Download OC 2/e 9 Alcohols

9
Organic
Chemistry
William H. Brown &
Christopher S. Foote
9-1
9
Alcohols
and
Thiols
Chapter 9
9-2
9 Structure - Alcohols
 The
functional group of an alcohol is
an -OH group bonded to an sp3
hybridized carbon
• bond angles about the hydroxyl oxygen
atom are approximately 109.5°
 Oxygen
is sp3 hybridized
• two sp3 hybrid orbitals form sigma bonds
to carbon and hydrogen
• the remaining two sp3 hybrid orbitals each
contain an unshared pair of electrons
9-3
9 Nomenclature-Alcohols
 IUPAC
names
• the longest chain that contains the -OH group is taken
as the parent
• the parent chain is numbered to give the -OH group the
lowest possible number
• the suffix -e is changed to -ol
 Common
names
• the alkyl group bonded to oxygen is named followed
by the word alcohol
9-4
9 Nomenclature-Alcohols
OH
CH3 CHCH 3
CH3 CH2 CH2 OH
1-Propanol
(Propyl alcohol)
2-Propanol
(Isopropyl alcohol)
OH
CH3 CH2 CH2 CH2 OH
1-Butanol
(Butyl alcohol)
CH3
CH3
CH3 CH2 CHCH3
CH3 COH
CH3 CHCH2 OH
CH3
2-Butanol
(sec-Butyl alcohol)
2-Methyl-1-propanol
(Isobutyl alcohol)
10
3 2
1
4
5
OH
6
cis-3-Methylcyclohexanol
9
8
7
1
6
2-Methyl-2-propanol
(tert- Butyl alcohol)
2
OH
3
4
5
Numbering of the
bicyclic ring takes
precedence over
the location of -OH
Bicyclo[4.4.0]decan-3-ol
9-5
9 Nomenclature of Alcohols
Problem: Write the IUPAC name for each alcohol.
OH
OH
(a)
(b)
(c)
(d) HO
HO
9-6
9 Nomenclature of Alcohols
 Compounds
containing more than one -OH group
are named diols, triols, etc.
CH2 CH2
OH OH
1,2-Ethanediol
(Ethylene glycol)
CH 3 CHCH2
HO OH
1,2-Propanediol
(Propylene glycol)
CH2 CHCH2
HO HO OH
1,2,3-Propanetriol
(Glycerol, Glycerine)
9-7
9 Nomenclature of Alcohols
 Unsaturated
alcohols
• the double bond is shown by the infix -en• the hydroxyl group is shown by the suffix -ol
• number the chain to give OH the lower number
HO
1
2 3
6
4 5
(E)-2-Hexene-1-ol
(trans-2-Hexen-1-ol)
9-8
9 Physical Properties
 Alcohols
are polar compounds
H
+
C
O
H
H
H
+
 They
interact with themselves and with other
polar compounds by dipole-dipole interactions
 Dipole-dipole interaction: the attraction between
the positive end of one dipole and the negative
end of another
9-9
9 Physical Properties
 Hydrogen
bonding: when the positive end of one
dipole is an H bonded to F, O, or N (atoms of high
electronegativity) and the other end is F, O, or N
• the strength of hydrogen bonding in water is
approximately 21 kJ (5 kcal)/mol
• hydrogen bonds are considerably weaker than
covalent bonds
• nonetheless, they can have a significant effect on
physical properties
9-10
9 Hydrogen Bonding
QuickTime™ and a
Photo - JPEG decompressor
are needed to see this picture.
9-11
9 Physical Properties
 Ethanol
and dimethyl ether are constitutional
isomers.
