Why does potassium permanganate oxidize formic acid and oxalic acid, but has no reaction with benzoic acid and acetic acid?

The reactions were heated in a water bath (~70-80 °C). Oxalic acid had turned colorless:

$$\ce{C2H2O4 + KMnO4 + 6 H+ -> CO2 + H2O + Mn^2+ + KOH}$$

The same reaction could be said for formic, but with the formation of a brownish precipitate:

$$\ce{HCOOH + KMnO4 + 6 H+ -> CO2 + H2O + MnO2 + KOH}$$

For both the benzoic and acetic acid, they remained purple. I do not know whether the structure could be the reason for not reacting at that certain temperature, or something. Could it be possible that the temperature alone is responsible for not allowing the redox reaction to occur for both acetic acid and benzoic acid?

I looked around online and read that alkanes and aromatic compounds don't react with $\ce{KMnO4}$ from one source (Experiment 5 – Reactions of Hydrocarbons, laney.edu). I was not aware of the fact that potassium permanganate could oxidize anything under certain conditions, thank you for that fact.

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    $\begingroup$ We have a policy which states that ‎you should show your thoughts, effort and attempts to answer your question yourself. It'll make us certain that ‎we aren't doing your homework for you, and that the Q/A is beneficial for broad audience. As "homework-like questions" are considered literal homeworks, self-study questions, puzzles, worked examples etc. Please edit in your full reasoning or thoughts on this. See Homework. Otherwise, the question may get closed as "low OP effort homework-like question." $\endgroup$
    – Poutnik
    Nov 14, 2019 at 14:24
  • $\begingroup$ Under the right conditions, KMnO4 will oxidize $any$ organic matter. You have to specify conditions and details. Also clarify the context of your question. $\endgroup$
    – AChem
    Nov 14, 2019 at 15:47
  • $\begingroup$ Your equations are basically wrong for two reasons. 1) It is simply impossible to produce a base like KOH from the reaction of an acid like HCOOH or H2C2O4. An acid cannot produce a base. If this would happen, the first molecule of the base will be immediately destroyed by the next molecule of acid. 2) The charges cannot disappear in chemistry. In your second equation, you have a total of 6 positive charges at the left-hand side, and no charges at all at the right-hand side. The same error occurs in your first equation ! $\endgroup$
    – Maurice
    Nov 14, 2019 at 17:50
  • $\begingroup$ Ah my apologies, I assumed this was right as my teacher had written these. But disregarding my equations, what could be the reason for the non-reaction of the two substances? Or could I not have gotten any conclusive results because of my equations? $\endgroup$ Nov 14, 2019 at 23:43


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