See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Esterification with Optically Pure Acid and Racemic Alcohol **Step 1: Analyze the reactants** - Acid: Enantiomerically pure (one stereoisomer only) - Alcohol: Racemic mixture (equal parts of R and S enantiomers) **Step 2: Understand ester formation mechanism** Esterification is a nucleophilic acyl substitution that does **not** affect the stereochemistry of either the acid or the alcohol component. The chiral centers remain intact. **Step 3: Determine products formed** When the optically pure acid reacts with the racemic alcohol: - Pure acid (say, R-configuration) + (R)-alcohol → Ester 1 (diastereomer) - Pure acid (R-configuration) + (S)-alcohol → Ester 2 (diastereomer) Two **diastereomeric esters** form in equal amounts. **Step 4: Assess optical activity** - Diastereomers have different physical properties (including specific rotations) - They are present in **unequal optical rotations** - The mixture is not racemic (would require equal but opposite rotations) - This is an **optically active mixture** with net rotation **Step 5: Eliminate other options** - **(B)** Pure enantiomer: ✗ Two diastereomers form - **(C)** Meso compound: ✗ Mixture, not a single compound - **(D)** Racemic mixture: ✗ Diastereomers don't have equal opposite rotations **Answer: (A) Optically active mixture**