See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify the reactants** - Acid: enantiomerically pure (one enantiomer only) - Alcohol: racemic mixture (50% R-enantiomer + 50% S-enantiomer) with one chiral carbon **Step 2: Analyze ester formation** Esterification is a simple acid-catalyzed condensation: $$\text{R-COOH} + \text{R'-OH} \rightarrow \text{R-COO-R'} + \text{H}_2\text{O}$$ The reaction does **not break the C-O bond** of the alcohol (Fischer esterification proceeds via acyl-oxygen cleavage). Therefore, the stereochemistry at the chiral carbon of the alcohol **remains unchanged**. **Step 3: Determine ester composition** - Pure acid enantiomer reacts with R-alcohol enantiomer → Ester enantiomer 1 - Same pure acid enantiomer reacts with S-alcohol enantiomer → Ester enantiomer 2 Since the alcohol was racemic (1:1 ratio), both ester enantiomers form in **equal proportions (1:1)**. **Step 4: Conclusion** The product is a **1:1 mixture of two enantiomeric esters**, which is the definition of a **racemic mixture**. **Answer: (D) racemic mixture** ✓