I have read that quinone systems are conjucated but not anti-aromatic. I had read the related questions:

  1. Is the aromaticity broken in some resonance structures of para-nitro-aminobenzene?
  2. Is a quinone ring aromatic?.

In one of those answers, I read a point which said that it was non aromatic:

[...] and you must be able to draw double bonds - internal to the ring - from each ring carbon atom. [...]

I don't get what this means as I had never heard of such a thing in Hückel's rule. I did read elsewhere that the system must be completely conjugated which is true in this case but due to cross conjugation the electrons are not delocalised throughout the system? I don't get why cross conjugation affects the delocalisation of the electrons this much? What is the proper explanation for this?


1 Answer 1


The reason you must be able to draw double bonds within the ring is the $4n+2$ rule applies rigorously only when the conjugated atoms are all in such a ring. If you are drawing double bonds to adjacent atoms off the ring then you don't have all the conjugated pi electrons in the ring anymore, and then there is no longer a $4n+2$ rule for aromaticity or a $4n$ rule for antiaromaticity.

In some systems it is possible to render contributing structures that have all the conjugated electrons in the ring and meet the $4n+2$ rule, in which case you can assert that this contribution is enhanced by aromaticity and the properties of the compound affected accordingly. For example, the aromatic contribution to cyclopropenones leads to enhanced polarity of the carbonyl group and increased basic character of the oxygen atom, compared with ordinary ketones.

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Illustration source: https://commons.m.wikimedia.org/wiki/File:C3H2OResonance.png. Drawn by WP user Smokefoot. Link to License

Quinones, however, cannot be given such a structure without leaving the highly electronegative oxygen short of an octet, so they lack any significant aromatic contribution.


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