A) \[C{{H}_{2}}C{{l}_{2}}\]
B) \[CHC{{l}_{3}}\]
C) \[C{{H}_{3}}Cl\]
D) \[CC{{l}_{4}}\]
Correct Answer: B
Solution :
\[C{{H}_{3}}N{{H}_{2}}+\underset{(X)}{\mathop{CHC{{l}_{3}}}}\,+3KOH\to 3KCl+C{{H}_{3}}NC+3{{H}_{2}}O\]
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