO4
CH3
H
Draw the products of each reaction.
a.
b.
H2
CH3CH2CH2CH CH2
Pd
Cl2
CH3CH2CH2CH CH2
H H
CH3CH2CH2C C H
H H
H H
CH3CH2CH2C C H
Cl Cl
c.
Br2
CH3CH2CH2CH CH2
Br Br
CH3CH2CH2C C H
H H
H H
d.
CH3CH2CH2CH CH2
H
Br
CH3CH2CH2C C H
Br H
e.
CH3CH2CH2CH CH2
H Cl
H H
CH3CH2CH2C C H
Cl H
f.
CH3CH2CH2CH CH2
H OH
H2SO4
HO H
CH3CH2CH2C C H
H H
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Chapter 13–8
13.20
Draw the products of the hydrogenation reactions.
H
H
a.
H H
H
C C
CH3CH2
C C
CH2
H
C C
CH2
CH2CH2CH2CH2CH2CH2CH2COOH
H2
Pd
C C
C C
CH2CH2CH2 CH2
CH3CH2
H
H
C C
CH3CH2
13.21
CH2CH2CH2CH2CH2CH2CH2COOH
H
H
C C
b.
H
H
H
H
CH2
CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOH
CH3CH2CH2CH2CH2CH2CH2 CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOH
To draw the polymers, draw three or more alkene molecules and arrange the carbons of the
double bonds next to each other. Break one bond of each double bond, and join the alkenes
together with single bonds. With unsymmetrical alkenes, substituents are bonded to every other
carbon. Use Example 13.7 as a guide.
Join these 2 C's.
CH3
a.
CH2
Join these 2 C's.
CH3
C
CH2
CH3
C
CH2CH3
CH2
H CH3 H CH3 H CH3
C
CH2CH3
C C
CH2CH3
C C
H CH2 H CH2 H CH2
CH3
CH3
Join these 2 C's.
CH3
b.
CH2
C
C
CN
CH2
C
CN
Cl
C
C C
C C
C C
H CN
H CN
H CN
Join these 2 C's.
H
CH2
H CH3 H CH3 H CH3
C
CN
H
c.
CH3
CH2
Join these 2 C's.
CH3
Join these 2 C's.
CH3
CH2
C C
H
CH2
Cl
C
H H
H H
H
H
C C
C C
C
C
H
H
H
Cl
Cl
Cl
Cl
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Chapter 13–9
13.22
Work backwards to determine what monomer is used to form the polymer.
Break these bonds to form the monomer.
H
H
H
CH2C
CH2C
CH2C
O
O
O
COCH3
COCH3
COCH3
H
formed from
CH2 C
O
COCH3
poly(vinyl acetate)
13.23
Name each aromatic compound as in Example 13.8. Name the substituents on the benzene ring.
With two groups, alphabetize the substituent names and use the prefix ortho, meta, or para to
indicate their location. With three substituents, alphabetize the substituent names, and number to
give the lowest set of numbers.
OH
CH2CH2CH3
OH on benzene ring = phenol
propyl
a.
m-butylphenol
c.
propylbenzene
CH3
CH2CH3
butyl
CH2CH2CH2CH3
ethyl
Br
b.
CH3 on benzene ring = toluene
bromo
d.
I
Cl
2-bromo-5-chlorotoluene
p-ethyliodobenzene
iodo
13.24
chloro
Draw the structure corresponding to each name.
NH2
a. pentylbenzene
!!!c. m-bromoaniline
CH2CH2CH2CH2CH3
Br
Cl
b. o-dichlorobenzene
CH2CH3
Cl
CH2CH3
!!!d. 4-chloro-1,2-diethylbenzene
Cl
13.25
Commercially available sunscreens contain a benzene ring. Therefore, compound (a) might be
found in a sunscreen since it contains two aromatic rings. Compound (b) does not contain any
aromatic rings.
13.26
Phenols are antioxidants because the OH group on the benzene ring prevents unwanted oxidation
reactions from occurring. Of the compounds listed, only curcumin (b) contains a phenol group
(OH on a benzene ring), making it an antioxidant.
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Chapter 13–10
13.26
Draw the products of each substitution reaction.
• Chlorination replaces one of the H’s on the benzene ring with Cl.
• Nitration replaces one of the H’s on the benzene ring with NO2.
• Sulfonation replaces one of the H’s on the benzene ring with SO3H.
Cl
Cl
Cl
Cl
Cl2
a.
SO3
c.
FeCl3
Cl
Cl
Cl
Cl
Cl
SO3H
Cl
Cl
NO2
HNO3
b.
H2SO4
H2SO4
Cl
13.27
Cl
Draw the products of the substitution reaction. The Cl can replace any of the H’s on the benzene
ring, giving three different products.
