Carbon atoms show `sp^(2)` hybridisation in compounds A and B. A decolourises alkaline potassium pemanganate solution whereas B cannot decolourise the solution. Then A and B are

Options

A. Propene and Ethene
B. Ethene and Benzene
C. Benzene and Ethene
D. Ethene and Propene

Answer

To solve the problem, we need to analyze the information given about compounds A and B, which both show `sp²` hybridization in their carbon atoms.

### Step-by-Step Solution:

1. **Identify the Characteristics of Compounds A and B**:
- Both compounds A and B have carbon atoms that are `sp²` hybridized.
- Compound A can decolorize alkaline potassium permanganate (KMnO₄) solution, while compound B cannot.

2. **Understanding the Reaction with KMnO₄**:
- Alkaline potassium permanganate is a strong oxidizing agent. It typically reacts with alkenes (which have carbon-carbon double bonds) to form diols, resulting in the decolorization of the purple KMnO₄ solution.
- Compounds that do not react with KMnO₄ are usually either saturated hydrocarbons (like alkanes) or compounds that do not have the necessary functional groups for oxidation.

3. **Analyzing Possible Compounds**:
- **Compound A**: Since it decolorizes KMnO₄, it is likely an alkene (like propene or ethene) that can undergo oxidation.
- **Compound B**: Since it does not decolorize KMnO₄, it is likely a compound that does not react with KMnO₄, such as benzene, which is an aromatic compound.

4. **Conclusion**:
- Based on the above analysis, we can conclude that:
- Compound A is likely ethene (C₂H₄) or propene (C₃H₆), as both are alkenes and can decolorize KMnO₄.
- Compound B is benzene (C₆H₆), which does not react with KMnO₄ and thus does not decolorize the solution.

5. **Final Answer**:
- Therefore, the answer is:
- A = Ethene (C₂H₄)
- B = Benzene (C₆H₆)