Hi! Good question. It is true that it goes through a halonium ion, but the “1,4” addition is still possible. What would happen here is that Cl(-) would attack the “4” position, and the double bond would move over to occupy the “2,3” positions with concurrent breakage of the halonium ion to give a product where Cl is attached to both the 1 and 4 positions, similar to the example in Section 5

The reaction rate is related to the energy of the highest-occupied pi molecular orbital of the diene. How does the HOMO of a conjugated diene compare to the HOMO of a non-conjugated diene?

Thank you for your excellent explanation!!! You may want to add it to the main text — it clarifies things nicely.

My confusion came because in the first example of “ion pairing” it seemed as if a positive bromonium ion might merely function as a “lighthouse” guiding over any old Br- that happened to float on by to preferentially attack the carbon (C2) directly attached to bromonium ion over the more distant C4.

But in the 2,5 dimethyl hexa-2-4-diene example a ‘lighthouse’ effect seemed less tenable as C-4 would seem to carry a higher partial positive charge than C-2.

Many thanks for clearing that up!

Hi George –
” So does ion pairing at low temp mean that the VERY SAME Br (which just lost its hydrogen and is now Br-) does not diffuse away and that it stays in the vicinity and it attacks C-2 (in other words, H and Br react almost simultaneously)?”

Yes, exactly! To clarify, the term “ion pairing” usually means “tight” ion pairing, where the opposite charges, once formed, stick together through electrostatic attraction.

[ It’s helpful to remember that these reactions don’t occur in a vacuum – they are surrounded by a shell of (usually inert) solvent, so at low temperatures where these is very little molecular motion, the solvent forms a kind of “cage” around each molecule that requires energy to break out of. One analogy is a “ball pit” that kids like to play in.]

When the Br- is formed it will be in the vicinity of a carbocation, and there will be an electrostatic attraction to this carbon. At low temperatures, “ion pairing” could occur and if the Br- doesn’t have enough kinetic energy to leave the solvent cage, it will just “rebound” to the carbocation, resulting in the 1,2 product.

At higher temperatures, Br- could diffuse away, leaving the resonance-stabilized cation. Then attack at either carbocation becomes possible, with the barrier to addition at “C4” being less due to the fact that it better stabilizes positive charge (and will have a greater partial positive charge).

Under this rationale, the 1,4 product is the “kinetic” product.

However we do *not* call the reaction of Br- with either C2 or C4 in this case, “ion pairing”, since the Br- broke out of the solvent shell.

However, for 2,5 dimethyl hexa-2-4-diene I get confused. First you get protonation on C1 by HBr. So does ion pairing at low temp mean that the VERY SAME Br (which just lost its hydrogen and is now Br-) does not diffuse away and that it stays in the vicinity and it attacks C-2 (in other words, H and Br react almost simultaneously)?

Because if Br- did have time to diffuse away, then I’d think resonance would quickly kick in, and since C4 has relatively more positive charge (than C2) that should attract the next Br- ion that floats on by (by “ion pairing”)??? So C4 would react by “ion pairing.”

Thanks!

No. Protonation at C2 or C3 would result in a non resonance-stabilized carbocation. Much higher activation energy.