Transcript Benzene and Its Derivatives Ch#4

Aromatic Compounds
Discovery of Benzene
• Isolated in 1825 by Michael Faraday who
determined C:H ratio to be 1:1.
• Synthesized in 1834 by Eilhard Mitscherlich
who determined molecular formula to be
C6H6.
• Other related compounds with low C:H ratios
had a pleasant smell, so they were classified
as aromatic.
Benzene Structures
• Proposed in 1866 by Friedrich Kekulé, shortly
after multiple bonds were suggested.
• Failed to explain existence of only one isomer
of 1,2-dichlorobenzene.
H
H C
C
H
C
C
C
H
C
H
H
Kekule Structure
Prismane Structure
Benzene Structures
• Each sp2 hybridized C in the ring has an
unhybridized p orbital perpendicular to the
ring which overlaps around the ring.
Resonance Structure
=>
Unusual Reactions
• Alkene + KMnO4  diol (addition)
Benzene + KMnO4  no reaction.
• Alkene + Br2/CCl4  dibromide (addition)
Benzene + Br2/CCl4  no reaction.
• With FeCl3 catalyst, Br2 reacts with benzene to
form bromobenzene + HBr
(substitution!). Double bonds remain.
Unusual Stability
Annulenes
• All cyclic conjugated
hydrocarbons were
proposed to be
aromatic.
• However,
cyclobutadiene is so
reactive that it
dimerizes before it can
be isolated.
• And cyclooctatetraene
adds Br2 readily.
[6]anulene
[4]anulene
[8]anulene
What Does It Take to Be Aromatic?
• Alternating double and single bonds
• A magic number of pi electrons
• Resonance structures must be able to move
the pi electrons in a circular manar.
• Non bonding electrons can also participate in
the resonance structures.
Hückel’s Rule
• Hückel’s Rule is used to generate the magic
number of pi electrons.
• If the compound has a continuous ring of
overlapping p orbitals and has 4N + 2 pi
electrons, it is aromatic.
• If the compound has a continuous ring of
overlapping p orbitals and has 4N electrons, it
is antiaromatic.
• N is any integer, starting at zero
Generation of Magic Pi Electrons
Value of N
0
1
2
3
4N + 2
2
6
10
14
[N]Annulenes
• [4]Annulene is antiaromatic (4N e-’s)
• [8]Annulene would be antiaromatic, but it’s
not planar, so it’s nonaromatic.
• [10]Annulene is aromatic except for the
isomers that are not planar.
• Larger 4N annulenes are not antiaromatic
because they are flexible enough to become
nonplanar.
Tropylium Ion
• The cycloheptatrienyl cation has 6 p electrons and an
empty p orbital.
• Aromatic: more stable than open chain ion.
H
H
OH
+
H , H2O
+
Which of the Following are Aromatic?
a.
e.
b.
f.
c.
g.
d.
h.
Disubstituted Benzenes
• The prefixes ortho-, meta-, and para- are
• commonly used for the 1,2-, 1,3-, and 1,4• positions, respectively.
=>
3 or More Substituents
• Use the smallest possible numbers, but
• the carbon with a functional group is #1.
OH
O 2N
NO2
NO2
1,3,5-t rinit robenzene
O 2N
NO2
NO2
2,4,6-t rinit rophenol
Common Names of Benzene Derivatives
OH
CH3
phenol
toluene
H
C CH2
styrene
OCH3
NH2
aniline
anisole
O
O
O
C
C
C
acetophenone
CH3
benzaldehyde
H
benzoic acid
OH
Phenyl and Benzyl
• Phenyl indicates the benzene ring
• attachment. The benzyl group has
• an additional carbon.
Br
phenyl bromide
CH2Br
benzyl bromide
Fused Ring Hydrocarbons
• Naphthalene
• Anthracene
• Phenanthrene
Larger Polynuclear
Aromatic Hydrocarbons
• Formed in combustion (tobacco smoke).
• Many are carcinogenic.
• Epoxides form, combine with DNA base.
pyrene
=>
Physical Properties
• Melting points: More symmetrical than
corresponding alkane, pack better into
crystals, so higher melting points.
• Boiling points: Dependent on dipole moment,
so ortho > meta > para, for disubstituted
benzenes.
