Formation of Bromohydrins from alkenes using water and Br2

by James

Description: Alkenes treated with bromine (Br2) in the presence of water will form bromohydrins.

Notes: Note that this reaction is Markovnikoff selective and delivers the trans-product. This is essentially the same reaction as with NBS.


Notes: Note how in examples 1 and 2 the oxygen is attached to the most substituted carbon (Markovnikoff). The trans-stereochemistry is evident in example 1. Finally in the third example this is an example of an intramolecular reaction – often gives students trouble!

Mechanism: The alkene attacks Br2 to form a bromonium ion (Step 1, arrows A and B) which is then attacked on the more substituted carbon by water (Step 2, arrows C and D). The positively charged oxygen is then deprotonated to give the neutral alcohol.

Notes: It’s probably more reasonable to show water as the base in step 3. Although not important for this molecule, the bromine and oxygen are in a trans orientation since water attacks the bromonium ion from the back.


{ 4 comments… read them below or add one }


Is it possible to produce a halogenoalkane as a side product in this reaction, that is, Br- attacks the bromonium ion in the sec step?


James Ashenhurst

yes, but keep in mind the concentration of solvent is significantly higher than the concentration of the counter-ion, so it will be very small [concentration of H2O in 1L of water is 55 M!] At very high concentrations it might become more significant.


Gabby Holguin

in the mechanism, the lone pair attacks back to the double bond (right?) and if so, does it matter which side it attacks?



Yes it does. Thanks for bringing this up. Both bonds to the Br will be on the same face (formation of the bromonium ion is “syn”).


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