2-Methyl butane reacting with { Br }_{ 2 } in sunlight mainly gives:(A.) 1-Bromo-2-methyl butane(B.) 2-Bromo-2-methyl butane(C.) 2-Bromo-3-methyl butane(D.) 1-Bromo-3-methyl butane

To determine the major product of the reaction between 2-methylbutane and \( \text{Br}_2 \) in the presence of sunlight, we need to consider the mechanism of the reaction, which is a free radical halogenation.

### Steps to Solve:

1. **Identify the Structure and Possible Products:**
- **2-Methylbutane** has the structure:

- When 2-methylbutane reacts with bromine (\( \text{Br}_2 \)) in sunlight, it undergoes a free radical halogenation process. The reaction involves the formation of bromine radicals, which abstract hydrogen atoms from the 2-methylbutane molecule to form bromoalkanes.

2. **Determine the Types of Hydrogen Atoms:**
- 2-methylbutane has three types of hydrogen atoms:
- Primary hydrogens (from the methyl groups attached to the main chain)
- Secondary hydrogens (from the 2-position carbon)
- Tertiary hydrogens (from the 3-position carbon)

- Specifically, in 2-methylbutane:
- There are 6 primary hydrogens at the methyl groups.
- There are 2 secondary hydrogens on the 2-position carbon.
- There is 1 tertiary hydrogen on the 3-position carbon.

3. **Analyze the Stability of the Radicals:**
- The stability of the resulting radicals determines the major product:
- Tertiary radicals are more stable than secondary radicals.
- Secondary radicals are more stable than primary radicals.

- The tertiary radical forms when the hydrogen from the 3-position carbon is replaced. Thus, the product formed with a tertiary radical will be more stable and is likely to be the major product.

4. **Determine the Major Product:**
- The major product is derived from the most stable radical, which is the tertiary radical.

- Hence, when the tertiary hydrogen (from the 3-position carbon) is replaced by bromine, the product is **2-Bromo-2-methylbutane**.

### Conclusion:

The main product of the reaction between 2-methylbutane and \( \text{Br}_2 \) in sunlight is **2-Bromo-2-methylbutane**.

So the answer is **(B.) 2-Bromo-2-methylbutane**.