an ester is boiled with KOH. The product is cooled and acidified with conc. HCl. A white crystalline acid separates . The ester is

Options

A. methyl acetate
B. ethyl acetate
C. ethyl formate
D. ethyl benzoate

Explanation

To solve the question regarding the ester that, when boiled with KOH and subsequently acidified with concentrated HCl, yields a white crystalline acid, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- When an ester is boiled with KOH, it undergoes saponification, which results in the formation of a carboxylate salt and an alcohol.

2. **Identifying the Ester**:
- The question provides four options: methyl acetate, ethyl acetate, ethyl formate, and ethyl benzoate. We need to determine which of these esters will yield a white crystalline acid upon acidification.

3. **Analyzing the Options**:
- **Methyl Acetate (C2H3O2CH3)**: Upon hydrolysis, it produces acetic acid (which is a liquid).
- **Ethyl Acetate (C2H5O2C2H5)**: Upon hydrolysis, it also produces acetic acid (which is a liquid).
- **Ethyl Formate (C2H5O2CHO)**: Upon hydrolysis, it produces formic acid (which is a liquid).
- **Ethyl Benzoate (C6H5COOC2H5)**: Upon hydrolysis, it produces benzoic acid (which is a solid at room temperature).

4. **Performing the Hydrolysis**:
- When ethyl benzoate is boiled with KOH, it forms potassium benzoate and ethanol:
\[ \text{C}_6\text{H}_5\text{COOC}_2\text{H}_5 + \text{KOH} \rightarrow \text{C}_6\text{H}_5\text{COOK} + \text{C}_2\text{H}_5\text{OH} \]

5. **Acidifying the Product**:
- When potassium benzoate is acidified with concentrated HCl, it produces benzoic acid:
\[ \text{C}_6\text{H}_5\text{COOK} + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{COOH} + \text{KCl} \]
- Benzoic acid is a white crystalline solid.

6. **Conclusion**:
- The only ester that produces a white crystalline acid upon acidification is ethyl benzoate. Therefore, the answer is **ethyl benzoate**.

### Final Answer:
The ester is **ethyl benzoate**.