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 :
\[NaOBr\]converts\[CON{{H}_{2}}\]group into\[N{{H}_{2}}\]group. \[\underset{ethanamide}{\mathop{C{{H}_{3}}CON{{H}_{2}}}}\,\xrightarrow[\begin{smallmatrix} -N{{a}_{2}}C{{O}_{3}} \\ -NaBr \\ -{{H}_{2}}O \end{smallmatrix}]{NaOBr}\underset{mehana\min e}{\mathop{C{{H}_{3}}N{{H}_{2}}}}\,\]
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