Question

How do alkenes react with bromine?

Answer 1

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.