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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 .

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  • $\begingroup$ 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

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