[LogicX]

So I get kind of confused sometimes about aromatic rings with substitutions on them, and how that effects the aromatic protons in an NMR.

Let's say that you have a phenol ring.  I know that this is an activating, ortho/para director because of additional stable resonance forms during electrophilic substitution.  You can stick the positive charge on oxygen and give everything an octet.  But when you aren't doing electrophilic substitution, you are just trying to determine which protons would be further downfield on an aromatic ring, I use different reasoning.  I would think that since oxygen is electronegative it withdraws electrons and thus puts a partial positive charge on the carbon it is bonded to.  So the carbon bonded to that carbon (in the ortho position) would have a partial negative charge.  And so on around the ring.  So the meta group would have additional positive charge and thus would be the furthest downfield since more (+) charge means more deshielding.

Is this a valid way to think about it?  Sometimes I don't know if I should think of oxygen as being an electron donating (activating) group, or as being electron withdrawing because it is very EN.

[LogicX]

Ok I just tried to type a better explanation that my original post.  Here we go again:

I can't figure out how aromatic substituents effect aromatic rings.  I can think of it in a couple ways:

Is the group ortho or meta directing?  When we learned about aromatic electrophilic substitution, we obviously learned about these groups.  When an aromatic proton is lost, it will be preferentially lost in a certain position due to the additional resonance of whatever group is attached to the ring.  So this tells you where you can find extra positive charge on the ring (and thus where it would be more deshielded).  For example, an ortho directing group would have more positive charge on the meta positions.  This reasoning seems to work when I am doing problems.

However you aren't actually pulling off protons, you are just looking at their chemical environments.  So take an ortho directing group like OH.  The O is electronegative, so it should have a partial negative charge.  If it has a partial negative charge then that means it drew electrons out of the aromatic system.  So the whole thing would be deshielded, but maybe slightly less deshielded the further away you get from oxygen.  This reasoning doesn't seem to work when doing problems.

So basically, if you take a neutral aromatic ring with a substituent on it such as an ortho/para director like -OH, is there actually deshielding occuring at the meta positions?  We only learned about it in terms of resonance that could be drawn after you pulled a proton off and the ring had an overall +1 charge. 

I guess this boils down to me not understanding when hyperconjugation applies and when resonance stabilization applies.

[Vidya]

O can show electron withdrawing effect through sigma bonds which is due to its electronegative nature.However lone pairs on O can show electron donating effect due to resonance or  delocalization of lone pair within the ring.Now which effect is dominating? It is the delocalization of electrons which will beat the EN efffect.This lead to the more negative charge on Ortho or para than on Oxygen atom.