`C CI_(3)CH=CH_(2) overset(Cl_(2)+H_(2)O)rarr P`,
Identify major product `P.`

Options

A.
B.
C.
D.

Explanation

To solve the problem of identifying the major product \( P \) when \( CCl_3CH=CH_2 \) reacts with \( Cl_2 + H_2O \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactant is \( CCl_3CH=CH_2 \), which has a double bond between the second and third carbon atoms and a trichloromethyl group (\( CCl_3 \)) attached to the first carbon.

2. **Understand the Reaction Conditions**: The reaction is carried out in the presence of \( Cl_2 \) and \( H_2O \). This indicates that we will have a halogenation reaction followed by hydration.

3. **Apply Markovnikov's Rule**: According to Markovnikov's rule, during the addition of \( Cl_2 \) and \( H_2O \) across the double bond, the halogen (\( Cl \)) will attach to the carbon with more alkyl substituents. In this case, the carbon with \( CCl_3 \) (first carbon) is more substituted than the second carbon (which has only \( H \)).

4. **Formation of Carbocation**: When \( Cl_2 \) adds to the double bond, it will first form a carbocation on the second carbon. The structure now looks like this:
\[ CCl_3 - C^+ - CH_2 - Cl \]
Here, \( C^+ \) indicates the carbocation formed at the second carbon.

5. **Nucleophilic Attack by Water**: The water molecule will then attack the carbocation. The hydroxide ion (\( OH^- \)) from water will attach to the carbocation, while a proton (\( H^+ \)) will be released.

6. **Final Product Formation**: The final product will have the structure:
\[ CCl_3 - C - CH_2 - Cl - OH \]
This means that one \( Cl \) is attached to the second carbon and \( OH \) is attached to the same carbon.

7. **Identify the Major Product \( P \)**: The major product \( P \) is therefore:
\[ CCl_3 - CH(Cl) - CH_2(OH) \]

### Conclusion:
The major product \( P \) formed from the reaction of \( CCl_3CH=CH_2 \) with \( Cl_2 + H_2O \) is \( CCl_3 - CH(Cl) - CH_2(OH) \).