`CH_(3)-overset(CH_(3))overset(|)CH-CH_(2)-CH_(3) overset(Cl_(2))underset("UV light")to underset("major")((A))`, The compound "A" is

Options

A. `CH_(3)-overset(CH_(3))overset(|)CH-underset(Cl)underset(|)CH-CH_(3)`
B. `CH_(3)-overset(CH_(3))overset(|)underset(Cl)underset(|)C-CH_(2)-CH_(3)`
C. `CH_(3)-overset(CH_(3))overset(|)CH-CH_(2)-CH_(2)Cl`
D. `CH_(3)-overset(CH_(2)Cl)overset(|)underset(H)underset(|)C-CH_(2)-CH_(3)`

Answer

To solve the problem, we need to determine the major product of the reaction of 2-methylbutane with chlorine (Cl2) in the presence of UV light. This reaction involves a free radical mechanism.

### Step-by-Step Solution:

1. **Identify the Compound**:
The compound given is 2-methylbutane, which has the structure:
\[ \text{CH}_3 - \text{C}(\text{CH}_3) - \text{CH}_2 - \text{CH}_3 \]
This can also be represented as:
\[ \text{CH}_3 - \text{CH}(\text{CH}_3) - \text{CH}_2 - \text{CH}_3 \]

2. **Understand the Reaction Mechanism**:
The reaction of alkanes with chlorine in the presence of UV light proceeds via a free radical substitution mechanism. This involves the formation of free radicals that can lead to the substitution of hydrogen atoms by chlorine atoms.

3. **Determine Possible Radicals**:
In 2-methylbutane, we can remove hydrogen atoms to form different radicals. The possible radicals that can be formed are:
- Removing a hydrogen from the terminal CH3 group (primary radical).
- Removing a hydrogen from the CH2 group (secondary radical).
- Removing a hydrogen from the CH group (tertiary radical).

4. **Identify Stability of Radicals**:
The stability of the radicals formed is crucial in determining the major product:
- Primary radical (least stable)
- Secondary radical (more stable)
- Tertiary radical (most stable)

The most stable radical that can be formed from 2-methylbutane is the tertiary radical, which can be represented as:
\[ \text{CH}_3 - \text{C}(\text{CH}_3) - \cdot - \text{CH}_2 - \text{CH}_3 \]
Here, the radical is formed by removing a hydrogen from the carbon that is bonded to two other CH3 groups.

5. **Formation of the Major Product**:
The chlorine radical (Cl•) will preferentially attack the most stable radical, which is the tertiary radical. This results in the substitution of a hydrogen atom with a chlorine atom, leading to the major product:
\[ \text{CH}_3 - \text{C}(\text{CH}_3) - \text{CH}_2 - \text{CH}_2 - \text{Cl} \]
This product can be represented as:
\[ \text{CH}_3 - \text{C}(\text{CH}_3) - \text{CH}_2 - \text{Cl} \]

6. **Final Product**:
The major product of the reaction is 2-chloro-2-methylbutane, which can be represented as:
\[ \text{CCl}(\text{CH}_3)(\text{CH}_3) - \text{CH}_2 - \text{CH}_3 \]

### Conclusion:
The major product (A) formed when 2-methylbutane is treated with Cl2 in the presence of UV light is 2-chloro-2-methylbutane.