A) \[C{{H}_{3}}C{{H}_{2}}CH=NNHCON{{H}_{2}}\]
B) \[C{{H}_{3}}-CH=\underset{C{{H}_{3}}}{\mathop{\underset{|\,\,\,\,\,\,\,\,\,\,}{\mathop{NNH}}\,}}\,CON{{H}_{2}}\]
C) \[C{{H}_{3}}CH=\underset{C{{H}_{3}}}{\mathop{\underset{|\,\,\,\,\,\,\,\,\,\,}{\mathop{NCO}}\,}}\,NHN{{H}_{2}}\]
D) \[C{{H}_{3}}C{{H}_{2}}CH-NCONHN{{H}_{2}}\]
Correct Answer: A
Solution :
\[\underset{\text{(A)}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}OH}}\,\xrightarrow{[O]}\underset{\text{(B)}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CHO}}\,\] \[C{{H}_{3}}-C{{H}_{2}}-\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,=O+{{H}_{2}}NNHCON{{H}_{2}}\xrightarrow{HCl}\] \[C{{H}_{3}}C{{H}_{2}}CH=N-NHCON{{H}_{2}}\]You need to login to perform this action.
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