20
Organic
Chemistry
William H. Brown &
Christopher S. Foote
20-1
20
Aromatics I
Chapter 20
20-2
20 Benzene - Kekulé
 The
first structure for benzene was proposed by
August Kekulé in 1872
H
H
C
C
H
C
C
H
C
C
H
H
H
H
C
C
H
C
C
H
C
C
H
H
 This
structure, however, did not account for the
unusual chemical reactivity of benzene
20-3
20 Benzene - MO Model
 The
concepts of hybridization of atomic orbitals
and the theory of resonance, developed in the
1930s, provided the first adequate description of
benzene’s structure
• the carbon skeleton is a regular hexagon, with all CC-C and H-C-C bond angles 120°
H
120°
H
C
109 pm
H
H
120°
C
120°
C
C
C
139 pm

sp2-sp2
C H
sp2-1s
H
20-4
20 Benzene - MO Model
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Photo - JPEG decompressor
are needed to see this picture.
20-5
20 Benzene - MO Model
 Combination
of six 2p atomic orbitals gives six
molecular orbitals
• three bonding MOs and
• three antibonding MOs
 In
the ground state, the six pi electrons of
benzene occupy the three bonding MOs
20-6
20 Benzene MO Model
6
six uncombined
2p orbitals, each
with one electron
4
5
2
3
1
• the stability of benzene results from the fact that these
three bonding MOs are much lower in energy than the
six uncombined 2p atomic orbitals
20-7
20 Benzene - Resonance
 We
often represent benzene as a hybrid of two
equivalent Kekulé structures
• each makes an equal contribution to the hybrid and
thus the C-C bonds are neither double nor single, but
something in between
20-8
20 Benzene - Resonance
 Resonance
energy: the difference in energy
between a resonance hybrid and the most stable
of its hypothetical contributing structures in
which electrons are localized on particular
atoms and in particular bonds
 One way to estimate the resonance energy of
benzene is to compare the heats of
hydrogenation of benzene and cyclohexene
20-9
20 Benzene
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20-10
20 Concept of Aromaticity
 The
underlying criteria for aromaticity were
recognized in the early 1930s by Erich Hückel,
based on MO calculations
 To be aromatic, a compound must
• be cyclic
• have one p orbital on each atom of the ring
• be planar or nearly planar so that there is continuous
or nearly continuous overlap of all p orbitals of the
ring
• have a closed loop of (4n + 2) pi electrons in the cyclic
arrangement of p orbitals
20-11
20 Frost Circles
 Inscribe
a polygon of the same number of sides
as the ring to be examined such that one of the
vertices is at the bottom of the ring
 The relative energies of the MOs in the ring are
given by where the vertices touch the circle
 Those MOs
• below the horizontal line through the center of the ring
are bonding MOs
• on the horizontal line are nonbonding MOs
• above the horizontal line are antibonding MOs
20-12
20 Frost Circles
 Following
are Frost circles describing the MOs
for monocyclic, planar, fully conjugated four-,
five-, and six-membered
rings
*
*
4
6
4*
3
2
1
5*
2
3
1
4*
5*
2
3
1
20-13
20 Aromatic Hydrocarbons
 Annulene:
a cyclic hydrocarbon with a
continuous alternation of single and double
bonds
 [10]Annulene: according to Hückel’s criteria, this
unsaturated hydrocarbon should be aromatic
• it is cyclic
• it has one 2p orbital on each carbon of the ring,
• it has 4(2) + 2 = 10 pi electrons
 It
is not aromatic, however, because it is not
planar
20-14
20 [10]Annulene
• nonbonded interactions between the two hydrogens
that point inward toward the center of the ring force
the ring into a nonplanar conformation in which
overlap of the ten 2p orbitals is no longer continuous
20-15
20 [14]Annulene
• this unsaturated hydrocarbon meets the Hückel
criteria and is aromatic
20-16
20 [18]Annulene
 This
unsaturated hydrocarbon is also aromatic
20-17
20 Antiaromatic Hydrocbn
 Antiaromatic
hydrocarbon: a monocyclic, planar,
fully conjugated hydrocarbon with 4n pi
electrons (4, 8, 12, 16, 20...)
