Chapter 13 Answers and Solutions
13
Answers and Solutions to Text Problems
13.1
a. An alkane has only sigma bonds.
b. An alkyne has a sigma bond and two pi bonds.
c. An alkene has the groups on the carbon atom arranged at 120°.
13.2
a. alkene
13.3.
a. An alkane has the general formula CnH2n+2
b. An alkene has a double bond.
c. An alkyne has a triple bond.
d. An alkene has a double bond.
e. A cycloalkene has a double bond in a ring.
13.4
a. cycloalkene
d. alkyne
13.5.
CH2=CHCH3 Propene (propylene)
b. alkane
b. alkyne
e. alkane
c. alkyne
c. alkene
CH3
Cyclopropane
13.6
CH3CH=CHCH3
CH2=CHCH2CH3
13.7
a. Propene contains three carbon atoms with a carbon-carbon double bond.
Propyne contains three carbon atoms with a carbon-carbon triple bond.
b. Cyclohexane is a six-carbon cyclic compound with all carbon-carbon single bonds.
Cyclohexene is a six-carbon cyclic compound with a carbon-carbon double bond.
13.8
a. 1-butyne is a four-carbon compound with a triple bond between the first and second carbons.
2-butyne is a four-carbon compound with a triple bond between the second and third
carbons.
b. 1-methylcyclohexene is a six-carbon cyclic alkene with a methyl attached to one of the
carbons in the double bond. 3-methylcyclohexene is a six carbon cyclic alkene with a methyl
attached to one of the carbons next to the double bond.
13.9
a. The two-carbon compound with a double bond is ethene.
b. 2-methyl-1-propene
c. 4-bromo-2-pentyne
d. This is a four-carbon cyclic structure with a double bond. The name is cyclobutene.
e. This is a five-carbon cyclic structure with a double bond and an ethyl group. You must
count the two carbons of the double bond as 1 and 2. The name is 4-ethylcyclopentene.
f. Count the chain from the end nearest the double bond: 4-ethyl-2-hexene
13.10
a. 1-hexene
d. 1,2-dimethylcyclobutene
13.11
a. Propene is the three-carbon alkene. H2C=CHCH3
b. 6-methyl-2-heptyne
e. 4,6-dichloro-1-heptene
c. 1,4-dimethylcyclohexene
f. 5,6-dichloro-2-heptyne
Unsaturated Hydrocarbons
b. 1-pentene is the five-carbon compound with a double bond between carbon1 and carbon 2.
H2C=CHCH2CH2CH3
c. 2-methyl-1-butene has a four-carbon chain with a double bond between carbon 1 and carbon 2
and a methyl attached to carbon 2.
CH3

H2C=CCH2CH3
d. 3-methylcyclohexene is a six-carbon cyclic compound with a double bond between carbon 1
and carbon 2 and a methyl group attached to carbon 3.
CH3
e. 2-chloro-3-hexyne is a six-carbon compound with a triple bond between carbon 3 and 4 and a
chlorine atom bonded to carbon 2.
Cl
|
CH3CH C≡CCH2CH3
CH3
13.12
a.
CH3

