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Which among the following compounds is oxidised most easily?

  1. 2-propanol ($\ce{CH3CHOHCH3}$)
  2. phenol ($\ce{PhOH}$)
  3. ethoxyethane ($\ce{Et-O-Et}$)
  4. 1-methyl cyclohexanol

The answer given in my workbook is option 1. It's obvious to me that options 3 and 4 can't be the right answer (ether, hindrance). But I am confused between options 1 and 2.

I have arguments that can support either of the 2 options. There's less hindrance on phenol so formation of an intermediate ester like chromate ester would be easier their as compared to secondary alcohol. However, phenol is losing aromaticity as well, therefore activation energy of oxidation should be higher as compared to secondary alcohol.

Since $E_\mathrm{a}$ matters more, we would say that secondary alcohol is oxidised more easily, is it?

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Hindrance has little to do with this. Loss of aromaticity is a big deal, so the oxidation of phenol to quinone takes a high energy reagent (Ceric Ammonium Nitrate is a good one). Oxidation of a secondary alcohol by contrast is fairly easy.

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  • $\begingroup$ Btw, what does "high energy reagent" mean? $\endgroup$ – Archer Jul 13 '18 at 19:29
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    $\begingroup$ Reagent in a high oxidation state e.g Cr(VI), Ce (IV), Mn (VII), hypervalent iodine salts; further reading here scielo.br/… $\endgroup$ – Waylander Jul 13 '18 at 19:32
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    $\begingroup$ Also, choice #4 is a tertiary alcohol. What would it be oxidized to? Even if it were unhindered, oxidation is going to require breaking C-C bonds. $\endgroup$ – Curt F. Jul 13 '18 at 22:03
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    $\begingroup$ There's a good case to be made that #2 is the right answer. If I leave isopropanol in one sealed container with O2 headspace at 80 C for a month, and phenol in another container under the same conditions, at the end of the month my guess is that the phenol would be substantially less pure than the isopropanol due to oxidative degradation. $\endgroup$ – Curt F. Jul 13 '18 at 22:04

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