What is the difference in the functioning of alkaline and acidified $\ce{KMnO4}$ as a reagent in organic chemistry?
E.g., in the oxidation of a primary alcohol to carboxylic acid, we use acidified $\ce{KMnO4}$ and not alkaline $\ce{KMnO4}$. Why?
Answer
Permanganate as oxidizing agent works most efficiently in acidic solution, because it is reduced to the greatest extent in this medium, from oxidation state +VII in $\ce{MnO4-}$ to +II in $\ce{Mn^2+}$.
$$\ce{8H+ + MnO4- + 5e- ->~ Mn^2+ + 4H2O}$$
Therefore, the number of electrons transferred from oxidized species per mole $\ce{KMnO4}$ (5 electrons) is the largest, and the least amount of $\ce{KMnO4}$ is needed for the reaction with a certain amount of alcohol. This makes oxidation in acidic medium the best choice for economic reasons.
The oxidation in neutral medium is less efficient, as manganese(VII) is only reduced to manganese(IV), yielding insoluble $\ce{MnO2}$ as a byproduct which needs to be removed from the reaction mixture during purification of the desired product (carboxylic acid).
$$\ce{2H2O + MnO4- + 3e- ->~ MnO2\downarrow + 4OH-}$$
Oxidation in strongly alkaline medium is least efficient in terms of transferred electrons per mole permanganate, as the latter is only reduced to manganate(VI).
$$\ce{MnO4- +e- ->~ MnO4^2-}$$
However, when the reaction is done under these conditions, a higher reaction speed and selectivity can be achieved, presumably because the alcohol will react more readily with $\ce{MnO4-}$ when deprotonated. Nevertheless, oxidation in acidic medium can be preferred for reasons stated above and substrates which are sensitive to strong bases.
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