Sciencemadness Discussion Board - Heterocyclic aromaticity

Heterocyclic aromaticity

Nerro - 8-12-2005 at 03:40

Can someone explain to me why molecules such as imidazole and oxazole are aromatic? And how would the resonance structures be that led to the conclusion that imidazole is indeed aromatic?

When I draw the molecule and all its sigma and pi bonds I get pi bonds on 4 out of 5 of the atoms in the molecule. That would mean the overlap an't span the entire molecule... So how do the electrons become delocalized?

I just of something. If the N in the N-H is not sp3 but sp2 hybridized that would leave a perpendicular pi orbital free for the formation of a delocalized pi-band. But is the formation of an aromatic system so much more stable that it would justify the energetically unfavorable formation of an sp2 orbital rather than an sp3 orbital?

[Edited on Thu/Dec/2005 by Nerro]

sparkgap - 8-12-2005 at 04:20

Only the electron pair in the pyrrole-like nitrogen of the imidazole ring participates in electron delocalization. The electron pair of the pyridine-like nitrogen doesn't. Hückel's rule still stands, since you have 6 electrons in total delocalized; three from the two nitrogen atoms, and three from the three carbon atoms. The situation for oxazole is similar.

Hope this helps.

sparky (~_~)

Nerro - 8-12-2005 at 04:43

Yes that makes sense and that's what I thought was logical but what I'm primarily interested in is whether or not it is energetically favourable to form an aromatic system rather than an sp3 hybridized orbital on the "pyrrole-like nitrogen". Because normally that N would be sp3 hybridized.

Does Hückels rule always apply? And is just the fact that there are pi orbitals on all atoms in the rings enough justification to say that all pi orbitals can overlap to form an aromatic system? Because that would contradict with Hückels rule... (an fully conjugated 8-ring is not aromatic according to Hückel.) Why would there nessessarily need to be 4n + 2 pi electrons?

sparkgap - 8-12-2005 at 05:20

I no longer remember the exact numbers

, but I do remember that it was calculated that the stability conferred to the imidazole ring by electron delocalization was by a wide margin greater than forming an sp3 orbital over the nitrogen.

IIRC, Hückel's rule was determined empirically, but there seems to have been a quantum-mechanical justification for it. I lost my notes on this, so don't take my word for it yet.

Cyclooctatetraene is indeed not aromatic; it behaves more like a polyene (undergoes addition rather than substitution with halogens, etc.). Cyclobutene is quite unstable; preferring to undergo Diels-Alder dimerization. So systems with 4n+2 electrons really do have special stability conferred upon them...

sparky (^_^)

[Edited on 8-12-2005 by sparkgap]

Nerro - 8-12-2005 at 05:38

I've looked it up and frankly Hückels rule seems like a bs rule to me.

It only applies to cases in which you add benzene rings to a first benzene ring in a linear fashion.

So benzene is aromatic (4*1 + 2 = 6)
And Naphtalene is aromatic (4*2 +2 = 10)
And Anthracene is aromatic (4*3 +2 = 14)

But that's just basically stating the (apparantly) obvious. Because there are plenty of forms of C4n+2H4n+2 that are not aromatic. It only seems to apply to cases such as the afformentioned and the next versions of that form. (a chain of benzene rings that share 2 C's with their neighbour.)

Off topic, not on heterocyclic aromatic compounds.

BromicAcid - 8-12-2005 at 06:59

Quote:

I've looked it up and frankly Hückels rule seems like a bs rule to me.

It only applies to cases in which you add benzene rings to a first benzene ring in a linear fashion

Wha wha wha? In my advanced orgo class we were using it to get a primary determination if many many systems were aromatic or not and it lead to us exploring the outer limits or aromaticity, give the book "Advanced Organic Chemistry" Part A: Structure and Mechanism (Carey and Sundberg) a try, we did frost diagrams and the like for beznene but we hardly used Huckels rule for the analysis of systems containing benzene rings because they really like to keep their electrons to themselves, for example if you look at two benzene rings connected so that they form a cyclobutane ring in the middle you see that it could be considered like a cyclobutadiene ring in the middle, which is incredibly unstable and you would expect the molecule to show a degree of instability, however you can heat it up to very high temps (>500C) before it starts to decompose, this is not the stability you would expect based on Huckels rule, you would expect it to be unstable (at least to some degree) but with the benzene rings they like to keep that electron density to themselves, so honestly we used Huckels rule for everything but benzene and compounds containing a benzene ring.

Remember for it to be aromatic the pi density has to be in a plane, if you can't force it planar then it really doesn't want to be aromatic. Take for example [10] annulene, if it were planar then it would likely be aromatic, there is one isomer the Z,E,Z,Z,E that is pretty to planar if it wasn't for an unfavorable steric intereaction between two hydrogen atoms being crammed into one another, but if you attach these two points with a methylene, the CH2 ends up above the ring and out of the plane, this holds it in the geometry where the annulene part of the ring will be planar and as a result the molecule is aromatic. Expecially interesting to me are the aromatic cations and ions that you really wouldn't expect, like squaric acid, very interesting stuff.

[Edited on 12/8/2005 by BromicAcid]

mick - 11-12-2005 at 08:14

Imidazole forms salts with metals and is acidic on the organic level because as a salt it is stabilized by aromaticity. Purines and pyrimidines in DNA might work in the same way.
mick

Nerro - 11-12-2005 at 11:57

"acidic"... The pKa = 7,00 at 20oC and 1atm. That's not very acidic is it...

unionised - 13-12-2005 at 10:47

It is for an amine.