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CHAPTER 20
UNSATURATED HYDROCARBONS
SOLUTIONS TO REVIEW QUESTIONS
1.
The sigma bond in the double bond of ethene is formed by the overlap of two sp 2 electron
orbitals and is symmetrical about a line drawn between the nuclei of the two carbon
atoms. The pi bond is formed by the sidewise overlap of two p orbitals which are
perpendicular to the carbon-carbon sigma bond. The pi bond consists of two electron
clouds, one above and one below the plane of the carbon-carbon sigma bond.
Cl
Cl
H
≈
√
C“C
√
≈
2.
H
Cl
≈
√
C“C
√
≈
H
cis-1-2-dichloroethene
Cl
H
trans-1,2-dichloroethene
H Cl
ƒ
ƒ
H¬C¬C ¬H
ƒ
ƒ
Cl H
1,2-dichloroethane
We are able to get cis-trans isomers of ethene because there is no rotation around the
carbon-carbon double bond, so the two structures shown are not the same. With
1,2-dichloroethane, there is free rotation of the carbon-carbon single bond. In the
structure shown, exchanging the chlorine on the right with either hydrogen atom on that
carbon appears to make a different isomer, but rotation makes any of the three positions
equivalent.
3.
Rubber products deteriorate rapidly in smog-ridden areas because ozone, which is present
in smog, causes oxidation of the carbon-carbon double bonds, which changes the
properties of the rubber.
4.
Acetylene presents two different explosion hazards:
(a) It can form an explosive mixture with oxygen or air;
(b) When highly compressed or liquefied, acetylene may decompose violently, either
spontaneously or from a slight shock.
5.
The small molecule released in an elimination reaction can be reacted with and added to
the main elimination product. For example, an alkene can be prepared by release of HX
from an alkyl halide. In turn, HX can be added to an alkene to form an alkyl halide.
6.
During the 10-year period from 1935 to 1945, the major source of aromatic hydrocarbons
shifted from coal tar to petroleum due to the rapid growth of several industries which
used aromatic hydrocarbons as raw material. These industries include drugs, dyes,
detergents, explosives, insecticides, plastics, and synthetic rubber. Since the raw material
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needs far exceeded the aromatics available from coal tar, another source had to be found,
and processes were developed to make aromatic compounds from alkanes in petroleum.
World War II, which occurred during this period, put high demands on many of these
industries, particularly explosives.
7.
Benzene does not undergo the typical reactions of an alkene. Benzene does not decolorize
bromine rapidly and it does not destroy the purple color of permanganate ions. The
reactions of benzene are more like those of an alkane. Reaction of benzene with chlorine
requires a catalyst. Benzene does not readily add Cl 2 , but rather a hydrogen atom is
replaced by a chlorine atom.
Fe "
C H Cl + HCl
C H + Cl
6
6
2
6
5
8.
The 11-cis isomer or retinal combines with a protein (opsin) to form the visual pigment
rhodopsin. When light is absorbed, the 11-cis double bond is converted to a trans-double
bond. This process initiates the mechanism of visual excitation which our brains perceive
as light or light forms.
9.
Polycyclic aromatic hydrocarbons are potent carcinogens.
10.
Cyclohexane carbons form bond angles of about 109° in a tetrahedron; this causes the
ring to be either a “boat” or a “chair” shape. Benzene carbons form bond angles of 120°
in a plane. Therefore, the benzene molecule must be planar.
11.
According to the mechanism, the addition of an unsymmetrical module such as HX adds
to a carbon-carbon double bond. In the first step, the H adds to the carbon of the carboncarbon double bond that has the most hydrogen atoms on it, according to Markovnikoff’s
Rule. The second step completes the addition by adding the more negative element, X of
the HX.
12.
“Cracking” means breaking into pieces, which, according to the pyrolysis products, is
what occurs in the reaction.
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CHAPTER 20
SOLUTIONS TO EXERCISES
1.
(a)
ethane
H H
H≠•
C≠•
C≠H
¶
¶
H H
(b)
propane
H H H
H≠•
C≠•
C≠•
C≠H
¶
¶
¶
H H H
(b)
2.
(a)
3.
