`CH_(3)-CH=CH_(2)overset(Cl_(2)//H_(2)O)to`
The correct product is

Options

A.
B. `CH_(3)-underset(Cl)underset("|")"CH"-underset(Cl)underset(|)"CH"_(2)`
C.
D. `H_(3)C-underset(OH)underset("| ")"CH"-underset(Cl)underset("|")"CH"_(2)`

Explanation

To solve the question regarding the reaction of propene (CH₃-CH=CH₂) with chlorine (Cl₂) and water (H₂O), we will follow the steps of the reaction mechanism:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are propene (CH₃-CH=CH₂) and the reagents Cl₂ and H₂O.

2. **Formation of Cyclic Halonium Ion**:
When Cl₂ is added to the double bond of propene, one of the chlorine atoms forms a bond with one of the carbon atoms in the double bond. This results in the formation of a cyclic halonium ion. The structure can be represented as:
\[ \text{CH}_3-\text{C}^+(\text{Cl})-\text{CH}_2 \]
Here, the positive charge is on the carbon that is bonded to the chlorine atom.

**Hint**: Look for the formation of a cyclic structure when halogens add to alkenes.

3. **Nucleophilic Attack by Water**:
Water (H₂O), which acts as a nucleophile, attacks the more substituted carbon of the cyclic halonium ion. The oxygen atom of water donates a pair of electrons to form a bond with the positively charged carbon, leading to the opening of the cyclic structure.

4. **Formation of Alcohol**:
The product formed after the nucleophilic attack is a chlorohydrin. The structure can be represented as:
\[ \text{CH}_3-\text{C}(\text{OH})-\text{CH}_2\text{Cl} \]
Here, the OH group from water adds to the carbon that was not bonded to chlorine.

5. **Final Product**:
The final product of the reaction is 2-chloropropanol (CH₃-CHOH-CH₂Cl), where the hydroxyl group (OH) and the chlorine atom (Cl) are added to adjacent carbons.

### Conclusion:
The correct product formed from the reaction of propene with Cl₂ and water is 2-chloropropanol (CH₃-CHOH-CH₂Cl).

### Final Answer:
The correct product is **2-chloropropanol**.