It's mentioned in Peter Sykes , A Guidebook To Mechanism In Organic Chemistry , 6th edition, 13th impression by Pearson, Pg. No. 185 , that on addition of $\ce{HBr}$ to bromoethene the only product obtained is 1,1-dibromoethane. It's due to the positve mesomeric effect of bromine to stabilize the carbocation in this case, however the rate is remarkably slower as compared to addition to ethene.

enter image description here1

Another reaction corresponding to addition of $\ce{Br2}$ to ethene proceeds with the formation of a bromonium ion intermediate.

enter image description here2

My question is, why can't the first reaction just proceed with a bromonium ion intermediate, like (28) above can form a bromonium ion and give 1,2-dibromoethane as a product (there may be a possibility that the carbocation may be formed away from bromine to counteract the strong negative inductive effect and in that process it may interact with the lone pairs of bromine to form bromonium ion)? And if bromonium ion is unstable (that is if it can't be formed in 1st reaction) why does it form at all in reaction 2? Why can't the Bromonium ion in the 2nd reaction just undergo a 1,2- hydride shift and convert into the "more stable" ion like 29b. It should also lead to formation of geminal dibromide .It's a circular argument.

It's clearly mentioned that bromonium ion and carbocation are canonical forms of each other , which means you expect them to convert into bromonium ion whenever there's a positively charged carbon near a bromine atom.

Don't concentrate too much on the attacking species. I've clicked it on between the reaction mechanism where it stops that Bromonium ion and $\beta $ carbocation are canonical forms.3

Here's a similar example of phenonium ion.

[4]: https://i.sstatic.net/0j5Ww.jpg .

The carbocation doesn't stay the way it is in structure 4. It rather forms 3. Notice the bond between phenyl ring and the carbon atom on the right. It's not a completely solid bond.

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    $\begingroup$ Surely, you can also formulate bromonium formation for the first reaction. But clearly, the carbocation in alpha position to the brom atom will be better stabilized due to the mesomeric stabilization. $\endgroup$ Commented Jan 22, 2018 at 17:02
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    $\begingroup$ If you then consider that formation of the carbocation is likely the rate determining step, it makes sense to assume that essentially only the pathway via the alpha-carbocation is populated. $\endgroup$ Commented Jan 22, 2018 at 17:03
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    $\begingroup$ Anything can be formulated only question if it is reasonable. In the second reaction though no alpha cation can be formed. $\endgroup$ Commented Jan 22, 2018 at 17:43
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    $\begingroup$ Now we are getting somewhere. So you propose a 1,2 hydride shift after the bromonium ion has been formed. Thats a good question. I dont know for sure why this doesnt happen. Maybe the bromide is just the better nucleophile. $\endgroup$ Commented Jan 22, 2018 at 17:53
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    $\begingroup$ After some consideration I would say that the bromonium ion is likely even more stable than the alpha-bromonium ion you proposed after a possible hydride shift, which would explain all observations. $\endgroup$ Commented Jan 22, 2018 at 18:32

2 Answers 2


You are not going to see the formation of a primary carbocation—the activation energy is too high. Even though it is true the following are resonance structures of one another:

bromonium resonance

the geometry in which the primary carbocation is formed does not allow for the simultaneous formation of the bromonium ion:

primary carbocation formation

The bromonium would only be able to form after a rotation of the $\ce{C-C}$ bond, so we need to consider the primary carbocation sans-bromonium as the real product of the protonation of the alkene. This is a very high energy product, so it will be slow to form at low temperatures and is out-competed by the other reaction. Since positive charge on carbon-1 can immediately be delocalized to the bromine, this product of protonation is more stable (meaning it forms faster).


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First of all, structure (9)-bromonium ion intermediate and structure (10)-the open intermediate aren't canonical forms of each other simply because "In most resonance, $\sigma$ bonds are not involved(except in hyper-conjugation but what we have here is not hyper-conjugation), and only the $\pi$ or unshared electrons are put in different ways and canonical forms don't really exist(its our imagination) but reaction intermediates like bromonium ion and structure (10)-type intermediates have been isolated under special conditions." Had structure(9) and (10) been canonical forms of the same molecule, we would've never been able to isolate structure(9) and (10). The positive charge in structure(10) might be stabilized by an attraction for Br but that does not involve a full bond.(see the image below)

enter image description here

The possibility that "the carbocation may be formed away from bromine to counteract the strong negative inductive effect and in that process it may interact with the lone pairs of bromine to form bromonium ion" sounds unreasonable to me because the carbocation will always form on the bromine side as the +M effect of bromine that stabilizes (29a) & (29b) is way more than its destabilizing -I effect.

In no way is bromonium ion intermediate less stable than structure (10).
Ab initio molecular orbital studies show that bromonium ion intermediate is more stable than the intermediate in reaction 1(structure-(10)).

In the addition of HBr to bromoethene, it follows the convention: "the electrophile attacks the substrate before the nucleophile does". Electrophile need not actually be a positive ion but can be the positive end of a dipole or an induced dipole, with the negative part breaking off either during the first step or shortly after. So, the formation of the carbocation after the attack of the proton is inevitable.

But due to the fact that in reaction 1, the electrophile is hydrogen ion, the reaction takes place through the mechanism involving structure (10) and not the one in reaction 2 involving the bromonium ion intermediate(9).

I hope you realize that (2) never forms (4). Your statement saying that (4) forms (3) interconvert among themselves is not correct rather (3) and (4) are separate species formed through different set of mechanisms. None of the structure is a canonical form of the other.

enter image description here

The above reaction clearly shows that (12) and (9a) are not really canonical forms of each other. I really don't get the sense in which that statement on pg. 180 has been written but it definitely doesn't specifically mean that (9) and (10) are canonical forms of each other.

Don't get confused by the reaction that I mentioned above, Peter Sykes says This neighbouring group participation by bromine does not ofcourse prove that addition to alkenes proceeds via cyclic bromonium ions, in the past,the reaction was just used to show that such species(cyclic bromonium ions) are not just mere assumptions. That's all!

  • $\begingroup$ Structure 10 or 28 and bromonium ion intermediate are not canonical forms of each other...didn't i say that earlier??...because two canonical structures of the same compound have the same number of sigma bonds...that's a basic thing dude... $\endgroup$
    – Carrick
    Commented Jan 21, 2018 at 17:46
  • $\begingroup$ Can you please tell me the problem that you are still facing $\endgroup$
    – Carrick
    Commented Jan 22, 2018 at 7:43
  • $\begingroup$ @gowreeshmago I've edited the answer, hope its understandable now. $\endgroup$
    – Carrick
    Commented Jan 22, 2018 at 16:45
  • $\begingroup$ Its mentioned in Peter sykes that they are canonical forms. It's not me saying that. $\endgroup$
    – user50247
    Commented Jan 22, 2018 at 16:52
  • $\begingroup$ Can you tell the page number. $\endgroup$
    – Carrick
    Commented Jan 22, 2018 at 16:53

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