JEE Main 2025 Jan 24 Shift 1, Chemistry Q20: Preparation of Amines

Trace the three steps applied to propyne, $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}\equiv \mathrm{CH}$:

Step (i) Hg${}^{2+}$/H${}_2$SO${}_4$: hydration of the terminal alkyne by Markovnikov's rule. Water adds with OH on the more substituted carbon, giving an enol that tautomerises to the more stable ketone, propan-2-one (acetone): $\mathrm{C}{\mathrm{H}}_3-\mathrm{CO}-\mathrm{C}{\mathrm{H}}_3$.

Step (ii) HCN: nucleophilic addition of cyanide to the ketone carbonyl gives the cyanohydrin: $\mathrm{C}{\mathrm{H}}_3-\underset{\underset{\mathrm{C}{\mathrm{H}}_3}{|}}{\stackrel{\stackrel{\mathrm{OH}}{|}}{\mathrm{C}}}-\mathrm{CN}$.

Step (iii) H${}_2$/Ni: catalytic hydrogenation reduces the nitrile to a primary amine ($-$CN $\to$ $-$CH${}_2$NH${}_2$), leaving the tertiary OH untouched.

Final product: $\mathrm{C}{\mathrm{H}}_3-\underset{\underset{\mathrm{C}{\mathrm{H}}_3}{|}}{\stackrel{\stackrel{\mathrm{OH}}{|}}{\mathrm{C}}}-\mathrm{C}{\mathrm{H}}_2-\mathrm{N}{\mathrm{H}}_2$, i.e. 1-amino-2-methylpropan-2-ol.

$\therefore$ Option (2).