Acetone reacts with `I_(2)` in presence of `NaOH` to from

Options

A. `C_(2)H_(5)I`
B. `C_(2)H_(4)I_(2)`
C. `CHI_(3)`
D. `CH_(3)I`

Explanation

To solve the question, we will analyze the reaction of acetone with iodine (I₂) in the presence of sodium hydroxide (NaOH). This reaction is known as the iodoform reaction, which is characteristic of methyl ketones like acetone.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are acetone (CH₃COCH₃) and iodine (I₂) in the presence of sodium hydroxide (NaOH).

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

3. **Identify the Products**:
- The major product formed is iodoform (CHI₃).
- Other products include sodium iodide (NaI), sodium acetate (CH₃COONa), and water (H₂O).

4. **Balance the Reaction**:
Ensure that the number of atoms of each element is balanced on both sides of the equation. In this case, the equation is balanced as written.

5. **Conclusion**:
The main product of the reaction between acetone and iodine in the presence of sodium hydroxide is iodoform (CHI₃).

### Final Answer:
The product formed when acetone reacts with I₂ in the presence of NaOH is **iodoform (CHI₃)**.

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