See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The equilibrium constant for the dissociation of a cyanohydrin into a carbonyl compound and HCN reflects how stable the cyanohydrin is relative to its carbonyl precursor. A less stable cyanohydrin (more tendency to dissociate) has a larger K. Step 1: Identify the two cyanohydrins. - Reaction 1: Acetone cyanohydrin [(CH3)2C(OH)CN] dissociates into acetone + HCN with equilibrium constant K1. - Reaction 2: Acetaldehyde cyanohydrin [CH3CH(OH)CN] dissociates into acetaldehyde + HCN with equilibrium constant K2. Step 2: Assess the stability of each cyanohydrin. The stability of a cyanohydrin depends on steric and electronic factors. Acetone cyanohydrin has two methyl groups on the carbon bearing OH and CN. This introduces greater steric strain (steric crowding around the quaternary-like carbon) compared to acetaldehyde cyanohydrin, which has only one methyl group and one hydrogen. Step 3: Relate stability to the equilibrium constant. The equilibrium constant K for dissociation = [carbonyl compound][HCN] / [cyanohydrin]. - A less stable (more strained) cyanohydrin lies more to the right (products), giving a larger K. - Acetone cyanohydrin is less stable due to greater steric hindrance from the two methyl groups, so it has a greater tendency to dissociate back into acetone + HCN. - Acetaldehyde cyanohydrin is more stable (less steric crowding), so it dissociates less, giving a smaller K. Step 4: Therefore K1 (acetone cyanohydrin dissociation) > K2 (acetaldehyde cyanohydrin dissociation). Why other options fail: - (a) K1 = K2: Incorrect, the two carbonyl compounds differ in steric bulk, so their cyanohydrins have different stabilities. - (c) K2 > K1: Incorrect, acetaldehyde cyanohydrin is more stable and dissociates less. - (d) K1 = K2 = 1: Incorrect, there is no reason these equilibrium constants equal 1. Therefore, the correct answer is B.