Aromatic Compound Theory E

8/21/2019 Aromatic Compound Theory E

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CHEMISTRY

AROMATIC COMPOUNDSINTRODUCTION

(1)There were a large number of compounds which were obtained from natural sources, e.g. resins, balsams,'aromatic' oils, etc., which comprised a group of compounds whose structures were arbitrarily classified asaromatic (Greek : aroma, fragrant smell) compounds.(2) These compounds contained a higher percentage carboncontent than the corresponding aliphatichydrocarbons, and that most of the simple aromatic compounds contained at least six carbon atoms.

It was shown that when aromatic compounds were subjected to various methods of treatment, they oftenproduced benzene or a derivative of benzene.(3) The benzene containing aromatic compounds are called benzenoid compounds, these are cyclic, buttheir properties are totally different from those of the alicyclic compounds.(4) Benzene was first synthesised by Berthelot (1870) by passing acetylene through a red-hot tube :

3C2H

2 C

6H

6+ other products

(5) It is mostly prepared by the decarboxylation of aromatic acids, e.g. by heating benzoic or phthalic acidwith calcium oxide / (soda lime).

COOH

)CaO(NaOH

)CaO(NaOH

1. Aromatic Character : [The Huckel (4n + 2) rule]The following three rules are useful in predicting whether a particular compoundis aromatic or nonaromatic.

1. Aromatic compounds are cyclic and planar.2. Each atom in an aromatic ring is sp2 or sp hybridised.3. The cyclic molecular orbital (formed by overlap of p-orbitals) must contain (4n + 2) electrons, i.e.,

2, 6, 10, 14 ........ electrons. Where n = an integer 0, 1, 2, 3,..............

Molecular orbital theory of aromaticity

According tomolecular orbital theorybenzene is a regular flat hexagon. Thus each carbon atom is in astate of sp2 hybridisation. Hence, in benzene, there are six CH bonds, six CC bonds and 3CC -bondssix 2p

zelectrons (one on each carbon atom) are present in 2p

zorbitals, which are all parallel the p-orbitals

are perpendicular to the plane of the ring. These electrons can be paired in two ways, both being equallygood (b and c). Each 2p

zelectron, however, overlaps its neighbours equally, and thereforeall six atomic

orbitals a single hexacentric molecular orbitals.

(a) (b) (c) (d)

In the ground state, the total energy of the three pairs of delocalised -electrons is less than that of three

pairs of localised -electrons (b) or (c), and hence the benzene molecule is stabilised by delocalisation(resonance).

2. Comparison of Aromatic compounds with alkenesBenzene is not as reactive as alkenes, which react rapidly with bromine at room temperature to give additionproducts.For example, cyclohexenereacts to givetrans-1, 2-dibromocyclohexane. This reaction is exothermicby about 29 kcal/mol (121 kJ/mol.)

+ Br2 H = 29 kcal ( 121 kJ)

The analogous addition of bromine to benzene is endothermic because it requires the loss of aromaticstability. The addition is not seen under normal circumstances. The substitution of bromine for a hydrogen

atom gives an aromatic product. The substitution is exothermic, but it requires a Lewis acid catalyst toconvert bromine to a stronger electrophile.


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CHEMISTRY

+ Br2

H = + 2 kcal (+ 8kJ)

+ Br2

3FeBr + HBr H = 10.8 kcal ( 45 kJ)

Examples of Aromatic Compounds : (Table )


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CHEMISTRY

3. Aromatic Electrophilic Substitution (ArSE

2) Reactions in Benzene Ring

Like an alkene, benzene has clouds of pi electrons above and below its sigma bond framework. Althoughbenzenes pi electrons are in a stable aromatic system still they are available to attack a strong electrophileto give a carbocation. This resonance-stabilized carbocation is called a sigma complex because theelectrophile is joined to the benzene ring by a new sigma bond.

Thesigma complex (also called an arenium ion) is not aromatic because the sp3 hybrid carbon atom interruptsthe ring of p orbitals. This loss of aromaticity contributes to the highly endothermic nature of thus first step.

