# Why does the trimethylammonium cation have a larger -I effect than the ammonium cation?

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?

• Are the groups in a gaseous phase or in a polar solvent..? The choice of phases in case of these groups changes everything. Feb 22 '17 at 8:27
• @BhavyaSharma nothing as such is mentioned in the question. The question just wants me to find the group with highest -I effect Feb 22 '17 at 8:41
• Methyl groups donate inductively only into empty-p-orbital cations (e.g. carbenium ions). Ammonium cations cannot profit from that stabilisation.
– Jan
Feb 22 '17 at 23:34
• @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? Feb 23 '17 at 8:54
• This question was answered a year ago at chemistry.stackexchange.com/questions/88185/…. The ring donates a pi electron to N and one CH3 group hyperconjugates (in reverse). The effect is demonstrated by the strong meta and deactivation effect typical of pi deactivators; NH3+ is only a weak deactivator and therefore not a pi actor. Apr 21 at 14:05

$$\ce{N(CH3)3+}$$ has more -I than $$\ce{NH3+}$$ , this irony happens because if you consider $$\ce{N-CH3}$$ bond and $$\ce{N-H}$$ bond, which is more polar?
Of course $$\ce{N-H}$$ bond will be more polar due to more difference in electronegativity value, so that implies electron density will be more on $$\ce{N}$$ in $$\ce{NH3+}$$ than $$\ce{N}$$ on $$\ce{N(CH3)3+}$$.
And therefore since electron density is less on $$\ce{N}$$ of $$\ce{N(CH3)3+}$$ , it will have greater -I effect.