CCL: Solvation Free Energy of Proton in THF



When I modeled protonated THF, I got more reasonable numbers. However, I observe a significant deviationÂÂfor the anionic system that makes me think about the anionic nature of the complex.

Thanks a lot for all suggestions.

Busra


On Wed, Aug 27, 2014 at 2:33 PM, Andreas Klamt klamt/./cosmologic.de <owner-chemistry(!)ccl.net> wrote:

Sent to CCL by: Andreas Klamt [klamt^^cosmologic.de]
Dear Busra,

obviously Mariusz is absolutely right. If you really should like to try to calculate absolute pKa, then you should get aware that the state of the proton in THF surely is not an isolated proton, but a protonated THF. As in water you have H3O+.

But I also would not expect that you can get an accurate prediction of the total free energy of the protonated H3O+ with a continuum model.

Best regards

Andreas

Am 27.08.2014 11:38, schrieb Mariusz Radon mariusz.radon[A]gmail.com:

Sent to CCL by: Mariusz Radon [mariusz.radon::gmail.com]
On 08/25/2014 08:23 PM, BUSRA DERELI bsradereli++gmail.com wrote:
Sent to CCL by: "BUSRAÂ DERELI" [bsradereli::gmail.com]
Hello,

I am trying to calculate pKa values of two systems, one is anionic and the
other is cationic copper systems. I got so weird numbers not close to the
experimental values when I computed the free energy of proton in THF which is
the solvent of interest. I use M11-L local functional with SDD basis set on Cu
atom and 6-31G(d) on all other atoms in the system. Thus, I want to compare the
computed solvation free energy of proton in THF with the experimental value. In
some literature reports, I have seen absolute solvation free energy of proton
in a couple of solvents but not in THF.

Thanks,
Busra Dereli

Dear Busra:

Not very strange that you get weird number for solvation energy of
proton which (unlike other ions) is just a point charge! As far as I
know, it is extremely difficult to reproduce solvation energy of proton
in such calculations.

If your goal is to compare acidity of two substances, you are probably
more interested in the difference of their pKa values, than in the
individual pKa values. The point is that when you compute the
difference, the (huge) error on the proton solvation energy cancels out
exactly. Alternatively, you can compute relative pKa values of both
substances with respect to some other chemical, treated as reference,
whose actual pKa you take from experiment. That is much easier than
trying to reproduce absolute pKa values.

Best regards,
Mariusz Radon



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