h bond acceptor count

Why is there a difference in the h bond acceptor count for the same compound when I used the function acceptorcount and acceptor in marvinview

for example, for this compound, i found that the function acceptor gives a h bond acceptor count of 1 while acceptorcount gives a h bond acceptor count of 0. What is the difference between the two functions and why is it giving different information for the same molecule. the smiles string is below

 

[H][C@@]24[C@]3(CCCC4)C1=CC(OC5=CC=CC=C5)=CC=C1C[C@H]2N(C)CC3

 

thanks.

Hi,

 

In the "mview"  I obtained the attached figure.

As you can see the only acceptor is the "N" atom (through the lone pair) in this molecule. The "ether type oxygen" atom  connected with two aromatic rings is not acceptor.

You must be obtained a similar figure in "mview".

 

Jozsi

 

i did not use mview since i have 50 compounds in a text file with the smiles string, I used the command line with the code

cxcalc acceptorcount -H 7.4 test.txt

or i used

cxcalc acceptor -s 0.5 test.txt

both these functions give me the h bond acceptor count but for some of the compounds the 
acceptor count is different from the two functions

i dont want to use mview since it would require inputing one compound at a time

Hi,

 

1.  "cxcalc acceptorcount -H 7.4 test.txt"

This command means that the major microspecies form generated  firstly at pH=7.4.

 

For example for this molecule:

"[H][C@@]12CCCC[C@@]11CCN(C)[C@@H]2CC2=CC=C(OC3=CC=CC=C3)C=C12"

 

the major microspecies form is this,

"[H][C@@]12CCCC[C@@]11CC[NH+](C)[C@@H]2CC2=CC=C(OC3=CC=CC=C3)C=C12"

 

acceptorcount of this major microspecies form is "0", since "N" is protonated and no lonepair is present.

 

2. "cxcalc acceptor -s 0.5 test.txt"

The "acceptorcount" calculated by this command always refers to the submitted molecule.

For example:

"[H][C@@]12CCCC[C@@]11CCN(C)[C@@H]2CC2=CC=C(OC3=CC=CC=C3)C=C12"

Acceptorcount is "1" for this molecule because of the lone pair of the nitrogen.

 

Additionally the avarage "acceptorcount - pH" distribution is also calculated by this (2.) command.

For example at pH=7.50 the acceptorcount is "0". This is in agreement with the acceptorcount of the (1.) command.

 

Jozsi

thank you very much

this makes sense now