Question

What happens when primary secondary and tertiary alcohols are oxidized separately by using acidified kmn o4?

Answer 1

Well, usually you cannot oxidize a tertiary alcohol.....

Explanation:

But we use acidified `KMnO_4`, a POTENT oxidant.....whose reduction reaction is given......

`underbrace(MnO_4^(-))_"deep purple" +8H^+ + 5e^(-) rarrMn^(2+) + 4H_2O` `(1)`

And primary alcohols are USUALLY oxidized straight up to the carboxylic acid....for instance...`C(-I)rarrC(+III)`:

`RCH_2OH(aq) +H_2O(l) rarr RCO_2H(aq) +4H^(+) + 4e^(-)` `(2)`

And we add the individual redox reactions in the usual way to eliminate the electrons....

`4xx(1) + 5xx(2):`

`5RCH_2OH(aq) +5H_2O(l) +4MnO_4^(-) +32H^+ + 20e^(-) rarr 4Mn^(2+) + 16H_2O+5RCO_2H(aq) +20H^(+) + 20e^(-)`

To give finally.....

`5RCH_2OH(aq) +4MnO_4^(-) +12H^+ rarr 4Mn^(2+) + 11H_2O+5RCO_2H(aq)`

The which (I think) is balanced with respect to mass and charge....And we see the strongly coloured permanganate ion dissipate to give pale, colourless `Mn^(2+)`....

Now of course secondary alcohols are oxidized up to the ketone, a 2-electron oxidation....

`RCH(OH)R rarr RC(=O)R+2H^+ +2e^(-)` `(3)`

I will let you try to balance the reduction with permanganate ion.....

Tertiary alcohols should be reasonably inert to oxidation.....