• an antiaromatic hydrocarbon is especially unstable
relative to an open-chain fully conjugated
hydrocarbon of the same number of carbon atoms
 Cyclobutadiene
is antiaromatic. In the groundstate electron configuration of this molecule,
• two electrons fill the 1 bonding MO
• the remaining two electrons lie in the 2 and 3
nonbonding MOs
20-18
20 Cyclobutadiene
• planar cyclobutadiene has two unpaired electrons,
which make it highly unstable and reactive
*4
3
2
Four uncombined
2p orbitals, each
with one electron
1
Ground-state electron
configuration of
planar cyclobutadiene
20-19
20 Cyclobutadiene
• cyclobutadiene is trapped within the cage of a larger
molecule called a hemicarcerand (to rotate this
molecule, see the accompanying CD)
20-20
20 Cyclooctatetraene
 Planar
cyclooctatetraene, if it existed, would be
antiaromatic; it would have two unpaired
electrons in 4 and 5 nonbonding MOs
8
eight uncombined
2p orbitals, each
with one electron
6
7
4
5
2
3
1
20-21
20 Heterocyclic Aromatics
 Heterocyclic
compound: a compound that
contains more than one kind of atom in a ring
• in organic chemistry, the term refers to a ring with one
or more atoms are other than carbon
 Pyridine
and pyrimidine are heterocyclic analogs
of benzene; each is aromatic.
4
4
5
3
5
6
2
6
N1
Pyridine
3
N
N1
2
Pyrimidine
20-22
20 Pyridine
• the nitrogen atom of pyridine is sp2 hybridized
• the unshared pair of electrons lies in an sp2 hybrid
orbital and is not a part of the six pi electrons of the
aromatic system
• pyridine has a resonance energy of 134 kJ (32
kcal)/mol, slightly less than that of benzene
•
•
•
•
this orbital is perpendicular
to the six 2p orbitals of the
pi system
•
•
N
this pair of electrons
is not a part of the
4n + 2 pi electrons
20-23
20 Furan
• the oxygen atom of furan is sp2 hybridized
• one unshared pairs of electrons on oxygen lies in an
unhybridized 2p orbital and is a part of the aromatic
sextet
• the other unshared pair lies in an sp2 hybrid orbital
and is not a part of the aromatic system
• the resonance energy of furan is 67 kJ (16 kcal)/mol
•
this pair of electrons
is a part of the
4n + 2 pi electrons
•
•
•
O
this pair of
electrons
is not
20-24
20 Other Heterocyclics
CH 2 CH 2 NH 2
HO
N
N
H
H
Serotonin
(a neurotransmitter)
Indole
N
N
N
H3 C
N
O
H
Purine
O
CH 3
N
N
CH 3
N
N
Caffeine
20-25
20 Aromatic Hydrcbn Ions
 Any
neutral, monocyclic unsaturated
hydrocarbon with an odd number of carbons
must have at least one CH2 group and, therefore,
cannot be aromatic
CH2
Cyclopropene
CH2
Cyclopentadiene
CH2
Cycloheptatriene
• cyclopropene, for example, has the correct number of
pi electrons to be aromatic, 4(0) + 2 = 2, but does not
have a closed loop of 2p orbitals
20-26
20 Cyclopropenyl Cation
• if, however, the CH2 group of cyclopropene is
transformed into a CH+ group in which carbon is sp2
hybridized and has a vacant 2p orbital, the overlap of
orbitals is continuous and the cation is aromatic
H
H
H
+
+ H
H
H
H
+
H
H
Cyclopropenyl cation represented as a hybrid
of three equivalent contributing structures
20-27
20 Cyclopropenyl Cation
 When
3-chlorocyclopropene is treated with
SbCl5, it forms a stable salt
H
+
Cl
3-Chlorocyclopropene
Sb Cl 5
Antimony(V)
chloride
(a Lewis acid)
+
H Sb Cl 6 -
Cyclopropenyl
hexachloroantimonate
• this chemical behavior is to be contrasted with that of
5-chloro-1,3-cyclopentadiene, which cannot be made
to form a stable salt
20-28
20 Cyclopentadienyl C+
H
+ A gBF4
Cl
5-Chloro-1,3cyclopentadiene
+
H BF4
-
+ Ag Cl
Cyclopentadienyl
tetrafluoroborate
• if planar cyclopentadienyl cation existed, it would
have 4 pi electrons and be antiaromatic
• note that we can draw five equivalent contributing
structures for the cyclopentadienyl cation. Yet this
cation is not aromatic because it has only 4 pi
electrons.
