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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?

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    $\begingroup$ An $\ce{sp}$ hybridized species, which has a bond angle of $180^{\circ}$ is more electronegative than an $\ce{sp^2}$ hybridized species, which has a bond angle of $120^{\circ}$ followed by $\ce{sp^3}$ which has a bond angle of $109.5^{\circ}$. So, as bond angle increases, electronegativity increases. So, as you said that 1 has a greater bond angle than 2, it should be more electronegative and hence it would show greater inductive effect on $\overset{+}{\ce{N}}$. $\endgroup$ May 17 at 10:28
  • $\begingroup$ Are you considering a theoretical comparison or approaches to develop experimental approaches to produce measurements? $\endgroup$ May 17 at 19:25
  • $\begingroup$ theoretical comparison $\endgroup$
    – Ashish
    May 18 at 10:01
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    $\begingroup$ @LightYagami that seems like you are relating two phenomenon's but why does that happen. Even If ur relations are true, I don't really understand how can steric effects/repulsions lead to higher electronegativity $\endgroup$
    – Ashish
    May 18 at 19:38
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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).


* 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."

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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? :)

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