b. HC≡CCHCH3
CH3
d.
CH3

c. H2C=CHCHCHCH3

CH3
CH3
e.
CH3
CH2CH3
13.13
Constitutional isomers have the same number of each atom but the atoms are connected
differently. Geometric isomers have different arrangements of the groups attached to a double
bond.
13.14
In cis-2-butene, two methyl groups are on the same side of the double bond; in trans-2-butene,
methyl groups are on opposite sides of the double bond.
13.15 There are four constitutional isomers with the molecular formula C3H5Cl, three are alkenes and
one is a cycloalkane.
CH3
Cl
|
CH3C=CH2
ClCH2CH=CH2
Cl
CH3CH=CHCl
13.16
CH3C≡CCH3
HC≡CHCH2CH3
H2C=CHCH=CH2
Chapter 13 Answers and Solutions
13.17
a. This compound cannot have cis-trans isomers since there are two identical hydrogen atoms
attached to the first carbon.
b. This compound can have cis-trans isomers since there are different groups attached to each
carbon atom in the double bond.
c. This compound cannot have cis-trans isomers since there are two of the same groups attached
to each carbon.
13.18
a. This compound has cis-trans isomers.
b. This compound does not have cis-trans isomers.
c. This compound does not have cis-trans isomers.
13.19
a. cis-2-butene This is a four-carbon compound with a double bond between carbon 2 and
carbon 3. Both methyl groups are on the same side of the double bond; it is cis.
b. trans-3-octene This compound has eight carbons with a double bond between carbon 3 and
carbon 4. The alkyl groups are on opposite sides of the double bond; it is trans.
c. cis-3-heptene This is a seven-carbon compound with a double bond between carbon 3 and
carbon 4. Both alkyl groups are on the same side of the double bond; it is cis.
13.20
a. cis-2-pentene
13.21
a. trans-2-butene has a four-carbon chain with a double bond between carbon 2 and carbon 3.
The trans isomer has two methyl groups on opposite sides of the double bond.
b. trans-2-heptene
CH3
c. trans-3-heptene
H
C
C
H
CH3
b. cis-2-pentene has a five-carbon chain with a double bond between carbon 2 and carbon 3. The
cis isomer has alkyl groups on the same side of the double bond.
CH3
CH3
CH2
C
C
H
H
c. trans-3-heptene has a seven carbon chain with a double bond between carbon 3 and carbon 4.
The trans isomer has the alkyl groups on opposite sides of the double bond.
CH3
CH2
H
C
C
H
13.22
a.
CH2 CH2 CH3
CH3CH2
CH2CH3
C
CH3CH2CH2
H
CH2CH2CH3
C
H
C
H
H
CH3
C
H
c.
b.
C
H
C
CH2CH3
Unsaturated Hydrocarbons
13.23
a.
CH3CH2CH2CH2CH3
pentane
Cl
|
1,2-dichloro-2-methylbutane
b. ClCH2CCH2CH3
|
CH3
c. The product is a four-carbon cycloalkane with bromine atoms attached to carbon 1 and carbon
2. The name is 1,2-dibromocyclobutane.
Br
Br
d. When H2 is added to a cycloalkene, the product is a cycloalkane. Cyclopentene would form
cyclopentane.
+ H2
Pt
cyclopentene
cyclopentane
e. When Cl2 is added to an alkene, the product is a dichloroalkane. The product is a four-carbon
chain with chlorine atoms attached to carbon 2 and carbon 3 and a methyl group attached to
carbon 2. The name of the product is: 2,3-dichloro-2-methylbutane
CH3
CH3


CH3 C=CHCH3 + Cl2 → CH3CCHCH3
 
Cl Cl
2-methyl-2-butene
2,3-dichloro-2-methylbutane
f. CH3CH2CH2CH2CH3
13.24
pentane
Br

a. BrCH2CHCH2CH3
1,2-dibromobutane
b.
cyclohexane
c. CH3CH2CH2CH3
CH3

d. CH3CCHCH2CH3
 
Cl Cl
2,3-dichloro-2-methylpentane
`
butane
CH3
Br
e.
Br
1,2-dibromo-3-methylcyclohexane
Chapter 13 Answers and Solutions
CH3 Cl Cl
|
| |
f. CH3CHCCH
| |
Cl Cl
13.25
1,1,2,2-tetrachloro-3-methylbutane
a. When HBr is added to an alkene, the product is a bromoalkane. In this case, we do not need
to use Markovnikov's rule.
Br

CH3CH2 CHCH3
b. When H2O is added to an alkene, the product is an alcohol. In this case, we do not need to use
Markovnikov's rule.
OH
c. When HCl is added to an alkene, the product is a chloroalkane. We need to use
Markovnikov's rule, which says that hydrogen adds to the carbon with the greatest number of
hydrogens, in this case that is carbon 1.
Cl

CH3CHCH2CH3
d. When HI is added to an alkene, the product is an iodoalkane. In this case, we do not need to
use Markovnikov's rule.
CH3 I


