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Q.

But -2- ene on reaction with Alkaline ${KMnO}_{4}$ at elevated temperature followed by acidification will give

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a

${CH}_{3} COOH + {CH}_{3} CHO$

b

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c

$2 mole of {CH}_{3} CHO$

d

$2 mole of {CH}_{3} COOH$

answer is A.

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Detailed Solution

${CH}_{3} - CH = CH - {CH}_{3} \to_{H^{+}}^{{KMnO}_{4}} 2 {CH}_{3} COOH$

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