JS
Registered: 10/06/09
Posts: 18
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Reply with quote | #1 |
Professor,
The explanation for why the molecule in practice problem 8 is not aromatic is that the oxygen has two double bonds, meaning that only one of the bonds can contribute electrons to the p orbital closed loop, as the other bond must be perpendicular. I understand that. For the purposes of this question, let me refer to the electrons that are in the closed loop of p orbitals as Pz electrons and the ones that are perpendicular as Py electrons. The explanation then proceeds to say that the molecule therefore has only 4 pi electrons in the Pz orbital loop (from the O-C and C-C pi bonds), and therefore cannot be planar.
However, if we look at the nitrogen, we see that it has two electrons in the Py orbital that is bonded to the oxygen. Would that not mean that its lone pair of electrons can exist in the Pz orbital, contributing to the pi electron count and giving the molecule a total of 6 pi electrons, making it aromatic? I understand that such would make the nitrogen sp, but if the molecule as whole can become aromatic, it most certainly would, correct?
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