Question

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?

Shouldn't an alcohol form? My teacher says there is no possibility of forming an alcohol.

Answer 1

You're teacher is quite correct, and there is no possibility of forming of alcohol (i.e. purely `R-OH`, rather than a bromo-substituted `(Br)R-OH` species).

Explanation:

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.

Answer 2

`"C"_2"H"_4"BrOH"` is an alcohol. The formula is more commonly written as `"C"_2"H"_5"BrO"`. It is called bromoethanol, and has two isomers.

Explanation:

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.

Answer 3

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"`.