A) \[B{{r}_{2}}/KOH\]
B) \[HCl{{O}_{4}}\]
C) \[HN{{O}_{2}}\]
D) \[N{{H}_{3}}\]
Correct Answer: C
Solution :
\[\underset{{{\text{1}}^{\text{o}}}}{\mathop{C{{H}_{3}}C{{H}_{2}}-N{{H}_{2}}}}\,+HN{{O}_{2}}\to \]\[\underset{\text{Alcohol}}{\mathop{C{{H}_{3}}C{{H}_{2}}-OH\,}}\,+{{N}_{2}}+{{H}_{2}}O\]\[\underset{{{2}^{o}}}{\mathop{{{(C{{H}_{3}}C{{H}_{2}})}_{2}}NH+HN{{O}_{2}}}}\,\to \]\[\underset{\text{Nitroso}\,\,\text{amine}}{\mathop{{{(C{{H}_{3}}C{{H}_{2}})}_{2}}N-N=O}}\,+{{H}_{2}}O\]
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