Halogenation of Alkynes
Halogenation adds halogen atoms across the C≡C triple bond of an alkyne. Step-by-step mechanism, reagents and practice.
What happens
Halogenation adds halogen atoms across the C≡C triple bond of an alkyne. Common halogens are Cl₂ and Br₂. With one equivalent, the reaction stops at a dihaloalkene. With two equivalents, addition happens again and produces a tetrahaloalkane.
Reaction summary
R–C≡C–R + X₂ (1 equivalent) → dihaloalkene; + 2 X₂ (2 equivalents) → tetrahaloalkane
Reagents and conditions
- Starting material: 2-Butyne (alkyne)
- Reagents: Br₂ (1 equivalent)
- Conditions: X₂ = Cl₂ or Br₂ (I₂ is much less commonly used); 1 equivalent X₂ → dihaloalkene; 2 equivalents X₂ → tetrahaloalkane
- Products: 2,3-Dibromobut-2-ene (dihaloalkene, generally trans)
Mechanism
- Alkyne Reacts with X₂: The alkyne reacts with X₂ (shown here with Br₂).
- Bridged Halonium-Type Intermediate: A bridged halonium-type intermediate forms.
- Halide Attacks from the Opposite Side: X⁻ attacks from the opposite side, giving a dihaloalkene.
- Second Equivalent of X₂: With another equivalent of X₂, the remaining C=C double bond reacts again, producing the tetrahaloalkane.
- Product: With 1 equivalent of X₂ the reaction stops at the dihaloalkene (generally trans, E). With 2 equivalents the double bond disappears and the final product is a saturated tetrahaloalkane.
Selectivity
- Stereoselectivity: With 1 equivalent of X₂, anti addition generally favors the trans (E) dihaloalkene. With 2 equivalents, the double bond disappears, so the final product is a saturated tetrahalide.
Common mistakes
- The amount of halogen matters: 1 equivalent X₂ → stop at the alkene; 2 equivalents X₂ → continue to the alkane. Don't automatically remove the entire triple bond when only one equivalent is given.
Memory trick
Think of the triple bond as having two rounds of addition: C≡C → C=C → C–C. 1 X₂: first round → 2 halogens. 2 X₂: second round → 4 halogens. So: 1 equivalent = 2 X, 2 equivalents = 4 X.