When benzene reacts with neopentyl bromide `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH_(2)Br` in presence of `AlBr_(3)`, - is formed

Options

A.
B.
C.
D. all the three are formed

Answer

To solve the problem of what product is formed when benzene reacts with neopentyl bromide in the presence of AlBr3, we can follow these steps:

### Step 1: Understand the Reactants
We have benzene (C6H6) and neopentyl bromide (C(CH3)3CH2Br). Neopentyl bromide is a primary alkyl halide, which means it can undergo a reaction to form a carbocation when treated with a Lewis acid like AlBr3.

### Step 2: Formation of the Carbocation
When neopentyl bromide reacts with AlBr3, the bromine atom leaves, and a carbocation is formed. The reaction can be represented as follows:
\[ \text{C(CH}_3)_3\text{CH}_2\text{Br} + \text{AlBr}_3 \rightarrow \text{C(CH}_3)_3\text{CH}_2^+ + \text{AlBr}_4^- \]
Here, the AlBr3 acts as a Lewis acid, accepting the bromine and generating a carbocation.

### Step 3: Carbocation Rearrangement
The initial carbocation formed is a primary carbocation, which is less stable. To stabilize it, a rearrangement occurs. The adjacent methyl group can shift to the carbocation center, leading to a more stable tertiary carbocation:
\[ \text{C(CH}_3)_3\text{CH}_2^+ \rightarrow \text{C(CH}_3)_3^+\text{C(CH}_3) \]
This results in a tertiary carbocation, which is more stable.

### Step 4: Electrophilic Attack on Benzene
The stable tertiary carbocation now acts as an electrophile and attacks the benzene ring. The reaction can be represented as:
\[ \text{C}_6\text{H}_6 + \text{C(CH}_3)_3^+ \rightarrow \text{C}_6\text{H}_5\text{C(CH}_3)_3 \]
This forms a new carbon-carbon bond between the benzene ring and the carbocation.

### Step 5: Deprotonation
The final step involves the loss of a proton (H+) from the carbon that was attacked, restoring the aromaticity of the benzene ring:
\[ \text{C}_6\text{H}_5\text{C(CH}_3)_3 \rightarrow \text{C}_6\text{H}_5\text{C(CH}_3)_3 + \text{H}^+ \]
The product formed is 1-tert-butylbenzene, where the tert-butyl group is attached to the benzene ring.

### Final Product
The final product of the reaction is:
\[ \text{C}_6\text{H}_5\text{C(CH}_3)_3 \]

### Summary of the Steps
1. **Identify reactants**: Benzene and neopentyl bromide.
2. **Carbocation formation**: Neopentyl bromide reacts with AlBr3 to form a carbocation.
3. **Rearrangement**: The primary carbocation rearranges to a tertiary carbocation for stability.
4. **Electrophilic attack**: The tertiary carbocation attacks the benzene ring.
5. **Deprotonation**: A proton is lost, restoring aromaticity.