Re: [Rdkit-discuss] Aromaticity question

[Peter S. Shenkin]

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 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 carbonyl or imine (open chain or in a partially saturated ring) group's
>> carbon atom provides (effectively) 0 pi electrons
>> 'Ordinary' aromatic carbons provide 1 pi-electron
>> Pyridine-type nitrogens (2-connections) or pyridinium (3-connections)
>> provide 1 pi-electron
>> Pyrrole-type nitrogens (3-connections) provide 2 pi-electrons.
>>
>> So for example 3, that's
>> [image: image.png]
>> 10 pi-electrons in total, so the *system* is aromatic even though the
>> electron counts look odd on a per-ring basis (not unlike azulene).
>>
>
> On this specific instance: This is exactly what the RDKit currently does.
> However, in this scheme the left ring, which has 7 pi electrons, is not
> aromatic even though both the right ring and the envelope are.
>
> -greg
>
>
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Re: [Rdkit-discuss] Aromaticity question

[Greg Landrum]

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; the input SMILES c1c2c1ccc2 gets converted to
> c1cc2cc-2c1, where the fusion bond is perceived as 'pure single' rather
> than aromatic as it should be, so we do indeed end up with two non-aromatic
> rings embedded in an aromatic envelope.
>
> Is there any way to define the aromaticity of an individual ring based on
> its mapping to an aromatic envelope? Are there any cases where that
> wouldn't be true?
>

No, there isn't currently a way to do this. In the RDKit's aromaticity
model (explained on the page I referenced earlier), the individual rings
for azulene are seen as containing 5 and 7 electrons. This is not an
aromatic electron count. The 10 electrons of the envelope gets you the
aromaticity.

If anyone is deeply upset by this well-documented behavior w.r.t. fused
rings, it's certainly possible to add your own aromaticity model to the
RDKit.

-greg
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Re: [Rdkit-discuss] Aromaticity question

[Peter S. Shenkin]

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. This ends up being related to
> Peter's example; the input SMILES c1c2c1ccc2 gets converted to
> c1cc2cc-2c1, where the fusion bond is perceived as 'pure single' rather
> than aromatic as it should be, so we do indeed end up with two non-aromatic
> rings embedded in an aromatic envelope.
>

Yes; and it's cool that crystallography shows that the fusion bond  is
considerably longer than the ring bonds. However, a common representation
of azulene is as a cyclopentadienyl anion fused to a tropylium cation. But
this picture consider the fusion bond to be aromatic.

> Is there any way to define the aromaticity of an individual ring based on
> its mapping to an aromatic envelope? Are there any cases where that
> wouldn't be true?
>

In naphthalene, the fusion bond and both rings would be considered
aromatic, if that's what you're getting at.

[image: image.png]
>>>
>>
This is a very cute example, because every bond and every atom is aromatic,
yet one ring composed of fully aromatic bonds and atoms is not (considered)
aromatic. (But if you "pyridonize" the left ring, it becomes aromatic, too.)

-P.
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Re: [Rdkit-discuss] Aromaticity question

[Chris Earnshaw]

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 c1c2c1ccc2 gets converted to
c1cc2cc-2c1, where the fusion bond is perceived as 'pure single' rather
than aromatic as it should be, so we do indeed end up with two non-aromatic
rings embedded in an aromatic envelope.

Is there any way to define the aromaticity of an individual ring based on
its mapping to an aromatic envelope? Are there any cases where that
wouldn't be true?

Chris

On Tue, 23 Oct 2018 at 17:35, Greg Landrum  wrote:

>
> 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 carbonyl or imine (open chain or in a partially saturated ring) group's
>> carbon atom provides (effectively) 0 pi electrons
>> 'Ordinary' aromatic carbons provide 1 pi-electron
>> Pyridine-type nitrogens (2-connections) or pyridinium (3-connections)
>> provide 1 pi-electron
>> Pyrrole-type nitrogens (3-connections) provide 2 pi-electrons.
>>
>> So for example 3, that's
>> [image: image.png]
>> 10 pi-electrons in total, so the *system* is aromatic even though the
>> electron counts look odd on a per-ring basis (not unlike azulene).
>>
>
> On this specific instance: This is exactly what the RDKit currently does.
> However, in this scheme the left ring, which has 7 pi electrons, is not
> aromatic even though both the right ring and the envelope are.
>
> -greg
>
>
>
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Re: [Rdkit-discuss] Aromaticity question

[Greg Landrum]