20-29
20 Cyclopentadienyl Anion
 To
convert cyclopentadiene to an aromatic ion, it
is necessary to convert the CH2 group to a CHgroup in which carbon becomes sp2 hybridized
and has 2 electrons in its unhybridized 2p orbital
•
H
H
H
H •
•
H
H
•
:
H
H
the origin of the 6 pi electrons
in the cyclopentadienyl anion
H
H
H
H
H
H
H
Cyclopentadienyl anion
(aromatic)
20-30
20 Cyclopentadienyl Anion
 The
pKa of cyclopentadiene is 16
• in aqueous NaOH, it is in equilibrium with its sodium
salt
H
H
CH2 + NaOH
pK a 16.0
:
H
H
H
Na+ + H2 O
pKa 15.7
• it is converted completely to its anion by very strong
bases such as NaNH2 , NaH, and LDA
20-31
20 Cycloheptatrienyl C+
 Cycloheptatriene
forms an aromatic cation by
conversion of its CH2 group to a CH+ group with
its sp2 carbon having a vacant 2p orbital
H
H
H
H
H
H
+
+
H
H
H
H
H
H
H
Cycloheptatrienyl cation
(tropylium ion)
(aromatic)
H
20-32
20 Nomenclature
 Monosubstituted
alkylbenzenes are named as
derivatives of benzene
• many common names are retained
Toluene
OH
Phenol
Ethylbenzene
N H2
Aniline
Cumene
CHO
Benzaldehyde
Styrene
COOH
Benzoic acid
OCH3
Anisole
20-33
20 Nomenclature
 Benzyl
and phenyl groups
CH3
Benzene
Phenyl group, Ph-
O
Toluene
CH2 Benzyl group, Bn-
O
H 3 CO
Ph
1-Phenyl-1pentanone
4-(3-Methoxyphenyl)2-butanone
(Z)-2-Phenyl2-butene
20-34
20 Disubstituted Benzenes
 Locate
the two groups by numbers or by the
locators ortho (1,2-), meta (1,3-), and para
(1,4-)
• where one group imparts a special name, name
the compound as a derivative of that molecule
CH3
N H2
COOH
N O2
Cl
Br
4-Bromotoluene
(p-Bromotoluene)
3-Chloroaniline
(m-Chloroaniline)
2-Nitrobenzoic acid
(o-Nitrobenzoic acid )
20-35
20 Disubstituted Benzenes
• where neither group imparts a special name, locate
the groups and list them in alphabetical order
CH2 CH3
4
N O2
2
3
N O2
Br
1
2
2
3
1
Cl
1-Chloro-4-ethylbenzene
(p-Chloroethylbenzene)
1
1-Bromo-2-nitrobenzene
(o-Bromonitrobenzene)
N O2
1,3-Dinitrobenzene
(m-Dinitrobenzene)
20-36
20 Polysubstituted Derivs
• if one group imparts a special name, name the
molecule as a derivative of that compound
• if no group imparts a special name, list them in
alphabetical order, giving them the lowest set of
numbers
1 2
N O2
OH
CH3
N O2
Br
6
1
2
4
Br
2
4
4
Cl
Br
4-Chloro-2-nitrotoluene
2,4,6-Tribromophenol
1
Br
CH2 CH3
2-Bromo-1-ethyl-4nitrobenzene
20-37
20 NMR Spectroscopy
 Hydrogens
bonded to a benzene ring appear in
the region  6.5 to 8.5
 Aryl hydrogens absorb further downfield than
vinylic hydrogens, which is accounted for by an
induced ring current
• the induced ring current has an associated magnetic
field which opposes the applied field in the middle of
the ring but reinforces it on the outside of the ring
• thus, hydrogens on the benzene ring come into
resonance at a lower applied field; that is, at a larger
chemical shift than vinylic hydrogens
20-38
20 NMR Spectroscopy
 There
are, of course, no hydrogens on the inside
of a benzene ring. But there are in larger
annulenes, for example [18]annulene
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
these six hydrogens on
the inside of the ring come
into resonance at  -3.00
the twelve hydrogens on
the outside of the ring come
into resonance at  9.3
H
20-39
20 Phenols
 The
functional group of a phenol is an -OH group
bonded to a benzene ring
