Feasibility of Electrophilic attack on the Nitrogen over the Phenyl ring of Aniline

Question

I have learnt that the '-NH2' substituent attached to the phenyl ring, owing to the substitutent's highly positive mesomeric effect, activates the phenyl ring quite a lot for Electrophilic Aromatic Substitution, so much so that it easily forms a tri-substituted product when Br+ is used as an electrophile with Aniline, and that a protecting group such as Acetyl (introduced using Acetic Anhydride) has to be introduced on the Nitrogen to deactivate the ring for just monobromination to occur.

This made me wonder that if such is the influence of the '-NH2' on the benzene ring then why would the Nitrosonium ion generated in diazotization prefer attacking the lone pair of Nitrogen instead of the electronically enriched benzene ring on the ortho or para position?

Initially I lazily (and wrongly) attributed it to perhaps some stronger bonds forming between similar atoms, i.e Nitrogen-Nitrogen bonds being stronger than the Carbon-Nitrogen bonds.

But, more interestingly I realized that this line of reasoning wouldn't apply to my above stated acetylation of the Nitrogen in Aniline, for in that case too there isn't an Aromatic Substitution taking place and there are no Nitrogen-Nitrogen bonds.

In short, what is the reason for the feasibility of Diazotization and Acetylation in the way they occur with Aniline? Why aren't electrophilic aromatic substitution reactions taking place? I may have erred in my basics, please help me out.

Answer 1

Electrophilic Aromatic substitution in Aniline will occur when the $\pi$-electron cloud of Benzene ring attacks the electrophile generated (mostly in-situ) instead of the lone pair on the Nitrogen atom. Thus for occurance of Electrophilic aromatic substitution, there always remains a competition between the $\ce{N}$ atom and the $\pi$-electron rich benzene ring. You may think that the lone pair of nitogen remains fully delocalised through the Benzene ring, but there always remains a high enough electronic charge density on the Nitrogen atom to behave as a nucleophile and in fact in greater extent the electronic density of the ring (not always for Oxygen though, being highly electronegative, its less nucleophilic, but Nitrogen is certainly more nucleophilic).

Thus,you will observe that it is difficult to perform electrophilic aromatic substitution reaction on unprotected Aniline. Thus, most of the reactions on Aniline are initiated by attacking of the elctrophile by Nitrogen atom instead of the ring. It is its general trend. Thus to do substitution in the ring, we first need to decrease electronic charge density on the Nitrogen atom and that's why we first treat the Aniline with $\ce{CH3COCl}$ in Pyridine medium to form acetanilide, and then perform any other reaction on the ring like Bromination, Chlorination, Alkylation etc.

You will also observe that Aniline doesn't undergo Friedel Crafts alkylation/acylation to a great extent. As at the moment you add Lewis acid, Nitogen atom behave as Lewis Base and forms the adduct, which make the ring highly electron deficient due to induced positive charge on Nitrogen.

Thus, the Diazotization and Acetylation are the normal/expected reaction of Aniline as a effect of higher nucleophilicity of $\ce{N}$(due to lesser elctronegativity and weaker basicity). Instead if you have to perform Electrophilic aromatic substiution in the ring, you have to think about how to perform it.

Answer 2

Adding to Soumik Das' great answer, The ring would have to be activated to a large extent to overcome the nucleophilicity of aniline's nitrogen if you would like to add the nitrosonium ion at the benzene, as the aromaticity of the ring would have to broken in the rate determining step, which would have significant activation energy demands.And indeed, it has been observed that for tertiary aryl amines, electrophilic aromatic substitution of the NO+ ion takes place majorly.

As a sidenote, if you consider diazotization in the case of tertiary amines, then according to the mechanism, we need to have atleast one hydrogen present on the nitrogen(of the amine) for the tautomeric step to take place. But due to the lack of H in tertiary amines, the addition of electrophile and it's removal from the amine will probably exist as a parasitic equilibrium