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 carbonyl or imine (open chain or in a partially saturated ring) group's
> carbon atom provides (effectively) 0 pi electrons
> 'Ordinary' aromatic carbons provide 1 pi-electron
> Pyridine-type nitrogens (2-connections) or pyridinium (3-connections)
> provide 1 pi-electron
> Pyrrole-type nitrogens (3-connections) provide 2 pi-electrons.
>
> So for example 3, that's
> [image: image.png]
> 10 pi-electrons in total, so the *system* is aromatic even though the
> electron counts look odd on a per-ring basis (not unlike azulene).
>

On this specific instance: This is exactly what the RDKit currently does.
However, in this scheme the left ring, which has 7 pi electrons, is not
aromatic even though both the right ring and the envelope are.

-greg
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Re: [Rdkit-discuss] Aromaticity question

[Chris Earnshaw]

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 sure I understand the
'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'
statement. The key thing is probably that the 'more electronegative atom'
has to be outside the conjugated, potentially aromatic ring-system.
Following this analysis means you don't need to consider the resonance
form:
A carbonyl or imine (open chain or in a partially saturated ring) group's
carbon atom provides (effectively) 0 pi electrons
'Ordinary' aromatic carbons provide 1 pi-electron
Pyridine-type nitrogens (2-connections) or pyridinium (3-connections)
provide 1 pi-electron
Pyrrole-type nitrogens (3-connections) provide 2 pi-electrons.

So for example 3, that's
[image: image.png]
10 pi-electrons in total, so the *system* is aromatic even though the
electron counts look odd on a per-ring basis (not unlike azulene).

I hope that makes sense!

Chris

On Tue, 23 Oct 2018 at 16:17, Greg Landrum  wrote:

>
> 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
> simple, so I'm guessing those examples are out there.
>
> -greg
>
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Re: [Rdkit-discuss] Aromaticity question

[Peter S. Shenkin]

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 illustrations, I can't think of any reason that the
ring on the left should not be aromatic in all cases.

Yes, we do need to say that the exocyclic double bond, to be considered,
needs to be to an electronegative element, certainly N or O, though this
reference seems to indicate that S (a little surprisingly to me) should
also qualify. But to paraphrase Freud, "Sometimes an edge case is just an
edge case."

You wrote:

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.


So the first atom in each of your two cases is the ring atom? So the first
point refers to the C=O carbon in pyridone and the second refers to the N
in pyridine? (Just asking for clarification here.)

Personally, the way I think of 2-pyridone aromaticity is more from the
point of view of the ring N. (The operative word may be "personally.)  The
C=O is willy-nilly stuck in sp2 hybrdization. In the VB structure, then, N,
with 3 single bonds to it, appears to be sp3-hybridized at a first glance.
But because of O's electronegativity, that C=O has significant C-[O-]
character, leaving a vacant p orbital.

Then, to achieve the thermodynamic stability that aromaticity provides, the
ring C-N(-H)-C moiety can hybridize to sp2, as C=[N+](-H)-C, where the N
has donated two electrons into the bond that has just become double. This
picture helps me understand, chemically, why the exocyclic double bond has
to be to an electronegative atom in order to confer aromaticity; namely, it
leaves the p-orbital on the C from which the double bond emanates vacant
(in that resonance structure), which allows the ring N to rehybridize. An
off-ring =CH2 would not do that; the corresponding C-[CH1-] resonance
structure would not be sufficiently stable to contribute materially.

Given this, all three structures in your illustration exhibit an off-ring
=N, so all should make the left ring aromatic. I'm not sure whether this
addresses the concerns you expressed about the 2nd and 3rd structures in
the illustration.

The kind of gotcha I am afraid of is a situation where, in two condensed
rings, a double bond to the right ring might make the left ring aromatic
but a double bond to the left ring might make the right ring aromatic, but
somehow they could not be simultaneously aromatic. Thus, then, atom
ordering might dictate the result, and if that's the case, what would the
right answer be, chemically? Might neither of them actually be aromatic?

-P.

On Tue, Oct 23, 2018 at 11:15 AM Greg Landrum 
wrote:

> 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 
> wrote:
>
>>
>>
>> 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 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 

Re: [Rdkit-discuss] Aromaticity question

[Greg Landrum]

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
simple, so I'm guessing those examples are out there.

