1

Chapter 14Chapter 14Chapter 14Chapter 14

AromaticityAromaticityAromaticityAromaticity

Reactions of Reactions of Reactions of Reactions of

BenzeneBenzeneBenzeneBenzene

Organic ChemistryOrganic ChemistryOrganic ChemistryOrganic Chemistry

4444thththth EditionEditionEditionEditionPaula Paula Paula Paula YurkanisYurkanisYurkanisYurkanis BruiceBruiceBruiceBruice

AROMATIC ?AROMATIC ?

WHERE DID THE TERM ORIGINALLY COME FROM ?

A LOT OF NICEA LOT OF NICE--SMELLING COMPOUNDS SMELLING COMPOUNDS (SPICES IN PARTICULAR) HAVE BENZENE RINGS(SPICES IN PARTICULAR) HAVE BENZENE RINGS

CHO

OMe

anisaldehyde

(anise)

CH CH CHOcinnamaldehyde

(cinnamon)

OH

thymol

(thyme)

2

A LOT OF NICEA LOT OF NICE--SMELLING COMPOUNDS SMELLING COMPOUNDS (SPICES IN PARTICULAR) HAVE BENZENE RINGS(SPICES IN PARTICULAR) HAVE BENZENE RINGS

Today chemists have a different definition of “AROMATIC”which we will discuss later in the chapter.

OH

CH2

OMe

CH CH2

eugenol(cloves)

CHO

cuminaldehyde

(cumin)

Representative Compounds

HO

OH

CHCH2NHCH3

OH

adrenaline

ephedrine

CHCHNHCH3

OH

CH3

Ephedrine is commonly used as a stimulant, appetite suppressant, concentration aid, decongestant and to treat hypotension associated with regional anaesthesia.

Representative Compounds

amphetamine

CH2CHNH2

CH3

methamphetamine

CH2CHNHCH3

CH3

speed

3

Other Sources of Aromatic Hydrocarbons

• From high temperature distillation of coal tar• Heating petroleum at high temperature and pressure

over a catalystCH3 CH3

H3C

CH3

CH3

CH3

CH3

benzene toluene xyleneortho para meta

indene naphtalene phenantrene anthracene

Naming Aromatic Compounds

• Many common names – toluene = methylbenzene – aniline = aminobenzene

• Monosubstituted benzenes systematic names as hydrocarbons with –benzene

Br

NO2 CH2CH2CH3

C6H5Br = bromobenzene

C6H5NO2 = nitrobenzeneC6H5CH2CH2CH3 = propylbenzene

OH

NH2

CHO

CN

COOH

CH(CH3)2

Name the following compounds

phenol

anilin

benzonitrile

benzoic acid

benzaldehyde 2-phenylpropane

4

The Phenyl Group

• When a benzene ring is a substituent, the term phenyl is used (for C6H5

�)

– You may also see “Ph” or “φφφφ”in place of “C6H5”

• “Benzyl ” refers to “C6H5CH2�”

R

CHCH2CH2CH2CH3

CH3

CH2Br

Phenyl group

2-phenyl hexanebenzyl bromide

Disubstituted Benzenes• Relative positions on a benzene ring

– ortho- (o) on adjacent carbons (1,2)– meta- (m) separated by one carbon

(1,3)– para- (p) separated by two carbons

(1,4)• Describes reaction patterns (“occurs

at the para position”)

Br

Br

CH3ortho

meta

paraCH3

H3C

CHO

Fortho-dibromobenzene1,2-dibromobenzene

meta-xylene1,3-dimethylbenzene

para-fluorobenzaldehyde4-fluorobenzaldehyde

Naming Benzenes With More Than Two Substituents

• Choose numbers to get lowest possible values• List substituents alphabetically with hyphenated

numbers• Common names, such as “toluene” can serve as

root name (as in TNT)

