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Sodium Metal when treated with ethyne forms mono and di sodiumethynide why doesn't lithium ? Sodium reacting with terminal alkynes

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    $\begingroup$ Related: Why are lithides not known? $\endgroup$ – andselisk Aug 5 '17 at 17:47
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    $\begingroup$ @andselisk I'm not asking about lithides. In mono sodium ethynide sodium has a positive charge $\endgroup$ – Serotonin Aug 6 '17 at 3:46
  • $\begingroup$ It looks like it does form monolithium ethynide, which should also react to form dilithium ethynide. onlinelibrary.wiley.com/doi/10.1002/047084289X.rl035.pub2/… $\endgroup$ – Tyberius Aug 6 '17 at 3:58
  • $\begingroup$ @Tyberius Yep! You're right en.m.wikipedia.org/wiki/Acetylide anyways I asked because my school textbook was saying lithium does not form Acetylide $\endgroup$ – Serotonin Aug 6 '17 at 4:10
  • $\begingroup$ @NilayGhosh Thanks for the explanation. But why lithium doesn't reacts directly with terminal alkynes ? $\endgroup$ – Serotonin Aug 12 '17 at 1:50
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Sodium alkynide was formed by reacting sodium and acetylene but in case of lithium alkynide, it can be formed by reacting acetylene with organolithium reagents like BuLi, Lithium amide, LiHMDS. From wikipedia:

Lithium amide, LiHMDS, or organolithium reagents, such as butyllithium, are frequently used form lithium acetylides:

$\ce{H-C#C-H + BuLi ->[THF][-78℃] Li -#-H + BuH}$

The above process is well described in this paper which @Tyberius linked in the comment section.

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Lithium actually can be reacted with acetylene to produce the ethynide:

In the laboratory samples may be prepared by treating acetylene with a solution of lithium in ammonia, on −40°C, with creation of addition compound of $\ce{Li2C2 • C2H2 • 2NH3}$ that decomposes in stream of hydrogen at room temperature giving white powder of $\ce{Li2C2}$.

$\ce{C2H2 + 2Li -> Li2C2 + H2}$

However,

Samples prepared in this manner generally are poorly crystalline.

Various alternatives are therefore often preferred, and (some of) these are also given in the Wikipedia article, and by Nilay Ghosh in his answer.

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