Heterocyclic chemistry is one of the most important fields of organic chemistry and biochemistry. Ap — Organic Chemistry Chemistry Question
Imidazole
Heterocyclic chemistry is one of the most important fields of organic chemistry and biochemistry. Approximately 55 % of publications in organic chemistry are related to the field, and the number of heterocyclic compounds recently found is far more than that of homocyclic compounds. The five-membered ring compounds with two heteroatoms are often present in many substances that are important for life. For example, imidazole ring is present in the essential amino acid, histidine, and its decarboxylation product, histamine. Histidine residues are found at the active sites of ribonuclease and of several other enzymes and play a vital part in the structure and binding functions of hemoglobin. Several drugs are based on the imidazole ring such as nitroimidazole, cimetidine, azomycin, metronidazole, midazolam.
Draw the structures of 1,3-diazole (imidazole, C3H4N2), imidazol-1-ide anion, imidazolium cation, 1,3-oxazole (oxazole, C3H3NO) and 1,3-thiazole (thiazole, C3H3NS). Which structure(s) can be considered aromatic?
Model Answer
Structure aromatic or not:
- Imidazole (C3H4N2): aromatic [VISUAL]
- Imidazol-1-ide anion (C3H3N2): aromatic [VISUAL]
- Imidazolium cation (C3H5N2): NH N H, aromatic [VISUAL]
- Oxazole (C3H3NO): aromatic [VISUAL]
- Thiazole (C3H3NS): N S, aromatic [VISUAL]
Arrange imidazole, 1,3-oxazole and 1,3-thiazole in decreasing order of melting and boiling points and justify your order.
Model Answer
Melting point: Imidazole > Thiazole > Oxazole
Justification: Imidazole is the first because of intermolecular hydrogen bonding. Thiazole is placed before oxazole because thiazole’s molecular mass and polarizability are larger than those of oxazole.
Boiling point: Imidazole > Thiazole > Oxazole
Justification: Imidazole is the first because of intermolecular hydrogen bonding. Thiazole is placed before oxazole because thiazole’s molecular mass and polarizability are larger than those of oxazole.
Using structural formulae, write down equations for the ionization of imidazole, oxazole, and thiazole in water. Arrange the substances in decreasing order of base strength and justify your answer.
Model Answer
Equation for the ionization: [VISUAL]
Base strength: Imidazole > Thiazole > Oxazole
Justification: Conjugate acid of imidazole is symmetrical delocalized, forms stronger hydrogen bonding with water, i.e. more stable, thus imidazole more basic than oxadiazole and thiazole. Atom O is more electronegative than N and S, it decreased electron density at N of oxazole, decreased stability of oxazole’s conjugate acid making oxazole less basic than thiazole.
Propose a reaction mechanism showing the catalytic behavior of imidazole in hydrolyzing RCOOR’ without a participation of OH–. Justify this behavior based on the structure of imidazole.
Model Answer
Reaction mechanism: [VISUAL]
Explanation: Atom N-3 (N at 3-position) is strong nucleophile; The positive charge is delocalized; The imidazole is good leaving group.
Propose a reaction mechanism for the formation of 1,1’-carbonyldiimidazole (C7H6N4O, CDI) from imidazole and phosgen (COCl2).
Model Answer
Reaction mechanism: [VISUAL]
Explain why the C=O IR stretching frequency in 1,1’-carbonyldiimidazole is 100 cm–1 higher than that of 1,1’-carbonyldipyrrolidine (CO(C4H8N)2).
Model Answer
In 1,1'-carbonyldiimidazole (CDI): The pair of electrons from N-1 and four electrons of the remaining four atoms form a sextet of π-electron of aromatic system. They do not conjugate with C=O, thus do not affect the bond order of C=O.
In 1,1’-carbonyldipyrrolidine: The pair of electrons from N-1 conjugate with C=O decreasing the bond order of C=O, hence decrease its IR stretching frequency.
Write down reaction equations for the preparation of CDI (a) using a mixture of 4 mol imidazole and 1 mol phosgene and (b) using a mixture of 2 mol imidazole, 1 mol phosgene, and 2 mol NaOH. Explain why reaction (a) is preferable.
Model Answer
Equations:
(a) 4 C3H4N2 + COCl2 → (C3H3N2)2CO + 2 [C3H5N2]Cl (1)
2 mol of imidazole react with 1 mol of phosgene to form 1 mol of CDI and 2 mol of HCl; the other 2 mol of imidazole are used to react with the HCl.
(b) 2 C3H4N2 + COCl2 + 2 NaOH → (C3H3N2)2CO + 2 NaCl + 2 H2O (2)
Explanation: In imidazolyl groups of CDI the pair of electrons from N-1 and four electrons of the remaining four atoms form a sextet of π-electron of aromatic system. They do not conjugate with C=O. Two electron-withdrawing imidazolyl groups make C=O more active, the imidazole is good leaving group, hence CDI readily reacts with water from reaction (2):
(C3H3N2)2CO + H2O → 2 C3H4N2 + CO2 (3)
Because CDI readily reacts with water, reaction (a) is preferable as it avoids the production of water that occurs in reaction (b).
CDI is often used for the activation of carbonyl group for the coupling of amino acids in peptide synthesis.
Use curly arrow mechanisms to complete the scheme below, showing the formation of the active compound G from CDI and Alanine.
[VISUAL]
Model Answer
[VISUAL]
R = CH3(NH2)CH
Propose a reaction mechanism for the formation of dipeptide Ala-Gly from G and Glycine.
Model Answer
[VISUAL]