AIEEE and Earlier JEE Main Questions: Monosaccharides

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Glucose and fructose share the same formula, C six H twelve O six. Glucose has an aldehyde group, the CHO group. So, does fructose have one too? Many students say yes at once, and an old exam question was built on exactly that slip. Today we solve two short A I triple E questions on monosaccharides, each with a trap inside. Ideas like these still appear in JEE Main questions, so let's begin.

0:34What a carbohydrate is: aldoses and ketoses

So, first, what exactly is a carbohydrate? The course defines carbohydrates as polyhydroxy aldehydes or ketones. Polyhydroxy simply means many O H groups, the hydroxyl groups, okay? Or, compounds that give such units on hydrolysis. Today we stay with the simplest ones, the monosaccharides. With me? Every monosaccharide has one carbonyl group, a carbon double bonded to an oxygen. If it is an aldehyde group, CHO, the sugar is an aldose, like glucose, the first Fischer projection in the figure, with CHO at the top. In its name, capital D gives the configuration, and plus in brackets means it turns plane-polarised light to the right. Beside it are L-glucose, its mirror image, and D-galactose. Clear? If it is a keto group, the sugar is a ketose, like fructose. C equals O on screen, with its two short lines, is that double bond. Now find fructose, the last structure, written D, minus in brackets, fructose. Which carbon carries its C double bond O, and is there a CHO group anywhere? Yes, exactly, the C double bond O sits on carbon two. C dash two simply means carbon two, and beside glucose you also see C dash five, for carbon five. And there is no CHO group at all, because the top carbon of fructose is a C H two O H group, with a small lowered two. Got it?

2:09The formula trap: hydrates of carbon

Now, one more trap before the questions. Many carbohydrates fit the formula on screen, C, then H two O in brackets, where the letters x and y, written low, are whole numbers. It looks like carbon plus water, hence the old name, hydrates of carbon. Okay? So let me ask you this. Ethanoic acid, on the badge, can be written as C two, then H two O in brackets, two. It fits the formula perfectly. So, is ethanoic acid a carbohydrate? It is not, and that is the trap. Ethanoic acid, C two H four O two, is a carboxylic acid, not a polyhydroxy aldehyde or ketone. Right? And it fails the other way round too. Rhamnose, C six H twelve O five, has one oxygen too few to fit, yet it is a carbohydrate. Clear? So the formula is not a test; the functional groups decide, and that is exactly what our first question checks. With me?

3:11The ring, and the anomeric carbon

For the second question, we need one more idea. In glucose, the O H group on carbon five adds to the CHO group on carbon one. Okay? That closes a six-membered ring, five carbons and one oxygen, called a cyclic hemiacetal. In the figure, it is the hexagon with an O in one corner, called a pyranose ring, as the names show; the small numbers one and four mark carbons one and four, and the thick edges are the front. Clear? In the open chain, carbon one was the CHO carbon, in a double bond, so it was not asymmetric. In the ring, it carries an O H group, a hydrogen, the ring oxygen and carbon two. So, is carbon one asymmetric now? Yes, exactly, it is a new asymmetric carbon, and we call it the anomeric carbon. So its O H group can point two ways. Got it? For a D-sugar, the alpha form, Greek letter alpha, has the O H on carbon one below the ring, and the beta form, Greek letter beta, has it above. In the figure, that bold O H points down on the left, and up on the right. The middle box, with its double-headed arrows, shows them changing into each other in water, through the open chain, and the slow change in rotation that this causes is called mutarotation. The lines under the box give the equilibrium mixture, thirty six percent alpha and sixty four percent beta, with a rotation of plus fifty two point five degrees. Right?

4:44Example 1: functional groups common to all carbohydrates

Time for Example one, from A I triple E two thousand and nine. Which two functional groups are present in a typical carbohydrate? The four options pair up CHO with COOH, C double bond O with O H, O H with CHO, and O H with COOH, okay? Pause the video and try it first. Done? Okay, let's check. Many students picture glucose, with its CHO and O H groups, and tick O H and CHO at once. So that is the answer, isn't it? Now be careful, because that is exactly the trap. Glucose, an aldose, does have a CHO group and five O H groups. But glucose is only one kind of carbohydrate. With me? Now think of fructose, the last structure in the figure: its C double bond O is on carbon two, and it has no CHO group. So O H and CHO cannot describe every carbohydrate. Okay? Now look at what both sugars share. Each has a carbonyl group, C double bond O, the aldehyde group in one and the keto group in the other, and each has O H groups. So the answer is C double bond O and O H, the second option. The COOH options, carboxylic acid groups, do not fit the definition at all. Clear?

6:17Example 2: anomers of glucose differ only at C-1

Now Example two, from A I triple E two thousand and six. To what does the term anomers of glucose refer? Pause the video and try it first. Done? Okay, let's check. Many students mix up anomers with the D and L forms of glucose, which are mirror images of each other. So, are alpha and beta glucose mirror images? They are not, and that is the trap here. Anomers of glucose are isomers that differ in configuration at carbon one only. Configuration simply means which way the groups point in space. Okay? Carbon one is the anomeric, or hemiacetal, carbon, which became asymmetric only when the ring closed. In the figure, only that bold O H changes, down in alpha and up in beta, and everything else stays the same. Got it? So the two anomers differ at only one of their five asymmetric carbons, which makes them diastereomers, not mirror images. The real mirror image of D-glucose is L-glucose, with every asymmetric carbon reversed. With me? Their properties differ too, as the figure shows. Alpha rotates light by plus one hundred and twelve degrees, the small circle meaning degrees, and melts at four hundred and nineteen kelvin; m p means melting point, and capital K is kelvin. Beta rotates only plus nineteen degrees and melts at four hundred and twenty three kelvin. In the figure, alpha in square brackets with a small D is the specific rotation, and the line under each gives the water temperature it crystallises at, three hundred and three kelvin for alpha and three hundred and seventy one kelvin for beta. Clear? Now one last link to Example one: in fructose, the anomeric carbon is carbon two, the carbon that carried its keto group. So, for glucose, anomers are isomers that differ in configuration at carbon one. Right?

8:18Quick revision: C=O and OH, formula trap, anomeric carbon

So, quick revision. First, a typical carbohydrate has a carbonyl group and O H groups, and a ketose has no CHO group. Second, the hydrates of carbon formula is not a test; the functional groups decide. And third, anomers of glucose differ only at carbon one, the anomeric carbon, which is carbon two in fructose. Now go and try the practice problems in this chapter. All the best!