Overview

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. You want a pure primary amine, so you heat an alkyl halide with ammonia. Simple, right? But the flask gives you a mixture: primary, secondary and tertiary amines, and even a quaternary salt. So how do chemists make a pure primary amine? Today we see the whole story of amines on one map. It comes up often in JEE, so let's begin!

0:27One map, three classes

So, here is the whole story on one page. In the centre is the primary amine, RNH two: R is an alkyl or aryl group, the dash is the bond, and N H two is the amino group. Seven arrows on the left show ways to make it; nine on the right show its reactions. Okay? The small raised circle, the degree sign, marks the class: RNH two is primary, R two N H is secondary, and R three N is tertiary. So, to classify an amine, what exactly do we count? We count the carbon groups on the nitrogen itself, yes, exactly! Not the carbons around the carbon carrying the nitrogen. Keep this for Example one, right?

1:09Making amines by reduction

Now the left side, the preparations. Reduce a nitro compound with tin and H C l, written S n slash H C l, the slash meaning with; or iron and H C l; or hydrogen over palladium, H two slash P d. Okay? Next, reduce a nitrile with lithium aluminium hydride, L i A l H four, or hydrogen over nickel. And the nitrile itself can come from an alkyl halide, R X, as the question on screen says. So, compared with R X, how many extra carbons does the amine have? Yes, exactly, the amine has one carbon more. R X, with potassium cyanide, KCN, over the arrow, gives the nitrile RCN, and reduction gives RCH two N H two. The cyanide brings its own carbon, and that is the one C more on screen. Got it?

1:58Amides: always count the carbons

Now the amides. Reduce an amide with lithium aluminium hydride, and the amine keeps the same number of carbons. Right? But treat an unsubstituted amide, RCONH two, with bromine and sodium hydroxide, B r two slash N a O H: the Hoffmann bromamide degradation. Now the amine has one carbon fewer! With me? So let me ask you this. Benzamide has seven carbons, and aniline has six. Which reagent turns benzamide into aniline? Yes, it is bromine with sodium hydroxide, the Hoffmann route, which takes seven carbons down to six. Lithium aluminium hydride would keep all seven. This mix-up is common, so always count the carbons, okay?

2:42Getting a pure primary amine

Now, back to our puzzle. R X plus N H three, called ammonolysis, gives a mixture of primary, secondary and tertiary amines, because the primary amine formed is itself a nucleophile and attacks more alkyl halide. Clear? So, look at the map. Which arrow gives a pure primary amine, with no secondary or tertiary amine at all? Yes, exactly, it is the Gabriel phthalimide synthesis. It gives primary amines free of secondary and tertiary ones. So that solves our opening puzzle, okay? But careful, there is a condition. Aryl halides do not undergo nucleophilic substitution by the phthalimide anion, so no aniline this way. Gabriel gives only primary aliphatic amines, right?

3:27Reactions: the lone pair at work

Now the right side: most reactions use the lone pair on nitrogen. With an acid H X, X being a halogen, the amine forms a salt, RNH three plus, X minus; the raised plus and minus are charges. That shows its basic character. Okay? With an alkyl halide R prime X, where the small dash, called prime, means another alkyl group, we get secondary and tertiary amines, then the quaternary salt: our puzzle's mixture. Right? With an acid chloride, R prime COC l, or an anhydride, R prime C O in brackets, two, then O, we get an amide. This is acylation: an acyl group replaces a hydrogen on nitrogen. Got it?

4:08Tests for primary amines

Next, the tests. Heat a primary amine with chloroform, CHC l three, and alcoholic KOH, and you get an isocyanide, RNC, with a very offensive smell. Okay? This is the carbylamine test. Now tell me, does a secondary amine give it too? No, a secondary amine does not give it, because only primary amines respond, aliphatic or aromatic. Secondary and tertiary amines give no isocyanide. Clear? Then Hinsberg's reagent, benzenesulphonyl chloride, C six H five S O two C l. With a primary amine, the product keeps one acidic N H, so it is soluble in alkali. With me?

4:48Nitrous acid: aliphatic or aromatic?

Now nitrous acid, HNO two. With a primary aliphatic amine, the diazonium salt loses nitrogen at once, giving an alcohol and nitrogen gas, N two. Okay? But a primary aromatic amine, like aniline, kept cold at two seventy three to two seventy eight kelvin, gives a diazonium salt, A r N two plus, C l minus, where A r stands for an aryl group. It is stable only in the cold. So remember both conditions: aromatic, and cold, right?

5:16Example 1: classify amines as primary, secondary, tertiary

Now Example one, an easy one, from the concept Classification and Nomenclature. Classify the five amines on screen as primary, secondary or tertiary. Pause the video and try it first. Done? Okay, let's check: count the carbon groups on nitrogen. Part a is C H three in brackets, two, then CHNH two. Nitrogen carries just one group, so it is primary. Okay? Part b is C H three in brackets, three, then CNH two. The carbon joined to nitrogen carries three methyl groups. So it must be a tertiary amine, right? Careful, because that is exactly the trap here. Look at the nitrogen, not the carbon: it carries only one group, so part b is primary. Got it? The rest are quick. Part c, C H three NHC two H five, has methyl and ethyl on nitrogen: secondary. Part d, C two H five in brackets, three, then N, has three ethyl groups: tertiary. And part e, C six H five NHCH three, has phenyl and methyl: secondary.

6:27Example 2: JEE Main 2021, statements on primary amines

Clear? And now Example two, a harder one: a JEE Main question from twenty twenty one, from the concept Past JEE Questions. Which of the four statements on screen is not correct for primary aliphatic amines? Pause the video and try it first. Done? Okay, let's check, starting with our map. Statement four: Gabriel gives pure primary aliphatic amines, so it is correct. Statement two: nitrous acid turns primary aliphatic amines into alcohols, as the map shows, and the statement sets methylamine aside. So that is correct too, right? Statement three is about basic strength: the smaller the p K b, where the small b stands for base, the stronger the base. Dimethylamine, C H three in brackets two, N H, has three point two seven, less than methylamine, C H three N H two, at three point three eight, as the less than sign shows. So the secondary amine is stronger: statement three is correct. Okay? So only statement one is left to check. Which has more N H bonds, the dash being the bond: a primary or a secondary amine? Yes, exactly, the primary amine has two, and the secondary has only one. More hydrogen bonding means more intermolecular association, not less. So statement one is not correct, and the answer is option one. Clear?

7:52Quick revision: carbon count, Gabriel, tests, diazonium

So, quick revision. Classify amines by counting the carbon groups on nitrogen. Count carbons: the nitrile route adds one, Hoffmann removes one, and Gabriel gives a pure primary aliphatic amine. The carbylamine test picks out primary amines, and a primary aromatic amine with nitrous acid at two seventy three to two seventy eight kelvin gives a diazonium salt. Now try the problems in the next sections of this chapter. All the best!