 Their boiling points are dramatically different
• ethanol forms intermolecular hydrogen bonds which
increase attractive forces between its molecules,
which result in a higher boiling point
CH3 CH2 OH
CH3 OCH3
Ethanol
bp 78°C
Dimethyl ether
bp -24°C
9-12
9 Physical Properties
 In
relation to alkanes of comparable size and
molecular weight, alcohols
• have higher boiling points
• are more soluble in water
 The
presence of additional -OH groups in a
molecule further increases solubility in water and
boiling point
9-13
9 Physical Properties
MW
bp
(°C)
Solubility
in Water
Structural Formula
Name
CH3 OH
Methanol
Ethane
32
30
65
-89
Infinite
Insoluble
CH3 CH2 CH3
Ethanol
Propane
46
44
78
-42
Infinite
Insoluble
CH3 CH2 CH2 OH
CH3 CH2 CH2 CH3
1-Propanol
Butane
60
58
97
0
Infinite
Insoluble
CH3 ( CH 2 ) 2 CH 2 OH
1-Butanol
Pentane
74
72
117
36
8 g/100 g
Insoluble
HOCH2 ( CH2 ) 2 CH2 OH 1,4-Butanediol 90
CH3 ( CH 2 ) 3 CH2 OH
1-Pentanol
88
230
138
69
Infinite
2.3 g/100 g
Insoluble
CH3 CH3
CH3 CH2 OH
CH3 ( CH 2 ) 3 CH3
CH3 ( CH 2 ) 4 CH3
Hexane
86
9-14
9 Acidity of Alcohols
 In
dilute aqueous solution, alcohols are weakly
acidic
CH3 O H + : O H
H
Ka =
–
CH3 O:
+ H
[ CH3 O - ] [ H3 O + ]
= 10
[ CH3 OH]
+
O H
H
- 15.5
pKa = 15.5
9-15
9 Acidity of Alcohols
Compound
Structural
Formula
pKa
Hydrogen chloride
HCl
-7
Acetic acid
CH3 COOH
Methanol
CH3 OH
15.5
Water
Ethanol
H2 O
15.7
CH3 CH2 OH
15.9
2-Propanol
( CH3 ) 2 CHOH
17
2-Methyl-2-propanol
( CH3 ) 3 COH
18
4.8
Stronger
acid
Weaker
acid
*Also given for comparison are pK a values for water,
acetic acid, and hydrogen chloride.
9-16
9 Acidity of Alcohols
 Acidity
depends primarily on the degree of
stabilization and solvation of the alkoxide ion
• the negatively charged oxygens of methanol and
ethanol are about as accessible as hydroxide ion for
solvation; these alcohol are about as acidic as water.
• as the bulk of the alkyl group increases, the ability of
water to solvate the alkoxide decreases, the acidity of
the alcohol decreases, and the basicity of the alkoxide
ion increases.
9-17
9 Reaction with Metals
 Alcohols
react with Li, Na, K, and other active
metals to liberate hydrogen gas and form metal
alkoxides
2 CH3 OH + 2 N a
2 CH3 O - N a + + H2
Sodium methoxide
(MeO-Na+)
9-18
9 Reaction with NaH
 Alcohols
are also converted to metal salts by
reaction with bases stronger than the alkoxide
ion
• one such base is sodium hydride
CH3 CH2 OH + Na+ HEthanol
Sodium
hydride
CH3 CH2 O - Na + + H2
Sodium ethoxide
9-19
9 Reaction with HX
• 3° alcohols react very rapidly with HCl, HBr, and HI
• low-molecular-weight 1° and 2° alcohols are unreactive
under these conditions
CH3
CH3
25°C
CH3 CCl + H2 O
CH3 COH + HCl
CH3
2-Chloro-2methylpropane
CH3
2-Methyl-2propanol
• 1° and 2° alcohols require concentrated HBr and HI to
form alkyl bromides and iodides
OH +
1-Butanol
HBr
H2 O
reflux
Br
+
H2 O
1-Bromobutane
9-20
9 Reaction with HX
• with HBr and HI, 2° alcohols generally give some
rearrangement
Br
OH
+ HBr
3-Pentanol
a product of
rearrangement
+ H2 O
+
heat
3-Bromopentane
(major product)
Br
2-Bromopentane
• 1° alcohols with extensive -branching give large
amounts of rearranged product
Br
OH + HBr
 
2,2-Dimethyl-1propanol
