The compound that decolourises cold, dilute alkaline $K M n O_{4}$ is:

C_{6} H_{6}

(Benzene)

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C_{6} H_{12}

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C_{6} H_{14}

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C C l_{4}

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Solution

The correct option is C $C_{6} H_{6}$ (Benzene)
$C_{6} H_{12}$ (Alkene) decolourises cold, dilute ${K M n O}_{4}$ (Alkaline). The $r \times n$ is called hydroxylation, pink colour of ${K M n O}_{4}$ goes off and brown ${M n O}_{2}$ precipitates.

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Carbon atom shows $s p^{2}$ - hybridization in compound A and B. A decolourises alkaline $K M n O_{4}$ solution, where as B cannot. A and B are

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