CH3
Cl2
CH3
CH3
CH3
Cl
FeCl3
Cl
Cl
o-chlorotoluene
m-chlorotoluene
p-chlorotoluene
Solutions to End-of-Chapter Problems
13.29
a. molecular formula: C10H12O
b. aromatic ring, alkene, ether
c. trans
d. Tetrahedral C's are indicated. All other C's are
trigonal planar.
aromatic ring
trans alkene
ether
O
CH3
tetrahedral
CH3
anethole
tetrahedral
13.30
a. molecular formula: C10H12O2
b. aromatic ring, ester
c. trans
d. Tetrahedral C is indicated. All other C's are
trigonal planar.
trans alkene
O
aromatic ring
O
CH3
tetrahedral
methyl cinnamate
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Chapter 13–11
13.31
Use the general formulas [saturated hydrocarbon (CnH2n + 2), alkene (CnH2n), and alkyne
(CnH2n – 2)] to determine the molecular formula for each compound with 10 C’s.
a. (10 × 2) + 2 = 22: molecular formula C10H22
b. 10 × 2 = 20: molecular formula C10H20
13.32
Draw the structure of the hydrocarbon fitting the required description.
a.
13.33
c. (10 × 2) – 2 = 18: molecular formula C10H18
HC CCH2CH2CH2CH3
b. CH2 CHCH2CH2CH CH2
c.
Draw three alkynes with molecular formula C5H8.
CH3
HC CCH2CH2CH3
CH3C CCH2CH3
HC C C H
CH3
13.34
13.35
Draw the five constitutional isomers of C5H10 that contain a double bond.
CH2 CHCH2CH2CH3
CH3CH CHCH2CH3
CH3
CH3C CHCH3
CH3
CH2 CHCHCH3
CH3
CH2 CCH2CH3
Label each carbon as tetrahedral, trigonal planar, or linear by counting groups.
tetrahedral
trigonal planar
tetrahedral
a.
c.
trigonal planar
tetrahedral
trigonal planar
linear
b.
C CH
all trigonal planar
13.36
Predict the indicated bond angles in falcarinol.
a
b OH
c
H C C C C C C C CH2
H H H
d
a. trigonal planar: 120°
C C
(CH2)6CH3
H H e
b. tetrahedral: 109.5°
c. inear: 180°
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Chapter 13–12
13.37
Give the IUPAC name for each compound.
a.
b.
1
2
1
2
3
3
4
4
5
6
6 C chain
hexene
2-ethyl
4 C chain
butene
13.38
2-hexyne
2-ethyl-1-butene
Give the IUPAC name for each compound.
1
a. HC C CH2CH2CH2CH2CH3
7 C chain
13.39
a.
CH2CH3
1- heptyne
5 carbon ring; ethyl at C2
2-ethylcyclopentene
Give the IUPAC name for each compound using the following steps, as in Example 13.2:
[1] Find the longest chain containing both carbon atoms of the multiple bond.
[2] Number the chain to give the multiple bond the lower number.
[3] Name and number the substituents and write the complete name.
CH2 CHCH2CH2C(CH3)3
CH3
1
CH2 CHCH2CH2CCH3
2
3
4
CH3
CH3CH2C
CH3
double bond at C1
2 methyls at C5
CHCHCH2CHCH3
CH2 C CH2CH3
CH2CH2CH2CH2CH3
7 C's in the longest chain
heptene
2 methyls at C5 and C7
ethyl at C3
CH3CH2
CH3
8 C's in the longest chain
octene
c.
Answer: 5,5-dimethyl-1-hexene
CH3
CHCHCH2CHCH3
CH3CH2
CH3
CH3
6 C's in the longest chain
hexene
(CH3CH2)2C
5
CH2 CHCH2CH2CCH3
CH3
b.
2
b.
CH3CH2C
1
2
CH3
CH3
CHCHCH2CHCH3
3
4 5
6
7 8
Answer:
3-ethyl-5,7-dimethyl-3-octene
double bond at C3
1
2
ethyl at C2
CH2 C CH2CH3
Answer: 2-ethyl-1-heptene
CH2CH2CH2CH2CH3
3
4
5
6
7
double bond at C1
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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 13–13
d. CH3C CCH2C(CH3)3
5
1
CH3
CH3C CCH2CCH3
2
3
CH3
Answer: 5,5-dimethyl-2-hexyne
CH3C CCH2CCH3
CH3
6 C's in the longest chain
hexyne
triple bond at C2
CH3 6
2 methyls at C5
2 methyls at C5 and C6
5
CH3 CH3
1
e. CH3C C CH2 CH C CH2CH3
2
3
6
CH3 CH3
4
8
CH3C C CH2 CH C CH2CH3
CH2CH3
CH2CH3
Answer:
6-ethyl-5,6-dimethyl-2-octyne
triple bond at C2
8 C's in the longest chain
octyne
ethyl at C6
3
CH3
f.
5 4
6
CH2 CHCH2 C CH=CH2
CH3
1
CH3
6 C's in the longest chain
hexadiene
13.40
CH3
CH2 CHCH2 C CH=CH2
Answer:
3,3-dimethyl-1,5-hexadiene
2 methyls at C3
double bonds at C1 and C5
Give the IUPAC name for each compound using the following steps, as in Example 13.2:
[1] Find the longest chain containing both carbon atoms of the multiple bond.
[2] Number the chain to give the multiple bond the lower number.
