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Question

The treatment of H3CCH=CHCH3 with NaIO4 (or) boiling KMnO4 produces :

A
CH3CHO only
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B
CH3COOH only
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C
CH3CHO and CH3COOH
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D
CH3COOH and HCOOH
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Solution

The correct option is B CH3COOH only
CH3CH=CHCH3+NaIO42CH3COOH

Alkene on oxidation in presence of an oxidizing agent such as boiling KMnO4 undergoes oxidation to form diol as a product which on further oxidation produces aldehyde/ketone as a product.

These aldehyde on further oxidation produces a carboxylic acid. Here the alkene is symmetrical thus produces carboxylic acid only.

Option B is correct.

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