Why does $\ce{-NH2}$ not leave in $\ce{RCONH2}$ while in $\ce{RCONR2}$ $\ce{NR2}$ leaves when treated with Grignard? I thought $\ce{NH2}$ should leave as there negative charge is more stable on $\ce{N}$ compared to $\ce{NR2}$ where the alkyl groups have a +I effect already.

  • $\begingroup$ I would appreciate answers instead of downvotes :) $\endgroup$ – Rits Jan 5 '18 at 16:51
  • 2
    $\begingroup$ Well, that's pretty obvious, you just don't use grigniards with compounds having even weakly acidic hydrogens. $\endgroup$ – Mithoron Jan 5 '18 at 17:39
  • $\begingroup$ Ikr but suppose if I use 2 eqt of grignard then why wouldnt NH2 leave ? Why does only NR2 leaves? $\endgroup$ – Rits Jan 5 '18 at 18:27
  • $\begingroup$ Grignard reagents with deprotonate amides and the RN- species is not a leaving group $\endgroup$ – Waylander Jan 5 '18 at 18:28

$\ce{RCONH_2}$ is a protic acid, and even though it's weak the Grignard reagent is a powerful enough proton base to react. You get deprotonation of the amide instead of nucleophilic attack on the carbonyl group.


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