AAMC Practice Exam 2 Question 6 on Chemical & Physical Foundations Section

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Sep 15, 2016
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How is it that the following product is formed? If electrons from both oxidized hydroxyls on both sides of compound 3 undergo resonance, N-N would end up with a triple bond, wouldn't they?

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Forgive me for oversimplifying it, but the way I look at it and would probably answer it on the MCAT:

2 (2 (-OH)) + O2 ---laccase---> 2 (=O) + 2 H2O.

That's where I stop. Resonance be damned.

If I try to do resonance, I start with the =O, pushing one of the double bonds up to the O to become a lone pair to make it -O (-), then that pulls bonds until a lone pair comes off of one of the N so that it becomes roughly:

(Benzene)-CH=N=N=CH-(Benzene) with each benzene having the 2 -OCH3 and 1 -O(-). No idea if that's right, but I also don't see how it's important. If someone can illuminate it for me, I'd love to learn something new.