Baeyer's reagent oxidises ethylene to

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Ethylene chlorohydrins

Ethyl alcohol

${CO}_{2} and H_{2} O$

Ethane – 1, 2 – diol

answer is D.

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Detailed Solution

The reaction in the image demonstrates the hydroxylation of ethene ( $C_2H_4$ $C$ $2$ $$ $H$ $4$ $$ ) using Baeyer's reagent, which is an alkaline solution of potassium permanganate ( $KMnO_4$ $KMnO$ $4$ $$ ). Here’s the breakdown of the solution:

Reactant: Ethene ( $C_2H_4$ $C$ $2$ $$ $H$ $4$ $$ ), an alkene.
Reagent: Baeyer's reagent ( $KMnO_4$ $KMnO$ $4$ $$ in alkaline conditions).
Product: Ethane-1,2-diol ( $CH_2OH-CH_2OH$ $CH$ $2$ $$ $OH$ $-$ $CH$ $2$ $$ $OH$ ), a diol.

Explanation of the Process:

Ethene reacts with Baeyer's reagent to undergo hydroxylation.
The double bond in ethene breaks, and each carbon of the double bond gets a hydroxyl group ( $-OH$ $-$ $OH$ ).
This reaction is used as a qualitative test for the presence of double bonds (unsaturation) since $KMnO_4$ $KMnO$ $4$ $$ gets decolorized in the reaction.