John,
Thanks. Your reply is actually very helpful. Yes, I just care whether there
is no overall charge. If the whole compound is neutral, I am supposed to do
nothing with it. The case mentioned is from a compound database. I need to
make sure that each compound is neutral before my program goes to the next
step. In your last example, I believe either output would be enough to me.
My knowledge of Inchi is very limited. Could you please help me to verify
that if a Inchi has no layer starting with /p or /q, this Inchi must be
neutral?
I am also confused by the following part:
My guess would be the correct way would be to order the neutralisation of
charges using p*K*a? In which case you need a p*K*a predictor, again, not
simple [1].
Firstly, what is pKa?
Secondly, what does [1] refer to?
Yes, I did try RDKit and got help from developers. It looks their solution
may also have problems on multiple charge cases. I hope the number of cases
with this problem will be reduced after I add a function by calculating the
overall charge.
Again, thank you very much.
Merry Christmas and Happy New Year!
Yingfeng
On Mon, Dec 23, 2013 at 12:49 PM, John May <john...@ebi.ac.uk> wrote:
> Hi Yingfeng,
>
> In short, no. I don’t think it’s easy to provide a comprehensive solution
> for neutralisation. However approximations such as the RDKit SMARTS you’ve
> tried offer a good approach for most cases.
>
> What might be easier is to understand why you need to neutralise the
> compounds?
>
> Anyways, I’m not a chemist but I’ll try my best to answer as to why it’s
> not simple. Firstly in an InChI string you can tell if there is a charge
> when a layer starts with /p or /q.
>
> InChI=1S/C5H9NO4/c6-3(5(9)10)1-2-4(7)8/h3H,1-2,6H2,(H,7,8)(H,9,10)/*p-1*
> /t3-/m0/s1
>
>
> You could also check the atoms where the formal charge is not 0. I guess
> what you really mean by charged is whether there is no overall charge.
>
> C[C@@H](O)[C@H]([NH3+])C([O-])=O uncharged
> C[C@@H]([O-])[C@H]([NH3+])C([O-])=O charged
>
> Again a simple procedure of summing all the charges in a connected
> structure will tell you this:
>
> int sum = 0;
> for (IAtom a : m.atoms())
> sum += a.getFormalCharge();
> boolean charged = sum == 0;
>
> As for neutralising, that’s more tricky. There may be something in a dusty
> corner of the CDK code but I’m not aware of it. The neutralisation of one
> atom is easily made by adding/removing protons or breaking/making bonds.
> However when there are multiple charges it is non-trival as it involves a
> decision. Considering the example from earlier.
>
> C[C@@H]([O-])[C@H]([NH3+])C([O-])=O charged
>
> how do we decide which neutralised form is correct, these are both have no
> overall charge:
>
> C[C@@H](O)[C@H]([NH3+])C([O-])=O uncharged
> C[C@@H]([O-])[C@H]([NH3+])C(O)=O uncharged
>
> My guess would be the correct way would be to order the neutralisation of
> charges using p*K*a? In which case you need a p*K*a predictor, again, not
> simple [1].
>
> Neutralisation reduces to finding the ionisation a given pH (i.e. find the
> pH where the compound is neutral). ChemAxon offer this functionality but I
> have been told of examples where given two ionisation states of the same
> compound (one > desired pH, one < desired pH) the tool produces different
> output.
>
> Sorry I can’t be of more help.
>
> Thanks,
> John
>
> [1] Lee and Crippen, Predicting pKa
> http://pubs.acs.org/doi/abs/10.1021/ci900209w
>
> On 21 Dec 2013, at 14:05, Yingfeng Wang <ywang...@gmail.com> wrote:
>
> I have a compound with Inchi
>
> InChI=1S/C5H9NO4/c6-3(5(9)10)
> 1-2-4(7)8/h3H,1-2,6H2,(H,7,8)(H,9,10)/p-1/t3-/m0/s1
>
> First of, is there is a way to know whether it is charged?
>
> Secondly, is CDK able to neutralize it if it is charged?
>
> Thanks.
>
> Yingfeng
>
>
>
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