aromatic compounds. discovery of benzene isolated in 1825 by michael faraday who determined c:h...
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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.
CC
CC
C
C
H
H
HH
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 4N + 2 0 2 1 6 2 10 3 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 OH
H+ , H2O
H
+
H H H H
Which of the Following are Aromatic?
a. b. c. d.
e. f. g. 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.
NO2
NO2
O2N
1,3,5-trinitrobenzene
NO2
NO2
O2N
OH
2,4,6-trinitrophenol
Common Names of Benzene Derivatives
OH OCH3NH2CH3
phenol toluene aniline anisole
CH
CH2 C
O
CH3C
O
HC
O
OH
styrene acetophenone benzaldehyde benzoic acid
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.
Chapter 17 21
Electrophilic Aromatic Substitution
Electrophile substitutes for a hydrogen on the benzene ring.
Chapter 17 22
MechanismStep 1: Attack on the electrophile forms the sigma complex.
Step 2: Loss of a proton gives the substitution product.
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 FeBr3
H
H
H
H
H
H
H
H
H
H
HH
Br+ + FeBr4
_+ -
Chapter 17 24
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 25
Energy Diagram for Bromination
Chapter 17 26
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+ HNO3 I21/2 I
+ NO2 H2O+ ++ +
Chapter 17 27
Nitration of BenzeneUse 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 withbenzene, loses H+ to form nitrobenzene.
Chapter 9 28
SulfonationSulfur trioxide, SO3, in fuming sulfuric
acid is the electrophile.
S
O
O OS
O
O OS
O
O OS
O
O O
+ + +
_
_ _
S
O
OO
H
S
O
O
OH
+
_S
HOO
O
benzenesulfonic acid
Chapter 17 29
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 30
Sigma Complex
Intermediate is more stable if nitration occurs at the ortho or para position.
=>
Chapter 17 31
Energy Diagram
Chapter 17 32
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
H
NO2
+
OCH3
H
NO2
+
Chapter 17 33
Substitution on Anisole
Chapter 17 34
The Amino GroupAniline, 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 35
Summary of Activators
Chapter 17 36
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 37
Ortho Substitutionon Nitrobenzene
Chapter 17 38
Para Substitution on Nitrobenzene
Chapter 17 39
Meta Substitution on Nitrobenzene
Chapter 17 40
Energy Diagram
Chapter 17 41
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 42
Summary of Deactivators
Chapter 17 43
More Deactivators
Chapter 17 44
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 45
Sigma Complex for Bromobenzene
Ortho and para attacks produce a bromonium ionand other resonance structures.
No bromonium ion possible with meta attack.
Chapter 17 46
Energy Diagram
Chapter 17 47
Summary of Directing Effects
Chapter 17 48
Multiple SubstituentsThe 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
+
Chapter 17 49
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 50
Examples ofCarbocation Formation
CH3 CH CH3
Cl
+ AlCl3
CH3C
H3C H
Cl AlCl3+ _
H2C CH CH3
HFH3C CH CH3
F+
_
H3C CH CH3
OHBF3
H3C CH CH3
OH BF3+
H3C CH CH3+
+ HOBF3
_
Chapter 17 51
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
+
-
Chapter 17 52
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 53
Mechanism of Acylation
C
O
R
+
H
C
H
O
R
+
Cl AlCl3
_C
O
R +
HCl
AlCl3
Chapter 17 54
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, 100°C
1000 psi3H2,
CH3
CH3
Chapter 17 55
Side-Chain OxidationAlkylbenzenes are oxidized to benzoic acid by
hot KMnO4 or Na2Cr2O7/H2SO4.
CH(CH3)2
CH CH2
KMnO4, OH-
H2O, heat
COO
COO_
_
Chapter 17 56
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.
CHCH2CH3
Br
hBr2, CH2CH2CH3
AROMATIC REVIEW
Give the shape of benzene.
a. Tetrahedralb.Bentc. Trigonal pyramidald.Planar
Answer
a. Tetrahedralb.Bentc. Trigonal pyramidald.Planar
All six carbons and six hydrogens are in the same plane.
Give the hybridization of each carbon in benzene.
a. spb. sp2
c. sp3
d. sp4
Answer
a. spb. sp2
c. sp3
d. 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. Aromaticb.Antiaromaticc. Nonaromaticd.Acyclic
Answer
a. Aromaticb.Antiaromaticc. Nonaromaticd.Acyclic
The compound gives a whole number for N in Hűckel’s rule (4N + 2 = 6, N = 1).
Classify .
a. Aromaticb.Antiaromaticc. Nonaromaticd.Acyclic
Answer
a. Aromaticb.Antiaromaticc. Nonaromaticd.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. Aromaticb.Antiaromaticc. Nonaromaticd.Acyclic
Answer
a. Aromaticb.Antiaromaticc. Nonaromaticd.Acyclic
This cyclic compound does not have a continuous, overlapping ring of p orbitals and is nonaromatic.
Name
a. Anthraceneb.Naphthalenec. Phenanthrened.Benzene
Answer
a. Anthraceneb.Naphthalenec. Phenanthrened.Benzene
Naphthalene contains two benzene rings fused together.
Name
a. Anthraceneb.Naphthalenec. Phenanthrened.Benzene
Answer
a. Anthraceneb.Naphthalenec. Phenanthrened.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
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
NH2
Br
Cl
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 NH2 is at position one.
