`CH_(3)COCH_(2)Cl overset(OH^(-),Cl_(2))rarr " Product P "` is

Options

A. `ClCH_(2)COCH_(2)Cl`
B. `CH_(3)COCHCl_(2)`
C. both a and b
D. `ClCH_(2)COOH+CH_(3)Cl`

Explanation

To solve the question, we need to analyze the reaction of CH₃COCH₂Cl with OH⁻ and Cl₂ step by step.

### Step 1: Identify the Reactants
The reactants are:
- CH₃COCH₂Cl (which is 2-chloropropan-2-one)
- OH⁻ (hydroxide ion)
- Cl₂ (chlorine gas)

### Step 2: Role of OH⁻
The hydroxide ion (OH⁻) acts as a base. Its primary role in this reaction is to abstract a proton (H⁺) from the molecule. We need to identify where the acidic protons are located in CH₃COCH₂Cl.

### Step 3: Identify Acidic Protons
In CH₃COCH₂Cl, there are two potential acidic protons:
1. The proton on the CH₃ group (methyl group).
2. The proton on the CH₂ group (methylene group).

### Step 4: Proton Abstraction
When OH⁻ abstracts a proton, it can do so from either of the two positions. However, the proton on the CH₂ group is more acidic due to the electron-withdrawing effect of the carbonyl (C=O) group. Thus, OH⁻ will preferentially abstract the proton from the CH₂ group.

After the abstraction, we get:
- CH₃C(=O)CH₂⁻ (a carbanion is formed at the CH₂ position)

### Step 5: Formation of the Carbanion
The carbanion formed (CH₃C(=O)CH₂⁻) is stabilized by the adjacent carbonyl group (C=O), which can help stabilize the negative charge through resonance.

### Step 6: Reaction with Cl₂
Now, the carbanion (CH₃C(=O)CH₂⁻) will react with Cl₂. The negative charge on the carbanion will attack one of the chlorine atoms in Cl₂, leading to the substitution of the chlorine atom.

### Step 7: Product Formation
The chlorine atom that is attacked will bond with the carbon atom that originally had the negative charge, resulting in the formation of:
- CH₃C(=O)CHCl₂ (dichloro compound)

### Final Product
Thus, the product P formed from the reaction is CH₃C(=O)CHCl₂.

### Summary of Steps:
1. Identify the reactants.
2. Recognize the role of OH⁻ as a base.
3. Identify the acidic protons in the molecule.
4. Determine which proton is abstracted by OH⁻.
5. Form the carbanion and stabilize it.
6. React the carbanion with Cl₂.
7. Conclude with the final product.