Isomeric iodobutenes, C4H 7I
H
H
≈
√
C“C
√
≈
CH 3CH 2
I
cis-1-iodo-1-butene
ethene
H H
•
C
C≠≠•
¶
¶
H H
(c)
propene
H
H≠•
C≠•
C≠≠•
C≠H
¶
H H H
(c)
H≠C≠≠≠C≠H
H
I
≈
√
C“C
√
≈
CH 3CH 2
H
trans-1-iodo-1-butene
CH 3CH 2CI “ CH 2
2-iodo-1-butene
CH 3CHICH “ CH 2
3-iodo-1-butene
CH 2ICH 2CH “ CH 2
4-iodo-1-butene
H
I
≈
√
C“C
√
≈
CH 3
CH 3
cis-2-iodo-2-butene
H
CH 3
≈
√
C“C
√
≈
CH 3
I
trans-2-iodo-2-butene
H
H
≈
√
C“C
√
≈
CH 3
CH 2I
cis-1-iodo-2-butene
H
CH 2I
≈
√
C“C
√
≈
CH 3
H
trans-1-iodo-2-butene
ethyne
CH 3
ƒ
CH 3C “ CHI
1-iodo-2-methylpropene
CH 3
ƒ
CH 2IC “ CH 2
3-iodo-2-methylpropene
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propyne
H
H≠•
C≠C≠≠≠C≠H
¶
H
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4.
(a)
(b)
C3H 5Cl
CH 3
H
≈
√
C“C
√
≈
H
Cl
trans-1-chloropropene
H
H
cis-1-chloropropene
CH 3CCl “ CH 2
2-chloropropene
CH 2ClCH “ CH 2
3-chloropropene
CH 3
Cl
≈
√
C“C
√
≈
chlorocyclopropane
H
Cl
5.
(a)
CH 3
CH 3
ƒ
ƒ
CH 3CHCH “ CHCHCH 3
(b)
2,5-dimethyl-3-hexene
CH 3
ƒ
CH 3
CHCH 3
≈
√
C“C
√
≈
H
H
cis-4-methyl-2-pentene
CH 3CH 2
(c)
CH ‚ CCH “ CHCH 3
(d)
H
≈
√
C“C
√
≈
H
trans-3-hexene
3-penten-1-yne
CH 2CH 3
CH3
(e)
CH ‚ CCHCH 2CH 3
ƒ
CH 3
(f)
3-methyl-1-pentyne
6.
(a)
CH 2CH 3
ƒ
CH 2 “ CHCCH 2CH 3
ƒ
CH 3
3-ethyl-3-methyl-1-pentene
CH3CHCHCH2CH2CH3
3-methyl-2-phenylhexane
H
H
(b)
√
≈
C“C
cis, 1,2-diphenylethene
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(c)
(d)
CH ‚ CCHCH3
3-phenyl-1-butyne
(e)
cyclopentene
(f)
CH3
1-methylcyclohexene
7.
(a)
(c)
8.
(a)
(b)
3-isopropylcyclopentene
CH 3
ƒ
CH 3CH 2C “ CH 2
Numbering was started from the wrong end of the structure.
Correct name: 2-methyl-1-butene
CH 3CH 2
(b)
CH(CH3)2
≈
√
C“C
√
≈
CH 3
H
H
Numbering was started from the wrong end of the structure.
Correct name: cis-2-pentene
CH 3CH 2
CH 3
≈
√
C“C
√
≈
H
CH 3
This compound does not have cis-trans isomers. Correct name: 2-methyl-2-pentene
CH 2 “ CHCHCH 3
ƒ
CH 2CH 3
Longest chain contains five carbon atoms. Correct name: 3-methyl-1-pentene
Cl
The C “ C bond in cyclohexene is numbered so that substituted groups have the
smallest numbers. Correct name: 1-chlorocyclohexene
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CH 3
(c)
H
≈
√
C“C
√
≈
H
CH 2CH 2CH 3
Numbering was started from the wrong end of the structure.
Correct name: trans-2-hexene
9.
(a)
(b)
(c)
trans-3-methyl-3-hexene
3-phenyl-1-propyne
4,5-dibromo-2-hexyne
10.
(a)
(b)
(c)
cis-4-methyl-2-hexene
2,3-dimethyl-2-butene
3-isopropyl-1-pentene
11.
