Master Organic Chemistry Reaction Guide

Oxidative cleavage of alkenes to ketones/carboxylic acids using KMnO4

Description: Alkenes treated with KMnO4 are cleaved into carbonyl compounds. The type of carbonyl compound depends on the substitution pattern of the alkene.

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16 thoughts on “Oxidative cleavage of alkenes to ketones/carboxylic acids using KMnO4

  1. 2 Q’s:

    1) How come my textbook likes to first add KMnO4 with OH-, then add acid later? Any difference?

    2) Can this reaction be done with H2CrO4? I notice that KMnO4 and H2CrO4 do the same job with other reactions, like oxidation of alcohols, aldehydes, or alkyl substituents of benzene rings, so why not with this reaction also?

    1. 1) Not sure. Usually KMnO4 is used with acid. However, there’s a second important reaction of KMnO4 – when you treat alkenes with COLD, DILUTE KMnO4, it will form the vicinal diol. Base is often used in the second step to get the alcohol.

      2)The reaction is generally not done with H2CrO4. I wouldn’t be surprised if it CAN be done, but H2CrO4 is not generally used – this may be because the reaction is slow.

      1. Thanks for the quick response! Is it possible to have email reminders when you give a response? Thanks!

        1. My software doesn’t currently allow for that – I’m concerned as well it will slow down my site. I’ll try to respond promptly – you can check “recent comments” on the main page

  2. Is the number of carboxylic acids formed in your product, directly dependent on the number of vinylic hydrogens? two vynil h’s = two carboxylic acids? one vynil h = carboxylic acid + ketone??

    1. Well, the maximum number of carboxylic acids would be two. If each carbon of the double bond had a C-H bond then there would be two carboxylic acids. However if one of the carbons has *two* vinylic hydrogens, then that carbon will turn into carbon dioxide.

  3. Alkene with kmno4 in acidic condition oxidized to carbonyll compound but in basic medium it oxidized to diol why and how? Plz mechanism any one know then email me

    1. The first step in the oxidation is a cycloaddition where the C-C pi bond is broken, and two new C-O bonds are formed. At low temperatures, in the presence of weak base, NaOH will attack Mn, liberating the newly-formed diol and preventing further oxidation. If the cycloadduct is allowed to warm up, oxidative cleavage results.

  4. In the mechanism, what happens to the double bond attached to the oxygen that is attacked by the alkene? Does it go to Mn?

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