How do steric effects affect inductive effect in compounds with tetravalent nitrogen?

Question

I want to compare the electron density on methyl group between tri‐tert‐butyl(methyl)azanium (1) and triisopropyl(methyl)azanium (2), i.e. essentially the inductive effects of $\ce{-\overset{+}{N}R3}$ groups.

$$\underset{\large\textbf{1}}{\ce{Me-\overset{+}{N}(t-Bu)3}}\qquad\underset{\large\textbf{2}}{\ce{Me-\overset{+}{N}(i-Pr)3}}$$

The only difference I see in these compounds is that due to steric effects the bond angle would be greater in compound 1 than in 2. Based on this info alone I am not able to judge which one would have greater electronegativity on $\ce{\overset{+}{N}},$ or so as to say greater inductive effect due to the $\ce{-\overset{+}{N}R3}$ group.

How should I go about comparing the electron density on methyl group in both of these species?

Answer 1

Since you are asking for a theoretical explanation, I believe a theoretical calculation might be helpful. It is possible to calculate the partial charges on atoms using electronic structure programs. I have calculated the partial charges at $\text{B3LYP/pcseg-1//GFN2-xTB}$ level (with GAMESS and xTB).

$\ce{Me-\overset{+}{N}(^iPr)3}$

We only need to look at the partial charges on the Me group to know the electron-withdrawing capacity of $\ce{-N+(^iPr)_3}$ group.

\begin{array}{cc} \hline \mathrm{atom} & \mathrm{partial\;charge\;(e)} \\ \hline \ce{C} & -0.4367\\ \ce{H} & \;\;\, 0.1954\\ \ce{H} & \;\;\, 0.1957\\ \ce{H} & \;\;\, 0.1966\\ \hline \end{array}

Total charge on $\ce{Me} = 0.151$.

$\ce{Me-\overset{+}{N}(^tBu)3}$

Again, we only need to look at the Me group.

\begin{array}{cc} \hline \mathrm{atom} & \mathrm{partial\;charge\;(e)} \\ \hline \ce{C} & -0.3087\\ \ce{H} & \;\;\, 0.1436\\ \ce{H} & \;\;\, 0.1430\\ \ce{H} & \;\;\, 0.1414\\ \hline \end{array}

Total charge on $\ce{Me} = 0.1193$

So, the partial charges would indicate that $\ce{-N+(^iPr)3}$ is more electron withdrawing than $\ce{-N+(^tBu)3}$.

Explanation?

I am not sure I have a definitive explanation for this. But it's possible to look at the differences between $\ce{^iPr}$ and $\ce{^tBu}$—

1. The cone angle formed by the three R groups for $\ce{-N+R3}$ would be larger for $\ce{R=^tBu}$ obviously, due to the steric bulk of that group. When the angles are the ideal $109.5^\circ$ each bond can be assumed to form from $\mathrm{N\;sp^3}$ orbitals. When N is $\mathrm{sp^2}$, the three bonds would be $120^\circ$ to each other. So, when the cone angle increases, the amound of p-character in those three N-R bonds can be assumed to go down. Conversely the amount of p-character in the 4th bond goes up which would mean less electron withdrawing capacity (p-orbitals are further away from nucleus). So, $\ce{-N+(^tBu)3}$ should be weaker EWG.

2. As there is no empty p-orbitals on N, so there should be no hyperconjugative effects for the compounds drawn. However, if instead of $\ce{Me}$, there is some other group that has a $\pi$ system attached to N, then there is a possibility of hyperconjugation.

3. The $\ce{CH3}$ groups have a very weak electron donating effect, so $\ce{-N+(^tBu)3}$ will be slightly weaker EWG than $\ce{-N+(^iPr)3}$. *

Notice that points 1 and 3 predict the same trend, which matches with the calculated trend. However, it is difficult to tell which factor is the most important (and whether these factors actually mean anything).

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* Curiously, the partial charge on $\ce{C}$ of the side chain goes up when its $\ce{H}$ is replaced by $\ce{CH3}$, which would suggest that $\ce{CH3}$ is actually electron-withdrawing. This does not match with the standard notion of methyl groups being electron donating. I am not really sure if there is any explanation for this discrepancy. But note that there have been some evidence that $\ce{-CH3}$ groups can act as EWG's in some circumstances. The linked paper specifically mentions in the abstract—"That methyl groups attached to carbon atoms are electron donors must not be generally assumed."

Answer 2

Both N are sp³ hybridized. Both N are quaternary nitrogens with a positive formal charge. The tBu group is more electron donating (think vectors of the N-C-C geometries) than the iPr group, but there is more of a steric penalty due to its bulk; on the other hand, the N-C-H linkage in the iPr ammonium cation will also have a slightly larger dipole than the N-C-C. So, in answer to your question about electron density, I would say the [MeN(tBu)₃]⁺ again because of the EDG effect of tBu > iPr.

Does this make sense? :)