How do we represent quantitatively the oxidation of isopropyl alcohol to acetone by permanganate ion?

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Answer 1 · 2018-01-16T16:30:44

Well, you gots the oxidation reaction of secondary alcohol to a ketone.....

And so.....

#H_3Cstackrel0CH(OH)CH_3 rarr H_3Cstackrel(+II)C(=O)CH_3+2H^+ + 2e^-# #(i)#

The oxidation numbers of the ipso carbons are #0# and #+II#...respectively...hence a two electron oxidation....

And permanganate is probably reduced to #Mn^(2+)#...a five electron reduction....

#underbrace(MnO_4^(-))_"deep purple"+8H^+ +5e^(-)rarrMn^(2+)+4H_2O(l)# #(ii)#

AND we add these half-equations together in such a way that the electrons are eliminated...i.e. #5xx(i)+2xx(ii)#....

#5H_3Cstackrel0CH(OH)CH_3 +2MnO_4^(-)+6H^+ rarr 2Mn^(2+)+8H_2O(l)+5H_3Cstackrel(+II)C(=O)CH_3 #

This reaction is self-indicating. Permanganate ion is deeply purple in colour....#Mn^(2+# is almost colourless....