+
H2 O
2-Bromo-2-methylbutane
(a product of rearrangement)
9-21
9 Reaction with HX
 Based
on
• the relative ease of reaction of alcohols with HX (3° >
2° > 1°) and
• the occurrence of rearrangements,
 Chemists
propose that reaction of 2° and 3°
alcohols with HX
• occurs by an SN1 mechanism, and
• involves a carbocation intermediate
9-22
9 Reaction with HX - SN1
Step 1: proton transfer to the OH group gives an
oxonium ion
:
CH3
CH3 - C- OH + H
CH3
+
O H
rapid and
reversible
H
CH3 H
CH3 - C O
+ :O H
+
CH3 H
H
Step 2: loss of H2O gives a carbocation intermediate
CH3 H
CH3 - C O
+
CH3 H
slow, rate
determining
SN 1
CH3
H
CH3 - C+ + : O
CH3
H
A 3° carbocation
intermediate
9-23
9 Reaction with HX - SN1
Step 3: reaction of the carbocation intermediate (a Lewis
acid) with halide ion (a Lewis base) gives the product
CH3
CH3 - C+ +
CH3
:Cl -
fast
CH3
CH3 - C- Cl
CH3
2-Chloro-2-methylpropane
(tert-Butyl chloride)
9-24
9 Reaction with HX - SN2
 1°
alcohols react with HX by an SN2 mechanism
Step 1: rapid and reversible proton transfer
:
RCH2 - OH + H
+
O H
H
rapid and
reversible
+
RCH2 - O
H
+ :O H
H
H
Step 2: displacement of HOH by halide ion
Br: - +
+
RCH2 - O
H
H
slow, rate
determining
SN 2
H
RCH2 - Br
+ :O
H
9-25
9 Reaction with HX
 For
1° alcohols with extensive -branching
• SN1 not possible because this pathway would require a
1° carbocation
• SN2 not possible because of steric hindrance created
by the -branching
 These
alcohols react by a concerted loss of HOH
and migration of an alkyl group
9-26
9 Reaction with HX
• Step 1: proton transfer gives an oxonium ion
:
CH3
+
CH3 -C- CH2 -O- H + H O H
CH3
H
2,2-Dimethyl-1propanol
rapid and
reversible
CH3
+ H
CH3 -C- CH2 O + : O H
H
CH3
H
An oxonium ion
• Step 2: concerted elimination of HOH and migration
of a methyl group gives a 3° carbocation
CH3
CH3 - C-CH2
CH3
+
O
H
H
slow and
rate determining
(concerted)
+
CH3 -C- CH2 -CH3 + : O H
CH3
H
A 3° carbocation
intermediate
9-27
9 Reaction with HX
Step 3: reaction of the carbocation intermediate (a Lewis
acid) with halide ion (a Lewis base) gives the product
+
CH3 - C-CH2 - CH3 + : Cl CH3
fast
Cl
CH3 - C-CH2 - CH3
CH3
2-Chloro-2-methylbutane
9-28
9 Reaction with PBr3
 An
alternative method for the synthesis of 1° and
2° alkyl bromides is reaction of an alcohol with
phosphorus tribromide
• this method gives less rearrangement than with HBr
OH
+
PBr 3
0°
2-Methyl-1Phosphorus
propanol
tribromide
(Isobutyl alcohol)
Br
+
H3 PO 3
1-Bromo-2-methyl- Phosphorous
propane
acid
(Isobutyl bromide)
9-29
9 Reaction with PBr3
Step 1: formation of a protonated dibromophosphite,
which converts H2O, a poor leaving group, to a good
leaving group
a good
leaving group
:
R- CH2 -O- H + Br
P Br
R-CH2
Br
+
O PBr2 + :Br H
Step 2: displacement by bromide ion
+ R-CH2
SN2
+
O PBr2
:
Br:
-
R- CH2 -Br + HO- PBr 2
H
9-30
9 Reaction with SOCl2
 Thionyl
chloride is the most widely used reagent
for the conversion of 1° and 2° alcohols to alkyl
chlorides
• a base, most commonly pyridine or triethylamine, is
added to catalyze the reaction and to neutralize the
HCl
OH +
1-Heptanol
SOCl 2
pyridine
Thionyl
chloride
Cl + SO + HCl
2
1-Chloroheptane
9-31
9 Reaction with SOCl2
