But-2-ene on reaction with alkaline \(\mathrm{KMnO_4}\) at elevated temperature followed by acidification will give :
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- one molecule of \(\mathrm{CH_3CHO}\) and one molecule of \(\mathrm{CH_3COOH}\)
- 2 molecules of \(\mathrm{CH_3COOH}\)
- 2 molecules of \(\mathrm{CH_3CHO}\)
The Correct Option is B
Solution and Explanation
Cold dilute alkaline \(\mathrm{KMnO_4}\) would only hydroxylate it to butane-2,3-diol (option 3), but at elevated temperature the strong oxidant cleaves the double bond.
Each \(\mathrm{CH_3CH=}\) carbon is oxidised to the carboxylate \(\mathrm{CH_3COO^-}\) (an aldehyde intermediate is oxidised further under these conditions); acidification gives \(\mathrm{CH_3COOH}\).
So one mole of but-2-ene gives 2 molecules of acetic acid.
Answer: (2) 2 molecules of \(\mathrm{CH_3COOH}\)
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