Why are aromatic compounds stable?

Question

Delilah Anderson · Accepted Answer

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.

Paisley Boney · Answer

undefined Aromatic compounds are stable due to the resonance stabilization provided by delocalized pi electrons within the ring structure, which distributes electron density evenly around the ring. This resonance leads to lower overall energy and increased stability compared to non-aromatic compounds.