• Density: More dense than nonaromatics, less
dense than water.
• Solubility: Generally insoluble in water.
Electrophilic Aromatic Substitution
Electrophile substitutes for a hydrogen on
the benzene ring.
Chapter 17
21
Mechanism
Step 1: Attack on the electrophile forms the sigma complex.
Step 2: Loss of a proton gives the substitution product.
Chapter 17
22
Bromination of Benzene
• Requires a stronger electrophile than Br2.
• Use a strong Lewis acid catalyst, FeBr3.
Br Br
FeBr3
Br
H
Br
H
Br
-
FeBr3
H
H
H
+
H
+
Br
FeBr3
H
H
Br
+
H
_
+ FeBr4
H
H
H
Br
+
Chapter 17
HBr
23
Comparison with Alkenes
• Cyclohexene adds Br2, H = -121 kJ
• Addition to benzene is endothermic, not
normally seen.
• Substitution of Br for H retains aromaticity, H
= -45 kJ.
• Formation of sigma complex is rate-limiting.
Chapter 17
24
Energy Diagram for Bromination
Chapter 17
25
Chlorination and Iodination
• Chlorination is similar to bromination. Use
AlCl3 as the Lewis acid catalyst.
• Iodination requires an acidic oxidizing
agent, like nitric acid, which oxidizes the
iodine to an iodonium ion.
+
H
+
+ HNO 3 + 1/2 I2
I
Chapter 17
+
NO 2 + H2O
26
Nitration of Benzene
Use sulfuric acid with nitric acid to form the
nitronium ion electrophile.
O
H O
S
O H
O
H O
H O N
H O N
+
O
O
O
H O
H O N
+
O
O
H2O +
N+
_
+ HSO4
NO2+ then forms a sigma complex with
benzene, loses H+ to form nitrobenzene.
O
Chapter 17
27
Sulfonation
Sulfur trioxide, SO3, in fuming sulfuric
acid is the electrophile.
O
_
O
O
S
S+
O
O
O
O
O
S+
_
O
O
_
O
S
O
O
H O
S O
+
O
H
_
S +
O
O
HO
O
S
O
benzenesulfonic acid
Chapter 9
28
Nitration of Toluene
• Toluene reacts 25 times faster than benzene.
The methyl group is an activating group.
• The product mix contains mostly ortho and para
substituted molecules.
Chapter 17
29
Sigma Complex
Intermediate is
more stable if
nitration
occurs at the
ortho
or para
position.
=>
Chapter 17
30
Energy Diagram
Chapter 17
31
Activating, O-, P-Directing Substituents
• Alkyl groups stabilize the sigma complex by
induction, donating electron density through
the sigma bond.
• Substituents with a lone pair of electrons
stabilize the sigma complex by resonance.
OCH3
+
OCH3
NO2
NO2
+
H
H
Chapter 17
32
Substitution on Anisole
Chapter 17
33
The Amino Group
Aniline, like anisole, reacts with bromine water
(without a catalyst) to yield the tribromide.
Sodium bicarbonate is added to neutralize
the HBr that’s also formed.
Chapter 17
34
Summary of Activators
Chapter 17
35
Deactivating Meta-Directing Substituents
• Electrophilic substitution reactions for
nitrobenzene are 100,000 times slower
than for benzene.
• The product mix contains mostly the meta
isomer, only small amounts of the ortho
and para isomers.
• Meta-directors deactivate all positions on
the ring, but the meta position is less
deactivated.
Chapter 17
36
Ortho Substitutionon Nitrobenzene
Chapter 17
37
Para Substitution on Nitrobenzene
Chapter 17
38
Meta Substitution on Nitrobenzene
Chapter 17
39
Energy Diagram
Chapter 17
40
Structure of Meta-Directing Deactivators
• The atom attached to the aromatic ring will
have a partial positive charge.
• Electron density is withdrawn inductively
along the sigma bond, so the ring is less
electron-rich than benzene.
Chapter 17
41
Summary of Deactivators
Chapter 17
42
More Deactivators
Chapter 17
43
Halobenzenes
• Halogens are deactivating toward
electrophilic substitution, but are ortho,
para-directing!
• Since halogens are very electronegative, they
withdraw electron density from the ring
inductively along the sigma bond.
