Formation of products from the Reaction of KMnO4 with Alkenes

Question

Please help draw or explain what will be the products that from the reaction of KMnO₄ with the following alkenes at an elevated temperature under acidic conditions:

a) propene

b) cis-2-butene

c) trans-2-butene

Answer 1

Reaction of KMnO₄ with alkenes at elevated temperature under acidic conditions leads to oxidative cleavage of the double bond to form a pair of carbonyl (ketone and/or aldehyde) products. Aldehyde products oxidize further.

If a ketone is formed it stays a ketone, for example

(H₃C)₂C=C(CH₃)₂ —→ (H₃C)₂C=O + O=C(CH₃)₂ (two molecules of acetone)

If an aldehyde is formed after oxidative cleavage it gets further oxidized to a carboxylic acid, for example

H₃C–CH=CH–CH₃ —→ [H₃C–CH=O + O=CH–CH₃] —→ H₃C–CO₂H + HO₂C–CH₃

The alkene oxidizes to a pair of aldehydes which each oxidize further to give a pair of carboxylic acids

If formic acid (H–CO₂H) is formed it gets further oxidized to CO₂ (H₂O is also formed). For example

H₂C=CH–CH₃ —→ [H–CH=O + O=CH–CH₃] —→ [H–CO₂H] + HO₂C–CH₃ —→ CO₂ + HO₂C–CH₃

Above I put molecules in brackets to indicate they are intermediates that react further.

So the answers are

a) propene

H₃C–CH=CH₂ —→ H₃C–CO₂H + CO₂ + H₂O

b) cis-2-butene and c) trans-2-butene give the same result

H₃C–CH=CH–CH₃ —→ 2 H₃C–CO₂H

When the double bond is cleaved it doesn't matter whether it started out cis or trans.

I hope this helps. Just add a comment if you have any questions.