This is a cute example. The left ring is one in which every atom and every
bond is aromatic, and yet the ring is not aromatic. Unlike azulene, in
which neither ring, alone, is aromatic
On Tue, Oct 23, 2018 at 12:36 PM Greg Landrum
wrote:
>
> I'll try later (likely tomorrow) to explain what I mea
On Tue, Oct 23, 2018 at 7:07 PM Chris Earnshaw wrote:
>
> This prompted me to see what happens with azulene, which is another case
> where the envelope is aromatic but neither of the individual rings are
> based on a simple neutral representation. This ends up being related to
> Peter's example;
On Tue, Oct 23, 2018 at 1:08 PM Chris Earnshaw wrote:
> Interesting - I do hope your idea works out!
>
> This prompted me to see what happens with azulene, which is another case
> where the envelope is aromatic but neither of the individual rings are
> based on a simple neutral representation. Th
Interesting - I do hope your idea works out!
This prompted me to see what happens with azulene, which is another case
where the envelope is aromatic but neither of the individual rings are
based on a simple neutral representation. This ends up being related to
Peter's example; the input SMILES c1c
I'll try later (likely tomorrow) to explain what I meant a bit better. Or
maybe I'll just implement it (since it seems like it could be fairly easy).
On Tue, Oct 23, 2018 at 6:13 PM Chris Earnshaw wrote:
>
> Following this analysis means you don't need to consider the resonance
> form:
> A carbo
Hi
I think my approach to this is - Is there a resonance form in which the
ring in question in unequivocally aromatic and the separated charge ends up
somewhere sensible? The 'electron stealing' concept is a sort of handy
shortcut for this.
For Greg's examples, I'd say:
[image: image.png]
I'm not
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 illustra
On Tue, Oct 23, 2018 at 5:14 PM Greg Landrum wrote:
>
> 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.
>
>
What I should have added to this: I don't think it can possibly be this
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 sa
On Tue, Oct 23, 2018 at 4:08 PM Peter S. Shenkin 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 r
This is just to note that pyridones are considered aromatic by all SMILES
kits I've seen (thought I've certainly not seen them all!), and pyridone
itself is cited in the Daylight Theory Manual as an example of an exocyclic
double bond which does not break aromaticity.
-P.
On Tue, Oct 23, 2018 at
Mea culpa - I hit Reply rather than Reply All and so only sent this to
Greg...
On Tue, 23 Oct 2018 at 13:53, Chris Earnshaw wrote:
> Hi Greg
>
> Apologies again, I'm not trying to stir things up here. As we can see from
> some of the the other discussion there's no clear view of what constitutes
Hi, Greg,
Thank you for being so open in your response, and I certainly agree with
everything you just said. Here are my thoughts.
- Easily understandable explanation:
- From the Daylight theory manual (and you've used similar
language): *exocyclic
double bonds do not break aromati
On Tue, Oct 23, 2018 at 3:00 PM Peter S. Shenkin wrote:
>
> It's difficult to fault RDKit for making the same mistake that everybody
> else blithely accepts; but it would be great, IMO, if it could do better
> than everyone else in this regard.
>
Again, I have no argument whatsoever with this. B
Hi,
I raised the same issue that Francis raised on the RDKit Slack channel on
Jan 14, 1917, with a different example (c1c[nH]c2nccc-2c1). With the same
response. Of course, breaking the non-aromatic ring causes the remaining
aromatic ring to be perceived as aromatic, as Greg's response would imply
Ian,
I think the idea is that the (out-of-plane) p orbital on the
carbonyl C is both part of the ring pi-system and the carbonyl
pi-system. However, both pi-electrons in the carbonyl 'belong to' the
oxygen because it's more electronegative, and they thus aren't counted
in the 4N+2.
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 f
Dissent is fine, but it's important to remember that there are *always*
going to be edge cases and that we're not trying to model something
physically observable here. The concept of aromaticity is primarily there
to make canonicalization easier. Section 3.4.2 here:
http://www.daylight.com/dayhtml/
Sorry about this, but I think that 'perhaps sub-optimal' should be replaced
by 'definitely wrong'. The 'quasi-aromatic' system in these two structures
is identical and should behave as such, but in practice one of them matches
a pyridine SMARTS pattern and the other doesn't. That shouldn't be affec
The current implementation requires "exocyclic" bonds to actually be
*non-ring* bonds in order to be recognized as such.
This is perhaps sub-optimal, but it's clearly defined and avoids arguments
about when exactly an "exocyclic" bond starts stealing electrons.
-greg
On Tue, Oct 23, 2018 at 12:46
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.MolFromSmile
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
b
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('n12c1=NCCC2'))
Out[2]: 'C1=CC2=NCCCN2C=C1'
In [3]: Chem.MolToSmiles(Chem.MolFromSmile
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