A) \[C{{H}_{3}}COOH\]
B) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}COOH\]
C) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,HCOOH\]
D) \[C{{H}_{3}}C{{H}_{2}}COOH\]
Correct Answer: D
Solution :
The organic compound 'A' is propanoic acid \[(C{{H}_{3}}C{{H}_{2}}COOH).\] \[\underset{\begin{smallmatrix} propanoic\,acid \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(A) \end{smallmatrix}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-COOH}}\,\xrightarrow[{}]{N{{H}_{3}}}\] \[\underset{\begin{smallmatrix} ammonium\,propanoate \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(B) \end{smallmatrix}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CO\overline{O}\overset{+}{\mathop{N}}\,{{H}_{4}}}}\,\] \[\xrightarrow[-{{H}_{2}}O]{\Delta }C{{H}_{3}}\underset{\begin{smallmatrix} propanamide \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,(C) \end{smallmatrix}}{\mathop{-C{{H}_{2}}-CON{{H}_{2}}}}\,\] \[\underset{ethyla\min e}{\mathop{C{{H}_{3}}-C{{H}_{2}}-N{{H}_{2}}}}\,\xleftarrow[\begin{smallmatrix} Hoffmann \\ bromamide \\ reaction \end{smallmatrix}]{KOH+B{{r}_{2}}}\]
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