[3] Name and number the substituents and write the complete name.
a. CH2 CHCH2CHCH2CH3
CH3
2
1
6 C's in the longest chain
hexene
b. (CH3)2C CHCH2CHCH2CH3
3
double bond at C1
double bond at C2
1
CH3
CH2CH2CH3
8 C's in the longest chain
octene
6
Answer: 4-methyl-1-hexene
CH3
CH2CH2CH3
CH3 C CHCH2CHCH2CH3
4 5
CH2 CHCH2CHCH2CH3
CH2 CHCH2CHCH2CH3
CH3
2
3
methyl at C4
ethyl at C5
5
CH3 C CHCH2CHCH2CH3
CH3
Answer: 5-ethyl-2-methyl-2-octene
CH2CH2CH3
6
8
methyl at C2
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Chapter 13–14
c. CH3C CCHCH3
triple bond at C2
methyl at C4
CH2CH2CH3
1
CH3C CCHCH3
2
4
Answer: 4-methyl-2-heptyne
CH3C CCHCH3
CH2CH2CH3
CH2CH2CH3
7
5
7 C's in the longest chain
heptyne
d. (CH3)3CC CC(CH3)3
triple bond at C3
CH3
CH3
CH3
CH3CC CCCH3
CH3
6
5
6 C's in the longest chain
hexyne
CH3 3 CH3 2
4 methyls at C2 and C5
CHCH3
double bond at C2
3
7
CH3CH2CH2CH2C CHCH3
butyl at C3
f. (CH3)2C CHCH2CH2CH2CH C(CH3)2
double bonds at C2 and C7
1
CHCH2CH2CH2CH CCH3
CH3
9 C's in the longest chain
nonene
13.41
2
CH3C
3
7
89
CHCH2CH2CH2CH CCH3
CH3
CH3
Answer:
2,8-dimethyl-2,7-nonadiene
methyls at C2 and C8
Give the IUPAC name for each compound using the steps in Answer 13.39 and Example 13.2.
double bond at C1
2
a. 1
Answer: 3-butyl-2-heptene
CH2CH2CH2CH3
7 C's in the longest chain
heptene
CH3
1
CH3CH2CH2CH2C CHCH3
CH2CH2CH2CH3
CH3C
Answer: 2,2,5,5-tetramethyl-3-hexyne
CH3CC CCCH3
CH3
e. (CH3CH2CH2CH2)2C
CH3
3
4
CH3
4-methylcyclohexene
b.
2
1
methyl at C4
6 carbon ring
cyclohexene
3
CH2CH3
CH2CH3
2 ethyl groups at C3
3,3-diethylcyclobutene
4
4 carbon ring
cyclobutene
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Chapter 13–15
13.42
Give the IUPAC name for each compound using the steps in Answer 13.39 and Example 13.2.
double bonds at C1 and C4
double bond at C1
a.
2
1
2
3
4
CH2CH2CH3
3
4
b. 1
4-propylcylcopentene
1,4-cycloheptadiene
5
7 carbon ring
cycloheptene
propyl at C4
5 carbon ring
cyclopentene
13.43
To draw the structure corresponding to each name, follow the steps in Example 13.3.
• Identify the parent name to find the longest carbon chain or ring, and then use the suffix to
determine the functional group; the suffix -ene = alkene and -yne = alkyne.
• Number the carbon chain or ring and place the functional group at the indicated carbon. Add
the substituents and enough hydrogens to give each carbon four bonds.
1 2
a. 3-methyl-1-octene
8 carbon chain
double bond at C1
4
5
6
7
8
CH2 CHCHCH2CH2CH2CH2CH3
CH3
CH3
methyl at C3
C C
b. 1-ethylcyclobutene
4 carbon ring
double bond at C1
CH2CH3
3
4
CH2CH3
ethyl at C1
C C
1 2
c. 2-methyl-3-hexyne
6 carbon chain
triple bond at C3
3
C C C C C C C C
5
H
6
C C C C C C
CH3 C C CCH2CH3
CH3
CH3
methyl at C2
1 2
CH2CH3
4
5
6
C C C C C C C
d. 3,5-diethyl-2-methyl-3-heptene
3
CH3
7 carbon chain
double bond at C3
1
2
3
4
5
6
8 carbon chain
double bond at C2
3
4
5
CH2CH3
7
C C C C C C C
1 2
CH3
2 ethyl groups
at C3 and C5
CH2 C C CHCH2CH2CH3
H H
7 carbon chain
double bonds at
C1 and C3
f. cis-7-methyl-2-octene
CH2CH3
H H
CH3 C C C CCH2CH3
CH2CH3
methyl at C2
!e. 1,3-heptadiene
H
7
6
7
8
C C C C C C C C
CH3
CH3
C C
H
CH2CH2CH2CHCH3
H
CH3
cis
methyl at C7
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Chapter 13–16
13.44
To draw the structure corresponding to each name, follow the steps in Example 13.3.
• Identify the parent name to find the longest carbon chain or ring, and then use the suffix to
determine the functional group; the suffix -ene = alkene and -yne = alkyne.