5

Naming Benzenes With More Than Two Substituents

CH3

CH3

Br

12

4

NO2

CH3

NO2O2N 1

2

4

6

Exercise

CH3

F

NH2

NO2

…provide IUPAC names for the molecules shown above

Exercise 2Cl

CH2CH3

NO2

CH2CH3

O2N

Cl

CH2CH3

Cl

NO2

…provide IUPAC names for the molecules shown above

6

Chemistry of Benzene

– Benzene does not undergo any of the regular reactions of alkenes such as addition or hydration

Br2/CCl4

XBr

Br

H2/ catalyst,

slow

HBr

XH

Br

heat of hydrogenation: compare experimentalvalue with "expected" value for hypothetical"cyclohexatriene"

∆H°= – 208 kJ

ThermochemicalThermochemical Measures of StabilityMeasures of Stability

+ 3H2

Pt

120 kJ/mol

231 kJ/mol

208 kJ/mol

360 kJ/mol

3 x cyclohexene

7

120 kJ/mol

360 kJ/mol

3 x cyclohexene"expected" heat of hydrogenation of benzene is 3 x heat of hydrogenation of cyclohexene

208 kJ/mol

360 kJ/mol

3 x cyclohexeneobserved heat of hydrogenation is 152 kJ/mol less than "expected"benzene is 152 kJ/mol more stable thanexpected152 kJ/mol is the resonance energy of benzene

hydrogenation of 1,3-cyclohexadiene (2*H2) gives off more heat than hydrogenation of benzene (3*H2)!

231 kJ/mol

208 kJ/mol

8

heat of hydrogenation =

heat of hydrogenation =

3H2

Pt

3H2

Pt

Cyclic conjugation versus noncyclic conjugation

compared to localized 1,3,5-cyclohexatriene

___ kJ/mol

compared to 1,3,5-hexatriene

___ kJ/mol

exact value of resonance energy of benzene depends on what it is compared to, but regardless of model, benzene is more stable than expected by a substantial amount

Resonance Energy of Benzene

Structure and Stability of Benzene

• Benzene reacts with Br2 , after adding a catalyst, to give bromobenzene (where Br replaces H)

• This is substitution rather than the rapid addition reaction common to compounds with C=C, suggesting that in benzene there is a higher barrier

Br2 Br

BrFe

Br

9

Dibromination of benzene

• Based on Lewis structures we expect four products.

Br 2

Fe

Br Br 2

Fe

Br

Br

Br

Br

Br

Br

Br

Br

1,2-dibromo derivatives

Br

Br

Br

Br

long bond short bond

• Based on Lewis structures we expect two different 1,2-dibromo derivatives.

• However, they are identical.

Kekule’s attempt to explain

• Kekule suggested a rapid equilibrium

• Today we know these are “__________________”

Br

Br

Br

Br

Br

Br

Br

Br

10

Benzene’s Unusual Structure• All its C-C bonds are the same length: 139 pm

— between single (154 pm) and double (134 pm) bonds• Electron density in all six C-C bonds is identical• Structure is planar, hexagonal• C–C–C bond angles 120°• Each C is sp2 and has a p orbital perpendicular to the

plane of the six-membered ring

Drawing Benzene and Its Derivatives

• The two benzene resonance forms can be represented by a single structure with a circle in the center to indicate the equivalence of the carbon–carbon bonds

• This does not indicate the ______ of π electrons in the ring but reminds us of the delocalized structure

• We shall use one of the ________________ to represent benzene for ease in keeping track of bonding changes in reactions

1

2

34

5

6

1

4

2

35

6

BENZENE π MOLECULAR ORBITALS

π1

π3π2

π4 π5

π6

six p’ssix mo’s

topview

ENERGY

11

Molecular Orbitals of the π-system

Recall: Key Ideas on Benzene

• Unusually stable - heat of hydrogenation 150 kJ/mol less negative than a cyclic triene

• Planar hexagon: bond angles are 120°, carbon–carbon bond lengths 139 pm

• Undergoes substitution rather than electrophilicaddition

• Resonance hybrid with structure between two line-bond structures

• One more important factor is the number of electrons in the cyclic orbital

AROMATICITY AROMATICITY THE HUCKEL RULETHE HUCKEL RULE

12

QUESTION: Are all fully-conjugated, cyclic systems aromatic?

