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SCHOLAR Study Guide SQA CfE Higher Chemistry Unit 2: Nature’s Chemistry Authored by: Emma Maclean Reviewed by: Diane Oldershaw Previously authored by: Peter Johnson Brian T McKerchar Arthur A Sandison Heriot-Watt University Edinburgh EH14 4AS, United Kingdom. First published 2014 by Heriot-Watt University. This edition published in 2014 by Heriot-Watt University SCHOLAR. Copyright © 2014 Heriot-Watt University. Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educational purposes within their establishment providing that no profit accrues at any stage, Any other use of the materials is governed by the general copyright statement that follows. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, without written permission from the publisher. Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the information contained in this study guide. Distributed by Heriot-Watt University. SCHOLAR Study Guide Unit 2: SQA CfE Higher Chemistry 1. SQA CfE Higher Chemistry ISBN 978-1-909633-21-6 Printed and bound in Great Britain by Graphic and Printing Services, Heriot-Watt University, Edinburgh. Acknowledgements Thanks are due to the members of Heriot-Watt University's SCHOLAR team who planned and created these materials, and to the many colleagues who reviewed the content. We would like to acknowledge the assistance of the education authorities, colleges, teachers and students who contributed to the SCHOLAR programme and who evaluated these materials. Grateful acknowledgement is made for permission to use the following material in the SCHOLAR programme: The Scottish Qualifications Authority for permission to use Past Papers assessments. The Scottish Government for financial support. All brand names, product names, logos and related devices are used for identification purposes only and are trademarks, registered trademarks or service marks of their respective holders. i Contents 1 Esters 1.1 Prior knowledge . . . 1.2 Introduction . . . . . . 1.3 Naming and structure 1.4 Drawing esters . . . . 1.5 Making esters . . . . . 1.6 Breaking esters . . . . 1.7 Summary . . . . . . . 1.8 Resources . . . . . . 1.9 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 3 4 7 10 13 15 19 19 20 2 Fats and oils 2.1 Introduction . . . . . . . . . . 2.2 Fat or oil? . . . . . . . . . . . 2.3 A healthy diet? . . . . . . . . 2.4 The structure of fats and oils 2.5 Hydrogenation . . . . . . . . 2.6 Summary exercise . . . . . . 2.7 Summary . . . . . . . . . . . 2.8 Resources . . . . . . . . . . 2.9 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 25 27 30 33 35 36 37 37 38 3 Proteins 3.1 Prior knowledge . . . . 3.2 Introduction . . . . . . . 3.3 The structure of proteins 3.4 Digestion of protein . . . 3.5 Summary . . . . . . . . 3.6 Resources . . . . . . . 3.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 43 44 44 51 56 57 58 4 Chemistry of cooking 4.1 Prior knowledge . . 4.2 Flavour . . . . . . . 4.3 Protein Structures . 4.4 Cooking . . . . . . . 4.5 Summary questions 4.6 Summary . . . . . . 4.7 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 69 70 78 82 85 88 88 . . . . . . . . . . . . . . ii CONTENTS 4.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 5 Oxidation of food 5.1 Prior knowledge . . . . . . . . . . . 5.2 Introduction . . . . . . . . . . . . . . 5.3 Structure of alcohols . . . . . . . . . 5.4 Oxidation of alcohols - practical . . . 5.5 Aldehydes and ketones . . . . . . . 5.6 Oxidation of aldehydes and ketones 5.7 Antioxidants . . . . . . . . . . . . . . 5.8 Oxygen and food . . . . . . . . . . . 5.9 Ion-electron equations . . . . . . . . 5.10 Estimating antioxidant levels . . . . . 5.11 Summary . . . . . . . . . . . . . . . 5.12 Resources . . . . . . . . . . . . . . 5.13 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 94 95 98 100 103 106 109 110 114 115 117 118 119 6 Soaps, detergents and emulsions 6.1 Prior knowledge . . . . . . . . 6.2 Soap . . . . . . . . . . . . . . . 6.3 Detergents . . . . . . . . . . . 6.4 Emulsions . . . . . . . . . . . . 6.5 Summary . . . . . . . . . . . . 6.6 Resources . . . . . . . . . . . 6.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 128 128 137 139 143 144 145 7 Fragrances 7.1 Prior knowledge 7.2 Essential oils . . 7.3 Terpenes . . . . 7.4 Summary . . . . 7.5 Resources . . . 7.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 149 150 153 160 160 161 . . . . . . . 163 165 166 166 173 179 179 180 . . . . . . . . . . . . . . . . . . . . . . . . 8 Skin care 8.1 Prior knowledge . . . . 8.2 Introduction . . . . . . . 8.3 Ultraviolet (UV) radiation 8.4 Free radical reactions . 8.5 Summary . . . . . . . . 8.6 Resources . . . . . . . 8.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 End of unit test 183 Glossary 190 Answers to questions and activities 1 Esters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Fats and oils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Proteins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 192 196 199 © H ERIOT-WATT U NIVERSITY CONTENTS 4 5 6 7 8 9 Chemistry of cooking . . . . . . . Oxidation of food . . . . . . . . . . Soaps, detergents and emulsions Fragrances . . . . . . . . . . . . . Skin care . . . . . . . . . . . . . . End of unit test . . . . . . . . . . . © H ERIOT-WATT U NIVERSITY iii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 207 215 218 221 223 1 Topic 1 Esters Contents 1.1 Prior knowledge . . . . . . . . . . . . 1.2 Introduction . . . . . . . . . . . . . . . 1.2.1 Uses for esters . . . . . . . . . 1.3 Naming and structure . . . . . . . . . 1.3.1 Naming and structure of esters 1.4 Drawing esters . . . . . . . . . . . . . 1.5 Making esters . . . . . . . . . . . . . . 1.5.1 Esterification . . . . . . . . . . 1.6 Breaking esters . . . . . . . . . . . . . 1.6.1 Identifying esters . . . . . . . . 1.7 Summary . . . . . . . . . . . . . . . . 1.8 Resources . . . . . . . . . . . . . . . 1.9 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 4 5 7 9 10 13 14 15 17 19 19 20 Prerequisite knowledge Before you begin this topic, you should already know or be able to do the following: • describe what is meant by the terms hydrocarbon, homologous series, saturated and unsaturated (National 5, Unit 2); • name and draw structural formulae for straight chain alkanes (C 1 - C8 ), alkenes (C2 - C6 ) and cycloalkanes (C3 - C6 ) (National 5, Unit 2); • use, appropriately, molecular formulae and full and shortened structural formulae (National 5, Unit 2); • identify and draw structural formulae for isomer. (National 5, Unit 2); • an ester can be made by reacting a carboxylic acid and an alcohol (National 5, Unit 2); • some uses of esters are in food flavouring, industrial solvents, fragrances and materials (National 5, Unit 2); • carboxylic acids can be identified by the carboxyl ending, the COOH functional group and the ‘-oic’ name ending (National 5, Unit 2); 2 TOPIC 1. ESTERS • straight-chained carboxylic acids can be identified and named from the structural formulae. Given the name of straight chained carboxylic acid the structural formulae can be drawn (National 5, Unit 2); • an alcohol is identified from the -OH group and the ending ‘-ol’ (National 5, Unit 2); • straight chain alcohols are named from the structure formulae. Given the names of straight-chain alcohols structural and molecular formulae can be written. (National 5, Unit 2). Learning Objectives At the end of this topic, you should know that: • an ester can be identified from the ester group and by the name containing the -‘yl-oate’ endings; • an ester can be named given the names of the parent carboxylic acid and alcohol or from structural formulae; • structural formulae for esters can be drawn given the names of the parent alcohol and carboxylic acid or the names of esters; • esters have characteristic smells and are used as flavourings and fragrances; • esters are also used as industrial solvents; • esters are formed by the condensation reaction between carboxylic acid and an alcohol. The ester link is formed by the reaction of a hydroxyl group and the carboxyl group; • in condensation reactions, the molecules join together with the elimination of a small molecule, in this case water; • esters can be hydrolysed to produce a carboxylic acid and alcohol; • given the name of an ester or its structural formula, the hydrolysis products can be named and their structural formulae drawn; • the parent carboxylic acid and the parent alcohol can be obtained by hydrolysis of an ester; • in a hydrolysis reaction, a molecule reacts with water breaking down into smaller molecules. © H ERIOT-WATT U NIVERSITY TOPIC 1. ESTERS 1.1 3 Prior knowledge Test your prior knowledge Q1: What is the name of the following alcohol? a) b) c) d) Methanol Ethanol Propanol Butanol .......................................... Q2: What is the name of the following carboxylic acid? a) b) c) d) Methanoic acid Ethanoic acid Propanoic acid Butanoic acid .......................................... Q3: Which of the following is a common use of esters? a) b) c) d) Fuels Vinegar Flavourings Ammonia .......................................... © H ERIOT-WATT U NIVERSITY 4 TOPIC 1. ESTERS 1.2 Introduction What do the following all have in common? • The fragrance of flowering jasmine • Nail varnish • The sex life of the melon fly • Chopping a pineapple • A lump of butter They all involve esters. The esters contained in butter will be considered in a later topic. Fragrances generally contain complex mixtures of chemicals. The fragrances associated with many fruits and flowers contain simple esters. As well as being sweetsmelling, these esters are sufficiently volatile to allow molecules to reach the receptors in the nose and trigger a response in the brain. Esters in fruit smells This image shows some common fruits and important esters found in their aromas. © H ERIOT-WATT U NIVERSITY TOPIC 1. ESTERS 5 Structures of esters found in fruit aromas Q4: Name the ester found in the smell of pineapple. .......................................... Q5: Name the ester found in the smell of pear. .......................................... 1.2.1 Uses for esters Without our sense of smell, we find it impossible to taste our food properly. This important feature of our enjoyment of food is considered in more detail in a later topic. Our senses of smell and taste are inextricably linked. Consequently, fragrant esters such as ethyl hexanoate and ethyl benzoate are often used as artificial food flavourings (see figure below). © H ERIOT-WATT U NIVERSITY 6 TOPIC 1. ESTERS Artificial food flavourings Ethyl benzoate is one of many esters used in perfumery, although the exact ingredients of most perfumes are well kept secrets. Humans are by no means the only animals to use scent to attract a mate. The male melon fly releases an ester (see figure below) which attracts the female. Melon fly pheromone Chemicals released by one animal to attract others or to warn others are called pheromones. Esters have other uses in cosmetics. Ethyl ethanoate and ethyl butanoate are commonly used as solvents for nail varnish. They dissolve non-polar solutes and evaporate easily, leaving a layer of the solute on the nail. For similar reasons, the same solvents are also used in car body paints, enamel paints and some glues. © H ERIOT-WATT U NIVERSITY TOPIC 1. ESTERS 7 More uses for esters .......................................... Key Point Esters have many uses including flavourings, perfumes and solvents. 1.3 Naming and structure The name of an ester is derived from the name of the parent carboxylic acid and parent alcohol. © H ERIOT-WATT U NIVERSITY 8 TOPIC 1. ESTERS Naming an ester .......................................... 1. Starting from the names of the parent acid and alcohol The name of an ester always consists of two parts. The first part comes from the name of the parent alcohol. Remove the '-anol' name ending and replace with '-yl'. Name of the parent alcohol Q6: What would be the first part of the name of an ester formed from methanol? .......................................... Q7: What would be the first part of the name of an ester formed from butanol? .......................................... The second part comes from the name of the parent carboxylic acid. Remove the '-ic acid' name ending and replace with '-ate'. Name of the parent acid Q8: What will be the second part of the name of an ester formed from ethanoic acid? .......................................... Q9: What will be the second part of the name of an ester formed from methanoic acid? .......................................... 2. Starting from the structural formula of the ester The formula is first broken into two fragments. © H ERIOT-WATT U NIVERSITY TOPIC 1. ESTERS 9 Structural formula of the ester The diagram shows full and shortened structural formulae for the same ester. The bond being broken is the C-O bond which was formed when the ester was made. The fragment containing the carbonyl group arises from the parent carboxylic acid. Q10: Name the parent acid. .......................................... The other fragment arises from the parent alcohol. Q11: Name the parent alcohol. .......................................... Having identified the parent acid and alcohol, the ester can be named using the procedure described earlier. Q12: What is the name of the ester in the above diagram? .......................................... 1.3.1 Naming and structure of esters Naming and structure of esters Q13: Name the ester formed by the condensation of ethanol with butanoic acid. .......................................... © H ERIOT-WATT U NIVERSITY 10 TOPIC 1. ESTERS Q14: Name the ester. .......................................... Q15: An ester found in the smell of pears is formed by the condensation of ethanoic acid with propanol. Name the ester. .......................................... Q16: Name the ester. HCOOCH3 .......................................... Q17: Pentanol and ethanoic acid react to form an ester which is used in flavouring for pear drops. Name the ester.. .......................................... Q18: Name the ester. .......................................... Key Point Esters can be identified from the ester functional group and from the '-oate' name ending. The name of an ester is derived from the names of the parent carboxylic acid and alcohol. 