• Number the carbon chain or ring and place the functional group at the indicated carbon. Add
the substituents and enough hydrogens to give each carbon four bonds.
1 CH3
1 CH3
a. 1,2-dimethylcyclopentene
methyls at C1 and C2
b. 6-ethyl-2-octyne
8 carbon chain
triple bond at C2
1
2
3
4 5
6
7
8
C C C C C C C C
ethyl at C6
1
c. 3,3-dimethyl-1,4-pentadiene
5 carbon chain
double bonds at C1 and C4
2
CH3 C CCH2CH2CHCH2CH3
CH2CH3
CH2CH3
CH3
CH3
4 5
C C C C C
CH2 CH C CH CH2
CH3
CH3
2 methyls at C3
5
6
C C C C C
C
1
d. trans-5-methyl-2-hexene
6 carbon chain
double bond at C2
e. 5,6-dimethyl-2-heptyne
7 carbon chain
triple bond at C2
2 CH
3
2 CH
3
5 carbon ring
double bond at C1
2
3
4
methyl at C5
1
2
3
4
5
C C C C C
CH3
H
CH3
C
C
H
CH2CHCH3
CH3
trans
6
7
C C
CH3 C C CH2
CH3 CH3
CH CHCH3
CH3 CH3
methyls at C5 and C6
1
f. 3,4,5,6-tetramethyl-1-decyne
10 carbon chain
triple bond at C1
13.45
3
4
5
6
C C C
2
C
C
C C C C C
7
8
9
10
CH C CH
CH3 CH3 CH3 CH3
CH3
CH CH CH(CH2)3CH3
CH3 CH3 CH3
methyls at C3, C4, C5 and C6
Correct each of the incorrect IUPAC names.
a. The name 5-methyl-4-hexene places the double bond at C4
instead of C2. Assign the lower number to the alkene: 2methyl-2-hexene.
b. The name 1-methylbutene makes the last carbon in the chain a
substituent. In addition, the location of the double bond is not
specified. There are five carbons in the chain (not a methyl
substituent): 2-pentene.
c. The name 2,3-dimethylcyclohexene starts numbering
substituents at C2 instead of C1. Number to put the C=C
between C1 and C2, and then give the first substituent the
lower number: 1,6-dimethylcyclohexene.
CH3
CH3C CHCH2CH2CH3
2-methyl-2-hexene
CH3
CH CHCH2CH3
2-pentene
CH3 1
CH3 6
2
3
4
5
1,6-dimethylcyclohexene
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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 13–17
1
d. The name 3-butyl-1-butyne does not name the longest chain.
Name the seven carbon chain: 3-methyl-1-heptyne.
13.46
3
2H
HC CCCH3
CH2CH2CH2CH3
3-methyl-1-heptyne
Correct each of the incorrect IUPAC names.
CH3
a. The alkene has two methyl groups off of C2. Therefore, it
cannot be cis.
b. The name 2-methyl-2,4-pentadiene places the double bonds at
C2 and C4 instead of C1 and C3. The methyl group should be
at C4.
c. The name 2,4-dimethylcyclohexene starts numbering
substituents at C2 instead of C1. Number to put the C=C
between C1 and C2, and then give the first substituent the
lower number: 1,5-dimethylcyclohexene.
CH3C CHCH2CH2CH3
2-methyl-2-hexene
CH3 C
4-methyl-1,3-pentadiene
CH3 1
6
13.47
2
3
4
5
CH3
1,5-dimethylcyclohexene
1
d. The name 1,1-dimethyl-2-cyclohexene does not number the
double bond between C1 and C2. The methyl groups should
be on C3.
CHCH CH2
CH3
6
2
5
3 CH3
4
CH3
3,3-dimethylcyclohexene
When the two alkyl groups are on the same side of the double bond, the compound is called the
cis isomer. When they are on opposite sides it is called the trans isomer.
trans
cis
H
CH3CH2CH2CH2CH2
C C
H
C
H
H
H
C
C
C
CH2
H
C
H
C
H
OH
CHCHCH2CH2CH2COOH
S CH2
CHCONHCH2COOH
NHCOCH2CH2CHCOOH
NH2
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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 13–18
13.48
Draw the structures using the proper cis or trans arrangement around the carbon–carbon double
bond.
cis
a. CH3(CH2)8CH
C C
H
(CH2)12CH3
H
13.49
b.
trans
C C
H
C H
H
O
Give the IUPAC name for the alkene. Use the definition in Answer 13.47 to determine if it is the
cis or trans isomer.
5
1
4
2
cis-4-methyl-2-pentene
3
cis
4-methyl
13.50
Give the IUPAC name for the alkene. Use the definition in Answer 13.47 to determine if it is the
cis or trans isomer.
CH3
4
1
H3C 2
3
2-methyl
13.51
6
7
CH3
5
trans
trans-2-methyl-3-heptene
Draw the cis and trans isomers for each compound, as in Example 13.4.
a.
