logP and HB acceptocount

User bff01cf213

29-05-2009 06:33:14

Dear All,

 I am using evaluator to calculate the logP value with the command:

 evaluate -e "logP()"  test.smi

 I have tried to calculate the relevant value for Tigecycline. The smiles format
 is shown below:


 The logp value I obtained is logP=-3.09

 This value is rather different to the one reported to pubchem, drugbank and
 experimental one as shown below:

 Pubchem(XlogP) = -0.2

 Drugbank (predicted)= 0.81 (alogps)
 Experimental= 0.8

 Could you please help me to this.

 In addition, when I am trying to calculate the HB acceptor atoms using the
 command: evaluate -e "acceptorcount()" test.smi

 For the three following smile files I get different values to the ones
 reported in pubchem as shown bellow:

 evaluateAcceptorcount = 7 (pubchem =6)

 evaluateAcceptorcount = 7 (pubchem =6)


 evaluateAcceptorcount = 10 (pubchem =8)


 Could you please also help me to this?

 Many thanks,


ChemAxon efa1591b5a

05-06-2009 08:57:35

Hi Andreas,

your request was moved to this area (Structure based prediction) and it has been assigned to the appropriate expert who will respond shortly.




User 851ac690a0

05-06-2009 13:08:18



Probably there is a pKa bug ,this is why you obtained a large error for the predicted 'logP'.

I calculated the logP value of the molecule you attached with the latest release which can also be reached  on the Chemaxon's site.



The logP value of the neutral form is  0.92 and the predicted logD value is -0.08. Since the molecule is ampholyte  we need to accept the logP=-0.08 as a predicted value.   

This predicted value is close to the experimental logP value  which is 0.8.

We are going to further improve the logP prediction in the next 5.2.3 release.



The 'acceptorcount' difference occures because we consider the "S" atoms as acceptor sites.

We are going to add a user adjustable option to this tool for switching on/off the "S" atom as an acceptor.