Oxidation of ketones involves carbon-carbon bond cleavage. Name the products formed on the oxidation of $2,5-$dimethylhexan$-3-$one.

(N/A) According to Popoff's rule,the oxidation of an unsymmetrical ketone occurs by the cleavage of the $C-C$ bond on either side of the carbonyl group,such that the smaller alkyl group stays with the carbonyl carbon.
For $2,5-$dimethylhexan$-3-$one $(CH_3-CH(CH_3)-CO-CH_2-CH(CH_3)-CH_3)$,cleavage can occur in two ways:
$1$. Cleavage between $C_2$ and $C_3$: This yields $2-$methylpropanoic acid $(CH_3-CH(CH_3)-COOH)$ and $2-$methylpropanoic acid $(CH_3-CH(CH_3)-COOH)$.
$2$. Cleavage between $C_3$ and $C_4$: This yields $3-$methylbutanoic acid $(CH_3-CH(CH_3)-CH_2-COOH)$ and propan$-2-$one $(CH_3-CO-CH_3)$.
Propan$-2-$one can undergo further oxidation to form ethanoic acid $(CH_3COOH)$ and methanoic acid $(HCOOH)$,which eventually oxidizes to $CO_2$ and $H_2O$.