The final products formed when acetylene reacts with $HCN & {CH}_{3} COOH$ separately, respectively are

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${CH}_{3} - CH {(CN)}_{2} & {CH}_{2} = CH ({OCOCH}_{3})$

${CH}_{2} - CH CN & {CH}_{2} - CH {({OCOCH}_{3})}_{2}$

${CH}_{3} - CH {(CN)}_{2} & {CH}_{3} - CH {({OCOCH}_{3})}_{2}$

${CH}_{2} = CHCN & {CH}_{2} = CH ({OCOCH}_{3})$

answer is C.

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Detailed Solution

Addition of hydrogen cyanide: Acetylene react with 1 mole of HCN in presence of barium cyanide or cuprous cyanide catalyst to form vinyl cyanide.

$HC \equiv CH + HCO \overset{{Cu}_{2} {(CN)}_{2}}{\to} H_{2} C = CHCN$

Addition of acetic acid: Acetylene combines with acetic acid in presence of mercuric sulphate catalyst. It first forms vinylacetate and then ethylidene acetate

$HC \equiv CH + {HOCCCH}_{2} \overset{{Hg}^{+ 2}}{\to} {CH}_{2} = {CHOCOCH}_{3}$ ${CH}_{2} = {CHOCOCH}_{3} + {HOCOCH}_{3} \overset{{Hg}^{+ 2}}{\to} {CH}_{3} - CH {({OCOCH}_{3})}_{2} \overset{∆}{\to} {CH}_{3} CO)_{2} O$