See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question
Question
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💡 Solution & Explanation
Step 1 - Identify key constraints: • Molecular formula C4H8O3 (degree of unsaturation = 1, consistent with one C=O, i.e., a carboxylic acid) • Evolves CO2 with NaHCO3 → compound is a carboxylic acid (acidic enough to react with NaHCO3) • Optically active → has at least one chiral centre • Reacts with LiAlH4 to give an ACHIRAL compound → reduction removes or destroys the chiral centre Step 2 - Check molecular formula for each option: (a) CH3CH2CH(OH)COOH = C4H8O3 ✓ (b) CH3CH(Me)COOH = (CH3)2CHCOOH = C4H8O2 ✗ (only 2 oxygens, no extra OH) (c) CH3CH(CH2OH)COOH = C4H8O3 ✓ (d) CH3CH(OH)CH2COOH = C4H8O3 ✓ Option (b) is eliminated by molecular formula. Step 3 - Check optical activity (chiral centre): (a) CH3CH2CH(OH)COOH: chiral centre at C3 (four different groups: H, OH, CH2CH3, COOH) ✓ (c) CH3CH(CH2OH)COOH: chiral centre at C2 (four different groups: H, CH3, CH2OH, COOH) ✓ (d) CH3CH(OH)CH2COOH: chiral centre at C3 (four different groups: H, OH, CH3, CH2COOH) ✓ Step 4 - Apply LiAlH4 reduction gives achiral product: LiAlH4 reduces -COOH to -CH2OH and -OH is unaffected. (a) CH3CH2CH(OH)COOH → LiAlH4 → CH3CH2CH(OH)CH2OH The product still has a chiral centre at C3 (groups: H, OH, CH2CH3, CH2OH - all four different) → CHIRAL. Eliminated. (c) CH3CH(CH2OH)COOH → LiAlH4 → CH3CH(CH2OH)CH2OH The product is CH3CH(CH2OH)CH2OH. The central carbon has: H, CH3, CH2OH, CH2OH → two identical CH2OH groups → NO chiral centre → ACHIRAL ✓ (d) CH3CH(OH)CH2COOH → LiAlH4 → CH3CH(OH)CH2CH2OH The product has chiral centre at C2 (groups: H, OH, CH3, CH2CH2OH - all different) → CHIRAL. Eliminated. Step 5 - Conclusion: Only option (c), CH3CH(CH2OH)COOH (2-(hydroxymethyl)propanoic acid), satisfies all conditions: correct molecular formula C4H8O3, carboxylic acid (reacts with NaHCO3 to give CO2), optically active (chiral centre), and reduction with LiAlH4 gives 2-methyl-1,3-propanediol which is achiral (has a plane of symmetry, meso-like, with two identical CH2OH groups on the central carbon). Therefore, the correct answer is C.