Ethanoic acid on heating with ammonia forms compound A which on t…

MCQ

Ethanoic acid on heating with ammonia forms compound $A$ which on treatment with bromine and sodium hydroxide gives compound $B$. Compound $B$ on treatment with $NaNO_2$ /dil. $HCl$ gives compound $C$. The compounds $A,\,B$ and $C$ respectively are

✓
ethanamide, methanamine, methanol
B
propanamide, ethanamine, ethanol
C
$N-$ ethylpropanamide, methaneisonitrile, methanamine
D
ethanamine, bromoethane, ethanediazonium chloride

✓

Answer

Correct option: A.

ethanamide, methanamine, methanol

a
$\underset{Ethanoic\,\,\,acid}{\mathop{C{{H}_{3}}COOH}}\,\,\xrightarrow[\Delta ]{N{{H}_{3}}}\,\underset{\begin{smallmatrix}
(A) \\
Ethanamide
\end{smallmatrix}}{\mathop{C{{H}_{3}}CON{{H}_{2}}}}\,$ $\xrightarrow{B{{r}_{2}}\,/\,NaOH}$

$\underset{\begin{smallmatrix}
\,\,\,\,\,\,\,\,\,\,\,(B) \\
Methana\min e
\end{smallmatrix}}{\mathop{C{{H}_{3}}N{{H}_{2}}}}\,\,\xrightarrow{NaN{{O}_{2}}/dil.HCl}\underset{\begin{smallmatrix}
\,\,\,\,\,\,\,\,(C) \\
Methanol
\end{smallmatrix}}{\mathop{C{{H}_{3}}OH}}\,$