`CH_3COCH_3+2(H) underset(H_2O)overset(Na+Hg) to A overset(PCl_3)to B` where B is :

Options

A. 1 chloro propane
B. 2 chloro propane
C. chloro ethane
D. chloro propane

Explanation

To solve the given question, we need to analyze the reactions step by step.

### Step 1: Identify the starting compound
The starting compound is **propanone (CH₃COCH₃)**, which is a ketone.

### Step 2: Reduction of propanone
The reaction involves the reduction of propanone using sodium (Na) and mercury (Hg) in the presence of water (H₂O). This is a typical reaction for reducing ketones to alcohols.

- **Reaction**:
\[ CH₃COCH₃ + 2(H) \xrightarrow{Na + Hg, H₂O} A \]
- **Product A**: The reduction of propanone will yield **isopropanol (propan-2-ol)**, which is a secondary alcohol. The structure of isopropanol is:
\[ CH₃CHOHCH₃ \]

### Step 3: Reaction of isopropanol with PCl₃
Next, we need to react the product A (isopropanol) with phosphorus trichloride (PCl₃). The reaction of alcohols with PCl₃ results in the formation of alkyl chlorides.

- **Reaction**:
\[ CH₃CHOHCH₃ \xrightarrow{PCl₃} B \]
- **Product B**: The product of this reaction will be **isopropyl chloride (2-chloropropane)**. The structure of isopropyl chloride is:
\[ CH₃CHClCH₃ \]

### Final Answer
Thus, the final product B is **isopropyl chloride (2-chloropropane)**.

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