Oxidation, Ozonolysis and Polymerisation

Transcript

0:00Introduction

Hello everyone! Let us begin with a small mystery. We have an unknown alkene, C seven H fourteen. We cut it with ozone, and we get just two pieces: propanal and butanone. From these two pieces, can you rebuild the whole molecule? Sounds like detective work, right? By the end of this video, you will do it in under a minute. Oxidation and ozonolysis of alkenes come up often in JEE, so let's begin.

0:31Cold, dilute KMnO₄: two OH groups add

First, oxidation with permanganate. On screen is K M n O four, potassium permanganate; the small four means four oxygen atoms. Cold, dilute and alkaline, it is called Baeyer's reagent. Now tell me, does it break the double bond, or add to it? It adds, yes, exactly! Two O H groups add to the same face of the double bond, which we call syn addition. The product, with O H groups on neighbouring carbons, is a vicinal diol, or glycol. Okay? And you can see it happen. The purple colour of permanganate is discharged, and brown M n O two, manganese dioxide, appears. Right? Now the stereochemistry. Since both groups add to the same face, cis but two ene gives the meso diol, and trans but two ene gives the racemic diol; each arrow means gives. Careful: the meso result needs a symmetrical alkene like this one. This is the reverse of the result with bromine, so do not mix them up. Clear?

1:35Hot KMnO₄: the double bond breaks

Now let us turn up the heat. With hot or acidified K M n O four, or acidic K two C r two O seven, potassium dichromate, the double bond itself, written C double bond C, is broken. What each end becomes depends on its hydrogens. Okay? An end like equals C H two, with two hydrogens, goes all the way to C O two and water. Now take an end with one hydrogen, equals CHR, where R is an alkyl group. It stops at an aldehyde, right? No, no, careful, hot permanganate does not stop at the aldehyde! The equals CHR end gives RCOOH, an acid. With me? And an end with no hydrogen, equals C R two, becomes R two C O, a ketone. So count the hydrogens: two hydrogens give carbon dioxide, one gives an acid, and none gives a ketone. Got it?

2:30Ozonolysis: ozone cuts the double bond

Next, ozonolysis. O three, ozone, adds to the double bond and forms an ozonide. In the figure, the small raised numbers on R just label the four groups, R one to R four. Okay? Then the ozonide is cleaved with Z n and H two O, zinc and water: the reductive work-up. Look at the dashed line in the figure: cut the double bond there, and put an O on each end. So equals C H two gives HCHO, methanal, equals CHR gives the aldehyde RCHO, and equals C R two gives a ketone. Right? Now suppose the work-up uses H two O two, hydrogen peroxide, instead. What happens to those aldehydes? They become acids, exactly! This is the oxidative work-up, and the last line of the figure shows it: the ends now behave just as with hot K M n O four. Clear?

3:24Working back: from the pieces to the alkene

Now the trick that solves our mystery. To go backwards, from the products to the alkene, remove the two O atoms, and join the two carbonyl carbons by a double bond. Okay? Now, what if the double bond sits inside a ring? We cut it once, so how many products do we get? Just one, correct! The ring only opens, so a cyclic alkene gives one chain with a carbonyl group at each end, one dicarbonyl compound. And a diene has two double bonds, so it uses two moles of ozone. Got it?

3:59Ethene: oxirane and polythene

Now two reactions of ethene, C H two double bond C H two. With oxygen, over silver, written A g above the arrow, at five hundred and twenty three kelvin, ethene gives oxirane, or ethylene oxide. Okay? And polymerisation. Ethene molecules, n of them, join into one long chain; the small n after the bracket means the unit repeats n times. This is polythene. There are low-density and high-density forms. Which is made at the higher pressure? High-density, surely? No, no, careful, it is the low-density one! It needs three hundred and fifty to five hundred and seventy kelvin, and one thousand to two thousand atmospheres, with a trace of oxygen or a peroxide. Right? The reaction goes by free radicals, and the chains are branched. With me? High-density polythene needs only three hundred and thirty three to three hundred and forty three kelvin, and six to seven atmospheres. The catalyst is T i C l four, titanium tetrachloride, with A l, then C two H five in brackets with a three outside, which is triethylaluminium. Clear? Together they make the Ziegler Natta catalyst. Here the chains are linear and closely packed, which fits the name, high density. Got it?

5:13Example 1: ozonolysis of but-1-ene and 2,3-dimethylbut-2-ene

Time for the first example. Give the products of ozonolysis, with zinc and water, of part a, but one ene, and part b, two three dimethyl but two ene. Pause the video and try it first. Done? Okay, let's check. But one ene is C H two, double bond, CHCH two C H three. Cut the double bond and put an O on each end. The C H two end gives HCHO, methanal, and the other end gives C H three C H two CHO, propanal. Okay? Now part b. Each carbon of its double bond carries two methyl groups, so both ends are alike. How many different products do we get? Only one, yes! Each end gives the same molecule, C H three in brackets, two, C O, which is propanone. And with two such ends, we get two moles of it; mol means mole. Got it?

6:09Example 2: C7H14 from propanal and butanone, then hot KMnO4

Now, remember our mystery from the start? Let's solve it. An alkene, C seven H fourteen, gives propanal and butanone on ozonolysis. Identify it, and say what it gives with hot K M n O four. Pause the video and try it first. Done? Okay, let's check. Write the two pieces with their oxygens facing each other. On the left is propanal, C H three C H two C H double bond O, with its oxygen on the right. On the right, butanone, O double bond C, with a C H three on it, then C H two C H three. Okay? Now remove the two O atoms, and join the two carbonyl carbons by a double bond. We get C H three C H two C H, double bond, C, with a C H three on it, and then C H two C H three. Right? Always check your answer. Three carbons plus four carbons gives seven. And the hydrogens, three, two, one, three, two and three, add up to fourteen. So, C seven H fourteen, perfect! Now the name. The longest chain has six carbons, and the double bond is at carbon three from either end. Numbering from the end nearer the methyl group gives three methyl hex three ene. Clear? Now hot K M n O four. The equals C H end has one hydrogen, so it becomes an acid, C H three C H two COOH, which is propanoic acid. With me? And the other end has no hydrogen, so it stays a ketone, butanone. Mystery solved, in under a minute! Got it?

7:46Quick revision: syn diol, hot KMnO4, ozonolysis, polythene

So, quick revision. Cold, dilute permanganate gives a syn diol; hot permanganate breaks the double bond. Then equals C H two gives C O two, equals CHR an acid, and equals C R two a ketone. In ozonolysis with zinc and water, cut the double bond and put an O on each end; join the carbonyl carbons to go back. And polythene: low density at high pressure, high density with Ziegler Natta. Now go and try the practice problems on this page. All the best!