Amides on treating with lithium aluminum hydride yield.......
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Primary amines
Lithium aluminium hydride (\( \ce{LiAlH4} \)) is a strong reducing agent. With an amide, it reduces the >C=O group to >\( \ce{CH2} \) and keeps the nitrogen, so an amine with the same number of carbon atoms is formed:
\( \displaystyle \ce{R-CONH2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{R-CH2-NH2} \)
An unsubstituted amide (\( \ce{R-CONH2} \)) gives a primary amine. (An N-substituted amide such as R–CONHR′ would give a secondary amine in the same way.)
Why the other options are incorrect
Option (1) Carbonyl compound — Incorrect. \( \ce{LiAlH4} \) does not stop at the carbonyl stage; the nitrogen stays in the product.
Option (3) Carboxylic acids — Incorrect. An acid is formed by hydrolysis of an amide, not by reduction.
Option (4) Nitriles — Incorrect. Nitriles are formed by dehydration of amides (e.g. with \( \ce{P4O10} \)); \( \ce{LiAlH4} \) in fact reduces nitriles to primary amines too.
NCERT (2026-27 edition):
Reduction of Amides
