July 02, 2022, 12:26:48 PM
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Topic: Oxidation to form o-semiquinone radical  (Read 263 times)

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Offline cairenrs

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Oxidation to form o-semiquinone radical
« on: January 30, 2022, 07:41:18 PM »
I'm trying to understand this figure from a paper I'm reading, but I'm stuck on this part of their proposed reaction mechanism (see green box in attached image). As someone who barely passed organic chemistry 6 years ago, I have two questions.
1) How come the reaction starts at the meta hydroxyl group, but then switches to the hydroxl at the para position?
2) The paper states that the autooxidation reaction forms an o-semiquinone radical that quickly turns into o-quinone. But I thought the ortho-quinone would look like the image I have on the right - how come the authors consider their product as o-quinone when it clearly has an additional oxygen?
Any explanations or insight would be much appreciated!

Offline Babcock_Hall

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Re: Oxidation to form o-semiquinone radical
« Reply #1 on: January 31, 2022, 10:45:59 AM »
I only have time for a short reply right now, but one thing to keep in mind is that frequently only one of two or more resonance forms is drawn out; the rest are implicit.

Offline natalie5933

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Re: Oxidation to form o-semiquinone radical
« Reply #2 on: February 03, 2022, 03:23:15 AM »
A) In order to form the semiquinone the meta and para position are both involved
B) Is is one of the resonance form of 0-quinone. if u google the resonance forms you will see what I am trying to say.

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