See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify functional group from tests: The compound gives a negative Tollens test, ruling out aldehydes (which would give a positive Tollens test). The compound gives a positive test with 2,4-dinitrophenylhydrazine (2,4-DNPH), which is positive for both aldehydes and ketones. Together, these two results confirm the compound is a KETONE. Step 2 - Eliminate aldehyde options: Option (c) is an aldehyde (H—C=O group), so it is eliminated because it would give a positive Tollens test. Step 3 - Check molecular formula: C6H12O. All remaining options (a), (b), and (d) are ketones with formula C6H12O. Step 4 - Check optical activity: The compound must be optically active, meaning it must have at least one chiral carbon (a carbon bonded to four different substituents). - Option (a): CH3—C(=O)—CH2—CH(CH3)—CH3. The carbon bearing CH3 has groups: CH3, H, CH2C(=O)CH3, CH3 — two CH3 groups are identical, so this carbon is NOT chiral. No chiral center → not optically active. - Option (b): CH3—C(=O)—CH(CH2CH3)—CH3. The carbon alpha to the carbonyl bears: H, CH3, CH2CH3, and C(=O)CH3. All four substituents are different → this IS a chiral center. Therefore, compound (b) is optically active. - Option (d): CH3—C(=O)—CH2—CH2—CH2—CH3. This is a straight-chain ketone (2-hexanone) with no chiral center → not optically active. Step 5 - Conclusion: Only option (b), which is methyl sec-butyl ketone (3-methyl-2-pentanone or methyl (1-methylpropyl) ketone), satisfies all conditions: ketone (negative Tollens, positive 2,4-DNPH) and optically active (chiral center present). Therefore, the correct answer is B.