Which of the following is Mendius reduction?

The correct answer is (4) \(\mathrm{R}-\mathrm{CN} \rightarrow \mathrm{Na} / \mathrm{CH}_3 \mathrm{OH}\) or \(\mathrm{LiAlH}_4 \rightarrow \mathrm{R}-\mathrm{CH}_2 \mathrm{NH}_2\).
This is the reaction for Mendius reduction, which involves the reduction of nitriles ( R CN ) to primary amines ( \(\mathrm{R}-\mathrm{CH}_2 \mathrm{NH}_2\) ). The reduction can be carried out using reducing agents like sodium in methanol \(\left(\mathrm{Na} / \mathrm{CH}_3 \mathrm{OH}\right)\) or lithium aluminum hydride \(\left(\mathrm{LiAlH}_4\right)\).
Explanation of the other options:
- (A) \(\mathrm{R}-\mathrm{CONH}_2+\mathrm{Br}_2+\mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{R}-\mathrm{NH}_2\) : This is not Mendius reduction. This is a reaction of an amide \(\left(\mathrm{R}-\mathrm{CONH}_2\right)\) with bromine and alkali, which leads to the formation of a primary amine via Hofmann's degradation.
- (B) \(\mathrm{R}-\mathrm{NH}_2+\mathrm{RX}\) (excess) \(\rightarrow \mathrm{R}-\mathrm{NH}-\mathrm{R}\) : This is a N -alkylation reaction, where an amine \(\left(\mathrm{R}-\mathrm{NH}_2\right)\) reacts with an alkyl halide \((\mathrm{RX})\) to form a secondary amine ( \(\mathrm{R}-\mathrm{NH}-\) R ), not related to Mendius reduction.
- (C) \(\mathrm{R}-\mathrm{NH}_2+\mathrm{X}\) (Moist Alkene) \(\rightarrow\) Alkene \(+\mathrm{R}-\mathrm{NH}_2\) : This is not related to Mendius reduction. This could describe an amine reacting with a halogen, but it's not a standard reduction reaction like Mendius.