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I was looking for the $\mathrm{p}K_\mathrm{a}$ of imidazole and found this $\mathrm{p}K_\mathrm{a}$ table [1]:

\begin{array}{ll} \text{Compound} &\mathrm{p}K_\mathrm{a} \\ \hline \text{Imidazolium} &7.38 \\ \text{2-Aminoimidazolium} &8.46 \\ \text{1-Methyl-2-aminoimidazolium} &8.65 \\ \text{4,5-Dimethyl-2-aminoimidazolium} &9.21 \end{array}

Wouldn't you predict totally inverse order of $\mathrm{p}K_\mathrm{a}$ values?

For example, $\ce{-NH2}$ is a donor by resonance, it will make the cation more stable than imidazole, as far as I know. So, how to understand those values of $\mathrm{p}K_\mathrm{a}$?

I upload the first two compound structures:

IMIDAZOLE

1H-imidazole

AMINOIMIDAZOLE

1H-imidazol-2-amine

  1. Storey, B. T.; Sullivan, W. W.; Moyer, C. L. The Journal of Organic Chemistry 1964, 29 (10), 3118–3120. DOI 10.1021/jo01033a537.

Edit

Only for completeness I should add that pka values are of the protonates species, I mean imidazolium, 2-aminoimidazolium, and so on..

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    $\begingroup$ My point was that it is inductively withdrawing. Give the nature of the cation, resonance donating is probably not super helpful. $\endgroup$
    – Zhe
    Commented Aug 11, 2017 at 21:49
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    $\begingroup$ Everything's all right. You've got confused but what you pointed out in your edit. Cation becomes weaker acid - "more stable" just like you say. Higher pKbH+ means weaker acid - stronger neutral base. $\endgroup$
    – Mithoron
    Commented Aug 11, 2017 at 23:30
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    $\begingroup$ I don’t understand where the problem lies. A higher $\mathrm pK_\mathrm{a}$ means less likely to get deprotonated. The more electron-donating groups, the more a positive charge is stabilised and the less likely it is for a proton to leave — exactly what the data tells you. Am I missing something? $\endgroup$
    – Jan
    Commented Aug 12, 2017 at 2:27
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    $\begingroup$ Ah. Do you want to write up a self-answer to your question? =) $\endgroup$
    – Jan
    Commented Aug 12, 2017 at 9:54
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    $\begingroup$ That’s what we’re here for =) $\endgroup$
    – Jan
    Commented Aug 12, 2017 at 10:00

1 Answer 1

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As pointed out by @Jan and @Mithoron in comments, part of the answer was on the OP. I will complete it.

If we compare acidity of imidazolium and 2-aminoimidazolium (the cations), we find that 2-aminoimidazolium is a weaker acid because of the bigger $\mathrm{p}K_\mathrm{a}$ value.

The reason is that the amino group ($\ce{-NH2}$) is donor by resonance, stabilizing the cation more than imidazolium is, because it doesn't have any extra donor group.

The same reasoning can be done for the remainder compounds of the table. The more inductive or resonance donors, the bigger stability of the cation and weaker acid (bigger $\mathrm{p}K_\mathrm{a}$).

One last thing

As @Zhe observed, the pKa values of all those compounds are really different of guanidine($\mathrm{p}K_\mathrm{a}$=13.6). In the paper below, authors make the same observation. I haven't found an easy answer for that. Maybe it can be done with computational calculations.

  1. Storey, B. T.; Sullivan, W. W.; Moyer, C. L. The Journal of Organic Chemistry 1964, 29 (10), 3118–3120. DOI 10.1021/jo01033a537.
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  • $\begingroup$ With regard to the comparison to guanidine, see "The pKa Values of Some 2-Aminomidazolium Ions"J. Org. Chem. 1964, 29, 10, 3118–3120 doi.org/10.1021/jo01033a537 $\endgroup$
    – Andrew
    Commented Aug 1, 2023 at 21:26

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