JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Chapter_pending

The substrate is 4-hydroxybutanal, carrying both a free hydroxyl group and an aldehyde group. Methylmagnesium bromide is destroyed by any acidic hydrogen, so the first equivalent is consumed in deprotonating the $-\mathrm{OH}$ group (liberating methane), and only a second equivalent then adds across the aldehyde carbonyl, which on acidic work-up gives the secondary alcohol. Two moles of $\mathrm{CH}{}_3\mathrm{MgBr}$ are therefore needed per mole of the hydroxy-aldehyde, so x = 2 and y = 1, and the ratio x/y is 2.