Transcript Electophilic Aromatic Substituion

16. Chemistry of
Benzene: Electrophilic
Aromatic Substitution
Substitution Reactions of Benzene
and Its Derivatives
 Benzene is aromatic: a cyclic conjugated compound with 6  electrons
 Reactions of benzene lead to the retention of the aromatic core
2
Why this Chapter?
 Continuation of coverage of aromatic compounds in
preceding chapter…focus shift to understanding
reactions
 Examine relationship between aromatic structure
and reactivity
 Relationship critical to understanding of how
biological molecules/pharmaceutical agents are
synthesized
3
16.1 Electrophilic Aromatic Substitution
Reactions (EAS):
 Reactions typical of addition to alkenes do not
work on aromatic double bonds.
 Need more electrophilic (more positive) halogen in
order to break an aromatic double bond.
4
EAS: Bromination
EAS occurs in two steps: Addition followed by Elimination
 FeBr3 acts as a catalyst to polarize the bromine reagent
and so make it more positive (more electrophilic)
 The  electrons of the aromatic ring act as a nucleophile
toward the now more electrophilic Br2 (in the FeBr3 complex)
Addition
The cationic addition intermediate is called a sigma complex
5
EAS: Bromination
EAS occurs in two steps: Addition followed by Elimination
 The cationic
addition
intermediate
(sigma complex)
transfers a
proton to FeBr4-
Addition
(from Br- and FeBr3)
 Aromaticity is
restored

(in contrast with
addition in alkenes
which is not followed
by elimination)
Elimination
6
EAS: Bromination
EAS occurs in two steps: Addition followed by Elimination
 The cationic
addition
intermediate
(sigma complex)
transfers a
proton to FeBr4(from Br- and FeBr3)
 Aromaticity is
restored

(in contrast with
addition in alkenes
which is not followed
by elimination)
Elimination is driven by the stability of becoming aromatic
7
16.2 Other Aromatic Substitutions
 Chlorine and iodine (but not fluorine, which is too reactive) can
produce aromatic substitution products
Chlorination
requires FeCl3
Iodine must be
oxidized (with
Cu+ or peroxide)
to form a more
powerful I+
species
8
Example:
A common Amino Acid
9
Aromatic Nitration
 The combination of nitric acid and sulfuric acid produces
NO2+ (nitronium ion)
The reaction with benzene produces nitrobenzene
10
Reduction of Nitration Product:
To put an amino (NH2) group on an aromatic ring
first Nitrate
then reduce
11
Aromatic Sulfonation
 Substitution of H by SO3 (sulfonation) with a mixture of
sulfuric acid and SO3
 Reactive species is sulfur trioxide or its conjugate acid
Reaction is Reversible:
Can remove sulfonyl group with H+,H2O, heat
12
Aromatic Hydroxylation
 Hard to directly add –OH to an aromatic ring in lab.