The sigma complex regains aromaticity either by a reversal of the first step (returning to the reactants) or byloss of the proton on the tetrahedral carbon atom, leading to the substitution product.

The overall reaction is thesubstitution of an electrophile

)E( for a proton

)H( on the aromatic ring: electrophilicaromatic substitution.

Step 1 : Attack of an electrophile on benzene ring forms the sigma complex

H E H E

Resonance hybrid[ complex]Arenium Ion

Step 2 : Loss of a proton gives the substitution product.

H E

:Nu

E

+ Nu H

4. Activating Groups or Electron Releasing Groups (ERG)All groups having one or more lone pair of electrons are activating groups because they release electronstowards the nucleus increasing electron density and hence energyof the system. Reaction rate is increaseddue to low energy of activation. Examples :

> > > > > OR > > Ar > R

All the groups which are electron donating etc. are orthopara directing and

facilitate electrophilic substitution in the benzene ring.

Ex.1 Compare the activating effects of the following o, p-directors and explain your order

(a) OH....

, O:....-

, OC CH3....

||O

(b) 2..

HN and

O||

CHCHN 3..

Sol. (a) The order of activation is O- >OH>OCOCH3.The O- , with a full negative charge, is best able todonate electrons, there by giving the very stable uncharged intermediate

E

H O

In 3OCCH||O

, the C of theC = O + -

has a positive charge and makes demands on the ..

..O for electron

density this cross-conjugation diminishes the ability of the too donate es to the arenium ion.

(b) The order is NH2> NHCOCH

3because of cross - conjugation in the amide,Ar N = C CH 3

| |

H :O:.. -

.


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CHEMISTRY

5. Deactivating Group or Electron Withdrawing Group (EWG)

Such groups have tendency to withdraw electrons from the benzene nucleus and thus decreasing itselectron density are known as deactivating groups.Due to decrease in electron density of the ring, the rate of electrophilic substitution is retarded. These groupsdevelop positive charge at ortho and para positions leaving the meta-positions as the point of relativelyhighelectron density and hence the electrophilic substitution occurs at mposition, not at oand ppositions.

eg.

6. Halogenation

(a) Bromination of Benzene:Bromination follows the general mechanism for electrophilic aromatic substitution. Bromine itself is notsufficiently electrophilic to react with benzene, but a strong Lewis acid such as FeBr

3catalyzes the reaction.

Step 1 :Formation of a stronger electrophile.

Br ::

: Br + FeBr 3

:

: : Br ::

: Br :

:

FeBr3

:

Step 2 :Electrophilic attack and formation of the sigma complex.

Br ::

: Br FeBr3

:

: :+H

HH

HH

H

+ FeBr4

Step 3 :Loss of a proton gives the products.

BrH

H

HH

H

bromobenzene

+ HBr + FeBr3

+

H

intermediate products

reactants

energy

reaction coordinate

BrFeBr4

Br + HBr+ FeBr3

T .S2

-10.8 kcal/mol

+ Br2

+ FeBr3

T .S1

Formation of the sigma complex is rate determining and the transition state leading to it occupies thehighest-energy point on the energy diagram. This step is strongly endothermic because it forms a non-

aromatic carbocation. The second step is exothermic because aromaticity is regained and a molecule ofHBr is evolved. The overall reaction is exothermic by 10.8 kcal/mol (45 kJ/mol.)


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CHEMISTRY

(b) Chlorination of benzene

Chlorination of benzene works much like bromination, except that aluminum chloride (AlCl3) is most often

used as the Lewis acid catalyst.

+ Cl2

3AlCl + HCl

Iodination of benzene

Iodination of benzene requires an acidic oxidizing agent such as nitric acid. Nitric acid is consumed in thereaction, so it is a reagent (an oxidant) rather than a catalyst.

+2

1I

2 + HNO

3 + NO

2+ H

2O

Iodination probably involves an electrophilic aromatic substitution with iodonium ion

)(I acting as the

electrophile. The iodonium ion results from oxidation of iodine by nitric acid.

