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 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.
2.
The restricted rotation around the carbon-carbon double bond in 1,2-dichloroethene
means two different structural isomers exist, one with two chlorines on the same side of
the double bond (cis) and when the chlorines are on opposite sides of the double bond
(trans). For 1,2-dichloroethyne, the triple bond also has restricted rotation. However, as
all atoms lie in a line, there is only one way to arrange the atoms for 1,2-dichloroethyne.
3.
1-pentene is a liquid at room temperature (25°C) because its boiling point is above room
temperature (30°C). 2-methylpropene is a gas at room temperature because its boiling
point (–7°C) is much lower than 25°C.
4.
1-butene has a much lower melting point (–185°C) than 2-methylpropene (–14°C)
although their boiling points are almost the same (1-butene, –6°C; 2-methylpropene,
–7°C). 1-butene has a much different molecular shape than 2-methylpropene. This shape
difference has a great effect on melting points because the molecules fit closely together
in a solid. In a liquid the molecules are rapidly moving and molecular shape differences
are much less important. Thus, the boiling points for these two molecules are almost
the same.
5.
Trans fats contain double bonds in the trans isomer form. In contrast, most naturally
occurring unsaturated fats have double bonds in the cis form. Our bodies can’t
metabolized the trans fats because they are the wrong shape. The trans fats accumulate
over time and increase the risk of cardiovascular disease.
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
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.
- 277 -
- Chapter 20 -
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
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.
- 278 -
- 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
•
•≠≠C
•≠H
H≠¶
C≠C
H H H
(c)
H≠C≠≠≠C≠H
H
I
≈
√
C“C
√
≈
H
CH 3CH 2
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
C H 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
- 279 -
propyne
H
•
H≠¶
C≠C≠≠≠C≠H
H
- Chapter 20 -
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
≈
√
C“C
√
≈
Cl
chlorocyclopropane
H
Cl
5.
(a)
CH 3
ƒ
CH CHCH
CH 3
ƒ
CHCHCH
(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)
3
(d)
≈
√
C“C
√
≈
H
trans-3-hexene
3-penten-1-yne
H
CH 2CH 3
CH3
(e)
ƒ
CH 3
2
(f)
3
3-methyl-1-pentyne
6.
(a)
3-methyl-2-phenylhexane
CH 2CH 3
ƒ
2
2
CH3CHCHCH2CH2CH3
H
3
ƒ
CH 3
3-ethyl-3-methyl-1-pentene
(b)
√
≈
C“C
H
cis, 1,2-diphenylethene
- 280 -
- Chapter 20 -
(c)
(d)
CH ‚ CCHCH3
cyclopentene
3-phenyl-1-butyne
(e)
(f)
CH3
1-methylcyclohexene
CH(CH3)2
3-isopropylcyclopentene
7.
(a) 23 sigma bonds, 1 pi bond; (b) 17 sigma bonds, 1 pi bond; (c) 10 sigma bonds,
3 pi bonds; (d) 17 sigma bonds, 1 pi bond; (e) 15 sigma bonds, 2 pi bonds; (f) 33 sigma
bonds, 3 pi bonds.
8.
(a) 23 sigma bonds, 1 pi bond; (b) 27 sigma bonds, 7 pi bonds; (c) 20 sigma bonds, 5 pi
bonds; (d) 13 sigma bonds, 1 pi bond; (e) 19 sigma bonds, 1 pi bond; (f) 21 sigma bonds,
1 pi bond.
9.
10.
(a)
CH 3
ƒ
CH 3CH 2C “ CH 2
Numbering was started from the wrong end of the structure.
Correct name: 2-methyl-1-butene
(b)
CH 3CH 2
(c)
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
(a)
≈
√
C“C
√
≈
CH 3
CH 2 “ CHCHCH 3
ƒ
CH 2CH 3
Longest chain contains five carbon atoms. Correct name: 3-methyl-1-pentene
- 281 -
- Chapter 20 -
(b)
(c)
Cl
The
bond in cyclohexene is numbered so that substituted groups have the
smallest numbers. Correct name: 1-chlorocyclohexene
CH 3
H
≈
√
C“C
√
≈
H
CH 2CH 2CH 3
Numbering was started from the wrong end of the structure.
Correct name: trans-2-hexene
11.
(a) trans-6-chloro-3-heptene; (b) 4,4-dimethyl-2-pentyne; (c) 2-ethyl-1-pentene.
