Thank you so much for your very helpful reply - that is exactly what I was looking for :)
I very much enjoyed reading the article on the NextMove blog, too. Best Uli On Mon, Aug 22, 2022 at 8:26 PM John Mayfield <john.wilkinson...@gmail.com> wrote: > Something to note from SMILES there are no radicals, only undervalent > atoms. Which means your formula works correctly but from other formats > (e.g. MDL) you get an explicit unpaired electron added to the container. > Simple rules will get you pretty far and there are utilities like the CDK > AtomTypeMatcher which provide a global model but I would write what you > need since different valence models exist for different formats and based > on surrounds e.g. oxide's things change. > > I will caution against "robo chemistry > <https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6086778/>" where you try to > guess what the correct answer is, for example to be a negative charge looks > more reasonable: > > [Na+].O=C1[N-]C=CC=2C=CC(Br)=CC12.FC(F)(F)CI > > Either way - I would probably avoid the getValency() field and instead > switch on the atomic number and guard against unusual charges: > > int explValence = atomContainer.getBondOrderSum(atom); > switch (atomicNum) { > case 6: > if (charge == 0) > max(4 - explValence, 0); > break; > case 7: > if (charge == 0 && explValence > 3) > max(5 - explValence, 0); > else if (charge == 0) > max(3 - explValence, 0); > break; > } > > Further reading: > > MDL valence model: > https://www.ics.uci.edu/~dock/manuals/oechem/pyprog/mdlvalence.html > Further Reading: > > https://nextmovesoftware.com/blog/2013/02/27/explicit-and-implicit-hydrogens-taking-liberties-with-valence/ > OPSIN has a good valence checker: > > https://github.com/dan2097/opsin/blob/c827542501516a70fb91dce09bc1b275ddddb80d/opsin-core/src/main/java/uk/ac/cam/ch/wwmm/opsin/ValencyChecker.java > > John > > On Mon, 22 Aug 2022 at 08:47, Uli Fechner <u...@pending.ai> wrote: > >> Hi, >> >> I came across an issue today that seemed straightforward at the >> beginning, but after a while ceased to appear that easily accessible. Well, >> I probably shouldn't be surprised - I guess that is just cheminformatics at >> its best :) >> >> The following smiles popped up in my workflow: >> [Na+].O=C1[N]C=CC=2C=CC(Br)=CC12.FC(F)(F)CI >> >> This translates to the sole nitrogen (valency = 3, SP2) being a radical >> with no implicit hydrogen and two neighboring carbon atoms both of which >> are connected by single bonds. >> >> Irrespective of how that radical got there I want to 'remove' it by just >> adding an implicit hydrogen to the nitrogen atom. >> >> This then led to the more general question of how to remove radicals for >> common organic elements (C, N, O, P, S seems like a good start). >> >> I came up with the following formula: >> >> int numberOfUnpairedElectrons = (int) (atom.getValency() - >> atomContainer.getBondOrderSum(atom) + atom.getFormalCharge() - >> atom.getImplicitHydrogenCount()); >> if (numberOfUnpairedElectrons % 2 != 0) { >> atom.setImplicitHydrogenCount(atom.getImplicitHydrogenCount() + 1); >> } >> >> As this is chemistry, I am sure there are a lot of exceptions - even if >> the elements of interest are very restricted. >> >> Is the formula above a reasonable simplification? Or am I oversimplifying >> this? >> >> Best >> Uli >> _______________________________________________ >> Cdk-user mailing list >> Cdk-user@lists.sourceforge.net >> https://lists.sourceforge.net/lists/listinfo/cdk-user >> >
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