Ethylbenzene on reaction with Cl2 in sunlight gives ‘A’ which on reaction with alc. KOH gives ‘B’.Compounds ‘A’ and ‘B’ respectively are

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Ethylbenzene on reaction with ${Cl}_{2}$ in sunlight gives ‘A’ which on reaction with alc. KOH gives ‘B’.Compounds ‘A’ and ‘B’ respectively are

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$C_{6} H_{5} {CH}_{2} {CH}_{2} {ClandC}_{6} H_{5} {CH}_{2} {CH}_{2} OH$

$C_{6} H_{5} {CH}_{2} {CH}_{2} Cl and C_{6} H_{5} CH = {CH}_{2}$

$C_{6} H_{5} CH (Cl) {CH}_{3} and C_{6} H_{5} CH (OH) {CH}_{3}$

$C_{6} H_{5} CH (Cl) {CH}_{3} and C_{6} H_{5} CH = {CH}_{2}$

answer is D.

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Detailed Solution

Ethylbenzene when react with chlorine in presence of sunlight it gives 1-chloro-1-phenylethanewhich is compound (A) ,after this when further reaction of compound A with alc. KOH then dehydrohalogenation takes place and alkene is formed. In KOH OH ions act as base and it abstract β hydrogen from alkyl halide and convert it into alkene.

$C_{6} H_{5} {CH}_{2} {CH}_{3} + C l_{2} \overset{sunlight}{\to} C_{6} H_{5} CH (Cl) C H_{3} \overset{alc . KOH}{\to} C_{6} H_{5} CH = {CH}_{2}$ .