Isoalloxazine is called aromatic in literature. It's a part of FAD cofactor. In some PDB structures it has flat geometry (3DK9) in some cases it has not (3GYJ).

I've applied Hückel's $4n+2$ rule for $\pi$ electrons to check aromaticity. I draw my current guess about $\pi$ electrons with red dots and lone pairs with blue clouds:

  • I've found a total of 20 $\pi$ electrons, which suggests non-aromaticity.
  • On the left ring, I've identified 6 $\pi$ electrons, indicative of aromaticity.
  • N10 is anticipated to be $sp^3$ hybridized disrupting aromatic system. If N10 is $sp^2$ hybridized it contributes 2 $\pi$ electrons.
  • N3 atoms are likely $sp^2$ hybridized due to the presence of 2 $\ce{C=O}$ groups.

Which part of isoalloxazine is aromatic?

Isoalloxazine electronic structure

  • 1
    $\begingroup$ This is a somewhat interesting and useful exercise, but ultimately doesn't give much result, as far as actual aromaticity is concerned. N10 is conjugated and you can draw an antracene-like mesomeric structure. This unfortunately doesn't prove anything - doesn't say how major a contributor it is. $\endgroup$
    – Mithoron
    Commented Aug 26, 2023 at 14:08

1 Answer 1


Whoever instructs students to use the $4n+2$ rule on a molecule such as this deserves to be placed on a rack and pulled until he confesses that the rule does not really apply.

First off, technically the rule should be applied only to single conjugated rings. If we fudge our way past that minor detail by ignoring the internal carbon-carbon bond, there are those pendant carbonyl double bonds. Pendant pi bonding also means you don't have a ring, and so can't properly apply the $4n+2$ rule.

If we are really willing/forced to stretch things:

As described here, if polarizing carbonyl groups to eliminate pendant pi bonds involving them ($\ce{C=O->\overset{+}{C}–\overset{-}{O}}$) results in an aromatic structure, then the molecular structure will in fact favor such a polarization. If we do that here, and if we count the remaining conjugated pi electrons properly, we get not $20$ but $14$:

  • $6$ electrons in the left ring

  • $2$ additional electrons from the $\ce{N–R}$ nitrogen in the center ring but not from the other nitrogen, whose lone pair is outward-pointing

  • $4$ additional electrons fem the pi bonds drawn in the middle and right rings

  • $2$ additional electrons from the nitrogen on the right ring.

We can argue that a chain of three conjugated rings with $14$ pi electrons like anthracene would be expected to show aromatic character, but we had to pull several muscles to get there. Better to do a more sophisticated molecular orbital analysis, preferably backed by experimental measurements such as nmr chemical shifts of hydrogen atoms bonded to the ring system, to know for sure.


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