All the hexynes, C6H 10
CH 3CH 2CH 2CH 2C ‚ CH
1-hexyne
CH 3
ƒ
CH 3CH 2CHC ‚ CH
3-methyl-1-pentyne
CH 3
ƒ
CH 3CHC ‚ CCH 3
4-methyl-2-pentyne
12.
CH 3CH 2CH 2C ‚ CCH 3
2-hexyne
CH 3CH 2C ‚ CCH 2CH 3
3-hexyne
CH 3
ƒ
CH 3CHCH 2C ‚ CH
4-methyl-1-pentyne
CH 3
ƒ
CH 3C ¬ C ‚ CH
ƒ
CH 3
3,3-dimethyl-1-butyne
All the pentynes, C5H 8
CH 3CH 2CH 2C ‚ CH
1-pentyne
CH 3CH 2C ‚ CCH 3
2-pentyne
CH 3
ƒ
CH 3CHC ‚ CH
3-methyl-1-butyne
13.
Only structure (b) will show cis-trans isomers. CH 3CH “ CHCl
14.
Only structure (c) will show cis-trans isomers. CH 2ClCH “ CHCH 2Cl
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15.
(a)
(b)
(c)
CH 3CH 2CH 2CH “ CH 2 + Br2 ¡ CH 3CH 2CH 2CHBrCH 2Br
I
ƒ
CH 3CH 2C “ CHCH 3 + HI ¡ CH 3CH 2CCH 2CH 3
ƒ
ƒ
CH 3
CH 3
CH 3CH 2CH “ CH 2 + H 2O
CH “ CH2
±
(d)
16.
H2
H+
" CH CH CHCH
3
2
3
ƒ
OH
Pt, 25°C
1 atm
H O
(e)
2
CH 3CHCHCH 3
CH 3CH “ CHCH 3 + KMnO4 99:
cold
ƒ ƒ
HO OH
(a)
CH 3CH 2CH 2CH “ CH 2 + H 2O
H+
(b)
" CH CH CH CHCH
3
2
2
3
ƒ
OH
CH 3CH 2CH “ CHCH 3 + HBr ¡
CH 3CH 2CHBrCH 2CH 3 + CH 3CH 2CH 2CHBrCH
(c)
CH 2 “ CHCl + Br2 ¡ CH 2BrCHClBr
CH “ CH2
CHClCH3
±
(d)
17.
CH2CH3
HCl
¡
H O
(e)
2
CH 2 “ CHCH 2CH 3 + KMnO4 99:
CH 2CHCH 2CH 3
cold
ƒ
ƒ
OH OH
(a)
ethyne
CH ‚ CH
2-butyne
CH 3C ‚ CCH 3
propyne
CH 3C ‚ CH
(b)
(c)
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18.
(a)
(b)
(c)
19.
1-butyne
CH ‚ CCH 2CH 3
propyne
CH ‚ CCH 3
2-pentyne
CH 3C ‚ CCH 2CH 3
(a)
(b)
CH 3C ‚ CCH 3 + Br2 (1 mole) ¡ CH 3CBr “ CBrCH 3
Two-step reaction:
HCl "
CH ‚ CH + HCl ¡ CH 2 “ CHCl
CH 3CHCl 2
(c)
CH 3CH 2CH 2C ‚ CH + H 2 (1 mole)
Pt "
25°C
CH 3CH 2CH 2CH “ CH 2
Pt, 25°C
CH 3C ‚ CH + H 2 (1 mol) 1 atm " CH 3CH “ CH 2
CH 3C ‚ CCH 3 + Br2 (2 mol) ¡ CH 3CBr2CBr2CH 3
Two step reaction:
HCl "
CH 3C ‚ CH + HCl ¡ CH 3CCl “ CH 2
CH 3CCl 2CH 3
20.
(a)
(b)
(c)
21.
When cyclohexene,
reacts with:
Br
(a)
Br2 , the product is
1,2-dibromocyclohexane
Br
I
(b)
HI, the product is
(c)
H 2O, H +, the product is
(d)
KMnO4(aq), the product is
iodocyclohexane
OH
cyclohexanol
OH
OH
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cyclohexene glycol or
1,2-dihydroxycyclohexane
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22.
When cyclopentene,
(a)
reacts with:
Cl
Cl 2 , the product is
1,2-dichlorcyclopentane
Cl
23.
Br
(b)
HBr, the product is
(c)
H 2 , Pt, the product is
(d)
H 2O, H +, the product is
(a)
Two possible alkenes will yield the same product, CH 2 “ C(CH 3)CH 2CH 3
(2-methyl-1-butene); and CH 3C(CH 3) “ CHCH 3 (2-methyl-2-butene)
CH 3C(CH 3)2CH “ CH 2 (3,3-dimethyl-1-butene)
CH 2 “ C(CH 3)2 (2-methyl-1-propene)
(b)
(c)
bromocyclopentane
cyclopentane
OH
cyclopentanol
CH3
(d)
H3C
24.
(a)
(b)
(c)
(d)
CH 3C ‚ CCH 2CH 2CH 3 2-hexyne
CH 3C(CH 3)2C ‚ CH 3,3-dimethyl-1-butyne
HC ‚ CCH 2C(CH 3)3 4,4-dimethyl-1-pentyne
CH 3C ‚ CCH 3 2-butyne
25.
Reactions to convert 2-butyne to:
(a) 2,3-dibromobutane
CH 3C ‚ CCH 3 + Br2 (1 mole) ¡ CH 3CBr “ CBrCH 3
Pt, 25°C "
CH 3CBr “ CBrCH 3 + H 2
CH 3CHBrCHBrCH 3
(3,6-dimethylcyclohexene)
1 atm
(b)
(c)
2,2-dibromobutane
CH 3C ‚ CCH 3 + HBr ¡ CH 3CH “ CBrCH 3
HBr
" CH CH CBr CH
3
2
2
3
2,2,3,3-tetrabromobutane
CH 3C ‚ CCH 3 + Br2 (2 moles) ¡ CH 3CBr2CBr2CH 3
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26.
(a)
CH 3C ‚ CCH 2CH 3 + Cl 2 (1 mole) ¡ CH 3CCl “ CClCH 2CH 3
Pt, 25°C
CH 3CCl “ CClCH 2CH 3 + H 2
(b)
1 atm
" CH CHClCHClCH CH
3
2
3
CH 3C ‚ CCH 2CH 3 + HCl ¡ CH 3CCl “ CHCH 2CH 3
CH 3CCl “ CHCH 2CH 3 + HCl ¡ CH 3CCl 2CH 2CH 2CH 3
Can also yield CH 3CH 2CCl 2CH 2CH 3 at the same time.
(c)
27.
(a)
28.
(a)
(c)
29.
(a)
CH 3C ‚ CCH 2CH 3 + 2 Cl 2 ¡ CH 3CCl 2CCl 2CH 2CH 3
(b)
(c)
COOH
CH3
(b)
CH2CH3
OH
(d)
NH2
(b)
Br
(d)
Cl
Br
Br
Br
1,3,5-tribromobenzene
(c)
C(CH3)3
o-bromochlorobenzene
(d)
CH3
p-xylene
tert-butylbenzene
30.
(a)
Cl
(b)
CH3
ƒ
CH
NO2
NO2
m-dinitrobenzene
NO2
Cl
1,3-dichloro-5-nitrobenzene
(c)
CH3
(d)
1,1-diphenylethane
CH “ CH2
phenylethene (styrene)
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31.
(a)
bromodichlorobenzenes
Cl
Cl
Cl
Cl
Br
Br
3-bromo-1,2-dichlorobenzene
4-bromo-1,2-dichlorobenzene
Cl
Cl
Br
Cl
Br
Cl
2-bromo-1,3-dichlorobenzene
4-bromo-1,3-dichlorobenzene
Cl
Br
Cl
Cl
Cl
Br
5-bromo-1,3-dichlorobenzene
(b)
2-bromo-1,4-dichlorobenzene
The toluene derivatives of formula C9H 12 :
CH3
CH3
CH3
CH3
CH3
CH3
CH3
1,2,3-trimethylbenzene
1,2,4-trimethylbenzene
H3C
1,3,5-trimethylbenzene
CH3
CH3
CH3
CH2CH3
CH2CH3
o-ethyltoluene
m-ethyltoluene
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CH3
CH2CH3
p-ethyltoluene
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32.
