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,
Kekulé suggested that two forms of benzene were in rapid equilibrium:
We now look at benzene using two different possible approaches to try to describe its stability.
Remember with resonance structures, neither of the extremes actually exists – the structure is somewhere in between.
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)
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!
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.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.
Monocyclic compounds with alternating single and double bonds are termed Annulenes.
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  annulene and  annulene
It was eventually made in 1965 but has a very short lifetime. It is highly unstable – more unstable than it is “Anti-aromatic”.
- If, on ring closure, the p electron energy of an open chain polyene (alternating single and double bonds) decreases the molecule is classified as aromatic.
- If, on ring closure, the p electron energy increases, the molecule is classified as antiaromatic.
- If, on ring closure, the p electron energy remains the same the molecule is classified as non-aromatic e.g. COT (just a polyene).
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?
Thus delocalised p electrons cause peripheral protons to absorb at very low magnetic field strengths.Used as a criterion for Aromaticity.
Consider  annulene (4n + 2 electrons with n = 4)
6 inner protons d -3.0 ppm
X-ray structure of  annulene shows that it is very nearly planar – no bond alternation (double / single) supports delocalisation.
Possible definition of Aromatic Compounds“Cyclic systems which exhibit diamagnetic ring current and in which all of the ring atoms are involved in a single conjugated system.”
An interesting non-benzenoid aromatic compound is Azulene, which has large resonance energy and a large dipole moment.
In electronic terms pyridine is related to benzene.
Note. Anthracene often undergoes normal SEAr reactions.
Thiophene has more aromatic character than furan.
(Protonation of pyridine further reduces reactivity.)