Acetone reacts with iodine `(I_(2))` to form iodoform in the presence of

Options

A. `CaCO_(3)`
B. `NaOH`
C. `KOH`
D. `MgCO_(3)`

Explanation

To solve the question regarding the reaction of acetone with iodine to form iodoform, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The question states that acetone reacts with iodine (I₂). Acetone is a ketone with the formula CH₃COCH₃.

2. **Understand the Reaction Conditions**: The formation of iodoform (CHI₃) from acetone requires a base in addition to iodine. The options provided are CaCO₃, NaOH, KOH, and MgCO₃.

3. **Recall the Iodoform Reaction**: The iodoform reaction occurs when a methyl ketone (like acetone) is treated with iodine in the presence of a strong base. The base facilitates the reaction by deprotonating the acetone, allowing it to react with iodine.

4. **Choose the Correct Base**: Among the options provided, NaOH (sodium hydroxide) is a strong base that is commonly used in the iodoform reaction. KOH (potassium hydroxide) could also work, but it is not listed as the correct answer in this context.

5. **Write the Reaction**: The overall reaction can be represented as follows:
\[ \text{CH}_3\text{COCH}_3 + 3 \text{I}_2 + 4 \text{NaOH} \rightarrow \text{CHI}_3 + 3 \text{NaI} + 3 \text{H}_2\text{O} + \text{CH}_3\text{COONa} \]

6. **Conclusion**: The correct answer to the question is NaOH, as it is the base required for the iodoform reaction with acetone.

### Final Answer:
The correct option is **NaOH**.

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