I'm trying to understand what halogens do in electrophilic aromatic substitution. Consider the following.

Bromobenzene is introduced to a carbocation. Bromobenzene will likely react, but at a slower rate than unsubstituted benzene, because the bromine is inductively withdrawing. The bromine is also not resonance donating/activating, but can stabilize the intermediate product through resonance donation.

The bromine is an ortho/para director, but not because it donates electron density to these positions. Rather, when bromobenzene attacks a carbocation, the meta positions are formally positive, and through resonance donation, bromine can complete an octet in one of the resonance depictions.

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Is this an accurate depiction of what a halogen does to benzene?


The bromine is also not resonance donating/activating

You can't have it both ways. If the bromine can resonance interact with the intermediate, then it can also resonance interact in the starting bromobenzene.

Let's back up to the starting bromobenzene.

Bromobenzene will likely react, but at a slower rate than plain benzene, because the bromine is inductively withdrawing

That's correct, but let's compare further. The oxygen in phenol is electron withdrawing (oxygen's electronegativity is even greater than bromine's), yet phenol is a very reactive aromatic compound towards electrophilic aromatic substitution (it will decolorize a solution of bromine, it doesn't even need a Lewis acid catalyst). The difference between bromobenzene and phenol is that the phenolic oxygen can interact strongly via resonance and donate electrons into the aromatic ring, while the bromine in bromobenzene also resonance interacts with the aromatic ring, but much less efficiently (the larger size of bromine's $\ce{4p}$ orbitals makes overlap much less efficient with carbon's 2p orbitals). Although the interaction is much less efficient, there still is some resonance interaction between the bromine substituent and the ring. Look at the analogous resonance structures for bromobenzene that you drew for the intermediate, in the contest between inductive deactivation and weak resonance activation, the ortho and para positions are the least deactivated.

enter image description here

When bromobenzene undergoes electrophilic aromatic substitution, the whole ring is deactivated because (unlike phenol, where the resonance effect overwhelms the inductive effect) the weak resonance effect doesn't offset the strong inductive effect. However, since the bromine substituent does interact with the ortho and para positions by resonance, these two positions are less deactivated that the meta position which cannot interact via resonance with the bromine substituent.

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  • $\begingroup$ So two competing effects? @ron - induction vs resonance? But still not enough induction to completely deactivate benzene from reacting, unlike with a nitro group (both inductively and resonance deactivating)? $\endgroup$ – Dissenter Sep 26 '14 at 6:18
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    $\begingroup$ @Dissenter That is exactly right, enough inductive deactivation to make all positions react slower than benzene, enough resonance activation to differentiate the ortho and para positions from the meta position. Nice analogy with nitrobenzene! $\endgroup$ – ron Sep 26 '14 at 13:33
  • $\begingroup$ thank you; so any substituted benzene will react with a halonium ion, but not any substituted benzene will react in a Fridel crafts alkylation or acylation? $\endgroup$ – Dissenter Sep 26 '14 at 13:36
  • $\begingroup$ No, I don't think so. Just as some substituted aromatics are to deactivated to react in a Friedel-Crafts reaction, I suspect there are also some very deactivated aromatics that won't react with halonium ions. $\endgroup$ – ron Sep 26 '14 at 13:40
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    $\begingroup$ @Dissenter Yes, that's correct. $\endgroup$ – ron Sep 26 '14 at 13:45

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