Benzene reacts with `CH_(3)Cl` in the presence of anhydrous `AlCl_(3)` to form

Options

A. Toluene
B. Chlorobenzene
C. Benzyl chloride
D. Xylene

Explanation

To solve the question, we need to analyze the reaction between benzene and methyl chloride (CH₃Cl) in the presence of anhydrous aluminum chloride (AlCl₃). This reaction is a classic example of Friedel-Crafts alkylation.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are benzene (C₆H₆) and methyl chloride (CH₃Cl).
- The catalyst used is anhydrous aluminum chloride (AlCl₃).

2. **Understand the Role of the Catalyst**:
- Anhydrous AlCl₃ acts as a Lewis acid. It helps in generating a more reactive species from CH₃Cl.
- In this case, AlCl₃ interacts with CH₃Cl to form a methyl cation (CH₃⁺) and AlCl₄⁻.

3. **Formation of the Methyl Cation**:
- The reaction can be represented as:
\[ CH₃Cl + AlCl₃ \rightarrow CH₃^+ + AlCl₄^- \]
- Here, the methyl cation (CH₃⁺) is the electrophile that will react with benzene.

4. **Electrophilic Aromatic Substitution**:
- The benzene ring, which is rich in π electrons, will attack the methyl cation (CH₃⁺).
- This results in the formation of a new carbon-carbon bond between the benzene and the methyl group.

5. **Formation of Toluene**:
- After the electrophilic attack, the intermediate formed will lose a proton (H⁺) to restore the aromaticity of the benzene ring.
- The final product of this reaction is toluene (C₆H₅CH₃).

6. **Conclusion**:
- Therefore, the product formed when benzene reacts with CH₃Cl in the presence of anhydrous AlCl₃ is toluene.

### Final Answer:
The product formed is **toluene (C₆H₅CH₃)**.

---