Alkaline `KMnO_4` converts ethylene into

Options

A. ethane
B. ethanol
C. methanol
D. ethylene glycol

Explanation

To solve the question "Alkaline `KMnO_4` converts ethylene into," we will go through the reaction step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant in this case is ethylene, which has the molecular formula `C2H4`. It can be represented as `CH2=CH2`.

2. **Identify the Reagent**:
- The reagent is alkaline potassium permanganate, commonly represented as `KMnO4` in an aqueous (water) solution. This reagent is known for its oxidizing properties.

3. **Understand the Reaction Conditions**:
- The reaction typically occurs in an alkaline medium and at a temperature range of approximately 298 to 303 Kelvin.

4. **Reaction Mechanism**:
- When ethylene reacts with alkaline `KMnO4`, it undergoes syn-dihydroxylation. This means that the double bond in ethylene is broken, and two hydroxyl (–OH) groups are added across the double bond.

5. **Formation of the Product**:
- The product formed from this reaction is ethylene glycol, which is also known as ethane-1,2-diol. Its structure can be represented as `CH2(OH)–CH2(OH)`.

6. **Conclusion**:
- Therefore, the product of the reaction between ethylene and alkaline `KMnO4` is ethylene glycol.

### Final Answer:
The correct answer is **ethylene glycol**.

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