another problem with LogD

Dear Sir:


About the following molecular structure as an example for the MarvinSketch help to calculate the LogD. In my calculation, the predication of LogD is totally different from the values in the help files. I wonder what is the reason.

Dear Rong,


thank your for your report, I will correct it.





Annamaria

Dear Sir:


Thank you for your reply, and about this dis-match, which one is correct, is the help file correct or the sofeware?


By the way, I also found in the same help file about LogP, the prediction of Logp of the structure above is different, either. My calculation is -0.56, which in the file it is -1.09.

The value of the calculated log P also depends on the ion concentrations (chloride, sodium/potassium). The values in the documentation were not calculated with the default ion concentration, that might be the reason.

Hi:


I just adapted the ion concentration, and found what you said is right. If all the ion concentration is zero, and the value of LogP is about -4.5.


But is the same reason for the prediction of LogD? Just as I mentioned above, is my calculation of LogD of Sample1 is right or not?


Thank you!





Sincerely


Rong

Calculating with the default values, your results are right for example 1. I haven't checked how the curve changes when setting different parameters, but as well the pictures could be mixed up and we put the wrong curve next to the example molecule.





regards


Annamaria

Hi:


Thank you for your reply.


I have a little problem for predicting the LogP of Aminophylline which structure shown below. Because the molecular structure is divided into 3 parts, and I don't know how they are be assembled into one complete group (it seems impossible to predict the exact structure of the molecular). Therefore. Is their any method to predict the LogP of Aminophylline.





Sincerely


Rong

Hi,








Aminophylline is a complex structure.





I calculated its logP with the next approximation. See the attached figure as well.





logP(aminophylline) = 2*logP(theophylline) + logP(ethanediamine) + (2*0.65) ~


2*(-0.1)+(-2.0) +1.3 = -0.90








The effect of the "hidrogen bond" on the logP considered to be (2*0.65).





logP of aminophylline is -0.90 (calculated)


logP of theophylline is -0.10 (exp. value)





Jozsi

Hi:


Thanks for your reply.


According to your method, I calculated the Log of Aminophylline again. However, the result is different from that you provided.


In my calculation, the LogP of theophylline is -1.15 (calculated)


the LogP of ethanediamine is -1.40 (calculated)


So, LogP(Aminophylline)=2*logP(theophylline) + logP(ethanediamine) + (2*0.65)=2*(-1.15)+(-1.40) +1.3 = -2.4





Is any problem with my calculation above?


in addition, can LogD be calculated using the same method?





Sincerely


Rong

Hi,

Quote:

Is any problem with my calculation above?

It is OK.


The logP of theophylline is -0.1 according to certain experimental results. I accepted this value in the calculation.

Quote:

in addition, can LogD be calculated using the same method?

Ethylenediamine have 3 protonation states. See the attached picture.





Distribution of the 3 species at given pH can be calculated with Marvin.





For example at pH=7.4 you will obtain this distribution.





0.4% (neutral )


72.0% (mono protonated)


27.6% (di-protonated )





logP (neutral) = 2*(-0.1) +(-2.0) + 1.3=-0.9


logP(momo protonated) = 2*(-0.1) +(-4.8) + 1.3=-3.7


logP(di-protonated) = 2*(-0.1) +(-8.2) +1.3=-7.10





logD can be approximated with next calculation.





logD =log((0.4*10^-0.9+72.0*10^-3.7+27.6*10^-7.10)/100)=-3.18








At pH=10.0


Distribution of the species:


67.0% (neutral )


33.0% (mono protonated)


0.0% (di-protonated )





logD =log((67.0*10^-0.9+33.0*10^-3.7+0.0*10^-7.10)/100)=-1.07








Jozsi