1.4 Drawing esters It is also important to be able to draw a structural formula given the name of an ester. © H ERIOT-WATT U NIVERSITY TOPIC 1. ESTERS 11 Drawing structural formulae for esters 1. Remember that the first part of the name is derived from the parent alcohol and the second part from the parent acid. Parent alcohol and parent acid 2. Draw full structural formulae for the acid and alcohol in such a way that the functional groups are adjacent. The diagram shows the acid to the left of the alcohol. The parent acid and alcohol 3. Remove the OH group from the acid and H from the alcohol as shown to form water. Joining up the remaining fragments gives the full structural formula of the ester. © H ERIOT-WATT U NIVERSITY 12 TOPIC 1. ESTERS Full structural formula of the ester 4. Note that there are other ways to draw structural formulae to represent the same ester, all of which are equally correct. Alternative structural formulae for the same ester .......................................... Drawing ester structures Q19: Draw the full structural formula for propyl butanoate. .......................................... Q20: Draw the full structural formula for the ester formed by the condensation of ethanoic acid and methanol. © H ERIOT-WATT U NIVERSITY TOPIC 1. ESTERS 13 .......................................... Q21: Draw a shortened structural formula for butyl methanoate. .......................................... Q22: Draw a shortened structural formula for the ester formed by the condensation of propanol and ethanoic acid. .......................................... Key Point Shortened and full structural formulae for esters can be drawn given the names of the parent acid and alcohol or the name of the ester. .......................................... 1.5 Making esters An important reaction of carboxylic acids is with alcohols to form compounds called esters. In this topic, we will concentrate on how esters are formed. How they are broken down is considered later. The reaction is carried out by mixing the acid and alcohol, adding a few drops of concentrated sulfuric acid and heating the solution for 10-15 minutes. The concentrated sulfuric acid acts as a catalyst. Making an ester © H ERIOT-WATT U NIVERSITY 14 TOPIC 1. ESTERS Q23: Look at the diagram above. Suggest why a paper towel soaked in cold water is wrapped round the thin glass tube. .......................................... After heating for several minutes, the contents are poured into sodium hydrogencarbonate solution. Look at diagram B. There is effervescence and a layer of ester, which is immiscible with water, forms on top. The ester can be recognised by its distinctive smell. Most simple esters have a sweet fruity smell. Q24: Sodium hydrogencarbonate solution is also used to counteract acid indigestion. What type of solution is sodium hydrogencarbonate solution? a) Acidic b) Neutral c) Basic .......................................... Q25: Suggest why a solution of sodium hydrogencarbonate solution was used rather than water. .......................................... Q26: Which gas is responsible for the effervescence? .......................................... .......................................... 1.5.1 Esterification Esterification The formation of an ester Q27: Apart from the ester, which other substance is produced? .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 1. ESTERS 15 This reaction is described as a condensation reaction, in which two molecules combine to form a larger molecule at the same time eliminating a small molecule such as water. In this case, the reaction between the hydroxyl group and the carboxyl group produces an ester link. Condensation reaction Key Point Esters are formed by the condensation reaction between a carboxylic acid and an alcohol. The ester link is formed by the reaction of a carboxyl group with a hydroxyl group. .......................................... 1.6 Breaking esters Esters are formed by a condensation reaction which is reversible. The reverse reaction is known as hydrolysis. The ester is broken down to reform the parent carboxylic acid and alcohol by reaction with water. This reaction can be catalysed either by acid or alkali. If alkali is used, the salt of the carboxylic acid, a carboxylate, is produced. © H ERIOT-WATT U NIVERSITY 16 TOPIC 1. ESTERS Hydrolysis of an ester .......................................... Identifying the parent acid and alcohol From the structure of the ester, it is possible to identify the parent acid and alcohol. Q28: In the diagram above, which acid was used to make the ester? .......................................... Q29: Which alcohol was used? .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 1. ESTERS 1.6.1 17 Identifying esters Identifying esters Using the grid above answer the following questions. Q30: Name the parent acid and alcohol used to make the ester A. .......................................... Q31: Name the parent acid and alcohol used to make the ester B. .......................................... Q32: Name the parent acid and alcohol used to make the ester C. .......................................... Q33: Name the parent acid and alcohol used to make the ester D. .......................................... Q34: Name the parent acid and alcohol used to make the ester E. .......................................... Q35: Name the parent acid and alcohol used to make the ester F. .......................................... © H ERIOT-WATT U NIVERSITY 18 TOPIC 1. ESTERS Q36: Name the parent acid and alcohol used to make the ester G. .......................................... Q37: Name the parent acid and alcohol used to make the ester H. .......................................... Q38: Name the parent acid and alcohol used to make the ester I. .......................................... Q39: Name the parent acid and alcohol used to make the ester J. .......................................... Key Point The formation and hydrolysis of an ester is a reversible reaction. The parent carboxylic acid and the parent alcohol can be obtained by hydrolysis of an ester. .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 1. ESTERS 1.7 Summary Summary Esters: • An ester can be identified from the ester group and by the name containing the -‘yl-oate’ endings. • An ester can be named given the names of the parent carboxylic acid and alcohol or from structural formulae. • Structural formulae for esters can be drawn given the names of the parent alcohol and carboxylic acid or the names of esters. • Esters have characteristic smells and are used as flavourings and fragrances. • Esters are also used as industrial solvents. Making & breaking Esters: • Esters are formed by the condensation reaction between carboxylic acid and an alcohol. The ester link is formed by the reaction of a hydroxyl group and the carboxyl group. • In condensation reactions, the molecules join together with the elimination of a small molecule, in this case water. • Esters can be hydrolysed to produce a carboxylic acid and alcohol. • Given the name of an ester or its structural formula, the hydrolysis products can be named and their structural formulae drawn. • The parent carboxylic acid and the parent alcohol can be obtained by hydrolysis of an ester. • In a hydrolysis reaction, a molecule reacts with water breaking down into smaller molecules. 1.8 Resources Texts Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson, ISBN 978-1444167528 © H ERIOT-WATT U NIVERSITY 19 20 TOPIC 1. ESTERS 1.9 End of topic test End of topic 1 test Q40: Rum flavouring is based on the compound shown below. It can be made from: a) b) c) d) ethanol and butanoic acid. butanol and ethanoic acid. propanol and propanoic acid. propanol and ethanoic acid. .......................................... Q41: Identify the ester from the structures below. a) b) c) d) .......................................... Q42: Which of the following is an ester? a) b) c) d) Ethylamine Ethyl chloride Ethyl ethanoate Ethylbenzene .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 1. ESTERS 21 Q43: Phenylethanol has the structure shown below. It has a smooth rose-like odour and is used in perfumes together with its propanoate ester. Which of the following represents the propanoate ester of phenylethanol? a) b) c) d) .......................................... Q44: Apart from perfumes, name one other use for esters. .......................................... © H ERIOT-WATT U NIVERSITY 22 TOPIC 1. ESTERS Q45: Name the ester formed when butanoic acid reacts with methanol. .......................................... Q46: What type of reaction is taking place? .......................................... Q47: An alcohol and acid, when warmed with a few drops of concentrated sulfuric acid, react to form the compound below. Write the systematic name for the acid. .......................................... Q48: Write the systematic name for the alcohol. .......................................... © H ERIOT-WATT U NIVERSITY 23 Topic 2 Fats and oils Contents 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Fat or oil? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 27 2.2.1 Testing with bromine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 A healthy diet? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 30 2.4 The structure of fats and oils . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.4.1 Hydrolysis of fats and oils . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Hydrogenation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 35 2.6 Summary exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 37 2.8 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 38 Prerequisite knowledge Before you begin this topic, you should already know that: • an ester can be identified from the ester group and by the name containing the -‘yl-oate’ endings; • an ester can be named given the names of the parent carboxylic acid and alcohol or from structural formulae; • structural formulae for esters can be drawn given the names of the parent alcohol and carboxylic acid or the names of esters; • esters have characteristic smells and are used as flavourings and fragrances; • esters are also used as industrial solvents; • esters are formed by the condensation reaction between carboxylic acid and an alcohol. The ester link is formed by the reaction of a hydroxyl group and the carboxyl group; • in condensation reactions, the molecules join together with the elimination of a small molecule, in this case water; • esters can be hydrolysed to produce a carboxylic acid and alcohol; 24 TOPIC 2. FATS AND OILS • given the name of an ester or its structural formula, the hydrolysis products can be named and their structural formulae drawn; • the parent carboxylic acid and the parent alcohol can be obtained by hydrolysis of an ester; • in a hydrolysis reaction, a molecule reacts with water breaking down into smaller molecules. Learning Objectives At the end of this topic, you should know that: • fats and oils are a concentrated source of energy; • they are essential for the transport and storage of fat-soluble vitamins in the body; • fats and oils are esters formed from the condensation of glycerol (propane-1,2,3triol) and three carboxylic acid molecules; • the carboxylic acids are known as ‘fatty acids’ and are saturated or unsaturated straight-chain carboxylic acids, usually with long chains of carbon atoms; • the lower melting points of oils compared to those of fats is related to the higher degree of unsaturation of oil molecules; • the low melting points of oils are a result of the effect that the shapes of the molecules have on close packing, hence on the strength of van der Waals’ forces of attraction. © H ERIOT-WATT U NIVERSITY TOPIC 2. FATS AND OILS 2.1 25 Introduction The fats and oils which we commonly find in our foodstuffs can be classified according to their origin as animal, vegetable or marine. Fats and Oils Vegetable Animal Lard (Pig) Dripping (Beef) Mutton Olive Sunflower Soya bean Corn Marine Cod liver Whale Sardine Fats and oils Venn diagram At room temperature fats are solids and oils are liquids, but fats can be made liquid by heating (think about the butter on your toast!) and oils can be made into solid fats by cooling. There are similarities in their chemical structures. In fact both fats and oils are esters. As nutrients the fats and oils supply the body with concentrated food energy at 37.6 kJ per gram (actually 9 kcal or 9 "food calories"), which is about 2.25 times as much as the energy content of carbohydrates. Animal, vegetable and marine fats are all part of energy sources in the diet. Processed foods carry a label revealing the energy quota for each nutrient. © H ERIOT-WATT U NIVERSITY 26 TOPIC 2. FATS AND OILS Animal, vegetable and marine fats Food label with traffic light colour coding The recent types of food labelling There is an increasing concern that the public should be aware that are several health issues concerned with foods. For example, we should not eat too much salt, sugar and saturated fats. All these have been associated with chronic conditions, for example, heart and circulatory problems, obesity and diabetes. The UK Food Standards Agency and this site in the NHS have guidance on labelling. There are two basic systems: • One lists the GDAs (recommended guideline daily amounts for adults) for sugars, fats and salt; • The other is a 'traffic light' colour coding indicating whether the amount (often per serving) is low (green), medium (amber) or high (red). Although less precise, this colour guide gives information 'at a glance' - the greener the indicators, the healthier the food. You should now see these on many packaged foods. Key Point Fats and oils from living organisms can be classified according to origin as animal, vegetable or marine. They are all ester molecules which supply energy in the diet in a more concentrated form than carbohydrates. © H ERIOT-WATT U NIVERSITY TOPIC 2. FATS AND OILS 2.2 27 Fat or oil? The ester molecules which make up fats and oils differ from each other in melting point. This suggests that although the chemical structures are similar, there must be some difference causing the trend in properties. The main factor is the amount of unsaturation. In general oils have more carbon to carbon double bonds per molecule than fat molecules of a similar molecular size and are therefore said to be unsaturated. If there is more than one double bond the term polyunsaturated is used. Percentage by weight Oil or fat Saturated Mono-unsaturated Poly-unsaturated Beef dripping 52 44 4 Olive 13 77 9 Sunflower 11 20 69 Lard 41 47 12 Animal and vegetable oils and fats Q1: Which of the oils or fats in the above table has the most carbon to carbon double bonds? .......................................... Q2: Which of the oils or fats in the above table is likely to have the lowest melting point? .......................................... The molecules which make up fats and oils have three long hydrocarbon chains on each molecule and are triesters. The chains can be similar or different and can be saturated or unsaturated. Packing in saturates If the hydrocarbon chains are saturated, the triester has a structure which allows close packing of the molecules, a large amount of intermolecular attraction and a higher melting point. Fats tend to have a higher amount of saturation than oils. © H ERIOT-WATT U NIVERSITY 28 TOPIC 2. FATS AND OILS When a hydrocarbon chain is unsaturated, the double bonds present cause the chain to bend. Packing in unsaturates The presence of carbon to carbon double bonds in hydrocarbon chains introduces kinking to the structure, making it more difficult to nestle close together. Fewer intermolecular attractions lead to lower melting points. Oils tend to have a higher amount of unsaturation. Q3: Which of the above two diagrams is most likely to represent olive oil? a) Packing in saturates b) Packing in unsaturates .......................................... 