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?




The inductive effect can be quantitatively measured by the [Hammett equation](https://en.wikipedia.org/wiki/Hammett_equation)


$$\sigma(\ce{X}) = \mathrm{p}K\_\mathrm{a}(\ce{H}) - \mathrm{p}K\_\mathrm{a}(\ce{X})$$


where $\mathrm{p}K\_\mathrm{a}(\ce{X})$ refers to the $\mathrm{p}K\_\mathrm{a}$ of $\ce{X}$-substituted benzoic acid and $\mathrm{p}K\_\mathrm{a}(\ce{H})$ to the $\mathrm{p}K\_\mathrm{a}$ of unsubstituted benzoic acid. A more electron-withdrawing group will lead to a more acidic carboxylic acid group, a lower $\mathrm{p}K\_\mathrm{a}(\ce{X})$, and hence a more positive $\sigma$ coefficient.


Values for a wide range of substituents in the *meta* and *para* positions have been tabulated in Table I of Ref. 1. Also included are Swain and Lupton's $F$ ($F$ for *field*) parameters which are designed to remove the influence of resonance effects on the original Hammett coefficients $\sigma$ (given in the same table; see Ref. 2 for original discussion).


$$\begin{array}{ccccc} \hline
\text{Entry number} & \text{Substituent} & \sigma\_\mathrm{m} & \sigma\_\mathrm{p} & F \\ \hline
\text{main text, p 167} & \ce{NMe3+} & 0.88 & 0.82 & 0.89 \\
58 & \ce{NH3+} & 0.86 & 0.60 & 0.92 \\
241 & \ce{SMe2+} & 1.00 & 0.90 & 0.98 \\
15 & \ce{F} & 0.34 & 0.06 & 0.45 \\ \hline
\end{array}$$


So, perhaps the answer should be $\ce{SMe2+}$. I don't have a good explanation for why, given that sulfur is less electronegative than nitrogen. I also do not think that the difference between $\ce{NMe3+}$ and $\ce{NH3+}$ can solely be attributed to the inductive effect of the methyl groups, as it is quite possible that there are solvation or hydrogen bonding effects that affect the $\mathrm{p}K\_\mathrm{a}$ values. The bottom line is that I don't think this can be rationalised as simply as the question seems to want.


**References**


1. Hansch, C.; Leo, A.; Taft, R. W. A survey of Hammett substituent constants and resonance and field parameters. *Chem. Rev.* **1991,** *91* (2), 165–195. [DOI: 10.1021/cr00002a004.](http://doi.org/10.1021/cr00002a004) (there is a non-paywall copy [here](https://faculty.missouri.edu/~glaserr/3700s10/SC_1991_Hansch.pdf))
2. Gardner Swain, C.; Lupton, E. C. Field and resonance components of substituent effects. *J. Am. Chem. Soc.* **1968,** *90* (16), 4328–4337. [DOI: 10.1021/ja01018a024.](http://doi.org/10.1021/ja01018a024)


 11

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.


 10

$\ce{-NH3+}$ should be the answer. Your thinking about the group having positive charge being more electron withdrawing is correct. Sulfur being less electronegative than nitrogen, will not withdraw electron more strongly, hence will have less $-I$ effect. Now the two nitrogen groups remain. The $\ce{-NMe3+}$ group has three methyl groups donating their electron density to the $\ce{N}$ atom, hence fulfilling some of its electronic needs. So it will have less $-I$ effect than $\ce{-NH3+}$. I hope the thought process is clear.


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