See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify functional group clues: - Positive 2,4-dinitrophenylhydrazine (2,4-DNP) test: The compound contains a carbonyl group (C=O), i.e., it is either an aldehyde or a ketone. - Negative Tollens' reagent test: The compound is NOT an aldehyde (aldehydes give positive Tollens'). Therefore, the compound must be a ketone. This eliminates option (b), which is an aldehyde (H-C=O group), since it would give a positive Tollens' test. Step 2 - Check for optical activity: Optical activity requires at least one chiral (stereocentre) carbon — a carbon bonded to four different groups. Option (a): CH3-C(=O)-CH2-CH2-CH2-CH3 (hexan-2-one) - The carbonyl carbon is not chiral. Examining all carbons: C1(CH3), C2(C=O), C3(CH2), C4(CH2), C5(CH2), C6(CH3). None of these carbons have four different substituents. No chiral centre. NOT optically active. Eliminated. Option (c): CH3-C(=O)-CH(CH3)-CH2-CH3 (3-methylpentan-2-one) - C1: CH3, C2: C=O (ketone carbon), C3: CH with substituents — bonded to C2(COCH3), CH3, CH2CH3, and H. The four groups on C3 are: H, CH3, CH2CH3, and COCH3 — all four are different. C3 is a chiral centre. This compound IS optically active. It is a ketone (positive 2,4-DNP, negative Tollens'). This fits all criteria. Option (d): CH3-CH2-C(=O)-CH(CH3)-CH3 (4-methylpentan-3-one... checking: actually this is 4-methylpentan-3-one) - The chiral carbon candidate is C4: CH bonded to C3(C=O-CH2CH3), CH3, CH3, and H. The two methyl groups and the two substituents: H, CH3, CH3, and COC2H5. Wait — C4 has H, CH3, CH3, and C(=O)CH2CH3. Two of the groups are CH3 (identical), so C4 is NOT a chiral centre. NOT optically active. Eliminated. Step 3 - Conclusion: Only option (c) — CH3-C(=O)-CH(CH3)-CH2-CH3 (3-methylpentan-2-one) — satisfies all three conditions: it is a ketone (positive 2,4-DNP, negative Tollens') and it is optically active (chiral centre at C3). Therefore, the correct answer is C.