Final answer:
Alkynes react with halogens in an addition reaction where the halogen atoms add across the alkyne's triple bond, resulting in dihaloalkenes or tetrahaloalkanes if excess halogen is present.
Step-by-step explanation:
Alkynes react with molecular halogens through an addition reaction where the halogen molecules add across the triple bond. The alkyne's carbon-carbon triple bond, consisting of one σ bond and two π bonds, makes it very reactive. When a halogen like bromine (Br₂) reacts with an alkyne such as acetylene, the halogen atoms attach to each carbon of the triple bond, converting the triple bond into a single bond if excess halogen is present. As a result, the alkyne could initially form a dihaloalkene and subsequently convert into a tetrahaloalkane in the presence of excess halogen.
For instance, the reaction of acetylene with bromine would first yield 1,2-dibromoethene and could further proceed to form 1,1,2,2-tetrabromoethane if excess bromine is used:
Acetylene + Bromine → 1,2-Dibromoethene (initial addition product)
1,2-Dibromoethene + Additional Bromine → 1,1,2,2-Tetrabromoethane
This type of addition reaction is quite common in alkenes as well but uses only half the amount of halogen compared to alkynes, given the presence of only one π bond in alkenes rather than two.