The treatment of benzene with isobutene in the presence of sulphuric acid gives

Options

A. Iso-butyl benzene
B. Tert- butyl benzene
C. n-butyl benzene
D. No reaction

Explanation

To solve the question regarding the treatment of benzene with isobutene in the presence of sulfuric acid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants:**
- The reactants in this reaction are benzene (C6H6) and isobutene (C4H8).
- Isobutene can be represented as CH2=C(CH3)2.

2. **Understand the Role of Sulfuric Acid:**
- Sulfuric acid (H2SO4) acts as a catalyst in this reaction. It protonates the isobutene, leading to the formation of a carbocation.

3. **Formation of Carbocation:**
- When isobutene is treated with sulfuric acid, it forms a tertiary carbocation. The reaction can be represented as follows:
\[ \text{CH}_2=\text{C}(\text{CH}_3)_2 + \text{H}_2\text{SO}_4 \rightarrow \text{(CH}_3)_2\text{C}^+ \text{H}_2 \]
- The resulting tertiary carbocation is (CH3)2C+H2, which is a stable intermediate.

4. **Electrophilic Attack by Benzene:**
- Benzene, being an aromatic compound, can act as a nucleophile and attack the carbocation. The reaction can be represented as:
\[ \text{C}_6\text{H}_6 + \text{(CH}_3)_2\text{C}^+ \rightarrow \text{C}_6\text{H}_5\text{C}(\text{CH}_3)_3^+ \]
- This forms a new compound with a positive charge on the carbon that was bonded to the carbocation.

5. **Deprotonation:**
- The positively charged intermediate will lose a proton (H+) to restore aromaticity. This can be represented as:
\[ \text{C}_6\text{H}_5\text{C}(\text{CH}_3)_3^+ \rightarrow \text{C}_6\text{H}_5\text{C}(\text{CH}_3)_3 + \text{H}^+ \]
- The final product formed is tertiary butylbenzene (C6H5C(CH3)3).

6. **Conclusion:**
- Therefore, the treatment of benzene with isobutene in the presence of sulfuric acid gives tertiary butylbenzene.

### Final Answer:
The product formed is **tertiary butylbenzene**.