Benzene reacts with n-propyl chloride in the presence of anhydrous `AlCl_(3)` to give predominantly

Options

A. `n-`Propylbenzene
B. Isopropylbenzene
C. `3-`Propyl`-1-`chlorobenzene
D. None of these

Explanation

To solve the problem of how benzene reacts with n-propyl chloride in the presence of anhydrous AlCl₃, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are benzene (C₆H₆) and n-propyl chloride (C₃H₇Cl or CH₃CH₂CH₂Cl).

2. **Understand the Role of AlCl₃**:
- Anhydrous aluminum chloride (AlCl₃) acts as a Lewis acid and is used as a catalyst in Friedel-Crafts alkylation reactions.

3. **Formation of Carbocation**:
- When n-propyl chloride reacts with AlCl₃, it forms a carbocation. The n-propyl chloride can form a primary carbocation (n-propyl cation, CH₃CH₂CH₂⁺), which is less stable.

4. **Carbocation Rearrangement**:
- The primary carbocation can undergo a rearrangement to form a more stable carbocation. In this case, a 1,2-hydride shift occurs, converting the n-propyl carbocation into an isopropyl carbocation (CH₃CH⁺CH₃), which is more stable.

5. **Electrophilic Attack on Benzene**:
- The isopropyl carbocation (CH₃CH⁺CH₃) acts as an electrophile and attacks the benzene ring, resulting in the formation of a sigma complex (arenium ion).

6. **Deprotonation**:
- The sigma complex loses a proton (H⁺) to restore the aromaticity of the benzene ring. This step regenerates the AlCl₄⁻ ion, which can react with H⁺ to form HCl and AlCl₃.

7. **Final Product**:
- The final product of this reaction is isopropylbenzene (C₆H₅C₃H₇), also known as cumene.

### Conclusion:
The predominant product of the reaction between benzene and n-propyl chloride in the presence of anhydrous AlCl₃ is **isopropylbenzene (cumene)**.

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