Glucose: Preparation and Open-Chain Structure

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Starch is just glucose units, joined in long chains. So eighty one grams of starch should give eighty one grams of glucose, right? Well, you actually get more glucose than starch! Hold that thought. Today we prepare glucose, then crack its open-chain structure from five simple reactions, like detectives. This comes up often in JEE, so let's begin.

0:28Preparing glucose: two routes

So, where does glucose come from? First, from sucrose, cane sugar. We boil it with dilute hydrochloric acid, H C l, or sulphuric acid, H two S O four, in alcoholic solution; the small numbers written low count the atoms. Now tell me, do we get only glucose? We do not; look at the equation. Sucrose, C twelve H twenty two O eleven, takes up one water, and the H plus above the arrow means acid is needed. Out come glucose and fructose, both C six H twelve O six, in equal amounts. Okay? Second, from starch, the commercial method. We boil it with dilute sulphuric acid at three hundred and ninety three kelvin, capital K, under two to three atmospheres, written a t m. Clear? Starch is C six H ten O five in brackets, with a small n outside, the number of units. Each unit takes up one water and gives one glucose. This water is the key to our puzzle. With me?

1:36Clues 1 to 3: the chain and the carbonyl group

Now let us do the detective work. Glucose, also called dextrose, is C six H twelve O six, and capital M, its molar mass, is one hundred and eighty. The picture links each reagent to what it proves; the answer is already in the glucose box. Okay? Clue number one comes from long heating with hydrogen iodide, H I, which gives n-hexane; the arrow means gives, and the slanted n means normal, unbranched. So the six carbons form a straight chain. Got it? In clue number two, glucose forms an oxime with hydroxylamine, N H two O H, and a cyanohydrin with HCN, hydrogen cyanide. So which group gives both these reactions? Yes, exactly, it is the carbonyl group, C double bond O, the two short lines being the double bond. Right? But is it an aldehyde or a ketone? Clue number three decides. Bromine water, B r two water, a mild oxidising agent, gives gluconic acid, still with all six carbons. So which is it? That is correct, it is an aldehyde: an acid with all six carbons means the carbonyl group was at the chain's end. So glucose is an aldohexose: aldo for aldehyde, hexose for six carbons. Clear?

3:01Clues 4 and 5: the OH groups

Clue number four counts the O H groups. Ethanoic anhydride, C H three C O in brackets, with a small two for two such groups, turns glucose into a pentaacetate. Penta means five, so five O H groups. Now, could two of them sit on one carbon? They cannot, because two O H groups on one carbon would lose water and give a carbonyl group. Glucose is stable, so they are on five different carbons. With me? In clue number five, nitric acid, HNO three, oxidises glucose to saccharic acid, a dicarboxylic acid, with an acid group at each end. Okay? Now the clever part. Gluconic acid, from clue three, also gives saccharic acid, and its aldehyde end is already an acid. So what did nitric acid oxidise at the other end? Yes, exactly, it turned the C H two O H group into the second acid group. So one O H group is primary, at the end of the chain. Got it? So the clues give a straight chain: C H two O H at one end, four CHOH groups in brackets, counted by the small four, and the aldehyde group at the other end. That is the open-chain structure, and Fischer worked out the exact arrangement of its O H groups. Clear?

4:27D and L versus (+) and (-)

These are Fischer projections: the vertical line is the carbon chain, and each horizontal line carries one carbon's H and O H groups. D and L describe configuration. Look at C five, the lowest asymmetric carbon: O H on the right means D, on the left means L, okay? Most natural sugars belong to the D series. The plus or minus in brackets is the observed rotation of plane-polarised light: plus for dextro, to the right, minus for laevo, to the left. The first drawing, D-glucose, is plus. So does D always mean plus? It does not, and that is the trap: the two labels are independent. Fructose, on the far right, has its C five O H on the right, so it is D, yet it turns light to the left. So it is D minus fructose. Got it? L-glucose, the second drawing, is the mirror image of D-glucose: on every asymmetric carbon, the H and O H swap sides. With me? Finally, open-chain glucose has four asymmetric carbons, so two to the power four, the small four written high, gives sixteen aldohexoses, and eight are in the D series. Clear?

5:43Example 1: mass of glucose from 81 g of starch

Now Example one, which settles our puzzle. State the conditions for the commercial preparation of glucose from starch, and find the mass of glucose from eighty one grams of starch; g means grams. Pause the video and try it first. Done? Okay, let's check. The conditions are dilute sulphuric acid, three hundred and ninety three kelvin, and two to three atmospheres. Now, starch is a polymer, and n is unknown. So how do we count its moles? We count units, not chains. One unit is C six H ten O five, and the cross means times. Six times twelve, plus ten times one, plus five times sixteen, gives one hundred and sixty two. Right? So eighty one over one hundred and sixty two, where the bar means divided by, gives zero point five mole of units; m o l is mole. Okay? Each unit gives one glucose, so zero point five times one hundred and eighty gives ninety grams. With me? And that solves our puzzle! The extra nine grams is the water taken up, zero point five mole times eighteen. Eighty one plus nine gives ninety, so no mass is created. Got it?

7:03Example 2: mass of pentaacetate from 9 g of glucose

Example two is a little harder, and it uses clue four. Find the molar mass of glucose pentaacetate, and the mass of it formed from nine grams of glucose. Pause the video and try it first. Done? Okay, let's check. In acetylation, the H of each O H group is replaced by an acetyl group, C H three C O, and its molar mass, in brackets, is forty three. So we simply add five times forty three to one hundred and eighty, right? That is the trap, so be careful here: the acetyl group comes in, but the H atom goes out. So each O H group gains forty three minus one, which is forty two. Okay? Five O H groups gain five times forty two, which is two hundred and ten. So capital M is one hundred and eighty plus two hundred and ten, which gives three hundred and ninety grams per mole; the g mol to the power minus one on screen means grams per mole. Right? Now the nine grams of glucose. Nine over one hundred and eighty gives zero point zero five mole. With me? One glucose gives one pentaacetate, so zero point zero five times three hundred and ninety gives nineteen point five grams. That is clue four doing all the work. Clear?

8:30Quick revision: preparation, five clues, D and L vs + and -

So, quick revision. Glucose comes from sucrose or starch, boiled with dilute acid. Hydrogen iodide shows a straight chain, the oxime a carbonyl group, and bromine water an aldehyde. The pentaacetate shows five O H groups, and nitric acid shows that one is primary. D or L comes from C five, and plus or minus from the rotation. Now try the practice problems on this page. All the best!