The reaction of an ester RCOOR' with an alcohol R'OH in the presence of an acid gives

Options

A. ROOCH
B. R'COOH
C. R"COOR
D. RCOOR"

Explanation

To solve the question regarding the reaction of an ester \( RCOOR' \) with an alcohol \( R'OH \) in the presence of an acid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
We have an ester \( RCOOR' \) and an alcohol \( R'OH \).

2. **Understand the Role of Acid**:
The presence of acid (like concentrated \( H_2SO_4 \)) acts as a catalyst in this reaction. It helps in protonating the ester, making it more electrophilic.

3. **Nucleophilic Attack**:
The alcohol \( R'OH \) acts as a nucleophile. The oxygen atom in the alcohol has a lone pair of electrons and can attack the carbonyl carbon of the ester \( RCOOR' \). This carbon is partially positive due to the electronegative oxygen atom in the carbonyl group.

4. **Formation of Tetrahedral Intermediate**:
Upon the nucleophilic attack, a tetrahedral intermediate is formed. This intermediate has the structure \( RCOO^-R'OH_2^+ \).

5. **Elimination of Alcohol**:
The tetrahedral intermediate will then collapse, leading to the elimination of one of the alkoxy groups (the \( R' \) from the ester). This results in the formation of a new ester.

6. **Final Product**:
The final product of this reaction will be a new ester \( RCOOR'' \), where \( R'' \) is the alkyl group from the alcohol that attacked the ester.

7. **Name of the Reaction**:
This reaction is specifically called **transesterification**, where one ester is converted into another ester by exchanging the alkoxy group.

### Final Answer:
The reaction of an ester \( RCOOR' \) with an alcohol \( R'OH \) in the presence of an acid gives a new ester \( RCOOR'' \) (where \( R'' \) is the alkyl group from the alcohol), and this process is known as transesterification.