H

HH H

HH

H

H

HH H

H

H

H

H

HH

H

AROMATICITYAROMATICITY

KNOWN AROMATIC

? ? ?

??

152 kJ/mole Res.en.

Do these other rings have thesame kind of stability as benzene?

HUCKEL 4n+2 RULEHUCKEL 4n+2 RULE....

Prediction: Compounds that have 4n+2 π electrons ina planar cyclic array will be aromatic.

AROMATICITYAROMATICITY

POLYCYCLIC AROMATIC COMPOUNDS

benzene naphthalene anthracene

6 10 14

4n+2 series = 2, 6, 10, 14, 18, 22, 26, 30 …….. etc.

The rule was derived by observation of

Aromatic Compounds Have Special Properties

Aromatic compounds:

1) Must be cyclic and fully conjugated

2) Must have 4n+2 π electrons in the system

3) Must have the entire system planar

Characteristic Properties:

1) Special chemical stability

2) Give substitution reactions instead of addition

3) Show a ring currrent in the NMR

4) Will have no unpaired electrons in

the ππππ system molecular orbitalsplanarsystem

13

Polycyclic Aromatic Rings

phenanthreneplanar, 7 pairs

anthraceneplanar, 7 pairs

chryseneplanar, 9 pairs

Naphthalene OrbitalsAll Bonds are not equivalent

• Three resonance forms and delocalized electrons

Secondary magnetic field

generated by circulating ππππelectrons deshields aromaticprotons

Circulating ππππ electrons

Ring Current in BenzeneRing Current in Benzene

Bo

Deshielded

H HA proton placed in themiddle of the ring wouldbe shielded!

14

H

HH H

HH

H

H

HH H

H

H

H

H

HH

H

inner hydrogens -1.8 ppmouter hydrogens 8.9 ppm

CH2CH2

CH2 CH2CH2

-1.0 ppm

2.0 ppm

CH2CH2

CH2 CH2

CH2 CH2C

H

H

-1.4 ppm

RING CURRENTS CAN BE SEEN IN THE NMR

AROMATIC - SHOWSA RING CURRENT

18 ππππ

HUCKEL MNEMONICCYCLIC π MOLECULAR ORBITAL ARRAYS

2) Inscribe the ring in the circle point down

3) Each point where the polygon (ring)touches the circle represents anenergy level.

1) Draw a circle

4) Place the correct number of electrons inthe orbitals, starting with the lowestenergy orbital first.

ENERGY

BENZENE

6π electrons

ENERGY

closed*shell

AROMATIC

Aromaticcompoundswill have allof the occupiedππππ M.O. levels completely filled with nounpaired electrons.

(*completedlevel)

15

4π electrons

CYCLOBUTADIENE

ENERGY

open*shell

ANTI-AROMATIC

Does not have a completedshell and has unpaired electrons.

Does not have 4n+2 ππππ electrons.

(*incompletelevel)

. .

DIRADICAL

•Cyclobutadiene is so unstable that it dimerizes•by a self-Diels-Alder reaction at low termperature

Instability of cyclobutadiene

1 2

34

a

b

a

b

1 2

3

4

12

34

a

b

12

34

a

b

Diels-Alder -78 0 C

8π electrons

CYCLOOCTATETRAENE

ENERGY

ANTI-AROMATIC

open*shell

Does not have a completedshell and has unpaired electrons.

Does not have 4n+2 ππππ electrons.

(*incompletelevel)

16

8π electrons

CYCLOOCTATETRAENECYCLOOCTATETRAENE

not planar

Cyclooctatetraene has four double bonds, reacting with Br2, KMnO4, and HCl as if it were four individual alkenes.