 Reaction
of an alcohol with SOCl2 in the
presence of a 3° amine is stereoselective;
proceeds with inversion of configuration
CH 3 ( CH2 ) 5
C OH + SOCl 2
H
H3 C
(S)-2-Octanol
Thionyl
chloride
3° amine
( CH2 ) 5 CH 3
Cl C
H
CH3
+ SO 2 + H Cl
(R)-2-Chlorooctane
9-32
9 Reaction with SOCl2
Step 1: nucleophilic displacement of chlorine
R1
:
O
C O H + Cl S Cl
H
R2
Thionyl
chloride
R1
H
+ :Cl C O+
S O
H
R2 Cl
Step 2: proton transfer to the 3° amine gives an alkyl
chlorosulfite
R1
R1
+
+ HN R3
:
H
+ :N R3
C O+
S O
H
R2 Cl
A 3° amine
C O
H
R2
S O
Cl
An alkyl chlorosulfite
9-33
9 Reaction with SOCl2
Step 3: backside displacement by chloride ion and
decomposition of the chlorosulfite ester gives the alkyl
chloride
R1
-
Cl:
+
R1
C
HR
2
O
S
O
SN 2
Cl
+ O
C
H
O
S
+ :Cl -
R2
Cl
9-34
9 Alkyl Sulfonates
 Sulfonyl
chlorides are derived from sulfonic
acids
• sulfonic acids are strong acids like sulfuric acid
O
R- S- Cl
O
A sulfonyl
chloride
O
R- S- OH
O
A sulfonic acid
(a very strong acid)
O
R- S- OO
A sulfonate anion
(a very weak base and
stable anion; a very
good leaving group
9-35
9 Alkyl Sulfonates
A
commonly used sulfonyl chloride is ptoluenesulfonyl chloride (Ts-Cl)
O
CH3 CH2 OH + Cl-S
CH 3
O
p-Toluenesulfonyl
Ethanol
chloride
pyridine
O
CH 3 CH 2 O-S
CH3 + HCl
O
Ethyl p-toluenesulfonate
(Ethyl tosylate)
9-36
9 Alkyl Sulfonates
 Another
commonly used sulfonyl chloride is
methanesulfonyl chloride (Ms-Cl)
OH
+
O
Cl-S- CH3
pyridine
O
Cyclohexanol
Methanesulfonyl
chloride
O
O-S-CH3 + HCl
O
Cyclohexyl
methanesulfonate
(Cyclohexyl mesylate)
9-37
9 Alkyl Sulfonates
 Sulfonate
anions are very weak bases (the
conjugate base of a strong acid) and are very
good leaving groups for SN2 reactions
 Conversion of an alcohol to a sulfonate ester
converts HOH, a very poor leaving group, into a
sulfonic ester, a very good leaving group
9-38
9 Alkyl Sulfonates
 This
two-step procedure converts (S)-2-octanol
to (R)-2-octyl acetate
Step 1: formation of a p-toluenesulfonate (Ts) ester
CH3 ( CH2 ) 5
C OH
H
+ Cl-T s
pyridine
CH3 ( CH2 ) 5
C OT s
+ HCl
H
CH3
(S)-2-Octanol
CH3
(S)-2-Octyl tosylate
9-39
9 Alkyl Sulfonates
Step 2: nucleophilic displacement of tosylate
CH3 ( CH2 ) 5
O
-
CH3 CO Na
+
+
C OT s
H
CH3
(S)-2-Octyl tosylate
SN 2
ethanol
O
CH 3 CO
( CH2 ) 5 CH3
C
+
+ N a OTs
H
CH3
(R)-2-Octyl acetate
9-40
9 Dehydration of ROH
 An
alcohol can be converted to an alkene by
elimination of H and OH from adjacent carbons (a
-elimination)
• 1° alcohols must be heated at high temperature in the
presence of an acid catalyst, such as H2SO4 or H3PO4
• 2° alcohols undergo dehydration at somewhat lower
temperatures
• 3° alcohols often require temperatures at or slightly
above room temperature
9-41
9 Dehydration of ROH
CH3 CH2 OH
OH
H2 SO 4
180°C
CH2 = CH2
H2 SO 4
+
140°C
Cyclohexanol
CH3
CH3 COH
+
H2 O
H2 O
Cyclohexene
H2 SO 4
CH3
2-Methyl-2-propanol
(tert- Butyl alcohol)
50°C
CH3
CH3 C= CH2 + H2 O
2-Methylpropene
(Isobutylene)
9-42
9 Dehydration of ROH
• where isomeric alkenes are possible, the alkene
having the greater number of substituents on the
double bond usually predominates (Zaitsev rule)
OH
CH3 CH2 CHCH3
2-Butanol
8 5 % H3 PO 4
heat
CH3 CH= CH CH 3 + CH3 CH2 CH= CH2 + H2 O
