# 1-bromo-2-fluorobenzene reaction with magnesium and furan

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?

• 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. – SendersReagent Apr 7 '16 at 8:24
• 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. – SendersReagent Apr 7 '16 at 8:29
• 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? – jerepierre Apr 8 '16 at 0:17
• Yes! You are correct @jerepierre – Dissenter Apr 8 '16 at 0:20

• Wouldn't the electrons in the $\mathrm{sp^2}$ orbital have a pretty hard time getting decent overlap with $\mathrm{\sigma ^*_{CF}}$? – SendersReagent Apr 8 '16 at 3:22