[alexmahone]

Q. On bromination, the electron rich phenoxide ion will be attacked most readily:
(A) on the negatively charged oxygen atom
(B) on the ortho and para carbon atom
(C) on the meta carbon atom
(D) on the ortho carbon atom

My attempt:Due to resonance, the ortho and para carbon atoms will gain a negative charge. They will therefore repel the negatively charged bromine atom. So my guess is (C) because the meta carbon atom has no negative charge. However, the correct answer (from the answer key) is (B).

Could someone explain why? Thanks in advance.

[macman104]

Bromination is an electrophilic process, with the ring electrons attacking the bromine, and kicking out a bromide ion.  Look at the process for addition of bromine to a double bond.

http://www.docbrown.info/page06/OrgMechs1b.htm#electrophilic

See how the electrons attack the bromine?

[alexmahone]

So the bromine molecule splits into a bromine cation and bromine anion, and the bromine cation attacks the electron rich sites?

[macman104]

So the bromine molecule splits into a bromine cation and bromine anion, and the bromine cation attacks the electron rich sites?

Well, it splits during the reaction.  Just like in the mechanism attached.  It's more that the electrons from the ring attack the bromine molecule, and then the electrons kick over to from the bromine anion.  You don't actually form the bromine cation per se, but conceptually yes.

[alexmahone]

Why doesn't the 'bromine cation' attack the negatively charged oxygen atom then?

[macman104]

Why doesn't the 'bromine cation' attack the negatively charged oxygen atom then?

Because electrons attack positive charges.  Nucleophiles attack electrophiles.  When you draw a mechanism, you draw from the site of negative to the site of positive.  It's just how things go...I actually can't at the time come up with a chemical explanation.  I could bs, and say that positive is really just the absense of electrons, you can't actually attack with anything.  However, a negative charge actually has something to donate to the relationship.  I have no clue, lol.

[alexmahone]

Thanks for your help. 

[nox]

Why doesn't the 'bromine cation' attack the negatively charged oxygen atom then?

Bond strength perhaps? Maybe a C-Br bond is stronger than an O-Br bond and thermodynamically more favorable?

And maybe also to do with the fact that oxygen is "hard" while bromine is "soft", and the aromatic pi system is more polarizable

[spirochete]

Why doesn't the 'bromine cation' attack the negatively charged oxygen atom then?

That's an excellent question.    Nox gave some good ideas, those would be the same things I'd guess.  Asking "why" questions like this is part of what real chemists do.

A small point about language:  organic chemists typically refer to the nucleophile as "attacking" and the electrophile as being attacked.   I think that's all that MacMan was referring to in his post.

[alexmahone]

So are these the final products?

[nox]

notice how the products you drew are no longer aromatic -- in fact those are intermediates in the bromination reaction, aromaticity is such a great thermodynamic driving force that the proton is actually lost on the brominated carbon to regain aromaticity, and the end products would be ortho/para substituted bromophenols


in reality though I'm not sure if it's possible to get monobromination with the phenoxide anion, I remember reading somewhere it's simply too activated and you'd end up with the tribrominated product (2,4,6-tribromophenol) whereas monobromination with neutral phenol is possible under carefully controlled conditions (low temperature with carbon disulfide as solvent)