• But halogens have lone pairs of electrons that
can stabilize the sigma complex by
resonance.
Chapter 17
44
Sigma Complex for Bromobenzene
Ortho and para attacks produce a bromonium ion
and other resonance structures.
No bromonium ion
possible with meta attack.
Chapter 17
45
Energy Diagram
Chapter 17
46
Summary of Directing Effects
Chapter 17
47
Multiple Substituents
The most strongly activating substituent will
determine the position of the next
substitution. May have mixtures.
OCH3
OCH3
SO3H
SO3
O2N
H2SO4
OCH3
+
O2N
O2N
SO3H
Chapter 17
48
Friedel-Crafts Alkylation
• Synthesis of alkyl benzenes from alkyl
halides and a Lewis acid, usually AlCl3.
• Reactions of alkyl halide with Lewis acid
produces a carbocation which is the
electrophile.
• Other sources of carbocations:
alkenes + HF, or alcohols + BF3.
Chapter 17
49
Examples ofCarbocation Formation
Cl
CH3
H2C
OH
H3C
CH CH3
CH CH3
+ AlCl3
HF
CH CH3
BF3
_
CH3 +
C Cl AlCl3
H3C H
_
F
+
H3C CH CH3
+ BF3
H O
H3C
CH CH3
Chapter 17
_
+
H3C CH CH3 + HOBF3
50
Formation of Alkyl Benzene
CH3
+C
H
H
+
CH3
H
F
H
+
CH(CH3)2
F
CH(CH3)2
-
B OH
CH3
F
CH
+
CH3
H
Chapter 17
HF
F
B OH
F
51
Friedel-CraftsAcylation
• Acyl chloride is used in place of alkyl
chloride.
• The acylium ion intermediate is resonance
stabilized and does not rearrange like a
carbocation.
• The product is a phenyl ketone that is less
reactive than benzene.
Chapter 17
52
Mechanism of Acylation
O
O
C
C+
R
+
H
R
Cl
_
AlCl3
O
C
HCl
R +
AlCl3
H
Chapter 17
53
Catalytic Hydrogenation
• Elevated heat and pressure are required.
• Possible catalysts: Pt, Pd, Ni, Ru, Rh.
• Reduction cannot be stopped at an
intermediate stage.
CH3
CH3
3 H2, 1000 psi
Ru, 100°C
CH3
Chapter 17
CH3
54
Side-Chain Oxidation
Alkylbenzenes are oxidized to benzoic acid by
hot KMnO4 or Na2Cr2O7/H2SO4.
CH(CH3)2
-
KMnO4, OH
CH CH2
H2O, heat
Chapter 17
_
COO
_
COO
55
Side-Chain Halogenation
• Benzylic position is the most reactive.
• Chlorination is not as selective as
bromination, results in mixtures.
• Br2 reacts only at the benzylic position.
Br
CH2CH2CH3
Br2, h
Chapter 17
CHCH2CH3
56
AROMATIC REVIEW
Give the shape of benzene.
a. Tetrahedral
b.Bent
c. Trigonal pyramidal
d.Planar
Answer
a. Tetrahedral
b.Bent
c. Trigonal pyramidal
d.Planar
All six carbons and six hydrogens are in
the same plane.
Give the hybridization of each carbon
in benzene.
a.
b.
c.
d.
sp
sp2
sp3
sp4
Answer
a.
b.
c.
d.
sp
sp2
sp3
sp4
Each carbon in benzene is sp2 hybridized.
Give the bond angle of the atoms in
benzene.
a. 45°
b.60°
c. 90°
d.109.5°
e.120°
Answer
a. 45°
b.60°
c. 90°
d.109.5°
e.120°
The carbons are 120o apart in benzene.
Classify
a. Aromatic
b.Antiaromatic
c. Nonaromatic
d.Acyclic
.
Answer
a. Aromatic
b.Antiaromatic
c. Nonaromatic
d.Acyclic
The compound gives a whole number for
N in Hűckel’s rule (4N + 2 = 6, N = 1).
Classify
a. Aromatic
b.Antiaromatic
c. Nonaromatic
d.Acyclic
.
Answer
a. Aromatic
b.Antiaromatic
c. Nonaromatic
d.Acyclic
The compound is cyclic and has
continuous delocalized electrons, but
does not give a whole number for
Hűckel’s rule (4N + 2 = 8, N = 3/2).
