[Edher]

Saludos,

       This whole deal about Aromatic Substitution Reactions is confusing the living neurons out of me.

My book says that to determine whether  a group is ortho, para or meta follow these instructions:

When two groups of the same type (both ortho, para, or both meta) are present, the stronger director wins out.

When two groups of different type (one ortho, para and one meta) the ortho-para guides the reaction.

This is was my attempt at the following:

[victor]

If me, I determine whether a group is an o,p-director or m-director by the tendency of electronegativity strength, free elctron pairs and some formal charges.
for benzaldehyde, you can see that aldehyde group is composed of O and H atoms attached into C atom so it would be H-C=O. We know that O atom has a high electronegativity, so it would draw electrons toward it and indirectly this action makes the C atom has a positive partial charge.
If another group enters benzaldehyde, it has three possibilities, ortho, meta or para. If this group enters in o,p-positions, it would lead to positive charge overlapping between benzenonium ion and C atom's positive partial charge. That's why meta position is preffered because it doesn't lead to that "overlapping"...

[movies]

This is a pretty subtle effect, but if you really think about it, this makes sense.

In general, we draw reaction mechanisms with electrons from a nucleophile attacking an electrophile, right?  So think about that in terms of aromatic substitution reactions: an excess of electrons (from an electron donating group, an o-p director) will make that initial attack on the electrophile faster.  However, meta directors, which are electron withdrawing groups, remove electron density from the aromatic ring.  The consequence of that is a less nucleophilic aromatic system.

Think of it this way: EWGs slow down the reaction, and especially slow it down at the o-p positions because of the resonance structures you can draw.  The meta position isn't as deactivated as the others, so reaction happens there.  In the case of EDGs, however, the o-p positions have a lot more electron density so they are very activated.  Even if there is another EWG on the ring, the EDG will increase the electron density at the o-p positions, and that's all that really matters for where the reaction takes place.

It's a little complicated, I know, and it took me a long time to figure out why this makes sense, but I think you can figure it out too!