Aromatic Compound Theory E

CHEMISTRY

AROMATIC COMPOUNDS INTRODUCTION (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 as aromatic (Greek : aroma, fragrant smell) compounds. (2) These compounds contained a higher percentage carbon content than the corresponding aliphatic hydrocarbons, 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 often produced benzene or a derivative of benzene. (3) The benzene containing aromatic compounds are called benzenoid compounds, these are cyclic, but their 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 : 3C2H2  C6H6 + other products (5) It is mostly prepared by the decarboxylation of aromatic acids, e.g. by heating benzoic or phthalic acid with calcium oxide / (soda lime). COOH NaOH (CaO)

 

1.

NaOH (CaO)

 

Aromatic Character : [The Huckel (4n + 2) rule] The following three rules are useful in predicting whether a particular compound is aromatic or non–aromatic. 1. 2. 3.

Aromatic compounds are cyclic and planar. Each atom in an aromatic ring is sp2 or sp hybridised. 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 to molecular orbital theory benzene is a regular flat hexagon. Thus each carbon atom is in a state of sp2 hybridisation. Hence, in benzene, there are six C–H bonds, six C–C bonds and 3C–C -bonds six 2pz electrons (one on each carbon atom) are present in 2pz orbitals, 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 equally good (b and c). Each 2pz electron, however, overlaps its neighbours equally, and therefore all 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 alkenes Benzene is not as reactive as alkenes, which react rapidly with bromine at room temperature to give addition products. For example, cyclohexene reacts to give trans-1, 2-dibromocyclohexane. This reaction is exothermic by 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 aromatic stability. 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 to convert bromine to a stronger electrophile.

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CHEMISTRY H° = + 2 kcal (+ 8kJ)

+ Br2

FeBr

3 + Br2   

Examples of Aromatic Compounds :

+ HBr H° = – 10.8 kcal (– 45 kJ)

(Table )

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CHEMISTRY 3.

Aromatic Electrophilic Substitution (ArSE2) Reactions in Benzene Ring Like an alkene, benzene has clouds of pi electrons above and below its sigma bond framework. Although benzene’s pi electrons are in a stable aromatic system still they are available to attack a strong electrophile to give a carbocation. This resonance-stabilized carbocation is called a sigma complex because the electrophile is joined to the benzene ring by a new sigma bond. The sigma complex (also called an arenium ion) is not aromatic because the sp3 hybrid carbon atom interrupts the 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 by loss of the proton on the tetrahedral carbon atom, leading to the substitution product. 



The overall reaction is the substitution of an electrophile (E ) for a proton (H) on the aromatic ring: electrophilic aromatic 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

E 

Nu :  

4.

+ Nu – H

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

>

>

>

>

> –OR >

All the groups which are electron donating facilitate electrophilic substitution in the benzene ring. Ex.1

Sol.

> –Ar > –R etc. are ortho–para directing and

Compare the activating effects of the following o, p-directors and explain your order .. .. .. .. .. (a) – OH , , (b) and  N H  C  CH3  N H – O: – OC – CH 2 3 .. .. .. || || O O (a) The order of activation is – O > – OH > – OCOCH3. The – O , with a full negative charge, is best able to donate electrons, there by giving the very stable uncharged intermediate

E H

O

O || .. +  In  OCCH3 , the C of the C = O has a positive charge and makes demands on the  O  for electron ..

density this cross-conjugation diminishes the ability of the too donate e–s to the arenium ion. (b) The order is – NH2 > – NHCOCH3 because 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 its electron density are known as deactivating groups. Due to decrease in electron density of the ring, the rate of electrophilic substitution is retarded. These groups develop positive charge at ortho and para positions leaving the meta-positions as the point of relatively high electron density and hence the electrophilic substitution occurs at m–position, not at o–and p–positions.

eg.

6.

Halogenation (a) Bromination of Benzene : Bromination follows the general mechanism for electrophilic aromatic substitution. Bromine itself is not sufficiently electrophilic to react with benzene, but a strong Lewis acid such as FeBr3 catalyzes the reaction. Step 1 : Formation of a stronger electrophile. :

:Br – Br : + FeBr3

: :

: :

: :

: :

   :Br – Br – FeBr3

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

+

: :

H

: :

H

:Br – Br: FeBr3 H

H





+ FeBr4¯

H

Step 3 : Loss of a proton gives the products.

