A) Ethyl cyanide
B) Ethyl isocyanide
C) Formic acid
D) An amide
Correct Answer: B
Solution :
\[CHC{{l}_{3}}+{{C}_{2}}{{H}_{5}}N{{H}_{2}}+3KOH\to \]\[\underset{\text{Ethyl}\,\,\text{isocyanide}}{\mathop{{{C}_{2}}{{H}_{5}}N\vec{=}C}}\,+3KCl+3{{H}_{2}}O\]
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