Answer:
Option D
Explanation:
$ CH_{3}CH_{2}CH_{2}OH$ $\xrightarrow[{160^{0}-180^{0}C}]{conc.H_{2}SO_{4}}$ $CH_{3}CH_{}=CH_{2}$ $\underrightarrow{Br_{2}} \underrightarrow{alc.KOH}$ $CH_{3}CH_{}Br-CH_{2}Br$
(Y)
$[CH_{3}C(Br)=CH_{2}+CH_{3}CH=CHBr] $ $\xrightarrow[{-HBr}]{NaNH_{2}}$ $CH_{3}C\equiv CH$
(A) (B) (Z)
Alcoholic KOH brings about dehydrobromination of Y and give a mixture of vinyl bromide (A and B) while
NaNH2 being a strong base than alc . KOH readily brings about dehydrobromination of less reactive vinyl bromide to give propyne $CH_{3}C\equiv CH$ i.e, (Z)