Pure enantiomeric acid+optically active alcohol having chiral `C` atom `rarr`?
The product will be:

Options

A. An optically active mixtrue
B. A meso compound
C. A racemic mixtrue
D. A pure enantiomer

Explanation

To solve the question, we need to understand the reaction between a pure enantiomeric acid and an optically active alcohol. Let's break it down step by step:

### Step 1: Identify the Reactants
We have two reactants:
1. **Pure Enantiomeric Acid**: This is an acid that exists as one specific enantiomer (e.g., S-2-methylbutanoic acid).
2. **Optically Active Alcohol**: This alcohol also has a chiral carbon and exists as one specific enantiomer (e.g., S-butane-2-ol).

### Step 2: Understand the Reaction
When these two reactants are mixed, they undergo a condensation reaction (esterification) where the acid and alcohol react to form an ester and water. The general reaction can be represented as:
\[ \text{Acid} + \text{Alcohol} \rightarrow \text{Ester} + \text{Water} \]

### Step 3: Determine the Product
Since both reactants are optically active (one is a pure enantiomer and the other is also optically active), the product formed will also be optically active. The specific product in this case would be an ester formed from the combination of the two reactants.

### Step 4: Analyze the Optical Activity
The resulting product will not be a racemic mixture (which would require both enantiomers to be present in equal amounts), nor will it be a meso compound (which has an internal plane of symmetry). Instead, the product will retain optical activity because it is derived from one specific enantiomer of the acid and one specific enantiomer of the alcohol.

### Conclusion
Thus, the product of the reaction between a pure enantiomeric acid and an optically active alcohol will be an **optically active compound**.

### Final Answer
The product will be: **An optically active mixture**.

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