Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Dissolve pure alpha-D-glucose in water, and its specific rotation is plus one hundred and twelve degrees. Add nothing, just wait, and the rotation slowly falls. When it reaches plus eighty two point two degrees, how much of the glucose has changed into the other form? The rotation has dropped by about twenty seven percent, so many students answer twenty seven percent. But that is not the answer! So where is the mistake? Hold that thought. This video opens the unit problem bank on biomolecules. We will see how to use it, and then solve two of its questions, the first from Part A, the second from Part B, okay? Questions of this kind are typical of JEE, so let's begin.

0:51How to use this chapter

So, first, how to use this chapter. It has three parts. Part A is at the JEE Main level, so start there. Okay? Part B is at the JEE Advanced level, so expect harder questions. Right? And Part C is the concept checks. It tests the ideas behind the facts: not only what is true, but why. Clear? Now, the most important line. Attempt each part as a test, and read the solutions only after that. That is how you find out what you really know. With me?

1:25Tool 1: what a disaccharide gives on hydrolysis

Now, tool number one. On screen are the three common disaccharides: maltose and lactose at the top, and sucrose at the bottom. All three have the formula C twelve H twenty two O eleven; the small lowered numbers count the atoms. So all three have the same molar mass, three hundred and forty two. Okay? Maltose, at the top left, is glucose plus glucose, two glucose units joined together. So on hydrolysis, one maltose gives two glucose. Got it? But sucrose, at the bottom, is different: a glucose ring on the left, joined to a fructose ring on the right. Sucrose plus water, H two O, with H plus over the arrow, which means acid, gives glucose plus fructose, one molecule of each. Remember that, because Example one depends on it. With me? And glucose itself is C six H twelve O six, with molar mass one hundred and eighty. Clear? Now a Part C question, from the concept checks. Glucose and fructose, on their own, are both reducing sugars. So why is sucrose, made of the two, a non-reducing sugar? The reason is that both anomeric carbons, carbon one of glucose and carbon two of fructose, are used up in the glycosidic linkage. The figure says it too: C one to C two, both reducing groups used. So neither ring can open to give a free carbonyl group, and sucrose is non-reducing and shows no mutarotation either. Got it?

2:58Tool 2: anomers and mutarotation

Now, tool number two. On the left of this figure is alpha-D-glucose, alpha being the Greek letter, with the O H on carbon one below the ring. Alpha in square brackets, with a small D below it, is the specific rotation, plus one hundred and twelve degrees, and the small circle means degrees. Okay? On the right is beta-D-glucose, with that O H above the ring. Its specific rotation is only plus nineteen degrees. Right? Now dissolve the alpha form in water and wait. Does it stay alpha forever? No, it does not. It opens into the open-chain form, in the box in the middle, and closes again either way. At equilibrium in water, we get thirty six percent alpha and sixty four percent beta. With me? And the equilibrium rotation is plus fifty two point five degrees. This slow change in rotation is called mutarotation. Clear? One more rule, and the course uses it in the Part B solutions. In a mixture, each form adds its fraction times its own rotation; the cross means times. Keep this ready for Example two.

4:07Example 1: mass of sucrose that gives 90 g of glucose

Got it? Now Example one, from Part A, at the JEE Main level. Find the mass of sucrose that gives ninety grams of glucose on complete hydrolysis; the small g after the number means grams. Pause the video and try it first. Done? Okay, let's check. First, the moles of glucose. The letter n, the number of moles, is ninety over one hundred and eighty; the fraction bar means divided by. That gives zero point five mole, and m o l is short for mole. Right? Now, sucrose is a disaccharide, two sugar units in one molecule. So half a mole of glucose needs only a quarter mole of sucrose, right? That is exactly the trap, so be careful here. Two glucose units is maltose. Sucrose gives one glucose and one fructose, so only one glucose for each sucrose, as we saw in tool one. So we need zero point five mole of sucrose. With me? Now the mass. Zero point five times three hundred and forty two gives one hundred and seventy one grams. If you used the quarter mole, you got only half of this answer. Got it?

5:19Example 2: percent alpha glucose changed at +82.2 degrees

Now Example two, from Part B, and it is our puzzle from the start. Read it with me. Pure alpha-D-glucose, plus one hundred and twelve degrees, is dissolved and left to stand. Find the percentage that has changed into the beta form, plus nineteen degrees, when the specific rotation has fallen to plus eighty two point two degrees. Pause the video and try it first. Done? Okay, let's check. The rotation has fallen from one hundred and twelve to eighty two point two, and that is a fall of twenty nine point eight degrees. Okay? So the fraction changed is twenty nine point eight out of one hundred and twelve, which is about twenty seven percent, right? That is exactly the trap from our puzzle, so be careful. The beta form does not rotate zero degrees; it still rotates plus nineteen. So let the letter x stand for the fraction changed. The alpha left, one minus x, gives one hundred and twelve times one minus x, in brackets. The beta formed gives nineteen times x. Together they give eighty two point two. With me? Open the bracket and collect the terms. One hundred and twelve, minus ninety three times x, equals eighty two point two. The double arrow means which gives, so ninety three times x equals twenty nine point eight. Clear? So x is twenty nine point eight over ninety three, which gives about zero point three two; the wavy equals sign means approximately. So thirty two percent has changed, not twenty seven! And there is our missing five percent. The complete change, alpha to beta, lowers the rotation by only ninety three degrees, not one hundred and twelve, so more glucose must change to give the same fall. Got it? One last check, using tool two. Thirty two percent is below sixty four percent, the share of beta at equilibrium. So equilibrium is not yet reached, and the rotation is still falling. That makes sense, right?

7:23Quick revision: disaccharide hydrolysis, mutarotation

So, quick revision. Sucrose gives one glucose and one fructose, while maltose gives two glucose. In water, glucose mutarotates to thirty six percent alpha and sixty four percent beta, at plus fifty two point five degrees. And for a mixture, add fraction times rotation for each form, remembering that beta still rotates plus nineteen degrees. Now open Part A, and take it as a test before you read the solutions. All the best!