See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1: Identify all functional groups in the molecule. From the structure, the compound contains: 1. Carboxylic acid (-COOH) 2. Ketone (C=O, the acetyl group on the left) 3. Aldehyde (-CHO, on the right side) 4. Ester (-COOMe, methyl ester at the bottom) Step 2: Determine what LAH (LiAlH4) reduces. LAH is a powerful reducing agent that reduces ALL carbonyl-containing functional groups: - Carboxylic acids → primary alcohols ✓ - Ketones → secondary alcohols ✓ - Aldehydes → primary alcohols ✓ - Esters → primary alcohols ✓ So LAH reduces all 4 functional groups. LAH count = 4. Step 3: Determine what SBH (NaBH4) reduces. NaBH4 is a mild reducing agent that selectively reduces only aldehydes and ketones (and sometimes imines). It does NOT reduce: - Carboxylic acids ✗ (not reduced by NaBH4) - Esters ✗ (not reduced by NaBH4 under normal conditions) - Aldehydes ✓ (reduced) - Ketones ✓ (reduced) So NaBH4 reduces only 2 functional groups (aldehyde + ketone). SBH count = 2. Step 4: Match with answer choices. LAH = 4, SBH = 2 → corresponds to option (d) 4, 2. Step 5: Why other options fail. (a) 4, 4: Incorrect because NaBH4 does not reduce carboxylic acids or esters. (b) 4, 3: Incorrect; NaBH4 reduces only 2 (aldehyde and ketone), not 3. (c) 3, 4: Incorrect; LAH reduces all 4 groups, not 3, and NaBH4 does not reduce 4. Therefore, the correct answer is D.