AIEEE and Earlier JEE Main Questions: Alkenes

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Ozone cuts a double bond into two pieces. So cut the double bond of a ring alkene, and you should get two molecules, right? But you get just one chain, with a carbonyl group at each end. One JEE Main paper turned this around: here is the chain, now rebuild the ring. We will do that in Example two. Today we go through the older entrance-exam questions on alkenes. Most of them test just four ideas, and once you see those four, these questions become really simple. So, let's begin.

0:36Idea 1: adding hydrogen

So, idea number one: adding hydrogen. H two, with a small two written low, a subscript, is hydrogen gas. With nickel, platinum or palladium, N i, P t or P d, it adds across the double bond, the two short lines between the carbons, one mole per double bond. Okay? Now a question from the two thousand three paper; the arrow means converts into. Which reagent converts but one ene into butane: tin with H C l, zinc with mercury, P d with H two, or zinc with H C l? The dashes and the slash just join the parts of each reagent. P d slash H two, yes, exactly! It is the only option with H two and one of our three metals, so but one ene becomes butane. Clear?

1:24Idea 2: Markovnikov's rule in hydration

Idea number two is Markovnikov's rule: the electrophile adds so as to give the more stable carbocation. In the figure, propene takes H plus, a proton. On the top path, the little plus on the middle carbon is the positive charge: a secondary carbocation, with the small circle read as degree. It is more stable, so this path is fast, and bromide, B r minus, gives two bromo propane, the major product. The lower path, through the less stable primary cation, gives only the minor product. With me? Now add water instead, with dilute sulphuric acid, H two S O four. Same rule, so propene gives propan two ol, with the O H on the middle carbon, not the end. Right? Now the two thousand five paper asked this. Acid-catalysed hydration of alkenes other than ethene leads to which class of alcohols? Secondary or tertiary alcohols, yes, exactly! Markovnikov addition never puts the O H on a terminal C H two, an end carbon. Two degree and three degree mean secondary and tertiary. Got it?

2:32Idea 3: HOCl follows the same rule

Idea number three is the same rule with a new reagent: HOC l, from C l two plus H two O, chlorine and water. The electrophile is C l plus, with the small plus up high for its charge, and it adds first to the carbon with more hydrogens. Okay? So the JEE Main two thousand sixteen paper asked this. The reaction of propene with HOC l proceeds through which intermediate? Chlorine goes to the end carbon, giving C H three, C H with a plus, C H two C l; the dashes are single bonds. So it is the secondary cation, the more stable one, as before. Clear?

3:10Idea 4: ozonolysis, cut and add oxygen

Now idea number four, the biggest in this set: ozonolysis. In the figure, O three, ozone, adds to the double bond to form an ozonide, and zinc and water break it into two carbonyl compounds. R one to R four are four groups; the raised numbers are labels. The dashed line marks the cut, so the rule is: cut the double bond, and put an O on each end. Okay? The next line of the figure gives the pieces: equals C H two gives HCHO, methanal; equals CHR gives RCHO, an aldehyde; equals C R two gives R two C O, a ketone. The last line is for an oxidising work-up with H two O two, or for hot K M n O four: there, equals CHR gives the acid RCOOH, equals C H two goes all the way to C O two, and the ketone stays a ketone. But our questions use zinc and water. Right? So, from the two thousand eight paper: but two ene is treated with ozone, and the product with zinc and water. What is formed? Two moles of ethanal, yes, exactly! Cut the double bond in the middle, and each half becomes C H three CHO, ethanal. Two identical halves, so two moles. Clear? Now the two thousand eleven paper: ozonolysis of a compound gives methanal as one of the products. What does this confirm? That is right, a terminal equals C H two carbon. Methanal comes from an equals C H two at the end of a double bond, so the compound has a vinyl group. Got it? And here is the trick for working backwards. Remove the two oxygen atoms, and join the two carbonyl carbons by a double bond. With me?

4:56Example 1: symmetrical alkene giving 2 moles of ethanal

Now, Example one, from the two thousand ten paper. One mole of a symmetrical alkene gives, on ozonolysis, two moles of an aldehyde of molar mass forty four. Identify the alkene. Pause the video and try it first. Done? Okay, let's check. Which aldehyde has molar mass forty four? Methanal, HCHO, is thirty. Ethanal, C H three CHO, has two carbons at twelve each, four hydrogens at one each, and an oxygen at sixteen. So twenty four plus four plus sixteen gives forty four. So it is ethanal, and two moles of it means two identical halves. Right? Now the trick: remove both oxygen atoms and join the carbonyl carbons. We get C H three C H, double bond, CHCH three. Clear? But one ene also has four carbons, so could the answer be but one ene? Oh no, careful, that is the trap! But one ene would give propanal and methanal, two different aldehydes, not two moles of ethanal. Got it? So the alkene is but two ene, the symmetrical one. With me?

6:06Example 2: which ring alkene gives 5-oxo-2-methylhexanal

Now Example two, the harder one, from the JEE Main two thousand fifteen paper, and it is our puzzle. Which dimethyl cyclopentene gives five oxo, two methyl hexanal on ozonolysis? Pause the video and try it first. Done? Okay, let's check. Hexanal is a six carbon chain, and the small numbers above the carbons count it. Carbon one is OHC, the aldehyde group. Two methyl is a C H three branch on carbon two, in brackets, and five oxo is a C O, a ketone, at carbon five. Okay? Same trick as before: carbon one, the aldehyde carbon, joins carbon five, the ketone carbon, by a double bond. Right? Carbons one to five close a five-membered ring with a double bond, a cyclopentene. Ozone cut that ring open into one chain, just as in our puzzle. Clear? Now, where do the two methyls go? It is tempting to put both on the double bond, and answer one, two dimethyl cyclopentene. Is that right? No, no, wait, that one gives heptane two, six dione, with a ketone at both ends and no aldehyde. Our product has an aldehyde end, so carbon one carried a hydrogen, not a methyl. Got it? So carbon six is a methyl on carbon five, a doubly bonded carbon, and the other methyl sits on carbon two, next to carbon one, the other doubly bonded carbon. With me? Now name the ring. The doubly bonded carbon with the methyl is ring carbon one, the other doubly bonded carbon is ring carbon two, and ring carbon three carries the second methyl. That gives one, three dimethyl cyclopentene, and our ring is rebuilt! Right?

7:50Quick revision: hydrogenation, Markovnikov, HOCl, ozonolysis

So, quick revision. First, hydrogen with nickel, platinum or palladium adds across the double bond, and Markovnikov's rule means the more stable carbocation. Second, acid hydration gives secondary or tertiary alcohols, and in HOC l, chlorine plus goes to the carbon with more hydrogens. Third, ozonolysis puts an O on each end of the cut double bond, and you work back by joining the carbonyl carbons. Now go back to the seven past-paper questions on this page and solve each one yourself, without looking at the solutions. You can do this. All the best!