OH
OH
OH
OH
OH
CH3
Phenol 3-Methylphenol
(m-Cresol)
OH
1,2-Benzenediol 1,4-Benzenediol
(Catechol)
(Hydroquinone)
20-40
20 Phenols
• hexylresorcinol is a mild antiseptic and disinfectant
• eugenol is used as a dental antiseptic and analgesic
• urushiol is the main component of the oil of poison ivy
OH
OH
OH
OH
OH
OH
14
Hexylresorcinol
Eugenol
Urushiol
20-41
20 Acidity of Phenols
 Phenols
are significantly more acidic than
alcohols, compounds that also contain the OH group
OH + H2 O
CH3 CH 2 OH + H2 O
O - + H3 O +
CH3 CH 2 O
-
+ H3 O +
pKa = 9.95
pKa = 15.9
20-42
20 Acidity of Phenols
• the greater acidity of phenols compared with alcohols
is due to the greater stability of the phenoxide ion
relative to an alkoxide ion
O:
O:
:
: O:
:
:
:
:
:O:
O:
:
:
:
These three Kekulé
structures are
equivalent
These three contributing structures
delocalize the negative charge onto
the carbons of the ring
20-43
20 Acidity of Phenols
 Alkyl
and halogen substituents effect acidities
by inductive effects
• alkyl groups are electron-releasing
• halogens are electron-withdrawing
OH
OH
OH
m-Cresol
pKa 10.01
OH
Cl
CH3
Phenol
pKa 9.95
OH
CH3
p-Cresol
pKa 10.17
Cl
m-Chloro
phenol
pKa 8.85
p-Chlorophenol
pKa 9.18
20-44
20 Acidities of Phenols
• nitro groups increase the acidity of phenols by both
an electron-withdrawing inductive effect and a
resonance effect
OH
OH
OH
NO2
Phenol
pKa 9.95
m-Nitrophenol
pK a 8.28
NO2
p-Nitrophenol
pKa 7.15
20-45
20 Acidities of Phenols
• part of the acid-strengthening effect of -NO2 is due to
its electron-withdrawing inductive effect
• in addition, -NO2 substituents in the ortho and para
positions help to delocalize the negative charge
O
O
:
delocalization of
negative charge
onto oxygen
O
+
N
O
O
N
+
O
20-46
20 Acidity of Phenols
 Phenols
are weak acids and react with strong
bases to form water-soluble salts
• water-insoluble phenols dissolve in NaOH(aq)
OH
Phenol
pKa 9.95
(stronger acid)
+
N aOH
Sodium
hydroxide
(stronger base)
O - N a+ +
Sodium
phenoxide
(weaker base)
H2 O
Water
pKa 15.7
(weaker acid)
20-47
20 Acidity of Phenols
• most phenols do not react with weak bases such as
NaHCO3; they do not dissolve in aqueous NaHCO3
OH
Phenol
pKa 9.95
(weaker acid)
+
N aHCO 3
Sodium
bicarbonate
(weaker base)
O - N a+ +
Sodium
phenoxide
(stronger base)
H2 CO 3
Carbonic acid
pKa 6.36
(stronger acid)
20-48
20 Alkyl-Aryl Ethers
 Alkyl-aryl
ethers can be prepared by the
Williamson ether synthesis
• but only using phenoxide salts and alkyl halides
• aryl halides are unreactive to SN2 reactions
X + RO - N a+

no reaction
The following two examples illustrate
• the use of a phase-transfer catalyst
• the use of dimethyl sulfate as a methylating agent
20-49
20 Alkyl-Aryl Ethers
OH + CH2 = CHCH2 Cl
Phenol
N aOH, H2 O, CH2 Cl 2
Bu 4 N + Br -
3-Chloropropene
(Allyl chloride)
OCH2 CH= CH2
Phenyl 2-propenyl ether
(Allyl phenyl ether)
O
N aOH, H2 O, CH2 Cl 2
+
OH
CH3 OSOCH3
Bu 4 N + Br O
Phenol
Dimethyl sulfate
OCH3 + N a2 SO 4
Methyl phenyl ether
(Anisole)
20-50
20 Kolbe Carboxylation
 Phenoxide
ions react with carbon dioxide to give
a carboxylic salt
-
OH
O Na
+
CO 2
N aOH
H2 O
Phenol
H2 O
Sodium
phenoxide
OH O
+
CO Na
HCl
OH O
COH
H2 O
Sodium salicylate
Salicylic acid
20-51
20 Kolbe Carboxylation
• the mechanism begins by nucleophilic addition of the
