4
$\begingroup$

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?

Improve this question
$\endgroup$
2
  • $\begingroup$ 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 ♦
    Commented May 10, 2019 at 5:45
  • 3
    $\begingroup$ Also, Hückel's rule is for one cycle. $\endgroup$
    – Ivan Neretin
    Commented May 10, 2019 at 6:47

1 Answer 1

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$

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