A) \[C{{H}_{3}}C{{H}_{2}}CN\]
B) \[NCC{{H}_{2}}-C{{H}_{2}}CN\]
C) \[BrC{{H}_{2}}-C{{H}_{2}}CN\]
D) \[BrCH=CHCN\]
Correct Answer: B
Solution :
\[{{C}_{2}}{{H}_{5}}I\xrightarrow[-HI]{Alc.\,\,KOH}C{{H}_{2}}=C{{H}_{2}}\xrightarrow{B{{r}_{2}}}\] \[BrC{{H}_{2}}-C{{H}_{2}}Br\xrightarrow{KCN}NCC{{H}_{2}}-C{{H}_{2}}CN\]
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