I came across this doubt while solving a problem. I can't understand to which path (1 or 2) I should give priority, as both sound equally good to me.

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Primary carbenium ions are very unstable. Typically, they will not form unless the positive charge is very well stabilised — so a benzyl cation ($\ce{Ph-CH2+}$) will form but an ethyl cation ($\ce{H3C-CH2+}$) will not form under ambient conditions.

Secondary carbenium ions are much more stabilised by double hyperconjugation, but they are still subject to rapid Wagner-Meerwein rearrangements to form more stable tertiary carbenium ions. However, if there is a sufficient possibility of stabilisation inherent to the molecule, forming them is no problem. Stabilisation could arise from:

  • a geminal heteroatom donating an electron pair to delocalise the cation onto said heteroatom
  • a vicinal group that can donate an electron pair in a similar manner, forming a three-membered ring
  • a neighbouring double bond that can resonate with the carbocation generating an allyl-type cation

In your case, the vicinal bromine is able to form a bromonium ion as in your path I. Thus, the secondary carbenium ion is reasonably stabilised and will be the major product.

Note that protonation of the double bond is reversible. Protonation leading to the primary carbocation will — if no immediate Wagner-Meerwein migrating hydride is observed — quickly eliminate back to the double bond, further favouring path I.

This observation is a classic example of Markovnikov’s rule. It states:

When adding $\ce{H-X}$ to a double bond, the hydrogen will add to that side of the double bond which already has a higher number of hydrogens.

Markovnikov’s rule is not strict (e.g. it fails completely for borane addition) but its basis is in the formation of the more stable carbenium ion as first addition step of the mechanism.


Without looking up data, I would firstly assume that the $\ce{-CHBrCH3}$ group acts like any other alkyl group, and that path I would be preferred (by Markovnikov's rule). Bromine is electron-withdrawing via the inductive effect, but only very weakly.

I did go and look up data for this particular substrate. The reference is J. Org. Chem. 1936, 1 (4), 393–404. On page 399 the authors describe the addition of $\ce{HBr}$ to the above substrate.

In the absence of air and peroxides, however, 3-bromo-1-butene gives mostly (60 per cent.) 2,3-dibromobutane.

2,3-Dibromobutane is the product arising from path I, which confirms my assumption.

  • $\begingroup$ I don't know whether Markovnik rule can be applied here, but another way to arrive at why 1 is preferred is due to the higher stability of the secondary carbocation than the primary due to inductive effect of alkyl groups and hyperconjugation $\endgroup$ – Amritansh Singhal Sep 14 '16 at 14:55
  • $\begingroup$ That is precisely the rationale for Markovnikov's rule, so you are saying the same thing I did, albeit in a more detailed fashion. I was rather lazy to type all that out because I assumed whoever was reading it would understand why the rule applies. $\endgroup$ – orthocresol Sep 14 '16 at 14:56
  • $\begingroup$ Might be. I had commented because the markovnikov rule I've read talks about the addition of negative part of the addendum while in this case there is no mention of the negative part of the addendum. Though I think OH- can be regarded as the negative part and then we can say that markovnikov rule is applicable. $\endgroup$ – Amritansh Singhal Sep 14 '16 at 15:00
  • $\begingroup$ But if HBr along with peroxides is supplied then what would be the pathway? Actually this doubt is a part of the problem where HBr + ROOR was added as reagents. $\endgroup$ – DIPANJAN Chowdhury Sep 14 '16 at 15:04
  • $\begingroup$ @DIP If that was part of your question you should have asked it at the start, then I would have known to answer it. ): The same article that I cited describes the addition with peroxides. The selectivity is reversed, as is expected from a radical addition, which proceeds with anti-Markovnikov regioselectivity. The product distribution is 80% 1,3-dibromobutane and 20% 2,3-dibromobutane. [...] $\endgroup$ – orthocresol Sep 14 '16 at 15:16

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