Ketones are relatively much harder to oxidise, but they do undergo oxidation reactions at extreme temperatures. Our teacher also told us about Popoff's rule which states that, on oxidation of unsymmetrical ketones, the $\ce{>C=O}$ group remains with the smaller alkyl group. Could you please give me a hint as to how this happens, and the probable reaction mechanism for this?

Also, please give the complete balanced equation when butan-2-one is oxidized to a carboxylic acid.

  • 6
    $\begingroup$ Your statement of extreme temperature is wrong. With proper oxidants, ketones are oxidized under mild condition. $\endgroup$ – Ian Fang Sep 6 '13 at 16:47
  • 3
    $\begingroup$ 2-Butanone to propionic acid via the haloform reaction - fast, at room temperature. You can use chlorine laundry bleach. doi:10.1021/cr60052a001 $\endgroup$ – Uncle Al Mar 11 '14 at 18:42
  • $\begingroup$ @IanFang Is this correct: Ketones undergo oxidation 'to form carboxylic acids' only at extreme temperatures and using a strong oxidising agent? Because ketones do undergo oxidation at mild conditions. For example, alpha-hydroxy ketones oxidises even using mild oxidation agent (Tollens' Reagent). Am I right? $\endgroup$ – Inquisitive Oct 31 '20 at 19:22
  • $\begingroup$ According to popoff's rule, the unsymmetrical ketone on oxidation, C-C bond cleavage and keto group goes with (CH3)2 - CH - group $\endgroup$ – Divyansh Verma Dec 4 '20 at 12:29

I am uncertain about the "Popoff's" rule you mention. There are two reactions that can oxidize ketones, and one seems to follow the behavior you are suggesting, but does not form a carboxylic acid. The other reaction does form carboxylic acids, but is more complex.

The Baeyer-Villager oxidation is an oxidation of ketones to esters using a peracid in the presence of a mild base:

$$\ce{CH3COCH3 + RCO3H->[\ce{Na2HPO4}] CH3CO-OCH3 + RCOOH}$$

The mechanism involves the fragmentation of one of the $\ce{C-CO}$ bonds. Unsymmetrical ketones fragment in a predictable pattern, but not always that the carbonyl remains with the smaller group. The fragment that would be a more stable carbocation (even though carbocations are not formed in this reaction) is the one to move. The migratory aptitude is tertiary alkyl > cyclohexyl > secondary alkyl, aryl > H > primary alkyl > methyl . For exampl, with 2-methyl-3-octanone, the isopropyl group moves because it is attached via a $2^\circ$ carbon:

$$\ce{(CH3)2CHCOCH2CH2CH2CH2CH3 + RCO-O-OH ->[\ce{Na2HPO4}] (CH3)2CH{\bf O}COCH2CH2CH2CH2CH3 + RCOOH}$$

Nitric acid chews ketones apart into two carboxylic acids. Concentrated $\ce{KMnO4}$ in acid also does this. These reactions are a little bit harder to find information on, since they tend to be considered uncontrollable.

The reaction goes through a series of oxidations from ketone to $\alpha$-hydroxyketone to cleaved acids through several enol intermediates.

$$\ce{RCH2COCH2R <=>[\ce{HNO3}] RCH=C(OH)CH2R ->[\ce{HNO3}] RCH(OH)COCH2R}$$ $$\ce{RCH(OH)COCH2R <=>[\ce{HNO3}] RC(OH)=C(OH)CH2R ->[\ce{HNO3}] RCO}$$

If the ketone is unsymmetrical, there is no guarantee that is will cleave predictably. For example, 2-butanone could cleave into propanoic acid and carbon dioxide or two equivalents of acetic acid.

$$\ce{CH3CH2COCH3 ->[\ce{HNO3}] n(CH3CH2COOH + CO2) + (1-n)(2CH3COOH)}$$

A balanced equation for the formation of acetic acid would look like: $$\ce{CH3CH2COCH3 + 3NO3^- + -> 2 CH3CO2H + 3NO2^-}$$

A balanced equation for the formation of propanoic acid and carbon dioxide would look like: $$\ce{CH3CH2COCH3 + 4NO3^- + -> CH3CH2COOH + CO2 + 4NO2^- + H2O}$$

The permanganate reactions are tougher to balance, since permanganate is a three electron oxidant.


When I wrote this answer, I had never heard of Popoff's rule. No textbook I own mentions this rule, and a Google search about it brings up this question as the top hit (and similar questions at other sites like ask.yahoo.com as the other hits). I now know that Aleksandr Popov published a paper in Liebigs Annalen in 1872 describing a variation of the reaction that would become the Baeyer-Village oxidation. This article is behind a paywall for me, and the first page preview confirms I would not be able to make much of it. I only know a small amount of German and the PDF sadly looks like a low quality copy of a copy of a copy.

However, from the title "Die Oxydation der Ketone als Mittel zur Bestimmung der Constitution der fetten Säuren und der Alkohole" I can parse what the paper was about. Roughly this paper was about "The oxidization of ketones - a means for determining the constitution of the fatty acids and alcohols". Thus, his method involved the oxidation to the ester and hydrolysis of said ester in one step. I don't know what reagents Popov used, but I'm pretty sure it was not a peroxyacid (since this reagent will not hydrolyze the ester). I have previously answered a question about migratory aptitude in Baeyer-Villager reactions. Since I cannot read Popov's paper, nor can I find any authoritative resource on his rule (on- or offline), I have to assume that the known migratory aptitude for the Baeyer-Village reaction is the same as Popov's rule.

  • $\begingroup$ Isn't the major product(2 acids) predicted by popoff's rule? $\endgroup$ – evil999man Mar 10 '14 at 13:17
  • 4
    $\begingroup$ Even if you spoke german, this is old language and hard to understand (and the quality does not help either). However, here is what he used to oxidise: $30~\mathrm{g}~\ce{K2Cr2O7}, 20~\mathrm{g}~\ce{H2SO4}, 100~\mathrm{g}~\ce{H2O}$ for $6.1~\mathrm{g}$ butyl phenyl ketone (I think), followed by azeotropic distillation. Apparently they form (after some more steps) benzoe-, isobutyric- and acedic acid. (However this might happen...) I did not read much further as it pained my eyes and brain. I hope this unfogs the mystery of the paper a little. (BTW: there was a previous version in russian.) $\endgroup$ – Martin - マーチン May 30 '14 at 8:02
  • 4
    $\begingroup$ Popoff's rule: just pop off the smaller substituent when oxidizing a ketone! $\endgroup$ – Dissenter Jan 1 '15 at 15:08
  • 1
    $\begingroup$ Confirming what @Mart said, although I have read enough of these papers to not find them hard (and I am fascinated by the old spelling variants). Popov states a rule at the beginning that if a phenyl or methyl (or sometimes ethyl) ketone is oxidised, then the oxidation breaks the $\ce{C-C}$ bond on the other side of the ketone; so propyl methyl ketone would give acetic acid and propionic acid. Not sure if that is the rule or not. $\endgroup$ – Jan Oct 21 '15 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.