2.2.1 Testing with bromine Testing with bromine One way to measure the amount of unsaturation is to use the bromine test for alkene double bonds. The more bromine solution that an oil or fat can decolourise, the more unsaturated it must be. Bromination of an alkene This activity consists of a simulation showing the titration of different fats and oils with bromine solution. The amount of unsaturation is thus measured and compared with the melting points of the samples. © H ERIOT-WATT U NIVERSITY TOPIC 2. FATS AND OILS The diagram shows the apparatus used and a table of results. Bromine titration apparatus and results Q4: Which fat or oil decolourised the largest volume of bromine solution? .......................................... Q5: Which fat or oil has the most carbon to carbon double bonds? .......................................... Q6: Which fat or oil is best described as polyunsaturated? .......................................... Q7: Look at the table showing the state at room temperature and decide which of these statements is true. a) The lower melting point of oils compared to fats is caused by the lower unsaturation of oil molecules. b) The lower melting point of oils compared to fats is caused by the higher unsaturation of oil molecules. c) The higher melting point of fats compared to oils is caused by the higher unsaturation of fat molecules. d) The lower melting point of fats compared to oils is caused by the higher unsaturation of fat molecules. .......................................... .......................................... © H ERIOT-WATT U NIVERSITY 29 30 TOPIC 2. FATS AND OILS Key Point Bromine solution can be used to test fats and oils for the degree of unsaturation. The higher the unsaturation levels the lower the melting point. The lower melting point of oils compared to fats is a result of the higher unsaturation of oil molecules. This affects the shape of the molecules and the extent of close packing, hence the strength of the London dispersion forces of attraction. 2.3 A healthy diet? The most important function of fats and oils in a healthy diet is to provide energy in the body. A limited amount of energy can be stored as carbohydrate in the form of glycogen. Surplus energy is mainly stored as fat. Becoming excessively overweight is called becoming obese. This can lead to high blood pressure and diabetes. A high intake of saturated fat pushes up the cholesterol level in the blood and fatty deposits (atheroma) in the arteries can lead to blockage and heart disease. Polyunsaturated fats found in fish and vegetable oils are considered to be less harmful. Omega-3 and omega-6 fatty acids Studies carried out in the 1970s on the Greenland Inuit tribe showed that despite consuming large quantities of fats from sea food there was very little incidence of cardiovascular disease. This was attributed to the high levels of omega-3 fatty acids in the diet. Omega-3, or ω-3, fatty acids have a double bond between the carbon atoms 3 and 4 from the methyl terminal carbon (the ω-carbon). (Note that this is different from the chemists' normal numbering of carbon atoms which start from the functional group.) An omega-3 fatty acid that is essential in the diet is alpha-linolenic acid. Alpha-linolenic acid Two omega-3 and omega-6 polyunsaturated fatty acids that are often quoted in dietary capsules are: eicosapentenoic and docosahexenoic acids. You can see why they are usually abbreviated to EPA and DHA respectively! The structures are shown below. You do not need to memorise these - just note that they have many double bonds, including those at carbons ω-3 and ω-6. © H ERIOT-WATT U NIVERSITY TOPIC 2. FATS AND OILS 31 Eicosapentenoic acid (EPA) Docosahexenoic acid (DHA) Omega and fish oil with EPA and DHA Fat-soluble vitamins A healthy diet requires a source of many different vitamins. Vitamins A and D are both found in nature dissolved in the fats and oils contained in butter and cheese. Margarines do not have these vitamins and are added to the product by the manufacturer. This helps prevent diseases like 'rickets' caused by vitamin deficiency. Fats and oils are thus an essential nutrient in a balanced diet. Polyunsaturated fish and vegetable oils are considered to be less harmful than saturated oils and fats. Margarines have vitamins added to them to prevent a deficiency in the diet. © H ERIOT-WATT U NIVERSITY 32 TOPIC 2. FATS AND OILS Polyunsaturated fish and vegetable oils Margarine ingredients Healthy diet exercise Q8: Fill in the blanks from the word bank. .......................................... Key Point Fats and oils are important in a balanced diet and supply the body with energy in a more concentrated form than carbohydrates. There is evidence of a link between a high intake of saturated fat in the diet and heart disease. © H ERIOT-WATT U NIVERSITY TOPIC 2. FATS AND OILS 2.4 33 The structure of fats and oils The molecules which make up fats and oils are all triesters of an alcohol called glycerol (propane-1,2,3-triol). Because glycerol has three alcohol groups per molecule it is a trihydric alcohol and three carboxylic acid molecules can form ester links with each glycerol giving a range of triglycerides. For example, glycerol tristearate is used in cocoa butter. Glycerol Glycerol tristearate The three carboxylic acids are sometimes called fatty acids and these may or may not be identical. In all the common fats and oils the fatty acids, both saturated and unsaturated, are straight chain carboxylic acids containing even numbers of carbon atoms ranging from C 4 to C24 , primarily C16 and C18 . Identical and non-identical fatty acids © H ERIOT-WATT U NIVERSITY 34 TOPIC 2. FATS AND OILS Evidence for the structure of oils and fats can be obtained when an oil or fat is hydrolysed by treatment with superheated steam as illustrated in the next activity. 2.4.1 Hydrolysis of fats and oils Hydrolysis of fats and oils Use the diagram to study the reaction and answer the questions. Hydrolysis of a triglyceride Q9: Which functional group on the oil or fat is being broken in the hydrolysis reaction? .......................................... Q10: How many molecules of fatty acid are produced for every one molecule of glycerol? .......................................... Q11: The number of carbon atoms in the fatty acids produced by hydrolysis is always: a) odd. b) even. c) the same. .......................................... Q12: The fatty acid molecules produced by hydrolysis: a) are always the same. b) are always different. c) may or may not be identical. .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 2. FATS AND OILS 35 Key Point The hydrolysis of triglycerides produces one molecule of glycerol (a trihydric alcohol) and three molecules of fatty acids which can be identical to or different from each other. The fatty acids produced can be saturated or unsaturated and always contain even numbers of carbon atoms C 4 to C24 , primarily C16 and C18 . .......................................... 2.5 Hydrogenation Vegetable oils are generally less expensive and are believed to be a healthier part of the diet than animal fats (like butter). Because vegetable oils are liquid few people would want to use them poured onto bread as an alternative to butter. Simple addition of hydrogen to the double bonds of a vegetable oil can make the liquid more solid, and more acceptable as a spread. Controlled hydrogenation produces spreads which can be used straight from the refrigerator. Hydrogenation of a double bond Hydrogenation of oils © H ERIOT-WATT U NIVERSITY 36 TOPIC 2. FATS AND OILS Reducing the unsaturation like this is called hardening by hydrogenation and is carried out with hydrogen using a nickel catalyst. Too much hydrogenation would cause the fat to become too hard to spread and the process is controlled to produce margarines which spread easily. Sunflower and olive oil are used extensively to produce polyunsaturated margarines and these are able to be described as polyunsaturated because the fatty acids still contain many alkene double bonds. Peanut oil can be partially hydrogenated to give peanut butter. Key Point The conversion of oils into hardened fats involves the partial removal of unsaturation by the addition of hydrogen. 2.6 Summary exercise Summary exercise Q13: Fill in the blanks from the word bank. .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 2. FATS AND OILS 2.7 Summary Summary Fats and oils • Fats and oils are a concentrated source of energy. • Fats and oils can be classified as animal, vegetable or marine. • Fats and oils are important in a balanced diet and supply the body with energy in a more concentrated form than carbohydrates. • There is evidence of a link between a high intake of saturated fat in the diet and heart disease. • Fats and oils are essential for the transport and storage of fat-soluble vitamins in the body. • The lower melting points of oils compared to those of fats is related to the higher degree of unsaturation of oil molecules. • The low melting points of oils are a result of the effect that the shapes of the molecules have on close packing, hence on the strength of van der Waals’ forces of attraction. • Fats and oils are esters formed from the condensation of glycerol (propane1,2,3-triol) and three carboxylic acid molecules. • The carboxylic acids are known as ‘fatty acids’ and are saturated or unsaturated straight-chain carboxylic acids, usually with long chains of carbon atoms. • Bromine solution can be used to test fats and oils for the degree of unsaturation. The higher the unsaturation levels the lower the melting point. • The hydrolysis of triglycerides produces one molecule of glycerol (a trihydric alcohol) and three molecules of fatty acids which can be identical to or different from each other. • The fatty acids produced can be saturated or unsaturated and always contain even numbers of carbon atoms C 4 to C24 , primarily C16 and C18 . • The conversion of oils into hardened fats involves the partial removal of unsaturation by the addition of hydrogen. 2.8 Resources Texts Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson, ISBN 978-1444167528 © H ERIOT-WATT U NIVERSITY 37 38 TOPIC 2. FATS AND OILS 2.9 End of topic test End of topic 2 test Q14: Fats and oils can be classified as: a) b) c) d) esters. polyesters. fatty acids. soaps. .......................................... Q15: Fats have higher melting points than oils because: a) b) c) d) fat molecules are more loosely packed. fats have more cross-links between molecules. fats have more hydrogen bonds. fat molecules are more saturated. .......................................... Q16: Liquid fats can be made solid by: a) b) c) d) hydrolysis. hydrogenation. hydration. dehydration. .......................................... Q17: Which of the following decolourises bromine solution least rapidly? a) b) c) d) Hex-1-ene Cod liver oil Mutton fat Palm oil .......................................... Q18: Which of the following is likely to be produced by hydrolysis of a fat? a) b) c) d) Ethanol Glycerol Glucose Ethanoic acid .......................................... Q19: When oil is converted into a fat which bonds are broken? a) Carbonyl bonds b) Hydroxyl bonds c) Ester bonds © H ERIOT-WATT U NIVERSITY TOPIC 2. FATS AND OILS d) Double bonds C C .......................................... Q20: In an experiment, more bromine solution is required to react completely with oil X than with oil Y. Compared to oil Y, oil X is likely to have: a) b) c) d) a higher melting point and more alkene double bonds. a higher melting points and less alkene double bonds. a lower melting point and less alkene double bonds. a lower melting point and more alkene double bonds. .......................................... Q21: Identify the name which could be applied to reaction Y. a) b) c) d) e) f) Hydration Addition Hydrolysis Oxidation Hydrogenation Condensation .......................................... Q22: Identify the two names from the above list which could be applied to reaction X. .......................................... Q23: Identify the true statements about fats and oils. a) Fats and oils in the diet can supply the body with energy. b) Fats are likely to have low melting points compared to oils. c) Fats are likely to have a higher degree of unsaturation than oils. d) Molecules in fats are packed more closely together than in oils. e) Fats and oils are a less concentrated source of energy than carbohydrates. .......................................... © H ERIOT-WATT U NIVERSITY 39 40 TOPIC 2. FATS AND OILS Q24: Identify the compound with the same functional group as a fat molecule. c) a) b) O CH3 C OC2H5 d) f) e) .......................................... Q25: Identify the two compounds from the six above which could be produced by the hydrolysis of an oil. .......................................... Q26: What name is given to this type of reaction? .......................................... Q27: How many moles of hydrogen are needed to react with one mole of reactant A? .......................................... Q28: Which two fatty acids would you notinclude in an attempt to illustrate that the melting point increases with chain length? a) Stearic: C17 H35 COOH b) Arachidonic: C19 H39 COOH c) Palmitoleic: C15 H29 COOH d) Linoleic: C17 H31 COOH e) Behenic: C21 H43 COOH f) Myristic: C13 H27 COOH .......................................... © H ERIOT-WATT U NIVERSITY 41 Topic 3 Proteins Contents 3.1 Prior knowledge . . . . . . . . . . . . 3.2 Introduction . . . . . . . . . . . . . . . 3.3 The structure of proteins . . . . . . . . 3.3.1 Amino acids and peptide links 3.3.2 Essential amino acids. . . . . . 3.4 Digestion of protein . . . . . . . . . . . 3.4.1 Identifying the amino acids . . 3.5 Summary . . . . . . . . . . . . . . . . 3.6 Resources . . . . . . . . . . . . . . . 3.