n-Butane reacts with `Br_(2)` at `130^(@)` to give more amount of

Options

A. `CH_(3)-CH_(2)-underset(Br)underset(|)CH-CH_(3)`
B. `CH_(3)CH_(2)-CH_(2)-CH_(2)Br`
C. `CH_(3)CH_(2)-CH_(2)-CH_(2)Br`
D. all in equal amounts

Answer

To solve the question regarding the reaction of n-butane with bromine at 130°C, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of n-Butane**:
- n-Butane is a straight-chain alkane with the molecular formula C₄H₁₀. Its structure can be represented as:
\[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3 \]

2. **Understand the Reaction Type**:
- The reaction of n-butane with bromine (Br₂) is a halogenation reaction that proceeds via a free radical mechanism. This means that the reaction involves the formation of free radicals.

3. **Initiation Step**:
- When bromine is heated, it undergoes homolytic cleavage to form two bromine radicals:
\[ \text{Br}_2 \xrightarrow{\text{heat}} 2 \text{Br}^\bullet \]

4. **Propagation Step**:
- One of the bromine radicals will abstract a hydrogen atom from n-butane, leading to the formation of a butyl radical and hydrogen bromide (HBr):
\[ \text{Br}^\bullet + \text{C}_4\text{H}_{10} \rightarrow \text{C}_4\text{H}_9^\bullet + \text{HBr} \]
- The butyl radical formed is a primary radical (1°).

5. **Rearrangement of the Radical**:
- The primary radical can undergo rearrangement to form a more stable secondary radical (2°). In this case, the hydrogen atom from the adjacent carbon can migrate to the carbon with the radical, resulting in a more stable secondary radical:
\[ \text{C}_4\text{H}_9^\bullet \rightarrow \text{C}_4\text{H}_9^{\text{(2°)}} \quad (\text{after rearrangement}) \]

6. **Termination Step**:
- The secondary radical will then react with another bromine radical to form the final product, which is 2-bromobutane:
\[ \text{C}_4\text{H}_9^{\text{(2°)}} + \text{Br}^\bullet \rightarrow \text{C}_4\text{H}_9\text{Br} + \text{Br}^\bullet \]

7. **Final Product**:
- The major product formed from this reaction is 2-bromobutane, which is represented as:
\[ \text{CH}_3-\text{CHBr}-\text{CH}_2-\text{CH}_3 \]

### Conclusion:
The major product formed when n-butane reacts with bromine at 130°C is **2-bromobutane**.