-greg
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Re: [Rdkit-discuss] Aromaticity question

[Greg Landrum]

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  wrote:

>
>
> 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 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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Re: [Rdkit-discuss] Aromaticity question

[Greg Landrum]

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 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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Re: [Rdkit-discuss] Aromaticity question

[Peter S. Shenkin]

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 10:12 AM Chris Earnshaw 
wrote:

> 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 aromaticity in these cases. I'm of the school which says that
>> pyridone is at least somewhat aromatic because, in crude terms, the
>> electronegative carbonyl oxygen 'steals' the electron from the carbon, the
>> carbon provides an empty p-orbital to the conjugated ring, and the ring
>> nitrogen provides a pair of electrons - hence 4n + 2 and aromatic.
>>
>> However, the thing that really worries me is that the 'iminopyridine'
>> ring in n12c1=NCCC2 *should* be perceived in the same way as in
>> n12c1=NC.CC2 but in practice that doesn't happen - one matches the
>> pyridine SMARTS c1n1 and the other doesn't. This seems to be
>> potentially dangerous. The question of 'aromatic or not' is interesting,
>> but I'm actually more concerned with the consequences for compound
>> searching and filtering.
>>
>> As an approach, rather than simply checking if the exocyclic bond is in
>> another ring (of any type), would it be possible to check if that other
>> ring is fully conjugated? If it is, then the Huckel rule could/should be
>> applied to the fused system to determine aromaticity. If not, then the fact
>> the substituents form a ring is irrelevant and the potential aromatic
>> should be treated in the same way as the non-fused analogue. This would
>> avoid the current inconsistency, but there would no doubt still be some
>> challenging edge cases...
>>
>> Best regards,
>> Chris
>>
>> On Tue, 23 Oct 2018 at 12:43, Greg Landrum 
>> wrote:
>>
>>> 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/doc/theory/theory.smiles.html has more
>>> info about this, as does the RDKit documentation:
>>> http://rdkit.org/docs/RDKit_Book.html#aromaticity
>>>
>>> I'm willing to change the current behavior, but someone would need to
>>> explain what it should be changed to in a way that is clear, unambiguous,
>>> and that would allow a human being looking at the structure to relatively
>>> easily figure out whether or not a given ring is aromatic.
>>>
>>>
>>> On Tue, Oct 23, 2018 at 1:17 PM Chris Earnshaw 
>>> wrote:
>>>
 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 affected by whether the saturated substituents form a ring or
 not. I do appreciate that it gets messy to deal with as fused rings may be
 fully conjugated, but the current behaviour seems to be disturbingly
 inconsistent. It would be suboptimal to say that no exocyclic bond is
 allowed to steal electrons, but that may be better than what's happening
 here.

 Apologies for the dissent!

 Chris Earnshaw



 On Tue, 23 Oct 2018 at 11:57, Greg Landrum 
 wrote:

> 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 PM Francis Atkinson 
> 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]1'))
>> Out[2]: 'O=c1[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 

Re: [Rdkit-discuss] Aromaticity question

[Chris Earnshaw]

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
> aromaticity in these cases. I'm of the school which says that pyridone is
> at least somewhat aromatic because, in crude terms, the electronegative
> carbonyl oxygen 'steals' the electron from the carbon, the carbon provides
> an empty p-orbital to the conjugated ring, and the ring nitrogen provides a
> pair of electrons - hence 4n + 2 and aromatic.
>
> However, the thing that really worries me is that the 'iminopyridine' ring
> in n12c1=NCCC2 *should* be perceived in the same way as in
> n12c1=NC.CC2 but in practice that doesn't happen - one matches the
> pyridine SMARTS c1n1 and the other doesn't. This seems to be
> potentially dangerous. The question of 'aromatic or not' is interesting,
> but I'm actually more concerned with the consequences for compound
> searching and filtering.
>
> As an approach, rather than simply checking if the exocyclic bond is in
> another ring (of any type), would it be possible to check if that other
> ring is fully conjugated? If it is, then the Huckel rule could/should be
> applied to the fused system to determine aromaticity. If not, then the fact
> the substituents form a ring is irrelevant and the potential aromatic
> should be treated in the same way as the non-fused analogue. This would
> avoid the current inconsistency, but there would no doubt still be some
> challenging edge cases...
>
> Best regards,
> Chris
>
> On Tue, 23 Oct 2018 at 12:43, Greg Landrum  wrote:
>
>> 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/doc/theory/theory.smiles.html has more
>> info about this, as does the RDKit documentation:
>> http://rdkit.org/docs/RDKit_Book.html#aromaticity
>>
>> I'm willing to change the current behavior, but someone would need to
>> explain what it should be changed to in a way that is clear, unambiguous,
>> and that would allow a human being looking at the structure to relatively
>> easily figure out whether or not a given ring is aromatic.
>>
>>
>> On Tue, Oct 23, 2018 at 1:17 PM Chris Earnshaw 
>> wrote:
>>
>>> 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 affected by whether the saturated substituents form a ring or
>>> not. I do appreciate that it gets messy to deal with as fused rings may be
>>> fully conjugated, but the current behaviour seems to be disturbingly
>>> inconsistent. It would be suboptimal to say that no exocyclic bond is
>>> allowed to steal electrons, but that may be better than what's happening
>>> here.
>>>
>>> Apologies for the dissent!
>>>
>>> Chris Earnshaw
>>>
>>>
>>>
>>> On Tue, 23 Oct 2018 at 11:57, Greg Landrum 
>>> wrote:
>>>
 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 PM Francis Atkinson 
 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]1'))
> Out[2]: 'O=c1[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+]12c1NCCC2
>
> [n+]12c1NC.CC2
>
> What does it make of those?
>
> 

