As the Glaser coupling reaction is usually done with terminal alkynes with the hydrogen being straight, I wonder if the same reaction can occur in terminal alkynes connected to the backbone of a metal alkyne complexe.

As an example, in the trimerization reaction of acetylene, which happens with the formation of a metal alkyne complete; the hydrogens are bent in a v-shape to the carbon; not straight. Alkyne Trimerisation.

Would it be possible? If it weren’t the case, what would impede this reaction?

  • $\begingroup$ I don't think the fact that the hydrogens are "straight" is relevant. I guess in a sense it is, because it gives information about the bonding. The fact that the hydorgen is not "straight" in a metal alkyne complex means that it is not really an alkyne, the bondig situation is completely different. Now, during the Glaser coupling, you will form a bond between the copper and the carbon of the alkyne - I am not sure you can do the same with an alkyne metal complex, because it's not really an alkyne. Maybe pre-dissociation would help? $\endgroup$
    – Szgoger
    Jul 13, 2022 at 8:28

1 Answer 1


The hydrogen being straight corresponds to its sigma bond with carbon having more carbon $2s$ character and less carbon $2p$ character, which makes the covalent overlap somewhat less and the bond more polar. Thus the hydrogen is susceptible to deprotonation by the strong base that initiates the reaction.

If the bond angle becomes bent due to the electronic contributions from the metal, we have lost some $2s$ character in the carbon-hydrogen bond in favor of more $2p$, and the required deprotonation could be inhibited.

Note that in the alkyne trimerization shown in the question, the first two acetylene molecules are effectively reduced before the third one displaces the metal. Such a reduction is not seen in the Glaser coupling.


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