H
H
C C
CH2CH2CH2CH2CH2CH3
CH3
cis-2-nonene
H
CH2CH2CH2CH2CH2CH3
C C
CH3
H
H
H
b. CH3CH
C C
CH2CH2CH3
CH3
cis-2-methyl-3-heptene
CH2CH2CH3
H
C C
CH3CH
H
CH3
trans-2-nonene
trans-2-methyl-3-heptene
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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 13–19
13.52
Draw the cis and trans isomers for each compound, as in Example 13.4.
a.
C C
CH3
CH3
CH2CH3
cis-4,4-dimethyl-2-hexene
CH3 CH2CH3
C
C C
CH3
CH3
H
CH2CH2CH3
H
C C
CH3CH2
C
CH3
cis-3-heptene
H
C C
b.
CH2CH2CH3
CH3CH2
H
H
H
H
H
trans-3-heptene
trans-4,4-dimethyl-2-hexene
13.53
Constitutional isomers have the same molecular formula, but have the atoms bonded to different
atoms. Stereoisomers have atoms bonded to the same atoms but in a different three-dimensional
arrangement.
13.54
Draw the possible stereoisomers for 2,4-hexadiene.
H
H
C C
C C
CH3
13.55
CH3
H
CH3
C C
C C
H
H
H
H
CH3
CH3
H
C C
C C
H
H
H
CH3
H
Determine if the molecules are constitutional isomers, stereoisomers, or identical.
a.
and
same molecular formula
same connectivity
identical
b.
and
same molecular formula
different connectivity
constitutional isomers
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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 13–20
13.56
Determine if the molecules are constitutional isomers, stereoisomers or identical.
CH3
a.
CH3
same molecular formula
same connectivity
identical
a.
CH3
H2
CH3CH2CH2CH2CH2CH3
c.
(CH3)2CHCH2CH2CH2CH3
d.
Pd
H2
b. (CH3)2C
a.
CH3
Draw the products of each reaction by adding H2 to the double bond.
CH2 CHCH2CH2CH2CH3
13.58
H C
3
and
same molecular formula
same connectivity
different arrangement in space
(cis and trans isomers)
stereoisomers
b. H3C
and
13.57
CH3
CH3
CH3
CHCH2CH2CH3
Pd
CH3
H2
Pd
H2
CH2
H
Pd
CH3
Draw the products of each reaction by adding Br2 to the double bond.
CH3
Br2
CH2 CHCH2CH2CH2CH3
CH2CHCH2CH2CH2CH3
Br2
c.
CH3
Br
Br
Br Br
Br2
b.
(CH3)2C CHCH2CH2CH3
(CH3)2C CHCH2CH2CH3
d.
Br2
CH2
Br
CH2Br
Br Br
13.59
Draw the products of each reaction by adding HCl to the double bond.
Cl
HCl
a.
b. (CH3)2C
13.60
C(CH3)2
c.
Cl H
HCl
(CH3)2C
d.
C(CH3)2
HCl
CH2 CHCH2CH(CH3)2
CH2
HCl
CH3
Cl
Draw the products of each reaction by adding H2O to the double bond.
OH
a.
Cl
CH3CHCH2CH(CH3)2
+ H2O
H2SO4
b. (CH3)2C C(CH3)2 + H2O
c. CH2 CHCH2CH(CH3)2 + H2O
H2SO4
OH H
(CH3)2C C(CH3)2
d.
CH2
+ H2O
H2SO4
H2SO4
OH
CH3CHCH2CH(CH3)2
CH3
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 13–21
13.61
Draw the products of each reaction by adding the specified reagent to the double bond.
H
CH2CH3
a.
H2
CH2CH3
Pd
H
CH2CH3
Cl
CH2CH3
HCl
d.
H
H
H
Cl
CH2CH3
Cl2
b.
CH2CH3
CH2CH3
e.
Br
CH2CH3
HBr
H
H
Cl
H
CH2CH3
c.
Br
CH2CH3
Br2
CH2CH3
f.
H
H2SO4
Br
H
13.62
OH
CH2CH3
H2O
H
Draw the products of each reaction by adding the specified reagent to the double bond.
a. CH3CH CHCH3
+ HCl
CH3CHCH2CH3
d. CH3CH CHCH2CH2CH3 + Br2
CH3CHCHCH2CH2CH3
Br Br
Cl
b.
+ H2
Pd
(CH3)2CCH2CH2CH3
e. (CH3)2C CHCH2CH3 + HBr
Br
CH3
c.
Cl2
CH3
13.63
a.
Cl
Cl
CH3
CH3
f.
CH3
CH3
+ H2O
H2SO4
CH3
CH3
CH3
OH
CH3
Work backwards to determine what alkene is needed as a starting material to prepare each of the
alkyl halides or dihalides.
CH2 CH2
HBr
CH3CH2Br
c.
CH3
Cl2
Cl
CH3
Cl
HCl
b.
Cl
d. CH2 CHCH2CH(CH3)2
Br2
BrCH2CHCH2CH(CH3)2
Br
13.64
Markovnikov’s rule must be followed when determining the starting materials. 2-Bromobutane
can be formed as the only product of the addition of HBr to 1-butene and 2-butene. 2Bromopentane can be formed as the only product of the addition of HBr to 1-pentene.
