# 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.

##### 3 Answers
Aug 23, 2015

You're teacher is quite correct, and there is no possibility of forming of alcohol (i.e. purely $R - O H$, rather than a bromo-substituted $\left(B r\right) R - O H$ species).

#### Explanation:

Ethylene, ${H}_{2} C = C {H}_{2}$, 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 $O {H}_{2}$, $B {r}^{-}$, $O {H}^{-}$, cyanide etc.

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

${H}_{2} C = C {H}_{2} + B {r}_{2} \rightarrow {H}_{2}^{+} \ast C \ast - C {H}_{2} B r + B {r}^{-}$

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 $B {r}_{2} \left(a q\right)$, 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.

Aug 24, 2015

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

#### Explanation:

The $\text{OH} -$ 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. Aug 24, 2015

Remember that aqueous means dissolved in water, which implies that water is present to react if its pKa is low enough (low pKa$\to$strong acid$\to$less stable), which it is. The conjugate acid of bromide is $\text{HBr}$, 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 $\text{C"-"Br}$ 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 $\text{C"-"Br}$ bonds, attaching in a $\text{trans-}$ 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, $\text{C"_2"H"_4"BrOH}$ IS an alcohol. It's $\text{2-bromoethanol}$.