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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