5
$\begingroup$

Why does bromofluorobenzene react with magnesium and furan to form a benzyne intermediate?

Does it have something to do with a Grignard being highly basic and a fluorine being electron withdrawing, thereby making elimination a likely pathway?

Then does a Diels-Alder type reaction commence, with furan acting as the diene and the benzyne acting as the dienophile?

How likely or favorable is this reaction in real life?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ I think it's more because fluorine makes the ring electron-poor, allowing the first step (formation of the grignard) to happen more quickly. I'm kind of using the comparison of $\mathrm{S_N1}$ and $\mathrm{E1}$ to guide me to the thought that $\mathrm{S_NAr}$ would have some parallels with aromatic $\mathrm{E1}$, so it could be wrong. To see why fluorine is used for $\mathrm{S_NAr}$, see the answer here. $\endgroup$ – SendersReagent Apr 7 '16 at 8:24
  • $\begingroup$ Of course, thinking about it, $\mathrm{S_NAr}$ is more like the opposite of $\mathrm{S_N1}$. So... We'll see how logical that thought is. $\endgroup$ – SendersReagent Apr 7 '16 at 8:29
  • 2
    $\begingroup$ I'm guessing that you mean that 1-bromo-2-fluorobenzene and magnesium form a benzyne intermediate, and then furan reacts with that benzyne intermediate. Can you confirm? $\endgroup$ – jerepierre Apr 8 '16 at 0:17
  • $\begingroup$ Yes! You are correct @jerepierre $\endgroup$ – Dissenter Apr 8 '16 at 0:20
7
$\begingroup$

The overall pathway described is shown below. Treatment of the aryl bromide with magnesium gives a Grignard, which is equivalent to a carbanion. The carbanion can eliminate fluoride to give benzyne. Being highly strained, benzyne is a fantastic dienophile and reacts with furan through a Diels-Alder reaction.

Mechanism

Most of the techniques to generate benzyne use the same strategy: generate a carbanion next to a leaving group. The classic method taught in introductory organic chemistry is by deprotonation of an aryl halide with a very strong base (sodium amide), which eliminates the halide. The strategy shown here generates the carbanion using the milder method of forming a Grignard.

The elimination is possible because carbanions are very good at kicking out leaving groups. It's commonly taught that fluoride is a poor leaving group, but the end of that sentence is that it's a poor leaving group for SN2 reaction or compared to chloride, bromide, and iodide. Later, it's taught that enolates (a reasonably stable class of carbanions) are reactive enough to eliminate alkoxides/hydroxide. So it makes sense that an unstabilized carbanion as here should be able to kick out a relatively poor (but not that terrible) leaving group.

The Stoltz group has a nice summary of benzyne's history, generation, and reactivity.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Wouldn't the electrons in the $\mathrm{sp^2}$ orbital have a pretty hard time getting decent overlap with $\mathrm{\sigma ^*_{CF}}$? $\endgroup$ – SendersReagent Apr 8 '16 at 3:22
  • 1
    $\begingroup$ @DGS it's a syn elimination which is less optimal than anti, but it is locked in the reactive conformation $\endgroup$ – jerepierre Apr 8 '16 at 5:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.