Hydrocarbon `(CH_3)_3CH` undergoes reaction with `Br_2` and `CI_2` in the presence of sunlight, if the reaction with Cl is highly reactive and that with Br is highly selective so no.of possible products respectively is (are)

Options

A. 2,2
B. 2,1
C. 1,2
D. 1,1

Explanation

To solve the problem, we need to analyze the reactions of the hydrocarbon \((CH_3)_3CH\) (which is tert-butane) with \(Br_2\) and \(Cl_2\) in the presence of sunlight.

### Step-by-Step Solution:

1. **Identify the Hydrocarbon**:
The hydrocarbon given is \((CH_3)_3CH\), which is known as tert-butane.

2. **Understand the Reaction Conditions**:
The reactions occur in the presence of sunlight, which indicates that these reactions will proceed via a free radical mechanism.

3. **Reactions with Bromine (\(Br_2\))**:
- The reaction with bromine is described as highly selective. This means that bromine will preferentially react with the most stable free radical.
- In tert-butane, there are three equivalent primary hydrogens and one tertiary hydrogen. The formation of a tertiary radical is more stable due to hyperconjugation.
- Thus, the bromination of tert-butane will yield one major product, which is tert-butyl bromide \((CH_3)_3CBr\).

4. **Reactions with Chlorine (\(Cl_2\))**:
- The reaction with chlorine is highly reactive, meaning it will react with all available hydrogen atoms, leading to multiple products.
- Chlorination can lead to the formation of:
- One product where a tertiary hydrogen is replaced: \((CH_3)_3CCl\) (tert-butyl chloride).
- Additionally, chlorination can also replace one of the three equivalent primary hydrogens, leading to the formation of products like \(CH_3CHClCH_3\) (isobutyl chloride).
- Therefore, two products can be formed from the chlorination of tert-butane.

5. **Count the Products**:
- From the bromination, we have **1 product**.
- From the chlorination, we have **2 products**.

6. **Final Answer**:
The number of possible products from the reactions with \(Cl_2\) and \(Br_2\) respectively is **2 and 1**.

### Conclusion:
Thus, the answer to the question is that the number of possible products from the reaction with \(Cl_2\) and \(Br_2\) is **2 and 1** respectively.