so I've been designing an experiment where I will explore the effect of isopropanol concentration on the colour change of a solution containing permanganate, isopropanol, and NaOH. I know that the isopropanol will be oxidized to propanone, but I'm not sure exactly what the permanganate will be reduced to?

From what I've read online, I've seen that it could either gain 1 electron and become a green manganate (VI), or it could be reduced to turn into a brown MnO2 precipitate. What exactly will the permanganate become? The sources are kind of conflicting for me.

Any help would be greatly appreciated!

  • 1
    $\begingroup$ It depends on pH. $\endgroup$
    – Mithoron
    Feb 9 at 16:09

2 Answers 2


Isopropanol has two successive reactions with the ion permanganate $\ce{MnO4-}$ in basic solution. First it reacts by gaining one electron and becoming manganate (VI) ion $\ce{MnO4^{2-}}$ in a reaction $$\ce{2 MnO4- + C3H7OH + 2 OH- -> 2 MnO4^{2-} + C3H6O + 2 H2O}$$ But a second (and even a third) reaction may immediately occur in solution ; $$\ce{MnO4^{2-} + C3H7OH -> MnO2 + C3H6O + 2 OH-}$$ The third reaction is due to the instability of the ion manganate(VI), which is not chemically stable in solution (except in highly concentrated basic solution). Usually a disproportionation happens spontaneously according to : $$\ce{3 MnO4^{2-} + 2 H2O -> MnO2 + 2 MnO4- + 4 OH-}$$ This equation can be added one, two, three or many more times to any of the two first equations cited above. It depends on the pH.

Finally there is no simple equation for explaining the reaction of an alcohol like isopropanol on permanganate ion. It depends on the importance of the third reaction, which itself depends on the pH.

  • $\begingroup$ I assume such oxidation is performed at very basic pH, possibly to expose RO- anions, where manganate was relatively stable. I remember some manganometric back titration of organics and the intermediate solution with performed oxidation was before acidification dark green. $\endgroup$
    – Poutnik
    Feb 9 at 10:09
  • $\begingroup$ I assume the reagent of the 2nd reaction is $\ce{C3H7OH}$ ( $\ce{(CH3)2CHOH}$ ) and not $\ce{C3H6O}$ ( $\ce{CH3COCH3}$ ). // Also see Pourbaix Mn diagram, suggesting $\ce{MnO4^2-}$ stability above pH 13. $\endgroup$
    – Poutnik
    Feb 9 at 12:51
  • $\begingroup$ @Poutnik. I am sorry. My second equation was wrong. I have now changed its writing and replaced propanone by isopropanol in the reagents. Fortunately, It does not change too much the whole mechanism. $\endgroup$
    – Maurice
    Feb 9 at 13:48

The quick answer and rough rule when using potasium permanganate is that it changes oxidation number from +7 to +2 which is more favourable. It is reducing itself and acts like an oxidation agent (a very strong one). Most often, the side product would be black tar ($\ce{MnO2}$).

Anyway, in your case under basic conditions I think what Maurice said is right.


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