CH3CHCHCH3
e. When HBr is added to an alkene, the product is a bromoalkane. We need to use
Markovnikov's rule, which says that hydrogen adds to the carbon with the greatest number of
hydrogens, in this case that is carbon 2.
Br

CH3CH2 CCH2 CH3

CH3
f. Using Markovnikov’s rule, the H from HOH goes to the carbon 2 in the cyclohexane ring,
which has more hydrogen atoms. The OH then goes to carbon 1.
CH3
OH
Unsaturated Hydrocarbons
13.26
CH3

a. CH3CCH2 CH3
|
Cl
OH

b. CH3CH2CH2 CHCH2CH3
CH3
|
c. CH3 CCH3

Br
CH3
Br
OH
d.
e.
H Cl
| |
f. CH3CCCH3
| |
H Cl
13.27
a. Hydrogenation of an alkene gives the saturated compound, the alkane.
CH3

Pt
CH2=CCH3 + H2 →
CH3

CH3CHCH3
b. The addition of HCl to a cycloalkene gives a chlorocycloalkane.
Cl
+ HCl →
c. The addition of bromine (Br2) to an alkene gives a dibromoalkane.
Br Br


CH3CH=CHCH2CH3 + Br2 → CH3 CHCHCH2CH3
d. Hydration (the addition of H2O) to an alkene gives an alcohol. In this case, we use
Markovnikov's rule and attach hydrogen to carbon 1.
OH
|
H+
CH2=CHCH3 + H2O → CH3CHCH3
Cl Cl
| |
e. CH3C≡CCH3 + 2Cl2 → CH3CCCH3
| |
Cl Cl
CH3
CH3
+
13.28
a.
+ H 2O
H
→
b. CH3CH2CH=CHCH2CH3
OH
Pt
+ H2 → CH3CH2CH2CH2CH2CH3
Chapter 13 Answers and Solutions
CH3
CH3