H + HNO3+

2

1I

2

ioniodonium

I + NO2+ H2O

7. Nitration

Nitration is brought about by the action of concentrated nitric acid or a mixture of concentrated nitric acid and

sulphuric acid often called nitrating mixture. HNO3alone is a weak nitrating agent where as the mixture isstrong nitrating mixture when concentrated HNO3and concentrated H2SO4are mixed to form the nitratingmixture, NO2

+ (Nitronium ion) is produced as follows :

OH3 +

HNO3+ 2H2SO4

Now, the NO2+ ion attacks the benzene nucleus and forms an intermediate cation (a benzenonium ion) which

loses a proton to yield the nitro derivative.

+NO2

(-complex)

H

NO2+

(- complex)

H+ +

8. Sulphonation

The electrophilic reagent, SO3, attacks the benzene ring to form the intermediate carbocation.

2H2SO4 SO3+ +


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CHEMISTRY

Sulphonation, isreversibleand takes place in concentrated sulphuric acid.

+ SO3 + H

+

+

H

+ H

Energy

progress of the reaction

++

SO H3SO H3

SO H3

transition state for the rate-determining step

in the forward direction and for therate-determining step in the reverse direction

9. Friedel Craft reaction

(a) Alkylation :The carbon atom of alkyl halides,

XR , is an electrophile. The presence of a Lewis acid

catalyst is also required. Anhydrous aluminium chloride. AlCl3, being a Lewis acid, accepts a lone pair ofelectrons from halogen (Chlorine atom) of R . This makes R (alkyl) group to be sufficiently polar so as to

act as an electrophile. The mechanism for Friedal Crafts reaction involves the following steps.

(i)

(ii)

(iii) + AlCl4

Nature of Lewis acid as catalyst

The order of effectiveness of Lewis acid catalyst has been shown to beAlCl

3 > FeCl

3 > BF

3 > TiCl

3 > ZnCl

2 > SnCl

4

Ex.2 What would be the major product of a friedal-Crafts alkylation reaction using the following alkyl halides ?(a)CH

3CH

2Cl (b) CH

3CH

2CH

2Cl (c) CH

3CH

2CH(Cl)CH

3

(d) (CH3)

3CCH

2Cl (e) (CH

3)

2CHCH

2Cl (f) CH

2= CHCH

2Cl

Sol. (a) Ethylbenzene (b) I sopropylbenzene (c) Sec-butylbenzene(d) Tert-pentylbenzene (e) Tert-butylbenzene (f) 3-Phenylpropene

(b) Acylation :Acylation of benzene may be brought about with acid chlorides or anhydrides in presence ofLewis acids.Mechanism

Step 1 :Formation of an acylium ion.

R C Cl AlCl3

||O: :

+ -....

complex

-AlCl4 +

Step 2 :Electrophilic attack.

O||C|R

+

+

O||C

RH

H

sigma complex


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CHEMISTRY

Step 3 :Loss of a proton. Complexation of the product.

+

-

O||C

RH

H

sigma complex

:Cl AlCl3....

e.g. + (1) AlCl3

(2) H O2

Note : Friedal - Crafts acylations are generallyfree from rearrangements and multiple substitution. They do

not go on strongly deactivated rings.

e.g.

10. Structure of Benzene

Analysis and molecular-weight determinations show that the molecular formula of benzene is C6H

6. I t istobe

expected that benzene would exhibit marked 'unsaturated reactions'. This is found to be so in practice, e.g.(i) Benzene adds on halogen,the maximum number of halogen atoms being six.

(ii) Benzene may be catalytically hydrogenated to cyclohexane at higher temperature (200C).

(iii) Benzene forms a triozonide C6H

6(O

3)

3.

(iv) In the absence of sunlight, benzene undergoes substitution when treated with halogen.

(v) Halogen acids do not add on to benzene.

So we conclude that

(1) Benzene contains three double bonds.

(2) All the six hydrogen atoms in benzene are equivalent consequently there is only one possible mono-

substituted derivative and there are three possible disubstitution products of benzene.

