# Major Product of Alkene in Bromine Water

At school, we are taught that in bromine water, the following equilibrium is present: $$\ce{Br2 + H2O \rightleftharpoons BrOH + HBr}$$ Therefore, when you add an alkene to bromine water, you don't get bromination, but rather the $\ce{BrOH}$ reacts instead and you get the following reaction (lets say for ethene for example): $$\ce{C2H4 + BrOH -> C2H5BrO}$$ However I have looked at a few organic textbooks such as Clayden and they don't mention anything about $\ce{BrOH}$ forming in bromine water and then adding across the double bond. It says that the product of the above reaction will be $\ce{C2H4Br2}$.

In reality, both reactions probably both occur to some extent (also, for the first reaction, wouldn't you expect some of the $\ce{HBr}$ to add across the double bond as well?), but which one is the most prominent, or in other words, what is the major product?

• I would say the alkene reacts with the bromine to form the bromonium ion, which is then intercepted by H2O. H2O is a lousy nucleophile compared to Br- but if the solvent is water then the sheer excess of water molecules would probably lead to the bromohydrin. At pH 0 the equilibrium constant for the first reaction you quoted is on the order of $10^{-9}$ (you can easily check this with standard reduction potential data). I strongly doubt HOBr is the electrophile. – orthocresol Jul 11 '16 at 22:23
• As orthocresol said, the reaction mechanism is that of a halohydrin formation. I am not sure whether $\ce{BrOH}$ would form at all, actually. – Jason Jul 11 '16 at 22:30
• @orthocresol thanks, that makes sense. So you wouldn't expect HBr to add across the double bond since the equilibrium constant is so low? – Nanoputian Jul 11 '16 at 22:31
• Yes, I would not expect there to be any appreciable amount of HBr. – orthocresol Jul 11 '16 at 22:52
• – Mithoron Jul 12 '16 at 1:16

@orthocresol mentioned the main points in his comment but I would just like to expand a little more. The main electrophile in bromine water is $\ce{Br2}$ and this is what does the initial attack on the double bond, forming a cyclic bromonium ion. The bromonium ion is a strong electrophile and so it will react with any nucleophiles that are around. Since the water molecules will far outnumber bromide ions in the solution then the major nucleophile is water, even though it is not as good a nucleophile as bromide. The nucleophile attacks in an $\mathrm{S_N2}$ reaction, leading to a halohydrin product, after loss of a proton. This mechanism also explains the trans stereochemistry observed in the product.
As @orthocresol mentioned, the equilibrium constant for the formation of $\ce{BrOH}$ will be very small and so $\ce{BrOH}$ will be negligible as an electrophile. This is explained in a bit more detail in @Jan's answer here.