Classify
a. Aromatic
b.Antiaromatic
c. Nonaromatic
d.Acyclic
.
Answer
a. Aromatic
b.Antiaromatic
c. Nonaromatic
d.Acyclic
This cyclic compound does not have a
continuous, overlapping ring of p
orbitals and is nonaromatic.
Name
a. Anthracene
b.Naphthalene
c. Phenanthrene
d.Benzene
Answer
a. Anthracene
b.Naphthalene
c. Phenanthrene
d.Benzene
Naphthalene contains two benzene
rings fused together.
Name
a. Anthracene
b.Naphthalene
c. Phenanthrene
d.Benzene
Answer
a. Anthracene
b.Naphthalene
c. Phenanthrene
d.Benzene
Anthracene contains three benzene rings
fused together.
Identify how carbon, diamond, and
graphite are related.
a. They are enantiomers of carbon.
b.They are diastereomers of carbon.
c. They are allotropes of carbon.
d.They have the same properties.
Answer
a. They are enantiomers of carbon.
b.They are diastereomers of carbon.
c. They are allotropes of carbon.
d.They have the same properties.
Name
Cl
NH2
Br
a. 4-Bromo-3chloroaniline
b.4-Bromo-3chlorophenol
c. 4-Bromo-3chloroanisole
d.1-Bromo-2-chloro-4aniline
e.1-Bromo-2-chloro-4phenol
Answer
a. 4-Bromo-3chloroaniline
b.4-Bromo-3chlorophenol
c. 4-Bromo-3chloroanisole
d.1-Bromo-2-chloro-4aniline
e.1-Bromo-2-chloro-4phenol
Aniline is the parent compound. The NH2 is
at position one.
CH3
Name
OH
a.
b.
c.
d.
e.
p-Methylphenol
m-Methylphenol
o-Methylphenol
4-Methylphenol
3-Methylphenol
Answer
a.
b.
c.
d.
e.
p-Methylphenol
m-Methylphenol
o-Methylphenol
4-Methylphenol
3-Methylphenol
The groups are on adjacent carbons,
which is ortho.
Name
O
C
NO2
H
a. 3-Amino-5benzaldehyde
b.5-Amino-3benzaldehyde
c. 3-Aminobenzaldehyde
d.5-Nitro-3benzaldehyde
e.3-Nitrobenzaldehyde
Answer
a. 3-Amino-5benzaldehyde
b.5-Amino-3benzaldehyde
c. 3-Aminobenzaldehyde
d.5-Nitro-3benzaldehyde
e.3-Nitrobenzaldehyde
Benzaldehyde is the parent compound.
Name
O
C
NO2
H
a. o-Aminobenzaldehyde
b. m-Aminobenzaldehyde
c. p-Aminobenzaldehyde
d. o-Nitrobenzaldehyde
e. m-Nitrobenzaldehyde
Answer
a. o-Aminobenzaldehyde
b. m-Aminobenzaldehyde
c. p-Aminobenzaldehyde
d. o-Nitrobenzaldehyde
e. m-Nitrobenzaldehyde
Benzaldehyde is the parent compound.
Name C6H5CH2CH2C≡CCH3.
a. 1-Phenyl-3-pentyne
b.5-Phenyl-2-pentyne
c. 4-Phenyl-2-pentyne
d.1-Phenyl-2-butyne
e.1-Phenyl-3-butyne
Answer
a. 1-Phenyl-3-pentyne
b.5-Phenyl-2-pentyne
c. 4-Phenyl-2-pentyne
d.1-Phenyl-2-butyne
e.1-Phenyl-3-butyne
Identify the slow step in
electrophilic aromatic substitution.
a. Formation of a stronger nucleophile.
b.Formation of the benzenonium ion.
c. Deprotonation to regain aromaticity.
d.Formation of a carbanion.
Answer
a. Formation of a stronger nucleophile.
b.Formation of the benzenonium ion.
c. Deprotonation to regain aromaticity.
d.Formation of a carbanion.
Benzene attacking the electrophile to form
the benzenonium ion is the slow step.
