IIT-JEE Questions (1981–1999): Reactions

Transcript

0:00Introduction

Hello everyone! Here is a detective puzzle from IITJEE, nineteen ninety three. There is a basic compound of nitrogen. Warm it with chloroform and alcoholic potash, and a gas with a terrible smell comes out. With nitrous acid, a colourless gas with no smell bubbles off. And the liquid distilled from that solution gives a yellow precipitate with iodine and alkali. So we have three test tubes and just one compound. Can you guess which compound it is? Today we revise the reactions of amines through older IITJEE questions, and by the end, you will name it yourself. These tests come up often, so let's begin.

0:43Carbylamine test: a foul-smelling isocyanide

On the right is the course's map of a primary amine, RNH two, where R is an alkyl or aryl group. First, the carbylamine test. The amine is heated with chloroform, CHC l three, and three KOH in alcohol. The small numbers written low count the atoms, and the arrow means gives. We get RNC, an isocyanide with a very offensive smell, plus three K C l and three water molecules. Okay? Now, IITJEE nineteen eighty eight asked for a test to tell ethylamine from diethylamine. So which of the two gives this terrible smell? Yes, exactly, only ethylamine, because its N H two group makes it a primary amine. The test works for primary amines only, aliphatic or aromatic. Clear? Diethylamine has two ethyl groups on its nitrogen, so it is secondary, and secondary and tertiary amines do not respond at all. Got it? And the reactive species is dichlorocarbene, C C l two, where the two dots in front are a lone pair of electrons on the carbon. Right?

1:52Nitrous acid, HNO₂, from NaNO₂ + HCl

Next, nitrous acid, HNO two, made from sodium nitrite, N a N O two, and hydrochloric acid, as the heading shows. With a primary aliphatic amine, we get the alcohol, ROH, plus nitrogen gas, N two, plus water. With me? And notice this: one mole of nitrogen gas comes out per mole of amine, and the course says this is used to estimate amines. Our puzzle needs it. Okay? Primary aromatic amines are different. In the cold, at two seventy three to two seventy eight kelvin, they give a stable diazonium salt, Ar N two plus, C l minus, where Ar is an aryl group and the raised signs are the charges. Right? Now, here is a question from IITJEE nineteen eighty one. Methylamine, ethylamine, diethylamine and triethylamine: which of these gives an oily nitrosoamine with nitrous acid in the cold? Diethylamine, that is correct, because it is the only secondary amine of the four. Secondary amines give a yellow oily N nitrosoamine, and the italic N tells us the nitroso group sits on the nitrogen.

3:06Hinsberg's test: telling the three classes apart

Clear? Now Hinsberg's test, shown on the right for all three classes, where one, two or three with a small raised circle means primary, secondary or tertiary. The reagent is benzenesulphonyl chloride: C six H five, a benzene ring, joined to S O two C l, the sulphonyl chloride group, used with aqueous KOH, right? With a primary amine, the chlorine is replaced by NHR, so the product still has one hydrogen on its nitrogen. And it dissolves in alkali. Okay? But why should this product dissolve in alkali at all? Because that N H is acidic, very good! The dash is the bond, and the sulphonyl group pulls electrons strongly, so the alkali can remove that hydrogen. Got it? With a secondary amine, the product has two R groups on the nitrogen and no N H left, so it is insoluble in alkali. With me? And a tertiary amine has no hydrogen on its nitrogen, so there is no reaction at all. Three classes, three results, so this test tells them apart, and can even separate a mixture of them. Clear?

4:16Example 1: C3H9N gives a Hinsberg solid insoluble in alkali

Time for Example one, from IITJEE nineteen ninety three. An amine, call it A, has the formula C three H nine N, three carbons, nine hydrogens and a nitrogen. It reacts with benzenesulphonyl chloride to give a solid that does not dissolve in alkali. Identify the amine. Pause the video and try it first. Done? Okay, let's check. It is insoluble in alkali, so it must be the tertiary amine, right? No, no, careful, that is exactly the trap in this question! But a tertiary amine does not react at all, and this one reacts and gives a solid, so it is not tertiary. Okay? And the solid does not dissolve in alkali, so no N H is left on its nitrogen. That makes our amine secondary. Right? The course counts four amines with this formula, and only one of them is secondary: an ethyl group and a methyl group on the same N H, as on screen. Clear? So the amine is N methyl ethanamine, where the italic N says the methyl group sits on the nitrogen. Got it?

5:24Example 2: amine from 112 mL N2 and the iodoform test

And now, our detective puzzle, also from nineteen ninety three. The compound gives a foul-smelling gas with chloroform and alcoholic potash. Then zero point two nine five grams of it, in aqueous H C l with sodium nitrite at zero degrees Celsius, gives one hundred and twelve millilitres of a colourless, odourless gas at STP, standard temperature and pressure. The distilled liquid, free of nitrogen, gives a yellow precipitate with alkali and iodine. Identify the compound. Pause the video and try it first. Done? Okay, let's check. The foul smell with chloroform and alcoholic KOH is the carbylamine test, so our compound is a primary amine. Okay? Next, the colourless, odourless gas from nitrous acid. Which gas is it? It is nitrogen, N two, yes, exactly! At STP, one mole of gas takes twenty two thousand four hundred millilitres, and the raised minus one means per mole. So one hundred and twelve over twenty two thousand four hundred gives zero point zero zero five moles of nitrogen, and at one mole per mole, the amine is zero point zero zero five moles too. Clear? So the molar mass, M, is zero point two nine five grams over zero point zero zero five moles, which gives fifty nine grams per mole. With me? Now, a saturated open-chain amine with one nitrogen is C n H two n plus three N, with molar mass fourteen times n, plus seventeen, where n counts the carbons. Set it equal to fifty nine, and the double arrow means so: n equals three, giving C three H nine N, our formula from Example one! But two of its four amines are primary. So which one is ours? The yellow precipitate is iodoform, CHI three, and this test needs an alcohol with a C H three group next to the carbon carrying the O H in brackets, the dash leading on to the rest of the molecule. Right? With three carbons, only propan two ol has that group. Taking this as the main product, with no rearrangement, the N H two sat on the middle carbon. Our compound is propan two amine, two methyl groups and the N H two on one carbon. So three test tubes, one compound, and our puzzle is solved. Got it?

7:55Quick revision: carbylamine, nitrous acid, Hinsberg test

So, quick revision. The carbylamine test, with its foul-smelling isocyanide, is given by primary amines only. With nitrous acid, a primary aliphatic amine gives the alcohol and one mole of nitrogen per mole. And in Hinsberg's test, the primary product dissolves in alkali, the secondary one does not, and a tertiary amine does not react. Now try the other past questions on this page yourself. All the best!