Hi Markus, Concatenating SMILES strings may not be the best way to solve this problem. I would just create a bond between the two atoms from Python and set the charges to zero, and adjust hydrogens (if necessary).
But since you asked...:-) $ obabel -:"O=C1CCC(=O)[N-]1" -xf 7 -osmi [N-]1C(=O)CCC1=O $ obabel -:"C1=C2CCCN3CCC(=C23)C=C1[CH+]C1=CC2=C3C(=C1)CCCN3CC2" -xl 13 -osmi c1c2CCCN3CCc(c23)cc1[C+](c1cc2c3c(c1)CCCN3CC2) With Python's string replace(), replace [N-] with N, and [C+] with C (assuming only a single instance of each in the strings), then concatenate: c1c2CCCN3CCc(c23)cc1C(c1cc2c3c(c1)CCCN3CC2)N1C(=O)CCC1=O You might be wondering where I got the numbers 7 and 13 from. Well, I just counted atoms in the SMILES string from the left starting at 1. If using Python, just get the Idx() of the atom that is a nitrogen and has a positive charge (and similarly for the C+). Regards, - Noel On Sun, 17 Nov 2019 at 16:58, Markus Grimm via OpenBabel-discuss < openbabel-discuss@lists.sourceforge.net> wrote: > Hello everyone, > > I'm trying to concatenate two smiles to get the product of some reactions. > In general, it is mostly clear where the bond is formed. An example: > > Nucleophile: O=C1CCC(=O)[N-]1 > Substrate: C1=C2CCCN3CCC(=C23)C=C1[CH+]C1=CC2=C3C(=C1)CCCN3CC2 > I want to rearrange the smiles always under some conditions. In this > example depending on where the charge is located respectively. So the bond > is formed between the [N-] and the [CH+]. I know openbabel can reorder the > smiles based on the atom number. But I'm clueless about how to use it. Is > there any way how to implement my problem in pybel? > > Best regards, > Markus > _______________________________________________ > OpenBabel-discuss mailing list > OpenBabel-discuss@lists.sourceforge.net > https://lists.sourceforge.net/lists/listinfo/openbabel-discuss >
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