A) \[C{{H}_{3}}Cl\]
B) \[C{{H}_{3}}CN\]
C) \[C{{H}_{3}}CC{{1}_{2}}N{{H}_{2}}\]
D) \[CH{{C}_{12}}CON{{H}_{2}}\]
Correct Answer: C
Solution :
\[C{{H}_{3}}CON{{H}_{2}}+PC{{l}_{5}}\xrightarrow{{}}C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}}\,-N{{H}_{2}}\]
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