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I was working on a question that basically boils down to deciding whether the nitro group ($\ce{-NO2}$) or the trimethylammonium group ($\ce{-N+(CH3)3}$) on a benzene ring, is a stronger electron withdrawing group (EWG).

The solution says that the trimethylammonium group is a stronger EWG since it has an overall positive charge, while the nitro group is overall neutral.

But I am getting confused here. Why should the nitro group be neutral? The $\ce{N}$ atom has 4 bonds (a single bond to an oxygen atom, a double bond to another oxygen atom, and another bond to the carbon of the ring), so the $\ce{N}$ atom in $\ce{NO2}$ should also have a positive charge?

Further I analysed that since $\ce{O}$ is more electronegative than $\ce{N}$, it tries to shift the $\pi$ electrons of the double bond towards itself, hence increasing the electrophilicity of the nitrogen of $\ce{NO2}$, while in the trimethyl amino group, the $\ce{N}$ being the more electronegative atom gets electrons donated to it due to the +I effect of the three $\ce{CH3}$ groups, thus reducing the electrophilicity of $\ce{N}$. So $\ce{-NO2}$ should be a stronger EWG than $\ce{-N+(CH3)3}$.

Where am I going wrong?

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    $\begingroup$ You are confusing the way the nitro group is drawn with a partial +ve charge on the N for an actual charge. It is not charged, this is just a convenient way of modelling it. The NMe3 group has an actual overall charge. $\endgroup$ – Waylander May 23 at 14:39
  • $\begingroup$ @Waylander, Sorry for the late reply, but does it mean that the $NMe_3$ group exists with the ring as an ion? Like $Na^+Cl^-$? $\endgroup$ – WTS May 25 at 6:07
  • $\begingroup$ Yes, there will be an anion associated with this such as Cl- $\endgroup$ – Waylander May 25 at 7:14

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