Question

Why are aromatic compounds stable?

Answer 1

They have a great deal of resonance stabilization, which is due to delocalization (the spreading out) of the `p` electrons in the conjugated `pi` system.

In general, the more spread out electron density is, the less interelectronic repulsion there is, and the less electrons are "stored" on average within a particular orbital, which decreases the overall energy of the molecule. Sometimes the resonance helps a little, sometimes a lot...

For instance, consider the hypothetical 1,3,5-cyclohexatriene, and compare its enthalpy of complete hydrogenation to that of benzene:

![)

(This would indicate how much energy input it requires to break all the `pi` interactions in the system.)

In theory, if benzene were just like cyclohexatriene (i.e. no resonance at all), its double bonds would absorb `"359.2 kJ/mol"` to break and form cyclohexane.

In reality, however, benzene has about `bb(-"150.7 kJ/mol")` of resonance stabilization, which makes its `pi` bonds about `ul(42%)` more stable than in cyclohexatriene.