Why does $\ce{N(CH_3)_3^+}$ have a larger $-I$ effect than $\ce{NH_3^+}$?

Since the methyl group is good at donating electrons it would stabilize the charge on nitrogen atom, decreasing its potential to withdraw electrons hence its $-I$ effect.

I can't figure out why its the other way around. Can someone help me understand this?

  • $\begingroup$ Are the groups in a gaseous phase or in a polar solvent..? The choice of phases in case of these groups changes everything. $\endgroup$ – Mitchell Feb 22 '17 at 8:27
  • $\begingroup$ @BhavyaSharma nothing as such is mentioned in the question. The question just wants me to find the group with highest -I effect $\endgroup$ – Osheen Sachdev Feb 22 '17 at 8:41
  • 1
    $\begingroup$ Methyl groups donate inductively only into empty-p-orbital cations (e.g. carbenium ions). Ammonium cations cannot profit from that stabilisation. $\endgroup$ – Jan Feb 22 '17 at 23:34
  • $\begingroup$ @Jan Oh I didn't know that, thank you! So basically inductive effect from methyl or hydrogen groups doesn't operate here. Then what is the factor that influences the stability of these ions? $\endgroup$ – Osheen Sachdev Feb 23 '17 at 8:54

The reason is that the methyl though gives electron via +I effect in case of H it is easy to cleave the N-H bond for nitrogen and release proton and take the electron pair from hydrogen but with C it being more electronegative than H the same is not possible .

  • $\begingroup$ But why is bond cleavage required? $\endgroup$ – Osheen Sachdev Feb 22 '17 at 8:55
  • $\begingroup$ to disperse the +ve charge from more electronegative atom to less electronegative atoms $\endgroup$ – Nitro phenol Feb 22 '17 at 9:00

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.