I am currently studying for an organic chemistry exam and I just couldn't come up with a solution to the following question:

Consider the reaction of 1-Propene + Br2 with water as solvent.

Why will the nucleophilic water molecule add on C-2 and rather than on C1 of the Bromonium-Ion? The answer seems to be, because in the transition state the bonding between Br and C will dissociate quicker than the bonding between O and C will form, which leads to a positive charge on that C-Atom in the transition state and therefore water will add on the more stable "carbokation-like" carbon. My question is: why does the Br-C bonding dissociate quicker than the O-C bonding forms? (I read that in the pearson organic chemistry textbook). Does it have anything to do with the Hammond-postulate? Or are there any other thermodynamic or kinetics reasons? So I am basically searching for an explanation why one bond would dissociate quicker than a different bond would associate, because I guess that's the reason why there is a partial positive charge that would then be more stable on the higher substituted carbon...

Unfortunately I haven't found an answer on the internet yet and also not in other textbooks. Maybe you can help :) Thanks!

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  • 1
    $\begingroup$ I learnt that halohydrin addition proceeds through a classical carbocation rather than a non-classical one (like this mechanism) and that the carbocation on the disubstituted C atom is more stable than the carbocation on the monosusbtituted C atom, which is why you get a secondary alcohol. $\endgroup$ Nov 26, 2020 at 14:06
  • $\begingroup$ What @AniruddhaDeb said. $\endgroup$
    – Waylander
    Nov 26, 2020 at 14:18
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    $\begingroup$ Why assume that the bromonium ion is symmetrical with all of the positive charge on the bromine? Why not consider it unsymmetrical with part of the positive charge residing on the secondary carbon.? $\endgroup$
    – user55119
    Nov 26, 2020 at 15:37
  • $\begingroup$ @user55119 Yeah, some explain it by electronegativity (inductive effect), but as I mentioned, the textbook explained it differently - that the partial positive charge in the transition state is due to different speed in bonding association and dissociation... which of course leads to different energy levels of the transition state (bc positive charge is more stable on the higher substituted carbon) and therefore the reaction speed is higher. $\endgroup$
    – Felix H.
    Nov 26, 2020 at 16:54
  • $\begingroup$ @AniruddhaDeb well, that would explain where the positive charge comes from... do you have any references (articles, textbook, etc.)? $\endgroup$
    – Felix H.
    Nov 26, 2020 at 16:56

1 Answer 1


This is not a 100% answer, but it was too detailed to fit in the comment section.

Following up from the comment, here's a reference from Inorganic Chemistry by Morrison and Boyd:

There is evidence, of a kind we are not prepared to go into here, that these compounds are not formed by addition of preformed hypohalous acid, $\ce{HOX}$, but by reaction of the alkene with, successively, halogen and water.

Sorry, but I'm too lazy to ChemDraw this atm :P

Halogen adds (Step 1) to form the halonium ion; this then reacts, in part, not with the bromide ion, but with water (Step 2) to yield the protonated alcohol.

Propylene gives the chlorohydrin in which chlorine is attached to the terminal carbon. This is typical behaviour for an unsymmetrical alkene; orientation follows markovnikov's rule, with positive halogen going to the same carbon that the hydrogen of a protic reagent would. Now, this orientation would be perfectly understandable if the intermediate were an open carbocation: the initial addition of halogen yields the more stable carbocation - secondary, in the case of propylene. But the stereochemistry indicates that the intermediate is not an open carbocation, but a cyclic halonium ion.

And that's all. Quite anticlimatic, as I was expecting a reason for the regioselectivity. Most other references and papers I have seen online also do not have a concrete explanation for the regioselectivity, and they resort to hand-wavy explanations involving resonance for the same. This reaction mechanism is also quite similar to Oxymercuration-demercuration, whose regioselectivity is again explained by the "three resonating structures". I could not find diagrams of these resonating structures, and while this question asks the same, the resonance structures or the intermediate are not properly explained in the answer.

My best suggestion for you would be to accept the explanation in your textbook as that is adequate for now. Until someone more knowledgable posts a better answer with references, that should suffice.

  • $\begingroup$ Well, thank you for your detailed comment. I really like studying theories and mechanisms in much detail, until I really understand what is going on and why. Unfortunately nobody seems to go into depth why the regioselectivity is the way it is here. Maybe the answer is found in quantum physics and thus not easy enough to be in a basic organic chemistry book, but let's see - maybe someone can come up with another explanation, although I doubt that, since I haven't even found studies/articles about it in chemistry databases... $\endgroup$
    – Felix H.
    Nov 27, 2020 at 16:09

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