2-Butene
1-Butene
(80%)
(20%)
9-43
9 Dehydration of ROH
 Dehydration
of 1° and 2° alcohols is often
accompanied by rearrangement
H2 SO 4
OH
+
140 - 170°C
3,3-Dimethyl2-butanol
2,3-Dimethyl2-butene
(80%)
2,3-Dimethyl1-butene
(20%)
• acid-catalyzed dehydration of 1-butanol gives a
mixture of three alkenes
OH
1-Butanol
H2 SO 4
+
140 - 170°C
trans-2-butene
(56%)
+
cis- 2-butene
(32%)
1-Butene
(12%)
9-44
9 Dehydration of ROH
 Based
on evidence of
• ease of dehydration (3° > 2° > 1°)
• prevalence of rearrangements
 Chemists
propose a three-step mechanism for
the dehydration of 2° and 3° alcohols
• because this mechanism involves formation of a
carbocation intermediate in the rate-determining step,
it is classified as E1
9-45
9 Dehydration of ROH
Step 1: proton transfer to the -OH group gives an
oxonium ion
HO:
CH3 CHCH2 CH3 + H
+
O H
rapid and
reversible
H
H +H
O
CH3 CHCH2 CH3 +
:O H
An oxonium ion
H
Step 2: loss of H2O gives a carbocation intermediate
H + H
O
CH3 CHCH2 CH3
slow, rate
determining
+
CH3 CHCH2 CH3 + H2 O:
A 2° carbocation
intermediate
9-46
9 Dehydration of ROH
Step 3: proton transfer from a carbon adjacent to the
positively charged carbon to water. The sigma
electrons of the C-H bond become the pi electrons of
the carbon-carbon double bond
H
+
CH3 -CH -CH- CH 3 + : O H
H
rapid and
reversible
CH3 -CH = CH -CH3
+
+ H O H
H
9-47
9 •Dehydration of ROH
alcohols with little -branching give terminal
alkenes and rearranged alkenes
 1°
• Step 1: proton transfer to OH gives an oxonium ion
:
O-H +
1-Butanol
+
H O H
rapid and
reversible
H
+
O-H + : O-H
H
H
• Step 2: loss of H from the -carbon and H2O from the
-carbon gives the terminal alkene
H O: +
H
+
O-H
H H H
E2
+
H O H +
1-Butene
H
9-48
9 Dehydration of ROH
Step 3: shift of a hydride ion from -carbon and loss of
H2O from the -carbon gives a carbocation
+
O-H
1,2-shift of a
hydride ion
H H H
+
+
H
A 2° carbocation
: O-H
H
Step 4: proton transfer to solvent gives the alkene
+
H
+ H2 O
E1
+
trans-2butene
+ H3 O +
cis- 2butene
9-49
9 Dehydration of ROH
 Dehydration
with rearrangement occurs by a
carbocation rearrangement
H+
+
OH - H2 O
3,3-DimethylA 2° carbocation
2-butanol
intermediate
+
A 3° carbocation
intermediate
H2 O
+ H3 O +
2,3-Dimethyl2-butene
H2 O
+ H3 O +
2,3-Dimethyl1-butene
9-50
9 Dehydration of ROH
 Acid-catalyzed
alcohol dehydration and alkene
hydration are competing processes
C
C
An alkene
 Principle
+ H2 O
acid
catalyst
C
C
H OH
An alcohol
of microscopic reversibility: the
sequence of transition states and reactive intermediates
in the mechanism of a reversible reaction must be the
same, but in reverse order, for the backward reaction as
for the forward reaction
9-51
9 Pinacol Rearrangement
 The
products of acid-catalyzed dehydration of a
glycol are different from those of alcohols
HO OH
CH3 -C- C-CH3
H2 SO 4
H3 C CH3
2,3-Dimethyl-2,3-butanediol
(Pinacol)
O CH3
CH3 -C- C-CH3 + H2 O
CH3
3,3-Dimethyl-2-butanone
(Pinacolone)
9-52
9 Pinacol Rearrangement
Step 1: proton transfer to OH gives an oxonium ion
:
HO O-H
+
CH3 -C- C-CH3 + H O H
H3 C CH3
H
rapid and
reversible
H
+
HO O-H
CH3 -C- C-CH3 + : O H
H
H3 C CH3
An oxonium ion
Step 2: loss of water gives a carbocation intermediate
