tertiary and quarternary carbon

Hello everybody,

I have just started working with Instant JChem so my question is very basic:

I would like to count tertiary and quarternary carbon atoms and thought that matchCount("[#6](~[#6])(~[#6])(~[#6])") would do that job. As far as I can tell, it works when there are only tertiary carbons in the molecule but will count querternary atoms as 4.

It would be great if somebody could give me a hint, how to change the formula!

Thanks a lot in advance!

Mieke

 

Ps: By the way: If I do a query by drawing the structure I am looking for; is there any possibility to translate this query into a SMARTS-string automatically?

 

Hi Mieke,

 

Unfortunately, it will not be appropriate, because your query will match substructures in quaternary carbon as well.

So you could subtract the number of quaternary carbons from this:

matchCount("[#6](~[#6])(~[#6])(~[#6])") - matchCount("[#6]~[#6](~[#6])(~[#6])~[#6]")

Ps: By the way: If I do a query by drawing the structure I am
looking for; is there any possibility to translate this query into a
SMARTS-string automatically?

Yes. In Marvin Sketch there are several ways, for example: Edit / Source and then View / SMARTS or Chemaxon Extended SMARTS. Or simply Edit / Copy as smiles (Ctrl-L) will copy SMARTS if the structure has query features.

 

Best regards,

Szabolcs

Hi Szabolcs,

thank you for the fast reply!

I was trying to count the sum of tertiary and quarternary atoms. I apologize for not making myself clear in this point. What confused me, was not that

matchCount("[#6](~[#6])(~[#6])(~[#6])")

counts quarternary carbon atoms as well, but that it counts 4 for every quarternary carbon atom and I don't understand why. Anyways, I adjusted your suggestion to substract the number of quarternary atoms to substracting their number thrice:

matchCount("[#6](~[#6])(~[#6])(~[#6])") - (3*matchCount("[#6]~[#6](~[#6])(~[#6])~[#6]"))

So, this is working now!

Thanks again for your help, have a good day!

Mieke

OK, I see.

The 4 times match count comes from the fact that a tertiary carbon has four 3-ligand "tertiary" substructure   match. (You can leave out each ligand in turn.)

 

Anyway - I am glad that you found the solution.

 

Best regards,

Szabolcs