Q.Complete the following reactions -

Reduction of Nitriles to Amines

Think of a nitrile as a molecule with a carbon atom triple-bonded to a nitrogen atom — that's the $-\mathrm{C}\equiv\mathrm{N}$ group. That triple bond is packed with electrons and is quite polarised: the nitrogen is more electronegative, so it pulls electron density toward itself, leaving the carbon slightly positive. This makes the carbon a good target for a nucleophilic attack by a hydride ion ($\mathrm{H}^-$).

When you treat a nitrile with a strong reducing agent like lithium aluminium hydride ($\mathrm{LiAlH}_4$) or hydrogen gas with a metal catalyst ($\mathrm{H}_2$/Ni, Pd, or Pt), you are essentially adding hydrogen atoms across that triple bond. The reaction does not stop at an imine (a $\mathrm{C}=\mathrm{N}$ intermediate) because the conditions are strongly reducing — it pushes all the way to a saturated $\mathrm{C}-\mathrm{N}$ single bond.

The key structural change: the nitrile carbon becomes a methylene ($-\mathrm{CH}_2-$) group, and the nitrogen becomes an amino ($-\mathrm{NH}_2$) group. So a nitrile $\mathrm{R}-\mathrm{C}\equiv\mathrm{N}$ becomes a primary amine $\mathrm{R}-\mathrm{CH}_2-\mathrm{NH}_2$.

Notice that the carbon chain has grown by exactly one carbon atom — the nitrile carbon is now part of the alkyl chain. This is a powerful method to extend a carbon skeleton by one unit while introducing an amine functionality.

The mechanism in brief (for $\mathrm{LiAlH}_4$)

1. A hydride ion ($\mathrm{H}^-$) attacks the electrophilic nitrile carbon, forming an imine anion intermediate: $\mathrm{R-C}^-=\mathrm{N}^-$.
2. A second hydride adds to the imine carbon, giving a dianion: $\mathrm{R-CH}_2-\mathrm{N}^{2-}$.
3. Aqueous work-up (adding water or dilute acid) protonates the nitrogen, yielding the free amine $\mathrm{R-CH}_2-\mathrm{NH}_2$.

Catalytic hydrogenation …