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STD 12 - 8.2 Carboxylic acids and their derivative — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 12 - 8.2 Carboxylic acids and their derivative4 Marks
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
$\underset{Ethanoic\,\,\,acid}{\mathop{C{{H}_{3}}COOH}}\,\,\xrightarrow[\Delta ]{N{{H}_{3}}}\,\underset{\begin{smallmatrix} 
 (A) \\ 
 Ethanamide 
\end{smallmatrix}}{\mathop{C{{H}_{3}}CON{{H}_{2}}}}\,$ $\xrightarrow{B{{r}_{2}}\,/\,NaOH}$

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

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