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I want to determine the carbons that are oxidatized in an organic reaction.

Is it an acceptable way to first figure out the oxidation number of every carbon in reactants and products and then compare the oxidation numbers (for example: $\ce{+I -> +II} =$ Oxidation; electron loss)?* Or is my attempt to work with oxidation numbers false (for example: an aldehyde and keton are on one oxidation level, but their carbon has different oxidation numbers?). Maybe I should go over the oxidation levels of functional groups given in many textbooks as tabulars (but my structures aren't that easy and have not the most common functional groups)?

*I have read in many posts here, that this is a legitimate way — but there is "my problem" with the keton and the aldehyd, as mentioned above.

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Yes, determining oxidation numbers and thereby finding out what is oxidised and what is reduced is a valid method, especially for reactions that include oxidations or reductions.

Your ‘problem’ with an aldehyde and a ketone is actually not a problem at all. Ketones have an oxidation state of $\mathrm{+II}$, which means that the only thing they could be oxidised to in a two-electron (i.e. non-radical) process would be $\ce{CO2}$ — hence why they are stable towards oxidation. Aldehydes, on the other hand, have an oxidation state of $\mathrm{+I}$, which means that they can easily be oxidised (to carboxylic acids, oxidation state $\mathrm{+III}$) by simplet two-electron processes.

Likewise, the different oxidation states of primary ($\mathrm{-I}$), secondary ($\mathrm{\pm 0}$) and tertiary ($\mathrm{+I}$) alcohols can be used to help understand why it makes sense to distinguish.

However, do note that many organic reactions — while definitely involving a change of oxidation states — do not fit into the classical redox paradigm well. Throughout the mechanism of a Horner-Wadsworth-Emmons reaction, the (tertiary) phosphonate carbon changes its oxidation state from $\mathrm{-II}$ to $\pm 0$ in two distinct, one-electron steps, while the aldehyde carbon goes from $\mathrm{+I}$ to $\mathrm{-I}$ also in two distinct steps — but nobody would go as far as calling it a redox reaction.

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