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.

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 $\text{2-bromoethanol}$.