`C_(6)H_(6)+Z overset(Anhy.AlCl_(3))to` Toluene
The compound Z is

Options

A. Acetic acid
B. Acetic anhydride
C. Acetone
D. Chloromethane

Answer

To solve the question, we need to identify the compound Z that reacts with benzene (C₆H₆) in the presence of anhydrous aluminum chloride (AlCl₃) to produce toluene (C₇H₈).

### Step-by-Step Solution:

1. **Identify the Reaction Type:**
The reaction described is a Friedel-Crafts alkylation reaction. In this type of reaction, an alkyl group is introduced to an aromatic ring (in this case, benzene).

2. **Recognize the Reactants:**
The reactant is benzene (C₆H₆), and it is reacting with a compound Z in the presence of anhydrous AlCl₃, which acts as a Lewis acid to facilitate the reaction.

3. **Determine the Product:**
The product of the reaction is toluene (C₇H₈). Toluene has a methyl group (CH₃) attached to the benzene ring.

4. **Identify Compound Z:**
Since the product is toluene, we need to identify what compound Z must be to provide the methyl group (CH₃) to the benzene. The most common source of a methyl group in Friedel-Crafts alkylation is chloromethane (CH₃Cl).

5. **Write the Reaction:**
The overall reaction can be represented as:
\[ C_6H_6 + CH_3Cl \overset{Anhyd. AlCl_3}{\rightarrow} C_7H_8 + HCl \]
Here, chloromethane (CH₃Cl) is the compound Z that reacts with benzene to form toluene.

6. **Conclusion:**
Therefore, the compound Z is chloromethane (CH₃Cl).

### Final Answer:
The compound Z is chloromethane (CH₃Cl).

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