Ethylbenzene on reaction with ${\mathrm{Cl}}_2$ in sunlight gives ‘A’ which on reaction with alc. KOH gives ‘B’.Compounds ‘A’ and ‘B’ respectively are

 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.

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2{\mathrm{CH}}_3+\mathrm{C}{\mathrm{l}}_2\stackrel{\mathrm{sunlight}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}\left(\mathrm{Cl}\right)\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{alc}.\mathrm{KOH}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}={\mathrm{CH}}_2$.