phenoxide ion to a carbonyl group of CO2
O:+
Sodium
phenoxide
O
C
O
O
(1)
O
C-O –
keto-enol
tautomerism
H
A cyclohexadienone
intermediate
(2)
OH O
C
O-
Salicylate anion
20-52
20 Oxidation to Quinones
 Because
of the presence of the electrondonating -OH group, phenols are susceptible to
oxidation by a variety of strong oxidizing agents
OH
O
H 2 Cr O 4
Phenol
O
1,4-Benzoquinone
(p-Quinone)
20-53
20 Oxidation of Phenols
O
OH
OH
O
K2 Cr2 O7
H2 SO4
1,2-Benzenediol
(Catechol)
1,2-Benzoquinone
(o-Quinone)
OH
O
K2 Cr2 O7
H2 SO4
OH
1,4-Benzenediol
(Hydroquinone)
O
1,4-Benzoquinone
(p-Quinone)
20-54
20 Quinones
 Perhaps
the most important chemical property
of quinones is that they are readily reduced to
hydroquinones
O
OH
Na 2 S 2 O4 , H 2 O
(reduction)
O
1,4-Benzoquinone
(p-Quinone)
OH
1,4-Benzenediol
(Hydroquinone)
20-55
20 Vitamin K
• both natural and synthetic vitamin K (menadione) are
1,4-naphthoquinones
O
CH3
Vitamin K 2
7
O
H
O
CH3
2-Methylnaphthalene
CrO 3
oxidation
CH3
O
2-Methyl-1,4-naphthoquinone
(Menadione)
20-56
20 Benzylic Oxidation
 Benzene
is unaffected by strong oxidizing
agents such as H2CrO4 and KMnO4
• halogen and nitro substituents are also unaffected by
these reagents
• an alkyl group with at least one hydrogen on its
benzylic carbon is oxidized to a carboxyl group
CH3
O2 N
COOH
H2 Cr O4
Cl
2-Chloro-4-nitrotoluene
O2 N
Cl
2-Chloro-4-nitrobenzoic acid
20-57
20 Benzylic Oxidation
 if
there is more than one alkyl group on the
benzene ring, each is oxidized to a -COOH group
H3 C
CH3
1,4-Dimethylbenzene
(p-xylene)
K2 Cr 2 O 7
H2 SO 4
O
HOC
O
COH
1,4-Benzenedicarboxylic acid
(terephthalic acid)
20-58
20 Benzylic Chlorination
 Chlorination
(and bromination) is by a radical
mechanism
CH3 + Cl 2
heat
or light
Toluene
CH2 CH3
Ethylbenzene
CH2 Cl + HCl
Chloromethylbenzene
(Benzyl chloride)
NBS
( PhCO 2 ) 2 , CCl 4
Br
CHCH3
1-Bromo-1-phenylethane
20-59
20 Benzylic Reactions
 Benzylic
radicals (and cations too) are easily
formed because of the resonance stabilization of
these intermediates
• the benzyl radical is a hybrid of five contributing
structures
C
C
C
C
C
20-60
20 Benzylic Halogenation
 Benzylic
bromination is highly regioselective
CH2 CH3
Ethylbenzene
NBS
(PhCO2 ) 2 , CCl4
Br
CHCH3
1-Bromo-1-phenylethane
(only product)
20-61
20 Benzylic Halogenation
 Benzylic
chlorination is also regioselective, but
less so than benzylic bromination
CH2 CH3
Ethylbenzene
 The
Cl 2
heat
or light
Cl
CHCH3 +
1-Chloro-1phenylethane
(90%)
CH2 CH2 Cl
1-Chloro-2phenylethane
(10%)
order of regioselectivity is
methyl < 1° < 2° < 3° < allylic = benzylic
Increasing radical stability
which parallels BDEs for the formation of radicals
20-62
20 Hydrogenolysis
 Hydrogenolysis:
cleavage of a single bond by H2
• among ethers, benzylic ethers are unique in that they
are cleaved under conditions of catalytic
hydrogenation
CH 2 O( CH2 ) 5 CH3 + H 2
Benzyl hexyl ether
Pd/ C
CH3
Toluene
+ HO( CH 2 ) 5 CH 3
1-Hexanol
20-63
20 Benzyl Ethers
 The
particular value of benzyl ethers is that they
can serve as protecting groups for the OH
groups of alcohols and phenols
• to carry out hydroboration/oxidation of this alkene,
the phenolic -OH must first be protected; it is acidic
enough to react with BH3 and destroy the reagent
CH2 CH=CH2
OH
2-(3-Propenyl)phenol
(2-Allylphenol)
CH2 CH2 CH2 OH
?
