Ethylbenzene is treated with $Cl X_{2}$ in sunlight (1 mol, monochlorination at the side chain, major product), then with alcoholic $KOH$ , and the resulting hydrocarbon is subjected to $O X_{3}$ followed by $Zn$ / $H X_{2} O$ . The carbonyl products are

Answer: A

Radical chlorination occurs mainly at the benzylic secondary carbon:

C X_{6} H X_{5} - CHCl - CH X_{3}

. Alcoholic KOH eliminates HCl to give styrene,

C X_{6} H X_{5} - CH = CH X_{2}

. Reductive ozonolysis gives

C X_{6} H X_{5} CHO

and

HCHO