Reaction: Bromine (Br₂) adds across the double bond of an alkene, forming a dibromoalkane product
The pi electrons of the double bond in the alkene attack one of the bromine atoms, breaking the Br-Br bond
One bromine atom forms a bond with each carbon atom, resulting in a bromonium ion intermediate
A bromide ion (Br⁻) attacks the bromonium ion, opening the ring and forming the dibromoalkane product