OH
2-(3-Hydroxy-1-propyl)phenol
20-64
20 Benzyl Ethers
• protect the phenolic -OH, carry out the
hydroboration/oxidation, and then remove the benzyl
protecting group by hydrogenolysis
2 . BH3 • THF
1 . ClCH2 Ph
(CH3 CH2 ) 3 N
OH
2-(2-Propenyl)phenol
(2-Allylphenol)
OH
O
Ph
O
H2
Pd/ C
Ph 3 . H2 O2 / NaOH
OH
OH
2-(3-Hydroxypropyl)phenol
20-65
20 Prob 20.15
State the number of p orbital electrons in each.
(a)
(b)
(c)
(e)
(f)
(g)
••
(d)
+
•
+
B
H
(h)
B H
(i)
(j)
••
••
20-66
20 Prob 20.23
From this spectral data, write a structural formula for
compound F, C12H16O.
1
H-NMR
0.83 (d, 6H)
2.11 (m, 1H)
2.30 (d, 2H)
3.64 (s, 2H)
7.2 - 7.4 (m, 5H)
13
C-NMR
207.82
134.24
129.36
128.60
126.86
50.88
50.57
24.43
22.48
20-67
20 Prob 20.24
From this spectral data, write a structural formula for
compound G, C10H10O.
1
H-NMR
2.50 (t, 2H)
3.05 (t, 2H)
3.58 (s, 2H)
7.1 to 7.3 (m, 4H)
13
C-NMR
210.19
136.64
133.25
128.14
127.75
126.82
126.75
45.02
38.11
28.34
20-68
20 Prob 20.29
Compound K, C10H12O2, is insoluble in water, 10% NaOH
and 10% HCl. Given this information and the following
spectral data, propose a structural formula for
compound K.
1
H-NMR
2.10 (s, 3H)
3.61 (s, 2H)
3.77 (s, 3H)
6.86 (d, 2H)
7.12 (d, 2H)
13
C-NMR
206.51 114.17
158.67 55.21
130.33 50.07
126.31 29.03
20-69
20 Prob 20.30
Propose a structural formula for each compound.
(a) C9 H9 Br O 2
1
H-NMR
1.39 (t, 3H)
4.38 (q, 2H)
7.57 (d, 2H)
7.90 (d, 2H)
(b) C8 H9 NO
13
C-NMR
165.73
131.56
131.01
129.84
127.81
61.18
14.18
1
13
H-NMR
C-NMR
2.06 (s, 3H) 168.14
7.01 (t, 1H)
139.24
7.30 (m, 2H) 128.51
7.59 (d, 2H) 122.83
9.90 (s, 1H) 118.90
23.93
(c) C9 H9 NO 3
1
H-NMR
2.10(s, 3H)
7.72 (d, 2H)
7.91 (d, 2H)
10.3 (s, 1H)
12.7 (s, 1H)
13
C-NMR
168.74
166.85
143.23
130.28
124.80
118.09
24.09
20-70
20 Prob 20.31
Each compound shows strong, sharp absorption
between 1700 and 1720 cm-1, and strong, broad
absorption over the region 2500-3500 cm-1.
