In my organic chemistry textbook, lithium is used to create a free radical of the alkynes, to then allow the anti addition of hydrogen to get an alkene.

"The Dissolving Metal Reduction of Alkynes" is a reaction that then uses $\ce{NH4Cl}$. Why put $\ce{NH4Cl}$?

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    $\begingroup$ I know that in the Birch reduction there is an alcohol (usually t-butanol) added to protonate the radical anion and allow the reaction to proceed, the ammonia solvent isn't sufficiently acidic to do that. I don't think you need to do that for alkynes. As far as I know the ammonium chloride is to quench excess sodium at the end of the reaction. $\endgroup$ – Abel Friedman Oct 20 '14 at 3:15
  • $\begingroup$ @AbelFriedman maybe jerepierre's answer is plausible? $\endgroup$ – yolo123 Oct 20 '14 at 3:29
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    $\begingroup$ @yolo123: Completely agree. $\endgroup$ – Abel Friedman Oct 20 '14 at 3:30
  • $\begingroup$ What about quenching sodium? $\endgroup$ – yolo123 Oct 20 '14 at 3:31
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    $\begingroup$ I think the ammonium chloride is used to protonated the amide ion. Protonation generates ammonia, which can easily be evaporated away. $\endgroup$ – Dissenter Oct 20 '14 at 4:31

Products from reaction step (1) are the trans-alkene and 2 equivalents of the amide $\ce{LiNHEt}$, a strong base. Step (2) is the workup of the reaction mixture with aqueous $\ce{NH4Cl}$ solution, which serves several purposes. First, it quenches the lithium amide and unreacted lithium metal.

$$\ce{LiNHEt + NH4Cl ->~ EtNH2 + NH3 + LiCl}$$ $$\ce{Li + H2O ->~ LiOH (aq) + \frac{1}{2} H2\uparrow}$$

Secondly, the trans-alkene is usually less water-soluble than ethyl amine and the lithium and ammonium salts, and can be separated from the workup mixture by precipitation or extraction with an organic solvent.


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