In Friedal craft reaction Toluene can be prepared by:

Options

A. `C_(6)H_(6) + CH_(3) CI`
B. `C_(6)H_(5)CI + CH_(4)`
C. `C_(6)H_(6) CH_(2) CI_(2)`
D. `C_(6)H_(6) + CH_(3) COCI`

Answer

To prepare toluene (methylbenzene) through a Friedel-Crafts reaction, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants needed for the Friedel-Crafts alkylation to synthesize toluene are benzene (C6H6) and methyl chloride (CH3Cl).

2. **Use of a Catalyst**:
In Friedel-Crafts reactions, a catalyst is required. Aluminum chloride (AlCl3) is commonly used as a Lewis acid catalyst. It helps in generating the electrophile from the alkyl halide.

3. **Formation of the Electrophile**:
When methyl chloride (CH3Cl) is treated with aluminum chloride (AlCl3), a complex is formed. The aluminum chloride accepts a pair of electrons from the chlorine atom in CH3Cl, leading to the formation of a methyl cation (CH3+), which is the electrophile.

\[ \text{CH}_3\text{Cl} + \text{AlCl}_3 \rightarrow \text{CH}_3^+ + \text{AlCl}_4^- \]

4. **Electrophilic Attack on Benzene**:
The generated methyl cation (CH3+) then acts as an electrophile and attacks the benzene ring. This results in the substitution of one hydrogen atom in the benzene ring with a methyl group.

\[ \text{C}_6\text{H}_6 + \text{CH}_3^+ \rightarrow \text{C}_6\text{H}_5\text{CH}_3 + \text{H}^+ \]

5. **Formation of Toluene**:
The final product of this reaction is toluene (C6H5CH3), where one hydrogen atom of benzene has been replaced by a methyl group.

### Final Reaction:
\[ \text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl} \]