(C) $1\%$ cold alkaline $KMnO_4$ is known as Baeyer's reagent. It reacts with ethylene $(CH_2=CH_2)$ to undergo syn-hydroxylation,resulting in the for

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(C) $1\%$ cold alkaline $KMnO_4$ is known as Baeyer's reagent. It reacts with ethylene $(CH_2=CH_2)$ to undergo syn-hydroxylation,resulting in the formation of ethylene glycol $(CH_2OH-CH_2OH)$. The reaction is: $CH_2=CH_2 + H_2O + [O] \xrightarrow{KMnO_4} CH_2OH-CH_2OH$.