-2
$\begingroup$

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.

|improve this question
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

$\ce{RCONH_2}$ 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.

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.