# Oxidation of alcohols by Ceric Ammonium Nitrate

I have realized that reaction with ceric ammonium nitrate essentially results in coordination of alcohol with $$\ce{Ce(IV)}$$ and my teacher referred that we should be able to proceed it to form ketone by dissociating $$\ce{C-H}$$ of the alcohol.I have the following doubts:

A) Precise mechanism as I am literally lost as to why the $$\ce{C-H}$$ being as strong bond is dissociated. what is the driving force here?

B) What is the final oxidation state of cerium as it seems that $$\ce{Ce(III)}$$ is produced but we initially had $$\ce{Ce(IV)}$$. Is there an electron transfer mechanism here?

• It is a redox reaction. Sep 20, 2020 at 15:42

In this reaction, the ion $$\ce{Ce^{4+}}$$ is a powerful oxidant. It needs one electron to be transformed into the more stable ion $$\ce{Ce^{3+}}$$. This electron may come from the central function $$\ce{-CHOH-}$$ of any secondary alcohol $$\ce{R-CHOH-R'}$$. The bond between $$\ce{O}$$ and $$\ce{H}$$ is able to loose $$1$$ electron and $$1$$ $$\ce{H+}$$ ion. Same thing happens with the bond $$\ce{C-H}$$ belonging to the same group $$\ce{-CHOH-}$$. As a consequence, two electrons are emitted with two $$\ce{H+}$$ ions, and the remaining organic molecule becomes a ketone $$\ce{R-CO-R'}$$. The two electrons react with two ions $$\ce{Ce^{4+}}$$. The two half-equations are : $$\ce{Ce^{4+} + e^- -> Ce^{3+}}$$ $$\ce{R-CHOH-R'-> R-CO-R' + 2 H+ + 2 e-}$$ so that the final equation is : $$\ce{2 Ce^{4+} + R-CHOH-R' -> 2 Ce^{3+} + R-CO-R' + 2 H^+}$$