6
$\begingroup$

I wish to ask where imidazole and aminoimidazole will be protonated on first protonation.

Imidazole:

Aminoimidazole:

I suppose that the nitrogen (without hydrogen) of the ring is the most probable to be protonated, but I am not sure about it. Do you know where it will be protonated first?

Any link to a study where the issue is solved will be good (I need the information for my thesis).

|improve this question
$\endgroup$
  • 1
    $\begingroup$ You can predict pKa values with the online webserver Chemicalize chemicalize.com $\endgroup$ – Jan Jensen Sep 28 '17 at 7:16
  • $\begingroup$ So you suggest to protonate the different nitrogens and see the result with that tool? $\endgroup$ – user43021 Sep 28 '17 at 7:24
  • $\begingroup$ No, just draw the molecule as you did above and Chemicalize will tell you the most probable de/protonated isomer as a function of pH $\endgroup$ – Jan Jensen Sep 28 '17 at 7:27
  • $\begingroup$ @JanJensen I don't know why it doesn't produces a value for -NH2 in 2-aminoimidazole, SMILES is c1cnc([nH]1)N . Maybe it just can't be protonated because of the resonance. $\endgroup$ – user43021 Sep 28 '17 at 7:33
  • 1
    $\begingroup$ Yes, Chemicalize doesn't consider all protonation states. Only the most probable. The $\ce{NH2}$ group won't be protonated for the reason you stated $\endgroup$ – Jan Jensen Sep 28 '17 at 7:42
1
$\begingroup$

For imidazole, the lone pair of pyrrole-type nitrogen (the one with hydrogen) is involved in aromaticity of the whole compound, as it contributes two π-electrons in addition to four of those from carbon-carbon or carbon-nitrogen bonds, which makes it 6, which is a classical sign of Huckel aromaticity. In fact, this NH-nitrogen tends more to act as an acidic center than as a basic one, as it loses its hydrogen when treated by strong bases, like $\ce{NaNH2}$.

The pyridine-type nitrogen has its lone pair out of ring conjugation, thus it can act as a acceptor of hydrogen ion, placing its two electrons into a vacant orbital of a $\ce{H+}$.

For reference, you can see any heterocyclic chemistry textbook. I prefer the one authored by Joule, Mills. Or this.

enter image description here

When it comes to 2-aminoimidazole, this publication (which is unarguably very old) says all three nitrogens are involved in protonation, because the structure is a derivative of guanidine, a strong organic base. Which seems only legit to me if it said that the contribution of pyrrole-type nitrogen is negligible, as aromaticity is ruined in that case. I could also predict that exo-aminogroup gets protonated more likely, as its basicity is somewhat higher: its lone pair is on a sp3-orbital, rather than the lone pair of endo-nitrogen (the one without H), where it is on a sp2-orbital, thus, the contribution of s-orbital is higher and it is closer to the nucleus of nitrogen atom, decreasing its availability to act as a covalent bond acceptor.

enter image description here

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ (Yes, I know that hybridization theory is false and is not really loved here, but why not use it, when it is a powerful teaching instrument and helps to explain stuff) $\endgroup$ – MEL Science Sep 28 '17 at 5:56
  • 1
    $\begingroup$ Oh no, in the case of organic compounds, hybridisation theory is actually rather well received because it does a good job on explaining this without having to solve the Hamiltonian. $\endgroup$ – Jan Sep 28 '17 at 10:33
  • $\begingroup$ However, the exo-amino group’s lone pair will very likely not be in an $\mathrm{sp^3}$ hybrid orbital until very high $\mathrm{pH}$ values because it takes part in resonance. (Compare with the amino group of aniline.) Also, that third structure you are drawing there in the series of equilibria deserves a downvote. $\endgroup$ – Jan Sep 28 '17 at 10:35
  • $\begingroup$ I do absolutely agree about not-hundred-percent availability of exo-aminogroup lone pair, but still the comparative basicity of those two nitrogens is still a good question, which need more references to be found. But the second scheme, as you can read from my answer, is not my creation, but an illustration for the paper I mentioned. $\endgroup$ – MEL Science Sep 28 '17 at 10:45
  • $\begingroup$ If the paper you are referring to is the one you linked (B. T. Storey, W. W. Sullivan, C. L. Moyer, J. Am. Chem. Soc. 1964, 29, 3118–3120. DOI: 10.1021/jo01033a537.) I can only suggest you actually read it since it supports none of your claims. $\endgroup$ – Jan Oct 29 '17 at 2:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy