Consider the addition of HBr (in absence of any radical initiators or environment that promotes radical mechanistic pathway), to 3-bromocyclopentene.


The first step is the attack of $\ce{H^+}$ On the exposed π electron cloud. But my query is whether it forms a classical carbocation like this one to oppose the negative inductive effects of the halogen atom:

(R)-bromocyclopentane-1-ylium cation

Or does it proceed with the formation of bromonium ion like this one to partially complete its octet using lone pairs on Br.

cyclopentene-bromonium ion

These two would lead to different products but I'm interested in the major one. The first one would lead to 1,3-dibromocyclohexane, while the second one would lead to 1,2-dibromocyclohexane.

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    $\begingroup$ Yeah, there should be bromonium here, but it's not what is called non-classical carbocation. $\endgroup$ – Mithoron Jan 19 '18 at 15:24
  • $\begingroup$ But sir I have been told that the inductive effect is so strong that it lead the the 1st pathway. But I feel it should be the one with bromonium ion. As in anchimeric effects during substitution reactions. Is there a strong justification for that? $\endgroup$ – user50247 Jan 19 '18 at 15:33

Since it is electrophilic addition, it can go via 2 ways. Either by forming classical carbocation or cyclic carbocation. Since the electrophile will be H+ it will go with classical carbocation. Therefore product will be 1,3 dibromo cyclopentane. Also taken in account the mesomeric effect of bromine atom

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  • $\begingroup$ Why would it go with the classical carbocation pathway? You have not given the explanation for that, please elaborate a bit. $\endgroup$ – user50247 Jan 19 '18 at 12:15
  • $\begingroup$ And it's not a cyclic carbocation btw. It's bromine that's carrying the positive charge (if you are referring to this). $\endgroup$ – user50247 Jan 19 '18 at 12:18

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