Fructose

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Aldehydes give a silver mirror with Tollens' reagent, and ketones usually do not, okay? Now fructose, the sugar in fruits and honey, is a ketone. So with Tollens' reagent, nothing should happen, right? But you actually get a shining silver mirror! How can a ketone do that? Fructose comes up often in JEE biomolecules, so let's begin.

0:28Fructose: a ketohexose

So, here is fructose: C six, H twelve, O six; the small lowered numbers count the atoms, okay? It has the same formula as glucose, but a keto group, and six carbons. So how would you classify it? Yes, exactly, it is a ketohexose: keto for the keto group, and hexose for six carbons. Glucose, with its aldehyde group, is an aldohexose. Clear? Fructose is the sweetest natural sugar, found in fruits and in honey. And we get it, along with glucose, when sucrose, or cane sugar, is hydrolysed. With me?

1:08Open chain: a keto group at C-2

Now let us look at the open chain. The figure shows Fischer projections: each horizontal line carries the H and the O H group of one carbon. Fructose is the last one; the first is D-glucose, and the middle two are L-glucose and D-galactose. The formula on screen has the C O group in second place, three CHOH groups in the bracket with the small three, and a C H two O H group at each end; the dashes are bonds. Okay? So the keto group sits on carbon two; C dash two means carbon number two. In the figure, it is the C double bond O near the top. And all six carbons form one straight chain. Got it? Natural fructose is D, minus, fructose, with the minus sign in brackets. The alpha in square brackets, with a small D below, is the specific rotation, minus ninety two point four degrees; the small circle means degrees. It turns plane-polarised light to the left, so it is laevorotatory, hence the name laevulose. Right? Now here is a trap. Many students think the capital D means dextrorotatory, turning light to the right. But fructose turns light to the left. So, is the D in its name a mistake? It is not a mistake at all. D tells us the configuration at carbon five, the lowest asymmetric carbon in the Fischer projection. In the figure, its O H group is on the right, just as in D-glucose, so fructose is a D sugar. Okay? And the minus sign in brackets tells us the direction of rotation, which is found by experiment. The two labels are independent, so D, minus, fructose is perfectly fine. Clear?

2:54Ring form: a five-membered furanose

Like glucose, fructose also forms a ring. The O H group on carbon five adds to the C double bond O of carbon two; C equals O on screen is that double bond. Now count with me: how many atoms are in the ring? Carbons two, three, four and five, plus the oxygen that joins them, so four carbons and one oxygen, five atoms. Five-membered rings are called furanose, after furan. And be careful here: glucose forms a six-membered pyranose ring, but fructose a five-membered furanose. Don't mix them up, okay? Carbon two is now the anomeric carbon. So there are two forms, alpha and beta D-fructofuranose, with the Greek letters alpha and beta on screen. They differ only at carbon two. Got it?

3:48Reactions: what each one tells us

Now the reactions, and what each one tells us. With HCN, hydrogen cyanide, fructose forms a cyanohydrin, and with N H two O H, hydroxylamine, an oxime; the arrow means gives. Both show that a carbonyl group is present. Okay? With H I, hydrogen iodide, it gives normal hexane; the small letter n, written in front, stands for normal. So the six carbons form a straight chain, just as in glucose. Right? Now bromine water, a mild oxidising agent. It oxidises glucose, but not fructose, because fructose has no aldehyde group. So bromine water tells glucose and fructose apart. With me? Now Tollens' and Fehling's reagents. Fructose is a ketone, and even mild bromine water could not oxidise it. So surely it will not reduce these either, right? It actually reduces both, and this is a classic trap: the silver mirror of our puzzle. Notice the words in brackets: both reagents are alkaline. Okay? Here is the reason. In alkali, fructose rearranges through an enediol: ene for a carbon carbon double bond, and diol for two O H groups. The enediol gives glucose and mannose, which are aldoses, and the reagent oxidises them; the double half arrows mean equilibrium. So that is how a ketone gives a silver mirror. Got it?

5:15Example 1: asymmetric carbons in fructose and its ring

Time for Example one. How many asymmetric carbon atoms are there in open-chain fructose, and in fructofuranose, the ring form? And how many ketohexoses of this structure are possible? Pause the video and try it first. Done? Okay, let's check. An asymmetric carbon carries four different groups. Carbons one and six have two hydrogens each, and carbon two is in the C O double bond, so they are out. That leaves carbons three, four and five, three asymmetric carbons; the double arrow on screen means so. Okay? Now let us look at the ring. When the ring closes, the oxygen on carbon five bonds to carbon two, and the C O double bond opens up. So what happens to carbon two now? That is right, carbon two becomes asymmetric too. It now carries an O H group, the ring oxygen, a C H two O H group and carbon three, four different groups. So the ring has four asymmetric carbons. With me? Now the number of ketohexoses. Four asymmetric carbons in the ring, so two to the power four, with the small raised four, gives sixteen, right? Be careful here, because that is the trap. The ring carbon only gives the alpha and beta forms of the same sugar, so we count from the open chain. Three asymmetric carbons, two ways each, give two to the power three, which is eight. Clear?

6:47Example 2: what glucose, mannose and fructose share

Now a harder one, Example two. D-glucose, D-mannose and D-fructose all give the same enediol in dilute alkali. What feature do they share? Pause the video and try it first. Done? Okay, let's check. In the enediol, carbons one and two are joined by a double bond, the ene part of the name. So the CHO group of glucose at carbon one, and the keto group of fructose at carbon two, both end up in that double bond. With me? Now, in glucose, carbon two is asymmetric, with its O H group on the right. In the enediol, it sits in a double bond. So what happens to its configuration? Yes, exactly, it is lost. Any carbon in a double bond cannot carry four different groups, so it cannot be asymmetric. So any difference at carbon two simply disappears. Okay? So if all three sugars give the very same enediol, which carbons must match in all of them? It is the rest of the chain: carbons three, four and five have the same configuration in all three. Check it in the figure, comparing D-glucose, the first, with D-fructose, the last: the O H group is on the left at carbon three, and on the right at carbons four and five, in both. That shared chain links all three in alkali. See how it all fits?

8:16Quick revision: keto at C-2, furanose ring, Tollens test

So, quick revision. Fructose is a ketohexose with its keto group at carbon two; it is D, minus, fructose, or laevulose. In the ring, the O H group on carbon five adds to carbon two, giving a five-membered furanose. And bromine water does not oxidise it, yet it reduces Tollens' reagent, through an enediol. Now try the practice problems on this page. All the best!