See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The equilibrium constant (K_eq) for HCN addition to an aldehyde (cyanohydrin formation) depends on the electrophilicity of the carbonyl carbon. A more electrophilic carbonyl carbon is attacked more readily by the nucleophile CN-, giving a larger K_eq. Electron-withdrawing groups increase electrophilicity of the carbonyl carbon, while electron-donating groups decrease it. Step 1: Identify the substituents and their electronic effects. - Compound I: 4-methoxybenzaldehyde — MeO is an electron-donating group (EDG) by resonance (lone pair donation into ring), which reduces the electrophilicity of the carbonyl carbon relative to benzaldehyde. - Compound II: Benzaldehyde — no substituent, serves as the reference. - Compound III: 4-(dimethylamino)benzaldehyde — NMe2 is a stronger electron-donating group than MeO (nitrogen donates more effectively than oxygen), which further reduces the electrophilicity of the carbonyl carbon. Step 2: Rank electrophilicity of carbonyl carbon. - II (no substituent) > I (moderate EDG: OMe) > III (strong EDG: NMe2) Step 3: Rank K_eq for HCN addition. - Higher electrophilicity → higher K_eq. - Therefore: K_eq(II) > K_eq(I) > K_eq(III) Step 4: Why other options fail. - (a) I > II > III: Wrong, because MeO reduces electrophilicity below benzaldehyde, so II should be higher than I. - (b) II > III > I: Wrong, NMe2 is a stronger EDG than OMe, so III should have lower K_eq than I, not higher. - (c) III > I > II: Wrong, EDG groups decrease K_eq, so III (strongest EDG) should have the lowest K_eq. Therefore, the correct answer is D.