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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).

Gaurang Tandon
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    You can predict pKa values with the online webserver Chemicalize https://chemicalize.com/ – Jan Jensen Sep 28 '17 at 07:16
  • So you suggest to protonate the different nitrogens and see the result with that tool? –  Sep 28 '17 at 07:24
  • 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 – Jan Jensen Sep 28 '17 at 07:27
  • @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. –  Sep 28 '17 at 07:33
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    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 – Jan Jensen Sep 28 '17 at 07:42
  • You already asked about it https://chemistry.stackexchange.com/questions/80504/protonation-of-compounds-with-two-or-three-neighboring-nitrogens – Mithoron Sep 28 '17 at 13:00
  • Possible duplicate of Protonation of Guanidine – Mithoron Sep 28 '17 at 13:11

1 Answers1

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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

Gaurang Tandon
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MEL Science
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    (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) – MEL Science Sep 28 '17 at 05:56
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    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. – Jan Sep 28 '17 at 10:33
  • 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. – Jan Sep 28 '17 at 10:35
  • 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. – MEL Science Sep 28 '17 at 10:45
  • 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. – Jan Oct 29 '17 at 02:45
  • @Jan I think you’re being somewhat unfair. OP already mentioned that the contribution of the pyrrole-type nitrogen is negligible as aromaticity is ruined, which is precisely the third structure you are complaining about (the meaning is that this structure contributes negligibly to the equilibrium mixture). It could perhaps be explained or drawn slightly clearer, but it is not wrong. BTW DMAP is a useful analogy to draw. Where does it get protonated? – orthocresol Dec 28 '17 at 00:40
  • Hückel aromaticity in a heterocycle; no. – Martin - マーチン Jan 26 '18 at 08:48