12.
(a) 3-phenyl-1-butyne; (b) 2-methyl-2-hexene; (c) cis-3,4-dimethyl-3-hexene.
13.
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
14.
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
15.
CH 3CH 2CH 2C ‚ CCH 3
2-hexyne
CH 3CH 2C ‚ CCH 3
2-pentyne
CH 3
ƒ
CH 3CHC ‚ CH
3-methyl-1-butyne
(a) 1-butyne (The smallest numbered carbon that is involved in the triple bond is used to
number the triple bond.); (b) 2-hexyne (The parent chain is the longest continuous carbon
chain that contains the triple bond.); (c) propyne (The triple bond has only one possible
location in propyne and no number is needed.)
- 282 -
- Chapter 20 -
16.
(a) 4-methyl-2-pentyne (The parent chain is numbered from the end closest to the triple
bond.); (b) 4-methyl-2-hexyne (The parent chain is the longest continuous carbon chain
that contains the triple bond.); (c) 3-bromo-1-butyne (The parent chain is numbered from
the end closest to the triple bond.).
17.
Cis-trans isomers exist for (c) only. Alkynes do not have cis-trans isomerism. Alkenes
have cis-trans isomerism only when each double-bonded carbon is attached to two
different groups.
18.
Cis-trans isomers exist for (b) only. Alkynes do not have cis-trans isomerism. Alkenes
have cis-trans isomerism only when each double-bonded carbon is attached to two
different groups.
19.
(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)
20.
±
H2
H+
" CH CH CHCH
2
3
3
ƒ
OH
Pt, 25°C
1 atm
CH2CH3
H O
(e)
2
CH 3CH “ CHCH 3 + KMnO4 99:
cold: CH 3CHCHCH 3
ƒ ƒ
HO OH
(a)
CH 3CH 2CH 2CH “ CH 2 + H 2O
(b)
(c)
" CH CH CH CHCH
2
2
3
3
ƒ
OH
CH 3CH 2CH “ CHCH 3 + HBr ¡
CH 3CH 2CHBrCH 2CH 3 + CH 3CH 2CH 2CHBrCH3
CH 2 “ CHCl + Br2 ¡ CH 2BrCHClBr
CH “ CH2
(d)
H+
±
HCl
¡
- 283 -
CHClCH3
- Chapter 20 -
21.
H O
(e)
2
CH 2 “ CHCH 2CH 3 + KMnO4 99:
cold: CH 2CHCH 2CH 3
ƒ
ƒ
OH OH
(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)
CH 3CH “ CH 2
1 atm
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
22.
(a)
(b)
(c)
23.
When cyclohexene,
reacts with:
Br
(a)
the product is
1,2-dibromocyclohexane
Br
I
(b)
HI, the product is
iodocyclohexane
(c)
H O H + the product is
(d)
KM O (
OH
cyclohexanol
OH
) the product is
OH
24.
When cyclopentene,
(a)
Cl
cyclohexene glycol or
1,2-dihydroxycyclohexane
reacts with:
Cl
the product is
1,2-dichlorcyclopentane
Cl
(b)
HBr, the product is
Br
bromocyclopentane
- 284 -
- Chapter 20 -
25.
(c)
H
(d)
H O 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
(2-methyl-2-butene)
(3,3-dimethyl-1-butene)
(2-methyl-1-propene)
(b)
(c)
P the product is
cyclopentane
OH
cyclopentanol
CH3
(d)
26.
H3C
(3,6-dimethylcyclohexene)
(a)
(b)
(c)
(d)
2-hexyne
3,3-dimethyl-1-butyne
4,4-dimethyl-1-pentyne
2-butyne
CH 3
27.
(a)
CH
C
CH
CH 3 + HBr
CH 2
Br
CH 3
C
CH
CH 3
(b)
CH
C
CH
CH 3 + 2 HBr
CH 3
Br
CH 3
C
CH
CH 3
CH 3
Br
CH 3
(c)
CH
C
CH
CH 3 + 2 Br2
CH3
28.
(a)
CH
C
C
CH2 CH3 + Cl 2
Br
Br
CH 3
CH
C
CH
Br
Br
CH 3
Cl
Cl
CH3
CH
C
C
CH3
CH3
- 285 -
CH2 CH3
- Chapter 20 -
CH3
(b)
CH
C
C
CH2
CH2 CH3 + HCl
Cl
CH3
C
C
CH3
CH3
CH3
(c)
CH
C
C
CH2 CH3 + 2 HCl
CH3
CH3
O
29.
