A gas formed by the action of alcoholic KOH on ethyl iodide, decolourises alkaline `KMnO_(4)` solution. The gas is

Options

A. `CH_(4)`
B. `C_(2)H_(6)`
C. `C_(2)H_(4)`
D. `C_(2)H_(2)`

Explanation

To determine the gas formed by the action of alcoholic KOH on ethyl iodide, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are ethyl iodide (C2H5I) and alcoholic KOH. Ethyl iodide is an alkyl halide, and KOH is a strong base.

### Step 2: Understand the Reaction Type
When alcoholic KOH is used with alkyl halides, it typically leads to an elimination reaction rather than a substitution reaction. This is because the alcoholic medium favors the formation of alkenes.

### Step 3: Write the Reaction
The elimination reaction of ethyl iodide with alcoholic KOH can be represented as follows:
\[ \text{C}_2\text{H}_5\text{I} + \text{KOH} \rightarrow \text{C}_2\text{H}_4 + \text{KI} + \text{H}_2\text{O} \]
Here, ethyl iodide loses a hydrogen atom and the iodine atom, resulting in the formation of ethylene (C2H4), which is an alkene.

### Step 4: Identify the Gas
The product of this reaction is ethylene (C2H4), which is a gas at room temperature.

### Step 5: Analyze the Decolourization of Alkaline KMnO4
The question states that the gas decolourizes alkaline KMnO4 solution. Alkaline KMnO4 is a strong oxidizing agent and is typically purple in color. Ethylene (C2H4) can undergo oxidation, which leads to the formation of ethylene glycol (C2H6O2) or other products, resulting in the decolourization of the solution.

### Conclusion
Thus, the gas formed by the action of alcoholic KOH on ethyl iodide, which also decolourizes alkaline KMnO4, is ethylene (C2H4).

### Final Answer
The gas is **C2H4 (ethylene)**.

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