A) \[Zn/HCl\]
B) \[Zn/N{{H}_{4}}Cl\]
C) \[Zn/NaOH\]
D) \[Sn/HCl\]
Correct Answer: B
Solution :
Nitromethane forms methyl hydroxylamine on reduction in neutral medium with \[Zn/N{{H}_{4}}Cl.\] \[\underset{\text{nitromethane}}{\mathop{C{{H}_{3}}N{{O}_{2}}}}\,+4[H]\xrightarrow[\Delta ]{Zn,N{{H}_{4}}Cl}\] \[\underset{\begin{smallmatrix} N-\text{methyl}\,\text{hydroxyl} \\ \,\,\,\,\,\,\,\,\text{amine} \end{smallmatrix}}{\mathop{C{{H}_{3}}NHOH}}\,+{{H}_{2}}O\]
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