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CHAPTER 22
ALCOHOLS, ETHERS, PHENOLS, AND THIOLS
SOLUTIONS TO REVIEW QUESTIONS
1.
The question allows great freedom of choice. These shown here are very simple examples
of each type.
(a) an alkyl halide
CH 3CH 2Cl
(b)
a phenol
(c)
an ether
(d)
an aldehyde
(e)
(f)
a ketone
a carboxylic acid
(g)
(h)
an ester
a thiol
OH
CH 3CH 2OCH 2CH 3
H
ƒ
CH 3C “ O
O
‘
CH 3CCH 3
CH 3COOH
O
‘
CH 3C ¬ OCH 2CH 3
CH 3CH 2SH
2.
Alkenes are almost never made from alcohols because the alcohols are almost always the
higher value material. This is because recovering alkenes from hydrocarbon sources in an
oil refinery (primarily catalytic cracking) is a relatively cheap process.
3.
Isopropyl alcohol is usually used for rubbing alcohol, rather than n-propyl alcohol,
because of price. The easy way to make the alcohol, by adding H 2O to propene, yields
isopropyl alcohol rather than n-propyl alcohol. Any process for producing n-propyl
alcohol would be much more expensive.
4.
Oxidation of primary alcohols yields aldehydes. Further oxidation yields carboxylic acids.
Examples:
O
‘
[O] "
CH 3CH 2OH
CH 3C ¬ H + H 2O
O
O
‘
‘
[O] "
CH 3C ¬ H
CH 3C ¬ OH
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5.
1,2-Ethanediol is superior to methanol as an antifreeze because of its low volatility.
Methanol is much more volatile than water. If the radiator leaks gas under pressure
(normally steam), it would primarily leak methanol vapor so you would soon have no
antifreeze. Ethylene glycol has a lower volatility than water, so it does not present this
problem.
6.
Glucose has five alcohol functional groups; these hydroxyl groups attract water by
hydrogen bonding and cause glucose to be very soluble in the blood stream.
7.
Physiological effects of:
(a) methanol: It is a poisonous liquid capable of causing blindness or death if taken
internally. Exposure to methanol vapors is also very dangerous.
(b) ethanol: It can act as a food, a drug, and a poison. The body can metabolize small
amounts of ethanol to produce energy; thus it is a food. It depresses brain functions
so that activities requiring skill and judgment are impaired; thus it is a drug. With
very large consumption, the depression of brain function can lead to
unconsciousness and death; thus it is a poison.
8.
±
benzene
O2
CH3CH “ CH2
H2SO4
propene
CH3
ƒ
¬ C¬ O ¬ O ¬ H
ƒ
CH3
dilute
H2SO4
10.
OH
±
phenol
cumene hydroperoxide
9.
CH3
ƒ
¬ CH
ƒ
CH3
cumene
(isopropylbenzene)
CH3CCH3
‘
O
acetone
Some common phenols include (a) hydroquinone, a photographic reducer and developer,
(b) vanillin, a flavoring, (c) eugenol, used to make artificial vanillin, (d) thymol, used as
an antiseptic in mouthwashes, (e) butylated hydroxytoluene (BHT) an antioxidant and, (f)
the cresols, used as disinfectants.
Low molar mass ethers present two hazards. They are very volatile and their highly
flammable vapors form explosive mixtures with air. They also slowly react with oxygen
in the air to form unstable explosive peroxides.
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11.
Ethanol (molar mass = 46.07) is a liquid at room temperature because it has a significant
amount of hydrogen bonding between molecules in the liquid state, and thus has a much
higher boiling point than would be predicted from molar mass alone. Dimethyl ether
(molar mass = 46.07) is not capable of hydrogen bonding to itself, so has low attraction
between molecules, making it a gas at room temperature.
12.
Fats are a major source of energy in the body. Some infants die from SIDS due to a lack
of energy because they are unable to degrade fats. One of the intermediate compounds in
the degradation of fats is an alcohol, which in the process is oxidized to a ketone.
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CHAPTER 22
SOLUTIONS TO EXERCISES
1.
2.