Cl
CH3
C
C
Cl
CH3
CH3
CH
(a)
(c)
CH3
NO 2
30.
(b)
OH
(d)
(a)
NO 2
(c)
CH2
CH
CH2
CH3
(b)
(d)
NO 2
Br
Cl
31.
(a)
I
Cl
(c)
CH3
(b)
CH
CH2 CH3
CH3
(d)
- 286 -
Br
NH2
CH2 CH3
Cl
32.
CH3
(a)
(c)
CH2 CH3
Cl
Cl
CH3
COOH
NO 2
(b)
(d)
Br
33.
(a)
NO 2
NO 2
bromodichlorobenzenes
Cl
Cl
Cl
Cl
Br
Br
3-bromo-1,2-dichlorobenzene
4-bromo-1,2-dichlorobenzene
Cl
Cl
Br
Cl
Cl
Br
2-bromo-1,3-dichlorobenzene
4-bromo-1,3-dichlorobenzene
Cl
Br
Cl
Br
Cl
Cl
2-bromo-1,4-dichlorobenzene
5-bromo-1,3-dichlorobenzene
- 287 -
- Chapter 20 -
(b)
The toluene derivatives of formula C9H 12 :
CH3
CH3
CH3
CH3
CH3
CH3
CH3
1,2,3-trimethylbenzene
H3C
1,2,4-trimethylbenzene
CH3
1,3,5-trimethylbenzene
CH3
CH3
CH3
CH2CH3
CH2CH3
CH2CH3
34.
(a)
p-ethyltoluene
m-ethyltoluene
o-ethyltoluene
trichlorobenzenes
Cl
Cl
Cl
Cl
Cl
Cl
Cl
1,2,3-trichlorobenzene
(b)
Cl
1,2,4-trichlorobenzene
Cl
1,3,5-trichlorobenzene
The benzene derivatives of formula C8H 10 :
CH3
CH3
CH3
CH3
m-xylene or 1,3-dimethylbenzene
o-xylene or 1,2-dimethylbenzene
CH2CH3
H3C
CH3
ethylbenzene
p-xylene or 1,4-dimethylbenzene
35.
(a)
(b)
(c)
p-chloroethylbenzene
propylbenzene
m-nitroaniline
(d)
(e)
- 288 -
p-bromophenol
triphenylmethane
36.
(a)
(b)
(c)
styrene
m-nitrotoluene
2,4-dibromobenzoic acid
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
(d)
(e)
isopropylbenzene
2,4,6-tribromophenol
Br
FeBr3
±
HBr
bromobenzene
(b)
CH3
CH3
± HNO3
NO2
H2SO4
CH3
± H2O
CH3
1,4-dimethyl-2-nitrobenzene
40.
(a)
±
(b)
AlCl3
CH3CHCH3
ƒ
Cl
CH3+KMnO4
CH3
ƒ
CHCH3+HCl
isopropylbenzene
H2O
COOH
benzoic acid
- 289 -
- Chapter 20 -
41.
CH 3
ƒ
When
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.
43.
Two tests can be used. (1) Baeyer test—hexene will decolorize
solution;
cyclohexane will not. (2) In the absence of sunlight, hexene will react with and decolorize
bromine; cyclohexane will not.
CH 3
C
C
H
CH 3
H
H
C
CH 3
CH 2 CH 3
CH 3
C
C
CH 3
H
C
CH 2 CH 3
CH 3
trans -3,4-dimethyl-2-hexene
cis-3,4-dimethyl-2-hexene
CH 3
H
CH 3 CH 3
C
C
H
CH 2 CHCH3
CH 3
trans-3,5-dimethyl-2-hexene
CH 3
H
C
H
C
CH 2 CHCH 3
cis-3,5-dimethyl-2-hexene
CH 3
H
CHCH 3
CH 3
C
CH
CH 3 CH 3
C
CH 3
C
H
CH 3
CH
CHCH3
C
CH 3
trans -4,5-dimethyl-2-hexene
cis-4,5-dimethyl-2-hexene
- 290 -
- Chapter 20 -
43.
(cont.)