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 44 44 45 49 51 54 56 57 58 Prerequisite knowledge You should already be able to: • describe what is meant by the terms hydrocarbon, homologous series, saturated and unsaturated (National 5, Unit 2); • explain how condensation polymers are made and identify the monomers from a section of the polymer and vice versa (National 5, Unit 2); • identify, and understand the chemistry of alcohols (National 5, Unit 2); • identify, and understand the chemistry of carboxylic acids (National 5, Unit 2); • identify, and understand the chemistry of esters (Higher, Unit 2, Topic 1). Learning Objectives By the end of this topic, you should be able to state that: • nitrogen is essential for protein formation by plants and animals. • proteins are the major structural materials of animal tissue; • proteins are also involved in the maintenance and regulation of life processes; • enzymes are proteins; 42 TOPIC 3. PROTEINS • amino acids, the building blocks from which proteins are formed, are relatively small molecules which all contain an amino group (NH 2 ), and a carboxyl group (COOH); • the body cannot make all the amino acids required for body proteins and is dependent on dietary protein for supply of certain amino acids known as essential amino acids; • proteins are made of many amino acid molecules linked together by condensation reactions; • in these condensation reactions, the amino group on one amino acid and the carboxyl group on a neighbouring amino acid join together, with the elimination of water; • the link which forms between the two amino acids can be recognised as an amide link (CONH) also known as the peptide link; • proteins which fulfil different roles in the body are formed by linking differing sequences of amino acids together; • during digestion, enzyme hydrolysis of dietary proteins can produce amino acids; • the structural formulae of amino acids obtained from the hydrolysis of proteins can be identified from the structure of a section of the protein. © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 3.1 43 Prior knowledge Test your prior knowledge Q1: What is the name of the following alcohol? a) b) c) d) Methanol Ethanol Propanol Butanol .......................................... Q2: What is the name of the following carboxylic acid? a) b) c) d) Methanoic acid Ethanoic acid Propanoic acid Butanoic acid .......................................... Q3: What is the name of the following ester? a) b) c) d) Methyl Ethanoate Ethyl Methanoate Methyl Methanoate Ethyl Ethanoate .......................................... © H ERIOT-WATT U NIVERSITY 44 TOPIC 3. PROTEINS 3.2 Introduction Proteins are involved in a great variety of essential functions in both plants and animals. They are naturally occurring condensation polymers which contain the elements carbon, hydrogen, oxygen and nitrogen. Proteins make up around 18% of the human body and are found in muscle tissue, hair and skin. Silk, feathers and scales are all made from proteins and of course, they are familiar to us in foodstuffs like meat, fish, eggs, peas, beans and wheat. The key element in protein formation is nitrogen. Proteins Only plants can make proteins from simple nitrogen-containing compounds taken in from the soil. Animals cannot do this and are dependent on eating plants or other animals as food to provide a protein source. This protein can then be digested and converted into the animals' own proteins. Foodstuff with an animal origin such as fish, eggs, meat and cheese are rich protein sources. Beans and pulses such as lentils are protein sources used in vegetarian diets. Key Point Nitrogen is essential for protein formation by plants and animals. 3.3 The structure of proteins Proteins are an essential part of every living cell. They are condensation polymers and the body requires specific proteins such as hormones and enzymes to perform specific tasks. Providing a complete protein diet to all of the world's population is a major health issue. Body builders wishing to bulk up their muscle protein sometimes add protein supplements to their diet. © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 45 Protein supplements This section deals with the monomer units which go to make the polymers, looks at how they are assembled and considers the diet necessary to provide an adequate healthy supply. In most people a balanced diet does not require any protein supplement. 3.3.1 Amino acids and peptide links Proteins are very large molecules which are formed by the condensation reactions of amino acids. There are between 20 - 30 different amino acids which occur in proteins and these can polymerise in almost any order directed by the cell. Each type of protein can have thousands of amino acids linked in one protein molecule. The potential variety of combinations is vast. The structure of the protein or the section of protein is based on the constituent amino acids. The amino acids found in proteins have the amino group and the carboxyl group attached to the same carbon atom. These are the α - amino acids (α = alpha). α being the first carbon atom from the carboxyl group. The structure of amino acids The structure of amino acids © H ERIOT-WATT U NIVERSITY 46 TOPIC 3. PROTEINS Amino acids contain both an amine group and a carboxylic acid group. These can react together forming an amide bond. Since amino acids contain two functional groups it is possible for them to link together into long chains in a condensation polymerisation. The protein molecules contain amide links. If amide links are present in living things then they are called peptide links. Therefore, in the protein molecules formed, the amide links are usually referred to as peptide links. Q4: Examples of amino acids Name the amino acid from the above examples in which hydrogen has replaced "R". .......................................... Q5: Name the amino acid from the above examples which would have the highest relative molecular mass. .......................................... Q6: Which amino acid from the above examples contains sulfur? .......................................... 3.3.1.1 Condensation of amino acids Condensation of amino acids The diagram shows three amino acids previously coded as A, B and D forming a product with two peptide links. Look at the diagram and answer the questions. Condensation of amino acids © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 47 Q7: Which small molecule is also a product of these condensation reactions? .......................................... Q8: What other name is sometimes given to the peptide link? .......................................... Q9: The combination of three amino acids shown is sometimes known as a tripeptide. Suggest what a combination of just two amino acids might be called? .......................................... Q10: The combination shown here is B-A-D. How many different combinations could there be of three amino acids using each only once? .......................................... Key Point Proteins are built up from the amino acids which have an amine functional group and a carboxyl functional group. Condensation of amino acids produces the amide (peptide) link. This is formed by the reaction of an amine group with a carboxyl group. The structure of a section of protein is based on the constituent amino acids. .......................................... 3.3.1.2 Condensation polymers Condensation polymers, such as polyesters and polyamides, are most often made from two different monomers, each with two functional groups which can react together producing a small molecule which can easily be removed. The characteristics of condensation polymerisation are: • in order for reacting molecules to produce large molecules (polymers) monomers require at least two reacting functional groups; • the polymerisation process also produces small molecules (often water). Note that this is different from addition polymerisation, where only one functional group (usually a double or triple bond) is involved, and the polymer is the only product. © H ERIOT-WATT U NIVERSITY 48 TOPIC 3. PROTEINS Condensation polymers Condensation polymers .......................................... Q11: The following questions relate to the figure below which shows the reaction between a carboxylic acid and an alcohol. As well as producing the -CO-O- linkage, name the small molecule which has also been produced. .......................................... Q12: What do you notice about the functional groups still present on the ester? .......................................... Q13: In the presence of more acid and alcohol, what would you expect to happen next? .......................................... This repeated ester formation will continue until one of the reactants is used up. © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 49 Polyester formation Polyester formation Q14: How many ester links are shown in the final molecule in the figure above? .......................................... Q15: How many water molecules have been formed in total? .......................................... Key Point Condensation polymers are made from monomers with two functional groups per molecule. A small molecule is also produced as condensation occurs. 3.3.2 Essential amino acids. Proteins are vital to the life of both plants and animals. Between 20 - 30 different amino acids are needed to make the proteins specific to the needs of the human body. The body cannot make all the amino acids required for body proteins and is dependent on dietary protein for supply of certain amino acids known as essential amino acids. If any are missing from the diet, important proteins cannot be produced and severe malnutrition can occur. Cereals can be deficient in some essential amino acids. For example, wheat is deficient in an amino acid called lysine. © H ERIOT-WATT U NIVERSITY 50 TOPIC 3. PROTEINS Wheat field Animal proteins like meat and fish provide the best supply of essential amino acids. Protein from plants vary widely in their amino acid content and vegetarians have to take care that their diet includes sufficient essential amino acids. Of the four amino acid structures, three, namely glycine, alanine and cysteine are non-essential and can be constructed in the human body by modifying other amino acids if absent from the diet. Simple amino acids Q16: Name an essential amino acid illustrated in the figure above. .......................................... Q17: Would wheat protein enhanced with glycine and alanine provide an adequate protein supply for a vegetarian? Explain your answer. .......................................... Q18: Which of these is the best source of essential amino acids? © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS a) b) c) d) 51 Wheat Carrots Beef Apples .......................................... Key Point Proteins specific to the body's needs are built up within the body. The body cannot make all the amino acids required for body proteins and is dependent on dietary protein for supply of certain amino acids known as essential amino acids. 3.4 Digestion of protein Digestion is the name we give to the breakdown of foodstuffs in the body. During digestion of protein foods the molecules are broken down by biological catalysts called enzymes and converted into the smaller units of amino acids. This takes place in a chemical reaction called hydrolysis. These amino acids are then passed from the small intestine into the blood and re-assembled into proteins specific to the body's needs. They become the feedstock for new proteins. If the amino acids are not required in that form they can sometimes be converted into ones which are needed or are broken down into waste products. In this way, "you are what you eat" and the protein-rich foods including meat, fish, eggs which you eat, are demolished and rebuilt into human muscle, skin, hair etc. Hydrolysis of protein The diagram shows three amino acids joined together in a tripeptide with a structure BA-D. The tripeptide is being hydrolysed. Look at the diagram and answer the questions. Hydrolysis of protein © H ERIOT-WATT U NIVERSITY 52 TOPIC 3. PROTEINS Q19: Which small molecule was added into the structure during the reaction? .......................................... Q20: What name is given to the link that is broken in this hydrolysis? .......................................... Q21: How many individual amino acids were formed by this hydrolysis? .......................................... Close examination of a section of protein should allow us to identify the structural formulae of the amino acids which make it up. Try this example. Q22: Identify the amino acid in the centre of this section of protein. .......................................... .......................................... The identification of the individual amino acids present in any one sample of protein is done by separation of the amino acid mixture produced by hydrolysis and comparison with known amino acids in a process known as chromatography. Some of the hydrolysed protein is spotted onto filter paper. Small spots of known amino acids are spotted alongside. The paper is placed in a solvent which rises up the paper carrying the amino acids at different rates. When the solvent is almost at the top, the paper is removed and the colourless amino acids are made visible by "developing" them with ninhydrin spray and warming the paper. Comparison of the positions of known and unknown spots allows identification to be made. Chromatography Look at the diagram showing the results of a chromatographic separation for the hydrolysed protein "X". The 'Simple amino acids' diagram in the 'Essential amino acids' topic will be needed to identify the amino acids present. © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 53 Protein hydrolysis analysis These questions refer to hydrolysed protein "X". Q23: How many amino acids are present in the hydrolysed protein "X"? Answer by entering the number. .......................................... Q24: Name any amino acid NOT present in X. .......................................... Q25: It is possible to conclude from the results that hydrolysed protein X contains a) b) c) d) B and C A and B only A, B and C A, B and D .......................................... There are many possible problem solving questions linked to these types of chromatogram. The question below refers to sample "X" and the on-line simulation has two other protein samples "Y" and "Z". Further questions on these can be found on the website. Q26: Use a piece of paper to draw your prediction of the chromatogram which would result if this protein section was hydrolysed and separated alongside samples of A, B, C and D. (Refer to the 'Simple amino acids' diagram in the 'Essential amino acids' topic.) .......................................... © H ERIOT-WATT U NIVERSITY 54 TOPIC 3. PROTEINS Q27: To answer this, and the other "further questions" it is necessary to access the on-line activity . Run the simulation for hydrolysed protein "Y". Explain fully your conclusions about the composition of Y from the results of the experiment. NOTE: There could be 3 marks for this answer in an exam setting. Try to give three pieces of information. Write an answer before revealing the possible model answer. .......................................... Q28: Run the simulation for hydrolysed protein "Z". How many amino acids are present in the hydrolysed protein "Z"? Answer by entering the number. .......................................... Q29: Name an amino acid which is not present in "Z" .......................................... Q30: Which amino acid seems to be present in the greatest amount? .......................................... .......................................... 