Chapter 9 : Alkynes

Summary

• The two p bonds of an alkyne can be formed using two consecutive elimination reactions.
• The leaving groups are usually halides (esp. Br or Cl)
• Reagent : usually NaNH₂ (a strong base, pKa = 35)
• Either geminal (1,1-) or vicinal (1,2-) dihalides can be used.
• Since 1,2-dihalides can be prepared by addition of X₂ to an alkene, an alkene can be converted into an alkyne in two steps.
• These reactions are typically E2 reactions and occur via an alkenyl halide.
• The strong base is needed in order to cause elimination of the alkenyl halide.

What is the alkyne product from the reactions of the following with NaNH2 :
(a) 2,2-dibromopropane		(c) 1,2-dibromohexane
(b) 1,1-dibromooctane		(d)  2,3-dibromohexane