Why Is the Amine Group in Procaine More Polar in the Para Position?

[nymbler_064]

I have just been trying to teach myself about the effect of activating and deactivating groups on electrophilic substitution. However, I am a bit confused as to why the amine group in procaine (anesthetic) is in the para position, when the aromatic ring is attached to a deactivating ester. What I am really trying to explain is why the molecule is more polar for having the amine group in the para position, as opposed to the ortho or meta positions.
The structure of procaine is available here:
http://upload.wikimedia.org/wikipedia/commons/thumb/4/4a/Procaine.svg/220px-Procaine.svg.png

Thankyou very much.

 

[QuasiP]

Activating and deactivating groups really only affect the reactions that you would perform to synthesize the molecule. In this case, the main product would likely depend on the order that one performs the reactions. The textbook method for generating aromatic amines is a nitration followed by a reduction. To generate esters, it is some kind of Friedel-Crafts acylation.

If one performs the acylation first, the ester group will be meta directing (and deactivating), so the desired product will not be preferred in the nitration.

If one performs the nitration/reduction first, the resulting amine group is both activating and ortho/para directing, so the acylation will be more likely to form the desired product.

Note, however, that these reactions are probably not how the synthesis is performed in practice. I think you would do it like this:

toluene --(nitration)--> p-nitrotoluene --(reduce, reduce)--> p-aminobenzoic acid --(SOCl2)--> acid chloride --(substitute)--> Procaine

The product which is energetically favourable overall is not always the product formed in reactions. That's the beauty of chemistry - it's all about tricking the molecules to do what you want.