The Friedel-Crafts reaction of benzene with `n-`butyl chloride at
`0^(@)C` produces :

Options

A. `C_(6)H_(5)-CH_(2)CH_(2)CH_(2)CH_(3)`
B. `C_(6)H_(5)overset(CH_(3))overset(|)(CH)-CH_(2)CH_(3)`
C. `C_(6)H_(5)-CH_(2)-underset(CH_(3))underset(|)(CH)-CH_(3)`
D. all of these

Answer

To solve the question regarding the Friedel-Crafts reaction of benzene with n-butyl chloride at 0°C, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants:**
The reactants in this reaction are benzene (C6H6) and n-butyl chloride (C4H9Cl).

2. **Understand the Friedel-Crafts Alkylation:**
The Friedel-Crafts reaction involves the alkylation of an aromatic ring. In this case, n-butyl chloride will react with aluminum chloride (AlCl3) to generate a carbocation.

3. **Formation of the Carbocation:**
When n-butyl chloride reacts with AlCl3, it forms a butyl carbocation (C4H9^+). The reaction can be represented as:
\[ \text{C4H9Cl} + \text{AlCl3} \rightarrow \text{C4H9}^+ + \text{AlCl4}^- \]

4. **Consider Carbocation Rearrangement:**
The butyl carbocation is a primary carbocation, which is less stable. To achieve greater stability, a hydride shift occurs, converting the primary butyl carbocation into a more stable secondary carbocation (isobutyl carbocation).

5. **Reaction with Benzene:**
The isobutyl carbocation then reacts with benzene. The electrophilic aromatic substitution occurs where the carbocation attacks the benzene ring, leading to the formation of an alkyl-substituted benzene.

6. **Formation of the Product:**
The product formed from this reaction is isobutylbenzene (or 1-phenyl-2-methylpropane). The reaction can be summarized as:
\[ \text{C6H6} + \text{C4H9}^+ \rightarrow \text{C6H5C4H9} + \text{HCl} \]

7. **Final Product:**
The final product of the Friedel-Crafts reaction of benzene with n-butyl chloride at 0°C is isobutylbenzene.

### Final Answer:
The product of the Friedel-Crafts reaction of benzene with n-butyl chloride at 0°C is **isobutylbenzene**.

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