`CH_(3)COCH_(3)` can be converted to ` CH_(3)CH_(2)CH_(3)` by the action of

Options

A. `HIO_(3)`
B. `HNO_(3) `
C. `HI + Red P`
D. `H_(3)PO_(3)` .

Explanation

To convert \( CH_3COCH_3 \) (acetone) to \( CH_3CH_2CH_3 \) (propane), we need to perform a reduction reaction. Here’s the step-by-step solution:

### Step 1: Identify the Reactants and Products
- **Reactant**: \( CH_3COCH_3 \) (acetone)
- **Product**: \( CH_3CH_2CH_3 \) (propane)

### Step 2: Understand the Type of Reaction
- The reaction involves the conversion of a ketone (acetone) into an alkane (propane). This requires the reduction of the carbonyl group (C=O) in the ketone to a methylene group (C-H).

### Step 3: Choose the Reducing Agent
- To achieve this reduction, we need a strong reducing agent. Common reducing agents for this type of reaction include lithium aluminum hydride (LiAlH4), but in this case, we will focus on the options provided in the question.

### Step 4: Analyze the Given Options
- The options provided are: \( HIO_3, HNO_3, HI, H_3PO_3 \).
- Among these, HI (hydroiodic acid) is known to act as a strong reducing agent when used with red phosphorus.

### Step 5: Apply the Reducing Agent
- When acetone reacts with HI, it undergoes reduction where the carbonyl group is converted into a methylene group, effectively transforming the ketone into an alkane.

### Step 6: Write the Final Reaction
- The reaction can be summarized as:
\[ CH_3COCH_3 + HI \rightarrow CH_3CH_2CH_3 + I_2 + H_2O \]
- Here, HI reduces the ketone to propane.

### Conclusion
- Therefore, the correct answer to the question is **HI**.

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