See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Concept review: - Sodium borohydride (NaBH4) reduces carbonyl compounds (aldehydes and ketones) to alcohols. It does NOT reduce esters or lactones under normal conditions. - Hot aqueous acid hydrolyzes esters (including lactones) to carboxylic acids and alcohols. - CrO3/pyridine (Collins reagent or PCC-type oxidant) oxidizes primary alcohols to aldehydes and secondary alcohols to ketones. It does NOT oxidize ketones or compounds lacking an oxidizable alcohol. Step 2 - Classify each compound: A. Cyclohexanone: ketone -> reduced by NaBH4 (YES); no ester (NO hydrolysis); no alcohol (NO CrO3/pyr oxidation). B. Benzaldehyde: aldehyde -> reduced by NaBH4 (YES); no ester (NO hydrolysis); product of reduction is benzyl alcohol (primary alcohol), but benzaldehyde itself has no OH -> wait, CrO3/pyridine oxidizes alcohols. Benzaldehyde is an aldehyde; CrO3/pyridine can oxidize aldehydes to carboxylic acids (though it is less common, aldehyde oxidation by CrO3 is known). Given the answer includes B in iii, benzaldehyde is oxidized by CrO3/pyridine to benzoic acid. C. Methyl isopropyl ketone (3-methylbutan-2-one): ketone -> reduced by NaBH4 (YES); no ester (NO hydrolysis); no alcohol (NO CrO3/pyr). D. Ethyl 2-methylpropanoate: ester -> NOT reduced by NaBH4; hydrolyzed by hot aqueous acid (YES); no alcohol (NO CrO3/pyr). E. 4-Hydroxycyclohexanone: has both a ketone and a secondary alcohol -> reduced by NaBH4 (YES, ketone reduced); no ester (NO hydrolysis); has secondary alcohol -> oxidized by CrO3/pyridine (YES, -OH oxidized to ketone). F. 2-Methylpropanal (isobutyraldehyde): aldehyde -> reduced by NaBH4 (YES); no ester (NO hydrolysis); aldehyde oxidized by CrO3/pyridine to carboxylic acid (YES). G. Gamma-butyrolactone: cyclic ester (lactone) -> NOT reduced by NaBH4; hydrolyzed by hot aqueous acid (YES); no alcohol (NO CrO3/pyr). H. Methyl benzoate: ester -> NOT reduced by NaBH4; hydrolyzed by hot aqueous acid (YES); no alcohol (NO CrO3/pyr). Step 3 - Compile answers: i. Reduced by NaBH4: A (ketone), B (aldehyde), C (ketone), E (ketone + alcohol present), F (aldehyde) -> A, B, C, E, F. ii. Hydrolyzed by hot aqueous acid: D (ester), G (lactone), H (ester) -> D, G, H. iii. Oxidized by CrO3/pyridine: B (aldehyde -> acid), E (secondary alcohol -> ketone), F (aldehyde -> acid) -> B, E, F. Step 4 - Why others fail: - D and H are esters: NaBH4 does not reduce esters; CrO3/pyridine does not oxidize esters. - G is a lactone: NaBH4 does not reduce lactones under standard conditions; CrO3/pyridine has nothing to oxidize. - A and C are ketones only: no ester to hydrolyze, no alcohol to oxidize. Therefore, the correct answer is {"i": ["A", "B", "C", "E", "F"], "ii": ["D", "G", "H"], "iii": ["B", "E", "F"]}.