3.4.1 Identifying the amino acids The structural formula of the amino acids obtained from the hydrolysis of proteins can be identified if these can be compared with a bank of standard samples. Identifying the amino acids does not tell us the order in which they occur in the protein section. This tripeptide forms three different amino acids on hydrolysis and shows their names and structures. Three simple amino acids Along with the structures in the 'Simple amino acids' diagram in the 'Essential amino acids' topic, this provides a bank of seven amino acid structures. © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 55 Example Questions Refer to this section of a protein. Identify the peptide links. Imagine the C-N bond breaking and join the -H from the water to the nitrogen end, and the -OH from the water to the carbonyl end of each fragment. Identify the individual amino acids from the 'Simple amino acids' and 'Three simple amino acids' diagrams. Q31: How many peptide links are shown? .......................................... Q32: How many amino acids are shown? .......................................... Q33: Name the amino acid on the left end of the protein section. .......................................... Q34: Name the amino acid on the right end of the protein section. .......................................... Key Point During digestion, the hydrolysis of proteins produces amino acids. The structural formulae of amino acids obtained from the hydrolysis of proteins can be identified from the structure of a section of the protein. © H ERIOT-WATT U NIVERSITY 56 TOPIC 3. PROTEINS 3.5 Summary Summary • Nitrogen is essential for protein formation by plants and animals. • Proteins are the major structural materials of animal tissue. • Proteins are also involved in the maintenance and regulation of life processes. • Enzymes are proteins. • The structure of a section of protein is based on the constituent amino acids. • Amino acids, the building blocks from which proteins are formed, are relatively small molecules which all contain an amino group (NH 2 ), and a carboxyl group (COOH). • The body cannot make all the amino acids required for body proteins and is dependent on dietary protein for supply of certain amino acids known as essential amino acids. • Proteins are made of many amino acid molecules linked together by condensation reactions. • Condensation polymers are made from monomers with two functional groups per molecule. A small molecule is also produced as condensation occurs. • In these condensation reactions, the amino group on one amino acid and the carboxyl group on a neighbouring amino acid join together, with the elimination of water. • The link which forms between the two amino acids can be recognised as an amide link (CONH) also known as the peptide link when in living things. • Proteins which fulfil different roles in the body are formed by linking differing sequences of amino acids together. • During digestion, enzyme hydrolysis of dietary proteins can produce amino acids. • The structural formulae of amino acids obtained from the hydrolysis of proteins can be identified from the structure of a section of the protein. • Chromatography can separate and identify these amino acids by comparison with a bank of known amino acids. © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 3.6 Resources Text • Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson, ISBN 978-1444167528 © H ERIOT-WATT U NIVERSITY 57 58 TOPIC 3. PROTEINS 3.7 End of topic test End of topic 3 test Q35: In which of the following kinds of compounds is nitrogen always present? a) Carbohydrates b) Enzymes c) Polyesters d) Oils .......................................... Q36: Essential amino acids can be defined as the amino acids which: a) are necessary for building proteins. b) plants cannot synthesise for themselves. c) are produced when any protein is hydrolysed. d) humans must acquire through their diet. .......................................... Q37: Animals can synthesise proteins specific to the body's needs from: a) digested plant proteins. b) fats and oils. c) nitrogen in the air. d) essential compounds in soil. .......................................... Q38: Identify the process which occurs when amino acids are converted into proteins. a) Hydration b) Addition c) Condensation d) Hydrogenation e) Hydrolysis f) Denaturing .......................................... Q39: Which of the following are α - amino acids? a) © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 59 b) c) d) e) f) .......................................... Q40: Which of the following is a β - amino acid? a) b) © H ERIOT-WATT U NIVERSITY 60 TOPIC 3. PROTEINS c) d) e) f) .......................................... Q41: Identify the molecules which could be produced when a protein is hydrolysed. a) b) © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 61 c) d) e) f) .......................................... Q42: Identify the molecule which would be present in the hydrolysed sample of the section of protein above. © H ERIOT-WATT U NIVERSITY 62 TOPIC 3. PROTEINS a) b) c) d) e) © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 63 f) .......................................... Q43: The artificial sweetener aspartame is a derivative of two amino acids. Select/highlight the part of the molecule which is the peptide (amide) bond. .......................................... Q44: Name the type of compound produced by the hydrolysis of a protein. .......................................... Q45: This question refers to the chromatographic analysis of hydrolysed protein section P. © H ERIOT-WATT U NIVERSITY 64 TOPIC 3. PROTEINS Name the amino acids which are present in the hydrolysed protein section P. .......................................... Q46: How many amino acids are present in hydrolysed protein section P? .......................................... Q47: From the results, it can be concluded that the hydrolysed protein section P contains: 1. A and B 2. A and B only 3. A, B and C 4. A, B, C and D .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 3. PROTEINS 65 Q48: Which of these protein sections could be P? 1. 2. 3. 4. .......................................... © H ERIOT-WATT U NIVERSITY 66 TOPIC 3. PROTEINS Q49: If protein section Q, shown above, were hydrolysed and run against A, B, C and D, which spots would be present? 1. A only 2. A and D 3. A, C and D 4. A, B, C and D .......................................... © H ERIOT-WATT U NIVERSITY 67 Topic 4 Chemistry of cooking Contents 4.1 Prior knowledge . . . . . . . . . . . . . . 4.2 Flavour . . . . . . . . . . . . . . . . . . . 4.2.1 The senses . . . . . . . . . . . . . 4.2.2 The sense of taste and smell . . . 4.2.3 Flavour molecules . . . . . . . . . 4.2.4 The solubility of flavour molecules 4.2.5 The Maillard reaction . . . . . . . . 4.3 Protein Structures . . . . . . . . . . . . . 4.3.1 Protein changes on cooking . . . . 4.3.2 Collagen . . . . . . . . . . . . . . 4.4 Cooking . . . . . . . . . . . . . . . . . . . 4.4.1 Cooking methods . . . . . . . . . . 4.5 Summary questions . . . . . . . . . . . . 4.6 Summary . . . . . . . . . . . . . . . . . . 4.7 Resources . . . . . . . . . . . . . . . . . 4.8 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 70 70 71 73 75 77 78 80 82 82 84 85 88 88 89 Prerequisite knowledge You should already know that: • many plants are used by chemist in the design and manufacture of many everyday products such as pharmaceuticals soaps, cosmetics, dyes, medicines, foods or food colourings (National 4, Unit 2); • plants are a source of carbohydrates and oils which can be used for food or fuel (National 4, Unit 2); • carboxylic acids can be identified by the carboxyl ending, the COOH functional group and the '-oic' name ending (National 5, Unit 2); • straight-chained carboxylic acids can be identified and named from the structural formulae. Given the name of straight chained carboxylic acid the structural formulae can be drawn (National 5, Unit 2); • an alcohol is identified from the -OH group and the ending '-ol' (National 5, Unit 2); 68 TOPIC 4. CHEMISTRY OF COOKING • straight chain alcohols are named from the structure formulae. Given the names of straight-chain alcohols structural and molecular formulae can be written (National 5, Unit 2); • an ester can be made by reacting a carboxylic acid and an alcohol (National 5, Unit 2); • some uses of esters are in food flavouring, industrial solvents, fragrances and materials (National 5, Unit 2). You should already know how to: • name and draw structural formulae for straight chain alkanes (C1 - C8), alkenes (C2 - C6) and cycloalkanes (C3 - C6) (National 5, Unit 2); • use, appropriately, molecular formulae and full and shortened structural formulae (National 5, Unit 2); • identify and draw structural formulae for isomers (National 5, Unit 2); • describe what is meant by the terms hydrocarbon, homologous series, saturated and unsaturated (National 5, Unit 2). Learning Objectives At the end of this topic, you should be able to state that: Flavour molecules • the olfactory and taste senses in humans can be described; • food flavours mainly excite the senses of taste and smell; • molecular size and functional groups present affect the volatility of food molecules; • flavour molecules can be water- or oil-soluble, consequently cooking methods can affect the quality of the food; • cooking methods might enhance or destroy the food's flavour; • cooking changes (denatures) proteins, in particular it can make tough collagen palatable; • different cooking methods would be appropriate for different foods. Proteins • within proteins, the long-chain molecules may be twisted to form spirals, folded into sheets, or wound around to form other complex shapes; • the chains are held in these forms by intermolecular bonding between the side chains of the constituent amino acids; • when proteins are heated, during cooking, these intermolecular bonds are broken allowing the proteins to change shape (denature). • these changes alter the texture of foods. © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING 4.1 Prior knowledge Test your prior knowledge Q1: What is the name of the following ester? a) b) c) d) Ethyl ethanoate Ethyl propanoate Propyl ethanoate Propyl propanoate .......................................... Q2: What is the name of the following carboxylic acid? a) b) c) d) Methanoic acid Ethanoic acid Propanoic acid Butanoic acid .......................................... Q3: Which of the following best describes the following molecule? a) b) c) d) Alkane, saturated Alkane, unsaturated Alkene, unsaturated Alkene, saturated .......................................... © H ERIOT-WATT U NIVERSITY 69 70 TOPIC 4. CHEMISTRY OF COOKING 4.2 Flavour Why do most of us enjoy foods so much? Food can stimulate all our senses, but mainly it excites our sense of taste and smell. Imagine the different tastes and aromas involved in a steak and chips meal. A steak and chips meal 4.2.1 The senses It is obvious that when we eat food we exercise our sense of taste, one of our 'chemical' senses. The taste sensors are located on the tongue but investigation shows that our taste buds can detect only five basic different types of taste: • sweet • sour • bitter • salty • savoury (umami) © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING Our main sense used with food is our sense of smell. This system is able to detect a vast variety of different odours. Just think how many different odours, pleasant and otherwise, that you experience each day. Your sense of smell is able to detect very subtle changes of odour, and our enjoyment of food is very closely tied to its aroma. Taste and smell experiment You can try an experiment with a friend to illustrate the limited range of taste compared with smell. Use two or more different jams. Get your friend to hold his or her nose so that no air can pass the olfactory system, and close his or her eyes so that you can choose a spoon of jam without your friend knowing which one. Your friend should be able to say that the substance on the spoon was sweet; but which jam is in his/her mouth? Releasing the grip on the nose should reveal the type of jam. You could reverse roles and use different flavoured crisps. .......................................... 4.2.2 The sense of taste and smell Chemoreceptors are found in the nasal cavity and upon the surface of the tongue. Chemoreceptors receive chemical molecules from the environment and convert or transduce the chemical messages into electrical impulses signalling gustation (taste) and olfaction (smell). Chemoreceptors on a tongue Taste is mediated by chemoreceptors in both the mouth area and nasal cavity; the latter will be discussed in the next section. Taste signals from the mouth area occur when food, or another soluble substance, is dissolved in saliva and penetrates the taste buds. Penetration causes a chemical reaction that leads to firing of the receptors cells. The tongue is covered in bumps called papillae and the walls of most papillae feature taste buds. © H ERIOT-WATT U NIVERSITY 71 72 TOPIC 4. CHEMISTRY OF COOKING Taste buds contain the receptor cells for gustation (about 50 to a bud) and are also found elsewhere apart from the tongue, such as in the throat, roof of the mouth and inner cheeks. There are about 6,000 taste buds in the average human mouth. Although taste receptors are the main cause of perception of a particular flavour, it is worth noting that the flavour of a given food is also associated with its temperature, consistency and texture. How we detect odour Olfaction occurs when an odour is present in the atmosphere and chemoreceptors in the nose process the odour. Airborne molecules from the odour are inhaled into the nasal cavity. After circulating around the nasal cavity, so removing dust and other particles, some of the molecules come into contact with the receptor cells. This contact produces a chemical reaction resulting in the receptor cells firing. As can be seen in the illustration above, the olfactory receptor cells are found on the olfactory epithelium at the top of the nasal cavity. They are termed bipolar nerve cells, terminating in cilia at one end and axons at the other, that connect to the olfactory bulb, a structure where the transduced odour information is processed. In humans there are an average of about 40-50 million olfactory sensory neurons. Our sense of smell is very specific. Since the molecules effecting the response in the olfactory system do this by affecting neurons on a 3D surface, the 3D structural arrangement of the molecule can elicit different responses. For example, one form of carvone tastes and smells of spearmint, whereas the other form resembles caraway seed - the smell elicited depends on the arrangement of groups around the lower carbon atom in the ring. © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING 4.2.3 73 L-carvone D-carvone Spearmint Caraway seed Flavour molecules In the previous topic on fruit flavours you looked at esters as a group of chemicals that have typically pleasant aromas that occur naturally in many fruits and are used as artificial flavourings. Here are some other common food aromas that are not due to esters. © H ERIOT-WATT U NIVERSITY 74 TOPIC 4. CHEMISTRY OF COOKING Chemical Odour Diacetyl buttery Benzaldehyde almond Cinnamaldehyde cinnamon Limonene orange 4-hydroxy-3-methoxy-benzaldehyde vanilla Oct-1-en-3-one mushroom Q4: Here are the formulae for these six compounds. What do you notice about their structures? Think of anything that might be relevant to the fact that these all have odours. Diacetyl Limonene Benzaldehyde Cinnamaldehyde 4-hydroxy-3-methoxyOct-1-en-3-one benzaldehyde .......................................... Q5: What property of molecules do you think is vital for food flavour molecules to possess? a) b) c) d) Volatile Soluble in water Soluble in oil Dense .......................................... Q6: What property of molecules determines their volatility? .