JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

The addition of $\mathrm{HCN}$ to the aldehyde needs the nucleophilic cyanide ion, which a base supplies, so $X$ is $\mathrm{NaOH}$; it generates $\mathrm{C}{\mathrm{N}}^{-}$ that attacks the carbonyl carbon to give the cyanohydrin. When the nitrile of that cyanohydrin is reduced by $\mathrm{LiAl}{\mathrm{H}}_4$, the group $-\mathrm{C}\equiv \mathrm{N}$ is converted to $-\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$, so the nitrogen ends up on a new carbon next to the one carrying the hydroxyl group. The product $Y$ is therefore a $1,2$-amino alcohol, with the $-\mathrm{OH}$ and $-\mathrm{N}{\mathrm{H}}_2$ on adjacent carbons rather than on the same carbon.