(a) C1 0 H 1 2 O 3
1
13
H-NMR
C-NMR
2.49 (t, 2H)
173.89
2.80 (t, 2H)
157.57
3.72 (s, 3H) 132.62
6.78 (d, 2H) 128.99
7.11 (d, 2H) 113.55
12.4 (s 1H)
54.84
35.75
29.20
(b) C1 0 H 1 0 O 2
1
H-NMR
2.34(s, 3H)
6.38 (d, 1H)
7.18 (d, 1H)
7.44 (d, 2H)
7.56 (d, 2H)
12.0 (s 1H)
13
C-NMR
167.82
143.82
139.96
131.45
129.37
127.83
111.89
21.13
20-71
20 Prob 20.35
Arrange the entries in each set in order of increasing
acidity.
(a)
OH
(b)
OH
(c)
C CH
CH3 COOH
OH
HCO 3
-
H2 O
OH
CH2 OH
20-72
20 Prob 20.36
Explain the trends in acidity of phenol and the
monofluoro derivatives of phenol.
OH
OH
OH
OH
F
F
pKa 10.0
pKa 8.81
pKa 9.28
F
pKa 9.81
20-73
20 Prob 20.38
From each pair, select the stronger base.
(a)
O - or
(c)
O-
or
OH
-
HCO 3
-
(b)
O-
or
O-
(d)
O-
or
CH3 COO
-
20-74
20 Prob 20.39
Describe a chemical procedure to separate a mixture of
these compounds and recover each in pure form.
CH2 OH
Benzyl alcohol
CH3
OH
o-Cresol
20-75
20 Prob 20.46
Account for the following order of reactivity under SN1
conditions.
R
CH2 Br + CH3 OH
methanol
R
Rate of SN 1 reaction: R = CH3 O > CH3
CH2 OCH3 + HBr
> H > NO 2
20-76
20 Prob 20.47
Propose a mechanism for this rearrangement. Account
for the fact that the product is 2-phenylpropanal rather
than its constitutional isomer, 1-phenyl-1-propanone.
OH
H
H2 SO 4
OH
1-Phenyl-1,2-propanediol
O
+ H2 O
2-Phenylpropanal
20-77
20 Prob 20.49
Synthesize these compounds from ethylbenzene.
(a)
O
COH
Br
(b)
CHCH3
OH
(d)
CHCH3
(e)
CH2 CH (h)
(j)
C CH
O
CCH3 (f)
CH= CH2
CH2 CH2 OH
O
O
(g)
(c)
CH2 COH
(k)
Br Br
(i)
CHCH2
C CCH2 CH= CH2
20-78
20 Prob 20.50
Show how to convert 1-phenylpropane to the following
compounds.
Br
(a) C6 H5
OH
O
(b) C6 H5
(c) C 6 H5
(e) C 6 H5
(f) C 6 H5
Cl
(d) C6 H5
Cl
OH
OH
(g) C6 H5
(h) C6 H5
(i) C 6 H5
Cl
OH
20-79
20 Prob 20.51
Given this retrosynthetic analysis, propose a synthesis
for cabinoxamide. Note that aryl bromides form Grignard
reagents much more readily than aryl chlorides.
CHO
Cl
O
Cl
+
N
N
N
+ Cl
N
Br
Carbinoxamine
20-80
20 Prob 20.52
Propose (1) a mechanism for formation of I, (2) a
structure for II and a mechanism for its formation, and (3)
a mechanism for the formation of cromolyn sodium.
HO
+
O
OH
2,6-Dihydroxyacetophenone
O
HO
Cl base
OO
OH
Et O- C- C-OEt
Et O - N a+
OH
Epichlorohydrin
O
O
O
O
I
O
II
+
N
a
N aOH, H2 O
-
O
O
O
O
O - N a+
OH
O
O
O
O
Cromolyn sodium
20-81
20 Prob 20.53
Describe the chemistry of each step.
O
OH
Br
1. BH3, THF
2. H2O2, NaOH
CH3ONa
2,4,6-Trimethylphenol
(A)
O
O
(B)
O
CrO 3, H2SO4
Br2, H2O
acetone
KBr
OH
(C) COOH
Br
KOH
O
O
THF
(D)
20-82
20 Prob 20.53 (cont’d)
describe the chemistry of each step
O
Br
O
KOH
O
O
O
Br2, H2O
THF
Br
O
O
KBr
O
O
(D)
COOH
(E)
(F)
20-83
20
Aromatics I
End of Chapter 20
20-84