See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Acid-catalyzed racemization of an alpha-hydroxy ketone proceeds via an enediol intermediate. Step 1 - Identify the compound: The starting material is (R)- or (S)-3-hydroxy-2-pentanone (CH3-CO-CH(OH)-CH2CH3). It has one stereocentre at C3. Racemization means loss of stereochemical information at C3. Step 2 - Mechanism of acid-catalyzed racemization: In aqueous acid, alpha-hydroxy ketones can tautomerize to an enediol intermediate. Protonation of the carbonyl oxygen, followed by deprotonation at the alpha carbon (C3, the stereocentre) gives an enediol: CH3-C(OH)=C(OH)-CH2CH3. This enediol is planar/achiral at the former stereocentre because the carbon involved is now sp2 (part of the C=C double bond). Step 3 - Why the enediol causes racemization: The enediol intermediate (option c) is achiral and flat at both carbons of the double bond. Re-protonation from either face of the enediol gives the two enantiomers of the starting alpha-hydroxy ketone in equal amounts, resulting in racemization. Step 4 - Evaluate options: - (a) is only the enol (retains C=O, only one OH), not a full enediol; does not explain loss of stereocentre via a symmetric intermediate. - (b) retains the stereocentre with defined wedge/dash bonds; still chiral, so cannot explain racemization. - (c) is CH3-C(OH)=C(OH)-CH2CH3, the enediol where C2 and C3 are both sp2; no stereocentre exists, allowing re-protonation from either face to give racemic product. This is correct. - (d) has multiple stereocentres and additional OH groups (gem-diol or triol), which is not the relevant intermediate for simple acid-catalyzed racemization. Therefore, the correct answer is C.