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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 ..

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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?

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

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    $\begingroup$ 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. $\endgroup$ – Martin - マーチン Sep 28 '14 at 7:52
  • $\begingroup$ But is 2 eq. the minimum needed? @Martin $\endgroup$ – Dissenter Sep 28 '14 at 7:58
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    $\begingroup$ @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. $\endgroup$ – permeakra Sep 28 '14 at 8:31
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I think the comments basically have the answer - if you use only 2 equivalents of sodium amide, you will use 1.33 equivalents creating the alkyne and the remaining 0.67 equivalent will react with the alkyne after it is formed (an acid-base reaction) to form the acetylide salt -- probably a much faster reaction than the dehydrobromination reactions. So about 1/3 of the original dibromide will remain unreacted. You need the third equivalent of sodium amide to drive the reaction to completion.

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  • $\begingroup$ Is the other picture wrong then? $\endgroup$ – Dissenter Sep 29 '14 at 14:13
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    $\begingroup$ It is not whether the dihalide is geminal vs vicinal, but rather whether you end up with a terminal alkyne or not (you will in the case of 1,2-dibromoethane). If not, you should only need 2 equiv. Does that help? $\endgroup$ – iad22agp Sep 29 '14 at 14:22
  • $\begingroup$ Wait what? Can you show me the mechanism for 3 eq? $\endgroup$ – Dissenter Sep 29 '14 at 14:24
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    $\begingroup$ I think the handwritten diagram is correct showing the first 2 steps to get to acetylene. As you point out (sorry for missing this the first time), the answer key erroneously shows 3 equiv of NaNH2 being used to make neutral acetylene and then another equivalent used to make acetylide ion. $\endgroup$ – iad22agp Sep 30 '14 at 16:44
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The third amide is used to deprotonate the alkyne after it is formed. It drives the formation of alkyne to the right, making a lot better yield of alkyne. That is why they also added water as a proton source in the second step to protonate the acetylide ion.

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  • $\begingroup$ I don't think this is correct. The issue here is that alkynes are relatively acidic, so if you don't add a third equivalent of the base, you will end up quenching some of the base by using it to deprotonate the product alkyne. $\endgroup$ – Zhe Feb 1 '17 at 21:57

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