Animated Solution for Chemistry - Organic Chemistry: In the reaction shown below, the major product(s) formed is / are :

The journey to mastering organic chemistry is paved with understanding the subtle, yet profound, differences in reactivity between functional groups. Today, we are going to dive deep into a classic battle of nucleophiles. Imagine a molecular arena where two nitrogen atoms are vying for the attention of a single electrophile. Who will win? Let's find out!

Our reactant is a fascinating molecule: a benzene ring adorned with two distinct nitrogen-containing groups. On one side, we have an aliphatic primary amine ($-{\text{CH}}_2-{\text{NH}}_2$). On the other, we have an amide ($-\text{C}(=\text{O})-{\text{NH}}_2$).

Into this arena enters our reagent: acetic anhydride ($({\text{CH}}_3\text{CO}{)}_2\text{O}$). Acetic anhydride is a renowned acetylating agent. It is an electrophile, meaning it is electron-deficient and actively seeks out electron-rich species to react with. In the world of organic chemistry, these electron-rich species are called nucleophiles.

The core question of this problem is a test of selectivity: Which of the two nitrogen atoms is the stronger nucleophile?

The Distracted Fighter

Let's first examine the amide group at the bottom left of our molecule. At first glance, the nitrogen atom has a lone pair of electrons, making it a potential nucleophile. However, we must look at its neighborhood.

This nitrogen is directly bonded to a carbonyl carbon ($\text{C}=\text{O}$). The carbonyl oxygen is highly electronegative, pulling electron density towards itself. This creates a powerful resonance effect. The lone pair on the amide nitrogen is not localized; instead, it is delocalized into the carbonyl pi system.

Because this lone pair is "busy" participating in resonance, it is significantly less available to reach out and attack an external electrophile like acetic anhydride. The amide nitrogen is, therefore, a very weak nucleophile.

The Focused Striker

Now, let's turn our attention to the top right group, the aliphatic primary amine ($-{\text{CH}}_2-{\text{NH}}_2$). Notice the crucial difference: the nitrogen atom is separated from the benzene ring by a $-{\text{CH}}_2-$ group.

This separation means the nitrogen's lone pair cannot participate in resonance with the aromatic ring. It is completely localized on the nitrogen atom. It is sitting there, highly concentrated, and ready to strike.

In the realm of nucleophilicity, a localized lone pair is vastly superior to a delocalized one. Therefore, the aliphatic amine is a much stronger nucleophile than the amide.

When acetic anhydride is introduced into the system, it is immediately attacked by the most reactive nucleophile available. The localized lone pair of the aliphatic amine launches a nucleophilic attack on one of the carbonyl carbons of acetic anhydride.

This leads to the cleavage of the anhydride bond, transferring an acetyl group ($-\text{C}(=\text{O}){\text{CH}}_3$) to the aliphatic nitrogen. The amide group, being a weak nucleophile, remains completely untouched during this process.

The final major product features an newly formed amide bond at the top position, while the original amide at the bottom remains intact. The byproduct of this acetylation is a molecule of acetic acid (${\text{CH}}_3\text{COOH}$).

Comparing our deduced product with the given options, we find a perfect match with Option (A). This problem is a beautiful reminder that in organic chemistry, structure dictates function, and the availability of electrons is the ultimate key to reactivity!