A) \[\underset{\begin{smallmatrix} | \\ N{{H}_{2}} \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,-CHO\]
B) \[C{{H}_{3}}CH=NOH\]
C) \[HCONH-C{{H}_{3}}\]
D) All of these
Correct Answer: C
Solution :
\[\underset{acetic\,acid}{\mathop{C{{H}_{3}}COOH}}\,\xrightarrow{N{{H}_{3}}}C{{H}_{3}}COON{{H}_{4}}\xrightarrow[-{{H}_{2}}O]{\Delta }\] \[C{{H}_{3}}CO\] \[\underset{acetamide}{\mathop{N{{H}_{2}}}}\,\] The isomers of\[C{{H}_{3}}CON{{H}_{2}}\]are \[\underset{(I)}{\mathop{N{{H}_{2}}C{{H}_{2}}CHO}}\,,\,\,C{{H}_{3}}-\underset{(II)}{\mathop{CH}}\,=NOH\] \[H-\underset{(III)}{\mathop{CONH}}\,-C{{H}_{3}}\]
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