Name
a. p-Methylphenolb. m-Methylphenolc. o-Methylphenold. 4-Methylphenole. 3-Methylphenol
CH3
OH
Answer
a. p-Methylphenolb. m-Methylphenolc. o-Methylphenold. 4-Methylphenole. 3-Methylphenol
The groups are on adjacent carbons, which is ortho.
Name
a. 3-Amino-5-benzaldehyde
b.5-Amino-3-benzaldehyde
c. 3-Amino-benzaldehyde
d.5-Nitro-3-benzaldehyde
e.3-Nitro-benzaldehyde
NO2CO
H
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.
Name
a. o-Amino-benzaldehyde
b. m-Amino-benzaldehyde
c. p-Amino-benzaldehyde
d. o-Nitro-benzaldehyde
e. m-Nitro-benzaldehyde
NO2CO
H
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.
Name C6H5CH2CH2C≡CCH3.
a. 1-Phenyl-3-pentyneb.5-Phenyl-2-pentynec. 4-Phenyl-2-pentyned.1-Phenyl-2-butynee.1-Phenyl-3-butyne
Answer
a. 1-Phenyl-3-pentyneb.5-Phenyl-2-pentynec. 4-Phenyl-2-pentyned.1-Phenyl-2-butynee.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.
a. Hexachlorobenzeneb.Hexachlorocyclohexanec. 5,6-Dichloro-1,3-cyclohexadiened.Chlorobenzene
Cl2
AlCl3
Answer
a. Hexachlorobenzeneb.Hexachlorocyclohexanec. 5,6-Dichloro-1,3-cyclohexadiened.Chlorobenzene
One chlorine atom substitutes on the benzene.
a. Nitrobenzeneb.Anilinec. Chlorobenzened.Benzenesulfonic acid
1. HNO3, H2SO4
2. Zn, aq. HCl
Answer
a. Nitrobenzeneb.Anilinec. Chlorobenzened.Benzenesulfonic acid
A nitro group is added, which is then reduced to an amino group.
a. Nitrobenzeneb.Anilinec. Chlorobenzened.Benzenesulfonic acid
SO3
Answer
a. Nitrobenzeneb.Anilinec. Chlorobenzened.Benzenesulfonic acid
A sulfonic acid group is added to the benzene.
a. 2- and 4-nitroethylbenzeneb.3-Nitroethylbenzenec. 2- and 4-ethylbenzenesulfonic acidd.3-Ethylbenzenesulfonic acid
CH2CH3
HNO3
H2SO4
Answer
a. 2- and 4-nitroethylbenzeneb.3-Nitroethylbenzenec. 2- and 4-ethylbenzenesulfonic acidd.3-Ethylbenzenesulfonic acid
The ethyl group is an ortho and para director.
Classify a bromide.
a. Meta, activating groupb.Meta, deactivating groupc. Ortho and para, deactivating groupd.Ortho and para, activating group
Answer
a. Meta, activating groupb.Meta, deactivating groupc. Ortho and para, deactivating groupd.Ortho and para, activating group
The electrons can delocalize into the bromide, making another benzenonium ion intermediate.
Classify a nitro group.
a. Meta, activating groupb.Meta, deactivating groupc. Ortho and para, deactivating groupd.Ortho and para, activating group
Answer
a. Meta, activating groupb.Meta, deactivating groupc. Ortho and para, deactivating groupd.Ortho and para, activating group
a. 3-Propylanisoleb.2- and 4-propylanisolec. 3-Isopropylanisoled.2- and 4-isopropylanisole
OCH3
AlCl3
CH3CH2CH2Cl
Answer
a. 3-Propylanisoleb.2- and 4-propylanisolec. 3-Isopropylanisoled.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. Carbocationb.Carbanionc. Radicald.Acylium ion
Answer
a. Carbocationb.Carbanionc. Radicald.Acylium ion
An R–C≡O+ is an intermediate in a Friedel–Crafts acylation.
a. Chlorobenzeneb.Benzoic acidc. Benzaldehyded.Benzene
CO, HCl
AlCl3, CuCl
Answer
a. Chlorobenzeneb.Benzoic acidc. Benzaldehyded.Benzene
The Gatterman–Koch formylation forms benzaldehyde.
a. 2-Bromo-propylbenzene
b.3-Bromo-propylbenzene
c. 4-Bromo-propylbenzene
d. -Bromo- propylbenzene
e. -Bromo-propylbenzene
CH2CH2CH3
Br2
light
Answer
a. 2-Bromo-propylbenzene
b.3-Bromo-propylbenzene
c. 4-Bromo-propylbenzene
d. -Bromo- propylbenzene
e. -Bromo-propylbenzene
Bromine substitutes on the benzylic carbon.
a. 2-Isopropylphenolb.3-Isopropylphenolc. 4-Isopropylphenold.2-Isopropylphenol and 4-isopropylphenole.2-Isopropylphenol and 3-isopropylphenol
OH
(CH3)2CHOH
HF
Answer
a. 2-Isopropylphenolb.3-Isopropylphenolc. 4-Isopropylphenold.2-Isopropylphenol and 4-isopropylphenole.2-Isopropylphenol and 3-isopropylphenol
Phenols are highly reactive substrates for electrophilic aromatic substitution.
Chapter 17 110
End of Chapter 9