Animated Solution for Chemistry - Organic Chemistry: The major product in the above reaction is

Have you ever wondered what happens when a molecule has two similar functional groups, but only enough reagent to react with one? It becomes a chemical competition! In this problem, we are faced with a fascinating scenario: a single molecule containing both an aliphatic amine and an aromatic amine, reacting with acetic anhydride.

Let's dive into the mechanics of this reaction and discover why one amine emerges victorious.

Our reactant is a benzene ring adorned with two amino groups. At one position, we have a primary aromatic amine ($-{\text{NH}}_2$) directly attached to the ring. At another position, we have a primary aliphatic amine ($-{\text{CH}}_2{\text{NH}}_2$), separated from the ring by a methylene ($-{\text{CH}}_2-$) bridge.

We are treating this molecule with acetic anhydride, $({\text{CH}}_3\text{CO}{)}_2\text{O}$. Acetic anhydride is a classic acetylating agent. It's hungry for electrons, meaning it acts as an electrophile. To react with it, our molecule needs to provide a nucleophile—an electron-rich center. Both of our amino groups have a lone pair of electrons on their nitrogen atoms, making them both potential nucleophiles.

But who will attack first?

To determine the major product, we must evaluate the relative nucleophilicity of the two amino groups. Nucleophilicity is all about how available and willing a lone pair is to be donated to an electrophile.

Let's look at the aromatic amine first. The nitrogen atom is directly bonded to the $s{p}^2$-hybridized carbon of the benzene ring. Because of this direct attachment, the lone pair on the nitrogen can participate in resonance with the $\pi$-electron system of the ring. This delocalization spreads the electron density around the ring, making the lone pair less available for donation. Consequently, the aromatic amine is a relatively weak nucleophile.

Now, let's examine the aliphatic amine. Here, the nitrogen is attached to an $s{p}^3$-hybridized carbon (the $-{\text{CH}}_2-$ group). This methylene bridge acts as an insulator, completely preventing the nitrogen's lone pair from interacting with the benzene ring's $\pi$-system. The lone pair remains strictly localized on the nitrogen atom. This high concentration of electron density makes the aliphatic amine significantly more basic and a much stronger nucleophile.

When the acetic anhydride is introduced, the highly reactive, localized lone pair of the aliphatic amine attacks the carbonyl carbon of the anhydride much faster than the delocalized lone pair of the aromatic amine.

As a result, selective acetylation occurs at the aliphatic position. The $-{\text{CH}}_2{\text{NH}}_2$ group is converted into an amide ($-{\text{CH}}_2{\text{NHCOCH}}_3$), while the aromatic $-{\text{NH}}_2$ group remains untouched.

This beautiful display of regioselectivity leads us straight to our major product, perfectly matching option (d). It's a perfect reminder that in organic chemistry, the availability of electrons dictates the flow of the reaction!