(a)
trichlorobenzenes
Cl
Cl
Cl
Cl
Cl
1,2,3-trichlorobenzene
(b)
Cl
Cl
Cl
1,2,4-trichlorobenzene
Cl
1,3,5-trichlorobenzene
The benzene derivatives of formula C8H 10 :
CH3
CH3
CH3
CH3
o-xylene or 1,2-dimethylbenzene
m-xylene or 1,3-dimethylbenzene
CH2CH3
H3C
CH3
p-xylene or 1,4-dimethylbenzene
33.
ethylbenzene
All isomers that can be written by substituting another chlorine in
o-chlorobromobenzene.
(a)
(b)
Cl
Cl
Br
Br
Cl
Cl
2-bromo-1,4-dichlorobenzene
2-bromo-1,3-dichlorobenzene
(c)
(d)
Br
Cl
Br
Cl
Cl
Cl
1-bromo-2,4-dichlorobenzene
1-bromo-2,3-dichlorobenzene
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34.
All isomers that can be written substituting a third chlorine atom on o-dichlorobenzene.
Cl
Cl
Cl
Cl
Cl
Cl
1,2,3-trichlorobenzene
1,2,4-trichlorobenzene
35.
(a)
(b)
(c)
p-chloroethylbenzene
propylbenzene
m-nitroaniline
(d)
(e)
p-bromophenol
triphenylmethane
36.
(a)
(b)
(c)
styrene
m-nitrotoluene
2,4-dibromobenzoic acid
(d)
(e)
isopropylbenzene
2,4,6-tribromophenol
37.
(a)
(b)
(c)
2-chloro-1,4-diiodobenzene
o-dibromobenzene
m-chloroaniline
38.
(a)
(b)
(c)
2,3-dibromophenol
1,3,5-trichlorobenzene
m-iodotoluene
39.
(a)
±
Br2
Br
FeBr3
±
HBr
bromobenzene
(b)
CH3
CH3
± HNO3
CH3
NO2
H2SO4
± H2O
CH3
1,4-dimethyl-2-nitrobenzene
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40.
(a)
±
(b)
AlCl3
CH3CHCH3
ƒ
Cl
CH3+KMnO4
CH3
ƒ
CHCH3+HCl
isopropylbenzene
COOH
H2O
benzoic acid
41.
CH 3
ƒ
When CH 2 “ CCH 2CH 2CH 3 reacts with HBr, two products are possible:
CH 3
CH 3
ƒ
ƒ
CH 3CBrCH 2CH 2CH 3 and CH 2BrCHCH 2CH 2CH 3
The first will strongly predominate. This is the product according to Markovnikov’s rule
and forms because the tertiary carbocation intermediate formed is more stable than a
primary carbocation.
42.
Two tests can be used. (1) Baeyer test—hexene will decolorize KMnO4 solution;
cyclohexane will not. (2) In the absence of sunlight, hexene will react with and decolorize
bromine; cyclohexane will not.
43.
Methylpentenes that show geometric isomerism.
CH 3
ƒ
CH 3CH “ CCH 2CH 3
3-methyl-2-pentene
CH 3CH “ CHCHCH 3
4-methyl-2-pentene
ƒ
CH 3
44.
(a)
+
CH 3
methyl carbocation
+
(b)
CH 3CH 2CH 2
propyl carbocation
(c)
CH 3
ƒ
CH 3C +
ƒ
CH 3
tert-butyl carbocation
(d)
CH 3CH 2CH 2CH 2CH 2
+
pentyl carbocation
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45.
(a)
Br
±
Br2
1,2-dibromocyclohexane
¡
Br
(b)
Cl
±
(c)
HCl
chlorocyclohexane
¡
CH3
CH3
± HCl ¡
Cl
1-chloro-1-methylcyclohexane
(d)
Cl
± HCl ¡
±
Cl
CH3
46.
CH3
CH3
2-chloro-1-methylcyclohexane
3-chloro-1-methylcyclohexane
The reaction mechanism by which benzene is brominated in the presence of FeBr3 :
(a) FeBr3 + Br2 ¡ FeBr 4- + Br +
Formation of a bromonium ion (Br +), an electrophile.
(b)
H
Br
±
±
Br ± ¡
The bromonium ion adds to benzene forming a carbocation intermediate.
(c)
H
Br
Br
±
± FeBr 4– ¡
± FeBr3 ± HBr
A hydrogen ion is lost from the carbocation forming the product bromobenzene.
