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

CH₄

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C₂H₆

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C₂H₄

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C₂H₂

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Solution

The correct option is C
C₂H₄

We know that,

Alkyl halide o reaction with alcoholic KOH or NaOH gives alkene.

$R - C H_{2} - C H_{2} - X + K O H (a l c) ⟶ R - C H = C H_{2} + K X + H_{2} O$

And

Hydroxylation using $K M n O_{4} : 1 %$ cold alkaline $K M n O_{4}$ solution (Baeyer's reagent) or $d i l . K M n O_{4}$ gives glycol

Note.

(a) It is an unsaturation test because the pink colour of solution disappears.

(b) It is a example of Syn-addition.

Therefore,

$C H_{3} C H_{2} l + K O H (a l c) ⟶ C H_{2} = C H_{2} + K l + H_{2} O$

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

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A g N O_{3}

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