Hinsberg's Test

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Imagine one flask holding a mixture of three amines: a primary, a secondary and a tertiary amine, all mixed together. Can a single reagent tell them apart, and even help us separate them? It can, and today you will see exactly how. This test comes up often in JEE questions on amines, so let's begin.

0:26Hinsberg's reagent

So, let us meet the reagent first. Hinsberg's reagent is C six H five S O two C l, benzenesulphonyl chloride. C six H five is the phenyl group, a benzene ring, S O two is the sulphonyl group, and C l is chlorine. The small numbers written low count the atoms. Okay? Another reagent is also used, p toluenesulphonyl chloride; the italic p stands for para. It is the same kind of reagent, with a methyl group on the ring at position four, opposite the sulphonyl chloride group. Right? And the method is really simple. Shake the amine with the reagent and aqueous KOH, that is potassium hydroxide, an alkali. Then we just watch what dissolves and what does not. With me?

1:13Three classes, three results

Now the three classes, one by one, and the figure shows all three. First, a primary amine, RNH two, where R is the carbon group. The small raised circle after the number one is read as degree, so one degree means primary. The arrow means gives: the benzenesulphonyl group, C six H five S O two, takes the place of one hydrogen on nitrogen, and we get C six H five S O two NHR, a sulphonamide. Okay? Now look at that product carefully. Is any hydrogen still left on the nitrogen? Yes, exactly, one is left, and the little dash in N H is the bond. Right next to it sits the sulphonyl group, S O two, which strongly pulls electrons away. So this last hydrogen becomes acidic. Right? And an acidic hydrogen meets the alkali. It forms a salt, and the salt dissolves, so the product from a primary amine is soluble in alkali. Got it? Next, a secondary amine, R two N H, with two carbon groups and one hydrogen on nitrogen. It reacts the same way and gives C six H five S O two N R two. So tell me, will this product dissolve in the alkali? It will not, and here is why: no hydrogen is left on the nitrogen. Nothing acidic means no salt, so the product stays insoluble in alkali. Okay? And a tertiary amine, R three N, has no hydrogen on nitrogen at all. There is nothing to replace, so there is no reaction. Clear?

2:54What you see in the test tube

Now, what do you actually see in the test tube? Look at the right side of the figure. For a primary amine, you get a clear solution. And on adding acid, the sulphonamide is set free from its salt and comes out as a solid. Okay? Now here is a classic JEE trap. In the figure, the secondary amine leaves something insoluble, and the tertiary amine also leaves an insoluble layer. So both give the same result, right? They do not, so be careful here, and add acid to tell them apart. The secondary amine gives a solid, or sometimes an oil, that dissolves in neither acid nor alkali. Right? But the tertiary amine never reacted. That layer is just the amine itself, and an amine is a base, so it dissolves on adding acid. With me? So fix this line in your mind: only the sulphonamide of a primary amine dissolves in alkali.

3:49Topper's trick: count the H on nitrogen

Clear? Now let me share a quick trick that toppers use. Just count the hydrogens on nitrogen: two for a primary amine, one for a secondary amine, and zero for a tertiary amine. The reagent replaces one of them. So how many are left in each case? One left for primary, zero left for secondary, and the tertiary amine had nothing to give, so no reaction. Right? So one hydrogen left means soluble in alkali, and none left means insoluble. That's it, the whole test in one line!

4:21Example 1: which amines give a product soluble in KOH

Now Example one, an easy one to warm up. Which of these give, with benzenesulphonyl chloride, a product soluble in KOH: propan one amine, N methylethanamine, aniline, or N N dimethylmethanamine? Pause the video and try it first. Done? Okay, let's check, using our trick. Propan one amine is a chain of three carbons ending in N H two; the one in its name says where the N H two sits. Two hydrogens on nitrogen, so it is primary, and its product is soluble. Okay? Next, N methylethanamine. Its name ends in ethanamine, so it is primary too, right? Careful here, because the capital italic N in front means the methyl group sits on the nitrogen. So the nitrogen holds an ethyl, a methyl and only one hydrogen. It is secondary, and its product is insoluble. Clear? Now aniline, C six H five N H two. Some students skip it because it is aromatic, but its nitrogen still has two hydrogens. So it is primary, and its product dissolves. With me? And N N dimethylmethanamine: the italic N, written twice, says that two methyl groups sit on the nitrogen, plus the methyl of methanamine. The bracket with a small three outside means three methyl groups, and no hydrogen on nitrogen, so it is tertiary and does not react. Right? So the answer is propan one amine and aniline, both primary, one aliphatic and one aromatic. Got it?

5:58Example 2: separate a mix of 1, 2 and 3 degree ethylamines

And now Example two, a harder one, and it is our puzzle from the start. Outline how a mixture of ethanamine, N ethylethanamine, and N N diethylethanamine is separated using Hinsberg's reagent. Pause the video and try it first. Done? Okay, let's check. First, classify them; the labels in round brackets give the class. Ethanamine, C two H five N H two, is primary. N ethylethanamine has two ethyl groups on nitrogen, so it is secondary, and N N diethylethanamine has three, so it is tertiary. Okay? Now, shake the whole mixture with benzenesulphonyl chloride and aqueous KOH, just as in the test. Which of the three comes out completely unchanged? Yes, exactly, it is the tertiary amine, because it cannot react. So it is simply distilled off, or extracted with acid. Right? The secondary amine became the N N diethylsulphonamide, which is insoluble in alkali. It stays as a solid, so we just filter it off. Now, where is the primary amine hiding? It is in the filtrate, as a soluble salt, that is correct. The plus sign means we add acid to the filtrate, and the N ethylsulphonamide is set free and comes out. Clear? Last, boil each sulphonamide with concentrated hydrochloric acid, H C l, which hydrolyses it, and then alkali frees the amine. So all three amines come back, separated, and our puzzle is solved: one reagent sorted out the whole flask! Got it?

7:40Quick revision: reagent, solubility in alkali, count H on N

So, quick revision. Hinsberg's reagent is benzenesulphonyl chloride, used with aqueous KOH, an alkali. Primary amines give a product soluble in alkali, secondary amines an insoluble one, and tertiary amines do not react. And to remember it, count the hydrogens on nitrogen: one left after the reagent means soluble. Now go and try the practice problems on this page. All the best!