Why is alkaline $\ce{KMnO4}$ used in the oxidation of toluene to benzoic acid? Can acidified or neutral $\ce{KMnO4}$ be used in this conversion?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Related: Mechanism of arene side chain oxidation by permanganate $\endgroup$– orthocresolMar 25, 2018 at 17:06
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
Here are the three equations describing the reduction of manganese (and concurrent oxidation of whatever substrate may be present) under basic, neutral and acidic conditions respectively.
$$\ce{Mn^{+7}O4- +e- ->~ Mn^{+6}O4^2-~~~~[basic]}$$ $$\ce{2H2O + Mn^{+7}O4- + 3e- ->~ Mn^{+4}O2 + 4OH-~~~~[neutral]}$$ $$\ce{8H+ + Mn^{+7}O4^{-} + 5e- ->~ Mn^{+2} + 4H2O~~~~[acidic]}$$
Acidic conditions are the most economical as 5 electrons are transferred per mole of manganese, but oxidation can be achieved under all 3 conditions. A problem with neutral conditions is the precipitation of insoluble $\ce{MnO2}$ and the need for its subsequent separation. A problem with all 3 conditions is that oxidation will only occur if the material to be oxidized is somewhat soluble in the reaction medium.
The milder basic conditions are generally preferred for the oxidation of aromatic alkyl side chains because with the milder basic conditions an organic co-solvent that is soluble in water can be employed to facilitate solubility of the organic substrate.
