[Caeldom]

My teacher used bromine water as an example of a halogenation addition reaction with cyclohexene, to produce dibromocyclohexane. However, I’m a bit confused since in my textbook, it says 3-hexene will react with bromine water in a halohydration reaction, forming 4-bromo-3-hexanol. Is it just because one is cyclic and the other acyclic that the reactions differ?

Furthermore, I found out that bromine water partially disassociates into the equilibrium (leaning heavily to the left):

Br_2(l) + H₂O_(l) HOBr_(aq) + HBr_(aq)

Which means that there are both hypobromous acid and hydrogen bromide present in the solution as well. Doesn’t that mean they will all react with the hydrocarbon to form three or even four different products?

Or does it depend on the state of the reactant, the length of chain, concentrations etc... because to my understanding bromine will cause a halogenation, water will cause hydration, hypobromous acid will cause halohydration and hydrogen bromide will cause hydrohalogenation.

I also found out that this would all be simplified to a halogenation reaction if it were bromine in a non-aqueous solution. My teacher didn’t go into this much detail at all, though I haven’t had a chance to ask him specifically yet.


Also, what will be the favoured product in halohydration with unsymmetrical alkenes?

In a simple example like the reaction of 2-pentene with hypobromous acid:

CH3-CH=CH-CH2-CH3_(g) + HOBr_(aq)

will it form 3-bromo-2-pentanol, or 2-bromo-3-pentanol?

I know there’s Markovnikov’s Rule when it comes to hydrogen, but is there one for hydroxyl groups? Or perhaps it will form equal amounts of both?


Cheers,
Dan

[Jorriss]

It's probably close to 50-50. That's what I'd bet.

[Caeldom]

As in for halohydration with unsymmetrical alkenes you mean?

[Jorriss]

I was actually just referring to your specific example there.

[eunChae]

In my humble opinion, I think the following,
     In halohydration of terminal alkenes, bromination occurs in terminal carbon and OH goes to beta carbon. However, in unsymmetrical alkenes we should look at how carbon chain differs on both side. I mean, in your example methyl and ethyl doesnt differ too much in electron donating or withdrawing kinda thing but if there exists, for example, an efficient electron donating group, OH attacks to that side (since it reduces the partial positive charge of beta carbon).

    If I'm wrong, please let me know.

[orgopete]

To original poster, I agree I think your professor erred. This is how I rationalize what should happen in these reactions.

First of all, as an acid-base problem, water is a stronger base than Br^-. This is equivalent to saying HBr is a stronger acid than H₃O⁺. Therefore, if an alkene reacts with Br₂, the resulting bromonium ion can react with H₂0 or Br^-. Since H₂0 is more basic, we should expect it to react at a faster rate to give a bromohydrin.

That lays out a reasonable principle. Obviously, it doesn't take concentration effects into account. As the concentration of H₂0 decreases or if NaBr is added, then Br- may compete more effectively.

Re: unsymmetrical alkenes
The opening of epoxides and bromonium ions can be thought of as paradoxical to the rules for SN1/SN2 reactions. Methylcyclohexene is often used as an example as the stereochemical effects can also be learned. The reaction of methylcyclohexene with Br₂/H₂O will give …
See Cyclohexene reaction with bromine and water.

For the example of 2-pentene, I would predict the 3-pentanol to be greater than the 2-pentanol product because an ethyl is a better electron donor than methyl. What would make this very interesting is if someone would provide the data for this actual reaction. Which alcohol is predominant and by how much?