c. CH3C=CHCH3 + HBr → CH3 CCH2CH3

Br
CH3

d. CH3 C=CCH2CH3 + Cl2 →

CH3
CH3 Cl
 
CH3 CCCH2CH3
 
Cl CH3
CH3
CH3
e.
Cl
+ HCl
13.29
A polymer is a long-chain molecule consisting of many repeating smaller units. These smaller
units are called monomers.
13.30 Monomers are the small carbon molecules which repeat to make polymers.
13.31
Teflon is a polymer of the monomer tetrafluoroethene.
3 F
F
F
C
C
F
F
F
F
F
F
F
C
C
C
C
C
C
F
F
F
F
F
F
13.32
3 CH2
13.33
CH
CH2
CH CH2
CH CH2
CH
1,1-difluoroethene is the two carbon alkene with two fluorine atoms attached to carbon 1.
F
H
F
H
F
C
C C
C
C C
F
H
H
F
F
H
H
CN
CN
CN
CN
|
|
|
|
CH2CHCH2CHCH2CH
3 CH2=CH →
13.35 Cyclohexane, C6H12 , is a cycloalkane in which six carbon atoms are linked by single bonds in a
ring. In benzene, C6H6, an aromatic system links the six carbon atoms in a ring.
13.36 In the Health Note “Aromatic Compounds In health and Medicine, the aromatic portion of each
molecule is the benzene ring.
13.34
Unsaturated Hydrocarbons
13.37
The six carbon ring with alternating single and double bonds is benzene. If the groups are in the
1,2 position, this is ortho (o), 1,3 is meta (m) and 1,4 is para (p)
a. 1-chloro-2-methylbenzene; o-chlorotoluene
b. ethylbenzene
c. 1,3,5-trichlorobenzene
d. m-xylene; m-methyltoluene; 1,3-dimethylbenzene
e. 1-bromo-3-chloro-5-methylbenzene; 3-bromo-5-chlorotoluene
f. isopropyl benzene
13.38
a. benzene
c. 1,3-dichlorobenzene; m-dichlorobenzene
e. benzyl bromide
13.39
a.
c.
b. methylbenzene; toluene
d. 1,4-dimethylbenzene;p-xylene
f. 4-bromo-1,2-dichlorobenzene
The prefix m means that the two
chloro groups are in the 1 and 3 position.
b.
CH3
d.
CH2CH3
The prefix p means that the two groups
are in the 1 and 4 position.
CH3
Cl
CH3
13.40
a.
b.
Cl
CH2CH2CH3
CH3
Cl
c.
d.
Cl
13.41
Benzene undergoes substitution reactions because a substitution reaction allows benzene
Cl to
retain the stability of the aromatic system.
13.42
The aromatic ring in toluene is resistant to halogenation in the presence of light. However, the
alkyl group does undergo halogenation as a substitution reaction in which a chlorine atom
replaces a hydrogen atom when light is present.
Cl
13.43 a. No reaction
b.
NO 2
c.
Chapter 13 Answers and Solutions
13.44
CH3
CH3
CH3
Br
a.
and
and
Br
Br
SO3H
b.
NO2
c.
13.45
Propane is the three-carbon alkane with the formula C3H8. All the carbon-carbon bonds in
propane are single bonds. Cyclopropane is the three-carbon cycloalkane with the formula C3H6.
All the carbon-carbon bonds in cyclopropane are single bonds. Propene is the three carbon
compound which has a carbon-carbon double bond. The formula of propene is C3H6. Propyne is
the three carbon compound with a carbon-carbon triple bond. The formula of propyne is C3H4.
13.46
Butane is the four-carbon alkane with the formula C4H10; the carbon-carbon bonds in butane are
single bonds. Cyclobutane is a four-carbon cycloalkane with the formula C4H8. All the carboncarbon bonds in cyclobutane are single bonds. Cyclopbutene is a four-carbon compound, which
has a carbon-carbon double bond. The formula of butene is C4H6. 2-Butyne is a four-carbon
compound with a carbon-carbon triple bond. The formula of propyne is C4H6.
13.47
a. This compound has a chlorine atom attached to a cyclopentane; the IUPAC name is
chlorocyclopentane.
b. This compound has a five carbon chain with a chlorine atom attached to carbon 2 and a
methyl group attached to carbon 4. The IUPAC name is: 2-chloro-4-methylpentane.
c. This compound contains a five-carbon chain with a double bond between carbon 1 and carbon
2 and a methyl group attached to carbon 2. The IUPAC name is: 2-methyl-1-pentene.
d. This compound contains a five-carbon chain with a triple bond between carbon 2 and carbon
3. The IUPAC name is: 2-pentyne
e. This compound contains a five-carbon cycloalkene with a chlorine atom attached to carbon 1.
The IUPAC name is: 1-chlorocyclopentene
f. This compound contains a five carbon chain with a double bond between carbon 2 and
carbon 3. The alkyl groups are on opposite sides of the double bond. The IUPAC name is:
trans-2-pentene
g. This compound contains a six-carbon ring with a double bond and chlorine atoms attached to
carbon 1 and carbon 3 The IUPAC name is: 1,3-dichlorocyclohexene
Br
c. CH3
a.
b. CH3C≡CCH2CH3
CH2CH2CH2CH3
13.48
Br
C
H
CH3 Cl
CH3CH2


d. CH3 CHCCH2CH3 e.
C

H
Cl
H
C
H
Br
f.
C
CH2CH3
Cl
Unsaturated Hydrocarbons
Cl

g. H2C=CCHCH3

Cl
13.49
13.50
a. These structures represent a pair of constitutional isomers. In one isomer, the chlorine is
attached to one of the carbons in the double bond; in the other isomer the carbon
bonded to the chlorine is not part of the double bond.
b. These structures are cis-trans isomers. In the cis isomer, the two methyl groups
are on the same side of the double bond. In the trans isomer, the methyl groups are
on opposite sides of the double bond.
c. These structures are identical and not isomers. Both have five carbon chains with a double
bond between carbon 1 and carbon 2.
d. These structures represent a pair of constitutional isomers. Both have the molecular formula
C7H16 .One isomer is a six-carbon chain with a methyl group attached, whereas the other is
a five-carbon chain with two methyl groups attached.
CH3
H2C=CHCH2CH3
cyclobutane
methylcyclopropane
CH3

