Reduction Methods

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. You have an alkyl halide with three carbon atoms, and you need an amine with four carbon atoms. So where will that extra carbon come from? And on paper, a nitrile and an isocyanide look almost the same. If you reduce both, do you get the same kind of amine? By the end of this video, you will answer both. Conversions like these come up often in JEE, so let's begin.

0:29Nitro compounds: six hydrogens per nitro group

So, the first route is to reduce a nitro compound. RNO two is the nitro compound, where R is an alkyl or aryl group, and the small numbers written low are subscripts; they count the atoms. The arrow means is reduced to, and RNH two is the primary amine. For the reducing agent, take tin, iron or zinc with hydrochloric acid, where the slash means with, or hydrogen gas over nickel, palladium or platinum. Okay? Now count with me. Both oxygen atoms have to leave as water, and the nitrogen needs two hydrogen atoms of its own. So how many hydrogen atoms does one nitro group use up? It uses six hydrogen atoms, yes, exactly! Four of them leave with the two oxygens as two molecules of water, and two stay on the nitrogen. The H in square brackets means hydrogen supplied by the reducing agent. Right? Now, in industry, the choice is iron scrap with only a little hydrochloric acid. The reason is simple: the ferrous chloride formed, F e C l two, is hydrolysed, and that gives the hydrochloric acid back. Clear? And now a special case. The italic m stands for meta, so meta dinitrobenzene has two nitro groups on positions one and three of the ring. Warm it with ammonium sulphide: N H four in brackets, with a two outside the bracket for two ammonium groups, and S for sulphur. Only one nitro group is reduced, and we get meta nitroaniline. Got it?

2:03Nitriles: one carbon more than the halide

Now, back to our first puzzle. Start with an alkyl halide, R X, where X stands for a halogen. Treat it with potassium cyanide, KCN, and you get a nitrile, RCN, where the C N group came from the cyanide. With me? Next, reduce the nitrile. Use lithium aluminium hydride, written L i A l H four, or hydrogen over nickel, or sodium in ethanol. RCN becomes RCH two N H two, a primary amine. Okay? Now compare the halide R X with this amine. Which carbon is new, and where did it come from? It came from the cyanide, very good! The carbon of the C N group becomes the C H two next to the nitrogen. So the amine has one carbon more than the halide: three carbons in, four carbons out. That is our first puzzle solved. Right?

2:56Isocyanides: the secondary-amine trap

Now the second puzzle. Here is the nitrile again, and it gives a primary amine. Okay? An isocyanide is RNC: the same atoms, but now R is joined to the nitrogen. So reducing it also gives a primary amine, right? Be careful, because this is exactly the trap. In RNC, the carbon is joined only to the nitrogen, so after reduction it becomes a methyl group, C H three, sitting on that nitrogen. So RNC gives RNHCH three. The nitrogen now carries two carbon groups, which makes it a secondary amine. Second puzzle solved. Got it?

3:38Amides keep the carbon count

Next, amides. RCONH two is an amide, a C O group joined to N H two. With lithium aluminium hydride, the C O becomes C H two, so we get RCH two N H two, again a primary amine. Clear? Now, the nitrile route gave us one carbon more than the halide. What about the amide: does the number of carbons change? It does not change at all, correct! No carbon is added and no carbon is removed; the C O just becomes C H two. So the amine has the same number of carbons as the amide. Right? And if the nitrogen of the amide already carries alkyl groups, we call it an N substituted amide; the italic N tells you the group sits on nitrogen. Those amides give secondary and tertiary amines. With me?

4:31Two more routes to primary amines

Two more routes, and both give primary amines. First, oximes, reduced with lithium aluminium hydride, or with sodium in ethanol. Okay? Second, reductive amination. Take an aldehyde or a ketone, add ammonia, N H three, and hydrogen over nickel, and you get a primary amine. Clear?

4:52Example 1: reducing a nitrile and two amides to amines

Now it is your turn. Lithium aluminium hydride reduces three compounds: propanenitrile, ethanamide, and N methyl ethanamide. Name the three products. Pause the video and try it first. Done? Okay, let's check the first one together. Propanenitrile is a nitrile, and nitriles give one carbon more, so the product is butan one amine, right? That is the trap, so slow down here. The extra carbon came earlier, from potassium cyanide, when the halide became the nitrile. Propanenitrile already has three carbons, counting the carbon of the C N group. Reduction adds no carbon, so we get propan one amine, with three carbons. Okay? The second one is ethanamide, with two carbons. The C O becomes C H two, so we get ethanamine, again two carbons, the same count. Right? And the third one is N methyl ethanamide, where the methyl group sits on the nitrogen. Again the C O becomes C H two, the methyl group stays where it is, and we get N methyl ethanamine. There are two carbon groups on the nitrogen, so it is a secondary amine, made from an N substituted amide. Clear?

6:07Example 2: aniline mass from 12.3 g nitrobenzene, 80% yield

Now a numerical one. Find the mass of aniline obtained from twelve point three grams of nitrobenzene, if the yield is eighty percent. Here is a hint: find the moles first. Pause the video and try it. Done? Okay, let's check. Nitrobenzene is C six H five N O two. With carbon twelve, hydrogen one, nitrogen fourteen and oxygen sixteen, capital M, the molar mass, is seventy two plus five plus fourteen plus thirty two, which gives one hundred and twenty three. So the moles are twelve point three over one hundred and twenty three, which gives zero point one mole. On screen, g means grams and mol means moles. Okay? Now, the nitro group becomes an amino group, and the rest of the molecule does not change. So one mole of nitrobenzene gives one mole of aniline, C six H five N H two, and its molar mass is ninety three. With me? With full conversion, zero point one mole of aniline is zero point one times ninety three, which gives nine point three grams. So the answer is nine point three grams, right? That is not the final answer yet, so read the question again. Nine point three grams is what we get only if all of it reacts, but the yield is eighty percent. So nine point three times zero point eight gives seven point four four grams. Never forget the yield. Got it?

7:36Quick revision: nitro, nitrile, isocyanide and amide routes

So, quick revision. Nitro compounds are reduced by tin, iron or zinc with hydrochloric acid, or by hydrogen over nickel, palladium or platinum, and each nitro group uses six hydrogen atoms. Okay? Going from a halide to a nitrile to an amine gives one carbon more, while an amide keeps the same number of carbons. Right? And an isocyanide gives a secondary amine, while oximes and reductive amination give primary amines. Now go and try the practice problems on this page. All the best!