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.