H Br

H

+ HBr + FeBr 3 H

H H bromobenzene

T .S1 reactants

T .S2

+ Br2 + FeBr3

H

energy

+

Br¯FeBr4

intermediate

products

-10.8 kcal/mol

Br + HBr + FeBr3

reaction coordinate

Formation of the sigma complex is rate determining and the transition state leading to it occupies the highest-energy point on the energy diagram. This step is strongly endothermic because it forms a nonaromatic carbocation. The second step is exothermic because aromaticity is regained and a molecule of HBr 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. AlCl3 + Cl2   

+ HCl

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

1 I + HNO3  2 2

+

+ NO2 + H2O



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 +

7.

1 I 2 2





+ NO2 + H2O

I iodonium ion

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. HNO3 alone is a weak nitrating agent where as the mixture is strong nitrating mixture when concentrated HNO3 and concentrated H2SO4 are mixed to form the nitrating mixture, NO2+ (Nitronium ion) is produced as follows :

H3 O  + 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.

H

+

NO2 ( -complex)

+

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, is reversible and takes place in concentrated sulphuric acid.

+ H

+ SO3

transition state for the rate-determining step in the forward direction and for the rate-determining step in the reverse direction

H

Energy

+

SO3H SO3H

+

+ H+

+ SO3H progress of the reaction

9.

Friedel Craft reaction





(a) Alkylation : The carbon atom of alkyl halides, R  X , 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 of electrons 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 Craft’s 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 be AlCl3 > FeCl3 > BF3 > TiCl3 > ZnCl2 > Ex.2

Sol.

SnCl4

What would be the major product of a friedal-Crafts alkylation reaction using the following alkyl halides ? (a) CH3CH2Cl (b) CH3CH2CH2Cl (c) CH3CH2CH(Cl)CH3 (d) (CH3)3CCH2Cl

(e) (CH3)2CHCH2Cl

(f) CH2 = CHCH2Cl

(a) Ethylbenzene (d) Tert-pentylbenzene

(b) Isopropylbenzene (e) Tert-butylbenzene

(c) Sec-butylbenzene (f) 3-Phenylpropene

(b) Acylation : Acylation of benzene may be brought about with acid chlorides or anhydrides in presence of Lewis acids. Mechanism Step 1 : Formation of an acylium ion.

:O: || .. + R – C – Cl .. – AlCl 3

AlCl4 +

complex

Step 2 : Electrophilic attack. O || C+ | R

+

O || C H H

R

sigma complex

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

+

O || C H H

R

.. :Cl .. – AlCl 3

sigma complex

e.g.

(1) AlCl3

+

(2) H2O

Note : Friedal - Crafts acylations are generally free 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 C6H6. It is to be 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 (200ºC). (iii) Benzene forms a triozonide C6H6(O3)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 monosubstituted derivative and there are three possible disubstitution products of benzene.

11.

Preparation of Arenes : (A) Benzene

(1)

By polymerisation (of Acetylene) : Re d hot iron tube 3HC  CH  

(2)

By decarboxylation (of Benzoic acid) :

NaOH   

NaOH    (CaO )

+ Na2CO3

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CHEMISTRY (3)

By catalytic reforming of n-Hexane : Pt, 873 K   

Pt, 873 K CH3 – (CH2)4 – CH3   

 3H2

 H2

(4)

By reduction (of Benzene diazonium Chloride) :

+ H3PO2 + H2O 

+ N2

(B) Toluene : (1)

By Friedel-Crafts reaction : + CH3Cl

AlCl

3  

+ HCl Toluene

(2)

By Wurtz fitting reaction :

dry ether

+ 2Na + CH3Br  Bromobenzene (3)

Toluene

From Grignard reagents :

+

(4)

+ 2NaBr

CH3Br 

+ MgBr2

By catalytic reforming of n-Heptane :

873 K, Pt   

873 K, Pt   

(  H2 )

(  3H2 )

Methylcyclohexane

Toluene

(C) Xylene : CH3 | CH3 X, AlCl3

  

CH3

+

o-Xylene

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CHEMISTRY 12.

Reactions of Benzene :

Ex.3

Complete the following KMnO

4      X

Sol.

