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Aromatic Compounds
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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.
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Kekul Structure
Proposed in 1866 by Friedrich Kekul, shortly
after multiple bonds were suggested.
Failed to explain existence of only one isomer
of 1,2-dichlorobenzene.
CC
CC
C
C
H
H
H
H
H
H
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Resonance Structure
Each sp2hybridized C in the ring has an
unhybridizedporbital perpendicular to the
ring which overlaps around the ring.
=>
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Unusual Reactions
Alkene + KMnO4diol (addition)
Benzene + KMnO4 no reaction.
Alkene + Br2
/CCl4
dibromide (addition)
Benzene + Br2/CCl4no reaction.
With FeCl3catalyst, Br2reacts with benzene to
form bromobenzene + HBr
(substitution!). Double bonds remain.
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Unusual Stability
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Annulenes
All cyclic conjugatedhydrocarbons wereproposed to bearomatic.
However,cyclobutadiene is soreactive that it
dimerizes before it canbe isolated.
And cyclooctatetraeneadds Br2readily.
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What Does It Take to Be Aromatic?
Alternating double and single bonds
A magic number of pi electrons
Resonance structures must be able to movethe pi electrons in a circular manar.
Non bonding electrons can also participate in
the resonance structures.
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Hckels Rule
Hckels Rule is used to generate the magic
number of pi electrons.
If the compound has a continuous ring of
overlappingporbitals and has 4N+ 2 pi
electrons, it is aromatic.
If the compound has a continuous ring of
overlappingporbitals and has 4N electrons, it
is antiaromatic.
Nis any integer, starting at zero
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Generation of Magic Pi Electrons
Value of N 4N+ 2
0 2
1 62 10
3 14
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[N]Annulenes
[4]Annulene is antiaromatic (4Ne-s)
[8]Annulene would be antiaromatic, but its
not planar, so its nonaromatic.
[10]Annulene is aromatic except for the
isomers that are not planar.
Larger 4Nannulenes are not antiaromatic
because they are flexible enough to become
nonplanar.
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Tropylium Ion
The cycloheptatrienyl cation has 6pelectronsand an emptyporbital.
Aromatic: more stable than open chain ion.
H OH
H+, H2O
H
+
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Which of the Following are Aromatic?
a. b. c. d.
e. f. g. h.
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Disubstituted Benzenes The prefixes ortho-, meta-, and para- are
commonly used for the 1,2-, 1,3-, and 1,4-
positions, respectively.
=>
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3 or More Substituents
Use the smallest possible numbers, but
the carbon with a functional group is #1.
NO2
NO2
O2N
1,3,5-trinitrobenzen
NO2
NO2
O2N
OH
2,4,6-trinitrophenol
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Common Names of Benzene
Derivatives
OH OCH3NH2CH3
phenol toluene aniline anisole
C
H
CH2 C
O
CH3C
O
HC
O
OH
styrene acetophenone benzaldehyde benzoic acid
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Phenyl and Benzyl
Phenyl indicates the benzene ring
attachment. The benzyl group has
an additional carbon.Br
phenyl bromide
CH2Br
benzyl bromide
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Fused Ring Hydrocarbons Naphthalene
Anthracene
Phenanthrene
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Larger Polynuclear
Aromatic Hydrocarbons
Formed in combustion (tobacco smoke).
Many are carcinogenic.
Epoxides form, combine with DNA base.
pyrene =>
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Physical Properties
Melting points: More symmetrical thancorresponding alkane, pack better intocrystals, so higher melting points.
Boiling points: Dependent on dipole moment,so ortho> meta>para, for disubstitutedbenzenes.
Density: More dense than nonaromatics, lessdense than water.
Solubility: Generally insoluble in water.
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Chapter 17 21
Electrophilic Aromatic Substitution
Electrophile substitutes for a hydrogen onthe benzene ring.
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Chapter 17 22
Mechanism
Step 1:Attack on the electrophile forms the sigma complex.
Step 2:Loss of a proton gives the substitution product.
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Chapter 17 23
Bromination of Benzene
Requires a stronger electrophile than Br2.
Use a strong Lewis acid catalyst, FeBr3.
Br
HBr+
Br Br FeBr3 Br Br FeBr3+ -
Br Br FeBr 3
HH
H
H
H
H
H
H
H
H
H H
Br+ + FeBr4
_+ -
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Chapter 17 24
Comparison with Alkenes
Cyclohexene addsBr2, 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.
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Chapter 17 25
Energy Diagram for Bromination
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Chapter 17 26
Chlorination and Iodination
Chlorination is similar to bromination. Use
AlCl3as the Lewis acid catalyst. Iodination requires an acidic oxidizing
agent, like nitric acid, which oxidizes the
iodine to an iodonium ion.