47.
¬ HCl
(a)
CH 3CHClCH 2CH 3
(b)
(c)
CH 2ClCH 2CH 2CH 2CH 3
Cl
" CH “ CHCH CH + CH CH “ CHCH
2
2
3
3
3
¬ HCl
" CH “ CHCH CH CH
2
2
2
3
–HCl
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48.
Yes, there will be a color change (loss of Br2 color). The fact that there is no HBr formed
indicates that the reaction is not substitution but addition. Therefore, C4H 8 , must contain
a carbon-carbon double bond. Three structures are possible.
CH 3
ƒ
CH 3CH 2CH “ CH 2
CH 3CH “ CHCH 3
CH 3C “ CH 2
49.
Baeyer test: Add KMnO4 solution to each sample. The KMnO4 will lose its purple color
with 1-heptene. There will be no reaction (no color change) with heptane.
50.
The carbon-carbon bonds in the three molecules are different. Ethane has a single bond
between carbon atoms formed from the overlap of two sp 3 hybridized orbitals. Ethene has
a double bond between carbon atoms. One bond is a sigma bond formed by the overlap of
two sp 2 hybridized orbitals, the other is a pi bond formed by the sidewise overlap of two
p orbitals. Ethyne has a triple bond between carbon atoms—one sigma bond formed from
the overlap of two sp hybridized orbitals and two pi bonds formed from the sidewise
overlap of p orbitals.
51.
Other possibilities are
cyclopentane
CH3
CH2CH3
CH3 CH3
(cis)
CH3
CH3
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CH3
CH3
(trans)
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52.
Benzene compounds of formula C8H 9Cl
CH2CH3
Cl
CH2CH3
CH2CH3
CHClCH3
CH2CH2Cl
Cl
Cl
CH3
CH3
CH3
CH3
CH3
CH2Cl
Cl
Cl
CH3
CH3
CH3
CH2Cl
Cl
CH3
CH3
Cl
CH3
Cl
CH3
CH2Cl
Cl
CH3
53.
(a)
(b)
CH3
CH 3
ƒ
CH 2 “ CH ¬ C ¬ CH 2 ¬ CH 3
ƒ
sp2 CH 3
sp2
¯˚˚˘˚˚˙
all sp 3
HC ‚ C ¬ CH 3
sp
sp sp 3
(c)
sp3
CH3
all sp2
H3C
sp3
- 327 -
CH3
HEINS20-311-328.v2.qxd
12/9/07
10:46 PM
Page 328
- Chapter 20 -
54.
(a)
Four isomers:
C4H8
(b)
One isomer
C5H8
CH 2 “ CHCH 2CH 3
1-butene
CH 2 “ C(CH 3)2
2-methyl-1-propene
CH 3CH “ CHCH 3
2-butene (cis and trans)
H
C
HC
H2C
CH2
CH2
cyclopentene
55.
(b) is the correct structure. (a) is an incorrect structure because the carbons where the two
rings are fused have five bonds, not four bonds to each carbon.
56.
Carbon-2 has two methyl groups on it. The configuration of two of the same groups on a
carbon of a carbon-carbon double bond does not show cis-trans isomerism.
57.
(a)
(b)
(c)
(d)
58.
Chemically distinguishing between benzene, 1-hexene, and 1-hexyne
Step 1. Add KMnO4 solution to a sample of each liquid. Benzene is the only one in
which the KMnO4 does not lose its purple color.
Step 2. To 0.5 mL samples of 1-hexene and 1-hexyne add bromine solution dropwise
until there is no more color change of the bromine (from reddish-brown to colorless).
1-hexyne (with a triple bond) will decolor about twice as many drops of bromine as
1-hexene. Thus the three liquids are identified.
59.
(a)
CH 3C ‚ CH
(b)
CH 3C ‚ CH
(c)
CH 3C ‚ CH
alkyne or cycloalkene
alkene
alkane
alkyne or cycloalkene
HCl
Br2
" CH CCl “ CH
3
2
HBr
" CH CBr “ CHB
3
r
2Cl 2
" CH CCl CHCl
3
2
2
- 328 -
HCl
" CH CClBrCH
3
3
" CH CClBrCH Br
3
2