Aromatic Compounds and Aromaticity
Solomons 6th Edition
Chapter 14 p 614 – 654
Chapter 15 p 655 – 703 (Reactions)
You will by now be familiar with the structure of benzene C6H6
Discovered in 1825 by Michael Faraday (RI).
Molecular formula deduced by Mitscherlich in 1834.
The fragrant odour of benzene and its derivatives led them to being classed as “aromatic”. This classification now has a chemical meaning – “aromaticity” is associated with a special stability resulting from structure.
Elucidation of the structure posed a problem – the molecular formula C6H6 indicated a highly unsaturated compound (double and/or triple bonds) but benzene does not show this behaviour.
Kekulé (1865) conceived a cyclic structure,
but this would imply alternating single and double bonds (C-C = 1.47Å, C=C = 1.34Å).
Kekulé suggested that two forms of benzene were in rapid equilibrium:
Later spectroscopic evidence showed all bond lengths to be equal and intermediate between single and double bond lengths (1.39 Å). It was also found that benzene was a flat (planar) molecule.
We now look at benzene using two different possible approaches to try to describe its stability.
Resonance hybrid, 2 canonical forms
-Try to represent both single and double bond character of each bond.
Remember with resonance structures, neither of the extremes actually exists – the structure is somewhere in between.
Further, all bond angles in benzene are 120º (revise 12.5), p electrons are delocalised.
Resonance theory states that if more than one resonance form can be drawn for a molecule, then the actual structure is somewhere in between them. Furthermore, the actual energy of the molecule is lower than might be expected for any of the contributing structures. If a molecule has equivalent resonance structures it is much more stable than either canonical would be – hence the extra stability of benzene (called resonance energy).
The bond angles of 120° in benzene suggests that C atoms are sp2 hybridised. An alternative representation therefore starts with a planar framework and considers overlap of the p orbitals (p electrons).
Mix n x p atomic orbitals 
np molecular orbitals!
Remember ethene? (p 26)
The exact calculation of their position (shown below) is beyond our discussion.
Each MO can accommodate 2 electrons, so for benzene we see all electrons are paired and occupy low energy MO’s (bonding MO’s). All bonding MO’s are filled. Benzene is therefore said to have a closed bonding shell of delocalised p electrons and this accounts in part for the stability of benzene.
There is a simple “trick” for working out the orbital energies (625):
Frost-Musulin diagrams - polygon in a circle. Draw the molecular framework of a cyclic system of overlapping p-orbitals, making sure you put an atom at the bottom. Atomic positions (positions of p-orbitals) then map on to the energy level diagram!
This leads to the very important
Hückel’s Rule: The (4n + 2) p Electron Rule
For monocyclic planar compounds in which each atom has a p orbital (as in benzene) Hückel showed that compounds with (4n + 2) p electrons, where n = 0, 1, 2, 3 etc, would have closed shells of delocalised p electrons and should show exceptional stability (high resonance energy º “aromatic”).
i.e. planar monocycles with 2, 6, 10, 14….delocalised p electrons should be “aromatic”.
i.e. p electrons are delocalised over the entire ring and the compound is thereby stabilised by the delocalisation.
Consider planar cyclooctatetraene (COT) (8 p electrons).
Firstly construct the ‘polygon in a circle’.
No closed shell and 2 unpaired electrons in each of 2 non-bonding orbitals! Molecules with unpaired electrons are typically unstable and reactive.
Therefore a planar form of COT should not be aromatic.
Because no stability is gained by becoming planar it assumes a tub shape.
COT is non-aromatic and in fact stability would be lost if it became planar.
Monocyclic compounds with alternating single and double bonds are termed Annulenes.
Thus: benzene is [6] annulene and COT is [8] annulene.
Remember Hückel’s rule predicts that annulenes
will be aromatic if
i)
they have (4n + 2) p electrons
ii)
they have a planar C skeleton
A study of annulenes has verified Hückel’s rule.
Consider [14] annulene and [16] annulene
What about [10] annulene? - predict it would be a stable aromatic compound. However, H’s interfere preventing planarity therefore it is not aromatic.
(Note: naphthalene. Not really a test of Hückel’s rule since it is bicyclic but we can regard it as a similar case if we look at periphery!)
What about [4] annulene (cyclobutadiene)?
(Draw polygon in a circle for yourself)
It was eventually made in 1965 but has a very short lifetime. It is highly
unstable – more unstable than 
it
is “Anti-aromatic”.
e.g.
NMR as a test for aromaticity. (p 627)
Key evidence for electron delocalisation is provided by NMR.
Fact: 
Has
a single unsplit signal for H at d 7.27 ppm. This tells us that
all H are equivalent.
Importantly the signal appears at a low field strength – so the nuclei are deshielded compared to normal alkene protons.
How is this explained/understood in terms of electron delocalisation?
Induced magnetic field tries to ‘oppose’ (‘neutralise’) applied filed B0. But (since magnetic lines of force are continuous) at the position of the protons of benzene the applied field is reinforced by the field produced by the circulation of p electrons.
This causes the H nuclei to be strongly deshielded – the protons sense the sum of the two fields and therefore the applied field B0 does not have to be as high (strong).
Thus delocalised p electrons cause peripheral protons to absorb at very low magnetic field strengths.
Used as a criterion for Aromaticity.
Consider [18] annulene (4n + 2 electrons with n = 4)
12 outer protons d 9.3
6 inner protons d -3.0 ppm
X-ray structure of [18] annulene shows that it is very nearly planar – no bond
alternation (double / single) supports delocalisation.
“Cyclic systems which exhibit diamagnetic ring current and in which all of the ring atoms are involved in a single conjugated system.”
Cyclopentadiene is unusually acidic (pKa 16)
In contrast, pKa of cycloheptatriene is 36. Loss of HYDRIDE is unusually easy, however, because it leads to an aromatic cation – tropyllium ion.
We have seen that benzene exhibits unusual stability compared to “cyclohexatriene” structure.
Also
Difference (357 – 207 = 150 kJ/mol) is called the “Resonance Energy” of benzene.
An interesting non-benzenoid aromatic compound is Azulene, which has large resonance energy and a large dipole moment.
So far we have only considered carbon skeleton compounds. However, many compounds we find in nature are cyclic compounds with an element other than carbon in the ring. These are called Heterocyclic compounds. Further, some are aromatic compounds - can be termed heteroaromatic.
However, the degree of aromaticity (extra stability) may vary as the heteroatom changes.
In electronic terms pyridine is related to benzene.
Pyrrole has electrons arranged differently – related to the cyclopentadienyl anion.
(Similar electronic configurations for furan and thiophene)
In the Diels Alder reaction a double bond adds to a 1,3 conjugated diene (4+2 cycloaddition) to give a 6-membered ring.
Favoured by electron withdrawing groups on the dienophile and electron donating groups on the diene e.g.
An indication of the stability of benzene over that indicated by 
is
that it does not undergo a Diels Alder reaction, despite the fact that
we can ‘locate’ a diene fragment in its structure.
But interestingly
Exhibits diene behaviour – note product still has 2 benzenoid rings.
Note. Anthracene often undergoes normal SEAr reactions.
Thiophene has more aromatic character than furan.
Electrophilic aromatic substitution in heteroaromatics compared to benzene.
We know
And that
Pyridine – contains electron withdrawing N in place of CH \less reactive towards electrophiles (E+) - p deficient system.
(Protonation of pyridine further reduces reactivity.)
The 5-membered heteroaromatics furan, pyrroles and thiophene can be regarded as p excessive systems (6 p electrons over 5 atoms).