Cl2
AlCl3
a. Hexachlorobenzene
b.Hexachlorocyclohexane
c. 5,6-Dichloro-1,3-cyclohexadiene
d.Chlorobenzene
Answer
a. Hexachlorobenzene
b.Hexachlorocyclohexane
c. 5,6-Dichloro-1,3-cyclohexadiene
d.Chlorobenzene
One chlorine atom substitutes on the
benzene.
1. HNO3, H2SO4
2. Zn, aq. HCl
a. Nitrobenzene
b.Aniline
c. Chlorobenzene
d.Benzenesulfonic acid
Answer
a. Nitrobenzene
b.Aniline
c. Chlorobenzene
d.Benzenesulfonic acid
A nitro group is added, which is then
reduced to an amino group.
SO3
a. Nitrobenzene
b.Aniline
c. Chlorobenzene
d.Benzenesulfonic acid
Answer
a. Nitrobenzene
b.Aniline
c. Chlorobenzene
d.Benzenesulfonic acid
A sulfonic acid group is added to the
benzene.
CH2CH3
HNO3
H2SO4
a. 2- and 4-nitroethylbenzene
b.3-Nitroethylbenzene
c. 2- and 4-ethylbenzenesulfonic acid
d.3-Ethylbenzenesulfonic acid
Answer
a. 2- and 4-nitroethylbenzene
b.3-Nitroethylbenzene
c. 2- and 4-ethylbenzenesulfonic acid
d.3-Ethylbenzenesulfonic acid
The ethyl group is an ortho and para
director.
Classify a bromide.
a. Meta, activating group
b.Meta, deactivating group
c. Ortho and para, deactivating group
d.Ortho and para, activating group
Answer
a. Meta, activating group
b.Meta, deactivating group
c. Ortho and para, deactivating group
d.Ortho and para, activating group
The electrons can delocalize into the
bromide, making another benzenonium
ion intermediate.
Classify a nitro group.
a. Meta, activating group
b.Meta, deactivating group
c. Ortho and para, deactivating group
d.Ortho and para, activating group
Answer
a. Meta, activating group
b.Meta, deactivating group
c. Ortho and para, deactivating group
d.Ortho and para, activating group
OCH3
AlCl3
CH3CH2CH2Cl
a. 3-Propylanisole
b.2- and 4-propylanisole
c. 3-Isopropylanisole
d.2- and 4-isopropylanisole
Answer
a. 3-Propylanisole
b.2- and 4-propylanisole
c. 3-Isopropylanisole
d.2- and 4-isopropylanisole
The methoxy group is an ortho, para
director. The propyl group rearranges.
Give the intermediate in a Friedel–
Crafts acylation.
a. Carbocation
b.Carbanion
c. Radical
d.Acylium ion
Answer
a. Carbocation
b.Carbanion
c. Radical
d.Acylium ion
An R–C≡O+ is an intermediate in a
Friedel–Crafts acylation.
CO, HCl
AlCl3, CuCl
a. Chlorobenzene
b.Benzoic acid
c. Benzaldehyde
d.Benzene
Answer
a. Chlorobenzene
b.Benzoic acid
c. Benzaldehyde
d.Benzene
The Gatterman–Koch formylation forms
benzaldehyde.
CH2CH2CH3
Br2
light
a. 2-Bromopropylbenzene
b.3-Bromopropylbenzene
c. 4-Bromopropylbenzene
d. -Bromopropylbenzene
e. -Bromopropylbenzene
Answer
a. 2-Bromopropylbenzene
b.3-Bromopropylbenzene
c. 4-Bromopropylbenzene
d. -Bromopropylbenzene
e. -Bromopropylbenzene
Bromine substitutes on the benzylic
carbon.
OH
(CH3)2CHOH
HF
a. 2-Isopropylphenol
b.3-Isopropylphenol
c. 4-Isopropylphenol
d.2-Isopropylphenol and 4-isopropylphenol
e.2-Isopropylphenol and 3-isopropylphenol
Answer
a. 2-Isopropylphenol
b.3-Isopropylphenol
c. 4-Isopropylphenol
d.2-Isopropylphenol and 4-isopropylphenol
e.2-Isopropylphenol and 3-isopropylphenol
Phenols are highly reactive substrates
for electrophilic aromatic substitution.
End of Chapter 9
Chapter 17
110