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Q.

4 - methyl - pent - 2-ene when heated with acidified KMnO₄ gives

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a

One mole each of carboxylic acid and aldehyde

b

Two moles of carboxylic acid

c

Two moles of ketone

d

One mole each of carboxylic acid and ketone

answer is C.

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Detailed Solution

${CH}_{3} - CH ({CH}_{3}) - CH = CH - {CH}_{3} \to_{∆}^{{KMnO}_{4} / H^{+}} {CH}_{3} - CH ({CH}_{3}) - COOH + HOOC - {CH}_{3}$

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