(a)
methanol
CH 3OH
(b)
(c)
3-methyl-l-hexanol
1,2-propanediol
CH 3
ƒ
CH 3CH 2CH 2CHCH 2CH 2OH
CH 3CHCH 2OH
ƒ
OH
(d)
1-phenylethanol
(e)
2,3-butanediol
(f)
2-propanethiol
(a)
2-butanol
(b)
2-methyl-2-butanol
(c)
2-propanol
(d)
cyclopentanol
(e)
1-pentanethiol
(f)
4-ethyl-3-hexanol
¬ CHCH3
ƒ
OH
CH 3CH ¬ CHCH 3
ƒ
ƒ
OH OH
CH 3CHCH 3
ƒ
SH
CH 3CHCH 2CH 3
ƒ
OH
CH 3
ƒ
CH 3CCH 2CH 3
ƒ
OH
CH 3CHCH 3
ƒ
OH
¬ OH
CH 3CH 2CH 2CH 2CH 2SH
CH 2CH 3
ƒ
CH 3CH 2CHCHCH 2CH3
ƒ
OH
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3.
Eight open chain isomeric alcohols with the formula C5H 11OH:
OH
OH
ƒ
ƒ
CH 3CH 2CH 2CH 2CH 2OH
CH 3CH 2CH 2CHCH 3
CH 3CH 2CHCH 2CH 3
1-pentanol
2-pentanol
3-pentanol
CH 3
CH 3
CH 3
ƒ
ƒ
ƒ
CH 3CH 2CCH 3
CH 3CHCH 2CH 2OH
CH 3CH 2CHCH 2OH
ƒ
OH
3-methyl-l-butanol
2-methyl-l-butanol
2-methyl-2-butanol
CH 3
CH 3
ƒ
ƒ
CH 3CH CHCH 3
CH 3CCH 2OH
ƒ
ƒ
OH
CH 3
3-methyl-2-butanol
2,2-dimethyl-l-propanol
4
(a)
C3H 8O
(b)
C4H 8O
Alcohols
CH 3CH 2CH 2OH
CH 3CH(OH)CH 3
Alcohols
CH 3CH 2CH 2CH 2OH
CH 3CH 2CH(OH)CH 3
(CH 3)2CHCH 2OH
(CH 3)3C ¬ OH
Ethers
CH 3OCH 2CH 3
Ethers
CH 3OCH 2CH 2CH 3
CH 3OCH(CH 3)2
CH 3CH 2OCH 2CH 3
5.
Of the eight compounds in question 3, the primary alcohols are: 1-pentanol; 3-methyl-1butanol; 2-methyl-l-butanol; and 2,2-dimethyl-l-propanol. The secondary alcohols are: 2pentanol; 3-pentanol; and 3-methyl-2-butanol. The tertiary alcohol is 2-methyl-2-butanol.
6.
The primary alcohols in question 4 are:
CH 3CH 2CH 2OH
CH 3CH 2CH 2CH 2OH
CH 3CH(OH)CH 3
The secondary alcohols are:
The tertiary alcohol is:
(CH 3)3C ¬ OH
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CH 3CH 2CH(OH)CH 3
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7.
8.
9.
10.
The names of the compounds are:
(a) 1-butanol (butyl alcohol)
(b) 2-propanol (isopropyl alcohol)
(c) 2-methyl-3-phenyl-l-propanol
(d) oxirane (ethylene oxide)
(e)
(f)
(g)
2-methyl-2-butanol
2-methylcyclohexanol
2,3-dimethyl-1,4-butanediol
The names of the compounds are
(a) ethanol (ethyl alcohol)
(b) 2-phenylethanol
(c) 3-methyl-3-pentanol
(d) 1-methylcyclopentanol
(e)
(f)
(g)
3-pentanol
1,2-propanediol
4-ethyl-2-hexanol
Chief product of dehydration
CH 3
ƒ
(a) CH 3C “ CHCH 3
(b) CH 3CH “ CHCH 2CH 3
2-methyl-2-butene
2-pentene
(c)
cyclohexene
Chief product of dehydration
CH3
(a)
1-methylcyclopentene
(b)
(c)
11.
