# How do alkenes react with bromine?

##### 1 Answer
Jul 17, 2016

Alkenes perturb the electron cloud in ${\text{Br}}_{2}$, polarizing it and allowing addition of ${\text{Br}}^{+}$ across the alkene.

The remaining ${\text{Br}}^{-}$ can backside-attack to form the vicinal dibromide product.

1-1. The $\pi$ bonding electrons from the alkene move towards one $\text{Br}$'s antibonding orbitals; that $\text{Br}$ becomes partially positive (${\delta}^{+}$), and the top $\text{Br}$ 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 $\text{Br"-"Br}$ bond, and the bonding electrons move into now-nonbonding orbitals.

This accounts for the upper arrow.

1-3. However, the bottom $\text{Br}$ 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 ${\text{Br}}^{-}$ 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 $\text{Br}$ on the final product.

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