11. Preparation of Arenes :(A) Benzene

(1) By polymerisation (of Acetylene) :

3HC CH tubeironhotdRe

(2) By decarboxylation (of Benzoic acid) :

NaOH )CaO(

NaOH

+ Na2CO3


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CHEMISTRY

(3) By catalytic reforming of n-Hexane :

CH3 (CH2)4 CH32

K873,Pt

H

2

K873,Pt

H3

(4) By reduction (of Benzene diazonium Chloride) :

+ H3PO2+ H2O + N2

(B) Toluene :

(1) By Friedel-Crafts reaction :

+ CH3Cl 3AlCl + HCl

Toluene

(2) By Wurtz fitting reaction :

+ 2Na + C H3Br etherdry

+ 2NaBr

Bromobenzene Toluene

(3) From Grignard reagents :

+ CH3Br + MgBr2

(4) By catalytic reforming of n-Heptane :

)H(

Pt,K873

2

)H3(

Pt,K873

2

Methylcyclohexane Toluene

(C) Xylene :

33 AlCl,XCH

CH3CH3

o-Xylene

|

+


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CHEMISTRY

12. Reactions of Benzene :

Ex.3 Complete the following

4KMnO X

Sol.COOH

Ex.4 Gives the products of the following reactions :

(a)

Cl

+ 3AlCl X (b) CH3 Cl +

OCH3

3AlCl Y

(c)

3

33

3

CHCl|| CHCHCCH

|CH

+

H C CH CH3 3

3AlCl

Z


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CHEMISTRY

Sol. X = phenylcyclohexane

Y =

OCH3CH3

OCH3

CH3O - methyl anisole p methyl anisole

Z = 1 - Isopropyl 4 (1, 1, 2 trimethylpropyl) benzene

CH

H CCCH3 3

H CCH3

CH3

CH3H C3

Ex.5 Predict the products (if any) of the following reactions

(a) excess +

3

333

CH|

AlClCHCCH|Cl

X

Isobutylchloride

(b) +

lotanbu1

BFOHCHCHCHCH 32223

Y

(c)

NO2

+

Cl|

AlClCHCHCH 333 Z

nitrobenzene (excess)

(d)

Benzene(excess)

+

3

23

3

CH|

HFCHHCCCH|CH

P

Sol. X =

CH C CH3 3

|CH3

ter-butyl benzene

Y = or

CH3

4-sec. butyltoluene

|H C CH3

|CH2|

CH2

Z = No reaction CCHCH3

CH3

CH3

CH3

P = (1, 1, 2-Trimethylpropyl) benzene


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CHEMISTRY

Ex.6 Outline a synthesis of biphenyl from benzene.

Sol. Fe/Br2

Br

Cu

(Ullmann reaction)

13. ARYL HALIDES

(A) Preparation of Aryl Halides1. Halogenation :

+ X2

acidLewis

+ HX

X2= Cl

2, Br

2

Lewis acid = FeCl3, AlCl

3, ZnCl

2, Zn etc.

2. Decarboxylation :

CaO/NaOH + CO

2

3. From Phenol :

5PCl

+ POCl3+ HCl

(B) Mechanism of bimolecular nucleophilic substitution (ArSN

2)

+ NuRDS

IStep

IIStep

)fast(X

Intermediate ion is stabilized by resonance.

A group that withdraws electrons tends to neutralize the negative charge of the ring and so to become morenegative itself; this dispersal of the charge stabilizes the carbanion.

G withdraws electrons : stabilizes carbanion, activates

( (CH3

)3

,NO2

, CN, SO3

H, COOH, CHO, COR, X)


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CHEMISTRY

A group that releases electrons tends to intensify the negative charge, destabilizes the carbanion, and thusslows down reaction.

G releases electrons : destabilizes carbanion, deactivates

(NH2, OH, OR, R)

Orientiation in nucleophilic aromatic substitution :

At para position :

At meta position :

At ortho position :

Note : If electron withdrawing group in present at ortho and para position it especially activates the aromaticnucleophilic substitution reaction.

