Working Problems on Glucose and Fructose

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take three point six grams of glucose, and acetylate it completely with ethanoic anhydride. The by-product, ethanoic acid, comes out heavier than all the glucose you started with. So how can there be more acid than sugar? Keep that question in mind, okay? Problems like these on glucose and fructose come up often in JEE, and they are quick once you know a few numbers. So, let's begin.

0:29Glucose or fructose? Read the tests

So, first, let us look at the tests. Some of them cannot tell glucose and fructose apart. Both sugars give normal hexane on heating with H I, hydrogen iodide; the small n in front stands for normal, a straight chain of six carbons. And both reduce Tollens' and Fehling's reagents, fructose too, even though it is a ketone. Okay? Bromine water, a mild oxidising agent, does tell them apart. It oxidises the CHO group of glucose, the aldehyde group. Fructose has no aldehyde group, so bromine water does not oxidise it. With me? Now try this problem from the course. Here is a compound, C six H twelve O six, where the small lowered numbers count the atoms. It gives normal hexane with H I and reduces Fehling's solution, but it does not decolourise bromine water. Which sugar is it? Yes, exactly, it is fructose. It has a straight chain and it is a reducing sugar, but bromine water leaves it alone, so there is no CHO group. If you said glucose, check the bromine water test again. Clear?

1:40One mole of glucose: how many moles react?

Now, counting moles. On the right is the map of glucose reactions from earlier in this chapter, with an arrow for each reagent. Glucose has one aldehyde group, the CHO at the end of its formula. So how many moles of HCN, hydrogen cyanide, add to one mole of glucose? Yes, exactly, it is one mole, because one aldehyde group adds one HCN, giving the cyanohydrin. In the same way, one mole of hydroxylamine, N H two O H, gives the oxime. The short form after each number means moles. Got it? Now ethanoic anhydride. On the map, it is two acetyl groups, C H three C O, in brackets with a small two, joined through one oxygen. How many moles of it react with one mole of glucose? Yes, exactly, it is five moles, because glucose has five O H groups, and each one takes one molecule of anhydride. That is why it forms a pentaacetate. Right? Now Tollens' reagent. The CHO group is oxidised, and each glucose loses two electrons. They go to two silver ions, written A g with a small plus for one positive charge, one electron each. So we get two moles of silver. Okay? With Fehling's solution, the red-brown precipitate is copper one oxide, C u two O, where C u is copper. Two moles of silver before, so two moles of copper one oxide now, right? That is the trap, because it is only one mole. The same two electrons go to two copper ions, one each, and both copper atoms sit in a single unit of copper one oxide, C u two O, as the formula shows. Clear?

3:30Molar masses: build them from 180

Next, molar masses. Glucose is one hundred and eighty grams per mole; the small minus one, written high, means per. When the CHO group becomes COOH, the acid group, what gets added to the formula? Yes, exactly, it is one oxygen atom, which is sixteen. So gluconic acid, from bromine water, is one hundred and eighty plus sixteen, which gives one hundred and ninety six. Okay? In saccharic acid, the C H two O H end also becomes an acid group. That end loses two hydrogens and gains one oxygen, which is fourteen more. So it is one hundred and ninety six plus fourteen, which gives two hundred and ten. Right? Reduction goes the other way. The CHO group gains two hydrogens and becomes C H two O H, giving sorbitol, one hundred and eighty two. With me? And the pentaacetate. An acetyl group, C H three C O, is forty three, but it replaces one hydrogen of an O H group, so each one adds forty two. There are five of them; the cross means times. Five times forty two is two hundred and ten, and one hundred and eighty plus two hundred and ten gives three hundred and ninety.

4:51Optical rotation: one formula, two units

Got it? Now optical rotation. The angle we observe equals the specific rotation, times l, times c. The specific rotation is the alpha in square brackets, a fixed number of degrees for each substance. Okay? Here, l is the length of the tube. In what unit do we take it? Centimetres, like most lengths in the lab? Careful now, we take it in decimetres, written d m, and one decimetre is ten centimetres. This is a common slip in exams. Right? And c, the concentration, is in grams per millilitre, not grams per litre. Mix them up, and your answer is off by a factor of a thousand. Clear?

5:35Example 1: ethanoic acid from acetylating 3.6 g glucose

Now, remember our puzzle from the start? Let us settle it. Find the mass of ethanoic acid formed as the by-product when three point six grams of glucose is completely acetylated with ethanoic anhydride. Pause the video and try it first. Are you done? Okay, let us check. Three point six grams over one hundred and eighty, the fraction bar on screen, gives zero point zero two moles of glucose. Okay? Now, how many moles of ethanoic acid come out? One for each glucose, so zero point zero two moles, right? That is exactly the trap, so be careful. Each of the five O H groups takes one molecule of anhydride, and each reaction releases one molecule of ethanoic acid, C H three COOH on screen. With me? So the ethanoic acid is zero point zero two times five, which gives zero point one zero moles. Right? Its molar mass is sixty grams per mole, so the mass is zero point one zero times sixty, which gives six point zero grams. So our puzzle is solved, because each glucose releases five molecules of acid. And if you got one point two grams, you used one mole instead of five. Got it?

6:55Example 2: sucrose concentration from observed rotation

Now let us try a harder one. Here is a sucrose solution with a specific rotation of plus sixty six point five degrees; the small circles are degree signs. In a one decimetre tube, it shows a rotation of plus seven point nine eight degrees. Find its concentration. Pause the video and try it first, and watch the units. Are you done? Okay, let us check. Put the numbers into our formula. Plus seven point nine eight degrees equals plus sixty six point five degrees, times one, times c. Okay? So c is seven point nine eight, over sixty six point five times one. That gives zero point one two, in grams per millilitre, the unit our formula uses. Clear? Many students now write zero point one two grams per litre. Is that right? Careful now, that is the unit trap we saw earlier. One litre is a thousand millilitres, so zero point one two times a thousand gives one hundred and twenty grams per litre. So from one measured angle, we found how much sugar is dissolved. Nice, isn't it?

8:05Quick revision: bromine water test, mole ratios, rotation

So, quick revision. Bromine water oxidises glucose but not fructose, while both reduce Tollens' and Fehling's reagents. One mole of glucose takes five moles of ethanoic anhydride, and gives two moles of silver, but only one mole of copper one oxide. And the observed rotation is the specific rotation times l times c, with l in decimetres and c in grams per millilitre. Now go and try the practice problems on this page. All the best!