# N-fluorosuccinimide: any scope?

All halogen derivatives (except fluorine) of succinimide exist and are used in various chemical processes. What makes N-fluorosuccinimde difficult to prepare? Still chemists have not been able to figure out the preparation of N-fluorosuccinimde and don't even bother to synthesize it since electrophilic/radical fluorination is accomplished by N-fluoro-o-benzenedisulfonimide (NFOBS), N-fluorobenzenesulfonimide (NFSI), and Selectfluor. Moreover, even if NFS is successfully synthesized, can it be used as a fluorinating agent in electrophilic/radical fluorination?

According to Wikipedia, NBS is made by reacting sodium hydroxide and bromine in an ice-water solution of succinimide. The NBS product precipitates and is collected by filtration. I am assuming that NCS and NIS are made in the similar fashion (using appropriate halogens). But, why NFS has still not been made through a similar way (using fluorine)? Is the reaction condition unsuitable for fluorine or is there any other reason?

I was reading a forum discussing the scope of NFS as a fluorinating agent. It says:

Does anyone know why is N-fluorosuccinimide (201281-58-5) such uncommon reagent for fluorination? I checked some major suppliers and none of them sells it.

In my knowledge, N-Halogenesuccinimide are obtained by reacting of Succinimide with the halogen itself. This is valid for Chlorine, Bromine, and Iodine. Fluorine is too reactive. For the fluorine derivative, iodinesuccinimide is treated with silver fluoride to exchange the halogen.

The other option is to treat N-Hydroxysuccinimide with Thionyl fluoride or Phosphorous trifluoride.

Is this true? I couldn't find any source confirming the above synthesis and I am skeptical about the preparation process. Can NFS be made through the above process?

• Nice. Also, I'd posit that since fluorine is the most electronegative halide, getting a compound $\ce{X-F}$ to act as a source of $\ce{F+}$ is harder than getting a compound $\ce{X-I}$ to act as a source of $\ce{I+}$. Hence, again, a more electron-withdrawing $\ce{X}$ moiety is needed. Apr 8, 2017 at 13:08