A) \[C{{H}_{3}}CN\xrightarrow{LiAl{{H}_{4}}}\]
B) \[C{{H}_{3}}NC\xrightarrow{LiAl{{H}_{4}}}\]
C) \[C{{H}_{3}}CON{{H}_{2}}\xrightarrow{LiAl{{H}_{4}}}\]
D) \[C{{H}_{3}}CON{{H}_{2}}\xrightarrow{B{{r}_{2}},\,\,NaOH}\]
Correct Answer: B
Solution :
\[C{{H}_{3}}NC+4H\xrightarrow{LiAl{{H}_{4}}}\underset{{{2}^{o}}\,\,a\operatorname{mine}}{\mathop{C{{H}_{3}}NHC{{H}_{3}}}}\,\]Isocyanides are reduced to \[{{2}^{o}}\] amines. \[C{{H}_{3}}NC+4H\xrightarrow{LiAl{{H}_{4}}}C{{H}_{3}}C{{H}_{4}}N{{H}_{2}}\] (\[{{1}^{o}}\]amine) \[C{{H}_{3}}CON{{H}_{2}}\xrightarrow{LiAl{{H}_{4}}}C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\] (\[{{1}^{o}}\]amine) \[C{{H}_{3}}CON{{H}_{3}}+B{{r}_{2}}+4KOH\xrightarrow{{}}RN{{H}_{2}}\] Hofmann bromamide reaction\[+{{K}_{2}}C{{O}_{3}}+2KBr+{{H}_{2}}O\]
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