I agree that are potential gotchas, and even if we can't think of them, someone else might, which is one of the reasons that I think that, even following any due diligence we are able to accomplish, the facility, if implemented, should be subject to a runtime flag.
In your three graphical illustrations, I can't think of any reason that the ring on the left should not be aromatic in all cases. Yes, we do need to say that the exocyclic double bond, to be considered, needs to be to an electronegative element, certainly N or O, though this reference seems to indicate that S (a little surprisingly to me) should also qualify. But to paraphrase Freud, "Sometimes an edge case is just an edge case." You wrote: Considering the Kekule form of a structure: - If a C atom is valence saturated and has a double bond to a "more electronegative atom" (let's agree that N and O meet this definition and then argue about other things later), it contributes zero pi electrons to whatever ring system it's in. - If a "more electronegative atom" is valence saturated and has a double bond to a C, then it contributes two electrons to whatever ring system it's in. So the first atom in each of your two cases is the ring atom? So the first point refers to the C=O carbon in pyridone and the second refers to the N in pyridine? (Just asking for clarification here.) Personally, the way I think of 2-pyridone aromaticity is more from the point of view of the ring N. (The operative word may be "personally.) The C=O is willy-nilly stuck in sp2 hybrdization. In the VB structure, then, N, with 3 single bonds to it, appears to be sp3-hybridized at a first glance. But because of O's electronegativity, that C=O has significant C-[O-] character, leaving a vacant p orbital. Then, to achieve the thermodynamic stability that aromaticity provides, the ring C-N(-H)-C moiety can hybridize to sp2, as C=[N+](-H)-C, where the N has donated two electrons into the bond that has just become double. This picture helps me understand, chemically, why the exocyclic double bond has to be to an electronegative atom in order to confer aromaticity; namely, it leaves the p-orbital on the C from which the double bond emanates vacant (in that resonance structure), which allows the ring N to rehybridize. An off-ring =CH2 would not do that; the corresponding C-[CH1-] resonance structure would not be sufficiently stable to contribute materially. Given this, all three structures in your illustration exhibit an off-ring =N, so all should make the left ring aromatic. I'm not sure whether this addresses the concerns you expressed about the 2nd and 3rd structures in the illustration. The kind of gotcha I am afraid of is a situation where, in two condensed rings, a double bond to the right ring might make the left ring aromatic but a double bond to the left ring might make the right ring aromatic, but somehow they could not be simultaneously aromatic. Thus, then, atom ordering might dictate the result, and if that's the case, what would the right answer be, chemically? Might neither of them actually be aromatic? -P. On Tue, Oct 23, 2018 at 11:15 AM Greg Landrum <greg.land...@gmail.com> wrote: > hmmm, thinking about this I believe I'm coming to a simpler (and > efficient) scheme for this after all... > > It's going to take me a bit to formalize, and I would want to test it on a > bunch of molecules, but I *think* this works. > > Considering the Kekule form of a structure: > - If a C atom is valence saturated and has a double bond to a "more > electronegative atom" (let's agree that N and O meet this definition and > then argue about other things later), it contributes zero pi electrons to > whatever ring system it's in. > - If a "more electronegative atom" is valence saturated and has a double > bond to a C, then it contributes two electrons to whatever ring system it's > in. > > That certainly handles the things we've discussed so far, as well as easy > cases like pyridine and quinone. Now I need to try and find some stuff that > breaks it. > > -greg > > > On Tue, Oct 23, 2018 at 5:08 PM Greg Landrum <greg.land...@gmail.com> > wrote: > >> >> >> On Tue, Oct 23, 2018 at 4:08 PM Peter S. Shenkin <shen...@gmail.com> >> wrote: >> >>> >>> - Easily understandable explanation: >>> - From the Daylight theory manual (and you've used similar >>> language): *exocyclic double bonds do not break aromaticity.* >>> - I'd alter this to *double bonds exocyclic to the ring in >>> question do not break aromaticity*. (I.e., even if they are in >>> other rings) >>> - Beyond this, conventional electron counting explains everything >>> in Francis's example and mine. >>> - >>> >>> You're close, but I think there's something missing. >> Exocyclic double bonds do not prevent an atom from being considered >> aromatic, but they *may* "steal" a pi-electron - e.g. the C that's double >> bonded to the O in pyridone contributes zero electrons to the aromatic >> ring. The challenge here is to define which exocyclic double bonds can do >> this. >> >> For example, you guys are agreeing that the N exocyclic bond next to the >> boxed C here: >> [image: image.png] >> does remove an electron. >> >> and, to go all the way in the other direction, what happens here: >> [image: image.png] >> And here: >> [image: image.png] >> Is that left ring aromatic in all cases? If not, why not? >> >> -greg >> >
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