Q68 / 180 · NEET 2026 (Re-NEET) · Chemistry
Identify the reactions which give aniline as the major product.

Choose the correct answer from the options given below.

Choose the correct answer from the options given below.
- AC and D only
- BA and B only
- Correct answerCB and D only
- DA and C only
Solution
Let us analyze each reaction to determine the major product:
Reaction A: Benzonitrile ($C_6H_5CN$) is reduced by $LiAlH_4$, a strong reducing agent, to form benzylamine.
$C_6H_5CN \xrightarrow{LiAlH_4} C_6H_5CH_2NH_2$
Reaction B: Benzamide ($C_6H_5CONH_2$) undergoes Hoffmann bromamide degradation in the presence of $KOH$ and $Br_2$ to form aniline.
$C_6H_5CONH_2 \xrightarrow{KOH, Br_2} C_6H_5NH_2$
Reaction C: Nitrobenzene ($C_6H_5NO_2$) is not reduced to aniline by $NaBH_4$. $NaBH_4$ is a mild reducing agent and generally does not reduce the $-NO_2$ group.
Reaction D: Acetanilide ($C_6H_5NHCOCH_3$) undergoes acid-catalyzed hydrolysis upon heating to yield aniline and acetic acid.
$C_6H_5NHCOCH_3 \xrightarrow{HCl, H_2O, \Delta} C_6H_5NH_2 + CH_3COOH$
Thus, reactions B and D give aniline as the major product.
Answer: B and D only
Reaction A: Benzonitrile ($C_6H_5CN$) is reduced by $LiAlH_4$, a strong reducing agent, to form benzylamine.
$C_6H_5CN \xrightarrow{LiAlH_4} C_6H_5CH_2NH_2$
Reaction B: Benzamide ($C_6H_5CONH_2$) undergoes Hoffmann bromamide degradation in the presence of $KOH$ and $Br_2$ to form aniline.
$C_6H_5CONH_2 \xrightarrow{KOH, Br_2} C_6H_5NH_2$
Reaction C: Nitrobenzene ($C_6H_5NO_2$) is not reduced to aniline by $NaBH_4$. $NaBH_4$ is a mild reducing agent and generally does not reduce the $-NO_2$ group.
Reaction D: Acetanilide ($C_6H_5NHCOCH_3$) undergoes acid-catalyzed hydrolysis upon heating to yield aniline and acetic acid.
$C_6H_5NHCOCH_3 \xrightarrow{HCl, H_2O, \Delta} C_6H_5NH_2 + CH_3COOH$
Thus, reactions B and D give aniline as the major product.
Answer: B and D only
Hence, the correct answer is option: 'C'.