-1
$\begingroup$

compound

On a test, I marked this as aromatic, but the answer key says it's wrong. My thought process:

  • It has 10pi electrons, so it follows Huckel's rule
  • All carbons are sp2 hybridized
  • It is cyclic
  • Every atom is conjugated

Everything points towards it being aromatic, is there something I'm missing?

Is there some sort of rule that says each individual ring must be aromatic for the whole molecule to be aromatic?

Any help would be greatly appreciated.

$\endgroup$
2
  • 3
    $\begingroup$ This is a carbene not carbanion, it changes a lot. $\endgroup$
    – Mithoron
    Feb 9 at 16:43
  • $\begingroup$ Because all of the hydrogens are implied at all other carbons not bearing the electron pair, it is fair to say that there is a hydrogen attached to the carbon in question. In which case the carbanion would be aromatic. $\endgroup$
    – user55119
    Feb 14 at 17:40

1 Answer 1

2
$\begingroup$

The electron pair on the right is meant to be outward pointing in the plane of the atoms, not part of the pi system. If you had the electrons in the pi system, then the carbon would be attached to a hydrogen atom and would have a negative formal charge; we would write this as $\ominus$ rather than as the electron pair.

So there are only eight conjugated pi electrons, and you can work from there.

$\endgroup$
2
  • 1
    $\begingroup$ Are there only eight electrons in the pie though? I don't think this is as easy as this. If there are 8 electrons in pi orbitals, this would make it anti-aromatic (in weird cases of thinking and extending Hückel's rules beyond the useful). But that couldn't be the ground state. So, there is probably a benzene ring with "intact aromaticity" and a fused second ring with a triplet carbene, which has some sort of delocalisation, 3c-3e bonding or so. There are means to look at this more thoroughly, but they don't fit a textbook narrative. It's a shame we don't have a citation for the exercise… $\endgroup$ Feb 9 at 23:57
  • $\begingroup$ @Martin you may be right, but I am rolling with the punch. I am guessing the OP is supposed to be applying the 4n+2 rule for aromaticity however inaccurate that may be for a polycycle. In any event your model could also be called antiaromatic in the sense that it spontaneously localizes groups of pi electrons rather than allowing a full delocalization. $\endgroup$ Feb 10 at 0:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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