CH2 CHCH2CH3
+ HBr
CH3CHCH2CH3
CH3CH CHCH3
1-butene
CH2 CHCH2CH2CH3 + HBr
2-bromobutane
+ HBr
CH3CHCH2CH3
Br
Br
2-butene
2-bromobutane
CH3CHCH2CH2CH3
Br
1-pentene
2-bromopentane
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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 13–22
13.65
Work backwards to determine what reagent is needed to convert 2-methylpropene to each
product.
a.
(CH3)2C=CH2
b.
(CH3)2C=CH2
c.
13.66
HCl
H2
Pd
H2O
(CH3)2C=CH2
(CH3)3CCl
d. (CH3)2C=CH2
(CH3)3CH
e. (CH3)2C=CH2
(CH3)3CBr
Br2
(CH3)2CCH2Br
Br
Cl2
f. (CH3)2C=CH2
(CH3)3COH
H2SO4
HBr
(CH3)2CCH2Cl
Cl
a. The addition of Br2 could be used to tell the difference between cyclohexane and cyclohexene.
Br2 is red in color. There is no reaction when Br2 is added to cyclohexane, so the solution
would remain red. The bromines will add across the double bond, though, when Br2 is added
to cyclohexene, thus yielding colorless 1,2-dibromocyclohexane.
b. The addition of Br2 could also be used to distinguish between cyclohexene and benzene. When
Br2 is added to cyclohexene, 1,2-dibromocyclohexane is formed and the red color of the Br2
disappears. When Br2 is added to a solution containing benzene, the red color will remain
because the benzene will not undergo a subsitution reaction, except in the presence of FeBr3.
13.67
To draw the polymer, draw three or more alkene molecules and arrange the carbons of the double
bonds next to each other. Break one bond of each double bond, and join the alkenes together with
single bonds. With unsymmetrical alkenes, substituents are bonded to every other carbon. Use
Example 13.7 as a guide.
Join these 2 C's.
H
CH2
H
C
CH2
CH2
C
COOH
COOH
CH2C
CH2C
CH2C
H
H
H
Draw the polymer using the steps in Example 13.7.
Join these 2 C's.
CH3
CH2
Join these 2 C's.
CH3
C
CH2
COOCH3
13.69
COOH COOH COOH
H
C
COOH
13.68
Join these 2 C's.
COOCH3 COOCH3 COOCH3
CH3
C
CH2
COOCH3
CH2C
C
CH2C
CH3
COOCH3
CH2C
CH3
CH3
Draw the polymers using the steps in Example 13.7.
Join these 2 C's.
CH2CH3
a. CH2
C
CH2CH3
CH2
CH2CH3
C
H
CH2
C
H
Join these 2 C's.
b. CH
2
Join these 2 C's.
Cl
C
C
CH3
CH2
CH2
CH2
H
H
H
Join these 2 C's.
Cl
CH2
CN
CH3
CH2C CH2C CH2C
H
Cl
CH2
CN
CH3
C
CN
Cl
Cl
Cl
CH2C
CH2C
CH2C
CN
CN
CN
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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 13–23
Join these 2 C's.
c.
Cl
CH2
Cl
C
CH2
CH2
C
Cl
Cl
Cl
CH2C CH2C
CH2C
Cl
Cl
Cl
Cl
Draw the polymers using the steps in Example 13.7.
Join these 2 C's.
a. CH2
Join these 2 C's.
OCH3
OCH3
CH2
C
CH2
CH2
CH2
Join these 2 C's.
c. CH2
C
CO2CH3
CH2
C
H
Cl
Cl
Cl
CH2C
CH2C
CH2C
CO2CH3
CO2CH3
CO2CH3
Join these 2 C's.
H
H
H
CH2C
CH2C
CH2C
H
C
CH2
C
NHCOCH3
NHCOCH3
CH2C
H
CO2CH3
H
H
CH2C
H
Cl
C
CO2CH3
OCH3
Join these 2 C's.
Cl
C
OCH3
CH2C
H
Join these 2 C's.
b. CH2
C
H
Cl
OCH3
OCH3
C
H
13.71
Cl
Cl
C
Cl
13.70
Join these 2 C's.
NHCOCH3
NHCOCH3 NHCOCH3
NHCOCH3
Work backwards to determine what monomer was used to form the polymer.
Each one of these units is from the monomer:
13.72
Br
Br
Br
CH2 C
CH2 C
CH2 C
Cl
Cl
Cl
Cl
CH2 C
Br
Work backwards to determine what monomer was used to form the polymer.
Each one of these units is from the monomer:
CH3
CH3
CH2 C
CH2 C
C O
CH2 C
C O
OCH2CH3
13.73
CH3
OCH2CH3
CH2 C
CH3
C O
C O
OCH2CH3
OCH2CH3
Draw two resonance structures by moving the double bonds.
