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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?

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    $\begingroup$ It is a redox reaction. $\endgroup$
    – Mathew Mahindaratne
    Sep 20, 2020 at 15:42

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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 : $$\ce{2 Ce^{4+} + R-CHOH-R' -> 2 Ce^{3+} + R-CO-R' + 2 H^+}$$

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  • $\begingroup$ ok i get the mechanism of this, the driver is the powerful oxidising nature of ce4+ right? however this mech makes me ask one more qn are you implying that 2 moles of CAN is required? my highschool book say it require 1 mol per alcohol. can you refer a book or a site which says otherwise? $\endgroup$
    – napstablook
    Sep 20, 2020 at 16:56
  • $\begingroup$ @ Napstablook. Your high school book is wrong. I am sorry to tell it this way. It is impossible to oxidize one mole alcohol with one mole Cerium(IV). $\endgroup$
    – Maurice
    Sep 20, 2020 at 20:26

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