See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: When an alcohol reacts with a carboxylic acid under acidic conditions (Fischer esterification), an ester is formed. When a chiral alcohol reacts with a chiral acid, the resulting esters are diastereomers if the two reactants have different (non-mirror-image) configurations at their stereocenters. Step 1: Identify the reactants. - A racemic mixture of 2-butanol is used: (R)-2-butanol and (S)-2-butanol. - (R,R)-tartaric acid is the chiral acid (a single enantiomer with two stereocenters, both R). Step 2: Determine the esterification products. - (R,R)-tartaric acid has two carboxylic acid groups. In acidic medium, esterification occurs at one or both -COOH groups with 2-butanol. - When (R)-2-butanol reacts with (R,R)-tartaric acid, the product is an ester with configuration (R) at the butanol-derived center and (R,R) at the tartaric acid centers → one stereoisomeric ester. - When (S)-2-butanol reacts with (R,R)-tartaric acid, the product is an ester with configuration (S) at the butanol-derived center and (R,R) at the tartaric acid centers → another stereoisomeric ester. Step 3: Compare the two ester products. - The two products differ in configuration at the 2-butanol-derived stereocenter: one is (R) and the other is (S), while both share the same (R,R) tartaric acid backbone. - These two products are NOT mirror images of each other (they would be mirror images only if the tartaric acid portion were also inverted). Since the tartaric acid portion is identical (R,R) in both, the two products are non-superimposable, non-mirror-image stereoisomers — i.e., diastereomers. Step 4: Why other options fail. - (a) Racemic: A racemic mixture consists of equal amounts of two enantiomers. The two products here are diastereomers, not enantiomers, so the mixture is not racemic. - (c) Meso: A meso compound has stereocenters but is achiral due to an internal plane of symmetry. The products here do not have the required internal symmetry to be meso. - (d) Pure enantiomer: A pure enantiomer would result if only one stereoisomer were formed. Since both (R)- and (S)-2-butanol are present and both react with the same (R,R)-tartaric acid, two distinct diastereomeric esters are formed, not a single pure enantiomer. Therefore, the correct answer is B.