H
+
HO O-H
CH3 -C- C-CH3
HO
+
CH3 -C- C-CH3
H3 C CH3
H3 C CH3
+ H2 O:
9-53
9 Pinacol Rearrangement
Step 3: a 1,2- shift of methyl gives a more stable
carbocation
H
O CH3
H +
O CH3
:
H
O CH3
CH3 -C- C-CH3
CH3 -C- C-CH3
+
CH3
CH3 -C- C-CH3
+
H3 C
CH3
Step 4: proton transfer to solvent completes the reaction
H +
O CH3
H 2 O: + CH3 -C- C-CH3
CH3
:O CH3
CH3 -C- C-CH3 + H3 O +
CH3
9-54
9 Oxidation: 1° ROH
A
primary alcohol can be oxidized to an aldehyde
or a carboxylic acid, depending on the
experimental conditions
OH
CH3 -C H
H
A primary
alcohol
[O]
O
CH3 -C- H
An aldehyde
[O]
O
CH3 -C- OH
A carboxylic
acid
• to an aldehyde is a two-electron oxidation
• to a carboxylic acid is a four-electron oxidation
9-55
9 Oxidation: 1° ROH
A
common oxidizing agent for this purpose is
chromic acid, prepared by dissolving
chromium(VI) oxide or potassium dichromate in
aqueous sulfuric acid
+ H2 O
Chromium(VI)
oxide
CrO 3
K2 Cr 2 O 7
Potassium
dichromate
H2 SO 4
H2 SO 4
H2 Cr 2 O 7
H2 Cr O 4
Chromic acid
H2 O
2 H2 Cr O 4
Chromic acid
9-56
9 Oxidation: 1° ROH
 Oxidation
of 1-octanol gives octanoic acid
• the aldehyde intermediate is not isolated
CH3 (CH2 ) 6 CH2 OH
1-Octanol
CrO3
H2 SO4 , H2 O
O
CH3 (CH2 ) 6 CH
Octanal
(not isolated)
O
CH3 (CH2 ) 6 COH
Octanoic acid
9-57
9 Oxidation: 1° ROH
 Pyridinium
chlorochromate (PCC): a form of
Cr(VI) prepared by dissolving CrO3 in aqueous
HCl and adding pyridine to precipitate PCC
N+
H
CrO 3 Cl -
• PCC is selective for the oxidation of 1° alcohols to
aldehydes; it does not oxidize aldehydes further to
carboxylic acids
9-58
9 Oxidation: 1° ROH
 PCC
oxidation of a 1° alcohol to an aldehyde
O
PCC
OH
Geraniol
H
Geranial
9-59
9 Oxidation: 2° ROH
 2°
alcohols are oxidized to ketones by both PCC
and chromic acid
OH
O
+ H2 Cr O 4
2-Isopropyl-5-methylcyclohexanol
(Menthol)
acetone
+ Cr
3+
2-Isopropyl-5-methylcyclohexanone
(Menthone)
9-60
9 Oxidation: 1° & 2° ROH
 The
mechanism of chromic acid oxidation of an
alcohol involves two steps
Step 1: formation of an alkyl chromate ester
O
OH
fast and
reversible
+ HO-Cr-OH
H
Cyclohexanol
O
O
O-Cr-OH
O
+
H2 O
H
An alkyl chromate
9-61
9 Oxidation: 1° & 2° ROH
Step 2: proton transfer to solvent and decomposition of
the alkyl chromate ester gives the product
O
O
H
slow, rate
Cr- OH determining
O
chromium(IV)
:
H
O
H
O +
Cyclohexanone
H3 O +
+
OCr- OH
O
9-62
9 Oxidation: 1° & 2° ROH
 In
chromic acid oxidation of a CHO group, it is
the hydrated form that is oxidized
O
R- C-H
+ H2 O
fast and
reversible
R- C-OH
H
R- C-OH
H
An aldehyde
hydrate
An aldehyde
OH
OH
O-Cr O 3 H
H2 Cr O4
R- C-OH
O
R- C-OH
H2 O:
H
An alkyl
chromate ester
A carboxylic
acid
9-63
9 Oxidation of Glycols
 Glycols
are cleaved by oxidation with periodic
acid, H5IO6 (or, alternatively HIO4•2H2O)
OH
+
OH
cis-1,2-Cyclohexanediol
HIO 4
Periodic
acid
CHO
CHO
Hexanedial
+ HIO 3
Iodic
acid
9-64
9 Oxidation of Glycols
• the glycol undergoes a two-election oxidation
C OH
C O
C OH
C O
+ 2H
+
+ 2e
-
• periodic acid undergoes a two-electron reduction
HIO 4
+ 2H
Periodic acid
+
+ 2e
-
HIO 3 + H 2 O
Iodic acid
9-65
9 Oxidation of Glycols
 The
mechanism of periodic acid oxidation of a
glycol is divided into two steps
