The treatment of H3C−CH=CH−CH3 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 BCH3COOH only CH3CH=CHCH3+NaIO4→2CH3COOH
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.