Acetylene on reaction with Alkaline $K M n O_{4}$ forms:

C H_{2} = C H_{2}

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oxalic acid

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C H_{3} C O O H

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C O_{2} + H_{2} O

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Solution

The correct option is A oxalic acid
Ethyne or acetylene on reaction with cold $K M n O_{4}$ adds 4 $O H^{-}$ to the triple bond. It is oxidised by a dilute aqueous solution of potassium permanganate to form oxalic acid.

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Similar questions

In this reaction alkaline $K M n O_{4}$ behaves as:

$C H_{3} - C H_{2} - O H a l k a l i n e K M n O_{4} + h e a t - -------------- \to {C H}_{3} - C O O H$

C H_{3} - C H_{2} - O H a l k a l i n e K M n O_{4} + H e a t - ------------- \to C H_{3} - C O O H

K M n O_{4}

K M n O_{4}

{CH}_{3} {CH}_{2} OH (aq) X \to {CH}_{2} = {CH}_{2} (g) + H_{2} O(l)

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