H
CH 3
C
C
H
C
H
H
CH 3
C
C
CH 3
CH 2 CH 3
CH 3
CH 3
C
CH 2 CH 3
trans-4,4-dimethyl-2-hexene
CH 3
cis-4,4-dimethyl-2-hexene
CH 3
H
CH 3 CH 3
C
C
H
CH 3 CH 3
CH 3
CH 2 CCH 3
C
C
CH 2 CCH 3
CH 3
CH 3
cis-3,5,5-timethyl-2-hexene
trans-3,5,5-trimethyl-2-hexene
CH 3
H3 C
CH 2 CH 3
C
C
CH 3 CH 2
C
CH 3 CH 2
CH 3
trans -3,4-dimethyl-3-hexene
H
CH 2 CH 3
H
CH 3
C
CH 3 CH
C
cis-3,4-dimethyl-3-hexene
CH 2 CH 3
C
CH 3
C
CH 3
C
CH 3 CH
CH 3
trans -2,4-dimethyl-3-hexene
CH 2 CH 3
CH 3
cis-2,4-dimethyl-3-hexene
CH 3
H
CHCH 3
C
CH 3 CH
H
H
C
C
H
CH 3 CH
CH 3
CH 3
C
CHCH 3
CH 3
cis-2,5-dimethyl-3-hexene
trans-2,5-dimethyl-3-hexene
- 291 -
- Chapter 20 -
43.
(cont.)
CH 3
CH 3
CH 3 CH 2
CCH 3
H
C
C
CH 3 CH 2
H
H
CH 3
C
C
H
(a)
CH 3CH 2
primary carbocation (a positively charged carbon bonded
to one other carbon)
(b)
CH3 CHCH3
secondary carbocation (a positively charged carbon
bonded to two other carbons)
(c)
CH 3
C
tertiary carbocation (a positively charged carbon bonded
to three other carbons)
CH 3
CH 3
45.
CH 3
cis-2,2-dimethyl-3-hexene
trans -2,2-dimethyl-3-hexene
44.
CCH 3
CH3
CH3
(a)
+
H 2O
OH
CH3
CH3
(b)
+ H 2O
OH
CH3
CH3
CH3
CH3
(c)
+ H 2O
OH
CH3
CH3
- 292 -
- Chapter 20 -
CH3
CH3
(d)
46.
CH3
CH3
OH
+ H 2O
The reaction mechanism by which benzene is brominated in the presence of FeBr3 :
(a) FeBr3 + Br2 ¡ FeBr 4- + Br +
+
Formation of a bromonium ion
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
48.
Yes, there will be a color change (loss of
color). The fact that there is no HBr formed
indicates that the reaction is not substitution but addition. Therefore, 4 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
solution to each sample. The
will lose its purple color
with 1-heptene. There will be no reaction (no color change) with heptane.
50.
I would not expect graphene to react easily with HCl in an addition reaction. Graphene is
composed of sheets of aromatic rings that do not easily undergo addition reactions.
- 293 -
- Chapter 20 -
51.
Br
Br
Br
Cl
Cl
Cl
Cl
Cl
Cl
Br
Cl
Cl
Cl
Br
Cl
Cl
Br
Cl
52.
53.
The double bonds in fatty acids are oxidized. This reaction adds oxygen and the double
bond converts to a single bond. This reaction is thought to start the process that eventually
results in “hardening of the arteries.”
(a)
(b)
CH 3
ƒ
CH 2 “ CH ¬ C ¬ CH 2 ¬ CH 3
ƒ
2
2
C
H3
sp
sp
¯˚˚˘˚˚˙
all sp 3
HC ‚ C ¬ CH 3
sp
sp sp 3
(c)
sp3
CH3
all sp2
H3C
sp3
54.
(a)
Four isomers:
C4H8
CH 2 “ CHCH 2CH 3
1-butene
CH 2 “ C(CH 3)2
2-methyl-1-propene
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CH 3CH “ CHCH 3
2-butene (cis and trans)
- Chapter 20 -
(b)
One isomer
C5H8
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
solution to a sample of each liquid. Benzene is the only one in
which the
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)
CH3C
(c)
CH 3C ‚ CH
alkyne or cycloalkene
alkene
alkane
alkyne or cycloalkene
CH
HCl
" CH CCl “ CH
3
2
Br2
2Cl 2
" CH CBr
3
HBr
CHBr
" CH CCl CHCl
3
2
2
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" CH CClBrCH
3
3
HCl
" CH3 CClBrCH2 Br