I want to do ab-initio calculation for an organic D-pai-A dye in presence of a solvent, and this molecule has a cyanoacetic group as an anchoring group, I am not sure about its neutral or anion form in solvent. Is the acidity of anchoring group (as well as basity of solvent) enough for predicting deprotonation of this group in ethanol?
Thanks in advance
Hi Yavar,
You should compute the pKa values to determine the protonation state. I suggest you reading, for instance, this paper: Phys. Chem. Chem. Phys., 2013, 15, 11875; where you can find a nice methodology to predict pKa in both ground and excited state, I guess something interesting for your dyes. Hope that helps.
Best,
Jose
Hi Yavar,
You should compute the pKa values to determine the protonation state. I suggest you reading, for instance, this paper: Phys. Chem. Chem. Phys., 2013, 15, 11875; where you can find a nice methodology to predict pKa in both ground and excited state, I guess something interesting for your dyes. Hope that helps.
Best,
Jose
In this moment I can not to answer your question because I do not know about what you are talking related with chemistry, I mean I need to know more about the process that you are studying.
with best regards
mleyva
Hi,
A difference in the pKa values of indicator and solvent is enough for predicting the deprotonation of indicator in a solvent. Very small difference (0-2) in their pKa values means dye will remain unaffected of the medium, however a difference higher than 3 is sufficient to deprotonate the dye in a given solvent.
I hope this will work.
Shashi Kant
Cyanoacetic acid (pKa=2.45) is a carboxylic acid somewhat stronger than acetic (pKa= 4.76). I'd expect that in water, some 10% of the acid would be ionized. Roughly, I'd expect an amount of this order in ethanol.
As regards a computational study, I'd suggest to use something similar to the
method given in J. Phys. Org.Chem. 2011 , 24, 1209-1211. Of course, this would apply to both the neutral and the anionic form. As regards, H+, I'd apply the same method to a protonated form of the solvent.
Good luck!
Thanks for your very good and instructive comments,
Dear Jose and dear Prof. Abbound referred papers is very good and I am reading them right now, thanks again.
Dear all,
If we have the pKa for both solute and solvent,
Is there any direct approach to calculate or estimate the fraction of deprotonated solutes?
Dear Professor Abboud,
At first thanks for your previous answers to my question,
I have a question (that may be so simple in chemistry)
If we have a organic molecule with pKa about 2.7 (by cyanoacetic group)
and a pure ethanol solvent ( I thinks its pKa is more than 7),
Is that possible to calculate exactly the fraction of deprotonated solutes?
Thanks for your time.
Yavar
Dear Jose,
Thanks for your reference,
as far as I know, It is possible to predict the fraction of deprotonation of a molecule
(for example cyanoacetic pKa~2.5) in different solvent with using their pka,
but I am not able its straight equation or approach......
thanks again for your consideration
Hi Yavar,
It is seen that for complete (100%) proton transfer between an acid and base/medium a pKa difference of 8 is required. Below this difference a combination of both ionic and molecular/neutral species exist. So if pKa difference of 8 is giving 100% ionic species, you can easily convert your pKa difference into the fraction of deprotonated species. For your system (cyanoacetic acid and etahnol) it should be around 57%.
Dear Shashi,
thanks for your consideration.
I did bad mistake about ethanol pKa,
pKa for ethanol is about 15.7 (wikipedia),
now, your rule is about ethanol or ethanol(H+)?
Kindly, If it is possible please write your reference.
Thanks again.
Dear Yavar,
This pKa difference works well for the synthesis of protic ionic liquids (between Bronsted acid and Bronsted base) of high ionic content/ionicity. Please see the attachment and read the section "Transport Properties and ionicity", there you will find the answer. Simply you can read the ionicity concept in books.
Hope this will be sufficient for you.
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