See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: NaHSO3 (sodium bisulphite) undergoes nucleophilic addition to carbonyl compounds. However, this reaction is subject to steric limitations. The bisulphite ion (HSO3-) is a relatively bulky nucleophile. The reaction proceeds readily with: 1. All aldehydes (unhindered carbonyl with at least one H) 2. Methyl ketones (CH3-CO-R) where one group is methyl 3. Cyclic ketones with 8 or fewer carbons It does NOT react with ketones where both groups on the carbonyl carbon are larger than methyl (i.e., ketones with two alkyl groups each containing 2 or more carbons), due to excessive steric hindrance. Now evaluating each option: (a) CH3CHO (acetaldehyde): This is an aldehyde. Aldehydes readily react with NaHSO3. Reacts. (b) CH3-CO-Et (butanone / methyl ethyl ketone): This is a methyl ketone (one side is CH3). Methyl ketones react with NaHSO3. Reacts. (c) Et-CO-Et (diethyl ketone / pentan-3-one): Both groups attached to the carbonyl are ethyl groups (each has 2 carbons). This creates sufficient steric bulk on both sides that the bulky HSO3- nucleophile cannot attack the carbonyl carbon effectively. Does NOT react. (d) Ph-CH2-CO-CH3 (phenylacetone): One group is CH3 (methyl), so it qualifies as a methyl ketone. Despite the PhCH2 group on the other side, since one substituent is methyl, NaHSO3 can react with it. Reacts. Therefore, diethyl ketone (Et-CO-Et, option c) does not react with NaHSO3 because both alkyl groups are ethyl, making the carbonyl too sterically hindered for the bulky bisulphite nucleophile. Therefore, the correct answer is C.