Aromatic Compound Theory E

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) NaOH(CaO)

1.

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. 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 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 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 equally good (b and c). Each 2p_zelectron, 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.

+ Br₂ H° = + 2 kcal (+ 8kJ)

+ Br₂ FeBr3 _{+ HBr H° = – 10.8 kcal (– 45 kJ)}

3.

Aromatic Electrophilic Substitution (ArS

_E

2) 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    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 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 >

> –Ar > –R

All the groups which are electron donating etc. are ortho–para 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 – CH} 3 .. .. || O (b) ₂ . . H N  and O || CH C H N ₃ . .   

Sol. (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 In OCCH₃ || O  , 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> – NHCOCH₃because of cross - conjugation in the amide, Ar – N = C – CH3 | | H :O:_..

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 FeBr₃catalyzes 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.

Br – : : : Br FeBr3 : : : + H H H H H H    + FeBr₄¯

Step 3 : Loss of a proton gives the products.

Br H H H H H bromobenzene + HBr + FeBr3 + H intermediate products reactants energy reaction coordinate Br¯FeBr4 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 the highest-energy point on the energy diagram. This step is strongly endothermic because it forms a non-aromatic carbocation. The second step is exothermic because non-aromaticity is regained and a molecule of HBr is evolved. The overall reaction is exothermic by 10.8 kcal/mol (45 kJ/mol.)

(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.

+ Cl_{2 }AlCl_3_ + 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.

+ 2 1

I

₂+ HNO₃  + 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+ 2 1

I

₂  ion iodonium 

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 is strong nitrating mixture when concentrated HNO₃and concentrated H₂SO₄are mixed to form the nitrating mixture, NO₂+_{(Nitronium ion) is produced as follows :}



O

H₃ +

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.

Sulphonation, is reversible and takes place in concentrated sulphuric acid. + SO₃ + H + + H + H Energy

progress of the reaction

++

SO H3

SO H3

SO H3

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

rate-determining step in the reverse direction

9.

Friedel Craft reaction

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

  _



 X

R , 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 > SnCl₄

Ex.2 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 Sol. (a) Ethylbenzene (b) Isopropylbenzene (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 of Lewis 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 R H H sigma complex

Step 3 : Loss of a proton. Complexation of the product. + -O || C R H H sigma complex :Cl – AlCl3 .. .. e.g. + (1) AlCl3 (2) H O2

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 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 Redhotirontube (2) By decarboxylation (of Benzoic acid) :

    NaOH ) CaO ( NaOH_    _{+ Na} 2CO3

(3) By catalytic reforming of n-Hexane : CH₃– (CH₂)₄– CH₃ 2 K 873 , Pt H     2 K 873 , Pt H 3    

(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 :

+ 2Na + CH₃Br dryether + 2NaBr

Bromobenzene Toluene

(3) From Grignard reagents :

+ CH₃Br  + MgBr₂

(4) By catalytic reforming of n-Heptane :

) H ( Pt , K 873 2     ) H 3 ( Pt , K 873 2     Methylcyclohexane Toluene

(C) Xylene :













CH3X,AlCl3 CH3 CH3 o-Xylene | +

12.

Reactions of Benzene :

Ex.3 Complete the following      KMnO4 X Sol. COOH

Ex.4 Gives the products of the following reactions :

(a) Cl + _{ }AlCl_3 X (b) CH 3– Cl + OCH3    AlCl3 Y (c) 3 3 3 CH Cl | | CH C HC C H | CH    + H C – CH – CH3 3     AlCl3 Z

Sol. X = phenylcyclohexane Y = OCH3 CH3 OCH3 CH3

O - methyl anisole p – methyl anisole

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

CH H C–C–CH3 3 H C–CH3 CH3 CH3 H C3

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

(a) excess + 3 3 3 3 CH | AlCl CH C C H | Cl     X

I

sobutylchloride (b) + l o tan bu 1 BF OH CH CH CH C H_{3 2 2 2 3}       _{ Y} (c) NO2 + Cl | AlCl CH CH C H₃   ₃ ₃  Z nitrobenzene (excess) (d) Benzene (excess) + 3 2 3 3 CH | HF CH HC C C H | 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 C–––CH–CH3 CH3 CH3 CH3 P = (1, 1, 2-Trimethylpropyl) benzene

Ex.6 Outline a synthesis of biphenyl from benzene. Sol. _Br2_/_Fe_ Br    Cu (Ullmann reaction)

13.

ARYL HALIDES

(A) Preparation of Aryl Halides

1. Halogenation :

+ X₂ Lewisacid + HX

X₂= Cl₂, Br₂

Lewis acid = FeCl₃, AlCl₃, ZnCl₂, Zn etc.

2. Decarboxylation :       NaOH/CaO _{+ CO} 2 3. From Phenol :    PCl5 + POCl₃+ HCl

(B) Mechanism of bimolecular nucleophilic substitution (ArS

_N

2)

+ Nu RDS I Step_     II Step ) 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 more

negative itself; this dispersal of the charge stabilizes the carbanion.

G withdraws electrons : stabilizes carbanion, activates

 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)

(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) (Major is oxidative cleavage of ring)

(4) _

(p-nitroso phenol)

(6) (7) (8) (9) (Fries Rearrangement) (10) (11) (polymer) (12) (13) (14) (15) (16)

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)

(8)

(9)

(10)

16.

ANILINE

(A) Prepartion of Aniline

(1)

(2)

(3)

(B) Chemical Reactions of Aniline

(1) (2)

(3)

(5) (6) (Diazotisation) (7) (8) (9) (10) (11) (12) (oxidation) (13) C₆H₅NC: (14)

17.

Benzenediazonium chloride

(1) (2) (3) (4) (5) (6) (7) (8) (9)

(10) (11) (12) (13) (14) (15) (16) (17)

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– CH₂– CH₂– Cl AlCl3

Ans.

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

3. (a) + _{ }H_

(b) +

(c)

Ans. (a) (b) (c)

4. + H₂SO_{4 } Ans. Sol. 2H₂SO₄ + + 5. Sol. + 6. Ans.

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

Sol.

8. _Br2_/Fe_

Ans.

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

9. (a) _Br_2/_H_2O_{ (b)}

Ans. (a) (b)

10. (a) _KMnO_4_/H__ (b) KMnO4/H

Ans. (a) (b)

Sol. (a & b) In aromatic hydrocarbons if at least one benzylic hydrogen is present then by oxidation benzoic acid is formed. 11. (a) _2.aq.KOH CHCl . 1 _3    (b) _2.aq.KOH CCl . 1 _4    (c) _(excess) O H / Br_{2 2}       Ans./Sol. (a) (b) (c) + 3HBr 13. 4 2 7 2 2 SO H O Cr Na_    Ans.