OH
H2O2
HSO3F
Formation of Electrophile
O
H O S
H O O H
O
F
H
H O O H
O H
13
Aromatic Hydroxylation
 Biological systems use Enzymes to hydroxylate aromatics
14
16.3 Alkylation of Aromatic Rings: The
Friedel–Crafts Reaction
Electrophilic C+ forms
 Alkylation (substitution
of Carbon compounds)
among most useful
electrophilic
aromatic subsitution
reactions
 Aromatic
substitution of R+ for
H+
 Aluminum chloride
promotes the
formation of the
carbocation
Addition
Elimination
15
Friedel-Crafts Alkylation: Limitations
 Only alkyl halides can be used (F, Cl, I, Br)
 Aryl halides (Ar-X) and vinylic
halides (CH2=CH-X) do not react
(their carbocations are too hard to form)
 Will not work with rings containing an amino group substituent or a
strongly electron-withdrawing group
16
Friedel-Crafts Alkylation: Control Problems
 Multiple alkylations can occur because the first
alkylation is activating (makes aromatic ring more nucleophilic
so more likely to add again)
17
Friedel-Crafts Alkylation: Control Problems
Carbocation Rearrangements
 Similar to
those that
occur during
electrophilic
additions to
alkenes
 Can involve H
or alkyl shifts
18
Example:
Cl
AlCl3
19
Friedel-Crafts Acylation
 Reaction of an acid chloride (RCOCl) and an aromatic ring
in the presence of AlCl3 introduces acyl group, COR
Benzene with acetyl chloride yields acetophenone
20
Friedel-Crafts Acylation: Mechanism
 Similar to alkylation
 Reactive electrophile: resonance-stabilized acyl cation
 An acyl cation does not rearrange
Electrophilic C+ forms
= Acylium ion
Addition
Elimination
21
EAS: Summary
22
16.4 Substituent Effects
+
Nucleophile
E+
E
Electrophile
 Substituents that donate electrons make ring more nucleophilic

Electron donating groups (edg) activate the ring toward EAS
 Substitutients that withdraw electrons make less nucleophilic

Electron withdrawing groups (ewg) deactivate the ring toward EAS
23
Substituent Effects
 Substituents influence by induction
edg activate the ring toward EAS
ewg deactivate the ring toward EAS
Halogens, C=O, CN, and NO2
withdraw electrons through s
bond connected to ring
Alkyl groups donate electrons
24
Substituent Effects
 Substituents influence by resonance
Halogen, OH, alkoxyl (OR), and amino substituents donate electrons
edg activate the ring toward EAS
25
Substituent Effects
 Substituents influence by resonance
C=O, CN, NO2 substituents withdraw electrons from the aromatic ring by resonance
ewg deactivate the ring toward EAS
26
Substituent Effects
 Substituents influence by resonance
edgs put negative character at o- and p- positions
(make o- and p- positions more nucleophilic)
Direct incoming electrophile into o- or p- spots
ewgs put positive character at o- and p- positions
(make o- and p- positions less nucleophilic)
Direct incoming electrophile into m- spots
27
28
Explanation of Substituent Effects
 Activating groups
donate electrons
to the ring,
stabilizing the
Wheland
intermediate
(carbocation)
 Deactivating
groups withdraw
electrons from the
ring, destabilizing
the Wheland
intermediate
29
16.5 Ortho- & Para-Directing
Activators: Alkyl Groups
 Alkyl groups activate: direct further substitution to positions ortho and
para to themselves
Alkyl group is most effective in
the ortho and para positions
30
Ortho- and Para-Directing Activators:
OH and NH2
 Alkoxyl, and amino groups have a strong, electron-donating
resonance effect
Most pronounced
at the ortho and
para positions
31
Ortho- & Para-Directing Deactivators:
Halogens
 Electron-withdrawing inductive effect outweighs weaker electron-
donating resonance effect
Resonance
effect is only at
the ortho and
para positions,
stabilizing
carbocation
intermediate
32
Meta-Directing Deactivators
 Inductive and resonance effects reinforce each other
 Ortho and para intermediates destabilized by deactivation of
carbocation intermediate
Resonance
cannot produce
stabilization
33
Summary:
34
Another
Summary:
35
16.6 Trisubstituted Benzenes:
Additivity of Effects
 If the directing effects of the two groups are the
same, the result is additive
36
Learning Check:
Br2, FeBr3
?
37
Solution:
Br2, FeBr3
?
