Animated Solution for Chemistry - Organic Compounds: Primary, secondary and tertiary amines can be separated using

Separating Amines

Imagine you are handed three test tubes containing clear liquids. You are told that one contains a primary ($1^{\circ }$) amine, another a secondary ($2^{\circ }$) amine, and the last a tertiary ($3^{\circ }$) amine. How do you tell them apart?

This is a classic qualitative analysis problem in organic chemistry. To solve it, we need a chemical "sorting hat"—a reagent that reacts uniquely with each class of amine based on their structural differences. Enter the Hinsberg Test.

The traditional hero of this test is benzene sulphonyl chloride ($C_6{H}_{5}S{O}_{2}Cl$). However, its close cousin, para-toluene sulphonyl chloride ($p-C{H}_3{C}_6{H}_{4}S{O}_{2}Cl$), is frequently used in modern laboratories because it is a solid and easier to handle. Both reagents operate on the exact same chemical principle: they seek out replaceable hydrogen atoms on the amine's nitrogen.

The Soluble Secret

Let's drop our reagent into the test tube containing the primary amine ($R-N{H}_2$). A reaction immediately occurs, forming an N-alkyl sulphonamide.

Now, here is the crucial catch: the nitrogen atom in this new product still has one hydrogen atom attached to it. Because this nitrogen is bonded directly to a highly electronegative sulphonyl group ($-S{O}_2-$), the electron density is pulled away from the $N-H$ bond. This makes the remaining hydrogen strongly acidic.

When we add a strong base like aqueous sodium hydroxide ($NaOH$) to the mixture, the base easily plucks off this acidic proton, forming a water-soluble sodium salt. The result is a perfectly clear, homogeneous solution.

The Stubborn Precipitate

Now, let's shift our focus to the secondary amine ($R_{2}NH$). It also reacts with the sulphonyl chloride, forming an N,N-dialkyl sulphonamide.

But notice the structural difference? There are absolutely no hydrogen atoms left on the nitrogen! Without an acidic proton, the $NaOH$ base has nothing to attack. Therefore, the N,N-dialkyl sulphonamide remains completely insoluble in the basic aqueous layer, forming a distinct, visible precipitate.

The Silent Bystander

Finally, what about the tertiary amine ($R_{3}N$)? For the initial sulphonamide formation to even occur, the amine must possess at least one replaceable hydrogen atom to eliminate a molecule of $HCl$ with the chloride from the reagent.

Tertiary amines have no such hydrogen. Therefore, there is absolutely no reaction with the Hinsberg reagent. The tertiary amine simply remains unreacted and insoluble in the basic mixture.

By simply observing the solubility of the reaction mixture in a strong base, we can confidently separate and identify these amines. The primary amine dissolves, the secondary forms a precipitate, and the tertiary does not react at all. The reagent that makes this elegant separation possible is para-toluene sulphonyl chloride.