# Alkyne formation from vicinal dihalide

Why must 3 equivalents of sodium amide be used here to form the alkyne? The part I specifically have issue with is in red.

From what I understand, we should just use 2 equivalents of sodium amide to get rid of the bromines and to form the alkyne. Another equivalent of sodium amide will form the acetylide anion. Removing two protons effectively forms the triple bond and eliminates the two halogens. Also, take a look at example 3 here ..

Why then does the book's solution manual want me to use 3 equivalents of sodium amide to get neutral acetylene, and another equivalent to get to the acetylide ion?

As you can see, my book's solution manual contradicts another book's discussion of obtaining alkynes from vicinal dihalides.

• It might be, that in some cases a terminal proton might be abstracted more readily, than the one necessary to form the triple bond. To ensure you get your alkyne you just add more base and reprotonate later. – Martin - マーチン Sep 28 '14 at 7:52
• But is 2 eq. the minimum needed? @Martin – Dissenter Sep 28 '14 at 7:58
• @Dissenter 2 eq. are an absolute minimum. In practice some excess is often useful. First, some excess helps to ensure that all vic-dihalogenalcane reacted. And $\ce{NaNH2}$ adsorbs water from air during storage/weighting, so it should be in excess to compensate. That NOT saying about deprotonation of terminaly alkynes. – permeakra Sep 28 '14 at 8:31