The comments on this answer seem to detail a trend in which $\ce{NR3+, NHR2+, NH2R+}$ all show meta-directing nature towards electrophilic substitution, with the slightly stronger para-direction of $\ce{NH3+}$ being the odd one out, as cited from March's Advanced Organic Chemistry.

The 2020 edition has this to say, on pages 617 and 618 under "Aromatic Substitution, Electrophilic" (emphasis mine);

Groups that lack an unshared pair on the atom connected to the ring are electron -withdrawing with a (+) charge or (δ+) dipole. In this category are, in approximate order of decreasing deactivating ability, $\ce{NR3+}$, .... and $\ce{NH3+}$. Also in this category are all other groups with a positive charge on the atom directly connected to the ring... and many groups with positive charges on atoms farther away, since these are often still powerful −I groups. The field-effect explanation predicts that these should all be meta directing and deactivating, and (except for $\ce{NH3}$ +) this is the case. The $\ce{NH3+}$ group is an anomaly, since this group directs para about as much as or a little more than it directs meta, often because the ammonium ion is in equilibrium with the amine. The $\ce{NH2Me+, NHMe2 +}$, and $\ce{NMe3+}$ groups all give more meta than para substitution, with the percentage of para product decreasing with the increasing number of methyl groups.

@ron's explanation coupled with the equilibrium argument makes a pretty good reason for why aniline behaves this way.

But why do the other groups meta-direct? The field effect mentioned above seems to refer to inductive effects, which should make them para-directing.


1 Answer 1


This does not answer your question exactly, but I hope this can shed some light on the anomalous behaviour of anilinium cations.

First of all, it is not true that the para-product is always favoured with the $\ce{NH3+}$ ion, it depends on the reaction conditions. There are two effects to consider here—1) kinetic/thermodynamic control, and 2) rate determining step of the reaction. If the reaction is under thermodynamic constrol, the most stable isomer will form regardless of the directing property of the substituent. If it is under kinetic control, then you have to look at the rate determining step of the reaction. The electrophilic aromatic substitution is a two step process in general, and if the first step is rate determining, then the nucleophilicity of each postion of the ring will be the main factor. If the second step is rate determining then the energy of the intermediate will also matter. Obviously the ortho and para intermediates will be unstable as the positive charge is delocalized next to the electron withdrawing group. So, it's not always true that inductive effect by itself cannot be meta-directing.

Anyway, back to your original question. Let's consider the nitration of aniline in presence of conc. $\ce{H2SO4}$.

nitration of aniline

Brickman and Ridd[1] has noted the relative ratios of products for different sulfuric acid concentrations. $$\begin{array} {|r|r|}\hline \text{conc. of }\ce{H2SO4}(\%) & 85 & 92.4 & 96.4 & 100 \\ \hline ortho(\%) & 6 & - & - & - \\ \hline meta(\%) & 34 & 53 & 58 & 64 \\ \hline para(\%) & 59 & 47 & 42 & 36 \\ \hline \end{array} $$ As you can see, the meta product becomes more favoured as the concentration of $\ce{H2SO4}$ increases.

What about the N-methylated anilines? In another paper, Brickman, Utley and Ridd[2] have mentioned the percentages of meta and para products (ortho does not form at high acid concentrations, probably due to the steric effect). $$\begin{array} {|r|r|}\hline \text{conc. of }\ce{H2SO4}(\%) & 90.9 & 96.2 & 99.8 \\ \hline ortho(\%) & - & - & - \\ \hline meta(\%) & 61 & 67.5 & 70.3 \\ \hline para(\%) & 39 & 32.5 & 29.7 \\ \hline \end{array}$$ $$\text{For N-methylanilinium ion }\ce{Ph-NH2Me+}$$ Here, we again see the same trend that the increasing sulfuric acid concentration increases the meta-product percentage. However, the meta-product is now the major product.

$\ce{Ph-NHMe2+}$ also shows similar results. The N,N,N-triemthylanilinium ion $\ce{Ph-NMe3+}$ shows $89\%$ m-product and $11\%$ p-product at $98\%$ conc. $\ce{H2SO4}$.

In the two papers, the authors mention that the analysis of acid concentration vs. rate plot seems to indicate that the substitution predominantly happens on the anilinium cation i.e. the protonated state, so there is no possibility of the para-product arising from the reaction of the deprotonated compound. Sulfuric acid is a pretty strong acid, so it's unlikely that there is much deprotonated aniline left at such high concentration. The p/m ratio of aniline nitration changes by a factor of less than $2$, when the acid concentration is changed from $90$ to $98\%$, which reduces the proportion of deprotonated form by a factor of $31$. So, it's not possible that the para-product is coming from aniline, because then the rate profile would have been much more sensitive to the concentration of acid.

They have also ruled out significant hyperconjugative effects, by comparing the rates of the benzylammonium variants of all of these anilines. So, the effect must be mostly from inductive effects.

After a lengthy discussion, it is noted that the $\ce{-NH3+}$ group is much more solvated than the $\ce{-NHMe2+}$, $\ce{-NMe3+}$ etc. groups. So, it is possible that the solvent forms hydrogen bonds to the $\ce{-NH3+}$ group and allows the positive charge on it to spread away. This means that the inductive effect from $\ce{-NH3+}$ is actually weaker than the methylated groups. So, the methylated anilinium ions are simply more strongly averse to reaction on ortho/para positions than anilinium.

However, the authors note after that[2]:

The above explanation accounts for the major differences in the substituent effects recorded in Table 3 but it is not completely satisfactory, for the extent of para-substitution in the anilinium ion is more than would be expected if the effect of replacing $\ce{-N+Me}$, by $\ce{-N+H}$, were simply to "scale down" the effect of the positive pole; such an effect would never cause the para-position to be more reactive than one meta-position. This anomaly is also illustrated by the comparison of the anilinium ion with the benzyltrimethyl-ammonium ion: the latter ion is more reactive by a factor of ca. 25 but gives less para- substitution. It appears that the protonated poles give significantly higher p/m ratios than expected from their reactivity relative to the methylated poles.

The reason for this is not entirely clear ...

So, it's not completely known why aniline gives more para-products than its N-methylated derivatives.

This is only considering the nitration reaction. If you consider other reactions, there are varied directing nature. For example the sulfonation reaction (done with $\ce{H2SO4}$ and $\ce{SO3}$) produces almost entirely the para-product[3] (sulfanilic acid), and this has been attributed to the fact that the sulfonation is reversible (thermodynamic control). I haven't been able to find any data on halogenation or other substitutions on the anilinium ion.

  1. M. Brickman, J.H. Ridd, J. Chem. Soc., 1965, 6845-6851
  2. M. Brickman, J.H.P. Utley, J.H. Ridd, J. Chem. Soc., 1965, 6851-6857
  3. J.J. Jacobs Jr., D.F. Othmer, A. Hokanson, Ind. Eng. Chem., 1943, 35, 3, 321–323

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