Why is a cationic trialkylammonium functional group inductively withdrawing in nature?

The alkyl groups on the nitrogen ion should donate electrons (alkyl groups are electron donating in nature). So, irrespective of the nitrogen ion being a cation (which is an electrophile), the alkyl groups should compensate the electrophilicity of the nitrogen cation, and the ion as a whole should be electron donating in nature.

Then why is $\ce{R3N+}$ considered to be electron withdrawing?

• The inductive effect is very weak; the inductive effect will reduce the charge on the nitrogen atom but it does not make it zero. Nitrogen is quite electronegative and it hates to be positively charged (even partially). It tries to pull electrons towards itself wherever possible. – Yashas Apr 3 '17 at 13:57
• I don't know the reason, but $\text{R}_3\text{N}^+$ is considered highly de-activating and meta-directing for electrophilic substitution on Benzene. I wanted to ask this question! – jonsno Apr 3 '17 at 15:15

By far, the overall positive charge of the $\ce{[-NR3]+}$ group is its dominating characteristic when considering inductive behavior. Bear in mind that the magnitude of the charge variations due to the inductive effect is much smaller than $q$, the electron charge. These variations are written as $\delta^+$ and $\delta^-$, after all, and it's implicit in the notation that $|\delta^\pm| \ll 1$.

So, no matter how electron-donating the alkyl chains might be relative to the hydrogens in a simple amine group, $\ce{[-NH3]+}$, the overall functional group remains strongly electron withdrawing.

It's worth pointing out here a common misconception about ammonium cations; quoting Fry$^\dagger$ (emphasis added):

[One must] appreciate that while the central nitrogen atom of the [tetra-n-butylammonium] cation is customarily written with a positive charge, quantum chemical computations at all levels of sophistication all agree that in tetraalkylammonium ions each of the hydrogen atoms actually carries a small positive charge and all of the carbons and the central nitrogen atom carry a negative charge.

Thus, even though we usually write tetraalkylammonium cations $\ce{(R4N+)}$ and trialkylammonium side groups $\ce{(-R3N+)}$ as though the positive charge is localized on the central nitrogen atom, the strongly-electronegative $\ce N$ will in fact always carry a $\delta^-$ charge. It would thus be more accurate to write these species as $\ce{[NR4]+}$ and $\ce{[-NR3]+}$, respectively.

$^\dagger$ Fry, A.J. Electrochem Soc Interface, Summer 2016 issue, pp. 37-40. doi:10.1149/2.F04162if
(free article through 9 Apr 2017!)

• In the first part of the answer, you've written that alkyl groups will direct some electron density towards the nitrogen atom (although very feebly). If that's the case then why $NR_3+$ is even more electron-withdrawing than $NH_3+$? Shouldn't alkyl groups, with their superior +I ability than H atoms, decrease the -I effect in $NR_3+$? – Aumkaar Pranav Feb 28 at 8:58
• @AumkaarPranav Very good question! I don't actually know the reason for that. I'd suggest asking it as a brand new question...it does seem to be a hole in the logic of my answer. – hBy2Py Feb 28 at 14:23

Nitrogen is a very electronegative element and the fact that nitrogen bears a positive charge in the ammonium ion makes it even more electrophilic(as you pointed out) . Alkyl groups are only weak electron donating groups and can't help much about nitrogen's positive charge.Hence the ammonium cation is inductively withdrawing in nature.