I have a feeling you may need to make two reactions.  Let's consider a dirt
simple case:

>>> rxn = AllChem.ReactionFromSmarts("[C:1][N:2]>>[C:1].[N:2]")

>>> prods = rxn.RunReactants([Chem.MolFromSmiles("CN")])

>>> Chem.MolToSmiles(prods[0][0])

'C'

>>> Chem.MolToSmiles(prods[0][1])

'N'

>>>

Note that this reaction is explicitly breaking a bond.  I think this is
what you are seeing with your example.

Note that similar to the "." on the reagent side meaning multiple reagents,
the "." on the right hand side means there will be multiple products.

Does this help at all?

Cheers,
 Brian

On Thu, Mar 30, 2017 at 12:07 PM, Stephen Roughley <s.rough...@vernalis.com>
wrote:

> Dear Greg/RDKitters,
>
>
>
> This may be user error, or misunderstanding of rSMARTS, so can anyone
> throw some light on the following behaviour?
>
>
>
> First example works as expected – there are 2× Ph in m4, so we end up with
> 2×2×2 copies of the expected product:
>
>
>
> rSMARTS4='([*:1]-&!@c1:[c&!H0]:[c&!H0]:[c&!H0]:[c&!H0]:[c&!
> H0]:1.[*:2]-&!@c1:[c&!H0]:[c&!H0]:[c&!H0]:[c&!H0]:[c&!H0]:1)
> >>([*:1]-!@c:1:c:c(-F):c:c:c1.[*:2]-!@c:1:c:c(-F):c:c:c1)' #Replace 2×
> Ph-* with 2× 3-Fl-C6H4-*
>
> rxn4=AllChem.ReactionFromSmarts(rSMARTS4)
>
> rxn4
>
> m4=Chem.MolFromSmiles('c1ccccc1CCOCc1ccccc1')
>
> m4
>
> prodsbi=rxn4.RunReactants((m4,))
>
> for prod in prodsbi:
>
>     Chem.SanitizeMol(prod[0])
>
> Draw.MolsToGridImage([prod[0] for prod in prodsbi],molsPerRow=4,
> subImgSize=(200,200))
>
>
>
> Now consider the following – the only difference I can think of is that
> the [*:1] and [*:2] atoms map to adjacent, directly bonded atoms – I cant
> see why that should matter…
>
>
>
> m3=Chem.MolFromSmiles('c1ccccc1COc1ccccc1')
>
> m3
>
> prodsbi=rxn4.RunReactants((m3,))
>
> for prod in prodsbi:
>
>     Chem.SanitizeMol(prod[0])
>
> Draw.MolsToGridImage([prod[0] for prod in prodsbi],molsPerRow=8,
> subImgSize=(200,200))
>
>
>
> Just to be sure this is as I think it looks..
>
> prodsbi[0][0]
>
>
>
> Any suggestions as to why this happens, and whether it is the expected
> behaviour? (And how to avoid it?!)
>
> Thanks,
>
> Steve
>
>
>
>
>
>
>
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