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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?
$\ce{N(CH3)3+}$ 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.
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 .
