enter image description here

I don't think it follows the $4n + 2$ rule. The double bonds give 8 π electrons. Then the lone pairs on the $\ce{NH}$ and $\ce{NH2}$ groups are delocalized (are they?) so they must contribute 4 π electrons. Thus in total there are 12 π electrons, which doesn't follow Hückel's rule. So, why is it aromatic?

  • $\begingroup$ Hückel's rule is for conjugated circuits, so $\ce{NH2}$ group is irrelevant, whereas $\ce{NH}$ is (lone pair on $\mathrm{2p}_z$ orbital), resulting in $n = 10$ and overall aromaticity. $\endgroup$
    – andselisk
    May 10, 2019 at 5:45
  • 3
    $\begingroup$ Also, Hückel's rule is for one cycle. $\endgroup$ May 10, 2019 at 6:47

1 Answer 1


The $4n+2$ rule applies only for a conjugated ring. If you assume that the amino group contributes a pair of pendant $\pi$ electrons into the ring system then perforce you do not have a conjugated ring. And even if you don't assume such a delocalization from the amino group, the cross-link within the ring system is still a stretch for the $4n+2$ rule.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.