JEE Advanced and IIT-JEE Questions: Monosaccharides

Transcript

0:00Introduction

Hello everyone! Here is a challenge from JEE Advanced, twenty fifteen. You are given the Fischer projection of D-glucose, and you must describe L-glucose. Now, D or L is decided by carbon five alone, so many students flip just carbon five, move on, and lose the marks. Today we see exactly why, with more past questions on glucose and fructose. Questions like these come up often, so let's begin.

0:26Fischer projections: D or L, plus or minus

So, the figure shows Fischer projections of D-glucose, L-glucose, D-galactose and D-fructose. The chain stands vertical, with CHO on top in the three aldoses and C H two O H at the bottom, where the small two written low means two hydrogens. Each O H points left or right, okay? The labels C dash two and C dash five mark carbons two and five. Now let us read D or L from these projections. Which carbon decides whether a sugar is D or L? Yes, exactly, it is the lowest asymmetric carbon, which is carbon five in a hexose. O H on the right means a D sugar, and O H on the left means an L sugar. In D-glucose, the O H on carbon five is on the right. With me? Its name has a plus sign in brackets, which means glucose turns plane-polarised light to the right. So tell me, is every D sugar plus? It is not, and that is a classic trap. Look at the last projection, D-fructose, with a minus sign in brackets; its C double bond O group sits on carbon two. The plus or minus is the observed rotation, dextro or laevo, found by experiment, and it does not follow from the D or L label. Clear? And L-glucose, the second projection, is the mirror image of D-glucose, its enantiomer. Which O H groups change sides is exactly our Example one, so keep this picture in mind. Got it?

2:02The ring: the anomeric carbon and anomers

Next, let us close the ring. The O H on carbon five adds to the CHO group on carbon one, closing a six-membered ring of five carbons and one oxygen, called a pyranose ring. In the figure, it is the hexagon with oxygen at one corner; the small numbers one and four label two of its carbons. Okay? Now think about carbon one. In the open chain it was the CHO carbon, and it was not asymmetric. What about in the ring? Yes, exactly, it becomes asymmetric, because it now carries four different groups: an O H, a hydrogen, the ring oxygen and carbon two. We call it the anomeric carbon. Clear? So its O H can point two ways, giving two forms, named with the Greek letters alpha and beta. In the figure, alpha, on the left, has the bold O H on carbon one below the ring, and beta has it above. The arrows link both rings through the open-chain form in the box; the numbers come from the mutarotation lesson. Right? Now a quick one from IITJEE, two thousand five. D-glucopyranose means D-glucose in its pyranose ring. What are its two forms, from a solution of D-glucose, called? Isomers, anomers, epimers or enantiomers? They are anomers, because they differ only at the anomeric carbon, carbon one. They are not enantiomers, since they are not mirror images of each other. Got it?

3:38Reducing tests: Fehling's and Tollens'

Now the reducing tests, with a statement question from IITJEE, two thousand seven. Statement one says that glucose gives a reddish-brown precipitate with Fehling's solution. Okay? Statement two says the reaction gives C u O and gluconic acid. Let me ask you this: is statement two true? If you said false, you are right, and here is the catch: the reddish-brown precipitate is C u two O, where the small two written low means two copper atoms. So statement one is true, and statement two is false. Clear? With Tollens' reagent, the CHO group is oxidised to the carboxylate, and a shiny silver mirror forms. Okay? Now bring in fructose, a ketose, with a keto group on carbon two and no CHO group at all. So does fructose give Tollens' test? Yes, it does, and it reduces Fehling's solution too. Both reagents are alkaline, and in alkali, fructose rearranges through an enediol into glucose and mannose, which get oxidised. Right? So all monosaccharides, aldoses and ketoses alike, are reducing sugars. That settles IITJEE, two thousand four, which asked for the pair where both give a positive Tollens' test: glucose and fructose. With me?

5:02Example 1: draw L-glucose from D-glucose

Now Example one, our JEE Advanced question from twenty fifteen. In the Fischer projection of D plus glucose, the O H groups on carbons two, three, four and five are on the right, left, right and right. Describe the Fischer projection of L minus glucose. Pause the video and try it first. Done? Okay, let's check. First write down D-glucose. From carbon two to carbon five, the O H groups are right, left, right, right. Okay? Now remember our hook. D or L is decided by carbon five, so we flip only carbon five and keep the rest, right? That is exactly the trap, so be careful here. L-glucose is the mirror image of D-glucose, and in a mirror image every asymmetric carbon is reversed, all four of them, not only carbon five. Clear? So in L-glucose, from carbon two to carbon five, the O H groups are left, right, left, left. The O H on carbon five is now on the left, so it is an L sugar. Got it?

6:14Example 2: stereoisomers of a D-pyranose with C-2 as CH2

Now a harder one, Example two, from IITJEE, twenty twelve. An aldohexose has the open chain on screen: CHO, then C H two, then three CHOH groups, and C H two O H at the end, where each short dash is a single bond. It exists in its D-configuration. Find the total number of stereoisomers in its pyranose form. Pause the video and try it first. Done? Okay, let's check. Number the chain from the CHO end: carbon one is CHO, and carbon two is C H two. Carbons three to five are CHOH, and carbon six is the C H two O H group. With me? Now, which carbons of this open chain are asymmetric? Careful, this sugar is not glucose. Yes, exactly, only carbons three, four and five are asymmetric. Carbon two carries two hydrogens, so it cannot be. Count it, as in glucose, and you go wrong right away. Clear? In the ring, carbon one also becomes asymmetric, the anomeric carbon. So four asymmetric carbons, and two to the power four, the small four written high, gives sixteen, right? Sixteen is the trap here, so be careful. The question says D-configuration, so carbon five is fixed, with its O H on the right in the Fischer projection. Only carbons three and four are free, with two ways each. Okay? Now add the ring, where carbon one is the anomeric carbon. It can be alpha or beta, which gives two more ways. Right? So the total is two times two times two, where the cross means times, giving eight stereoisomers. Sixteen counts the L forms too, but the question fixed the D-configuration. Got it?

8:16Quick revision: L-glucose, anomers, reducing sugars

So, quick revision. First, L-glucose is the mirror image of D-glucose, with every asymmetric carbon reversed. Second, in the ring, carbon one becomes the anomeric carbon, and the alpha and beta forms are anomers. And third, Fehling's solution gives C u two O, and all monosaccharides, fructose too, are reducing. Now go over the past questions on this page, and try each one yourself before reading the solution. All the best!