H2C=C−CH3
CH3
1-butene
CH3
C
C
C
C
H
CH3
trans-2-butene
H
H
cis-2-butene
2-methyl-1-propene
H
CH3
CH3
13.51 The structure of methylcyclopentane is
.
It can be formed by the hydrogenation of four cycloalkenes.
CH3
CH3
CH3
CH2
13.52
When light fall on a molecule of cis-2-retinal, which is attached to a protein, the cis isomer
changes to trans. The trans isomer will not fit the protein and separation from the protein
generates a signal, which the brain interprets as light.
13.53
a.
CH3
CH2CH3
C
H
CH3
C
H
C
H
cis-2-pentene both alkyl groups are on
the same side of the double bond
H
C
CH2CH3
trans-2-pentene both alkyl groups are on
opposite sides of the double bond
Chapter 13 Answers and Solutions
CH3CH2
b.
CH3CH2
CH2CH3
C
C
H
C
H
CH3
CH3
CH3
C
CH3
trans-2-butene both alkyl groups are on
opposite sides of the double bond
CH2CH2CH3
C
CH3
C
H
H
a. CH3CH2 CH2CH3
d.
H
C
H
cis-2-hexene both alkyl groups are on
the same side of the double bond
13.54
Br
C
H
H
cis-2-butene both alkyl groups are on
the same side of the double bond
CH3
H
C
C
H
d.
C
H
CH2CH3
trans-3-hexene both alkyl groups are on
opposite sides of the double bond
cis-3-hexene both alkyl groups are on
the same side of the double bond
c.
H
C
CH2CH2CH3
trans-2-hexene both alkyl groups are on
opposite sides of the double bond
Br

c. CH3CH2CHCH3
b.
Cl
Cl


e. CH3CHCHCH3
f. no reaction
OH
CH3 Cl
 
h.
i. CH3CH−CCH3
|
CH3
a. The reaction of H2 in the presence of a Ni catalyst changes alkenes into alkanes. The reactant
must be cyclohexene.
OH

g. CH3CH2CHCH3
13.55
b. Br2 adds to alkenes to give a dibromoalkane. Since there are bromine atoms on carbon 2 and
carbon 3 the double bond in the reactant must have been between carbons 2 and 3.
CH3CH=CHCH2CH3
c. HCl adds to alkenes to give a chloroalkane. The product has three carbons and the double
bond must be between carbons 1 and 2.
CH2=CHCH3
d. An alcohol is formed when H2O adds to an alkene in the presence of acid (H+). The alkene
which adds water to form
OH
H+
+ H2O
Unsaturated Hydrocarbons
13.56
13.57
2 HC≡CH
13.59
13.60
+ 2 H2O + heat
CN

H
C
CH
Styrene is 2
and acrylonitrile is H2C=CH. A section copolymer of styrene and
acrylonitrile would be:
H
13.58
+ 5 O2 → 4 CO2
H
CN H
H
CN H
H
CN
C
C
C
C
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
H
O
||
COCH3 CH3
CH3
|
|
|
CH2CCH2CCH2C
|
|
|
O=CCH3 CH3 O=CCH3
a. methylbenzene; toluene
b. 1-chloro-2-methylbenzene; o-chlorotoluene (1,2 position is ortho, o)
c. 1-ethyl-4-methylbenzene; p-ethyltoluene (1,4 position is para,p)
d. 1,3-diethylbenzene; m-diethylbenzene (1,3 position is meta,m)
CH3
Cl
CH2CH3
CH3
a.
b.
c.
CH3
d.
Cl
CH3
CH3
b. o-bromotoluene, m-bromotoluene,
p-bromotoluene
d. No products
13.61
a. Chlorobenzene
c. Benzenesulfonic acid
13.62
a. Benzene and HNO3 in the presence of sulfuric acid, H2SO4
b. Benzene and sulfur trioxide in the presence of sulfuric acid, H2SO4
c. Benzene and bromine in the presence of iron (III) bromide