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.

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+ , H2O

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.

1,3,5-trinitrobenzene

2,4,6-trinitrophenol

Common Names of Benzene Derivatives

OH OCH3NH2CH3

phenol toluene aniline anisole

styrene acetophenone benzaldehyde benzoic acid

Phenyl and Benzyl

• Phenyl indicates the benzene ring

• attachment. The benzyl group has

• an additional carbon.

phenyl bromide

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 Br FeBr3 Br Br FeBr3+ -

Br Br FeBr3

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 S O H

+ HSO4

_H O N

H2O + N

+ then forms a sigma complex withbenzene, loses H+ to form nitrobenzene.

Chapter 9 28

SulfonationSulfur trioxide, SO3, in fuming sulfuric

acid is the electrophile.

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.

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.

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

+ AlCl3

Cl AlCl3+ _

H2C CH CH3

HFH3C CH CH3

H3C CH CH3

OH BF3+

H3C CH CH3+

+ HOBF3

Chapter 17 51

Formation of Alkyl Benzene

CH(CH3)2+

CH(CH3)2

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

Cl 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.

CH3Ru, 100°C

1000 psi3H2,

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

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

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

The compound gives a whole number for N in Hűckel’s rule (4N + 2 = 6, N = 1).

Classify .

Answer

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 .

Answer

This cyclic compound does not have a continuous, overlapping ring of p orbitals and is nonaromatic.

a. Anthraceneb.Naphthalenec. Phenanthrened.Benzene

Answer

Naphthalene contains two benzene rings fused together.

Answer

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.

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

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.

a. p-Methylphenolb. m-Methylphenolc. o-Methylphenold. 4-Methylphenole. 3-Methylphenol

Answer

a. p-Methylphenolb. m-Methylphenolc. o-Methylphenold. 4-Methylphenole. 3-Methylphenol

The groups are on adjacent carbons, which is ortho.

a. 3-Amino-5-benzaldehyde

b.5-Amino-3-benzaldehyde

c. 3-Amino-benzaldehyde

d.5-Nitro-3-benzaldehyde

e.3-Nitro-benzaldehyde

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.

a. o-Amino-benzaldehyde

b. m-Amino-benzaldehyde

c. p-Amino-benzaldehyde

d. o-Nitro-benzaldehyde

e. m-Nitro-benzaldehyde

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

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 nitro group is added, which is then reduced to an amino group.

Answer

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

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

The electrons can delocalize into the bromide, making another benzenonium ion intermediate.

Classify a nitro group.

Answer

a. 3-Propylanisoleb.2- and 4-propylanisolec. 3-Isopropylanisoled.2- and 4-isopropylanisole

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

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

(CH3)2CHOH

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

Aromatic Compounds. Discovery of Benzene Isolated in 1825 by Michael Faraday who determined C:H...

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