IIT-JEE Questions (1979–2005): Alcohols

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Two bottles have lost their labels: one holds propan one ol, and the other holds propan two ol. Both have the same formula, C three H eight O, okay? One student says: oxidise both with a strong oxidant, add Fehling's solution, and the aldehyde will show itself. Will that work? Hmm, hold that thought. Today we solve two questions from old IITJEE papers, from two thousand and one and nineteen ninety six. Questions like these come up often, and each needs just a few ideas from this chapter. So, let's begin.

0:41Idea 1: primary or secondary?

So, idea number one: is the alcohol primary, secondary or tertiary? Count the carbons joined to the carbon that carries the O H group. The small circle written high is read as degree: one degree is primary, two degree secondary, three degree tertiary. Okay? Propan one ol is C H three, C H two, C H two, O H; the small numbers written low count the hydrogens. Its O H carbon is joined to just one carbon, so it is primary. Right? Now propan two ol, with the O H group on the middle carbon. Is it primary, secondary or tertiary? It is secondary, yes, exactly! C H three, then C H with O H in brackets, then C H three. The middle carbon is joined to two carbons. Clear?

1:31Idea 2: oxidation, and where it stops

Now idea number two: oxidation. Oxidation takes a primary alcohol to an aldehyde, then on to a carboxylic acid; each arrow means goes to. With me? Now let me ask you this. With a strong oxidant, does a primary alcohol stop at the aldehyde, or go on to the acid? It goes on to the acid! Acidified potassium permanganate, K M n O four, or potassium dichromate, K two C r two O seven, carries it all the way. Remember this; it is the trap in Example one. Got it? Next, a secondary alcohol gives a ketone, and under these conditions the ketone is not oxidised further. Okay? And here is a cleaner method: hot copper, C u, at five hundred and seventy three kelvin; the capital K is kelvin. It removes H two, a molecule of hydrogen, so this is dehydrogenation: a primary alcohol gives an aldehyde, and a secondary one gives a ketone. Clear?

2:31Idea 3: Fehling's solution

Idea number three is a test: Fehling's solution. Take an aldehyde and a ketone. Which one reduces it? The aldehyde, very good! An aliphatic aldehyde reduces it, and a red-brown precipitate of C u two O appears. Right? But a ketone shows no change at all. With me? So, to tell a primary alcohol from a secondary one, oxidise it only as far as the aldehyde or ketone, then test the product. An aldehyde means primary, and a ketone means secondary. Okay?

3:05Idea 4: the iodoform test

Now idea number four, the iodoform test. Treat the compound with iodine, I two, and with N a O H, sodium hydroxide. If the test is positive, we get a yellow precipitate of iodoform, CHI three. Clear? Which alcohols give it? Ethanol, and any alcohol of the type C H three, C H with O H in brackets, R, where R is the rest of the molecule. So the O H carbon must carry a hydrogen and a methyl group, C H three. Got it? Now, iodoform has just one carbon. Where does that carbon come from? From the methyl group, exactly! That carbon leaves as iodoform. The rest of the molecule stays in the alkaline solution as the salt of an acid with one carbon fewer. Right?

3:57Example 1: tell propan-1-ol from propan-2-ol with hot Cu

Time for Example one, from the IITJEE paper of two thousand and one. How are propan one ol and propan two ol best distinguished? Option a: oxidation with alkaline K M n O four, then Fehling's solution. Option b: oxidation with acidic dichromate, then Fehling's solution. Option c: heating with copper, then Fehling's solution. And option d: concentrated sulphuric acid, H two S O four, then Fehling's solution. Pause the video and try it first. Done? Okay, let's check. Propan one ol is primary, and propan two ol is secondary. So we want an aldehyde from one and a ketone from the other; then Fehling's solution can tell them apart. With me? Now look at option b, acidic dichromate. It is a strong oxidant, and it oxidises both alcohols. So option b is the answer, right? No, no, wait, that is the trap! Remember idea two: a strong oxidant does not stop at the aldehyde. It takes propan one ol on to C H three C H two COOH, which is propanoic acid. Alkaline permanganate in option a does the same. So no aldehyde is left for Fehling's solution to find. Okay? And option d? Concentrated sulphuric acid removes water and gives an alkene; the short form conc on screen means concentrated. No aldehyde forms, so Fehling's solution finds nothing. Right? Now option c, heating with copper. It dehydrogenates both alcohols cleanly. Propan one ol gives propanal, C H three C H two CHO, an aldehyde, and propan two ol gives propanone, C H three COCH three, a ketone. Clear? Only propanal reduces Fehling's solution and gives the red-brown precipitate; propanone does not. So option c is the answer. And our puzzle is solved: that student's strong oxidant would spoil the test.

6:03Example 2: identify C8H10O as 1-phenylethanol by iodoform

Got it? Now the harder one, Example two, from the nineteen ninety six paper. Compound D, C eight H ten O, gives a yellow precipitate with alkaline iodine. Acidifying the filtrate gives a white solid, compound E, C seven H six O two. Identify both compounds. Here is a hint: think of idea four. Pause the video and try it first. Done? Okay, let's check. The yellow precipitate is iodoform, so D has the unit C H three, C H with O H in brackets; the dash at the end joins it to the rest of the molecule. Okay? Now count the carbons. D has eight, and E has seven. So exactly one carbon has gone, the methyl carbon, as iodoform. Right? Now, E is the acid left behind, with the formula C seven H six O two. Which acid is it? It is benzoic acid, very good! C six H five is the phenyl group, a benzene ring, and COOH is the carboxylic acid group: seven carbons in all. Clear? Now, careful. Two phenylethanol has the same formula, C eight H ten O, so could D be two phenylethanol? No, and that is the trap! In two phenylethanol, C six H five C H two C H two O H, the O H sits on a C H two at the end of the chain. There is no C H three, C H with O H unit, so it gives no iodoform. With me? So the carbon of COOH in E was the O H carbon of D, and its methyl group left as iodoform. Put the methyl back: D is C six H five, C H with O H, C H three, which is one phenylethanol. Eight carbons, ten hydrogens and one oxygen, so the formula matches. And E is benzoic acid. Got it?

8:09Quick revision: oxidants, hot Cu, Fehling's and iodoform

So, quick revision. Strong oxidants take a primary alcohol all the way to the acid, while hot copper stops at the aldehyde. With Fehling's solution, an aliphatic aldehyde gives the red-brown precipitate, and a ketone does not. And in the iodoform test, the C H three, C H with O H unit gives yellow iodoform, and leaves an acid with one carbon fewer. Now go and try the practice problems on this page. All the best!