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According to "Organic Chemistry as a secondary language", during elimination-addition, formation of benzyne involves 2 steps (as shown in the bottom photo). I wonder during 2nd step, it is just electron transfer and 'kick out' the Cl as leaving group.

Can the 2 step process be simplified as 1 step? (as shown in the upper photo) The only difference is that I used the electrons from C-H bond to form the 3rd C-C bond, where the book suggest a deprotonated anion is first formed then C-C is subsequently formed. I wonder what causes the difference? Thank you.

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    $\begingroup$ The loss of HX from chloro- and bromobenzene is concerted with amide ion while fluorobenzene loss of HF is stepwise. J. D. Roberts, et al., J. Am. Chem. Soc., 78, 611 (1956). $\endgroup$
    – user55119
    Jun 16, 2020 at 18:10
  • $\begingroup$ But I would really like to know why it must be a 2- instead of 1-step process, is the (central) anion more stable? What caused the electron to stay at 1 point but not immediately transferred to form 3rd C-C bond? Thanks $\endgroup$
    – 234ff
    Jun 17, 2020 at 1:14
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    $\begingroup$ look carefully at user55119's comment and you will realize that it is satisfying the criteria you have yourself stated. The anion formed in the fluorobenzene case will be quite a bit more stable than that in chlorobenzene and bromobenzene, hence theformer goes through a stepwise process, while the other halobenenes adopt a concerted path $\endgroup$ Jun 17, 2020 at 3:52
  • $\begingroup$ OK thanks. So in this case, forming benzyne from chlorobenzene is actually a concerted process rather than 2-step process as the book says? (since the anion is much less stable) $\endgroup$
    – 234ff
    Jun 17, 2020 at 3:58
  • $\begingroup$ Yes, benzyne to chlorobenzene is concerted $\endgroup$ Jun 17, 2020 at 4:32

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