(a)
CH 3CH “ CHCH 3
CH 3
ƒ
CH 3C “ CHCH 2CH 3
CH 3CH 2CH 2CHCH 2CH 2CH 3
ƒ
OH
OH
(b)
(c)
CH3
2-butene
2-methyl-2-pentene
4-heptanol
cyclohexanol
3-methylcyclobutanol
HO
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12.
(a)
(b)
(c)
13.
¬ OH
cyclopentanol
ƒ¬ CH3
OH
CH 3 CH 3
ƒ
ƒ
CH 3CH CHCHCH 3
ƒ
OH
1-methylcyclopentanol
2,4-dimethyl-3-pentanol
Oxidation of a primary alcohol
CH3CH2
O
ƒ
‘
CH3CH2CH2CHCH2 ¬ C ¬ H
[O]
CH2CH3
ƒ
CH3CH2CH2CHCH2CH2OH
aldehyde
[O]
O
CH3CH2
ƒ
‘
CH3CH2CH2CHCH2 ¬ C ¬ OH
carboxylic acid
14.
Oxidation of a secondary alcohol
CH 3
CH 3
ƒ
ƒ
[O] "
CH 3CCHCH 2CH 3
CH 3CHCHCH 2CH 3
ƒ
‘
OH
O
ketone
15.
(a)
(b)
(c)
H+
CH 3CH “ CH 2 + H 2O 9: CH 3CHCH 3
ƒ
OH
H+
CH 3CH 2CH “ CH 2 + H 2O 9: CH 3CH 2CHCH 3
ƒ
OH
H+
CH 3CH 2CH “ CHCH 3 + H 2O 9: CH 3CH 2CHCH 2CH 3 + CH 3CH 2CH 2CHCH 3
ƒ
ƒ
OH
OH
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16.
(a)
(b)
(c)
17.
(a)
(b)
(c)
18.
(a)
(b)
(c)
19.
H+
CH 3CH “ CHCH 3 + H 2O 99: CH 3CH 2CHCH 3
ƒ
OH
H+
CH 3CH 2CH 2CH “ CH 2 + H 2O 99: CH 3CH 2CH 2CHCH 3
ƒ
OH
OH
ƒ
H+
CH 3C “ CHCH 3 + H 2O 99: CH 3CCH 2CH 3
ƒ
ƒ
CH 3
CH 3
CH 3CHBrCH 3
2-bromopropane
bromocyclohexane (cyclohexyl bromide)
Br
CH 3
ƒ
CH 3CHCH 2CH 2Br
1-bromo-3-methylbutane
CH 3CHCH 2CH 3
ƒ
OH
CH 2CH 3
ƒ
CH 3CHCHCH 2CH 3
ƒ
OH
2-butanol
3-ethyl-2-pentanol
¬ OH
cyclopentanol
140°C "
CH 3CH 2OCH 2CH 3 + H 2O
diethyl ether
(a)
2 CH 3CH 2OH + H 2SO4
(b)
CH 3CH 2CH 2OH + H 2SO4
(c)
O
‘
K2Cr2O7
C
CH 3CH(OH)CH 2CH 3 99:
CH
3 CH 2CH 3 + H 2O
H SO
2
180°C "
CH 3CH “ CH 2 + H 2O
propene
4
2-butanone
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(d)
O
‘
H+
CH 3CH 2C ¬ OCH 2CH 3 99:
CH 3CH 2COOH + CH 3CH 2OH
H O
2
propanoic acid
20.
(a)
2 CH 3CH 2OH + 2 Na ¡ 2 CH 3CH 2O -Na+ + H 2
(b)
O
‘
K2Cr2O7
CH 3CH 2CH 2CH 2OH 99:
CH
CH
CH
C
¬H
3
2
2
H SO
2
¬ CH “ CH2
4
H2O
(d)
¬ CHCH3
ƒ
OH
O
O
‘
‘
CH 3CH 2C ¬ OCH 3 + NaOH ¡ CH 3CH 2C ¬ O -Na+ + CH 3OH
(a)
o-methylphenol
(c)
21.
ethanol
H2SO4
(b)
m-dihydroxybenzene
OH
OH
CH3
OH
(c)
4-hydroxy-3-methoxybenzaldehyde
H¬C“O
OCH3
OH
22.