Step 1: formation of a cyclic periodic ester
O
O
C OH
+ O
C OH
I
C
O
OH
I
C
O
OH + H 2 O
O
O
A cyclic periodic ester
Step 2: redistribution of electrons within the fivemembered ring
O
C
O
I
C
O
OH
C O
+ HIO 3
C O
O
9-66
9 Thiols: Structure
 The
functional group of a thiol is an SH (sulfhydryl) group bonded to an
sp3 hybridized carbon
 The bond angle about sulfur in
methanethiol is 100.3°, which
indicates that there is considerably
more p character to the bonding
orbitals of divalent sulfur than there
is to oxygen
9-67
9 Nomenclature
 IUPAC
names:
• the parent is the longest chain that contains the -SH
group
• change the suffix -e to -thiol
• as a substituent, it is a sulfanyl group
 Common
names:
• name the alkyl group bonded to sulfur followed by the
word mercaptan
CH3
CH3 CH2 SH
CH3 CHCH 2 SH
HSCH 2 CH2 OH
Ethanethiol
(Ethyl mercaptan)
2-Methyl-1-propanethiol
(Isobutyl mercaptan)
2-Sulfanylethanol
(Mercaptoethanol)
9-68
9 Thiols: Physical Properties
 The
difference in electronegativity between S
(2.5) and H (2.1) is 0.4. Because of the low
polarity of the S-H bond, thiols
• show little association by hydrogen bonding
• have lower boiling points and are less soluble in water
than alcohols of comparable MW
Thiol
bp (°C)
Methanethiol 6
Ethanethiol
35
1-Butanethiol 98
bp (°C)
Alcohol
Methanol 65
Ethanol
78
1-Butanol 117
9-69
9 Thiols: Physical Properties
 Low-molecular-weight
thiols = STENCH
• the scent of skunks is due primarily to these two thiols
CH3
CH3 CHCH 2 CH2 SH
3-Methyl-1-butanethiol
CH3 CH= CHCH 2 SH
2-Butene-1-thiol
9-70
9 Thiols: preparation
 The
most common preparation of thiols, RSH,
depends on the very high nucleophilicity of
hydrosulfide ion, HS+
CH3 ( CH2 ) 8 CH2 I + N a SH
1-Iododecane
SN 2
Sodium
hydrosulfide
+ -
CH3 ( CH2 ) 8 CH2 SH + N a I
1-Decanethiol
9-71
9 Thiols: acidity
 Thiols
are stronger acids than alcohols
CH3 CH2 SH + H2 O
pKa = 8.5
pKa = 15.9
CH3 CH2 OH + H2 O
CH3 CH2 S
-
+ H3 O+
+
CH3 CH2 O + H3 O
9-72
9 Thiols: acidity
 When
dissolved an aqueous NaOH, they are
converted completely to alkylsulfide salts
+
CH3 CH2 SH + N a OH
-
pKa = 8.5
Stronger
Stronger
acid
base
-
CH3 CH2 S N a
Weaker base
+
+
H2 O
pKa = 15.7
Weaker acid
9-73
9 Thiols: oxidation
 Thiols
are oxidized to disulfides by a variety of
oxidizing agents, including O2.
• they are so susceptible to this oxidation that they must
be protected from air during storage
2 RSH +
A thiol
1 O
2 2
RSSR + H2 O
A disulfide
• the most common reaction of thiols in biological
systems in interconversion between thiols and
disulfides, -S-S-
9-74
9 Prob 9.22
From each pair of compounds, select the one more
soluble in water.
(a) CH2 Cl 2
o r CH3 OH
O
(b) CH3 CCH3 or
CH2
CH3 CCH3
(c) CH3 CH2 Cl o r N aCl
(d) CH3 CH2 CH2 SH or CH3 CH2 CH2 OH
(e)
OH
CH3 CH2 CHCH2 CH3
O
or CH3 CH2 CCH 2 CH3
9-75
9 Prob 9.24
From each pair of compounds, select the one more
soluble in water.
(a) CH2 Cl2
or
CH3 CH2 OH
(b) CH3 CH2 OCH2 CH3
O
(c) CH3 CCH3
or
or CH3 CH2 OH
CH3 CH2 OCH2 CH3
(d) CH3 CH2 OCH2 CH3
or CH3 ( CH2 ) 3 CH3
9-76
9 Prob 9.25
Calculate the percent of each isomer present at
equilibrium. Assume a value of DG° (equatorial to axial)
for cyclohexanol is 4.0 kJ (0.95 kcal/mol).
