A) \[{{P}_{2}}{{O}_{5}}\]
B) \[NaOBr\]
C) \[LiAl{{H}_{4}}/{{H}_{2}}O\]
D) \[Na(Hg)/{{C}_{2}}{{H}_{5}}OH\]
Correct Answer: B
Solution :
\[\text{NaOBr}\]converts\[-\text{CON}{{\text{H}}_{\text{2}}}\] group into \[-\text{N}{{\text{H}}_{\text{2}}}\]group. \[\underset{\text{ethanamide}}{\mathop{C{{H}_{3}}CON{{H}_{2}}}}\,\xrightarrow[\begin{smallmatrix} -N{{a}_{2}}C{{O}_{3}} \\ \,\,-NaBr \\ \,\,-{{H}_{2}}O \end{smallmatrix}]{NaOBr}\underset{\text{methanamine}}{\mathop{C{{H}_{3}}N{{H}_{2}}}}\,\]
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