When ethyl alcohol and KI reacted in presence of `Na_(2)CO_(3)` yellow crystals of ……. Are formed

Options

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

Answer

To solve the question regarding the reaction of ethyl alcohol with KI in the presence of Na2CO3, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this reaction are ethyl alcohol (C2H5OH) and potassium iodide (KI), with sodium carbonate (Na2CO3) acting as a catalyst.

2. **Understand the Reaction Type**: This reaction is a classic example of the iodoform reaction. It involves the conversion of a secondary alcohol into iodoform (CHI3) when treated with iodine in the presence of a base.

3. **Reaction Mechanism**:
- Ethyl alcohol (C2H5OH) is first converted to its corresponding alkyl halide, which in this case is ethyl iodide (C2H5I) when it reacts with KI.
- The presence of Na2CO3 helps in the deprotonation of the alcohol, facilitating the substitution reaction.
- The ethyl iodide can then undergo further reactions leading to the formation of iodoform.

4. **Formation of Iodoform**: The key product of this reaction is iodoform (CHI3). This compound is characterized by its yellow crystalline form.

5. **Conclusion**: Therefore, when ethyl alcohol reacts with KI in the presence of Na2CO3, the yellow crystals formed are iodoform (CHI3).

### Final Answer:
The yellow crystals formed are iodoform (CHI3).

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