Animated Solution for Chemistry - Organic Chemistry: Carbylamine test is used to detect the presence of primary amino group in an organic compound. Which of the following compound is formed when this test is performed with aniline?

Unmasking Primary Amines with the Carbylamine Test

In the vast and intricate world of organic chemistry, qualitative analysis is like detective work. Chemists rely on specific, observable reactions to identify the functional groups hidden within a molecule. Among these chemical interrogations, the Carbylamine Test (also known as the Hofmann isocyanide synthesis) stands out—not just for its chemical elegance, but for its notoriously unforgettable, foul odor.

Imagine you are handed a vial containing an unknown organic compound. You suspect it might be a primary amine, but you need proof. This is where the carbylamine test comes into play. The test is remarkably specific: it only works for primary amines ($1^{\circ }$ amines), whether they are aliphatic (like methylamine) or aromatic (like aniline).

In our specific problem, we are testing aniline ($C_6{H}_{5}N{H}_2$). Aniline is an aromatic primary amine where the amino group ($-N{H}_2$) is directly bonded to a benzene ring. To perform the test, we introduce two crucial reagents: chloroform ($CHC{l}_3$) and ethanolic potassium hydroxide ($KOH$).

The magic happens when we apply heat to this mixture. But what exactly is going on at the molecular level? The reaction is driven by the generation of a highly reactive, electron-deficient intermediate called dichlorocarbene ($:CC{l}_2$).

When the strong base (ethanolic $KOH$) reacts with chloroform, it strips away a proton, leading to the loss of a chloride ion and the formation of the dichlorocarbene. This carbene is hungry for electrons. The nitrogen atom of our primary amine, armed with its lone pair of electrons, acts as a nucleophile and attacks the carbene.

Through a series of proton transfers and the elimination of two more chloride ions, the nitrogen and carbon atoms forge a strong triple bond. The original $-N{H}_2$ group is completely transformed into an isocyanide group ($-NC$).

The overall balanced equation for aniline is:

A Scent to Remember

The product of this reaction is phenyl isocyanide ($C_6{H}_{5}NC$). While the chemical equation is neat and balanced, the physical reality of this reaction is an assault on the senses. Isocyanides (or carbylamines) are infamous for their incredibly foul, pungent, and nauseating odor.

In a laboratory setting, the sudden emission of this terrible smell is the definitive, positive confirmation that the original compound was indeed a primary amine.

You might wonder, why don't secondary ($2^{\circ }$) or tertiary ($3^{\circ }$) amines give this test? The answer lies in the mechanism. To form the triple bond of the isocyanide group ($-N\equiv C$), the amine nitrogen must lose two protons during the reaction.

Secondary amines only have one proton attached to the nitrogen, and tertiary amines have none. Therefore, they simply cannot complete the transformation. If you were to test N-methylaniline (a secondary amine) with chloroform and $KOH$, the reaction would fail, and you would be spared the foul smell.

Returning to our problem, when aniline is subjected to the carbylamine test, the $-N{H}_2$ group is converted into an $-NC$ group. The resulting compound is phenyl isocyanide, making option (c) the correct answer. It is a beautiful example of how a profound structural change at the atomic level can result in a dramatic, macroscopic observation.