Consider the following reaction `CH_3-underset((A))(CH_2)-O-H+CH_3-underset((B))overset(O)overset(||)C-OHrarrCH_3-overset(O)overset(||)C-O-CH_2-CH_3+HOH` In the above reaction ester formation takes place by

Options

A. Breaking of O - H bond of A and O - H bond of B
B. Breaking of C - O bond of A and C - O bond of B
C. Breaking of O - H bond of A and C - O bond of B
D. Breaking of C - O bond of A and O - H bond of B

Answer

To solve the problem, we need to analyze the given reaction step by step. The reaction involves the formation of an ester from an alcohol and a carboxylic acid.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are:
- An alcohol (CH₃-CH₂-OH)
- A carboxylic acid (CH₃-C(=O)-OH)

2. **Understand Ester Formation**:
Ester formation occurs through a condensation reaction, where an alcohol reacts with a carboxylic acid to form an ester and water (H₂O).

3. **Mechanism of Reaction**:
- The reaction typically requires an acid catalyst (like H₂SO₄) to facilitate the process.
- The acid donates a proton (H⁺) to the alcohol, making it a better nucleophile.
- The alcohol then attacks the electrophilic carbon of the carboxylic acid.

4. **Bond Breaking**:
- In the alcohol, the O-H bond breaks, releasing a water molecule (H₂O).
- In the carboxylic acid, the C-O bond breaks, allowing the alcohol to bond with the carbonyl carbon.

5. **Formation of the Ester**:
- After the nucleophilic attack, the intermediate formed will lose a water molecule (H₂O), resulting in the formation of the ester.
- The final product is an ester (R-C(=O)-O-R').

6. **Conclusion**:
- In this reaction, the breaking of the O-H bond in the alcohol and the C-O bond in the carboxylic acid leads to the formation of the ester and water.

### Final Answer:
The ester formation takes place through the breaking of the O-H bond in the alcohol and the C-O bond in the carboxylic acid.