The major product of reaction between n-butane and bromine at `130^(@)C` is

Options

A. `CH_(3)CH_(2)CH_(2)CH_(2)Br`
B. `CH_(3)CH_(2)underset(CH_(3))underset(|)(CH)Br`
C. `CH_(3)-underset(CH_(3))underset(|)(C)HCH_(2)Br`
D. `CH_(3)-underset(CH_(3))underset(|)(C)-Br_(2)`

Answer

To determine the major product of the reaction between n-butane and bromine at 130°C, we can follow these steps:

### Step 1: Identify the Reactants
We start with n-butane, which has the molecular formula \( \text{C}_4\text{H}_{10} \) and can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \]

### Step 2: Understand the Reaction Conditions
The reaction is taking place in the presence of bromine (\( \text{Br}_2 \)) and at a high temperature of 130°C. This suggests that a free radical halogenation reaction will occur.

### Step 3: Mechanism of the Reaction
In free radical halogenation, the reaction proceeds through three main steps: initiation, propagation, and termination. At high temperatures, bromine can homolytically cleave to form bromine radicals.

### Step 4: Formation of Free Radicals
When n-butane reacts with bromine, free radicals are formed. The hydrogen atoms in n-butane can be replaced by bromine atoms, leading to the formation of different alkyl bromides.

### Step 5: Stability of Free Radicals
The stability of the free radicals formed during the reaction is crucial in determining the major product. The possible free radicals that can form from n-butane are:
- Primary free radical (1°) from the terminal carbon
- Secondary free radical (2°) from the second carbon

Since secondary free radicals are more stable than primary free radicals, the reaction will preferentially lead to the formation of the more stable secondary radical.

### Step 6: Identify the Major Product
The major product will be formed when bromine replaces a hydrogen atom from the carbon that forms the more stable secondary radical. The major product of this reaction will be:
\[ \text{CH}_3\text{CH}_2\text{CHBr}\text{CH}_3 \]
This can also be represented as 2-bromobutane.

### Conclusion
Thus, the major product of the reaction between n-butane and bromine at 130°C is 2-bromobutane.

### Final Answer
The major product is \( \text{CH}_3\text{CH}_2\text{CHBr}\text{CH}_3 \) (2-bromobutane).

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