3
$\begingroup$

When $\ce{RMgX}$ (Grignard reagent) reacts with chloramine ($\ce{NH2Cl}$) it forms $\ce{RNH2}$.

But why doesn't it form $\ce{RH}$ by taking a hydrogen atom from chloramine as the negative charge so developed on nitrogen will be more stable as nitrogen has high electronegativity and $\ce{Cl}$ being a electron withdrawing group further stabilise the anion.

So where am I wrong? Can anybody suggest a mechanism?

|improve this question
$\endgroup$
  • $\begingroup$ On nitrogen, chlorine is no longer electron-withdrawing. In $\ce{NCl3}$, chlorine actually bears the partial positive charge (demonstrated by reacting with chloride ions to form chlorine gas). $\endgroup$ – Jan Nov 19 '17 at 14:05
  • $\begingroup$ Still may be more electron wothdrawing/less electron releasing than hydrogen, though. $\endgroup$ – Oscar Lanzi Nov 19 '17 at 20:53
5
$\begingroup$

The $\ce{N-H}$ protons are relatively weakly acidic and this makes deprotonation kinetically less favored if an alternative nucleophilic attack mechanism is available. With $\ce{NH3}$ no such alternative is there, due to lack of a decent leaving group, so deprotonation ultimately prevails; but $\ce{NH2Cl}$ allows displacement of a chloride ion and thus enables nucleophilic attack.

|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.