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I wonder what the product of the above reaction is, that is, bromination of the given alkene in non polar solvent carbon tetrachloride (proceeds through cyclic bromonium ion intermediate).

Once the cyclic bromonium ion is formed, can the ring expand as shown or is the usual product formed? Five remembered rings happily expand to six membered ones if possible - so what's going on here?

It'd be great if you could quote sources with your answer - and help me understand whether or not species involving cyclic bromonium ions can rearrange in a reaction, and why?

Thanks!

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  • $\begingroup$ I did a quick search, and it appears to be possible, at least for norbonyl type structures, but I haven't seen other examples yet. For example: pubs.acs.org/doi/abs/10.1021/jo401888f $\endgroup$ – Zhe Jul 19 '18 at 13:21
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This is apparently not a likely rearrangement, even in more strained circumstances. There are a number of reactions out there in the literature like the one below.

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This result makes sense if we consider the relative stability of the bromonium ion vs. your proposed rearrangement. I've redrawn the mechanism to show all the formal charges. The bromonium should be much more stable (positive charge delocalized over two carbon atoms, all atoms have full octet) compared to a secondary carbocation.

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Bromonium rearrangements are not entirely impossible though, check out this fun example (J. Org. Chem., 1998, 63, 2646 DOI: 10.1021/jo9722055)

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As with all chemical reactions, the product(s) is/are determined by either kinetic or thermodynamic control.

Under thermodynamic control (where the transition-state activation energies are relatively low and all reactions are reversible) the products equilibrate to the lowest energy product. Under kinetic control (where the rate determining step is effectively irreversible), the difference in activation energies determines the product ratio. You might properly guess this is a kinetically-controlled reaction, since 1,2-dibromoalkanes do not spontaneously form bromonium ions and Br- in CCl4.

If you draw out the elementary steps of the reaction mechanism, following the rules for arrow-pushing (described, for example, here) you should be able to identify the step which determines the product:

Bromonium ion rearrangement

The rearrangement product requires going through a non-stabilized secondary carbocation transition state (4) versus direct opening of the bromonium ion as in (1). For a deeper understanding, consider what changes to the structure or reaction conditions might favor the rearranged product.

Ref: Advanced Organic Chemistry: Part A: Structure and Mechanisms. Francis A. Carey, Richard J. Sundberg, pp. 362-371

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  • $\begingroup$ What's your source, or is this just a guess? $\endgroup$ – arya_stark Jul 17 '18 at 1:31
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First of all, the product you drew containing only one bromine atom is not correct since you added bromine to it, so it should have 2 bromine atoms ( there are some stages in this addition, but to skip to the most important ones, the bromine molecule suffers a heterolytic cleavage to form bromonium and bromide, the bromonium adds to the double bond according to Markovnikov’s rule. Now, the bromonium ion you drew where bromine is bonded to 2 carbon atoms is the result of the carbocation “attacking” one electron pair of bromine in an attempt to stabilise itself. Basically, once bromonium reacted with the double bond, you are left with a positive charge which will be offset by the bromide ion ). Under no circumstance can you have an organic compound with one bromine atom when you add bromine to an alkene. Probably you forgot to put a second atom of bromine there. Anyway, in every reaction where you encounter carbocations, there will be rearrangements occuring, but there are some exceptions such as this one. Normally, if you added hidrobromic acid, you would have rearrangements because there will not be any cyclic ions forming. In this case, like I told you, a carbocation will be formed when bromonium ion reacts with the double bond. That carbocation, however, will not rearrange itself because it has found another method to stabilise itself: it forms a bond with the bromine atom, generating a cyclic ion where the positive charge is more spread and thus more stable. And then the bromide ion attacks from the opposite side.

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  • $\begingroup$ I haven't forgotten to put the second bromine atom, it's simply not there. If the reaction proceeds via ring expansion as proposed by me, there's no step in which the bromide ion would possibly attack and thus remains in solution. $\endgroup$ – arya_stark Jul 13 '18 at 14:46
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    $\begingroup$ @schrodinger_16, AndrewB is correct. There should be a carbocation remaining after the rearrangement, which could be attacked by bromide. That said, I think you have still asked an interesting question. I hope to have time to provide an answer. $\endgroup$ – jerepierre Jul 13 '18 at 15:58
  • $\begingroup$ If the compound with a cyclohexane ring has only one bromine atom, according to the rearrangement you proposed, there should be a positive charge residing on the tertiary carbon in the ring, so the compound is not electrically neutral ( if you said that the bromide stays in solution ). $\endgroup$ – AndrewB Jul 14 '18 at 16:56
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Here is a reference. http://research.cm.utexas.edu/nbauld/teach/alkenes3.html#bromination Read the part “electrophilic addition of bromine and other halogens”. Technically, to refine my argument, the positive charge is mainly on the bromine.

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  • $\begingroup$ Why don't you merge the answers? You can edit to do so. $\endgroup$ – Avnish Kabaj Jul 16 '18 at 6:44

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