Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take benzylamine and four methylaniline. Both have the formula C seven H nine N, both are primary amines, and both carry a benzene ring. Yet the Gabriel phthalimide synthesis can make only one of them. So which one, and why? Hold that thought, we will solve it in Example one. This chapter gives you three sets of mixed problems from the whole unit, all in the style of recent JEE Main papers. So let us sharpen the tools they need. Let's begin.

0:36How this chapter works

So, what is inside this chapter? Three sets of fifteen original problems each, in the style of recent JEE Main papers, and each set mixes all the chapters of the unit. Okay? Give yourself twenty five minutes for each set, and only then check the solutions. So tell me, how much time is that for each problem? One hundred seconds, yes, exactly! Twenty five minutes is fifteen hundred seconds, with s for seconds, and fifteen hundred over fifteen gives one hundred; the division sign means over. So the tools must be ready in your head. Right?

1:14Tool 1: the class of an amine, and two tests

Now, tool one. The class of an amine is the number of carbon groups on N, the nitrogen. RNH two is primary, R two N H secondary, and R three N tertiary, where R is a carbon group. The small numbers written low count atoms, and in brackets, one, two or three with a small raised circle is short for primary, secondary and tertiary. Got it? Now take tert butylamine, whose nitrogen sits on a carbon carrying three methyl groups. Is it a tertiary amine? It is not tertiary, it is primary, and many students lose marks here. C H three in brackets, with a small three outside, means three methyl groups on one carbon, and the dash joins that carbon to N H two. Nitrogen carries only one carbon group, so the amine is primary. Clear? Next, the carbylamine test: heat the amine with chloroform, CHC l three, plus alcoholic KOH, potassium hydroxide. Only primary amines respond, aliphatic or aromatic, giving an isocyanide with a very offensive smell. With me? And Hinsberg's reagent, C six H five S O two C l, is benzenesulphonyl chloride. The product from a primary amine dissolves in alkali, the one from a secondary amine does not, and a tertiary amine does not react. Okay?

2:36Tool 2: making amines, count the carbons

Now, tool two: making amines. Always count the carbons. Reducing a nitrile gives one carbon more than the alkyl halide, R X, you started from. Reducing an amide with lithium aluminium hydride, L i A l H four, keeps the same carbons. Right? But in the Hoffmann bromamide reaction, the amide RCONH two, with bromine and four N a O H, sodium hydroxide, gives RNH two; the arrow means gives. That amine has one carbon fewer than the amide. Clear? Next, the Gabriel phthalimide synthesis gives pure primary amines. Now, aniline is a primary amine, so can Gabriel make it? It cannot, and that is the trap. We would need an aryl halide, and aryl halides do not undergo nucleophilic substitution by the phthalimide anion. So the nitrogen must sit on an s p three carbon, with the three written small and raised. And not just any such carbon: a tertiary halide gives elimination instead, so even tert butylamine is out. Got it?

3:39Tool 3: diazonium salts

Now, tool three, diazonium salts. The map shows the preparation on the left, and on the right, eight reactions that replace nitrogen and two coupling reactions that keep it, giving azo dyes. Here, A r N H two is an aromatic amine, with A r for an aryl group. With sodium nitrite and two H C l, at two seventy three to two seventy eight kelvin, where capital K is kelvin, it is diazotised. Okay? Will a primary aliphatic amine, like ethanamine, also give a stable diazonium salt? It will not, and that is a classic trap. An aliphatic diazonium salt loses nitrogen at once, giving mainly the alcohol. So we need a primary aromatic amine, kept cold, and the salt is used at once. Got it? Then come the replacements, where nitrogen is lost. Copper salts, C u C l, C u B r or C u C N, give the Sandmeyer reaction; copper powder with H C l or H B r, written H X, gives the Gattermann reaction. The slash means with. Right? And if you warm the salt with water, you get the phenol, A r O H, while nitrogen gas, N two, bubbles off. Keep this in mind for Example two. Clear?

4:51Example 1: which amines can Gabriel synthesis make

Now, Example one, a medium problem from Set one. How many of these can be prepared by the Gabriel phthalimide synthesis: methanamine, propan one amine, aniline, N methylaniline, benzylamine, and four methylaniline? Pause the video and try it first. Done? Okay, let's check. Methanamine and propan one amine are primary, with nitrogen on an s p three carbon, so both can be made. With me? Aniline and four methylaniline have nitrogen on the ring, so they would need an aryl halide. And in N methylaniline, the italic capital N puts a methyl group on nitrogen, so it is secondary. All three are out. Okay? Now benzylamine. It also has a benzene ring, so it is out too, right? It is not out, it can be made, and this is our opening puzzle! In benzylamine, C six H five C H two N H two, the nitrogen sits on the C H two carbon, an s p three carbon, so it behaves as an aliphatic amine. Benzylamine yes, four methylaniline no. So the answer is three, and the puzzle is solved. Got it?

5:58Example 2: N2 volume and phenol mass from 4-methylaniline

And now Example two, a hard one from Set three. Two point one four grams of four methylaniline is diazotised, and the solution warmed until the reaction is complete. Find the volume of nitrogen at STP, and the mass of the phenol formed. Pause the video and try it first. Done? Okay, let's check. C H three, C six H four, N H two has seven carbons, nine hydrogens and one nitrogen. So capital M, the molar mass, is one hundred and seven grams per mole; the raised minus one means per. Then n, the moles, is two point one four over one hundred and seven, which gives zero point zero two mole. Right? Diazotisation gives the salt, A r N two plus, C l minus, with the raised signs as charges. Warm water turns it into the phenol, with nitrogen and H C l, giving one mole of nitrogen per mole of amine. So we get zero point zero two mole of nitrogen. Clear? One mole of gas at STP occupies twenty two thousand four hundred millilitres. So capital V, the volume, is zero point zero two times twenty two thousand four hundred, which gives four hundred and forty eight millilitres; the cross means times. Okay? Now the mass. The phenol is just ordinary phenol, C six H five O H, right? Be careful here, it is not, because the methyl group stays on the ring. The product is four methylphenol, C H three, C six H four, O H, with molar mass one hundred and eight. So its mass is zero point zero two times one hundred and eight, which gives two point one six grams. Got it?

7:40Quick revision: amine class, carbon count, diazotisation

So, quick revision. Classify by the carbon groups on nitrogen; the carbylamine test is for primary amines only, and in Hinsberg's test only the primary product dissolves in alkali. Count carbons: a nitrile adds one, Hoffmann removes one, and Gabriel gives primary amines with nitrogen on an s p three carbon. Diazotise aromatic amines at two seventy three to two seventy eight kelvin; warm water gives the phenol and one nitrogen per amine. Now take Set one, start a twenty five minute timer, and check the solutions only at the end. All the best!