H+
HNO3 I21/2 I+
NO2 H2O+ ++ +
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Chapter 17 27
Nitration of Benzene
Use sulfuric acid with nitric acid to form the
nitronium ion electrophile.
H O N
O
O
H O S O H
O
O
+ HSO4
_H O N
OH
O+
H O N
OH
O+
H2O + N
O
O
+
NO2+then forms a sigma complex with
benzene, loses H+to form nitrobenzene.
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Chapter 9 28
Sulfonation
Sulfur trioxide, SO3, in fuming sulfuric
acid is the electrophile.
S
O
O O
S
O
O O
S
O
O O
S
O
O O
+ + +
_
_ _
S
O
O O
H
S
O
O
OH
+
_
S
HOO
O
benzenesulfonic a
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Chapter 17 29
Desulfonation
All steps are reversible, so sulfonic
acid group can be removed by heating in
dilute sulfuric acid.
This process is used to place deuterium in
place of hydrogen on benzene ring.
Benzene-d6
D
D
D
D
D
D
D2SO4/D2O
large excess
H
H
H
H
H
H
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Chapter 17 30
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.
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Chapter 17 31
Sigma Complex
Intermediate is
more stable if
nitrationoccurs at the
ortho
orpara
position.
=>
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Chapter 17 32
Energy Diagram
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Chapter 17 33
Activating, O-, P-Directing Substituents
Alkyl groups stabilize the sigma complex byinduction, donating electron density throughthe sigma bond.
Substituents with a lone pair of electrons
stabilize the sigma complex by resonance.
OCH3
H
NO2
+
OCH3
H
NO2
+
b l
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Chapter 17 34
Substitution on Anisole
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Chapter 17 35
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 thats also formed.
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Chapter 17 36
Summary of Activators
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Chapter 17 37
Deactivating Meta-Directing Substituents
Electrophilic substitution reactions fornitrobenzene are 100,000 times slowerthan for benzene.
The product mix contains mostly the meta
isomer, only small amounts of the orthoand para isomers.
Meta-directors deactivate all positions onthe ring, but the meta position is lessdeactivated.
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Chapter 17 38
Ortho Substitutionon Nitrobenzene
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Chapter 17 39
Para Substitution on Nitrobenzene
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Chapter 17 40
Meta Substitution on Nitrobenzene
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Chapter 17 41
Energy Diagram
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Chapter 17 42
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.
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Chapter 17 43
Summary of Deactivators
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Chapter 17 44
More Deactivators
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Chapter 17 45
Halobenzenes
Halogens are deactivating toward
electrophilic substitution, but are ortho,
para-directing!
Since halogens are very electronegative, they
withdraw electron density from the ringinductively along the sigma bond.
But halogens have lone pairs of electrons that
can stabilize the sigma complex byresonance.
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Chapter 17 46
Sigma Complex for Bromobenzene
Ortho and para attacks produce a bromonium ion
and other resonance structures.
No bromonium ion
possible with meta attack.
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Chapter 17 47
Energy Diagram
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Chapter 17 48
Summary of Directing Effects
l i l S b i
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Chapter 17 49
Multiple Substituents
The most strongly activating substituent will
determine the position of the next
substitution. May have mixtures.
OCH3
O2N
SO3
H2SO4
OCH3
O2N
SO3H
OCH3
O2N
SO3H
+
F i d l C f Alk l i
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Chapter 17 50
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.
l fC b i i
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Chapter 17 51
Examples ofCarbocation Formation
CH3 CH CH3
Cl
+ AlCl3
CH3C
H3C H
Cl AlCl3+ _
H2C CH CH3
HF
H3C CH CH3
F+
_
H3C CH CH3
OHBF3
H3C CH CH3
OH BF3+
H3C CH CH3+
+ HOBF3
_
F i f Alk l B
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Chapter 17 52
Formation of Alkyl Benzene
C
CH3
CH3
H+
H
H
CH(CH3)2+
H
H
CH(CH3)2
B
F
F
F
OHCH
CH3
CH3
+HF
BF
OHF
+
-
Li it ti f F i d l C ft
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Chapter 17 53
Limitations of Friedel-Crafts
Reaction fails if benzene has a substituent that
is more deactivating than halogen.
Carbocations rearrange. Reaction of benzene
with n-propyl chloride and AlCl3produces
isopropylbenzene.
The alkylbenzene product is more reactive
than benzene, so polyalkylation occurs.
F i d l C ft A l ti
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Chapter 17 54
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.