Al[ OCH( CH3 ) 2 ) ] 3
HO
acetone
A
OH
B
9-77
9 Prob 9.26
Complete each acid-base reaction. Use curved arrows to
show the flow of electrons.
+
(a) CH3 CH2 OH + HOH
H
(b) CH3 CH2 OCH2 CH3
O
+ HOSOH
O
(c) CH3 CH2 CH 2 CH2 CH2 OH + HI
9-78
9 Prob 9.26 (cont’d)
Complete each acid-base reaction. Use curved arrows to
show the flow of electrons.
O
O
(d) CH3 CH2 CH 2 COH + HOSOH
O
(e)
OH + BF3
+
(f) CH3 CH= CH CH CH 3
+ HOH
9-79
9 Prob 9.27
From each pair, select the stronger acid and write a
structural formula for its conjugate base.
(a) H2 O or H2 CO3
(b) CH3 OH or CH3 COOH
(c) CH3 CH2 OH or CH3 C CH
(d) CH3 CH2 OH or CH3 CH2 SH
9-80
9 Prob 9.28
From each pair select the stronger base. Write a
structural formula for its conjugate acid.
(a) OH-
or
CH3 O- (each in H2 O)
(b) CH3 CH2 O-
or
CH3 C C-
(c) CH3 CH2 S-
or
CH3 CH2 O-
(d) CH3 CH2 O
or
NH2 -
9-81
9 Prob 9.29
In each equilibrium, label the stronger acid and base, and
the weaker acid and base. Estimate the position of
equilibrium.
(a) CH3 CH2 O - +
CH3 C CH
(b) CH3 CH2 O + HCl
(c) CH3 COOH + CH3 CH2 O -
CH3 CH2 OH + CH3 C C–
CH3 CH2 OH +
Cl
-
CH3 COO- + CH3 CH2 OH
9-82
9 Prob 9.32
Complete each equation, but do not balance
(a)
(b)
(c)
(d) HO
OH + H2 Cr O4
OH + SOCl 2
OH
+ HCl
OH +
HBr
(excess)
9-83
9 Prob 9.32 (cont’d)
Complete each equation, but do not balance
OH
(e)
+ H2 Cr O4
OH
+ HIO 4
(f)
OH
(g)
(h)
1 . Os O4 , H2 O 2
2 . HIO 4
OH
+
SOCl 2
9-84
9 Prob 9.34
When A or B is treated with HBr, racemic 2,3dibromobutane is formed. When C or D is treated with
HBr, meso 2,3-dibromobutane is formed. Explain.
CH3
HO
CH3
H
H
Br
CH3
A
H
Br
CH3
B
CH3
CH3
OH
H
OH
HO
H
H
H
Br
Br
H
CH3
CH3
C
D
9-85
9 Prob 9.36
Show how to bring about each conversion.
(a)
OH
(c)
OH
(b)
OH
OH
OH
O
(d)
OH
H
O
(e)
CH2
CH2 Cl
9-86
9 Prob 9.36 (cont’d)
Show how to bring about each conversion.
(f)
CHCH3
(g) CH3 ( CH2 ) 6 CH 2 OH
(h)
O
CCH3
OH
O
CH3 ( CH2 ) 6 CH
OH
OH
(i)
CH 2
O
COH
9-87
9 Prob 9.37
Propose a mechanism for the following pinacol
rearrangement.
HO
O
OH
BF3 • Et 2 O
+ H2 O
9-88
9 Prob 9.40
Propose a mechanism for this reaction.
H
O CH2 OH
Tetrahydrofurfuryl
alcohol
A rSO3 H
+
H2 O
O
Dihydropyran
9-89
9 Prob 9.43
Show how to bring about this conversion.
OH
O
O
9-90
9 Prob 9.44
Propose a structural formula for the product of this
reaction and a mechanism for its formation.
OH
NaOH
C7 H1 2 O
OT s
9-91
9 Prob 9.45
Propose a mechanism for the formation of the products
of this solvolysis.
H2 O
T sO
DMSO
Chrysanthemyl tosylate
+
OH
Artemisia alcohol
OH
Yomogi alcohol
9-92
9 Prob 9.46
Show how to convert cyclohexene to each compound.
O
(a)
OH
(d)
(c)
(b)
O
OCH3
O
H
(e) H
O
9-93
9
Alcohols
and
Thiols
End of Chapter 9
9-94