38
39
Substituents with Opposite Effects
 If the directing effects of two groups oppose each other,
the more powerful activating group decides the principal
outcome
 Usually gives mixtures of products
40
Meta-Disubstituted Compounds
 The reaction site is too hindered
 To make aromatic rings with three adjacent substituents,
it is best to start with an ortho-disubstituted compound
41
16.7 Nucleophilic Aromatic Substitution
Aryl halides with ewg’s o- and p- react with nucleophiles
Sanger’s reagent: For labeling proteins in biochemistry
42
Nucleophilic Aromatic Substitution
43
Nucleophilic Aromatic Substitution
44
Nucleophilic Aromatic Substitution
ewg’s o- and p- stabilize the intermediate
Intermediate
Meisenheimer complex
is stabilized by electronwithdrawal
Halide ion is lost to give
aromatic ring
45
16.8 Benzyne
 Phenol is prepared on an industrial scale by treatment of chlorobenzene
with dilute aqueous NaOH at 340°C under high pressure
Elimination reaction gives a triple bond intermediate called benzyne
46
Evidence for Benzyne
 Bromobenzene with
14C*
only at C1 gives substitution product with
label scrambled between C1 and C2
Reaction proceeds through a symmetrical intermediate in which C1 & C2
are equivalent— must be benzyne
47
Structure of Benzyne
 Benzyne is a highly distorted alkyne
 The triple bond uses sp2-hybridized carbons, not the
usual sp
 The triple bond has one  bond formed by p–p
overlap and another by weak sp2–sp2 overlap
48
16.9 Benzylic Oxidation
 Alkyl side chains with a C-H next to the ring can be
oxidized to CO2H by strong reagents such as



KMnO4 and
Na2Cr2O7
O2, Co(III)
 Converts an alkylbenzene into a benzoic acid,
 ArR  ArCO2H
49
Examples:
50
Benzylic Bromination
 Reaction of an alkylbenzene with N-bromo-succinimide
(NBS) and benzoyl peroxide (radical initiator) introduces
Br into the side chain
51
Mechanism of NBS (Radical) Reaction
 Abstraction of a benzylic hydrogen atom by a bromine radical
generates an intermediate benzylic radical which reacts with Br2 to
yield product
 Br· radical cycles back into reaction to carry chain
52
Mechanism of NBS (Radical) Reaction
 Benzylic radical is stabilized by resonance
53
16.10 Aromatic Reductions
 Aromatic rings are inert to catalytic hydrogenation under
conditions that reduce alkene double bonds
 Can selectively reduce an alkene double bond in the
presence of an aromatic ring
54
Aromatic Reductions
 Reduction of aromatic ring requires more powerful reducing conditions
•high pressure
•rhodium catalysts
55
Reduction of Aryl Alkyl Ketones
 Aromatic ring activates neighboring carbonyl group toward reduction
Ketone is
converted into an
alkylbenzene by
cat H2/Pd
56
Examples:
57
Learning Check:
 Synthesize para-chlorobenzoic acid from benzene.
 Synthesize meta-chlorobenzoic acid from benzene.
58
Solution:
 Synthesize para-chlorobenzoic acid from benzene.
 Synthesize meta-chlorobenzoic acid from benzene.
59
16.11 Synthesis of Trisubstituted Benzenes
 plan a sequence of reactions in right order is valuable to
synthesis of substituted aromatic rings
60
Synthesis of Trisubstituted Benzenes
Work Backwards
61
Synthesis of Trisubstituted Benzenes
Work Backwards
62
Synthesis of Trisubstituted Benzenes
63
Learning Check:
 Synthesize 4-chloro-2-propylbenzenesulfonic acid from benzene.
SO3H
Cl
CH2CH2CH3
64
Solution:
 Synthesize 4-chloro-2-propylbenzenesulfonic acid from benzene.
SO3H
Cl
CH2CH2CH3
65
Learning Check:
Which of the following groups is an activator?
O
O
I
A
C
B
CH 3
C
C
O
N(CH3)2
SO 3H
D
O
C
E
CH3
Solution:
Which of the following groups is an activator?
O
O
I
A
C
B
CH 3
C
C
O
N(CH3)2
SO 3H
D
Generally and except for the halogens, groups
with lone pairs of electrons adjacent to the ring
are activators.
O
C
E
CH3
Learning Check:
Which of the following groups is an ortho- para- director?
A.
B.
C.
