The product of reacting an alkene with #"Br"_2(aq)# is #"C"_2"H"_4"BrOH"#. What is its name? Why is the product NOT a geminal dibromide?

Ethylene, #H_2C=CH_2#, is an electron-rich species, a nucleophile that has electrons to donate, however, after it has donated its pi electrons, the intermediate will undergo reaction with nucleophilic species, such as #OH_2#, #Br^-#, #OH^-#, cyanide etc.

On the other hand, molecular bromine is polarizable, and can be represented as #Br^(delta-)-Br^(delta+)#; that is the bromine nucleophile becomes polarized, and individual bromine atoms acquire partial negative or partial positive charges. The partially positive #Br# atom reacts with the electron-rich ethylene. We can represent this as:

#H_2C=CH_2 + Br_2 rarr H_2^+ **C** -CH_2Br + Br^-#

The highlighted carbon is now cationic, and will react with any nucleophile present: this could be water, or cyanide, or bromide anion, whatever is in solution. Since we used #Br_2(aq)#, i.e. bromine solution in water, water will certainly be present in abundance, and you are likely to form a bromo-alcohol.

If, however, the olefin was treated with #H^+#, the first step of the reaction is to form a #C-H# bond; the subsequent carbocation is likely to react with water to form an alcohol.

The #"OH"-# group is a hydroxyl group that in this case forms an alcohol functional group.

The following image is the structural formula for 1-bromoethanol.

The following image is the structural formula for 2-bromoethanol.

Remember that aqueous means dissolved in water, which implies that water is present to react if its pKa is low enough (low pKa#->#strong acid#->#less stable), which it is. The conjugate acid of bromide is #"HBr"#, whose pKa is about #-9#, so bromide is a very weak base, and additionally not a good nucleophile (electron-donor).

It's quite true that bromine liquid reacting with ethene gives dibromoethane. That is an early mechanism you would have learned in the first few weeks of a first-year organic chemistry class.

However, water disrupts this mechanism in the middle. What happens is:

• In the first step, bromine makes the cyclopropane-analog bonds.
• Then the carbon centers of each #"C"-"Br"# becomes electrophilic (electron-acceptor). Water, a stronger nucleophile than bromide, can attack either one (due to symmetry) from behind to bond and break one of the #"C"-"Br"# bonds, attaching in a #"trans-"# addition.

  (normally, it would be the other bromide attacking from behind, not water)

  • Water finishes the mechanism by deprotonating the attached water to form hydronium and bromide in solution.

    So yes, there IS the possibility of forming an alcohol. In fact, #"C"_2"H"_4"BrOH"# IS an alcohol. It's #"2-bromoethanol"#.