Organic Chemistry III - Heterocyclic Compounds PDF

ORGANIC CHEMISTRY III HETEROCYCLIC COMPOUNDS Dr. Khulood H. Oudah  5-Membered rings with two heteroatoms Imidazole and Pyrazole  Introduction ❖Replacing a CH group in the pyrrole ring with a nitrogen atom can give rise to two compounds : pyrazole and imidazole...

ORGANIC CHEMISTRY III HETEROCYCLIC COMPOUNDS Dr. Khulood H. Oudah  5-Membered rings with two heteroatoms Imidazole and Pyrazole  Introduction ❖Replacing a CH group in the pyrrole ring with a nitrogen atom can give rise to two compounds : pyrazole and imidazole. ❖ Only one nitrogen atom can contribute two electrons to the aromatic sextet. It is the nitrogen with the hydrogen (red) and it is described as pyrrole-like nitrogen. ❖While the second nitrogen which has no hydrogen (blue) is described as pyridine-like.  Resonance structures of imidazole ❖The lone pair on pyrrole-like nitrogen is delocalized round the ring while that on the pyridine-like nitrogen is localized in sp2 orbital on nitrogen. Thus these compounds have properties intermediate between those of pyrrole and pyridine.  Physical Properties of Imidazole and Pyrazole  Imidazole and pyrazole are water soluble solids and insoluble in aprotic solvent.  They have very much higher boiling point:256 and 187 °C respectively, this difference is due to imidazole has an extensive hydrogen bonding than pyrazole thus imidazole molecules can exist as oligomers, consequently more energy is required to break these bonds to bring the molecules from one phase to another.  On the other hand pyrazole molecules can form dimers only thus lesser energy is required to break these molecules. imidazole molecules as a polymer pyrazole molecules as a dimer N H N N H N N N N H N N H N  N- subsituted imidazole and pyrazole have lower boiling and melting points than the unsubstituted compounds due to inability to form H-bonds. 1-(4-chlorobenzyl)-1H-imidazole Basicity  Imidazole is a stronger base than pyrazole or pyridine and of course pyrrole. Thus imidazole and pyrazole are more stabilized than pyrrole in acidic medium.  Basicity order: Imidazole > Pyridine > Pyrazole > pyrrole ❑ This can be explained as follows: i) Pyrrole is not basic because the lone pair on the only nitrogen is needed to complete the aromatic pi system and protonation if occurs at all occurs at carbon rather than on nitrogen and the resulting cation is not aromatic. ii) Imidazole has two nitrogen atoms and on protonation the positive charge can be delocalized over them but although in pyridine its N is available for protonation as it is not involved in maintaining aromaticity, it has only has one nitrogen on which to stabilize the positive charge thus it is less basic than imidazole. iii) Pyrazole is much weaker base than imidazole and pyridine although it also have two nitrogen atoms as in case of imidazole. This difference is due to the fact that the positive charge in pyarzolium ion is less stabilized than in the imidazolium ion and pyridinium ion. So imidazole is a stronger base than pyridine whereas pyrazole is a weaker base than pyridine:  Effect of substitution on basicity ❖ Generally E.D.G groups on the ring increase the basicity while E.W.G. decrease it. ❖ N-methyl imidazole is more basic than imidazole itself. ❖ However, N-methylpyrazole is less basic than pyrazole which can be attributed to steric hindrance effect which cause difficulty in accessing the lone pair of electron by the proton. N N Basicity > N < N N N N N H H CH3 CH3  Acidity  Imidazole unsubstituted in the 1-position are weak acids.  Its acidity is greater than that of pyrrole and equals that of pyrazole.  Thus imidazole and pyrazole are amphoteric substances. Acidity N > N N H H  Electrophilic Aromatic Substitution  Diazoles are less reactive than 5-membered heterocycles with one heteroatom (pyrrole and its analogs) in electrophilic aromatic substitution due to the inductive electron-withdrawing effect of the second heteroatom.  However, they are more reactive than pyridine due to delocalization of the lone pair of electrons on the N-atom make the C- atoms bear negative charges while in pyridine the N- atom exerts inductive electron withdrawing effect only.  The orientation in pyrazole, is at the 4-position due to the deactivation effect of the pyridine-like nitrogen.  E+ E N N N H N H Pyrazole 4-substituted  The orientation in imidazole, is at 5-position, due to the additional N-atom deactivates its vicinal positions N-like pyridine deactivate its vicinal positions 4 N3 5 2 N1 H most strongly activated position  However, if the position 5 is occupied the electrophiles will be directed to 4-position. Electrophilic substitution: reactivity: reactive towards many electrophiles (E+); >benzene but C5 (for NR systems; if NH then C4 and C5 are identical) pyrazole: C4 - less reactive than imidazole NB. electron donating substituents enhance reactivity towards electrophiles  Electrophilic Substitution 1) Nitration 2) Sulfonation HO3S H2SO4 / SO3 N N N 100° C N H H 3) Halogenation Diazocoupling:  Since it occurs with electron rich aromatic compounds it occurs only with imidazole and activated pyrazole ( with EDG) in alkaline medium. N Ph N=N +Cl- N N N N=NPh H HO HO N=N-Ph PhN2+ Cl N N N N H H Lithiation N N N BuLi RX N Li R N N ph ph ph CH3 Li CH3 N BuLi N N N R R Acylation N N Ac2O N N H COCH3 Alkylation CO2CH3 CO2CH3 N CH2N2 N H3C N H3C N H CH3 Reaction with oxidizing reagents MeO N O2 / MeOH NH N MeO N O H H  SYNTHESIS OF IMIDAZOLE AND PYRAZOLE Pyrazole From 1,3 - Dicarbonyl Compounds and Hydrazines.  SYNTHESIS OF IMIDAZOLE AND PYRAZOLE Imidazole From α-Hydroxy ketone and Foramide. R O R R OH R NHCHO N -2H2O + 2 HCONH2 1 R1 N R OH R1 OH Foramide R1 NHCHO  -Hydroxy ketone H  R O R NHCHO R N + HCONH2 R1 N R1 OH Foramide R1 NHCHO  -Hydroxy ketone H N C N BuLi CH3CN + PhCN Ph N Ph N H