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

Options

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

Answer

To solve the question regarding the reaction of acetone with iodine (I₂) in the presence of sodium hydroxide (NaOH), we can follow these steps:

### Step-by-Step Solution:

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

2. **Understand the Reaction Type**: This reaction is known as the iodoform reaction, which occurs when a methyl ketone (like acetone) reacts with iodine in the presence of a base (NaOH).

3. **Write the Balanced Reaction**: The balanced chemical equation for the reaction is:
\[ CH₃COCH₃ + 3I₂ + 4NaOH \rightarrow CHI₃ + 3NaI + CH₃COONa + 3H₂O \]
- Here, CHI₃ is iodoform, NaI is sodium iodide, CH₃COONa is sodium acetate, and H₂O is water.

4. **Identify the Main Product**: The main product formed in this reaction is iodoform (CHI₃).

5. **Identify Byproducts**: The byproducts of this reaction are sodium iodide (NaI), sodium acetate (CH₃COONa), and water (H₂O).

6. **Conclusion**: Therefore, when acetone reacts with iodine in the presence of NaOH, the main product formed 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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