Question

$C_6H_5$ CO $NHCH_3$ can be converted into $C_6H_5CH_2NHCH_3$ by

• A. $H_2-Pd/C$
• B. Zn-Hg/HCl
• C. $NaBH_4$
• D. $LiAlH_4$

Answer 1

Answer: (D)

D) $LiAlH_4$

Answer 2

The reagent required must reduce the amide carbonyl (–CO–) in $C_6H_5CO–NHCH_3$ to a $CH_2$ group. Among the listed reagents, $LiAlH_4$ is a strong hydride donor that effectively reduces amides to amines.

$LiAlH_4$ reduces N-methylbenzamide ($C_6H_5CO–NHCH_3$) to N-methylbenzylamine ($C_6H_5CH_2–NHCH_3$), whereas the other reagents are either too mild ($NaBH_4$), require catalytic hydrogenation conditions ($H_2–Pd/C$), or are used for Clemmensen reductions (Zn–Hg/HCl) which are not effective for amide groups.