The conversion of 1-butene to 2-butyne can be achieved through which of the following reaction sequences?
- HBr; 2. NaC≡CH
- Br₂/CCl₄; 2. NaNH₂ (excess), heat
- H₂O, H₂SO₄; 2. H₂SO₄, heat; 3. Br₂
- Br₂/CCl₄; 2. KOC(CH₃)₃ (excess)
Explanation: To convert an alkene to an internal alkyne, a common strategy is addition of a halogen followed by a double elimination. The double elimination may also involve isomerization of the alkyne. Route B is correct. Step 1: Br₂ adds across the double bond of 1-butene to form 1,2-dibromobutane. Step 2: An excess of a very strong base, sodium amide (NaNH₂), is used to perform a double dehydrohalogenation. This initially forms 1-butyne. However, in the presence of hot NaNH₂, the terminal alkyne will isomerize to the more thermodynamically stable internal alkyne, 2-butyne. Route A is incorrect. Step 1 gives 2-bromobutane. The acetylide anion in Step 2 would act as a base, causing E2 elimination to give 2-butene. Route C is futile. It converts 1-butene to 2-butanol, then back to 2-butene (the major product). Route D is incorrect. While KOC(CH₃)₃ is a strong base, it is generally not strong enough to effectively perform the second elimination (from the intermediate vinyl bromide) to form the alkyne in high yield.