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Is there any reason why this grignard reagent wouldn't work? My best guess is that perhaps the nitrogen could act as a second nucleophilic "head", thereby not giving a good yield on the intended product. Or maybe this isn't stable enough?

proposed piperidine grignard

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  • $\begingroup$ The variables regarding why the reagent may or may not be effective for certain processes depends a lot on what your using it for? $\endgroup$ – Technetium Nov 25 '15 at 8:10
  • $\begingroup$ I'm really just wondering if there's a way it could deactivate itself, like how you shouldn't make a grignard reagent that can reach a carboxyl group on itself. Originally I intended to use it to attack benzoyl chloride. Can the grignard really deprotonate itself? It seems to me like the nitrogen's proton is too close? $\endgroup$ – timaeus222 Nov 25 '15 at 8:30
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    $\begingroup$ I'm almost sure it can and will deprotonate and thus deactivate itself. Whether or not the nitrogen's proton is close to the Grignard site, does not matter at all. $\endgroup$ – Ivan Neretin Nov 25 '15 at 8:38
  • $\begingroup$ @Ivan and the OP—Please (self-)answer this question so it wouldn't show up in the 'unanswered' list. i.e. close the case. $\endgroup$ – M.A.R. Nov 25 '15 at 18:41
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I guess it wouldn't even exist. Grignard's reagent is basic enough to easily take off an imine proton, so as soon as you try to make such a thing, it will turn upon itself.

Probably they have a suitable protecting group for this, but here I'm stepping outside my expertise.

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  • $\begingroup$ Why are you calling the amine an imine? $\endgroup$ – electronpusher Mar 31 '17 at 18:45
  • $\begingroup$ Obviously because I mistook the secondary amine's $\ce{-NH-}$ for imine's $\ce{=NH}$. Who cares. Grignard would instantly destroy both all the same. $\endgroup$ – Ivan Neretin Mar 31 '17 at 21:21
  • $\begingroup$ Just wondering, thought maybe I was missing something. $\endgroup$ – electronpusher Mar 31 '17 at 21:47

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