(a)
p-nitrophenol
(b)
2,6-dimethylphenol
OH
OH
H3C
NO2
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(c)
o-dihydroxybenzene
OH
OH
23.
(a)
(b)
(c)
2-methyl-l-propanol
OH
ƒ
CH 2 ¬ CH ¬ CH 3
ƒ
CH 3
3-methyl-2-butanol
OH
ƒ
CH 3 ¬ CH ¬ CH ¬ CH 3
ƒ
CH 3
cyclopentanol
¬ OH
24.
(a)
(b)
(c)
Methanol, CH 3 ¬ OH
CH 3
ƒ
2,2-dimethyl-l-propanol CH 3 ¬ C ¬ CH 2 ¬ OH
ƒ
CH 3
CH 3
ƒ
3,3-dimethyl-1-pentanol CH 3 ¬ CH 2 ¬ C ¬ CH 2CH 2 ¬ OH
ƒ
CH 3
25.
(a)
(b)
(c)
methanol
2-methyl- l-propanol
cyclohexanol
26.
(a)
(b)
(c)
cyclopentanol
2-propanol
2-methyl-2-propanol
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27.
(a)
(b)
phenol
m-methylphenol
(c)
(d)
2-ethyl-5-nitrophenol
4-bromo-2-chlorophenol
28.
(a)
(b)
p-dihydroxybenzene (hydroquinone) (c)
2,4-dimethylphenol
(d)
29.
Order of increasing solubility in water [c (lowest), a, b, d (highest)]
(c) CH 3CH 2CH 2CH 2CH 3 (1)
(b) CH 3CH(OH)CH 2CH 2CH 3 (3)
(a) CH 3CH 2OCH 2CH 2CH 3 (2)
(d) CH 3CH(OH)CH(OH)CH 2CH 3 (4)
30.
Order of decreasing solubility in water: [a (highest), d, c, b (lowest)]
(a) CH 3CH(OH)CH(OH)CH 2OH (1)
(c) CH 3CH 2CH 2CH 2OH (3)
(d) CH 3CH(OH)CH 2CH 2OH (2)
(b) CH 3CH 2OCH 2CH 3 (4)
31.
Fourteen isomeric ethers; C6H 14O
2,4-dinitrophenol
m-hexylphenol
CH 3OCH 2CH 2CH 2CH 2CH 3
methyl-n-pentyl ether (1-methoxypentane)
CH 3
ƒ
CH 3OCHCH 2CH 2CH 3
2-methoxypentane
CH 2CH 3
ƒ
CH 3OCHCH 2CH 3
3-methoxypentane
CH 3
ƒ
CH 3OCH 2CHCH 2CH 3
1-methoxy-2-methylbutane
CH 3
ƒ
CH 3OCH 2CH 2CHCH 3
1-methoxy-3-methylbutane
CH 3 CH 3
ƒ
ƒ
CH 3OCH ¬ CHCH 3
2-methoxy-3-methylbutane
CH 3
ƒ
CH 3OCH 2C ¬ CH 3
ƒ
CH 3
1-methoxy-2,2-dimethylpropane
CH 3CH 2OCH 2CH 2CH 2CH 3
n-butyl ethyl ether (1-ethoxybutane)
CH 3
ƒ
CH 3CH 2OCHCH 2CH 3
sec-butyl ethyl ether
(2-ethoxybutane)
CH 3
ƒ
CH 3CH 2OCH 2CHCH 3
ethyl isobutyl ether
(1-ethoxy-2-methylpropane)
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32.
