What would the mechanism for the reaction between 2-(chloromethyl)furan and potassium cyanide be, as detailed below?

Reaction scheme

I know that for 6-membered rings like pyridine, a $\mathrm{S_NAr}$ reaction can occur. However, I'm not sure as to how the mechanism differs between the two in a way that the cyanide group is added at the 5-position instead of the methyl group.

  • 2
    $\begingroup$ google "vinylogous SN2 reaction" $\endgroup$
    – ron
    Commented Apr 25, 2017 at 18:46
  • $\begingroup$ I had a look and I'm not sure how to relate it to the mechanism in question $\endgroup$
    – Noktis
    Commented Apr 25, 2017 at 19:47
  • $\begingroup$ Probably SN1 not SN2, as outlined in Blaise's answer. $\endgroup$ Commented May 25, 2017 at 23:47

2 Answers 2


Although the answer was already appropriately provided, for anyone interested in further reading, the reaction is covered in detail by Divald and coworkers:1

Scheme II

Essentially, the proposed mechanism starts from the resonance structures of 2-chloromethyl furan, where the chlorine bond is treated as ionic and the electron-donating property of oxygen stabilizes the shared positive charge similar to a benzyl cation. The 5' position is therefore electrophilic (and is presumably further activated by the inductive effects of a neighboring oxygen), resulting in a 60/40 yield of 5/2-methyl substitution.

It is however mentioned early in the paper that 5' substitution is only seen in protic solvents. This could be explained in the proposed mechanism, where after addition to the 5' position, elimination of the geminal hydrogen and tautomerization proceed through exchange with a protic solvent.


  1. Divald, S.; Chun, M. C.; Joullié, M. M. Chemistry of 2-(chloromethyl)furans. Reaction of 2-(chloromethyl)furans with aqueous potassium cyanide and other nucleophiles. J. Org. Chem. 1976, 41 (17), 2835–2846. DOI: 10.1021/jo00879a010.

Cyanide ion attacks the 5 position of the furan. The 4,5 double bond migrates to 3,4. The 2,3 double bond migrates to the Me group kicking out Chloride and giving the exomethylene (double bond outside tbe ring) intermediate. Then the bonds reorganise to re-aromatise giving the product shown.

I hope this makes sense. I have no access to a drawing package.


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