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JEE CHEMISTRY MCQ: HYDROCARBONS

1 Hard

1 of 1

Analyzing +4 | -1

1,3-Butadiene is treated with hot

\mathrm{KMnO}_4 / \mathrm{NaIO}_4

, the product(s) formed is(are)

View answer and explanation

Correct option: C

2 \mathrm{CO}_2+\mathrm{HOOC}-\mathrm{COOH}

The correct answer is C (

2 \mathrm{CO}_2+\mathrm{HOOC}-\mathrm{COOH}

) because hot KMnO4/NaIO4 strongly oxidizes 1,3-butadiene to carbon dioxide and oxalic acid.

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