# Mechanism of acetylene formation from iodoform and silver

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?

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

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.
• 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? – Justanotherchemist Mar 20 '18 at 7:54