Carbon atom shows ${sp}^{2} - hybridization$ in compounds A and B. A decolourises alkaline potassium permanganate solution whereas B cannot decolourise the solution. Then A and B are

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Propene and Ethene

Benzene and Ethene

Ethene and Benzene

answer is B.

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Detailed Solution

Carbon atom in both the compounds shows ${sp}^{2}$ hybridization but due to Aromaticity Benzene cannot decolourises the alkaline potassium permanganate solution.

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