Compounds With 4n π Electrons Are Not Aromatic (May be Antiaromatic)

• Planar, cyclic molecules with 4 n πelectrons are much less stable than expected (anti-aromatic)

• They will distort out of plane and behave like ordinary alkenes

• 4- and 8-electron compounds are not delocalized (single and double bonds)

cyclooctatetraene

c y c lo b u ta d ie n e

12 π

14 π = 4(3) + 2

18 π = 4(4) + 2

HH 10 π = 4(2) + 216 π

20 π

SOME CYCLIC POLYENES

AROMATIC

AROMATIC

ANNULENES

[10]-annulene

[12]-annulene

[14]-annulene

[16]-annulene

[18]-annulene

[20]-annulene

AROMATIC ?

17

The Annulenes

• The [10]annulenes below should be aromatic but none of them can be planar

• Steric interaction of the indicated hydrogens• Large angle strain in the flat molecules

Steric repulsion between opposing hydrogensForces the ππππ-system out of planarity

4

5 / 6

18

N N

H

O S

HETEROCYCLIC COMPOUNDSHETEROCYCLIC COMPOUNDS

..

.. ..

..

..

..

pyridine pyrrole furan thiophene

All have 6ππππ electrons in a cyclic array.

The unshared pairs are in thecyclic pi system (one pair in each case).

These compounds have reactions similar to benzene, rather than to alkenes. They will give substitution reactionsunder conditions similar to those for benzene.

pair not inπ π π π system

pair not inπ π π π system

pair not inπ π π π system

Aromatic Heterocycles: Pyridine and Pyrrole

• Aromatic compounds can have elements other than carbon in the ring

• Heterocyclic compounds contain elements other than carbon in a ring, such as N,S,O,P

• There are many heterocyclic aromatic compounds and many are very common

• Cyclic compounds that contain only carbon are called carbocycles (not homocycles)

• Nomenclature is specialized

Pyridine

• A six-membered heterocycle with a nitrogen atom in its ring

• π electron structure resembles benzene (6 electrons)• The nitrogen lone pair electrons are not part of the

aromatic system (perpendicular orbital)• Pyridine is a relatively weak base compared to normal

amines but protonation does not affect aromaticity

19

Pyrrole

• A five-membered heterocycle with one nitrogen• π electron system similar to that of cyclopentadienyl anion• Four sp2-hybridized carbons with 4 p orbitals

perpendicular to the ring and 4 p electrons• Nitrogen atom is sp2-hybridized, and lone pair of

electrons occupies a p orbital (6 π electrons)• Since lone pair electrons are in the aromatic ring,

protonation destroys aromaticity, making pyrrole a very weak base

Pyrrol, Furan

N H

N

H

Lone pair part of π-system,N, and O are sp2-hybridized

O

O

Second lone-pairLocated in sp2-hybridizedorbital

PyrrolResonance contributors

N

HN

H

N

H

N

H

N

H

pos. charge

Note: there is a pos.charge on the most eletronegative atom !!

20

Furan

O O O

O O

Resonance contributors

Exercise

• Draw the orbital pictures, and energy diagram of furan, to explain why it is aromatic.

• Imidazole is an important heterocycle. Explain which nitrogen is pyrrole like, which is pyridine like

O

N

N

H

QuinolineQuinoline

NN••••

NN ••••

IsoquinolineIsoquinoline

21

Polycyclic Aromatic Compounds

• Aromatic compounds can have rings that share a set of carbon atoms (fused rings)

• Compounds from fused benzene or aromatic heterocycle rings are themselves aromatic

benz[a]pyrene coronene

Why Aromatic ?

Benz[a]pyrene coronene

20 ππππ electrons 24 ππππ electrons

Huckel: …18, 22, 26 !!

One pair doesn’t participate !

