Cyclic Structure, Anomers and Mutarotation

Transcript

0:00Introduction

Hello everyone! Let's start with a small mystery. Dissolve pure alpha-D-glucose in water, and its optical rotation reads plus one hundred and twelve degrees. Add nothing, just wait, and the reading slowly falls to plus fifty two point five degrees. So what is changing inside that solution? And glucose has an aldehyde group, yet it fails some aldehyde tests. One small ring explains both, and questions on it come up often in JEE, so let's begin.

0:32What the open chain cannot explain

So, recall the open-chain structure of glucose, with an aldehyde group, CHO, on carbon one. An aldehyde should give Schiff's test, right? So tell me, does glucose give it? No, it does not, and that surprises many students. It also forms no addition product with sodium bisulphite, N a HSO three, with a small subscript three. With me? Next, glucose pentaacetate, with all five O H groups acetylated, does not react with hydroxylamine, N H two O H, which turns a carbonyl group into an oxime. So it has no free CHO group. Okay? And third, glucose exists in two crystalline forms, named with the Greek letters alpha and beta. The open chain explains none of these three facts, so we need a better structure.

1:28The ring: a cyclic hemiacetal

Clear? Here is the idea that fixes everything. Inside the same molecule, an O H group adds to the aldehyde group and closes a ring of six atoms, five carbons and one oxygen. So which carbon must carry that O H? Yes, exactly, it is the O H on carbon five, adding to the CHO group on carbon one; C dash five on screen means carbon five. Count with me: carbons one to five, plus the oxygen, make six atoms. The product is a six-membered cyclic hemiacetal, the ring in the figure. Got it? In the open chain, carbon one, the CHO carbon, was not asymmetric. But in the ring it becomes a new asymmetric carbon, called the anomeric carbon. Right? So the O H on this carbon can point two ways, and two forms are possible, alpha and beta. Those are exactly the two crystalline forms we could not explain! We will compare the two forms in a minute.

2:30Anomers and the Haworth structure

Got it? These two cyclic forms are called anomers. They differ only in the configuration at the hemiacetal carbon, okay? For aldoses like glucose, the hemiacetal carbon is carbon one, and for ketoses like fructose, it is carbon two. With me? These are Haworth structures: the ring lies flat, with its thick edge nearest to you. In the alpha form on the left, the bold O H on carbon one points down, below the ring. Now, in the beta form on the right, where does it point? Yes, exactly, it points up, above the ring. So in the Haworth structure of a D-sugar, alpha has the O H on the anomeric carbon below the ring, and beta has it above; for glucose, that is carbon one. Easy to remember, because beta and above both have the letter b in them. Clear? The six-membered ring is called pyranose, after pyran, and a five-membered ring is called furanose, after furan. So the left one is alpha-D-glucopyranose, and the plus sign in brackets gives the sign of its rotation. Got it?

3:35Two anomers, and mutarotation

Alpha-D-glucose melts at four hundred and nineteen kelvin; m p means melting point, and capital K means kelvin. Alpha in square brackets, with a small D below it, is the specific rotation, plus one hundred and twelve degrees, and the little circle means degrees. It crystallises from a concentrated solution at three hundred and three kelvin, okay? Beta-D-glucose melts at four hundred and twenty three kelvin, rotates only plus nineteen degrees, and crystallises from a hot saturated solution at three hundred and seventy one kelvin. Got it? Now back to our mystery solution, where the rotation of alpha falls. So the alpha ring simply flips straight into the beta ring, right? That is exactly the trap, because the change is not direct. In water, the ring first opens into the open-chain form, and then closes again, either way. The double half-arrows on screen mean an equilibrium that runs both ways. With me? This slow change in rotation is called mutarotation. Starting from alpha, the rotation falls from plus one hundred and twelve degrees; starting from beta, it rises from plus nineteen degrees. Both reach the same value, plus fifty two point five degrees, at equilibrium. We will find its exact composition in Example two. Okay?

4:57Why Tollens' works but Schiff's fails

Now here is one last puzzle. Glucose does reduce Tollens' and Fehling's reagents, and for glucose those reagents need a free aldehyde group. But the ring has no free CHO group. So how does glucose manage it? Yes, exactly, it works through the open-chain form in our equilibrium. The box in the middle of the figure says the open-chain form is only about zero point zero two percent. Only a small amount, right? That small amount is enough for Tollens' and Fehling's reagents, because they use it up for good, and the equilibrium keeps supplying more. Clear? But Schiff's reagent reacts reversibly, so this small amount is too little for it, and for sodium bisulphite too. Okay? Finally, look at the pentaacetate. Its O H on carbon one is acetylated too, so the ring cannot open to give a free aldehyde, and no oxime forms. So one ring explains all three facts. Beautiful, isn't it?

5:56Example 1: asymmetric carbons in open-chain and ring glucose

Now the first example. How many asymmetric carbon atoms are there in the open-chain form of glucose, and how many in alpha-D-glucopyranose? Pause the video and try it first. Done? Okay, let's check. An asymmetric carbon carries four different groups. Carbon one, the CHO carbon, and carbon six, the C H two O H carbon, do not. Carbons two to five do, so the open chain has four. Clear? Now for the ring, closing it adds no new carbon atom, so the answer is still four, right? No, and that is exactly the trap, because carbon one has changed. In the ring it carries an O H, a hydrogen, the ring oxygen and carbon two, four different groups. So four plus one gives five asymmetric carbons. Got it?

6:52Example 2: percent alpha and beta glucose at +52.5 degrees

Now a harder one, and it finishes our mystery. The specific rotations of alpha- and beta-D-glucose are plus one hundred and twelve and plus nineteen degrees, and the equilibrium mixture shows plus fifty two point five degrees. Find the composition of the mixture. Pause the video and try it first. Done? Okay, let's check. Let the letter x be the fraction of alpha in the mixture. Then the fraction of beta is one minus x, okay? Each form adds its own rotation times its fraction. So one hundred and twelve times x, plus nineteen times one minus x, equals fifty two point five. Right? Open the bracket and collect terms: one hundred and twelve times x, minus nineteen times x, is ninety three times x, and fifty two point five minus nineteen is thirty three point five. So ninety three times x equals thirty three point five. The double arrow means which gives, so x is thirty three point five over ninety three, which gives about zero point three six; the wavy equals sign means approximately. So the mixture is thirty six percent alpha and sixty four percent beta. That solves our mystery: alpha slowly turns into beta, through the open chain, until beta is the majority, and the rotation falls. Clear?

8:18Quick revision: anomeric carbon, alpha vs beta, mutarotation

So, quick revision. First, the O H on carbon five adds to the CHO group, and carbon one becomes the anomeric carbon. Second, for a D-sugar, alpha has the O H on the anomeric carbon below the ring, and beta has it above. And third, mutarotation goes through the open chain to thirty six percent alpha and sixty four percent beta, at plus fifty two point five degrees. Now go and try the practice problems on this page. All the best!