The reaction of acetylene (C₂H₂) with bromine (Br₂) is a halogenation reaction. It occurs in two steps, and the products depend on the amount of bromine used.

Step 1: Addition of one bromine molecule

C₂H₂ + Br₂ → CHBr=CHBr (1,2-dibromoethene)

Step 2: Addition of a second bromine molecule

CHBr=CHBr + Br₂ → CHBr₂CHBr₂ (1,1,2,2-tetrabromoethane)

Overall reaction:

C₂H₂ + 2Br₂ → CHBr₂CHBr₂

Explanation:

* Step 1: The bromine molecule adds across the triple bond of acetylene, breaking the triple bond and forming a double bond. This results in the formation of 1,2-dibromoethene.

* Step 2: If enough bromine is present, it will react further with the double bond of 1,2-dibromoethene, breaking the double bond and forming a single bond. This results in the formation of 1,1,2,2-tetrabromoethane.

Note: The reaction conditions can influence the outcome. If only a limited amount of bromine is present, the reaction will stop at the formation of 1,2-dibromoethene.