A) acetamide
B) ammonium acetate
C) methyl cyanide
D) aniline
Correct Answer: C
Solution :
\[\underset{Acetic\text{ }acid}{\mathop{C{{H}_{3}}COOH}}\,\xrightarrow[\Delta ]{N{{H}_{3}}}\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,\,\,\,\,(A) \\ Ammonium\text{ }acetate \end{smallmatrix}}{\mathop{C{{H}_{3}}COON{{H}_{4}}}}\,\]\[\xrightarrow[\Delta ,-{{H}_{2}}O]{{{P}_{2}}{{O}_{5}}}\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,(B) \\ Acetamide \end{smallmatrix}}{\mathop{C{{H}_{3}}CON{{H}_{2}}}}\,\xrightarrow[{{P}_{2}}{{O}_{5}}/\Delta ]{-{{H}_{2}}O}\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,\,(C) \\ Methyl\text{ }cyanide \end{smallmatrix}}{\mathop{C{{H}_{3}}CN}}\,\]
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