Reaction of benzene with `Me_3COCl` in the presence of anhydrous `AlCl_3` gives

Options

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Explanation

To solve the question regarding the reaction of benzene with `Me3COCl` in the presence of anhydrous `AlCl3`, we can follow these steps:

### Step 1: Understand the Reactants
We have benzene (C6H6) and `Me3COCl` (tert-butyl chloride). The `Me3CO` part is a tert-butyl group, which is a tertiary alkyl group.

### Step 2: Role of Anhydrous AlCl3
Anhydrous aluminum chloride (AlCl3) acts as a Lewis acid. It interacts with `Me3COCl` to form a more reactive species. The AlCl3 coordinates with the chlorine atom in `Me3COCl`, facilitating the generation of a carbocation.

### Step 3: Formation of the Carbocation
When `Me3COCl` reacts with AlCl3, the chlorine atom leaves, and a tert-butyl cation (`Me3C+`) is formed. The reaction can be summarized as follows:
- `Me3COCl + AlCl3 → Me3C+ + AlCl4-`

### Step 4: Electrophilic Aromatic Substitution
The tert-butyl cation (`Me3C+`) is a strong electrophile. It will react with benzene through an electrophilic aromatic substitution mechanism. The benzene ring donates one of its π electrons to the carbocation, forming a sigma complex (also known as an arenium ion).

### Step 5: Deprotonation
The sigma complex formed will lose a proton (H+) to restore aromaticity. This step regenerates the aromatic system of benzene and leads to the final product.

### Step 6: Final Product
The final product of the reaction is tert-butylbenzene. The structure can be represented as:
- C6H5-C(CH3)3

### Conclusion
The reaction of benzene with `Me3COCl` in the presence of anhydrous `AlCl3` gives tert-butylbenzene.

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