CH 3
ƒ
CH 3CH 2OC ¬ CH 3
ƒ
CH 3
t-butyl ethyl ether
(2-ethoxy-2-methylpropane)
CH 3CH 2CH 2OCH 2CH 2CH 3
di-n-propyl ether
(1-propoxypropane)
CH 3
ƒ
CH 3CH 2CH 2OCHCH 3
isopropyl-n-propyl ether
CH 3 CH 3
ƒ
ƒ
CH 3CHOCHCH 3
diisopropyl ether
Six isomeric ethers: C5H 12O
CH 3
ƒ
CH 3OCHCH 2CH 3
sec-butyl methyl ether
(2-methoxybutane)
CH 3OCH 2CH 2CH 2CH 3
n-butyl methyl ether
(1-methoxybutane)
CH 3
ƒ
CH 3OCCH 3
ƒ
CH 3
t-butyl methyl ether
(2-methoxy-2-methylpropane)
CH 3
ƒ
CH 3OCH 2CHCH 3
isobutyl methyl ether
(1-methoxy-2-methylpropane)
CH 3
ƒ
CH 2CH 2OCHCH 3
ethyl isopropyl ether
(2-ethoxypropane)
CH 3CH 2OCH 2CH 2CH 3
ethyl n-propyl ether
(1-ethoxypropane)
¢"
33.
CH 3CH 2CHCH 3 + 6 O2
ƒ
OH
4 CO2 + 5 H 2O
34.
CH 3CH 2OCH 2CH 3 + 6 O2
35.
Possible combinations of reactants to make the following ethers by the Williamson
synthesis:
CH 3ONa + CH 3CH 2Cl or
(a) CH 3CH 2OCH 3
CH 3CH 2ONa + CH 3Cl
¢"
4 CO2 + 5 H 2O
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(b)
CH2ONa
CH2OCH2CH3
±
CH3CH2Cl
±
CH3CH2ONa
CH2Cl
or
(c)
O ¬ CH2CH3
ONa
±
36.
CH3CH2Cl
Possible combinations of reactants to make the following ethers by the Williamson
synthesis:
CH 3CH 2CH 2ONa + CH 3CH 2CH 2Cl
(a) CH 3CH 2CH 2OCH 2CH 2CH 3
CH 3
ƒ
(b) HCOCH 2CH 2CH 3
ƒ
CH 3
CH 3
ƒ
HCONa + CH 3CH 2CH 2Cl
ƒ
CH 3
(c)
O
cannot be made by the Williamson synthesis. Cannot use
secondary RX.
37.
IUPAC names
(a) 2-methyl-2-butanethiol
(b) cyclohexanethiol
(c) 2-methyl-2-propanethiol
(d) 2-methyl-3-pentanethiol
38.
IUPAC names
(a) 2-methylcyclopentanethiol
(b) 2-propanethiol
(c) 2,3-dimethyl-2-pentanethiol
(d) 3-ethyl-2,2,4-trimethyl-3-pentanethiol
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39.
(a)
CH 3CHCH 3
ƒ
SH
(b)
CH3CH2 CHCH2CH2CH2SH
CH
CH3 CH3
(c)
SH
CH3
40.
(a)
SH
CH2CH3
41.
(b)
CH 3CH 2CHCH 2CH 3
ƒ
SH
(c)
CH 3
ƒ
CH 3CH 2CCH 2SH
ƒ
CH 3
Structures and correct names
(a)
(b)
CH 3CH 2CHCHCH 3
ƒ ƒ
H 3C OH
3-methyl-2-pentanol
m-methylphenol
OH
CH3
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42.
(c)
CH 3
ƒ
CH 3CH 2CH 2CH 2 ¬ O ¬ CHCH 3
isopropoxybutane
(d)
CH 3CH 2CH 2SH
1-propanethiol
Structure and correct names
(a)
CH 3
ƒ
CH 3CH 2CHCHCH 3
ƒ
SH
3-methyl-2-pentanethiol
(b)
CH 3CH 2CH ¬ OCH 3
ƒ
CH 3
2-methoxybutane
(c)
HOCH 2CH 2OH
1,2-ethanethiol
(d)
3-ethyl-2-methylphenol
OH
CH3
CH2CH3
43.
OH
4-hexylresorcinol or 4-hexyl-1,3-dihydroxybenzene
OH
CH2(CH2)4CH3
44.
OH
OH OH
cis-1,2-cyclopentanediol
OH
trans-1,2-cyclopentanediol
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45.
Increasing acidity
CH2OH
CH3
HO
< H2O <
46.
The common phenolic structure in the catecholamines is catechol, o-dihydroxybenzene.
OH
OH
47.