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING 75 Volatile or not? Q7: From your knowledge of intermolecular forces, decide whether these molecules are volatile or not: 1. Casein - the main protein in milk. 2. Anisole (C6 H5 OCH3 ) - found in aniseed 3. Hex-3-ene-1-ol - in freshly cut grass 4. Cellulose - carbohydrate polymer found in plants 5. Lignin - polymer found in woody plants 6. Menthol ((CH3 )2 CHC6 H9 (CH3 )OH) - from peppermint .......................................... 4.2.4 The solubility of flavour molecules Our sense of smell can detect an enormous number of different types of molecules. Two examples; toluene, an aromatic hydrocarbon named because of its pleasant odour and methanol, both have distinct odours. The former is almost insoluble in water; the later will mix with water in all proportions (because they can form hydrogen bonds with each other easily). Chemical analysis of the volatile compounds in Asparagus showed a majority of these were alcohols and aldehydes. Hexanal, hexenal and oct-1-en-3-ol predominated. These three compounds are all fairly soluble in water, so what do you imagine will happen to the flavour of asparagus if it is cooked by boiling in water? The cooked asparagus will loose its essential taste, which will be thrown out with the cooking water. How do you get round this problem? Substances that are soluble in water are not usually soluble in oils, so heat the asparagus in cooking oil or butter and the flavour will remain with the vegetable. On the other hand the flavour components of, say, broccoli are not very soluble in water, so boiling or steaming are appropriate cooking methods. Asparagus © H ERIOT-WATT U NIVERSITY Broccoli 76 TOPIC 4. CHEMISTRY OF COOKING Which cooking method? For the four compounds shown decide whether they will be soluble in water or in oil, and whether they will be volatile or not. Both of these properties depend on the intermolecular forces between molecules, so focus on that aspect of the compounds. Q8: .......................................... Q9: .......................................... Q10: .......................................... Q11: .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING 77 You might like to discuss with members of your class, or using the SCHOLAR discussion board, some foods that you particularly like or dislike, trying to give some reasons for your choice. .......................................... 4.2.5 The Maillard reaction As well as destroying flavours by employing incorrect cooking methods, most cooking enhances the flavours of food. One common reaction occurs when you make toast or roast meats. It was studied by the French scientist, Louis Camille Maillard, and carries his name. Amino acids and carbohydrates react in a series of complex reactions resulting in 'browning' of the food and the production of many flavour compounds. Each food has a different set of reactions resulting in characteristic flavours. White bread Toast Two compounds produced by the Maillard reaction are furfural and 2-acetylpyrroline. Furfural 2-Acetylpyrroline Q12: Are these two compounds likely to be flavour molecules? a) Yes b) No .......................................... Q13: Do you think these molecules will be soluble in water? © H ERIOT-WATT U NIVERSITY 78 TOPIC 4. CHEMISTRY OF COOKING a) Yes b) No .......................................... Q14: Do you think furfural and 2-acetylpyrroline are likely to be flavour molecules? a) Yes b) No .......................................... 4.3 Protein Structures The primary structure of long protein molecules is the linear sequence of amino acids held together by peptide links. This can bend and flex, in some cases leading to intramolecular bonding (usually hydrogen bonding) and a helix or a pleated sheet shape. Many fibrous proteins have this secondary structure. Helix structure H C O H C O N O C N C C O N O C C C N C R N N C C O R N H O HC R N H C O C R R CH N C N H C HC H O CH H O C Sheet structure C O C C N N H H H C H C C O O N hydrogen C N bonds C C C N C C H N O H N CH hydrogen bonds HC R Sheet and helix structures Fibrous proteins are long and thin and are the major structural materials of animal tissue. Examples like keratin (in hair, nails, and feathers), collagen (connective tissue between cells), actin and myosin (in muscles) and fibroin (in silk) are insoluble in water and resistant to acids and alkalis. Bending of the secondary structure can result in the tertiary shape taken up by globular proteins which have their spiral chains folded into compact units. Globular © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING proteins are involved in the maintenance and regulation of life processes and, due to their solubility, can move around the body in the bloodstream. Examples are haemoglobin, antibodies, all enzymes and many hormones e.g. insulin. Some enzymes and haemoglobin are made from assemblies of several protein chains. The way these fit together is called the quaternary structure. Fibrous and globular proteins Fibrous and globular proteins .......................................... Q15: What type of intramolecular bonding is usually responsible for the secondary structure of proteins? .......................................... © H ERIOT-WATT U NIVERSITY 79 80 TOPIC 4. CHEMISTRY OF COOKING 4.3.1 Protein changes on cooking The way most proteins work relies heavily on the three dimensional structure. For example, proteins called enzymes have important areas on their surface called the active site. This is where the chemical function of the enzyme takes place. The enzyme and the molecule on which it acts (the substrate) have to fit together like a lock and key. Only when this close fit is achieved can the substrate be changed into products, which can then leave the active site. Denaturing If the secondary, tertiary or quaternary structure of the enzyme is lost it is said to be denatured. This happens because the relatively weak hydrogen bonds holding the different chains together are very sensitive to changes in temperature or pH, and can be easily broken. The important shape of the active site is damaged and the 'lock' no longer fits the 'key'. Denaturing of a protein involves physical alteration of the molecular shape as a result of temperature or pH changes. These changes can be seen below where an egg is heated (poached) and when vinegar is added to milk to cause curdling. Natural protein Denatured protein Raw egg Poached egg Bottles of milk Curds - the start of making cheese © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING As you should know, protein structure is complex and can have up to four different components. Try these questions to make sure you are familiar with the terminology. Q16: The primary structure is the: a) b) c) d) aggregation of several polypeptide chains into a functional unit. interaction of amino acid side chains to give an overall folding to the protein. sequence of amino acids in the polypeptide chain. folding of the backbone of the chain into helices and pleated sheets. .......................................... Q17: The secondary structure is the: a) b) c) d) aggregation of several polypeptide chains into a functional unit. interaction of amino acid side chains to give an overall folding to the protein. sequence of amino acids in the polypeptide chain. folding of the backbone of the chain into helices and pleated sheets. .......................................... Q18: The tertiary structure is the: a) b) c) d) aggregation of several polypeptide chains into a functional unit. interaction of amino acid side chains to give an overall folding to the protein. sequence of amino acids in the polypeptide chain. folding of the backbone of the chain into helices and pleated sheets. .......................................... Q19: The quaternary structure is the: a) b) c) d) aggregation of several polypeptide chains into a functional unit. interaction of amino acid side chains to give an overall folding to the protein. sequence of amino acids in the polypeptide chain. folding of the backbone of the chain into helices and pleated sheets. .......................................... Q20: Which of these types of protein structure is likely to remain intact during denaturation? a) b) c) d) Primary Secondary Tertiary Quaternary .......................................... © H ERIOT-WATT U NIVERSITY 81 82 TOPIC 4. CHEMISTRY OF COOKING 4.3.2 Collagen Without the protein collagen you would literally fall apart! Collagen is the main structural component of connective tissue in the body; involved in skin, bone, tendons, ligaments and other parts. It forms 25 - 40% of the protein in the human body. In keeping with its function as 'structural scaffolding' collagen is a fibrous protein. It has long helical subunits intertwined into a triple helix resulting in great strength. Unusually for proteins it contains large quantities of two amino acids, glycine and proline, together with hydroxyproline and hydroxylysine, which are produced by enzymic hydroxylation on the initial proto-collagen. These amino acids in the backbone allow the chain to deform readily and produce a large number of hydrogen bonds resulting in a very strong structure. The structure of collagen The three chains are shown in different colours in the image below. Since collagen is so widespread in animal tissue it is always present to varying degrees in cuts of meat. Its presence in the living organism is vital, but when we come to eat it, collagen makes the meat very tough to chew and difficult to enjoy. So how do we make a tough cut into a tender morsel? We cook it. 4.4 Cooking Cooking can involve several different methods: roasting and boiling (stewing) are typical. The actual procedure, for example the time and temperature, will depend on the starting material and the desired result. When collagen is heated to about 60 ◦ C the quaternary structure starts to break down and the rigid triple helix will unwind to single strands. These strands of α-helices will, in turn, again lose the bonding which holds them in place becoming free protein chains. These chains form a mesh which absorbs water. These are no longer tough, rigid material, but soft and pliable. The original connective tissue has completely changed © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING 83 character. When carried to a considerable extent the end-product is gelatine, used to make jellies! Gristly meat Soft jelly Q21: Collagen is an example of: a) b) c) d) fat. oil. protein. carbohydrate. .......................................... Q22: When cooked, which protein structure in collagen is broken down? a) b) c) d) Primary Secondary Tertiary Quaternary .......................................... Other advantages of cooking Collagen is not the only protein in foods. All the other proteins will undergo denaturation to a greater or lesser extent, depending on the method and duration of cooking. Q23: Can you think of an advantage of denaturing protein in foods before eating (other than making it more palatable)? Hint: Think of what happens to food once eaten. .......................................... Q24: There is also some 'chemical' sense to having, say, pineapple with gammon, in addition to it tasting good! Do you know what it is? .......................................... Q25: Is there another benefit of cooking food? .......................................... © H ERIOT-WATT U NIVERSITY 84 TOPIC 4. CHEMISTRY OF COOKING 4.4.1 Cooking methods Two different cuts of meat: juicy steak - almost free of connective tissue and poorer cut with gristle. Prime steak Poor cut of steak Here is what they look like when you eat them. Prime steak cooked Poor cut of steak cooked The prime steak was placed under a hot grill for 10 minutes; in contrast, the poor cut of steak was simmered (kept almost at boiling point) for 2.5 hours. Q26: Which steak's proteins will be denatured and hydrolysed most? a) Prime steak cooked b) Poor cut of steak cooked .......................................... Q27: Which of the cooked states will be the softest and easiest to digest? © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING a) Prime steak cooked b) Poor cut of steak cooked .......................................... 4.5 Summary questions As stated at the beginning of this topic, this is not a cookery course, so these questions are essentially trivial in terms of a genuine chef's knowledge. Summary questions Q28: Capsaicin (see below) is the compound which gives chilli peppers their 'hotness'. There are several functional groups present in this molecule, how many can you find? .......................................... Q29: Look at the molecular structure again. If you decided to stir-fry your peppers, do you think that the capsaicin molecules would be volatile enough to be lost from the peppers when they were heated? a) Volatile - lost from the stir-fry b) Non-volatile - will remain in the cooked product .......................................... Q30: How did you decide the answer to the above question? .......................................... Q31: You may have noticed that when you boil cabbage, broccoli, sprouts and other greens the water turns distinctly green. But when you boil carrots the water does not turn orange. Here are the structures of the green pigment chlorophyll and the orange pigment βcarotene. © H ERIOT-WATT U NIVERSITY 85 86 TOPIC 4. CHEMISTRY OF COOKING chlorophyll β-carotene Which two non-metallic elements are present in chlorophyll but not in β-carotene? .......................................... Q32: Explain why chlorophyll is soluble in water whereas β-carotene is not. .