D.
E.
-Cl
-CHO
-CN
-NO2
None of the
above
Solution:
Which of the following groups is an ortho- para- director?
A.
B.
C.
D.
E.
-Cl
-CHO
-CN
-NO2
None of the
above
Although –Cl is a deactivator, it’s
lone pairs of electrons stabilize
electrophilic aromatic substitution
reactions and it is an o- pdirector.
Learning Check:
Which of the following groups most strongly activates an
aromatic ring toward Friedel-Crafts acylation?
A.
B.
C.
D.
E.
-NH2
-OCH3
-O-C(=O)CH3
NO2
H
Solution:
Which of the following groups most strongly activates an
aromatic ring toward Friedel-Crafts acylation?
A.
B.
C.
D.
E.
-NH2
-OCH3
-O-C(=O)CH3
NO2
H
Although the –NH2 group is
a stronger activator it
complexes with AlCl3 to
form a strongly deactivating
ammonium group.
Consequently, the methoxy
is a better activator.
Learning Check:
Which compound is the major product of the chlorination
shown below?
O
NH
A
Cl2
C
O
Cl
NH
C
FeCl3
Cl
B
NH
O
C
O
NH
C
C
Cl
Cl
O
D
NH
C
O
E
NH
C
Cl
Solution:
Which compound is the major product of the chlorination
shown below?
The nitrogen is an activator and o- p- director.
The C=O is a deactivator.
O
NH
A
Cl2
C
O
Cl
NH
C
FeCl3
Cl
B
NH
O
C
O
NH
C
C
Cl
Cl
O
D
NH
C
O
E
NH
C
Cl
Learning Check:
Which step in the following reaction will cause the proposed synthesis to fail?
A
B
Cl
NBS
Br
light
AlCl3 cat.
C
(H3C)3C
O
E
MnO2
HO
D
1. Hg(OAc)2, H 2O
2. NaBH4
O-
Solution:
Which step in the following reaction will cause the proposed synthesis to fail?
A
B
Cl
NBS
Br
light
AlCl3 cat.
C
(H3C)3C
O
E
MnO2
HO
O-
D
1. Hg(OAc)2, H 2O
2. NaBH4
Addition of water in step “D” occurs with Markovnikov selectivity.
What is the major product of the reaction of
nitrobenzene with Br2/FeBr3?
1.
2.
20%
20%
20%
20%
3
4
20%
3.
4.
5.
1
2
5
Which of the following compounds would react fastest
with HNO3/H2SO4?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
nitrobenzene (PhNO2)
toluene (PhCH3)
bromobenzene (PhBr)
anisole (PhOCH3)
benzoic acid (PhCOOH)
1
2
5
Which structure is a major intermediate formed in the
electrophilic nitration of chlorobenzene?
1.
2.
20%
20%
20%
20%
3
4
20%
3.
4.
5.
1
2
5
Which series of reactions will convert benzene into pnitrobenzoic acid?
1.
2.
3.
4.
5.
CH3Br/AlBr3 followed by
HNO3/H2SO4 followed by
KMnO4/H2O
HNO3/H2SO4 followed by
KMnO4/H2O followed by
CH3Br/AlBr3
KMnO4/H2O followed by
CH3Br/FeBr3 followed by
HNO3/H2SO4
CH3Br/AlBr3 followed by
KMnO4/H2O followed by
HNO3/H2SO4
HNO3/H2SO4 followed by
CH3Br/AlBr3 followed by
KMnO4/H2O
20%
1
20%
2
20%
20%
3
4
20%
5
Select the major product of the following reaction.
KMnO4
20%
20%
20%
20%
3
4
20%
H2O
1.
2.
3.
4.
5.
1
2
5
What is the electrophile in the following aromatic
substitution?
Cl
20%
AlCl3
20%
20%
20%
3
4
20%
2.
1.
3.
4.
5.
1
2
5
What is the best sequence of reactions to synthesize the desired
product (Pr = propyl)?