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 13–24
13.74
The structures A and B are the same compound even though A has the two Cl atoms on the same
double bond and B has the two Cl atoms on different double bonds, because the two structures are
resonance structures. They differ in the placement of the electrons, but the placement of the atoms
is the same.
13.75
Name each aromatic compound as in Example 13.8. Name the substituents on the benzene ring.
With two groups, alphabetize the substituent names and use the prefix ortho, meta, or para to
indicate their location. With three substituents, alphabetize the substituent names, and number to
give the lowest set of numbers.
b.
a.
bromo
fluoro
chloro
ethyl
p-chloroethylbenzene
13.76
o-bromofluorobenzene
Name each aromatic compound as in Example 13.8. Name the substituents on the benzene ring.
With two groups, alphabetize the substituent names and use the prefix ortho, meta, or para to
indicate their location. With three substituents, alphabetize the substituent names, and number to
give the lowest set of numbers.
a.
OH
propyl
b. H2N
bromo
aniline
phenol
m-propylphenol
13.77
Br
p-bromoaniline
Name each aromatic compound as in Example 13.8 and Answer 13.75.
Cl
nitro
NO2
a.
butyl
m-chloronitrobenzene
c.
o-butylethylbenzene
ethyl
chloro
Cl
b.
H2N
NO2
nitro
p-nitroaniline
OH
OH on benzene ring = phenol
Cl
2,5-dichloro
d.
2,5-dichlorophenol
NH2 on benzene ring = aniline
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Chapter 13–25
13.78
Name each aromatic compound as in Example 13.8. Name the substituents on the benzene ring.
With two groups, alphabetize the substituent names and use the prefix ortho, meta, or para to
indicate their location. With three substituents, alphabetize the substituent names, and number to
give the lowest set of numbers.
a. CH3(CH2)3
(CH2)3CH3
H2N
c.
NH2 on benzene ring = aniline
Br
ethyl
CH2CH3
b.
Br
bromo
CH3
o-bromoethylbenzene
bromo
ethyl
m-ethylaniline
p-dibutylbenzene
2 butyl groups
CH2CH3
I
iodo
d.
2-bromo-3-iodotoluene
CH3 on benzene ring = toluene
13.79
Draw and name the three isomers with Cl and NH2 as substituents. Recall that a benzene ring
with an NH2 group is named aniline.
NH2
NH2
NH2
Cl
Cl
o-chloroaniline
13.80
m-chloroaniline
Cl
p-chloroaniline
Draw the structure of 2,4,6-trinitrotoluene (TNT).
CH3
O2N
NO2
NO2
13.81
Work backwards to draw the structure from the IUPAC name.
NO2
nitro
CH3
a.
p-nitropropylbenzene
CH2CH2CH2CH3
chloro
Cl
butyl
OH on benzene ring = phenol
iodo
2-bromo-4-chlorotoluene
4
m-dibutylbenzene
b.
CH3 on benzene ring = toluene
bromo
2 Br
3
butyl
CH2CH2CH2CH3
I
d.
propyl
CH2CH2CH3
OH
1
NH2
1
e. I
6
5
NH2 on benzene ring = aniline
2 Cl
2-chloro-6-iodoaniline
3
4
iodo
chloro
o-iodophenol
c.
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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 13–26
13.82
Work backwards to draw the structure from the IUPAC name.
NO2
m-ethylnitrobenzene
a.
CH2CH3
1,3,5-trinitrobenzene
d.
O 2N
ethyl
fluoro
F
F
NO2
OH on benzene ring = phenol
OH
Br
o-difluorobenzene
b.
nitro
NO2
nitro
e.
bromo
Br
CH3
CH3 on benzene ring = toluene
p-bromotoluene
c.
bromo
Br
13.83
2,4-dibromophenol
Draw the products of each reaction.
SO3H
Cl
a. CH3
Cl2
CH3
FeCl3
CH3
c. CH3
CH3
CH3
SO3
H2SO4
CH3
CH3
NO2
b. CH3
13.84
HNO3
CH3
H2SO4
CH3
CH3
Determine the reagents for the reactions in the sequence.
A
B
Cl
Cl2
HNO3
H2SO4
FeCl3
Cl
O 2N
13.85
SO3H
Cl
D
O2N
C
SO3
H2SO4
Cl
H2
Pd
H2N
SO3H
Draw the three products formed in the reaction of bromobenzene.
Br
HNO3
H2SO4
Br
Br
NO2
Br
NO2
NO2
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Chapter 13–27
13.86
Draw the structures that result from the substitution of a chloro group onto the benzene ring.
Br
Br
FeCl3
+ Cl2
Br
Cl
Br
Br
+ Cl2
Br
Br
FeCl3
+
Br
Cl
Br
Br
Cl
A
C
Br
Br
+ Cl2
Br
Br
Cl
FeCl3
+
Br
+
Br
Br
B
Cl
Br
Cl
13.87
Vitamin E is an antioxidant because of the phenol, which has an OH bonded to the benzene ring.
13.88
BHA is an ingredient in some breakfast cereals and other packaged foods because it is a synthetic
antioxidant and can prevent oxidation and spoilage.
