[maadijk]

I've read that when the "bridge" is formed by the halogen (electrophilic attack) on an unsymmetrical alkene, most of the positive charge goes to the more substituted carbon. I'm not sure how to determine which one it is, since both carbons are directly bonded to 2 carbons and 1 hydrogen each. There is a phenyl group on Carbon #3's side and a carboxylic group on carbon #2's side. Does it have to do with electronegativity??

[discodermolide]

Look to see where the positive charge can best be stabilised (resonance).

[maadijk]

Alright, so to the right of Carbon #2 there are 2 oxygens, which are more electronegative than carbon, and they fulfil the octet (I think? with the bonds) so they provide high electron density. To the left of Carbon #3 there are just carbons and hydrogens. So, Resonance in this case has to do with spreading out the positive charges... I don't know how to continue

[discodermolide]

Draw the intermediate bridged compound you mentioned. Then open the bridge so that a positive charge appears on the carbon next to the benzene ring.
Now see if you can draw any resonance forms which may help stabilise this charge. Try the same the other way around and see what you get.

[maadijk]

Okay, so I see that in the first case (+ close to the ring), the double bonds int he ring can move around but the + charge would eventually just end up in the original place or move around in circles (?). In the other case I'm not sure if the oxygens can aid in anyway... I don't really understand

[discodermolide]

Well the only oxygen that could help is the OH one, giving a very strained resonance structure.
Moving the positive charge around the ring produces resonance stabilised structures which help the stability of any positive charge accumulating on the benzyllic carbon. Therefore the result of the reaction will reflect this.

[maadijk]

So, The carboxylic group is an electron withdrawing group, while the phenyl is an electron donating group. The phenyl provides resonance stabilization.

If the carbon closest to the ring is the best at stabilizing the positive charge, does it mean that it is also the most substituted one? So most of the + charge from the halonium ion would go to it?

[discodermolide]

It does not mean it is the most substituted one. It means that most of the positive charge would be expected to reside there.

[maadijk]

Thank you!