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.
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-Villiger 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.
Update:
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-Villiger 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-Villiger 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-Villiger reaction is the same as Popov's rule.