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING Q33: What substance in your kitchen cupboard would you expect to dissolve βcarotene? .......................................... Q34: You have to cook squid for a long time to make them tender. What do you think that tells you about the composition of squid muscles? .......................................... Q35: How many strands of collagen form the quaternary structure of this tough protein? a) b) c) d) 1 2 3 4 .......................................... Q36: When collagen is denatured by boiling for a period, which of the protein structures are destroyed? a) b) c) d) Quaternary Quaternary and tertiary Quaternary, tertiary and secondary Quaternary, tertiary, secondary, and primary .......................................... Q37: What type of bonds are involved in maintaining the secondary structure? a) b) c) d) Intermolecular Hydrogen bonds Peptide links Disulfide bridges .......................................... Q38: What type of bonds are involved in maintaining the primary structure? a) b) c) d) Intermolecular Hydrogen bonds Peptide links Disulfide bridges .......................................... Q39: What is the name of the chemical process that can destroy the primary structure of proteins? a) b) c) d) Oxidation Condensation Polymerisation Hydrolysis .......................................... © H ERIOT-WATT U NIVERSITY 87 88 TOPIC 4. CHEMISTRY OF COOKING 4.6 Summary Summary Flavour molecules • the olfactory and taste senses in humans can be described; • food flavours mainly excite the senses of taste and smell; • molecular size and functional groups present affect the volatility of food molecules; • flavour molecules can be water- or oil-soluble, consequently cooking methods can affect the quality of the food; • cooking methods might enhance or destroy the food's flavour; • cooking changes (denatures) proteins, in particular it can make tough collagen palatable; • different cooking methods would be appropriate for different foods. Proteins • within proteins, the long-chain molecules may be twisted to form spirals, folded into sheets, or wound around to form other complex shapes; • the chains are held in these forms by intermolecular bonding between the side chains of the constituent amino acids; • when proteins are heated, during cooking, these intermolecular bonds are broken allowing the proteins to change shape (denature). • these changes alter the texture of foods. 4.7 Resources Texts Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson, ISBN 978-1444167528 © H ERIOT-WATT U NIVERSITY TOPIC 4. CHEMISTRY OF COOKING 4.8 End of topic test End of topic test Q40: Much of the flavour of food depends on our sense of smell. In order to excite this sense, molecules must: a) b) c) d) contain nitrogen atoms be volatile be large have strong intramolecular bonds .......................................... Q41: Flavour molecules can be water-soluble or oil-soluble. Which of the following would you expect to be oil-soluble? a) b) c) d) Ethanoic acid Ethanol Beta-carotene Methyl methanoate .......................................... Q42: When considering the boiling point of molecules, which of the following properties are important? 1. Solubility in water 2. Molecular mass 3. Presence of unpaired electrons 4. Type of intermolecular bonding .......................................... Q43: In general, volatile molecules are: a) b) c) d) small with strong bonds between molecules large with strong bonds between molecules large with weak bonds between molecules small with weak bonds between molecules .......................................... Q44: Put the four molecules below in order of increasing boiling point: 1. Ethanol (C2 H5 OH) 2. Ethane (C2 H6 ) 3. Ethanal (CH3 CHO) 4. Ethanoic Acid (CH3 COOH) .......................................... Q45: Which type of bonding is important in deciding the order you have chosen? a) Ionic bonding © H ERIOT-WATT U NIVERSITY 89 90 TOPIC 4. CHEMISTRY OF COOKING b) Hydrogen bonding c) London dispersion forces d) Covalent bonding .......................................... Q46: In order to retain the oil-soluble flavour of a vegetable, it should be cooked for a: a) b) c) d) short time in boiling water long time in boiling water long time in oil short time in oil .......................................... Q47: Which definition applies to the primary structure in proteins? a) b) c) d) The presence of alpha-helical and beta-pleated sheets in the structure. The overall shape of the complete protein molecule. The first bonds to form after synthesis. The sequence of amino acids in the chain. .......................................... Q48: Heating of proteins causes denaturation. What is denaturation? .......................................... Q49: Which type of bonding is broken during denaturation? a) b) c) d) Covalent Disulfide Polar covalent Intermolecular .......................................... Q50: Which types of protein structure are disrupted during denaturation? a) b) c) d) Primary and quaternary Primary and secondary Primary, secondary and tertiary Secondary, tertiary and quaternary .......................................... Q51: Using your knowledge of proteins, describe why different methods are used to cook a stewing steak and a prime fillet steak. .......................................... © H ERIOT-WATT U NIVERSITY 91 Topic 5 Oxidation of food Contents 5.1 Prior knowledge . . . . . . . . . . . . . . . . . . . . 5.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Oxidation and reduction . . . . . . . . . . . . 5.3 Structure of alcohols . . . . . . . . . . . . . . . . . . 5.4 Oxidation of alcohols - practical . . . . . . . . . . . . 5.5 Aldehydes and ketones . . . . . . . . . . . . . . . . 5.5.1 From name to structure . . . . . . . . . . . . 5.6 Oxidation of aldehydes and ketones . . . . . . . . . 5.6.1 Oxidising agents . . . . . . . . . . . . . . . . 5.6.2 What are the compounds? . . . . . . . . . . 5.7 Antioxidants . . . . . . . . . . . . . . . . . . . . . . . 5.8 Oxygen and food . . . . . . . . . . . . . . . . . . . . 5.8.1 The antioxidant mechanism of ascorbic acid . 5.8.2 Other methods of preventing oxidation . . . . 5.9 Ion-electron equations . . . . . . . . . . . . . . . . . 5.10 Estimating antioxidant levels . . . . . . . . . . . . . . 5.11 Summary . . . . . . . . . . . . . . . . . . . . . . . . 5.12 Resources . . . . . . . . . . . . . . . . . . . . . . . 5.13 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 95 96 98 100 103 105 106 107 107 109 110 112 113 114 115 117 118 119 Prerequisite knowledge You should already know that: Flavour Molecules • the olfactory and taste senses in humans can be described; • food flavours mainly excite the senses of taste and smell; • molecular size and functional groups present affect the volatility of food molecules; • flavour molecules can be water- or oil-soluble, consequently cooking methods can affect the quality of the food; • cooking methods might enhance or destroy the food's flavour; 92 TOPIC 5. OXIDATION OF FOOD • cooking changes (denatures) proteins, in particular it can make tough collagen palatable; • different cooking methods would be appropriate for different foods; Proteins • within proteins, the long-chain molecules may be twisted to form spirals, folded into sheets, or wound around to form other complex shapes; • the chains are held in these forms by intermolecular bonding between the side chains of the constituent amino acids; • when proteins are heated, during cooking, these intermolecular bonds are broken allowing the proteins to change shape (denature); • these changes alter the texture of foods. Learning Objectives At the end of this topic, you should be able to state that: Alcohols • for compounds with no more than eight carbon atoms in their longest chain; • branched-chain alcohols can be named from structural formulae; • given the names of branched-chain alcohols, structural formulae can be drawn and molecular formulae written; • primary alcohols are oxidised, first to aldehydes and then to carboxylic acids; • secondary alcohols are oxidised to ketones; • when applied to carbon compounds, oxidation results in an increase in the oxygen to hydrogen ratio; • in the laboratory, hot copper(II) oxide or acidified dichromate(VI) solutions can be used to oxidise primary and secondary alcohols; • tertiary alcohols cannot be oxidised. Carbonyl compounds • branched-chain carboxylic acids, with no more than eight carbon atoms in their longest chain, can be named from structural formulae; • given the names of branched-chain carboxylic acids, structural formulae can be drawn and molecular formulae can be written; • aldehydes and ketones can be identified from the ‘-al’ and ‘-one’ name endings respectively; © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD • straight-chain and branched-chain aldehydes and ketones, with no more than eight carbon atoms in their longest chain, can be named from structural formulae; • given the names of straight-chain or branched-chain aldehydes and ketones, structural formulae can be drawn and molecular formulae can be written; • aldehydes, but not ketones, can be oxidised to carboxylic acids; • Fehling’s solution, Tollens’ reagent and acidified dichromate solution can be used to differentiate between an aldehyde and a ketone. Antioxidants • oxygen reacts with edible oils giving the food a rancid flavour; • antioxidants are molecules which will prevent these oxidation reactions taking place; • ion-electron equations can be written for the oxidation of many antioxidants. © H ERIOT-WATT U NIVERSITY 93 94 TOPIC 5. OXIDATION OF FOOD 5.1 Prior knowledge Test your prior knowledge Q1: Which of the following properties would you look up to determine how volatile a molecule is? a) b) c) d) Melting point Boiling point Density Atomic number .......................................... Q2: The following structure is limonene, present in oranges. Limonene is likely to be: a) b) c) d) volatile and soluble in water. volatile and insoluble in water. not volatile and soluble in water. not volatile and insoluble in water. .......................................... Q3: a) b) c) d) Which definition applies to the primary structure in proteins? The sequence of amino acids in the chain. The presence of alpha-helical and beta-pleated sheets in the structure. The overall shape of the complete protein molecule. The first bonds to form after synthesis. .......................................... Q4: a) b) c) d) When cooking meats, which protein structure is broken down? Primary Secondary Tertiary Quaternary .......................................... .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD 5.2 Introduction This topic revises and extends the chemistry of oxidation before studying the oxidation of alcohols and aldehydes. It then considers the role of antioxidants, particularly with regard to food preservation. Oxidation has previously been defined in terms of loss of electrons by a reactant whereas reduction was gain of electrons by a reactant. Oxidation and reduction occur simultaneously (a redox reaction). One cannot happen without the other. Even with carbon compounds, it is possible to write ion-electron half equations for oxidation and reduction but these are generally complex. When dealing with carbon compounds, we are more interested in what happens to the carbon compound than what happens to the oxidising agent. Balanced redox equations are rarely used. Instead, only the reacting carbon compound and the product carbon compound are shown. The oxidising agent is often shown by [O] written above the arrow in the equation, e.g. Oxidation of ethanol In the context of carbon compounds, oxidation and reduction can be defined in a different way. In Figure 2.1 above, the initial reaction involves the loss of two hydrogen atoms from the ethanol molecule. The second reaction, involves the gain of an oxygen atom. Both reactions are regarded as oxidation. What they have in common is an increase in the oxygen to hydrogen ratio. The following activity shows how to calculate this ratio and hence work out whether or not a reaction is an oxidation. © H ERIOT-WATT U NIVERSITY 95 96 TOPIC 5. OXIDATION OF FOOD Calculating the O:H ratio It can be seen that, when applied to carbon compounds, oxidation reactions result in an increase in the oxygen to hydrogen ratio. Since reduction is always the opposite of oxidation, reduction of a carbon compound must involve a decrease in the oxygen to hydrogen ratio. 5.2.1 Oxidation and reduction Oxidation and reduction For the first three questions, use this reaction: Q5: What is the oxygen to hydrogen ratio in the reactant? .......................................... Q6: What is the oxygen to hydrogen ratio in the product? .......................................... Q7: What type of reaction is this? a) Oxidation b) Reduction c) Neither .......................................... Q8: What type of reaction is this? © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD a) Oxidation b) Reduction c) Neither .......................................... Q9: What type of reaction is this? a) Oxidation b) Reduction c) Neither .......................................... Q10: What type of reaction is this? a) Oxidation b) Reduction c) Neither .......................................... © H ERIOT-WATT U NIVERSITY 97 98 TOPIC 5. OXIDATION OF FOOD Q11: What type of reaction is this? a) Oxidation b) Reduction c) Neither .......................................... Q12: What type of reaction is this? a) Oxidation b) Reduction c) Neither .......................................... Key Point When applied to carbon compounds, oxidation reactions result in an increase in the oxygen to hydrogen ratio. When applied to carbon compounds, reduction reactions result in a decrease in the oxygen to hydrogen ratio. .......................................... 5.3 Structure of alcohols Alcohols can be classified as either primary, secondary or tertiary. The classification depends on the position of the hydroxyl (OH) group in the molecule. In order to be able to be oxidised, the alcohol must have at least one hydrogen atom on the carbon atom which is attached to the OH group. © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD 99 Primary alcohols The hydroxyl group is joined to the end of the carbon chain. Examples include: Ethanol Butan-1-ol Oxidation of primary alcohols Primary (and secondary) alcohols oxidise by losing two H atoms as shown here, forming a carbonyl (C=O) group. The hydroxyl group of the alcohol is joined to an end carbon in the chain which also has a hydrogen atom attached. For example: Ethanol Ethanal When a primary alcohol is oxidised, the product is an aldehyde. This is because the carbonyl group (C=O) is formed at the end of the carbon chain. Aldehydes can also be identified by their names, which end in “al”. Secondary alcohols The hydroxyl group is joined to an intermediate carbon in the chain. ie. The OH group is in the middle of the carbon chain. Examples include: Propan-2-ol Butan-2-ol Oxidation of secondary alcohols Secondary alcohols also oxidise by losing two H atoms as shown here, forming a © H ERIOT-WATT U NIVERSITY 100 TOPIC 5. OXIDATION OF FOOD carbonyl (C=O) group. The hydroxyl group of the alcohol is joined to an intermediate carbon in the chain which also has a hydrogen atom attached. Propan-2-ol Propanone When a secondary alcohol is oxidised, the product is a ketone. This is because the carbonyl group (C=O) is formed within the carbon chain. Ketones can also be identified by their names, which end in “one”. Tertiary alcohols The hydroxyl group of the alcohol is joined to an end carbon in the chain which also has a branch attached. Examples include: 2-methylpropan-2-ol A tertiary alcohol cannot easily oxidise because there is no H atom on the C atom attached to the hydroxyl group (OH). 