[Sealion]

Hi guys,

I want to understand what is happening in the picture below


I think what the diagram is depicting is that the electron(s) do not stay in one place, and move around the benzene ring and -OH functional group.  The electron spends more time in the ortho and para positions much more so than the meta position, explaining the δ^- areas.

(Please correct me if I am wrong)

With this in mind, why does it spend more time in ortho, why not meta? I would expect the electronegativity to move away from the -OH group, due to the negative charge of the oxygen.

Also, why is the -OH δ⁺? Shouldn't it be electronegative (due to the oxygen?)

[Corribus]

Benzene, or any aromatic for that matter, is an electron deficient system.  Oxygen has a lot of extra unpaired electrons hanging around.  This is a general explanation for why benzene is partially negative and oxygen is partially positive.  True oxygen is fairly electronegative in general, but benzene is moreso.

As to why certain positions are partially negative and others (meta positions aren't), the resonance structures provided already give you at least a rudimentary explanation.  The presence of an oxygen provides additional resonance structures, but I challenge you to draw one where the negative charge is at the meta position.  This is why meta position doesn't have a partially negative charge.  The nature of resonance structure does not allow the formal charge to be located at this position - hence no partial negative charge.

A better way to see this would be to do a modified but simple Huckel treatment of phenol.  Then you would probably see quite clearly that electron density tends to localize in the ortho and para positions but not meta.  (This also explains why reactions with a substituted aromatic ring tend to happen at some positions but not others.)

[Sealion]

Thank you, that makes sense