Animated Solution for Chemistry - Organic Chemistry: Hinsberg's reagent is

Imagine you are working in an organic chemistry laboratory, and you are handed three unlabelled test tubes. You are told that one contains a primary amine, another a secondary amine, and the last one a tertiary amine. They all look like clear liquids and smell somewhat similar (usually like fish or ammonia). How do you tell them apart?

This is where a brilliant piece of chemical detective work comes into play, known as the Hinsberg Test. The hero of this test is a specific chemical compound called Hinsberg's Reagent.

Benzene Sulphonyl Chloride

Hinsberg's reagent is chemically known as benzene sulphonyl chloride. Its molecular formula is ${\text{C}}_6{\text{H}}_5{\text{SO}}_2\text{Cl}$.

Let's break down its structure to understand why it is so effective. At its core, it has a stable benzene ring (${\text{C}}_6{\text{H}}_5-$). Attached to this ring is the highly reactive sulphonyl chloride group ($-{\text{SO}}_2\text{Cl}$). In this group, the central sulfur atom is double-bonded to two highly electronegative oxygen atoms and single-bonded to a chlorine atom.

Because oxygen and chlorine are electron-greedy, they pull electron density away from the sulfur atom. This leaves the sulfur atom with a strong partial positive charge, making it an excellent electrophile (electron lover). When an amine, which has a lone pair of electrons on its nitrogen atom, comes near, it acts as a nucleophile and eagerly attacks this electron-deficient sulfur atom.

When a primary amine ($R-{\text{NH}}_2$) reacts with benzene sulphonyl chloride, the nitrogen attacks the sulfur, and the chlorine atom is kicked out as a leaving group. The resulting product is an N-alkylbenzene sulphonamide.

Here is the crucial part: a primary amine starts with two hydrogen atoms attached to the nitrogen. After the reaction, one hydrogen is lost to form $\text{HCl}$, but one hydrogen remains attached to the nitrogen in the sulfonamide product.

Because the adjacent sulphonyl group is strongly electron-withdrawing, it makes this remaining hydrogen highly acidic. If you add an aqueous alkali (like $\text{NaOH}$) to the test tube, this acidic hydrogen is easily removed, forming a water-soluble sodium salt.

Observation: The initial precipitate dissolves in alkali, confirming a primary amine.

Now, let's look at a secondary amine ($R_2\text{NH}$). It also attacks the sulfur atom, kicking out the chlorine to form an N,N-dialkylbenzene sulphonamide.

However, a secondary amine only starts with one hydrogen atom on the nitrogen. This single hydrogen is lost during the formation of the sulfonamide. As a result, the final product has no acidic hydrogen left on the nitrogen atom.

When you add aqueous alkali to this test tube, nothing happens. The sulfonamide cannot be deprotonated, so it remains as an insoluble solid.

Observation: A precipitate forms that does not dissolve in alkali, confirming a secondary amine.

Finally, what about a tertiary amine ($R_3\text{N}$)? A tertiary amine has zero hydrogen atoms attached to its nitrogen.

While the lone pair on the nitrogen can temporarily attack the sulfur atom, the intermediate formed cannot lose a proton to stabilize itself. Because there is no hydrogen to lose, the reaction reverses, and no stable sulfonamide product is formed.

Observation: No reaction occurs (the amine remains unreacted and insoluble in the aqueous mixture, though it will dissolve if you add acid later), confirming a tertiary amine.

By understanding the molecular structure of benzene sulphonyl chloride (${\text{C}}_6{\text{H}}_5{\text{SO}}_2\text{Cl}$) and how the number of hydrogen atoms on the amine dictates the solubility of the final product, we can flawlessly identify the class of any aliphatic amine. Therefore, the correct option for Hinsberg's reagent is indeed (c).