Imidazole is a five-membered heterocyclic compound with formula C3H4N2. Is an aromatic compound, indeed suits the Huckel - Von Dering empiric rule (a molecule is aromaticif contains 4n+ 2 π electrons). In order to simplify the discussion around imidazole (but is exactly the same with nitrogen-based heterocycles with more than one nitrogen), we refer to "pyridine-like" and "pyrrole-like" nitrogen. The pyrrole-like nitrogen is sp2 hybridized and has a lone pair on the perpendicular p orbital that therefore concurs to the aromaticity of the ring. Just like the nitrogen in a pyrrole is therefore not basic. On the other side the pyridine-like nitrogen is also sp2 hybridized but the lone pair lies on an sp2 orbital, "externally" to the molecule and is then basic and nucleophile.
We will see afterwards that reactivity of imidazole is to some extent intermediate compared with pyrrole and pyridine.
One peculiarity of imidazole is the impossibility to distinguish the two nitrogen atoms in solution. The hydrogen moves according to a tautomeric equilibria (that is exactly 50% of each form) from one nitrogen to the other:
To differentiate the two nitrogen atoms at least one substituent is necessary (N.B: the tautomeric equilibria is anyway present). For what concerns nomenclature, -1 position is that corresponding to the pyrrole nitrogen. Then we have to number in ascending order in the direction of the pyridine-like nitrogen.
Unfortunately sometimes the tautomerism makes unpossible to use imidazole in pharmaceutical preparations.
Acid base properties
Imidazole is an amphoteric substance.
As we mentioned previously, imidazole can act as a base considering that has a pyridine-like nitrogen. Actually imidazole is even more basic than pyridine, because is stabilized by mesomeric effect; the proton can be exchanged between the two nitrogens that are then indistinguishable.
pKa = 7 (pKb = 7) in acqueos solution is 50% protonated
On the other side imidazole can act as an acid and the proton on N-1 can be removed by a strong base (pKa relative to this proton is 14.5)
The acid - base behaviour of imidazole is very interesting, because it is not just an amphoteric substance, thanks pyrrole-like and pyridine-like nitrogen but is also consistently more basic than pyridine (pKa of the conjugated acid 5.3) and more acidic than pyrrole (pKa 17.5). It all depence on the symmetry of the nitrogen atoms, that can equally stabilize either the positive (a proton) or the negative charge.
For a good comprehension of imidazole behaviour in different reactions the best tool is still the π electron density map:
As you could have imagined, most of the electron density is localized on nitrogen atoms, but position -4 and -5 are electron rich as well. The carbon atom in position -2, then located between two nitrogen atoms is electron deficient instead, right because is in the middle between two much more electronegative atoms. We quickly get to an important conclusion:
towards electrophilic aromatic substitution position -4 and -5 (equivalent) are reactive (in spite of in general imidazole, just like pyrazole, is less reactive than pyrrole and more reactive than benzene)
Resuming: C4 and C5 are electron rich, C2 is electron deficient
- electrophilic aromatic substitution
- reactivity in alkaline environment (N - metallation, C- metallation)
THe proton more easily removable is the one bound to the pyrrole-like nitrogen, N–H, because nitrogen (heteroatoms in general) stands charges much better than carbon. But what if the pyrrole-like nitrogen is substituted, for example alkylated? Which is now the more acidic proton? Again, if you consider the electron density map you'll probably figure out which is the carbon that tends more to lose a proton to balance the electronic situation, that's C2, the electron deficient carbon.
As result of this considerations, here are the two class of reactions with imidazole as nucleophile:
1) We can generate by deprotonation a nitrogen anion (six electrons in the outer shell), to get N-substituted imidazoles:
2) Or we can generate a carbanion on C2 deprotonating N-substituted imidazoles. The carbanion can afterwards react with an electrophile:
- synthesis from amidine and α - haloketone
Is the most common synthetic approach. Amidine is obtainable by reaction of an amine and a nitrile. The logic standing behind the reaction is not that difficult. You have to focus on the atoms that you have to choose to synthesize the desired ring.
- - reaction mechanism (with amidine synthesis)
On one side we have an amidine, molecule that resembles an amide, that shows two nitrogen atoms each with a lone pair and therefore both virtually nucleophilic. On the other side, we have an α- haloketone, carrying two electrophilic centers instead: the carbonyl carbon and the adjacent carbon (α), considering that is bound to an halogen (higher electronegativity). Whatever is the reaction mechanism, the overall result is that we have both nitrogen atoms attacking respectively the two electrophilic centers. Once the ring is formed, the tendency for aromatization is that strong (low energy) that drives the reaction, and a proton is quickly removed.
- Synthesis starting from imine and ketone/aldehyde
In this case, reaction mechanism is even easier. Once the double imine is offhand synthesized (1,2 diketone and ammonia) the nucleophilic addition to the aldehyde (in red) is the equivalent of an aminal synthesis.