Re: [Rdkit-discuss] Aromaticity question

[Peter S. Shenkin]

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 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.
   - I've heard it asserted that introducing this modification would
   significantly slow down aromaticity perception. That is a real
   consideration which of course you can evaluate better than I.
   - This would be a major change and I would recommend being able to turn
   it on via a run-time flag before making it the default behavior.
  - This would allow user testing for performance and for the emergence
  of hidden gotchas at either the chemical or the computational level.

I hope this is helpful, at least as a starting point for discussion.

-P.



On Tue, Oct 23, 2018 at 9:08 AM Greg Landrum  wrote:

>
>
> 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. But a half-assed fix is
> worse than doing nothing. So in order for me to do something about it I
> need:
> "someone to explain how things should be changed to in a way that is
> clear, unambiguous, and that would allow a human being looking at the
> structure to relatively easily figure out whether or not a given ring is
> aromatic."
>
>
>
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Re: [Rdkit-discuss] Aromaticity question

[Greg Landrum]

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. But a half-assed fix is
worse than doing nothing. So in order for me to do something about it I
need:
"someone to explain how things should be changed to in a way that is clear,
unambiguous, and that would allow a human being looking at the structure to
relatively easily figure out whether or not a given ring is aromatic."
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Re: [Rdkit-discuss] Aromaticity question

[Peter S. Shenkin]

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.
Chemically, this is not an edge case.

I agree with Chris that this is just wrong. However, it might be some
consolation (if that's the word I want) that back when I had access to
other packages, I tried the same example in all of them and got the same
result in every case. Even Daylight's web-site depiction scheme gave the
same result. I would love to know what OpenEye does, and in any case, there
are a lot more SMILES implementations out there now than there were then,
and I'ld also love to know whether any of the others get it (to my mind,
and Chris's, and Francis's) right.

The usual response quotes Dave Weininger's comment in the Daylight Theory
manual:

It is important to remember that the purpose of the SMILES aromaticity
detection algorithm is for the purposes of chemical information
representation only! To this end, rigorous rules are provided for
determining the "aromaticity" of charged, heterocyclic, and electron
deficient ring systems. The "aromaticity" designation as used here is not
intended to imply anything about the reactivity, magnetic resonance
spectra, heat of formation, or odor of substances.

There are at least two reasonable responses:

1. What do we use SMARTS for? Is it not to locate compounds with common
substructures? Is that not predicated on the notion that at least to some
extent, common substructures will be correlated with common activity? An
implementation that fails to see these examples as aromatic will fail to do
find the aromatic substructure these compounds when an aromatic SMARTS is
used for the search.

2. If, per the quotation above, the only purpose is really
canonicalization, why do we to bother to aromatize pyrrole in SMILES? After
all, it's already unambiguous without doing so. Would an implementation
that did not perceive pyrrole as aromatic be considered acceptable? If not,
then why is this example acceptable? Perhaps because such examples are
rare. But are we not, in fact, usually looking for unusual examples of
compounds containing a desired substructure?


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.

-P.


