organic chemistry - Strongest negative inductive effect group between trimethylammonium, ammonium and dimethylsulfonium groups

Which of the following group exerts the strongest -I effect?

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


2. $\ce{-NH3+}$


3. $\ce{-S(CH3)2+}$


4. $\ce{-F}$

My idea is that positive species exerts more -I effect than neutral species. thus, 4) is neglected. Moreover, as nitrogen and sulfur have almost the same electronegativity 3) has two methyl groups and 2) none so 2) should exert more -I effect. However, the answer key says it to 1). Why is this?

Answer

Methyl groups are great at stabilising carbenium ions via an inductive effect — which should actually be considered a resonance effect — known as hyperconjugation. This effect, which is actually due to neighbouring $\ce{C-H}$ bonds, not indiscriminately due to neighbouring carbon atoms, is electronic in nature. The cationic carbon has an empty p orbital, which can line up with the $\sigma_{\ce{C-H}}$ orbitals of the neighbouring carbon. These two orbitals can then interact, stabilising the $\sigma_{\ce{C-H}}$ orbital and destablising the empty p orbital for an overall energy gain of the system. It can be understood in resonance terms by the following mesomeric structures:

$$\ce{H-CR2-CR2+ <-> H+\bond{...}CR2=CR2}\tag{1}$$

For this effect to be helpful, it is vital for the positive charge to derive from an empty p orbital.

In ammonium derivatives, this is not the case. Nitrogen is $\mathrm{sp^3}$ configured with four bonds to neighbouring atoms; the positive charge stems only from an extraneous proton in nitrogen’s nucleus; it cannot be attributed to any specific (empty) orbital. In fact, all valence and bonding orbitals are fully occupied; only antibonding and far-removed orbitals are vacant. Therefore, hyperconjugation cannot serve to stabilize ammonium cations.

Instead, we need to base our discussion on truly inductive effects based solely on the electronegativity of the bonding partners. I assume that this is the analysis your answer key wants you to perform. The first step is to eliminate fluorine which is not a cation. The second step is to distinguish between sulfur and nitrogen — the latter has a higher electronegativity and thus features a less stable cation.

The final step is then comparing an $\ce{-NH3+}$ group to an $\ce{-NMe3+}$ group. Here, we need to compare secondary effects: the electronegativity of carbon is higher than that of hydrogen. Thus, a nitrogen bound to three hydrogens experiences a greater partial negative charge from these three bonds than a corresponding nitrogen bound to three carbons. Therefore, the $\ce{-NH3+}$ cation is slightly more stabilised thanks to hydrogen’s lower electronegativity — and thus $\ce{-NMe3+}$ exhibits the greater -I effect.