Re: [Ambit-users] [URGENT] Issues with my SMIRKS reaction. Need help

Dear Yannick

If you need to transform/remove/add H atoms in Ambit-SMART it can be 
done only treating them as explicit H atoms. The SMIRKS must define 
explicitly what happens with each H atom.

The indicated implicit H atoms are used only for defining logical 
expression in order to match the substructure (target fragment) that is 
to be transformed according to the SMIRKS.

Here I suggest to you a working solution based on explicit H atoms:

(1) First you need two different SMIRKS for both cases since the H atom 
configuration is different (It is impossible with explicit H managing to 
handle both cases with a single SMIRKS):

(2) Secondly your target molecule must be set with explicit H atoms

(3) here are the SMIRKS for both cases:

[#6;A;H2X4:7]([H:99])([H])!@-[#6;A;H2X4:6]!@-[#8;X2:5][P;X4:4]([#8;X2:2])([#8;A;X2:3][H])=[O;X1:1]>>[#8;X2:2][P;X4:4]1(=[O;X1:1])[#8;A;X:3][#6;A;H1X4:7]([H:99])!@-[#6;A;H2X4:6]!@-[#8;X2:5]1

[#6;A;H2X4:7]([H:99])([H])!@-[#6;A;H2X4:6]!@-[#8;X2:5][P;X4:4]([#8;X2:2])([#8;A;X1-:3])=[O;X1:1]>>[#8;X2:2][P;X4:4]1(=[O;X1:1])[#8;A;X:3][#6;A;H1X4:7]([H:99])!@-[#6;A;H2X4:6]!@-[#8;X2:5]1

Notice that also the explicit H atoms are treated on atom 7 
[#6;A;H2X4:7]([H:99])([H])  because when you close the ring one of the H 
atoms must be removed (i.e. this is the unmapped H atom) the other H 
atom stays (i.e. this is the mapped H atom: [H:99])


With best regards
Nick




On 9/12/16 22:29, Yannick .Djoumbou wrote:
>
>
> On Sun, Jul 10, 2016 at 7:57 PM, Yannick .Djoumbou 
> <y.d...@gm... <mailto:y.d...@gm...>> wrote:
>
>     Hi all,
>
>     I have some issues with my SMIRKS manager.
>     have written the SMIKRS below that I am applying on one compounds:
>
>     [#6;A;H2X4:7]!@-[#6;A;H2X4:6]!@-[#8;X2:5][P;X4:4]([#8;X2:2])([#8;A;X2H1,X1-:3])=[O;X1:1]>>[#8;X2:2][P;X4:4]1(=[O;X1:1])[#8;A;X:3][#6;A;H1X4:7]!@-[#6;A;H2X4:6]!@-[#8;X2:5]1
>
>
>     I am trying to join two atoms here to form a cycle, The C7 and the
>     O3. In  the reactant,O3 can be linked to a Hydrogen atom (case 1)
>      or can be negatively charged (case 2). The problem is that in the
>     product, the O3 has a valence of 3 (with an implicit H) (case 1),
>     or a valence of 2 (case 2).
>
>     The examples are illustrated in the attached file.
>
>
>     Could anyone help?
>
>
>     Thanks.
>
>
>     Regards,
>
>