`C_2H_5 I` + alcoholic potash `to X` , Here , X is

Options

A. acetylene
B. methylene
C. ethylene
D. propylene

Explanation

To solve the problem, we need to analyze the reaction between C₂H₅I (ethyl iodide) and alcoholic potash (alcoholic KOH).

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is C₂H₅I, which is ethyl iodide.
- The reagent is alcoholic potash, which is a solution of potassium hydroxide (KOH) in alcohol.

2. **Understand the Reaction Type**:
- The reaction involves a nucleophilic substitution where the iodide ion (I⁻) from ethyl iodide is replaced by a hydroxyl group (OH⁻) from KOH.
- However, since the reaction is in alcoholic medium, it can also lead to elimination reactions.

3. **Determine the Product**:
- In the presence of alcoholic KOH, C₂H₅I undergoes dehydrohalogenation, leading to the formation of an alkene.
- The elimination of HI (hydrogen iodide) occurs, resulting in the formation of ethylene (C₂H₄), which can be represented as CH₂=CH₂.

4. **Byproducts**:
- Along with ethylene, the byproducts of this reaction are potassium iodide (KI) and water (H₂O).

5. **Conclusion**:
- Therefore, the product X formed from the reaction of C₂H₅I with alcoholic potash is ethylene (C₂H₄).

### Final Answer:
X is ethylene (C₂H₄).

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