22

Aromatic Ions

• The 4n + 2 rule applies to ions as well as neutral species

• Both the cyclopentadienyl anion and the cycloheptatrienyl cation are aromatic

• The key feature of both is that they contain 6 πelectrons in a planar ring of continuous p orbitals

cyclopentadienyl anion cycloheptatrienyl cation

H H H Cl

H H

X

+.. -

Cl-

NaOEtEtOH

The anion forms readily.The cation does notform at all.

6ππππ 4ππππ

CYCLOPENTADIENYL ANION AND CATION

AROMATIC ANTI-AROMATIC

The methylenehydrogens areacidic.

This compounddoes not dissolvein water.

•Relatively acidic (pKa = 16) •because the anion is stable

HH

H HH

-H+ -H-

-H.

6 ππππ 5ππππ 4ππππelectrons

anion radical cation

23

Electron Delocalization in Electron Delocalization in CyclopentadienideCyclopentadienide AnionAnion

HH HH

HH HH

HH

••••––

Electron Delocalization in Electron Delocalization in CyclopentadienideCyclopentadienide AnionAnion

HH HH

HH HH

HH

••••––

HH HH

HH HH

HH

••••––

Electron Delocalization in Electron Delocalization in CyclopentadienideCyclopentadienide AnionAnion

HH HH

HH HH

HH

••••––

HH HH

HH HH

HH

••••––

HH HH

HH HH

HH

•••• ––

24

Electron Delocalization in Electron Delocalization in CyclopentadienideCyclopentadienide AnionAnion

HH HH

HH HH

HH

••••––

HH HH

HH HH

HH

••••––

HH HH

HH HH

HH

•••• ––

HH HH

HH HH

HH

–– ••••

Electron Delocalization in Electron Delocalization in CyclopentadienideCyclopentadienide AnionAnion

HH HH

HH HH

HH

••••––

HH HH

HH HH

HH

••••––

HH HH

HH HH

HH

HH HH

HH HH

HH

•••• ––

HH HH

HH HH

HH

–– ••••

••••

Compare Acidities ofCompare Acidities of

CyclopentadieneCyclopentadiene and and CycloheptatrieneCycloheptatriene

HH HH

HH HH

HH HH

ppKKaa = 16= 16

KKaa = 10= 10--1616

ppKKaa = 36= 36

KKaa = 10= 10--3636

HH HH

HHHHHH

HH HH

HH HH

25

HH HH

HH HH

HH

••••––

Compare Acidities ofCompare Acidities of

CyclopentadieneCyclopentadiene and and CycloheptatrieneCycloheptatriene

Aromatic anion6 π electrons

Anion not aromatic8 π electrons

HH HH

HHHH

HH HH

HH

••••––

Cycloheptatriene• Cycloheptatriene has 3

conjugated double bonds joined by a CH2

• Removal of “H-” leaves the cation

• The cation has 6 electrons and is aromatic

H H H Cl

H H

X

+..-

Cl-

NaOEtEtOH

8ππππ 6ππππ

CYCLOHEPTATRIENYL ANION AND CATION

ANTI-AROMATIC AROMATIC

This compound ionizes easily in water.

The methylenehydrogens arenot acidic.

Doesn’t formeasily.

Dissolves in water.

26

Azulene

• Azulene, C10H8.

• Deep blue color, large dipole moment

• Is it aromatic ?• Explain dipole moment.

Azulene

• Is it aromatic ?• Explain dipole moment.

10 π electrons, ring system, planar� Hϋckel aromat

Azulene

• _____________ resonance structures

• ___________________ resonance structure

27

Azulene_____________________ resonance structure

One end: cyclopentadienyl anion likeOther end: cycloheptatriene cation like

cyclopentadienyl anion cycloheptatrienyl cation

Azulene

+

δδδδ+ δδδδ+

δδδδ+

δδδδ+δδδδ+

δδδδ-

δδδδ-

δδδδ-

- +

The contribution from the __________ _________ form leads to the presence of a large dipole moment . That dipole moment is associated with the color.