M h i f A l ti
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Chapter 17 55
Mechanism of Acylation
C
O
R
+
H
C
H
O
R
+
Cl AlCl3
_C
O
R +
HCl
AlCl3
C t l ti H d ti
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Chapter 17 56
Catalytic Hydrogenation
Elevated heat and pressure are required.
Possible catalysts: Pt, Pd, Ni, Ru, Rh.
Reduction cannot be stopped at an
intermediate stage.
CH3
CH3Ru, 100C
1000 psi3H2,
CH3
CH3
Sid Ch i O id ti
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Chapter 17 57
Side-Chain Oxidation
Alkylbenzenes are oxidized to benzoic acid by
hot KMnO4or Na2Cr2O7/H2SO4.
CH(CH3)2
CH CH2
KMnO4, OH-
H2O, heat
COO
COO_
_
Sid Ch i H l ti
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Chapter 17 58
Side-Chain Halogenation
Benzylic position is the most reactive.
Chlorination is not as selective as
bromination, results in mixtures.
Br2reacts only at the benzylic position.
CHCH2CH3
Br
hBr2,
CH2CH2CH3
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AROMATIC REVIEW
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Give the shape of benzene.
a. Tetrahedral
b. Bent
c. Trigonal pyramidal
d. Planar
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Answer
a. Tetrahedral
b. Bent
c. Trigonal pyramidal
d. Planar
All six carbons and six hydrogens are inthe same plane.
Give the hybridization of each carbon
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Give the hybridization of each carbon
in benzene.
a. sp
b. sp2
c. sp3
d. sp4
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Answer
a. sp
b. sp2
c. sp3
d. sp4
Each carbon in benzene is sp2hybridized.
Give the bond angle of the atoms in
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Give the bond angle of the atoms in
benzene.
a. 45
b. 60
c. 90
d. 109.5
e. 120
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Answer
a. 45
b. 60
c. 90
d. 109.5
e. 120
The carbons are 120oapart in benzene.
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Classify .
a. Aromatic
b. Antiaromaticc. Nonaromatic
d. Acyclic
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Answer
a. Aromatic
b. Antiaromatic
c. Nonaromatic
d. Acyclic
The compound gives a whole number forNin Hckels rule (4N+ 2 = 6, N= 1).
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Classify .
a. Aromatic
b. Antiaromaticc. Nonaromatic
d. Acyclic
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Answer
a. Aromatic
b. Antiaromatic
c. Nonaromatic
d. Acyclic
The compound is cyclic and hascontinuous delocalized electrons, butdoes not give a whole number forHckels rule (4N+ 2 = 8, N= 3/2).
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Classify .
a. Aromatic
b. Antiaromatic
c. Nonaromatic
d. Acyclic
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Answer
a. Anthracene
b. Naphthalene
c. Phenanthrene
d. Benzene
Naphthalene contains two benzenerings fused together.
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Name
a. Anthracene
b. Naphthalenec. Phenanthrene
d. Benzene
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Answer
a. Anthracene
b. Naphthalene
c. Phenanthrene
d. Benzene
Anthracene contains three benzene ringsfused together.
Identify how carbon diamond and
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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.
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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.
NH2Cl
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Name
a. 4-Bromo-3-
chloroanilineb. 4-Bromo-3-
chlorophenol
c. 4-Bromo-3-chloroanisole
d. 1-Bromo-2-chloro-4-
anilinee. 1-Bromo-2-chloro-4-
phenol
Br
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Answer
a. 4-Bromo-3-
chloroaniline
b. 4-Bromo-3-
chlorophenol
c. 4-Bromo-3-
chloroanisole
d. 1-Bromo-2-chloro-4-
aniline
e. 1-Bromo-2-chloro-4-
phenol
Aniline is the parent compound. The NH2isat position one.
CH3
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Name
a. p-Methylphenol
b. m-Methylphenol
c. o-Methylphenol
d. 4-Methylphenol
e. 3-Methylphenol
OH
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Answer
a. p-Methylphenol
b. m-Methylphenol
c. o-Methylphenol
d. 4-Methylphenol
e. 3-Methylphenol
The groups are on adjacent carbons,
which is ortho.
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Name
a. 3-Amino-5-
benzaldehydeb. 5-Amino-3-
benzaldehyde
c. 3-Amino-benzaldehyde
d. 5-Nitro-3-
benzaldehydee. 3-Nitro-
benzaldehyde
NO2C
O
H
A
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Answer
a. 3-Amino-5-benzaldehyde
b. 5-Amino-3-
benzaldehyde
c. 3-Amino-
benzaldehyde
d. 5-Nitro-3-benzaldehyde
e. 3-Nitro-
benzaldehyde
Benzaldehyde is the parent compound.
