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

That conversion was somewhat as follows:

$$\ce{CHI3->[Ag] HC#CH ->[NaNH2] NaC#CH ->[CH3Br] CH3C#CH}$$

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

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    $\begingroup$ What happens when "chloroform reacts with silver powder". Putting this in here so that future google searches redirect here, since this a popular question. $\endgroup$ Commented Feb 2, 2018 at 13:19

3 Answers 3


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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    $\begingroup$ How does C-C bond formation occur and how do you end up with a triple bond? Can you provide a mechanism? $\endgroup$
    – bon
    Commented Jun 4, 2016 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
    Commented Jun 5, 2016 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
    Commented Jun 5, 2016 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? $\endgroup$ Commented Mar 20, 2018 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.


It might be undergoing the same mechanism as the Wurtz reaction. Which is through free radical mechanism (Some say that Wurtz reaction also takes place through the ionic reactions).

Firstly the $\ce{C-I}$ bond breaks from both the iodoforms to form $\ce{CHI2}$ free radical which causes them to fuse forming $\ce{I2HC-CHI2}$. The same step repeats causing further breakng of $\ce{C-I}$ bond leading to formation of alkenes and alkynes later.

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    $\begingroup$ It is always best to have literature backing up your mechanisms. Opinion-based answers aren't very helpful, especially when you give no rationale behind why that mechanism might be considered correct. $\endgroup$ Commented May 10, 2020 at 16:15

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