Methyl alcohol is converted to formaldehyde by an oxidation reaction:
O
‘
[O] "
CH 3OH
HCH
48.
(a)
OH
O
2ƒ
‘
Cr2O7
CH 3CHCH 3 999:
CH
C
CH 3
3
H+
H+
(b)
CH 3CH 2CH 2CH “ CH 2 99: CH 3CH 2CH 2CHCH 3
H2O
ƒ
OH
(c)
2 CH 3CH 2OH + 2 Na (metal) ¡ 2 CH 3CH 2ONa + H 2
(d)
(e)
2-
Cr2O7
H+
CH 3CH 2CH “ CH 2 99: CH 3CH 2CHCH 3 999:
CH 3CH 2CCH 3
H+
H2O
ƒ
‘
OH
O
H2SO4
CH 3CH 2CH 2CH 2OH 99: CH 3CH “ CHCH 3 + CH 3CH 2CH “ CH 2
¢
HCl "
CH 3CH 2CHCH 3
ƒ
Cl
O
‘
NaOH "
CH 3CH 2CH 2Cl
CH
CH
C
¬H
CH 3CH 2CH 2OH 999:
3
2
+
H
2-
(f)
Cr2O7
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O– Na±
OH
± NaOH ¡
49.
CH3
± H2O
CH3
CH 3CH 3 + Cl 2
light "
CH 3CH 2Cl + HCl
O– Na±
OCH2CH3
± CH3CH2Cl ¡
CH3
± NaCl
CH3
50.
A simple chemical test to distinguish between:
(a) ethanol and dimethyl ether. Ethanol will react readily with potassium dichromate
and sulfuric acid to make acetaldehyde. Visibly, the orange color of the dichromate
changes to green. Ethanol reacts with metallic sodium to produce hydrogen gas.
Dimethyl ether does not react with either of these reagents.
(b) 1-pentanol and 1-pentene. 1-pentene will rapidly decolorize bromine as it adds to
the double bond. 1-pentanol does not react.
(c) p-methylphenol has acidic properties so it will react with sodium hydroxide.
Methoxybenzene does not have acidic properties and will not react with sodium
hydroxide.
51.
Differentiation between phenols and alcohols:
Phenols are acidic compounds; alcohols are not acidic.
Phenols react with NaOH to form salts; alcohols do not react with NaOH.
The OH group is bonded to a benzene ring in phenols and to a nonbenzene carbon atom
in alcohols.
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52.
Isomers of C8H 10O
¬ CH2CH2OH
¬ CHCH3
ƒ
OH
CH2OH
OH
CH2OH
CH3
CH2CH3
¬ OCH2CH3
CH2OH
CH2CH3
H3C
OH
CH3
¬ CH2OCH3
OCH3
OCH3
CH3
OCH3
H3C
CH2CH3
HO
CH3
53.
Only compound (a) will react with NaOH.
O – Na±
OH
± NaOH ¡
54.
± H2O
Order of increasing boiling points.
1-pentanol
1-octanol 6 1,2-pentanediol
6
138°C
194°C
210°C
All three compounds are alcohols. 1-pentanol has the lowest molar mass and hence the
lowest boiling point. 1-octanol has a higher molar mass and therefore a higher boiling
point than 1- pentanol. 1,2-pentanediol has two ¬ OH groups and therefore forms more
hydrogen bonds than the other two alcohols which causes its higher boiling point.
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55.
(a)
The primary carbocation that is formed first is unstable.
CH 3
ƒ
CH ¬ CH ¬ CH ¬ CH
2
2
3
Thus, the primary carbocation shifts to a much more stable, tertiary carbocation
C H3
ƒ
CH 3 ¬ C ¬ CH 2 ¬ CH 3
It is this intermediate that goes on to form the major product, 2-methyl-2-butene.
(b)
CH 3
ƒ
CH 3 ¬ C “ CH ¬ CH 3
Hydration will not form 2-methyl- l-butanol because the double bond lies between
the middle two carbons in 2-methyl-2-butene. By Markovnikov’s Rule, the ¬ H
will add to the double bonded carbon that already has a hydrogen. Thus, 2-methyl2-butanol will be the major product.
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