A gas formed by the action of alcoholic KOH on ethyl iodide decolourises alkaline $K M n O_{4}$ . The gas is :

C_{2} H_{6}

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C H_{4}

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C_{2} H_{2}

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C_{2} H_{4}

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Solution

The correct option is D $C_{2} H_{4}$
In presence of alcoholic $K O H$ , ethyl iodide undergoes dehydrohalogenation, to form ethylene $C_{2} H_{4}$ . Ethylene decolourises alkaline $K M n O_{4}$ .

Alkaline $K M n O_{4}$ is called Baeyer's reagent and is reduced to $M n O_{2}$ by ethylene.

$C_{2} H_{5} - I a l c . K O H - ------------ \to D e h y d r o h a l o g e n a t i o n C_{2} H_{4}$

$C H_{2} = C H_{2} + a l k a l i n e K M n O_{4} \to C H_{2} (O H) - C H_{2} - O H$

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Similar questions

K M n O_{4}

K M n O_{4}

A gas formed by the action of alcoholic KOH on ethyl iodide, decolorizes alkaline KMnO₄ solution. The gas is

C_{2} H_{6}, C_{2} H_{4}

C_{2} H_{2}

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