Francis
Sorry yes you're right, the C with the exocyclic d.b. doesn't contribute
its p electron to the pi system, but then doesn't that break the
aromaticity since a continuous ring of contributing p orbitals is surely a
requirement?
I would say that 2-pyridone should not be classed as aromatic for the same
reason but its tautomer 2-hydroxypyridine Oc1ccccn1 clearly is. In
2-pyridone the ring C-C bond lengths alternate between conjugated single
(1.45) and double (1.34) whereas in 2-hydoxypyridine they are all around
the aromatic C-C length (1.39).
I guess it all depends on how you define 'aromatic' but as I understand it
there are 4 necessary conditions:
1. Must be cyclic.
2. Every atom in the ring must be conjugated (i.e. contributes a p orbital
to the pi system).
3. Must have 4n+2 pi electrons.
4. Ring must be planar (i.e. any stereochemical distortion breaks the
aromaticity even if the other 3 conditions are fulfilled).
You could add that bond lengths between like atom types should be about
equal, but that follows from the other conditions.
Cheers
-- Ian
On Tue, 23 Oct 2018 at 11:45, Francis Atkinson <fran...@ebi.ac.uk> wrote:
> Ian,
>
> I make it 6 electrons: two from the N, none from the C double bonded
> to the exocyclic N, and one each from four other carbons in the ring. It's
> isoelectronic with *e.g.* pyridone, which is aromatic in RDKit...
>
> In [1]: from rdkit import Chem
>
> In [2]: Chem.MolToSmiles(Chem.MolFromSmiles('O=c1[nH]cccc1'))
> Out[2]: 'O=c1cccc[nH]1'
>
> The protonated/tautomerised version are indeed aromatic
> (interconverting bewteen these species was actually how I came across this
> issue), but I still reckon the unprotonated bicyclic should be aromatic
> too...
>
> Francis
> On 23/10/2018 11:18, Ian Tickle wrote:
>
>
> Hi, it seems to me that neither is aromatic since the N-substituted hetero
> ring breaks the Huckel rule by having 7 e- (2 from the N and 1 each from
> the 5 Cs). If you remove 1 e- from the N (so it's [n+]) and also make the
> external double bond into a single (picking up a proton on the other N) it
> becomes pyridinium which is certainly aromatic.
>
> [n+]12ccccc1NCCC2
>
> [n+]12ccccc1NC.CC2
>
> What does it make of those?
>
> Cheers
>
> -- Ian
>
>
> On Tue, 23 Oct 2018 at 10:37, Francis Atkinson <fran...@ebi.ac.uk> wrote:
>
>> Hello,
>>
>> In the following pair of molecules, the bicyclic is non-aromatic,
>> whereas the 'ring-opened' analogue is aromatic...
>>
>> In [1]: from rdkit import Chem
>>
>> In [2]: Chem.MolToSmiles(Chem.MolFromSmiles('n12ccccc1=NCCC2'))
>> Out[2]: 'C1=CC2=NCCCN2C=C1'
>>
>> In [3]: Chem.MolToSmiles(Chem.MolFromSmiles('n12ccccc1=NC.CC2'))
>> Out[3]: 'CCn1ccccc1=NC'
>>
>> Notebook version:
>>
>> https://nbviewer.jupyter.org/gist/flatkinson/b88eb42510a79594a9e37042eeb7e224
>>
>> This seems counter-intuitive to me: I don't see why the pyridine in the
>> first molecule shouldn't be aromatic, just as it is in the second.
>>
>> Am I missing something here?
>>
>> Thanks,
>>
>> Francis
>>
>> --
>> Dr Francis L Atkinson
>>
>> Chemogenomics Group
>> European Bioinformatics Institute (EMBL-EBI)
>> European Molecular Biology Laboratory
>> Wellcome Genome Campus
>> Hinxton
>> Cambridge CB10 1SD
>> United Kingdom
>>
>> (01223) 494473
>>
>>
>>
>> _______________________________________________
>> Rdkit-discuss mailing list
>> Rdkit-discuss@lists.sourceforge.net
>> https://lists.sourceforge.net/lists/listinfo/rdkit-discuss
>>
> --
> Dr Francis L Atkinson
>
> Chemogenomics Group
> European Bioinformatics Institute (EMBL-EBI)
> European Molecular Biology Laboratory
> Wellcome Genome Campus
> Hinxton
> Cambridge CB10 1SD
> United Kingdom
>
> (01223) 494473
>
>
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