Naphtalene versus Azulene

28

Spectroscopy of Aromatic Compounds

• IR: Aromatic ring C–H stretching at 3030 cm−−−−1 and peaks 1450 to 1600 cm−−−−1(See Figure 15-13)

• UV: Peak near 205 nm and a less intense peak in 255-275 nm range

• 1H NMR: Aromatic H’s strongly deshielded by ring and absorb between δ 6.5 and δ 8.0– Peak pattern is characteristic of positions of

substituents

3000

-C-H=C-H

31003300

=C-H=

2900 2850 2750

-CH=O(weak)

increasing CH Bond Strength

sp3-1ssp2-1ssp-1s

increasing frequency (cm-1)

aldehyde

increasing s character in bond

increasing force constant K

STRONGER BONDS HAVE LARGER FORCE CONSTANTSAND ABSORB AT HIGHER FREQUENCIES

CH BASE VALUE = 3000 cm-1

The C=C stretching region

• C=C double bond at 1650 cm-1 is often weak or not even seen.

• C=C benzene ring shows peak(s) near 1600 and 1400 cm-1 , one or two at each value - CONJUGATION

LOWERS THE VALUE.

• When C=C is conjugated with C=O it isstronger and comes at a lower frequency.

29

TolueneToluene

CH3

AROMATIC

benzene

oops

benzeneC-H

C=C

H

H

H

OUTOUT--OFOF--PLANE BENDINGPLANE BENDING

above

below

PLANE

(OOPS)

H

ALKENES

BENZENES

also with

10 11 12 13 14 15

1000 900 800 700 cm-1

Monosubstituted

Disubstituted

ortho

meta

para

Trisubstituted

1,2,4

1,2,3

1,3,5

BENZENES µ

s s

s

s s

s

s

s

s

m

m

m

mOOPS

RING H’s

combination bands

30

1H-NMR of Aromatic Compounds

8.00 7.90 7.80 7.70 7.60 7.50 7.40 7.30 7.200.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

N+

O-

O

CH 3

C H 3

1H-NMR of Aromatic Compounds

8.30 8.20 8.10 8.00 7.90 7.80 7.70 7.60 7.50 7.40 7.30 7.20 7.10 7.00 6.90 6.800.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

N+

O-

O

CH3

1H-NMR of Aromatic Compounds

8.50 8.40 8.30 8.20 8.10 8.00 7.90 7.80 7.70 7.60 7.50 7.40 7.30 7.200.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

N+

O-

O

31

1H-NMR of Aromatic Compounds

8.70 8.60 8.50 8.40 8.30 8.20 8.10 8.00 7.90 7.80 7.70 7.60 7.50 7.40 7.30 7.20 7.10 7.000.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

OH

N+

O-

O

OHO

13C NMR of Aromatic Compounds

• Carbons in aromatic ring absorb at δ 110 to 140 ppm

• Shift is distinct from alkane carbons but in same range as alkene carbons

170 165 160 155 150 145 140 135 1300

10

20

30

40

50

60

70

80

90

100

110

120

129.01(5;3)130.08(6;2)

133.76(4)

168.57(7)

OH9

6

1

5

2

4

3

7

O8

32

170 165 160 155 150 145 140 135 130 125 120 1150

10

20

30

40

50

60

70

80

90

100

110

120

115.86(6;2)

122.35(4)

132.49(5;3)

162.00(1)168.97(8)

OH10

5

4

6

3

1

2

8

O9

OH7

The Effect of Aromaticity on the pKaValues of Some Compounds

CH3CH3 CH3CH2-

+ H+pKa = 50

H

H

pKa = 15H

Phenols

• The functional group of a phenol is a -OH group bonded to a benzene ring

1,2-Benzenediol (Catechol)

1,4-Benzenediol (Hydroquinone)