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Name
a. o-Amino-
benzaldehydeb. m-Amino-
benzaldehyde
c. p-Amino-benzaldehyde
d. o-Nitro-
benzaldehydee. m-Nitro-
benzaldehyde
NO2C
O
H
A
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Answer
a. o-Amino-benzaldehyde
b. m-Amino-
benzaldehyde
c. p-Amino-
benzaldehyde
d. o-Nitro-benzaldehyde
e. m-Nitro-
benzaldehyde
Benzaldehyde is the parent compound.
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A
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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
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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.
A
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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 formthe benzenonium ion is the slow step.
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a. Hexachlorobenzene
b. Hexachlorocyclohexanec. 5,6-Dichloro-1,3-cyclohexadiene
d. Chlorobenzene
Cl2
AlCl3
A
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Answer
a. Hexachlorobenzene
b. Hexachlorocyclohexane
c. 5,6-Dichloro-1,3-cyclohexadiene
d. Chlorobenzene
One chlorine atom substitutes on thebenzene.
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a. Nitrobenzene
b. Anilinec. Chlorobenzene
d. Benzenesulfonic acid
1. HNO3, H2SO4
2. Zn, aq. HCl
Answer
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Answer
a. Nitrobenzene
b. Aniline
c. Chlorobenzene
d. Benzenesulfonic acid
A nitro group is added, which is thenreduced to an amino group.
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a. Nitrobenzene
b. Anilinec. Chlorobenzene
d. Benzenesulfonic acid
SO3
Answer
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Answer
a. Nitrobenzene
b. Aniline
c. Chlorobenzene
d. Benzenesulfonic acid
A sulfonic acid group is added to thebenzene.
CH CH
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a. 2- and 4-nitroethylbenzene
b. 3-Nitroethylbenzenec. 2- and 4-ethylbenzenesulfonic acid
d. 3-Ethylbenzenesulfonic acid
CH2CH3
HNO3
H2SO4
Answer
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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 paradirector.
Classify a bromide
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Classify a bromide.
a. Meta, activating group
b. Meta, deactivating group
c. Ortho and para, deactivating groupd. Ortho and para, activating group
Answer
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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 thebromide, making another benzenoniumion intermediate.
Classify a nitro group
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Classify a nitro group.
a. Meta, activating group
b. Meta, deactivating group
c. Ortho and para, deactivating groupd. Ortho and para, activating group
Answer
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Answer
a. Meta, activating group
b. Meta, deactivating group
c. Ortho and para, deactivating group
d. Ortho and para, activating group
OCH3
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a. 3-Propylanisole
b. 2- and 4-propylanisolec. 3-Isopropylanisole
d. 2- and 4-isopropylanisole
AlCl3
CH3CH2CH2Cl
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Give the intermediate in a Friedel
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Crafts acylation.
a. Carbocation
b. Carbanion
c. Radicald. Acylium ion
Answer
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Answer
a. Carbocation
b. Carbanion
c. Radical
d. Acylium ion
An RCO+is an intermediate in aFriedelCrafts acylation.
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a. Chlorobenzene
b. Benzoic acid
c. Benzaldehyde
d. Benzene
CO, HCl
AlCl3, CuCl
Answer
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Answer
a. Chlorobenzene
b. Benzoic acid
c. Benzaldehyded. Benzene
The GattermanKoch formylation formsbenzaldehyde.
CH2CH2CH3
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a. 2-Bromo-
propylbenzeneb. 3-Bromo-
propylbenzene
c. 4-Bromo-propylbenzene
d. -Bromo-
propylbenzenee. -Bromo-
propylbenzene
Br2
light
Answer
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Answer
a. 2-Bromo-propylbenzene
b. 3-Bromo-
propylbenzenec. 4-Bromo-
propylbenzene
d. -Bromo-propylbenzene
e. -Bromo-
propylbenzene
Bromine substitutes on the benzylic
carbon.
OH
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a. 2-Isopropylphenol
b. 3-Isopropylphenolc. 4-Isopropylphenol
d. 2-Isopropylphenol and 4-isopropylphenol
e. 2-Isopropylphenol and 3-isopropylphenol
(CH3)2CHOH
HF
Answer
8/13/2019 122 Ch 4 Aromatic
111/112
Answer
a. 2-Isopropylphenol
b. 3-Isopropylphenol
c. 4-Isopropylphenold. 2-Isopropylphenol and 4-isopropylphenol
e. 2-Isopropylphenol and 3-isopropylphenol
Phenols are highly reactive substrates
for electrophilic aromatic substitution.
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End of Chapter 9
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