For the reaction `C_(6)H_(6)+(CH_(3))_(2)CHCH_(2)Cl overset("Anhy".AlCl_(3))rarr` The final product is

Answer

To solve the given reaction of benzene (C₆H₆) with isobutyl chloride ((CH₃)₂CHCH₂Cl) in the presence of anhydrous AlCl₃, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are benzene (C₆H₆) and isobutyl chloride ((CH₃)₂CHCH₂Cl). Anhydrous AlCl₃ is used as a catalyst.

2. **Formation of the Carbocation**:
When isobutyl chloride reacts with AlCl₃, it forms a carbocation. The chlorine atom leaves, and the isobutyl group forms a carbocation:
\[ (CH₃)₂CHCH₂^+ \]
This carbocation is a primary carbocation, which can rearrange to form a more stable tertiary carbocation through a hydride shift.

3. **Hydride Shift**:
The primary carbocation undergoes a 1,2-hydride shift to form a more stable tertiary carbocation:
\[ CH₃C^+(CH₃)₂ \]
This is the tertiary carbocation that will react with benzene.

4. **Electrophilic Attack**:
The benzene ring, which is rich in electrons, will attack the positively charged tertiary carbocation. This forms a sigma complex (also known as an arenium ion):
\[ C₆H₅^+ - C(CH₃)₃ \]

5. **Deprotonation**:
The sigma complex loses a proton (H⁺) to restore aromaticity. The proton can be abstracted by the AlCl₄⁻ ion formed during the reaction:
\[ C₆H₅C(CH₃)₃ + HCl \]

6. **Final Product**:
The final product of this reaction is tertiary butyl benzene, which can be represented as:
\[ C₆H₅C(CH₃)₃ \]

### Final Answer:
The final product of the reaction is **tertiary butyl benzene**.

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