Pr
1
2
3
20%
20%
20%
20%
2
3
4
20%
NH2
Step 1
Step 2
Step 3
1 HNO3/H2SO4
1.
X
SO4
2 HNO
2. 3/H2X
3.
X
3 AlCl3/PrCl
4.
X
SO4
4 HNO
5. 3/H2X
PrMgBr/H3O+
H2/Pd
AlCl3/PrBr
H2/Pd
HNO3/H2SO4
H2/Pd
H2/Pd
PrMgBr/H3O+
5 AlCl3/CH3CH2COCl
HNO3/H2SO4
H2/Pd
1
5
What are the reagents necessary for step 3 in the
following transformation?
1
2
20%
3
20%
20%
20%
3
4
20%
SO3H
1.
2.
3.
4.
5.
CH3COCl/AlCl3
CH3CH2Br/AlBr3
SO3/H2SO4
H2/Pd
SOCl2/AlCl3
1
2
5
Which of the following carbons of the benzyl radical
has the smallest unpaired electron density?
H
1
20%
H
5
20%
20%
20%
3
4
20%
2
3
1.
2.
3.
4.
5.
1
2
3
4
5
4
1
2
5
What would be the main product of nitration of
benzenesulfonic acid?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
o-nitrobenzenesulfonic
acid
p-nitrobenzenesulfonic
acid
2-nitrobenzoic acid
m-nitrobenzenesulfonic
acid
m-nitrobenzoic acid
1
2
5
Fluorine is less deactivating than chlorine in the aromatic
electrophilic substitution reactions. What is the main reason for
this reactivity trend?
1.
2.
3.
4.
5.
Fluorine forms stronger π
bonds than chlorine,
providing more resonance
stabilization.
Chlorine is more
electronegative than fluorine.
Because of it size chlorine
has stronger inductive
influence than fluorine.
Fluorine is smaller, making
the entry of electrophiles
easier.
Chlorine forms complexes
with electrophiles,
diminishing their reactivity.
20%
1
20%
2
20%
20%
3
4
20%
5
Which of the following represent the intermediate formed in the
reaction between p-chloronitrobenzene and hydroxide ion?
1.
2.
20%
20%
20%
20%
3
4
20%
3.
4.
5.
1
2
5
Based on the electronic structure, what kind of substituent effect
would you expect from the nitroso group?
25%
N O
1.
2.
3.
4.
25%
25%
25%
nitroso group
o,p-directing, deactivating
o,p-directing activating
m-directing, activating
m-directing, deactivating
1
2
3
4
Which one is not a limitation of Friedel-Crafts
alkylations?
1.
2.
3.
4.
5.
carbocation rearrangements
may occur
polyalkylation products are
possible
only substrates with
selected activating groups
can be used
vinyl halides cannot be used
to generate electrophiles
the halogen in the aluminum
halide must match one in
the alkyl halide
20%
1
20%
2
20%
20%
3
4
20%
5
What is the major product of the following reaction?
CH3
HOOC
Br2
20%
20%
20%
20%
3
4
20%
FeBr3
1.
2.
3.
4.
5.
1
2
5
What is the result of the reaction between mbromotoluene and sodium amide?
NaNH2
Br
1.
2.
3.
4.
5.
20%
20%
20%
20%
3
4
20%
NH3(l)
2-aminotoluene
3-aminotoluene
4-aminotoluene
all of these
none of these
1
2
5
Aromatic compounds can be oxidized to their radical cations by
removal of just one electron. Which of the following will be the
easiest to oxidize to a radical cation?
1.
2.
20%
20%
20%
20%
3
4
20%
3.
4.
5.
1
2
5
Which of the following is not a practical method to generate an
electrophile for aromatic substitution reaction?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
HNO3/H2SO4
Cl2/FeBr3
I2/CuCl2
SO3/H2SO4
CH3COCl/AlBr3
1
2
5