When `n-` butane is heated iin the presence of `AlCl_(3)HCl` it will be converted into

Options

A. (a) Ethane
B. (b) Propane
C. (c) Butane
D. Isobutane

Explanation

To solve the question of what n-butane converts into when heated in the presence of AlCl3 and HCl, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We start with n-butane, which has the structure CH3-CH2-CH2-CH3.

2. **Add the Catalysts**:
- The catalysts used are AlCl3 (aluminum chloride) and HCl (hydrochloric acid). AlCl3 is a strong Lewis acid that can accept electron pairs.

3. **Formation of Carbocation**:
- When n-butane is heated with AlCl3 and HCl, the hydrogen atom from one of the terminal carbons can break its bond, leading to the formation of a carbocation.
- The reaction can be represented as:
\[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3 \xrightarrow{\text{AlCl}_3, \text{HCl}} \text{CH}_3-\text{CH}^+-\text{CH}_2-\text{CH}_3 + \text{AlCl}_3 \]

4. **Carbocation Rearrangement**:
- The formed carbocation is not stable. To stabilize it, a methyl group (CH3) can shift from one of the adjacent carbons to the positively charged carbon, leading to a more stable tertiary carbocation.
- This rearrangement can be illustrated as:
\[ \text{CH}_3-\text{CH}^+-\text{CH}_2-\text{CH}_3 \rightarrow \text{CH}_3-\text{C}^+-\text{CH}_3-\text{CH}_3 \]

5. **Final Product Formation**:
- The stable carbocation can then react with a chloride ion (Cl-) from AlCl3 to form isobutane (also known as methylpropane).
- The final product can be represented as:
\[ \text{CH}_3-\text{C}(\text{CH}_3)^+-\text{CH}_3 + \text{Cl}^- \rightarrow \text{(CH}_3)_2\text{CHCl} \]

6. **Conclusion**:
- The final product of heating n-butane in the presence of AlCl3 and HCl is isobutane (C4H10), which is an isomer of n-butane.

### Final Answer:
When n-butane is heated in the presence of AlCl3 and HCl, it is converted into isobutane.