Q4(a). Sydnone and imidazole; imidazolium and imidazolide ions
How is sydnone different from imidazole? Imidazole upon reaction with an acid gives two products. Comment on the aromaticity of the products.
Answer in brief
A sydnone is a mesoionic 6π ring with separated charges and no basic nitrogen; imidazole is a neutral, amphoteric 6π heterocycle. Imidazole gives the imidazolium cation (protonation at N-3) and the imidazolide anion (loss of N–H); both keep a planar sextet and are aromatic.
Key points
- Imidazolium ion (with an acid). The proton goes to the pyridine-type N-3, whose lone pair lies in the ring plane and is not part of the sextet. Protonation therefore leaves the 6π system intact: the cation is aromatic. It has…
- Imidazolide ion (imidazole acting as an acid, e.g. with NaH or NaNH₂). The N–H proton is removed; the negative charge is shared by both N atoms, each of which keeps one lone pair in the plane. The ring still holds six π…
- protonation of pyrrole occurs on carbon and uses the sextet, so the pyrrolium ion is not aromatic. Imidazole is protonated without loss of aromaticity because it has a pyridine-type N. A proton on C-2, C-4 or C-5 would make…
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. Both are five-membered heteroaromatic rings with a 6π sextet, but they are of different kinds. Imidazole (1,3-diazole) is a neutral heteroaromatic compound that can be written with a normal covalent structure. A sydnone (a 1,2,3-oxadiazolium-5-olate) is mesoionic: no uncharged covalent structure can be drawn for it.
Reading of the question: “reaction with an acid” is read as the acid–base behaviour of imidazole: it is protonated by an acid to the imidazolium ion, and, being amphoteric, it can itself act as an acid and give the imidazolide ion. Both ions are discussed.
Sydnone versus imidazole
Scheme: Sydnone carries separated charges in every structure; imidazole does not.
| Feature | Sydnone | Imidazole |
| Ring atoms | O-1, N-2, N-3, C-4, C-5; exocyclic O on C-5 | N-1 (NH), C-2, N-3, C-4, C-5 |
| Bonding type | mesoionic: + spread over the ring, - on the exocyclic O | neutral; one pyrrole-type N, one pyridine-type N |
| Source of the 6π | ring O lone pair, N-2, N-3, C-4, C-5 (+ delocalised) | N-1 lone pair (2 e) + four C/N p electrons |
| Polarity | large dipole moment (about 6 D) | moderately polar; strong N–H…N hydrogen bonding |
| Acid–base | no basic N; hydrolysed by hot acid (ring opens) | amphoteric: base (pKa of conjugate acid 7.0) and weak acid (pKa 14.5) |
| Typical reaction | 1,3-dipolar cycloaddition (to pyrazoles); electrophilic substitution at C-4 | electrophilic substitution at C-4/C-5; N-alkylation and N-acylation |
The two ions of imidazole
Scheme: Protonation at N-3 and deprotonation at N-1 both give symmetrical ions.
- Imidazolium ion (with an acid). The proton goes to the pyridine-type N-3, whose lone pair lies in the ring plane and is not part of the sextet. Protonation therefore leaves the 6π system intact: the cation is aromatic. It has two equivalent resonance structures, with the + charge shared equally by both N atoms; the two N–H are equivalent. This symmetry is why imidazole (conjugate acid pKa 7.0) is a much stronger base than pyridine or pyrrole.
- Imidazolide ion (imidazole acting as an acid, e.g. with NaH or NaNH₂). The N–H proton is removed; the negative charge is shared by both N atoms, each of which keeps one lone pair in the plane. The ring still holds six π electrons: the anion is aromatic and is an ambident N-nucleophile.
- Compare pyrrole: protonation of pyrrole occurs on carbon and uses the sextet, so the pyrrolium ion is not aromatic. Imidazole is protonated without loss of aromaticity because it has a pyridine-type N. A proton on C-2, C-4 or C-5 would make that carbon sp³ and break the cycle: such a cation is not aromatic and is not formed.
With an acid chloride. Imidazole and RCOCl give the same pair of aromatic species in another form: the N-acylimidazole and imidazolium chloride (the second equivalent of imidazole takes up HCl).
Conclusion. A sydnone is a mesoionic 6π ring with separated charges and no basic nitrogen; imidazole is a neutral, amphoteric 6π heterocycle. Imidazole gives the imidazolium cation (protonation at N-3) and the imidazolide anion (loss of N–H); both keep a planar sextet and are aromatic.
Application. The imidazole of histidine switches between these forms near pH 7 and acts as the acid–base catalyst in enzymes such as chymotrypsin; imidazolium salts are the parent of N-heterocyclic carbenes and ionic liquids.
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All parts of Question 4
- 4(a) Sydnone and imidazole; imidazolium and imidazolide ions15 marks
- 4(b) 'Etard reaction, Clemmensen reduction, Michael addition15 marks
- 4(c)(i) Ethyl levulinate with excess methylmagnesium bromide10 marks
- 4(c)(ii) Reimer–Tiemann reaction of p-cresol: normal and abnormal products10 marks