AIEEE and Earlier JEE Main Questions: Preparation

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Bromine and sodium hydroxide take an amide and hand back an amine with one carbon fewer, so one carbon simply walks out of the molecule! Now tell me, for one mole of that amine, how many moles of sodium hydroxide are used up? Is it one mole, or two moles? Many students guess too low. Today we solve two past JEE Main questions on making amines, and you will count that answer yourself, atom by atom. Questions like these come up often, so let's begin.

0:36Count the carbons, then pick the route

So, first our tools. The figure is the course's map of a primary amine, RNH two, where R is an alkyl or an aryl group, and the small two written low is a subscript that counts atoms. We need only its left side, the ways to make it, and to choose among them we count carbons; on screen, C means carbon. If the amine keeps the same number of carbons as the alkyl halide, use the Gabriel synthesis, or ammonolysis, which is heating the halide with ammonia. Okay? One carbon more than the halide? Then use KCN, potassium cyanide, first, and reduce the nitrile you get. The cyanide brings its own carbon. Right? The same number of carbons as the amide? Reduce the amide with lithium aluminium hydride, L i A l H four. Every carbon stays. Clear? Now, what if the amine must have one carbon fewer than the amide? Which route will you pick? Yes, exactly, it is the Hoffmann bromamide degradation! That is bromine, B r two, with sodium hydroxide, N a O H, and the slash means together with. Both of today's questions sit on this one line, so let's open it up. With me?

1:48Inside the Hoffmann bromamide reaction

Here is the reaction. We start with an unsubstituted amide, RCONH two, where C O is the carbonyl group. The reagents, bromine and sodium hydroxide, sit above the arrow, and the arrow means gives. Out comes RNH two, a primary amine with one carbon fewer. Okay? Now let us look inside. In step one, bromine replaces one hydrogen on the nitrogen, giving the N bromoamide, named in brackets; the italic N says the bromine sits on nitrogen. So only one bromine atom goes onto the nitrogen, and the other leaves as bromide, B r minus, where the raised minus is its charge. Keep that in mind. Right? In step two, hydroxide, O H minus, takes the second hydrogen from the nitrogen, giving an anion. The bromine is still sitting on the nitrogen, for now. Clear? Now the bromine on the nitrogen leaves as bromide, and at the same moment the group R moves from the carbon to the nitrogen, which would otherwise be short of electrons. So that makes two bromide ions in all. We get the isocyanate, RNCO; each equals sign is a double bond, and the small dash is a single bond. Got it? Look at the isocyanate. The old carbonyl carbon now sits all alone, between the nitrogen and the oxygen. So where does it finally go? Perfect, it leaves as carbonate. In step three, water adds, then hydroxide breaks the molecule into RNH two and carbonate, C O three, where the raised two minus is its charge. That is the carbon that walked out in our puzzle. Okay? And one condition. Steps one and two each took a hydrogen from the nitrogen, so the amide needs two hydrogens there. Only RCONH two reacts; an N substituted amide does not. Right?

3:45Acid to amide: ammonia, then heat

One more tool, and then the questions. Where does the amide come from? Take a carboxylic acid, RCOOH, and add ammonia, N H three: it simply forms a salt, RCOONH four, an ammonium salt. Clear? Now we heat this salt strongly. Which small molecule do you think it loses? Very good, it loses water. One molecule of H two O goes, and the salt becomes the amide, RCONH two. With me? And see the carbons: the acid, the salt and the amide all have the same number. In this whole chain, only the Hoffmann step removes a carbon. Remember this for Example two. Okay?

4:31Example 1: moles of NaOH and Br2 per mole of amine

Now Example one, from JEE Main twenty sixteen. In the Hoffmann bromamide degradation, how many moles of sodium hydroxide, and how many moles of bromine, are used for every mole of amine produced? Pause the video and try it first. Done? Okay, let's check, starting with bromine. One bromine molecule per amide: one atom left as bromide at once, and the one on nitrogen left later. So one mole of B r two gives two moles of bromide ions, and with sodium, two moles of sodium bromide, N a B r. Okay? Now the carbon that walked out. It leaves as carbonate, and with sodium that is sodium carbonate, N a two C O three. With me? So, how much sodium hydroxide? Two sodium bromides need two sodiums, so the answer is two moles, right? That is the trap, so careful here: the sodium carbonate also needs two sodiums. Two plus two gives four, so we need four moles of sodium hydroxide. Clear? Now write it all together, exactly as the course gives it. One amide, one bromine and four sodium hydroxide give the amine, sodium carbonate, two sodium bromide and two waters. The two waters balance the hydrogen and the oxygen, with five oxygen atoms on each side. Okay? So the answer is four moles of sodium hydroxide and one mole of bromine, for every mole of amine. That solves our puzzle: people guess too low because they forget the carbonate. Got it?

6:05Example 2: identify the acid that gives ethanamine

And now Example two, a harder one, from JEE Main twenty thirteen. Compound A reacts with ammonia to give compound B, which on heating gives compound C, and compound C, with bromine in the presence of KOH, potassium hydroxide, gives the amine at the end of the question. Your job is to identify compound A, okay? Pause the video and try it first. Done? Okay, let's check. We work backwards, from the end. The amine on screen is ethanamine, an ethyl group carrying N H two, and it has two carbons. With me? Now, the amine has two carbons, so the acid should have two carbons as well. That makes compound A ethanoic acid, right? That is exactly the trap, so be careful here. The last step is bromine with KOH, the Hoffmann reaction, and Hoffmann removes one carbon. So compound C, the amide, has one carbon more, three carbons in all. Clear? Three carbons, with the amide group at the end: an ethyl group joined to CONH two. So compound C is propanamide. Right? Now, which compound gives propanamide on heating? Think back to our last tool. Yes, exactly, it is the ammonium salt. So compound B is ammonium propanoate, the salt that loses water on strong heating. Okay? And the salt came from the acid and ammonia, with no change in carbons. So compound A is propanoic acid, the acid with three carbons. Check it forwards once: acid, then salt, then amide, then ethanamine. Everything fits. Got it?

7:51Quick revision: Hoffmann moles and working backwards

So, quick revision. Hoffmann: an amide, with bromine and sodium hydroxide, gives a primary amine with one carbon fewer. For each mole of amine, four moles of sodium hydroxide and one mole of bromine; remember the carbonate. And in a chain, an acid gives an ammonium salt and then an amide, so work backwards and add back the carbon Hoffmann removed. Now go and try the other past JEE questions in this chapter. Count the carbons, and they will feel easy. All the best!