5
$\begingroup$

The documentation for the indigo module can be found here

http://ggasoftware.com/opensource/indigo/api#inputoutput

So for instance if I have a molecule object for some SMILES string, e.g. "[C](=[O])", I wish to calculate the valency of each atom, for instance here the desired output would be [atom=C, unbound_electrons=2],[atom=O, valency=0]

If I consider the atom "[C]" Can anyone explain why is this code printing [atom=C, unbound_electrons=0] not [atom=C, unbound_electrons=4]

from indigo import *
indigo = Indigo()

mol=indigo.loadMolecule("[C]")

print(mol.grossFormula(),"\n")

for atom in mol.iterateAtoms():
        print([atom.symbol(),atom.radicalElectrons()])

EDIT: I could work it out if I could generate a list of the types of bonds on the atom in conjuction with atom.atomicNumber(). E.g. if I could say [C] has a double bond I could take it's atomic number - 2 (second shell) - 2 (double bond)

EDIT#2: This might be useful for visualising what i'm talking about

from indigo_renderer import *
renderer = IndigoRenderer(indigo)
renderer.renderToFile(mol,"mol.png")

EDIT#3: I am not a chemist, so might have got some concepts wrong

$\endgroup$
5
$\begingroup$

I was looking into this, and the results I was getting seemed very anomalous. Consider this iteration from a single carbon atom at -4 through +4, this is the output I receive (python code at the bottom of the post):

[C+4]
C valence= 4 radicalE= 0 charge= 4 implicitH= 0
[C+3] |^1:0|
C valence= 4 radicalE= 1 charge= 3 implicitH= 0
[C+2] |^3:0|
C valence= 4 radicalE= 2 charge= 2 implicitH= 0
[C+]
C valence= 0 radicalE= 0 charge= 1 implicitH= 0
[C]
C valence= 0 radicalE= 0 charge= 0 implicitH= 0
[C-]
C valence= 0 radicalE= 0 charge= -1 implicitH= 0
[C-2] |^3:0|
C valence= 4 radicalE= 2 charge= -2 implicitH= 0
[C-3] |^1:0|
C valence= 4 radicalE= 1 charge= -3 implicitH= 0
[C-4]
C valence= 4 radicalE= 0 charge= -4 implicitH= 0

I printed them using the canonicalSmiles format. After I realized something strange was going on (neutral, and +/- 1 charges were 0, but everything else was fine), I dug into SMILES documentation to find out that currently it appears only up to divalent radicals are supported. Something like a lone carbon atom has technically a tetra-valent radical, so it defaults to nothing, because it cannot be interpreted by SMILES.

I pulled this from ChemAxon which explains what the notation after the atom means. I'm going to assume that |^1:0| is the notation for a mono-valent system. The 0 indicates the index of the atom, which is always zero for this example.

Radical numbers
Atom indexes with:
- divalent radical center are written after "^2:"
- divalent singlet radical center are written after "^3:"
- divalent triplet radical center are written after "^4:"
- trivalent radical center are written after "^5:"

And from the Indigo website itself it states under the supported extensions

Radical numbers: monovalent, divalent singlet, and divalent triplet

So, I don't think you've done anything wrong, you've just simply hit a limitation of the SMILES system and the interpreter of the indigo software. Here is the code I used below:

from indigo import *
indigo = Indigo()

from indigo_renderer import *

renderer = IndigoRenderer(indigo)
mols = []
mols.append(indigo.loadMolecule("[C+4]"))
mols.append(indigo.loadMolecule("[C+3]"))
mols.append(indigo.loadMolecule("[C+2]"))
mols.append(indigo.loadMolecule("[C+1]"))
mols.append(indigo.loadMolecule("[C]"))
mols.append(indigo.loadMolecule("[C-]"))
mols.append(indigo.loadMolecule("[C-2]"))
mols.append(indigo.loadMolecule("[C-3]"))
mols.append(indigo.loadMolecule("[C-4]"))

i = 0
for mol in mols:
    print mol.canonicalSmiles()
    for atom in mol.iterateAtoms():
      print atom.symbol(),"valence=",atom.valence(),"radicalE=",atom.radicalElectrons(),"charge=",atom.charge(),"implicitH=",atom.countImplicitHydrogens()
    filename = 'mol'+str(i)+'.png'
    renderer.renderToFile(mol,filename)
    i += 1
$\endgroup$
  • $\begingroup$ Thanks so much for the level of detail here! This is extremely helpful. This confirms my worries, I've developed a work around that examines the bond network and counts bonds per atom. It's a bit messy and roundabout, but it seems to work! $\endgroup$ – Freeman Jul 22 '13 at 16:11
4
$\begingroup$

I have never used the indigo framework, but there's more than one way to process SMILES strings in Python. Open Babel and Pybel might be an alternative:

import pybel

# SMILES for flurazepam, taken from  
# http://chemspider.com/chemical-structure.3276
flurazepam = 'CCN(CC)CCN1C(=O)CN=C(C2=C1C=CC(=C2)Cl)C3=CC=CC=C3F'

# SMILES for trimethylamine, taken from 
# http://chemspider.com/chemical-structure.1114
trimethylamine = 'CN(C)C'

# SMILES for nitromethane, taken from 
# http://chemspider.com/chemical-structure.6135
nitromethane = 'C[N+](=O)[O-]'

mol = pybel.readstring('smi', flurazepam)

for atom in mol.atoms:
    print '{:<5} {:3} {:3} {:3}'.format(atom.type, atom.formalcharge, \
        atom.implicitvalence, atom.valence)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.