Set 2

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle from Set two. Two unlabelled bottles: one holds pentan two one, the other pentan three one. Both are ketones, with exactly the same formula. So which one simple test can tell them apart? Keep that in mind. By the end, the same test will also sort out two alcohols made from one alkene. Problems like these come up often in JEE Main, so let's begin.

0:28The iodoform test: what it looks for

First, the test itself. The iodoform test uses I two plus N a O H, that is iodine with sodium hydroxide; the small two written low means two iodine atoms, and the plus sign means together with. When the test is positive, we see a yellow precipitate of CHI three, which is iodoform. Okay? Now, iodoform, CHI three, has just one carbon atom. So where in the molecule being tested does that carbon come from? From a methyl group sitting right next to a C O group, yes, exactly! That is C H three C O, and the dash at the end is the bond to the rest of the molecule. The arrow means gives: the methyl carbon leaves as CHI three, and what remains is the salt of an acid with one carbon fewer. So the test is given by methyl ketones, and by ethanal. Right? And for alcohols, the group is C H three C H, then O H in brackets, so the O H hangs on the carbon next to the methyl group. That covers ethanol, and alcohols of the type C H three CHOHR, where R is a carbon group. Clear? Now, a trap. Ethanoic acid also contains the C H three C O group. So does ethanoic acid give the iodoform test? No, it does not, and here is the reason. The group must be joined to a carbon or to a hydrogen; that is what C or H on the screen means. In ethanoic acid it is joined to an O H group instead, so ethanoic acid, its esters and its amides do not respond. But ethanol and ethanal both do. Got it?

2:15Water on an alkene: two routes, two alcohols

Now, the second idea: adding water across a double bond. The first route is oxymercuration demercuration. First mercuric acetate, H g, then O A c in brackets with a small two, where O A c is short for acetate, together with water; the slash means together. Then sodium borohydride, N a B H four. With me? Take propene, with its double bond at the end of the chain. By this route, which carbon gets the O H group? The middle carbon, exactly! This route is Markovnikov. On the screen, the dashes are single bonds, and the equals sign is the double bond. The hydrogen goes to the end carbon, which already has more hydrogens, and the O H goes to the middle carbon, so propene gives propan two ol. Okay? The second route is hydroboration oxidation. First diborane, B two H six, or B H three in THF, the solvent tetrahydrofuran. Then hydrogen peroxide, H two O two, with hydroxide, O H minus; the small minus written high is its charge. Same propene: this time, which carbon gets the O H group? The end carbon, yes, and that is why this route is called anti-Markovnikov. Here we get C H three, C H two, C H two, O H, which is propan one ol. Same alkene, a different alcohol, just by changing the reagent. Right? And one more point: neither of these routes rearranges, while acid-catalysed hydration goes through a carbocation and can rearrange. Keep both routes in mind for Example two. Clear?

4:00Example 1: test to tell pentan-2-one from pentan-3-one

Now, Example one, our puzzle from the start. Which test distinguishes pentan two one from pentan three one? The number in each name tells you which carbon carries the C O group. Pause the video and try it first. Done? Okay, let's check. First, write the structures. Pentan two one is C H three, then C O, then C H two C H two C H three: five carbons, with the carbonyl group on carbon two. Okay? Pentan three one has the carbonyl in the middle, on carbon three: C H three C H two, then C O, then C H two C H three. Right? Now, many students reach for two four DNP, or for Tollens' reagent. Will either of these tell the two ketones apart? No, neither of them will. Both are C five H ten O ketones; the small numbers written low count the atoms. Two four DNP, two four dinitrophenylhydrazine, gives a precipitate with any aldehyde or ketone, so both respond, and that precipitate can be yellow too, so do not mistake it for iodoform. Tollens' reagent picks out aldehydes, so both fail. Clear? So find a group that only one of them has. Pentan two one has C H three C O dash, so it is a methyl ketone. Pentan three one has an ethyl group on each side of the carbonyl, and no methyl right next to it. So, the iodoform test it is. Got it? With iodine and sodium hydroxide, pentan two one gives a yellow precipitate of iodoform, and pentan three one gives none. That's it, our two bottles are sorted!

5:45Example 2: which alcohol from pent-1-ene gives iodoform

Now a harder one, Example two. Pent one ene is converted into an alcohol, in part a by oxymercuration followed by sodium borohydride, and in part b by hydroboration followed by alkaline hydrogen peroxide. Which of the two alcohols gives the iodoform test? Pause the video and try it first. Done? Okay, let's check. Pent one ene is C H two, double bond, C H, then C H two C H two C H three. The double bond is between carbon one and carbon two. With me? Part a first, which is oxymercuration. So does the O H group go to carbon one, or to carbon two? On the badge, C dash one and C dash two are short for these two carbons. Carbon two, exactly, because oxymercuration is Markovnikov! We get C H three, C H, O H in brackets, then C H two C H two C H three, which is pentan two ol, with no rearrangement. Okay? For part b, hydroboration is anti-Markovnikov, so the O H goes to carbon one, the end carbon. That gives C H three, then three C H two groups, then C H two O H, which is pentan one ol. Right? Now, a trap. Both alcohols have a C H three group at one end. So both give the iodoform test, don't they? No, no, wait, look at where the O H sits! In pentan two ol, the O H is on carbon two, right next to the methyl group. That is the C H three CHOH group, joined to carbon, so the alcohol from part a gives the yellow precipitate. Clear? In pentan one ol, the O H is on the end carbon, and the methyl group is at the far end of the chain. There is no C H three CHOH group, so there is no iodoform. Got it? So only the alcohol from part a, pentan two ol, gives the iodoform test. Same alkene, but the reagent you choose decides the result!

7:59Quick revision: iodoform, 2,4-DNP, Tollens, hydration routes

So, quick revision. The iodoform test needs a C H three C O group, or a C H three CHOH group, joined to carbon or hydrogen. Two four DNP responds to any aldehyde or ketone, while Tollens' reagent picks out aldehydes, and simple ketones do not respond. And on an alkene, oxymercuration puts the O H the Markovnikov way, and hydroboration the anti-Markovnikov way. Now try all fifteen problems of Set two in twenty five minutes, then check the solutions. All the best!