Iodoform is formed on warming `I_2` and `NaOH` with:

Options

A. Acetophenone
B. `CH_(3)OH`
C. `HCOOH`
D. `C_(6)H_(6)`

Answer

To determine which compound forms iodoform when warmed with iodine (I₂) and sodium hydroxide (NaOH), we need to understand the conditions under which the iodoform reaction occurs.

### Step-by-Step Solution:

1. **Understanding the Iodoform Reaction**:
The iodoform reaction occurs with compounds that have a methyl ketone group (RCOCH₃) or secondary alcohols that can be oxidized to methyl ketones. The presence of the methyl group adjacent to the carbonyl group is crucial for the formation of iodoform (CHI₃).

2. **Analyzing the Options**:
We have four options to consider:
- **Option 1: Acetophenone (C₆H₅COCH₃)**: This compound has a methyl ketone functional group (RCOCH₃). Therefore, it can react with iodine and sodium hydroxide to form iodoform.
- **Option 2: Methanol (CH₃OH)**: This is a primary alcohol and does not have a methyl ketone structure. Thus, it cannot undergo the iodoform reaction.
- **Option 3: Formic Acid (HCOOH)**: This is a carboxylic acid and does not contain a methyl ketone group. Therefore, it cannot undergo the iodoform reaction.
- **Option 4: Benzene (C₆H₆)**: This is an aromatic hydrocarbon and does not have a methyl ketone or secondary alcohol structure. Hence, it cannot undergo the iodoform reaction.

3. **Conclusion**:
Based on the analysis, the only compound that can react with iodine and sodium hydroxide to form iodoform is **Acetophenone**.

### Final Answer:
**Acetophenone** is the compound that forms iodoform when warmed with iodine and sodium hydroxide.

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