The reagent which can be used to distinguish acetopheone from benzophenone is :

Options

A. 2,4-dinitrophenyl hydrazine
B. aqueous solution of `NaHSO_3`
C. Benedict reagent
D. `I_(2) and Na_(2)CO_(3)`

Explanation

To distinguish between acetophenone and benzophenone, we can use the iodoform test. Here’s a step-by-step solution to the question:

### Step 1: Identify the Structures
- **Acetophenone** has the structure:
\[ \text{CH}_3\text{C}(=O)\text{C}_6\text{H}_5 \]
- **Benzophenone** has the structure:
\[ \text{C}_6\text{H}_5\text{C}(=O)\text{C}_6\text{H}_5 \]

### Step 2: Understand the Iodoform Test
The iodoform test is used to identify the presence of a methyl ketone (a ketone with the structure RCOCH3) or a secondary alcohol that can be oxidized to a methyl ketone.

### Step 3: Apply the Iodoform Test
- **Acetophenone** contains the methyl group (CH3) adjacent to the carbonyl group (C=O). When acetophenone is treated with iodine (I2) in the presence of sodium hydroxide (NaOH), it undergoes the iodoform reaction to produce iodoform (CHI3), which is a yellow precipitate.

Reaction:
\[ \text{CH}_3\text{C}(=O)\text{C}_6\text{H}_5 + \text{I}_2 + \text{NaOH} \rightarrow \text{CHI}_3 \text{ (yellow ppt)} + \text{other products} \]

- **Benzophenone**, on the other hand, does not have a methyl group adjacent to the carbonyl group. Therefore, it will not react with iodine in the presence of sodium hydroxide and will give no precipitate.

Reaction:
\[ \text{C}_6\text{H}_5\text{C}(=O)\text{C}_6\text{H}_5 + \text{I}_2 + \text{NaOH} \rightarrow \text{No reaction} \]

### Step 4: Conclusion
Since acetophenone gives a yellow precipitate of iodoform while benzophenone does not react, the reagent that can be used to distinguish between acetophenone and benzophenone is **I2 + NaOH**.

### Final Answer:
The reagent which can be used to distinguish acetophenone from benzophenone is **I2 + NaOH**.

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