Acetone reacts with iodine to form iodoform in the presence of

Options

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

Explanation

To solve the question about the reaction of acetone with iodine to form iodoform in the presence of a base, we can break it down into several steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is acetone (C3H6O), which has the structure CH3-CO-CH3.
- The other reactants are iodine (I2) and sodium hydroxide (NaOH).

2. **Understand the Reaction Conditions**:
- The reaction requires a basic medium, which is provided by sodium hydroxide (NaOH). This base will help in the deprotonation of acetone.

3. **Deprotonation of Acetone**:
- In the presence of NaOH, one of the acidic hydrogen atoms (H) from the methyl group adjacent to the carbonyl (C=O) in acetone is removed. This forms an enolate ion:
\[ CH3-C(=O)-CH3 \xrightarrow{NaOH} CH3-C(=O)^{-}-CH2 \]

4. **Formation of Enol**:
- The enolate ion can tautomerize to form an enol:
\[ CH3-C(=O)^{-}-CH2 \rightarrow CH2=CH-CO-CH3 \]

5. **Iodination of the Enolate**:
- The enolate ion reacts with iodine (I2). The iodine adds to the enolate, replacing the hydrogen atoms:
\[ CH2=CH-CO-CH3 + I2 \rightarrow CHI2-CO-CH3 \]
- This step occurs three times, replacing three hydrogen atoms with iodine atoms.

6. **Formation of Iodoform**:
- The final product after the complete reaction will be iodoform (CHI3) and a carboxylate ion:
\[ CHI3 + CH3COO^{-} \]

7. **Final Products**:
- The products of the reaction are iodoform (CHI3) and sodium acetate (CH3COO^-).

### Conclusion:
The reaction of acetone with iodine in the presence of sodium hydroxide leads to the formation of iodoform (CHI3).

### Answer:
The correct answer is that acetone reacts with iodine to form iodoform in the presence of sodium hydroxide (NaOH).

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