Complete the following :
`CH_(3)COOHoverset(CH_(3)OH)underset(H_(2)SO_(4))rarr`

Explanation

To complete the reaction of acetic acid (CH₃COOH) with methanol (CH₃OH) in the presence of concentrated sulfuric acid (H₂SO₄), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Reactants**: The reactants are acetic acid (CH₃COOH) and methanol (CH₃OH). The presence of concentrated sulfuric acid (H₂SO₄) indicates that we are performing an esterification reaction.

2. **Esterification Reaction**: In the presence of H₂SO₄, acetic acid reacts with methanol to form an ester. The general reaction for the esterification of a carboxylic acid with an alcohol can be represented as:
\[ \text{RCOOH} + \text{R'OH} \xrightarrow{\text{H}_2\text{SO}_4} \text{RCOOR'} + \text{H}_2\text{O} \]
Here, R is the acyl group from the acid (in this case, CH₃CO-) and R' is the alkyl group from the alcohol (in this case, CH₃-).

3. **Write the Reaction**: For our specific case:
\[ \text{CH}_3\text{COOH} + \text{CH}_3\text{OH} \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_3\text{COOCH}_3 + \text{H}_2\text{O} \]
The product formed is methyl acetate (CH₃COOCH₃) and water (H₂O).

4. **Final Products**: The final products of the reaction are:
- Methyl acetate (CH₃COOCH₃)
- Water (H₂O)

### Complete Reaction:
\[ \text{CH}_3\text{COOH} + \text{CH}_3\text{OH} \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_3\text{COOCH}_3 + \text{H}_2\text{O} \]