Reaction of benzene with `Me_(3)C COCl` in the presence of anhydrous `AlCl_(3)` gives

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Explanation

To solve the question regarding the reaction of benzene with `Me3C COCl` in the presence of anhydrous `AlCl3`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have benzene (C6H6) and `Me3C COCl` (which is tert-butyl carbonyl chloride).

2. **Understand the Role of Anhydrous AlCl3**:
- Anhydrous AlCl3 acts as a Lewis acid and facilitates the formation of an electrophile from `Me3C COCl`.

3. **Formation of the Electrophile**:
- When `Me3C COCl` reacts with `AlCl3`, it forms a complex that generates a tert-butyl cation (`Me3C+`) and a chloride ion (Cl-). The carbonyl group (C=O) remains attached to the AlCl4- complex.

4. **Electrophilic Aromatic Substitution**:
- The generated tert-butyl cation (`Me3C+`) is a strong electrophile. It will react with benzene through electrophilic aromatic substitution.

5. **Formation of the Product**:
- The benzene ring donates a pair of electrons to the tert-butyl cation, forming a sigma complex (arenium ion). After deprotonation, the product formed is tert-butylbenzene.

6. **Final Product**:
- The final product of the reaction is tert-butylbenzene (C6H5C(CH3)3).

### Summary of the Reaction:
The reaction of benzene with `Me3C COCl` in the presence of anhydrous `AlCl3` yields tert-butylbenzene.

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