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So I've read that during the formation of the Grignard reagent, The R-X bond cleaves homolytically forming free radicals. My question is how is this happeneing? Due to the high electronegativity difference between Carbon and Halide group shouldn't it result in R+ and X- ? To be clear, I have no problem following through how the electrons move around. My only doubt is how exactly the Alkyl-Halide bond resulted in a free radical instead of an ion.

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  • $\begingroup$ related chemistry.stackexchange.com/questions/67601/… $\endgroup$
    – Mithoron
    Jan 13, 2019 at 19:43
  • $\begingroup$ To be clear, I have no problem following through how the electrons move around. My only doubt is how exactly the Alkyl-Halide bond resulted in a free radical instead of an ion. $\endgroup$ Jan 13, 2019 at 21:39

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The electronegativity difference between carbon and the halogen used for Grignard reagents is actually not all that high. Alkyl fluorides are generally un-reactive towards magnesium, and other halogen whose alkyl compounds are more reactive have electronegativity ranging from only a few tenths above carbon to a few tenths below carbon.

In any event the neutral alkyl halide molecule does not directly cleave into either radicals or ions. According to the University of Texas, the first stage of the reaction involves a one-electron reduction of the alkyl halide to a radical anion; the magnesium is oxidized to a radical cation. The alkyl halide anion, not the original molecule, cleaves off its halide ion to form an alkyl radical. Thus, in effect, after the initial electron transfer you get "a little bit of both" -- the alkyl group is a radical but the halogen (even with its modest electronegativity) is an anion. The alkyl radical, the magnesium radical cation, and the previously displaced halide ion then combine to get the Grignard reagent molecule (the combination process finishes off the oxidation of magnesium to the +2 state). At no point in the given mechanism do you have a free radical halogen atom.

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