COOH Ex.4

Gives the products of the following reactions :

OCH3

Cl (a)

+

AlCl3   X

(b) CH3 – Cl +

AlCl3   Y

CH3 H3C – CH – CH3 | AlCl 3    Z (c) H3C  HC  C  CH3 + | | Cl CH3

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CHEMISTRY

Sol.

X=

phenylcyclohexane OCH3

OCH3 CH3

Y= CH3

O - methyl anisole

p – methyl anisole H3C

CH3 CH

Z = 1 - Isopropyl – 4 – (1, 1, 2 – trimethylpropyl) benzene H3C–C–CH3 H3C–CH CH3

Ex.5

Predict the products (if any) of the following reactions Cl | + H3 C  C  CH3  AlCl3 | CH3

(a) excess

 X

Isobutylchloride + H3 C  CH2  CH2  CH2  OH  BF3

(b)



Y

1  butanol

NO2 H3C  CH  CH3  AlCl3  | Cl nitrobenzene (excess)

(c)

+

(d) Benzene (excess)

CH3 | + H3 C  C  HC  CH2  HF  | CH3

Z

P

CH3 | CH3 – C – CH3

Sol.

X=

CH3

Y=

or | H3C – CH | CH2 | CH2

ter-butyl benzene

CH3

Z = No reaction

4-sec. butyltoluene

C–––CH–CH3 P = (1, 1, 2-Trimethylpropyl) benzene CH3 CH3

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CHEMISTRY Ex.6

Outline a synthesis of biphenyl from benzene. Br

Sol.

Cu 

Br2 / Fe

 

   

(Ullmann reaction)

13.

ARYL HALIDES (A) Preparation of Aryl Halides

1.

Halogenation :

Lewis acid

 + X2 

+ HX

X2 = Cl2, Br2 Lewis acid = FeCl3, AlCl3, ZnCl2, Zn etc. 2.

Decarboxylation :

NaOH / CaO

 

3.

+ CO2

From Phenol :

PCl

5  



+ POCl3 + HCl

(B) Mechanism of bimolecular nucleophilic substitution (ArSN2)

Step I

+ Nu   RDS

 X  ( fast )

   Step II

Intermediate ion is stabilized by resonance.



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

G withdraws electrons : stabilizes carbanion, activates (– (CH3)3, –NO2, –CN, –SO3H, –COOH, –CHO, –COR, –X)

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CHEMISTRY 

A group that releases electrons tends to intensify the negative charge, destabilizes the carbanion, and thus slows 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 aromatic nucleophilic 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. 182°C, which turns pink on exposure to air and light. It is moderately soluble in cold water, but is readily soluble in concentrated sulphuric acid (Liebermann reaction), when phenol is dissolved in concentrated sulphuric acid and a few drops of aqueous sodium nitrite added, a red colour is obtained on dilution and turns blue 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)

(4)

(Major is oxidative cleavage of ring)



(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. = 211°C (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)

(13)

(oxidation)

C6H5 N   C:

(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 : + HClO4 

(b)

+ 2 AgClO4 

1.

(a)

Ans.

(a)

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

(b)

+ AgCl

AlCl3 + CH3 – CH2 – CH2 – Cl  

2.

Ans. Sol.

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

3.

(a)

+

(b)

+

H 

(c)

Ans.

(a)

(b)

Sol.

Since the fridel craft acylation is faster than alkylation.

(c)

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CHEMISTRY

+ H2SO4 

4.

Ans.

Sol.

2H2SO4

+

+

5.

Sol.

+

6.

Ans.

Sol.

7.

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

Br / Fe 2  

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CHEMISTRY

Sol.

Br / Fe 2  

8.

Ans.

Sol.

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

9.

(a)

Ans.

(a)

Sol.

(a)

Br / H O 2 2 

(b)

(b)

ion increases the reactivity towards electrophile due to increases the electron density in ring.

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CHEMISTRY

KMnO / H   4  

(b)

KMnO / H   4  

10.

(a)

Ans.

(a)

Sol.

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

11.

(a)

1. CHCl3     2. aq.KOH

(b)

1. CCl4     2. aq.KOH

(c)

Br / H O 2 2 (excess)

(b)

Ans./Sol.

(a)

13.

(b)

(c)

+ 3HBr

Na2Cr2O7     H2SO4

Ans.

Sol.

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

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