The products formed in the reaction are :
`C_(6)H_(5)-overset(O)overset(||)C-OH+CH_(3)overset(18)OHunderset("Heat")overset(H_(2)SO_(4))rarr`

Options

A. `C_(6)H_(5)-overset(O)overset(||)C-overset(18)OCH_(3)` and `H_(2)O`
B. `C_(6)H_(5)-overset(O)overset(||)C-OCH_(3)` and `H_(2)overset(18)O`
C. `C_(6)H_(5)-overset(O)overset(||)C-CH_(2)overset(18)OH` and `H_(2)O`
D. `C_(6)H_(5)overset(18)OCH_(3),CO` and `H_(2)O`

Answer

To solve the given reaction, we will break down the steps involved in the reaction between benzoic acid (C6H5C(=O)OH) and methanol (CH3OH) in the presence of sulfuric acid (H2SO4) and heat. The methanol contains an isotope of oxygen (O-18), which will be incorporated into the product.

### Step-by-Step Solution:

1. **Identify the Reactants:**
- The reactants are benzoic acid (C6H5C(=O)OH) and methanol (CH3OH) with O-18 isotope.

2. **Protonation of Benzoic Acid:**
- In the presence of sulfuric acid (H2SO4), the hydroxyl group (-OH) of benzoic acid is protonated, forming a protonated carboxylic acid intermediate (C6H5C(=O)OH2+).

3. **Nucleophilic Attack:**
- The methanol (CH3OH) acts as a nucleophile. The oxygen atom of the methanol (which is O-18) attacks the carbonyl carbon of the protonated benzoic acid, leading to the formation of a tetrahedral intermediate.

4. **Formation of Tetrahedral Intermediate:**
- The tetrahedral intermediate consists of C6H5C(OH)(OCH3) with the O-18 from methanol now attached to the carbon.

5. **Proton Transfer:**
- A proton transfer occurs within the tetrahedral intermediate, stabilizing the structure and preparing it for the next step.

6. **Elimination of Water:**
- The tetrahedral intermediate collapses, leading to the elimination of a water molecule (H2O), resulting in the formation of the product.

7. **Formation of the Final Product:**
- The final product is methyl benzoate with the O-18 isotope incorporated, represented as C6H5C(=O)OCH3 (with O-18).

### Final Product:
- The final product formed in the reaction is O-18 labeled methyl benzoate (C6H5C(=O)OCH3 with O-18).