Set 2

Transcript

0:00Introduction

Hello everyone! Let us begin with two small puzzles from Set two. First, take seventeen point one grams of sucrose, ordinary cane sugar, and hydrolyse it completely. It splits into two sugars, glucose and fructose. Yet the two products together weigh more than the sucrose we started with. Where does the extra mass come from? Second, fructose has no aldehyde group, yet it passes Benedict's test, while sucrose, ordinary table sugar, fails it. It sounds strange, doesn't it? By the end of this video, you will settle both. Problems like these come up often in JEE Main, so let's begin.

0:40Disaccharides: two units, one oxygen bridge

So, first a quick recap from chapter two. The figure shows three disaccharides: maltose and lactose on top, and sucrose, which is cane sugar, at the bottom. Each ring is one sugar unit, and the capital O between two rings is an oxygen bridge, the glycosidic linkage; the small numbers beside it name the carbons it joins. In COC on screen, the dashes are bonds. And when the bridge forms, a molecule of water, H two O, is lost; the small lowered two counts the hydrogen atoms. Okay? Now let me ask you this. Water left when the bridge formed. So when we break the bridge, what must come back in? Yes, exactly, the water comes back in. Dilute acid, or an enzyme, breaks the glycosidic linkage with a molecule of water, and that is hydrolysis. Keep this in mind, because it is the key to our first puzzle. Right? Now, all three disaccharides share one formula, C twelve H twenty two O eleven, where the small lowered numbers count the atoms. Under each structure, the Greek letters alpha and beta tell us which anomer of each unit is used. With me? So all three have the same molar mass, capital M, three hundred and forty two grams per mole. The small g is grams, and the raised minus one on mol means per mole. Clear?

1:59Inversion of cane sugar

Now, the reaction behind our first puzzle. Sucrose plus water gives D-glucose plus D-fructose. Over the arrow you see H plus, the hydrogen ion of an acid, or the enzyme invertase; either one does the job. The capital D in front of each name gives the configuration, as in chapter one. Okay? So tell me, one mole of sucrose gives how many moles of fructose? Yes, exactly, one mole of fructose, and one mole of glucose too. The equation is balanced as it stands, so the two sugars come out in equal moles. Got it? Glucose and fructose have the same formula, C six H twelve O six. Six carbons give seventy two, twelve hydrogens give twelve, and six oxygens give ninety six. So both have a molar mass of one hundred and eighty grams per mole. Right? And one more fact. Sucrose is dextrorotatory: it turns the plane of polarised light to the right. Fructose turns it to the left, and more strongly than glucose turns it to the right. So this equimolar mixture is laevorotatory: the sign of rotation is inverted, and the mixture is called invert sugar. With me?

3:12Reducing sugars and Benedict's test

Now let us take up the second puzzle. Benedict's test and Fehling's test both use an alkaline copper complex. C u two plus is the copper ion, and the small raised two plus is its charge. The arrow means gives: the sugar reduces it to C u two O, which comes down as a red precipitate. Okay? And which sugars give this red colour? Only the reducing sugars, the sugars that can reduce this alkaline copper reagent. Right? Now here is the trap. Fructose is a ketohexose. Its carbonyl is a keto group at carbon two, not an aldehyde group. So does fructose give a positive Benedict's test? Yes, it does, and many students get this wrong. All monosaccharides are reducing sugars, the aldoses and the ketoses alike. Clear? And here is why. In alkali, fructose rearranges through an enediol into glucose and mannose, and those reduce the reagent. Benedict's solution is alkaline, so fructose passes. Got it? Now let us come to the disaccharides. Such a sugar is reducing only if one of its two units still has a free hemiacetal group, which can open to an aldehyde. Keep this rule ready, because we need it in Example two. With me?

4:29Example 1: fructose mass from hydrolysis of 17.1 g sucrose

Now, Example one, our first puzzle in numbers. Find the mass of fructose obtained by the complete hydrolysis of seventeen point one grams of sucrose. Pause the video and try it first. Done? Okay, let's check. First, the moles of sucrose. Small n of sucrose, on screen, is the number of moles of sucrose. Seventeen point one over three hundred and forty two gives zero point zero five mole. Right? Now, watch out for a trap here. Sucrose splits into two units, fructose and glucose. So fructose is just half of seventeen point one, that is eight point five five grams, right? No, that is the trap, because water joins in. So count moles, not grams. One mole of sucrose gives one mole of fructose, so we get zero point zero five mole of fructose. Okay? So the mass of fructose is zero point zero five times one hundred and eighty, which gives nine grams. The cross means times. Clear? Now let us check it, and settle our first puzzle. Water has a molar mass of eighteen grams per mole, so the water used is zero point zero five times eighteen, which gives zero point nine grams. With me? Seventeen point one plus zero point nine gives eighteen grams of products, which is nine grams of glucose and nine grams of fructose. So the extra mass came from the water. Got it?

5:56Example 2: how many sugars give a positive Benedict's test

Now a harder one, Example two, and it settles our second puzzle. How many of these give a positive Benedict's test: fructose, sucrose, lactose, starch and maltose? Pause the video and try it first. Done? Okay, let's check them one by one. Fructose is a monosaccharide, and all monosaccharides are reducing, thanks to that enediol. So fructose gives a positive test. Okay? Next comes sucrose, ordinary cane sugar. It is a sugar, and it is sweet, so it must give a positive test, right? No, it does not, and this is the classic trap. Look at sucrose at the bottom of the figure. C dash one on screen means carbon one, and carbon one of glucose is joined to carbon two of fructose. Both reducing groups are used up in the linkage, so sucrose is non-reducing. Clear? Now lactose, at the top right of the figure. Carbon one of its glucose unit is still a free hemiacetal, which can open to an aldehyde. So lactose is reducing, and the answer is yes. Got it? Now let us look at starch. Starch is made only of glucose units, and glucose is reducing. So starch is reducing too, right? Careful here, because the answer is no. Starch is a polysaccharide, a polymer of D-glucose, and polysaccharides like starch are non-reducing. So starch gives no red precipitate. With me? And maltose, at the top left of the figure. Carbon one of its second glucose unit is still free, so maltose is reducing, and the answer is yes again. Right? So the positive ones are fructose, lactose and maltose, and the answer is three. And that settles our second puzzle: fructose passes, and sucrose fails. Clear?

7:44Quick revision: sucrose hydrolysis and reducing sugars

So, quick revision. Hydrolysis adds water, so one mole of sucrose gives one mole of glucose and one mole of fructose. Always count moles. All monosaccharides reduce Benedict's solution, even fructose, which is a ketose. Sucrose and starch are non-reducing, while maltose and lactose are reducing. You have done really well today. Now try the practice problems of Set two on this page, then check the solutions. All the best!