3-Methylphenol (m-Cresol)

Phenol

OH OHOH OH

OHCH3

OH

33

Acidity of Phenols

• Phenols are significantly more acidic than alcohols, compounds that also contain the -OH group

Phenol: pK a = 9.95

Ethanol: pK a = 15.9

OH O-

CH3 CH2 O -CH3 CH2 OH

H2 O+

H2 O+

H3 O++

H3 O++

Acidity of Phenols

• We account for the increased acidity of phenols relative to alcohols in the following way• delocalization of the negative charge on a phenoxide

ion stabilizes it relative to an alkoxide ion• because a phenoxide ion is more stable than an

alkoxide ion, phenols are stronger acids than alcohols

• Note that while this reasoning helps us to understand why phenols are more acidic than alcohols, it does not give us any way to predict how much stronger they are

Acidity of Phenols

•• ••

O

•••• •• ••

••••

••

These three contributing structures delocalize the negative charge onto

carbon atoms of the ring

H

H

H

OO

•• •••• •• ••••

These 2 Kekulé structures are equivalent

OO

34

Acidity of Phenols

• Ring substituents, particularly halogen and nitro groups, have marked effects on the acidity of phenols

pKa 9.18

Phenol

pKa 9.95

OH OH

Clp-Chloro-

phenolpKa 7.15

OH

NO2p-Nitro-phenol

ACIDITY Res forms

Acidity of Phenols• Phenols are weak acids and react with strong

bases to form water-soluble salts• water-insoluble phenols dissolve in NaOH(aq)

Phenol Sodium hydroxide

Sodium phenoxideWater

pKa = 9.95

pKa = 15.7

(stronger acid)

(weaker acid)

(stronger base)

(weaker base)

OH

O- Na +

+ NaOH

+ H2 O

35

Acidity of Phenols

• most phenols do not react with weak bases such as NaHCO3; they do not dissolve in aqueous NaHCO3

(Weaker base)(Weaker acid)

pKa = 6.36Carbonic acid

Sodium phenoxide(Stronger base)

(Stronger acid)

pKa = 9.95

Sodium bicarbonatePhenol

OH

O- Na +

+ NaHCO3

+ H2 CO3

Electron withdrawal decreases reactivity toward electrophilic substitution and increases acidity

Electron donation increases reactivity toward Electrophilic substitution and decreases acidity

36

Reactions of Aromatic Sidechains

• Benzylic bromination using NBS• Benzylic oxidation using KMnO4

Oxidation with neutral MnO4-

CH3

CH2CH2CH3

MnO4-

MnO4-

MnO4-

COOH

COOH

COOH

Oxidation with neutral MnO4-

CH3

CH2CH2CH3

MnO4-

MnO4-

MnO4-

COOH

COOH

COOH

37

NBS BrominationCH3

CH2CH2CH3

NBS/CCl4

NBS/CCl4

NBS/CCl4

NBS BrominationCH3

CH2CH2CH3

NBS/CCl4

NBS/CCl4

NBS/CCl4

CH2 Br

CH2CH2CH3

Br

Br

NBS Bromination

O

O

O

O

hνννν O

O

N

O

O

Br

Br

CH2

H

CH2

N

O

O

BrCH2

Br

38

Birch Reduction

Li/NH3

Birch Reduction

H3C

O

CH3

Li/NH 3

Li/NH 3

Li/NH 3

H3C

O

CH3

With substituted aromaticcompounds, the Birch reduction will proceed to place the double bondon a ring carbon which is attached to electron donating substituents…

Birch Reduction

H3C

O

CH3

Li/NH 3

Li/NH 3

Li/NH 3

H3C

O

CH3

39

Birch Reduction : Mechanism

Na

H2N H

H

HNaH

HH2N HH

HH

H

Homework• Predict the outcome of the following reactions:

NBS/CCl4

RO-OR

MnO4-/H2O

Li/NH 3