On Tue, Oct 23, 2018 at 8:15 AM Francis Atkinson  wrote:

> 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.
>
> I am sure that explanation would pain a theoretical chemist, but, as
> Greg has pointed out, this is as much an informatics issue as a chemistry
> issue.
>
> The RDKit aromaticity perception is quite an inclusive one: others (
> *e.g.* Biovia's) are less so and wouldn't count pyridone as aromatic.
>
> Francis
> On 23/10/2018 12:48, Ian Tickle wrote:
>
>
> 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
> requirementI would say that 2-pyridone should not be classed as aromatic
> for the same reason but its tautomer 2-hydroxypyridine Oc1n1 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  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]1'))
>> Out[2]: 'O=c1[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 

Re: [Rdkit-discuss] Aromaticity question

[Francis Atkinson]

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.


    I am sure that explanation would pain a theoretical chemist, but, 
as Greg has pointed out, this is as much an informatics issue as a 
chemistry issue.


    The RDKit aromaticity perception is quite an inclusive one: others 
(/e.g./ Biovia's) are less so and wouldn't count pyridone as aromatic.


        Francis

On 23/10/2018 12:48, Ian Tickle wrote:


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 requirementI would say that 2-pyridone should not 
be classed as aromatic for the same reason but its tautomer 
2-hydroxypyridine Oc1n1 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 > 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]1'))
Out[2]: 'O=c1[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+]12c1NCCC2

[n+]12c1NC.CC2

What does it make of those?

Cheers

-- Ian


On Tue, 23 Oct 2018 at 10:37, Francis Atkinson mailto: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('n12c1=NCCC2'))
Out[2]: 'C1=CC2=NCCCN2C=C1'

In [3]: Chem.MolToSmiles(Chem.MolFromSmiles('n12c1=NC.CC2'))
Out[3]: 'CCn1c1=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



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Chemogenomics Group
European Bioinformatics Institute (EMBL-EBI)
European Molecular Biology Laboratory
Wellcome Genome Campus
Hinxton
Cambridge CB10 1SD
United Kingdom

(01223) 494473


--
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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Re: [Rdkit-discuss] Aromaticity question

[Greg Landrum]

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/doc/theory/theory.smiles.html has more info
about this, as does the RDKit documentation:
http://rdkit.org/docs/RDKit_Book.html#aromaticity

I'm willing to change the current behavior, but someone would need to
explain what it should be changed to in a way that is clear, unambiguous,
and that would allow a human being looking at the structure to relatively
easily figure out whether or not a given ring is aromatic.


On Tue, Oct 23, 2018 at 1:17 PM Chris Earnshaw  wrote:

> 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 affected by whether the saturated substituents form a ring or
> not. I do appreciate that it gets messy to deal with as fused rings may be
> fully conjugated, but the current behaviour seems to be disturbingly
> inconsistent. It would be suboptimal to say that no exocyclic bond is
> allowed to steal electrons, but that may be better than what's happening
> here.
>
> Apologies for the dissent!
>
> Chris Earnshaw
>
>
>
> On Tue, 23 Oct 2018 at 11:57, Greg Landrum  wrote:
>
>> 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 PM Francis Atkinson 
>> 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]1'))
>>> Out[2]: 'O=c1[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+]12c1NCCC2
>>>
>>> [n+]12c1NC.CC2
>>>
>>> What does it make of those?
>>>
>>> Cheers
>>>
>>> -- Ian
>>>
>>>
>>> On Tue, 23 Oct 2018 at 10:37, Francis Atkinson 
>>> 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('n12c1=NCCC2'))
 Out[2]: 'C1=CC2=NCCCN2C=C1'

 In [3]: Chem.MolToSmiles(Chem.MolFromSmiles('n12c1=NC.CC2'))
 Out[3]: 'CCn1c1=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



 ___
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>>> --
>>> 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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>>> Rdkit-discuss@lists.sourceforge.net
>>> https://lists.sourceforge.net/lists/listinfo/rdkit-discuss
>>>
>> ___
>> Rdkit-discuss mailing list
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>> 

Re: [Rdkit-discuss] Aromaticity question

[Greg Landrum]

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 PM Francis Atkinson  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]1'))
> Out[2]: 'O=c1[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+]12c1NCCC2
>
> [n+]12c1NC.CC2
>
> What does it make of those?
>
> Cheers
>
> -- Ian
>
>
> On Tue, 23 Oct 2018 at 10:37, Francis Atkinson  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('n12c1=NCCC2'))
>> Out[2]: 'C1=CC2=NCCCN2C=C1'
>>
>> In [3]: Chem.MolToSmiles(Chem.MolFromSmiles('n12c1=NC.CC2'))
>> Out[3]: 'CCn1c1=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
>>
>>
>>
>> ___
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>> 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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>
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