13.89
Methoxychlor is more water soluble than DDT. The OCH3 groups can hydrogen bond to water.
This increase in water solubility makes methoxychlor more biodegradable.
13.90
2,4-D is soluble in water because it contains an –OCH2COOH group. This group can hydrogen
bond to water through the oxygen bound to the benzene ring and the two oxygens on the carboxy
group (COOH). DDT has no groups that are able to hydrogen bond to water, so DDT is insoluble
in water.
13.91
H H
H
C C
a.
CH3CH2CH2CH2CH2
H
C C
CH2
CH2CH2CH2CH2CH2CH2CH2COOH
H
H
H2, Pd
C C
CH2CH2CH2CH2CH2CH2CH2COOH
CH3CH2CH2CH2CH2CH2CH2CH2
H
H
C C
CH3CH2CH2CH2CH2
b.
+
Partial hydrogenation adds hydrogen
to one of the double bonds.
CH2 CH2CH2CH2CH2CH2CH2CH2CH2CH2COOH
Complete hydrogenation adds hydrogen to both of the double bonds, forming:
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOH
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Chapter 13–28
c.
CH2CH2CH2CH2CH2CH2CH2COOH
H
one possibility:
C C
H
CH3CH2CH2CH2CH2CH2CH2CH2
trans
13.92
Draw the structures and determine which will have the higher melting point.
a. CH3(CH2)2CH2
b. CH3(CH2)2CH2
H
H
C C
H
C C
H
c.
13.93
(CH2)7COOH
H
H
H
C C
H
C C
H
H
C C
H
H
C C
(CH2)7COOH
The all-trans isomer will have the higher melting point because it has a more linear shape
than the isomer with the cis double bond. As a result, the all-trans isomer packs more closely
together in the solid, and thus requires more energy to separate upon melting.
Recall from Section 13.12 that many phenols are antioxidants.
a.
b.
CH2CH2CH2OH
an alcohol
not an antioxidant
c. HO
OCH3
an ether
not an antioxidant
OCH3
This compound could be an
antioxidant because it has an OH
group bonded to the aromatic ring.
13.94
All commercial sunscreens contain a benzene ring. The structure in part a contains two benzene
rings and therefore might be an ingredient in a commercial sunscreen. The structure in part b
contains two cyclohexane rings and therefore would not be an ingredient in a commercial
sunscreen.
13.95
When benzene is oxidized to phenol, it is converted to a more water-soluble compound that can
then be excreted in the urine.
13.96
A PAH is a polycyclic aromatic hydrocarbon, a compound that contains two or more benzene
rings that share carbon–carbon bonds. The structure of anthracene, a PAH mentioned in Section
13.10D, is shown below.
13.97
H
H
a.
CH3(CH2)5
(CH2)7COOH
cis
palmitoleic acid
13.98
(CH2)7COOH
H
b.
C C
CH3(CH2)5
H
trans
stereoisomer
H
H
c.
C C
C C
CH3(CH2)4
(CH2)8COOH
constitutional isomer
one possibility
Polyethylene is a long chain hydrocarbon. Water and carbon dioxide are formed when
polyethylene undergoes combustion.
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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 13–29
13.99
All the carbons in benzene are trigonal planar with 120° bond angles, resulting in a flat ring. In
cyclohexane, all the carbons are tetrahedral, so the ring is puckered.
13.100 p-Dichlorobenzene is a nonpolar molecule but o-dichlorobenzene is a polar molecule because the
p-dichlorobenzene molecule is symmetrical and therefore does not have a net dipole moment. oDichlorobenzene, on the other hand, has a net dipole.
Cl
Cl
Cl
Cl
nonpolar
the dipoles cancel
polar
13.101
a. CH2=CH(CH2)4CH3
7 C chain
1-heptene
b. CH2=CH(CH2)4CH3
c. CH2=CH(CH2)4CH3
d. polymerization:
H2
CH3(CH2)5CH3
H2O
CH3CH(OH)CH2CH2CH2CH2CH3
R
R
R
CH2C
CH2C
CH2C
H
H
H
R = (CH2)4CH3
13.102
a. CH2=CH(CH2)7CH3 10 C chain
1-decene
b. CH2=CH(CH2)7CH3
c. CH2=CH(CH2)7CH3
d. polymerization:
H2
CH3(CH2)8CH3
H2O
CH3CH(OH)CH2CH2CH2CH2CH2CH2CH2CH3
R
R
R
CH2C
CH2C
CH2C
H
H
H
R = (CH2)7CH3
13.103 cis-2-Hexene and trans-3-hexene are constitutional isomers because the double bond is located in
a different place on the carbon chain (C2 vs. C3).
CH3
CH2CH2CH3
C C
H
H
CH2CH3
C C
H
cis-2-hexene
CH3CH2
H
trans-3-hexene
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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 13–30
13.104 Determine the two monomer units for Saran.
H
Cl
H
Cl
CH2 C
CH2 C
CH2 C
CH2 C
Cl
Cl
Cl
Cl
A
B
A
B
H
CH2 C
A
Cl
+
CH2 C
Cl
Cl
B
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.