Site of first protonation of imidazole and aminoimidazole

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


**Imidazole:**


![](https://i.stack.imgur.com/PowBv.png)


**Aminoimidazole:**


![](https://i.stack.imgur.com/Q3QWF.png)


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).




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](http://web.pdx.edu/~wamserc/CH332S96/12notes.htm).


[![enter image description here](https://i.stack.imgur.com/jcath.png)](https://i.stack.imgur.com/jcath.png)


When it comes to 2-aminoimidazole, [this publication](http://pubs.acs.org/doi/abs/10.1021/jo01033a537)  (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](https://i.stack.imgur.com/C4ith.png)](https://i.stack.imgur.com/C4ith.png)


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