My book mentions a synthetic route for turning iodoform into propyne.

That conversion was somewhat as follows:


The first step is the conversion of iodoform into acetylene with silver. I am usure how that reaction works. Could someone provide a mechanism?

  • $\begingroup$ What happens when "chloroform reacts with silver powder". Putting this in here so that future google searches redirect here, since this a popular question. $\endgroup$ – Gaurang Tandon Feb 2 '18 at 13:19

What happens is that the C-I bond being weak, the iodide group is attacked by the silver atom (similar to what Na does in Wurtz reaction). As, there are 3 I-atoms and 2 molecules of iodoform usually react, 6 silver atoms are used up in the reaction and they precipitate out 6 molecules of AgI. The 2 carbon atoms form a triple bond to form acetylene.

EDIT- $$\ce{2CHI3 + 6Ag-> C2H2 + 6AgI}$$ $$\ce{C-I + 2Ag +I-C->I\bond{...}Ag + Ag\bond{...}I + C\bond{...}C}$$ As there are 3 C-I bonds, 6 Ag atoms form bonds with I-atoms, Hence, triple bond is formed between the 2 C-atoms. I hope that explains it.

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  • 3
    $\begingroup$ How does C-C bond formation occur and how do you end up with a triple bond? Can you provide a mechanism? $\endgroup$ – bon Jun 4 '16 at 18:35
  • $\begingroup$ What @bon said and additionally can you expand on how "the iodide group is attacked by the silver atom"? $\endgroup$ – K_P Jun 5 '16 at 8:55
  • $\begingroup$ While this still isn't what I'd call mechanism, the reaction is redox on surface of Ag powder, so exact mechanism may be elusive. $\endgroup$ – Mithoron Jun 5 '16 at 19:36
  • $\begingroup$ Perhaps something related to the Ullmann-reaction? Forming an alkyl-silver-iodide, which reacts with a second silver to form silver iodide and alkyl-silver. Another iodoform enters the reaction, the c-c-bond is formed and silver iodide leaves. That would create something like $I_2HC-CHI_2$ and perhaps this further eliminates iodides with the help of silver in a similar fashion? $\endgroup$ – Justanotherchemist Mar 20 '18 at 7:54

Here the iodine atoms of iodoform are the leaving groups, which creates a precipitate of silver iodide by reacting with silver. But, there are no nucleophile to occupy the vacant places of the leaving groups. So it creates a triple bond with other molecule. I think that is the process.

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