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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?

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    $\begingroup$ 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, 2016 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, 2016 at 8:29
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    $\begingroup$ 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, 2016 at 0:17
  • $\begingroup$ Yes! You are correct @jerepierre $\endgroup$
    – Dissenter
    Apr 8, 2016 at 0:20

1 Answer 1

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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.

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  • $\begingroup$ 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, 2016 at 3:22
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    $\begingroup$ @DGS it's a syn elimination which is less optimal than anti, but it is locked in the reactive conformation $\endgroup$
    – jerepierre
    Apr 8, 2016 at 5:57
  • $\begingroup$ Don't forget the magnesium. When it us incorporated into the Grignard reagentbit becomes electrophilic and can attach to the fluorine. This facilitates the nucleophilic attack on the carbon-fluorine group by making a better leaving group. $\endgroup$
    – Oscar Lanzi
    Jul 10 at 23:40

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