See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: When alpha-hydroxy acids are heated, they undergo intermolecular esterification (condensation) to form cyclic diesters called lactides (cyclic dilactones), rather than simple dehydration or decarboxylation products. Step 1: Identify the starting material. The compound is lactic acid (2-hydroxypropanoic acid): CH3CH(OH)COOH. It has both a hydroxyl group (-OH) and a carboxylic acid group (-COOH) on the alpha carbon. Step 2: Determine the reaction pathway upon heating. Simple alpha-hydroxy acids when heated do NOT simply dehydrate to acrylic acid (that would require specific elimination conditions) nor do they decarboxylate to give pyruvic acid or reduce to ethanol. Instead, two molecules of lactic acid undergo mutual esterification: the -OH of one molecule reacts with the -COOH of another, and vice versa, forming a six-membered cyclic diester ring (a 1,4-dioxane-2,5-dione skeleton), releasing two molecules of water. This product is known as lactide (3,6-dimethyl-1,4-dioxane-2,5-dione). Step 3: Confirm the structure of option (a). Option (a) shows a six-membered ring with two ester linkages (-C(=O)-O-) and two CH(CH3) groups alternating in the ring, which is exactly the lactide structure formed from two lactic acid molecules. Step 4: Eliminate other options. - Option (b) CH2=CHCO2H (acrylic acid): This would result from dehydration (elimination of water) across the alpha carbon, but simple heating of lactic acid does not favor this pathway preferentially. - Option (c) CH3C(=O)CO2H (pyruvic acid): This would require oxidation, not just simple heating. - Option (d) CH3CH2OH (ethanol): This would require both decarboxylation and reduction, not feasible under simple heating. Therefore, the correct answer is A.