How do alkenes react with bromine?

1 Answer
Jul 17, 2016

Alkenes perturb the electron cloud in "Br"_2, polarizing it and allowing addition of "Br"^(+) across the alkene.

The remaining "Br"^(-) can backside-attack to form the vicinal dibromide product.

1-1. The pi bonding electrons from the alkene move towards one "Br"'s antibonding orbitals; that "Br" becomes partially positive (delta^(+)), and the top "Br" becomes partially negative (delta^(-)).

This accounts for the right-hand arrow.

1-2. Polarizing a bond by perturbing the electron cloud weakens the bond, and in this case, it was enough to break the "Br"-"Br" bond, and the bonding electrons move into now-nonbonding orbitals.

This accounts for the upper arrow.

1-3. However, the bottom "Br" also donates electron density into the alkene's antibonding orbitals, thus making a bridging connection between the left and right carbons.

This accounts for the left-hand arrow.

That was the complicated part. The rest is not too crazy.

2. The remaining "Br"^(-) is the nucleophile, attacking one of the carbons from the rear (a backside-attack), and breaking one of the bridging bonds (which are weak already).

This accounts for both arrows.

The backside-attack is what generates an anti-addition product. This is reflected in the trans relationship of both "Br" on the final product.

In terms of Markovnikov or anti-Markonivkov addition, it doesn't really matter because both "Br" added are identical atoms.