Addition of HOBr across C=C

Question

Source: Carey Advanced Organic Chemistry B, problem 4.1 b

Looking for interesting problems about addition to C=C, I came across this in the back of Carey. It looks like a simple addition across the alkene, but using HOBr which is something I've never came across in the lab (and perhaps not seen used as an example since I was an undergrad...).

The most obvious way to dissociate HOBr is to lose a proton, leaving the OBr anion behind, Wikipedia confirms this:

[...] hypobromous acid partially dissociates into the hypobromite anion OBr− and the cation H+.

So far so good. And based on this fact, you might expect the product to be as drawn below. However the original literature points towards a different product, also shown below.

Obviously, the product I've drawn as expected looks a little odd, with the unstable Br-O bond present. However, I cannot rationalise how the actual product has formed.

Ref.: JOC, 1968, 33, 3953.
The original literature from which the problem was taken

Three possibilities:

1. The HOBr doesn't dissociate as I'd imagined. There is the possibility of the Br forming the bromonium ion which is then opened by hydroxide, however to get the product as isolated, you have to attack in the anti-Markovnikov sense (i.e. the product would look like the one formed if you'd attacked a primary carbocation with hydroxide).

2. The product forms as I drew it, but rearranges to give the product they isolate. This is what I consider to be the most likely explanation, but I'm unable to see how this happens.

3. The secondary carbocation forms upon protonation, but this rearranges. Again, struggling to see how this would work as the secondary cation is the most stable one that can form in this system, as there is no possibility of a tertiary carbocation.

Any suggestions to point in the right direction? Even the name of a rearrangement would be helpful.

Answer 1

The 'expected' product comes from analogy of addition of strong acids like HBr or acid (often sulfuric acid) catalyzed hydration. However, The pKa of HOBr is 8.7 (ref), much lower than those conditions, so it's not reasonable to make that analogy. Put another way, the base in those reactions is an alkene, which is very weak. Only a very strong acid will protonate an alkene.

In some quick SciFinder searching, I haven't found any overwhelmingly convincing evidence for the regioselectivity observed in the product, but it does seem reasonable to me by a mechanism similar to what Aditya Dev suggests.

I look at HOBr as an oxygen with a leaving group (analogous to Br-Br). Attack by the alkene on the oxygen releases bromide and gives the tertiary carbocation. Kick back by the oxygen closes a protonated epoxide, which is analogous to the bromonium. In the bromonium case, these are considered resonance structures. I don't know if that's definitely the case here (oxygen is much smaller than bromine), but I've written it that way to continue the analogy. In any case, protonated epoxides are typically attacked at the site best able to stabilize positive charge, which is the tertiary carbon. The resulting bromohydrin is the 'observed' product.

Answer 2

Traynham and Pascual¹ were first to investigate the reaction of hypohalous acids with methylenecyclohexane 1 and that of the epoxide 3 of this olefin with HBr. While both regioisomeric chlorohydrins were reported, only one bromohydrin was obtained, regioisomer 2 (mp.82^o-83^o). The structural assignments were made without the aid of nmr spectroscopy.

Sisti² reinvestigated these reactions and provided convincing evidence with the aid of nmr that the structure 2 is the other regioisomer 8 (mp.82^o-83^o and undepressed mixture melting point). Predictable oxymercuration of olefin 3 led to mercurial 6 which was readily converted to bromohydrin 8 and also reduced with NaBH₄ to form tertiary cyclohexanol 7. Preparation of bromohydrin 2 was achieved in part by reduction of bromoester 4 with LiAlH₄. The non-ring methylene groups were distinguished by nmr: 2 (CH₂OH, δ 3.70); 8 (CH₂Br, δ 3.37).

To answer this inquiry properly, it is important to have the correct structure of the product before providing a mechanism for its formation.

1) Traynham, J. G.; Pascual, O. S. Tetrahedron, 1959, 7, 165.
2) Sisti, A. J. J. Org. Chem., 1968, 33, 3953.

Answer 3

Sorry to dig up an old question but I think some important points are missing:

The experimental result is easily explained by steric hindrance and polarization of a bromonium ion formed due to addition of $\ce{Br+}$ to the alkene (three membered ring). SN2 attack of water occurs (hydroxide is not very common in aqueous acidic media!) at the bromonium ion. Nucleophilic attack at the cyclohexyl ring is more difficult (slow) than at the exposed site (fast). Additionally, the cyclohexyl ring delivers a +I effect, deactivating the cyclohexyl ring carbon for nucleophilic attack.

Answer 4

Also, I wanted to add it's not actually possible to obtain pure HOBr. These species exist in complicated equilibria in aqueous solution. They are known to disproportionate to more stable oxidation states (including bromide) and they exist in an equilibrium with bromine and water. Similar to other electrophilic halogenating agents such as N-bromosuccinimide they supply a small concentration of the molecular halogen to the solution. You can treat this as the reactive entity in your reaction.