14. PHENOLS (ArOH) :

(A) Preparation of ArOH

(1) (2)

(3) (4)


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CHEMISTRY

(5) (6)

(7)

(B) Properties of phenol

These are characteristic of monohydric phenols. Phenol is a colourless crystalline solid, m.p. 43, b.p.182C, which turns pink on exposure to air and light. It is moderately soluble in cold water, but is readilysoluble in concentrated sulphuric acid (Liebermann reaction), when phenol is dissolved in concentratedsulphuric acid and a fewdrops of aqueous sodium nitrite added, a redcolour is obtained on dilution andturnsblue when made alkaline with aqueous sodium hydroxide.

Phenol is used as an antiseptic and disinfectant and in the preparation of dyes, drugs, bakelite, etc.

(C) Chemical Reactions of Phenols

(1)

(2)

(3) (Major is oxidative cleavage of ring)

(4)

(p-nitroso phenol)

(5)


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CHEMISTRY

(6)

(7)

(8)

(9) (Fries Rearrangement)

(10)

(11) (polymer)

(12)

(13) (14)

(15) (16)


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CHEMISTRY

15. Nitrobenzene

(A) Preparation

+ HNO3(conc.)

(B) General properties of nitrobenzene

(i) Yellow liquid(ii) Denser than water. Thus insoluble in water but soluble in organic solvents(iii) b.p. = 211C(iv) steam - volatile

(C) Chemical Reactions of Nitrobenzene

(1)

(2)

(3)

(4)

(5)

(6)

(7)


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CHEMISTRY

(8)

(9)

(10)

16. ANILINE

(A) Prepartion of Aniline

(1)

(2)

(3)

(B) Chemical Reactions of Aniline

(1) (2)

(3)

(4)


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CHEMISTRY

(5)

(6) (Diazotisation)

(7)

(8)

(9)

(10)

(11)

(12) (oxidation)

(13) C6H5 :CN (14)


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CHEMISTRY

17. Benzenediazonium chloride

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)


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CHEMISTRY

(10)

(11)

(12)

(13)

(14)

(15)

(16)

(17)


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CHEMISTRY

MISCELLANEOUS SOLVED PROBLEMS (MSPS)

Complete the following reactions :

1. (a) + HClO4

(b) + 2 AgClO4

Ans. (a) (b) + AgCl

Sol. Aromatic compounds are -(i) cyclic (ii) planer (iii) contains (4n + 2) no of-electrons where n = o , 1,2,3 .......(iv) does cyclic resonance between (4 n + 2) - electrons

2. + CH3 CH

2 CH

2 Cl 3

AlCl

Ans.

Sol. Aromatic compounds undergo electrophilic substitution reaction, and if possible then there is rearrangementof carbocation occurs.

3. (a) +

H

(b) +

(c)

Ans. (a) (b) (c)

Sol. Since the fridel craft acylation is faster than alkylation.


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CHEMISTRY

4. + H2SO

4

Ans.

Sol. 2H2SO

4 + +

5.

Sol. +

6.

Ans.

Sol. Incase of biphenyl one ring is electron donor and other is electron acceptor.

7. Fe/Br2


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CHEMISTRY

Sol.

8. Fe/Br2

Ans.

Sol. Incase of electrophilic substitution reaction stable carbonium ion intermediate is formed.

9. (a) OH/Br 22 (b)

Ans. (a) (b)

Sol. (a) ion increases the reactivity towards electrophile due to increases the electron density in ring.


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CHEMISTRY

10. (a)

H/KMnO4 (b) H/KMnO4

Ans. (a) (b)

Sol. (a & b) In aromatic hydrocarbons if at least one benzylic hydrogen is present then by oxidation benzoic acidis formed.

11. (a) KOH.aq.2

CHCl.1 3

(b) KOH.aq.2

CCl.1 4

(c) )excess(

OH/Br 22

Ans./Sol.

(a) (b) (c) + 3HBr

13.

42

722

SOH

OCrNa

Ans.

Sol. Oxidation of phenol with cromic acid produces a conjugated diketone known as benzoquinone.

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Aromatic Compound Theory E

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