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Neil Morrison

Hock rearrangement of hydroperoxides

Darla  Follow

The main parts of the mechanism that will affect the rate of the reaction are the following steps (full mechanism on Wikipedia): abstraction of a hydrogen to form a benzylic radical and the rearrangement of the of the protonated peroxide.These steps will be slower as they involve the formation of the higher energy species when ethylbenzene is used.

benzylic radical formation

Secondary benzylic hydrogen have a higher bond bond dissociation energies ($357.5\: \mathrm{kJ\: mol^{-1}}$) than tertiary benzylic hydrogen ($353.3\: \mathrm{kJ\: mol^{-1}}$)$^{[1]}$. This will diminish the rate at which the initiation step proceeds, though a difference of $4.2\: \mathrm{kJ\: mol^{-1}}$ shouldn't make or break the reaction.

carbocation formation

The formation of a primary carbocation would otherwise be highly unfavorable, except for that it is bonded to an oxygen atom. A lone pair on the oxygen can therefore be shared with the carbocation, thereby stabilizing it. This step will have a higher activation energy, but likely all that will be necessary for this reaction to occur is a little more heat and time.


[1] Baciocchi, E.; D'acunzo, F.; Galli, C.; Lanzalunga, O. Tertiary : Secondary : Primary C–H Bond Relative Reactivity in the One-Electron Oxidation of Alkylbenzenes. A Tool to Distinguish Electron Transfer from Hydrogen Atom Transfer Mechanisms. J. Chem. Soc., Perkin Trans. 2. 1996, 133–140.

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Grady Green  Follow
of the Hydroperoxide? Im wondering if, once formed, the hydroperoxide of EB would split up or not via the Hock Rearrangement.More
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Jakob Persson  Follow
Thanks! But isnt this addressing the More
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Finn Murray  Follow
@curious_cat I think the second half covers what youre looking for specifically.More
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Joann Doe  Follow
formationMore
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Joel Harvey  Follow
@curious_cat The hydroperoxide will form nearly instantly once the benzylic cation is formed. It is not part of the rate-determining step.More
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Angel Wallace  Follow

If the image below is correct (I believe it is), then higher substitution = more stable carbocation = more likely to react this way.

hock rearrangement/cumene process

That being said, a carbocation adjacent to an oxygen atom like that is pretty stabilized, so it may not require a tertiary carbon. I'm not sure what side reactions would be promoted by a non-tertiary carbocation, though.

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Doug Robb  Follow
Actually, this picture is wrong. It applies, but not perfectly. Will fix when Im back to my computer.More
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