How can the major product of the addition reaction of 2-bromo-3-chloro-2-butene with hydroiodic acid be predicted, since Markovnikov's rule fails to distinguish it?



1 Answer 1


TL;DR - Sometimes you need to do the experiment.

Like all questions involving Markovnikov's rule, you should compare the structures of the two carbocations:

At C2 (Bromo)

$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-CH(Cl)-C+(Br)-CH3}$$

The two issues to consider are resonance and induction.


You could draw a resonance structure to show that the bromo group is stabilizing the carbocation:

$$\ce{R2C+-Br <-> R2C=Br+}$$

However, since the bromine atom is much larger than the carbon atom, it cannot form as effective $\ce{p}-\pi$ overlap with carbon.


The carbocation has two stabilizing alkyl groups. What about that bromo group? Bromine is more electronegative than carbon, but not by much (2.96 vs 2.55 on the Pauling scale). Plus, bromine has a lot of electron density around it to donate, if it could. Bromine thus is a mild electron-withdrawing group by induction.

At C3 (chloro)

$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-C+(Cl)-CH(Br)-CH3}$$

We once again examine resonance and induction.


$$\ce{R2C+-Cl <-> R2C=Cl+}$$

Chlorine is smaller than bromine and closer in size to carbon. Thus, chlorine is a better resonance stabilizer than bromine.


The same two alkyl groups are present, so we will discount them. Chlorine is more electronegative than bromine (3.16 on Pauling), and it has less electron density around it to begin with. Chlorine is more powerful electron-withdrawer by resonance than bromine.


                  Chlorine              Bromine
Resonance         Stronger donor        Weaker donor
Induction         Stronger withdrawer   Weaker withdrawer

In many cases, resonance usually wins, so I might expect carbocation formation preferentially at C3 with the chloro group, but I would expect the other product to form also.

Now... what else might happen.

Regardless of the initial regioselectivity, the carbocation also can be stabilized by forming a halonium ion. Bromine is better at this since it is larger (and less electronegative so it can stabilize the positive charge better).

This halonium would be attacked at the less hindered position (C2), resulting in transfer of the bromine to C3.

So in effect, you can get three products:

$$\ce{CH3-CHCl-CIBr-CH3}$$ $$\ce{CH3-CICl-CHBr-CH3}$$ $$\ce{CH3-CClBr-CHI-CH3}$$

You would then want to actually do the experiment to determine which of these is really the major product.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.