JEE Advanced and IIT-JEE Questions: Di- and Polysaccharides

Transcript

0:00Introduction

Hello everyone! Let us start with a real question from JEE Advanced twenty sixteen. It gives L-glucose a specific rotation of minus fifty two degrees, and L-fructose, plus ninety two degrees. Invert sugar is a mixture of glucose and fructose, so you take the average and get plus twenty degrees. But the correct statement in that question says minus twenty degrees! So who made the mistake, you or the paper? Today we solve past JEE questions on disaccharides and polysaccharides with just four ideas. So, let's begin.

0:37Idea 1: the glycosidic linkage

So, let us start with idea one. In a disaccharide, two sugar units are joined through an oxygen atom, and a molecule of water is lost. This is the glycosidic linkage, C, O, C on screen, where each small dash is a bond. Chemically, it is an acetal linkage. Okay? And how do we break it? With dilute acid, or with an enzyme. Remember this, right? Now look at the figure. Top left is maltose, and top right is lactose. In both, C one of the first unit is joined to C four of the second; C one means carbon number one, and the small numbers in the figure mark these carbons. With me? At the bottom is sucrose. Here C one of glucose is joined to C two of fructose. This difference decides everything in idea two. Clear?

1:25Idea 2: reducing or non-reducing?

Now comes idea two. When C one still carries a free hemiacetal group, the O H on screen, the ring can open up to CHO, an aldehyde group. That aldehyde reduces Tollens' and Fehling's reagents, so the sugar is a reducing sugar. Okay? In maltose and lactose, C one of the second unit is still free. So both are reducing sugars. With me? Now tell me this. Glucose is a reducing sugar, and fructose is a reducing sugar too. Sucrose is made of glucose and fructose, so sucrose must also be a reducing sugar, right? The answer is no, and that is the trap many students fall into. In sucrose, the reducing group of glucose at C one and that of fructose at C two are both used up in the linkage. Nothing is left to open up, so sucrose is non-reducing. Got it? This idea answers an IITJEE question from two thousand and five. Two glucose units joined C one to C four keep a free hemiacetal group, so they reduce Tollens' reagent. Joined C one to C one, both anomeric carbons, the C one carbons, are locked in the linkage, so they do not. Clear?

2:37Idea 3: two labels in a sugar's name

Now idea three, the one behind our puzzle. The full name of a sugar carries two labels. Capital D or L gives the configuration, by comparison with glyceraldehyde. Okay? The plus or minus sign in brackets tells you the direction of rotation of plane-polarised light. Plus means to the right, and minus means to the left. Right? And the two labels are independent. Glucose is D and plus, but fructose is D and minus. So never guess the sign from the letter. With me? Now, an L sugar is the mirror image of the D sugar, with every asymmetric carbon reversed; the two are enantiomers. So tell me this. D-fructose has a specific rotation of minus ninety two point four degrees; the small raised circle means degrees. What does its mirror image, L-fructose, show? It shows plus ninety two point four degrees, yes, exactly. The mirror image turns the light by the same amount, the other way, so only the sign flips. Okay? And one more fact: most natural sugars, glucose and fructose among them, belong to the D-series. Keep this in mind for our puzzle. Clear?

3:46Idea 4: bromine water or nitric acid?

Now comes idea four, oxidation. Bromine water, B r two water on screen, where the small two means two atoms, oxidises aldoses only. So glucose changes into gluconic acid; the arrow means changes into. Okay? Gluconic acid has one COOH, that is, one carboxylic acid group. And fructose? It is a ketose, so bromine water leaves it unchanged. Right? Nitric acid oxidises glucose at both ends of the chain, giving saccharic acid, with two COOH groups. So bromine water gives gluconic acid, and nitric acid gives saccharic acid. Got it?

4:23Example 1: products of acid hydrolysis of sucrose

Now Example one, an easy one from IITJEE two thousand. Give the products A and B of the reaction of sucrose with H plus in water; H plus, with its raised plus sign, simply means acid. Pause the video and try it first. Done? Okay, let's check. The acid, H plus over the arrow, breaks the glycosidic linkage, and one water molecule, H two O, is used. Sucrose, C twelve H twenty two O eleven, gives two molecules of C six H twelve O six; the small lowered numbers count atoms. So both products are glucose, right? The answer is no, because two glucose units would be maltose. Sucrose is one glucose joined to one fructose, as we saw in idea one. With me? So, A equals D-glucose, and B equals D-fructose, one mole of each. Both are natural sugars from the D-series. Okay? And this equimolar mixture, with equal moles of the two sugars, is called invert sugar. Keep it in mind, because the next question is all about it. Got it?

5:33Example 2: invert sugar statements, JEE Advanced 2016

Now Example two, the harder one, from JEE Advanced twenty sixteen: our puzzle. The rotations given: sucrose is plus sixty six degrees, maltose is plus one hundred and forty, L-glucose is minus fifty two, and L-fructose is plus ninety two. Which statements about invert sugar are correct? Statement a says it is made by acid hydrolysis of maltose. Statement b says it is an equimolar mixture of D plus glucose and D minus fructose. Statement c says its specific rotation is minus twenty degrees. And statement d says that with bromine water it gives saccharic acid. Pause the video and try it first. Done? Okay, let's check, starting with statement a. Maltose hydrolyses to glucose only, and invert sugar comes from sucrose. So statement a is false. Okay? Next, statement b. Sucrose gives one mole each of D plus glucose and D minus fructose, as in Example one. So statement b is true. Clear? Now let us look at statement c. The question gives minus fifty two degrees for glucose and plus ninety two for fructose. Average them: minus fifty two plus ninety two is plus forty, and forty over two gives plus twenty degrees. So statement c is false, right? Careful now, because this is the trap from our puzzle. Those values are for the L-sugars, the mirror images, but invert sugar is made of natural D-sugars. So flip both signs: D-glucose is plus fifty two degrees, and D-fructose is minus ninety two degrees. With me? The alpha in square brackets is the symbol for specific rotation. One mole of each, with the same molar mass, so we take the simple mean. Plus fifty two plus minus ninety two gives minus forty, and minus forty over two gives minus twenty degrees. So statement c is true. The paper was right, and our puzzle is solved! Clear? Last, statement d. Bromine water oxidises only the glucose, to gluconic acid; saccharic acid needs nitric acid. So statement d is false. Got it? So the correct statements are b and c, okay? Notice that plus sixty six and plus one hundred and forty, for sucrose and maltose, were never needed. Extra data like this should not distract you. Right?

7:57Quick revision: reducing sugars, invert sugar, bromine water

So, quick revision. Free hemiacetal O H means a reducing sugar; sucrose has both ends locked, so it is non-reducing. Sucrose with acid gives D-glucose and D-fructose: invert sugar, at minus twenty degrees. Rotations given for L-sugars? Flip the signs. And bromine water gives gluconic acid. Now try the other past JEE questions on this page yourself. All the best!