See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Fischer esterification is the reaction of a carboxylic acid with an alcohol in the presence of a strong acid catalyst (conc. H2SO4) to form an ester and water. Step 1 - Identify the reactive groups: Lactic acid (2-hydroxypropanoic acid) contains two functional groups: a carboxylic acid (-COOH) and a secondary hydroxyl group (-OH at the alpha carbon). Allyl alcohol (prop-2-en-1-ol, CH2=CH-CH2-OH) is a primary alcohol. Step 2 - Selectivity of esterification: Under Fischer esterification conditions (conc. H2SO4, heat), the reaction preferentially occurs between the carboxylic acid group of lactic acid and the hydroxyl group of allyl alcohol. The carboxylic acid is far more reactive toward esterification than the secondary alcohol under these conditions. The alpha-hydroxyl group of lactic acid is a secondary alcohol and does not preferentially undergo esterification with another alcohol under these mild acid-catalyzed conditions. Step 3 - Product formation: The -COOH of lactic acid reacts with the -OH of allyl alcohol to form an ester linkage (-COO-CH2-CH=CH2), releasing water. The alpha-OH of lactic acid remains intact. Step 4 - Product structure: The product is allyl lactate: CH3-CH(OH)-COO-CH2-CH=CH2, which matches option (b). Step 5 - Why other options fail: - Option (a): Shows etherification at the alpha-OH with free carboxylic acid. Etherification of an alcohol with another alcohol does not readily occur under Fischer esterification conditions; the carboxylic acid is far more reactive. - Option (c): Shows an ethyl ester, but the alcohol used is allyl alcohol (not ethanol), so an ethyl ester cannot form. - Option (d): Shows a rearranged ketone/addition product, which is not expected under simple Fischer esterification conditions. Therefore, the correct answer is B.