5.4 Oxidation of alcohols - practical Two of the most common oxidising agents used to oxidise alcohols are copper(II) oxide and acidified potassium dichromate solution. © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD 101 Using copper(II) oxide as an oxidising agent Using acidified dichromate solution as an oxidising agent The copper(II) oxide is heated strongly and the alcohol vapour is passed over it. When the alcohol is oxidised, the black copper(II) oxide is reduced to pinkish brown copper. The alcohol is added to the orange solution. On warming the solution, the orange dichromate ions are reduced to green Cr3+ ions. Primary, secondary and tertiary alcohols behave differently towards these oxidising agents. Primary alcohols, such as ethanol, are oxidised in two stages (see figure below). With care, the oxidation can be stopped after stage 1. Oxidation of ethanol Q13: Name the product of the first stage oxidation. .......................................... Q14: Name the final product. .......................................... Oxidation of a primary alcohol involves the removal of two hydrogen atoms, one from the hydroxyl group and one from the carbon attached to the hydroxyl group. The result is formation of a C=O bond. Q15: What name is given to the C=O group? .......................................... Secondary alcohols, such as propan-2-ol, also have a hydrogen atom attached to the carbon atom carrying the OH group. A similar oxidation again results in the formation of a C=O group (see figure below). © H ERIOT-WATT U NIVERSITY 102 TOPIC 5. OXIDATION OF FOOD Oxidation of propan-2-ol Q16: Name the product of this reaction. .......................................... Unlike primary and secondary alcohols, tertiary alcohols have no hydrogen atom attached to the carbon carrying the OH group (see figure below). Ease of oxidation of alcohols Consequently, tertiary alcohols are difficult to oxidise. Useful oxidations Vinegar is made by the action of oxidising enzymes present in some strains of microorganisms to change the ethanol in a variety of wines, beers and cider into ethanoic acid. The name vinegar derives from the French 'vinaigre', sour wine. There is an enormous range of vinegars, the product depending on the source of material containing the ethanol, the time allowed for the oxidation, whether an aging process is included and the addition of substances such as herbs. Whisky contains a variety of aldehydes which impart characteristic flavours to the product. This slow oxidation takes place when the whisky is maturing in casks. The atmospheric oxygen slowly diffuses through the wood into the liquid and causes partial oxidation of the ethanol and other alcohols present. © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD Key Point Primary and secondary alcohols can be oxidised by a number of oxidising agents, including copper(II) oxide and acidified potassium dichromate. • Primary alcohols are oxidised first to aldehydes and then to carboxylic acids. • Secondary alcohols are oxidised to ketones. • Tertiary alcohols are resistant to oxidation. 5.5 Aldehydes and ketones Aldehydes and ketones are less familiar, but no less important, than alcohols. The two classes of compound are generally considered together since they are structurally similar. The functional group in both is the carbonyl group, C=O (see figure below) where R, R1 and R2 are alkyl groups. Aldehydes and ketones The difference is that aldehydes have the carbonyl on the end carbon atom whereas ketones have the carbonyl group between two carbon atoms. Alkanals are a homologous series of aldehydes based on the corresponding alkanes. The family name ending is '-al' and they are named in exactly the same way as alkanols, except that there is no need to number the position of the carbonyl group since it must be on the end carbon atom (C 1 ). Naming alkanals Q17: Copy the following diagram, or use a photocopy, and complete it using the items in the formula bank. Then compare your answer with the correct answer. © H ERIOT-WATT U NIVERSITY 103 104 TOPIC 5. OXIDATION OF FOOD Naming alkanals table .......................................... Alkanones are a homologous series of ketones based on the corresponding alkanes. They can be recognised by the family name ending '-one', e.g. • testosterone, one of the male sex hormones, is a complex ketone • progesterone, one of the female sex hormones, is a complex ketone Alkanones are also named in a similar way to alkanols. The position of the carbonyl group is shown by inserting the number of the carbon atom in front of the family name ending. This is only necessary when there is more than one possible position for the carbonyl group. Q18: How many carbon atoms are there in the simplest alkanone (please enter using digit(s))? .......................................... Q19: Draw the structure and name the simplest alkanone. .......................................... Q20: What is the simplest alkanone for which it is necessary to number the position of the carbonyl group? Draw structural formulae and name the two isomers. .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD 5.5.1 From name to structure From name to structure From name to structure For each of the following questions, draw the full structural formula for the given compound before revealing the correct answer. Q21: 2-methylpropan-1-ol .......................................... Q22: 3-methylbutanone .......................................... Q23: 2,3-dimethylbut-2-ene .......................................... © H ERIOT-WATT U NIVERSITY 105 106 TOPIC 5. OXIDATION OF FOOD Key Point Alkanals and alkanones are homologous series of aldehydes and ketones respectively, identified by the presence of the carbonyl functional group and the name endings '-al' and '-one' respectively. They are named in a similar way to alkanols. .......................................... 5.6 Oxidation of aldehydes and ketones Structure of aldehydes and ketones Many of the flavour molecules you saw in an earlier topic are aldehydes. Ketones are also found in flavourings. Both of these classes of compounds contain the carbonyl (C=O) group (see figure below). This topic will study these classes of compounds further. Aldehyde and ketone functional groups In a similar way to alcohols, aldehydes and ketones can be distinguished by a difference in their ease of oxidation. Q24: Which class of alcohols cannot be oxidised easily? a) Primary b) Secondary c) Tertiary .......................................... Q25: How many hydrogen atoms are bonded to the -C-OH carbon in tertiary alcohols? a) 2 b) 1 c) 0 .......................................... Q26: By analogy with alcohols, which of these will be difficult to oxidise? a) Aldehydes © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD 107 b) Ketones .......................................... 5.6.1 Oxidising agents In aldehydes, there is a hydrogen atom bonded to the carbonyl group. Consequently, aldehydes can be oxidised easily by mild oxidising agents such as acidified potassium dichromate. Benedict's solution (or the very similar Fehling's solution) and Tollens' reagent can also be used to oxidise aldehydes to the corresponding carboxylic acids. Using Benedict's solution as an oxidising agent Using Tollens' reagent as an oxidising agent From the figure above, you can see that when an aldehyde is heated with Benedict's solution for a few minutes in a hot water bath, the blue solution slowly produces a redorange precipitate of copper(I) oxide. Cu 2+ ions are reduced to Cu + ions. When an aldehyde is heated with Tollens' reagent for a few minutes in a hot water bath, the colourless solution slowly produces a silver mirror on the inside of the test tube. Ag+ ions are reduced to metallic silver. In ketones, the carbonyl group is always flanked by two carbon atoms. Ketones resist further oxidation since this would involve breaking a carbon to carbon bond. Key Point Aldehydes, but not ketones, can be oxidised by a number of oxidising agents, including Benedict's solution, to carboxylic acids. 5.6.2 What are the compounds? What are the compounds? A series of reactions is described which can be used to identify an unknown substance. A colourless liquid (A) had a pleasant aroma of pears. A had molecular formula C5 H10 O2 . When A was refluxed with aqueous sodium hydroxide solution one of the products was a volatile liquid B (formula C 3 H8 O). © H ERIOT-WATT U NIVERSITY 108 TOPIC 5. OXIDATION OF FOOD Gentle treatment of B with acidified potassium dichromate solution produced compound C (formula C3 H6 O), with a powerful pungent odour. Compound C produced a silver mirror with Tollens' reagent. The solution from the reaction of A with sodium hydroxide solution was acidified with hydrochloric acid and distilled. The vapour condensed to a colourless liquid, D, which turned pH paper red. D had empirical formula CH 2 O. When excess copper(II) carbonate was added to D, a gas was evolved and on filtering off the excess carbonate and evaporation, dark bluish-green crystals of substance E were produced. The exercise is to describe the reactions and compounds in this scheme and thereby identify substance A. You can try it yourself and compare your answer with the display answer for the last question in this activity. If you would like help, follow the trail of questions that will guide you to identifying the substances A to E. Q27: It is usual to start with the later substances in these schemes. Compound C reacts with Tollens' reagent. It is likely to be: a) b) c) d) e) f) hydrocarbon alcohol carboxylic acid ester aldehyde ketone .......................................... Q28: The molecular formula of C is C 3 H6 O. What is a possible structure for C? .......................................... Q29: C was prepared from B by gentle dichromate oxidation. What class of compound is B likely to be? a) b) c) d) e) f) hydrocarbon alcohol carboxylic acid ester aldehyde ketone .......................................... Q30: B has formula C 3 H8 O, what structures are possible? .......................................... Q31: Is there any hint as to which of these it is? .......................................... Q32: Now look at compound D. It is produced when the reaction products from A are acidified and 'turns pH paper red'. It is likely to be: a) hydrocarbon © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD b) c) d) e) f) alcohol carboxylic acid ester aldehyde ketone .......................................... Q33: D has empirical formula CH 2 O. What are possible structures? .......................................... Q34: Is there any help as to which it might be? .......................................... Q35: We are told that A has a pleasant aroma and reacts with sodium hydroxide solution to yield B and D. You know the chemical classes to which B and D belong. What class is A likely to be? a) b) c) d) e) f) hydrocarbon alcohol carboxylic acid ester aldehyde ketone .......................................... Q36: The formula of A is C 5 H10 O2 . Assuming what you have answered about the compounds B and C is correct, what is a structure for A? .......................................... Q37: What are the crystals of compound E? .......................................... Q38: The structures and names are all shown in the Display answer. .......................................... 5.7 Antioxidants All living organisms have to invest considerable energy in preventing their own decay. Once life processes cease, these decay processes take over to the benefit of the environment as a whole. These decay, degradation processes are essentially oxidations. With organic material becoming finally carbon dioxide, water and minerals. These can then be used by plants to make new tissue, which in turn will form new animal tissue when eaten. So the cycle continues. The 'carbon cycle' is illustrated next. Most of the cycle involves processes occurring in living organisms: photosynthesis and respiration. The decay of dead tissue (detritus) is recycled by decomposers. © H ERIOT-WATT U NIVERSITY 109 110 TOPIC 5. OXIDATION OF FOOD Healthy forest Decaying trees When the process is not supplied with sufficient oxygen anaerobic decay occurs with the formation, for example, of foul-smelling hydrogen sulfide (H 2 S) instead of the innocuous, fully oxidised water (H2 O) and sulfate ion (SO4 2- ). Consequently the environment is awash with micro-organisms ready to oxidatively degrade whatever is around! One way that plants particularly protect their tissues is by producing antioxidants. Examples of natural antioxidants include: • polyphenols (found in fruits, vegetables and chocolate); • unsaturated hydrocarbons (alpha-carotene - the orange colour in carrots and lycopene - the red colour in tomatoes); • flavonoids (a group of yellow polyphenols). Antioxidants can function in several ways, for example, as: • reducing agents; • free-radical scavengers; • complexing agents for metals; • quenchers of singlet oxygen formation. 5.8 Oxygen and food The most common reaction of oxygen with foods is when edible oils and fats turn 'rancid'. The lipids produce short-chain carboxylic acids, aldehydes and ketones all of which can confer strong pungent flavours and odours. The fats can also taste 'soapy'. When this occurs in milk or cream it is unpleasant; when similar processes occur in cheese maturation, these products enhance the taste. What is happening to cause these changes? One mechanism involves photo-oxidation. © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD 111 The initial reaction produces a peroxide at a C=C double bond. Lipid hydroperoxide formation The peroxide can then break the hydrocarbon chain to produce several products. Final products Q39: We are all being encouraged to eat less saturated fat and more unsaturated and polyunsaturated oils. Can you explain why this is likely to produce more rancid food? .......................................... We can reduce this undesirable oxidation by adding an antioxidant to the food. Some of these are shown in the table below. Antioxidant Natural/synthetic E number Ascorbic acid Natural E300 α-tocopherol Natural E307 BHT Synthetic E321 Propyl gallate Synthetic E310 Some antioxidants © H ERIOT-WATT U NIVERSITY 112 TOPIC 5. OXIDATION OF FOOD Q40: The E numbers from 300 to 340 are mostly antioxidants. Have a look around your food cupboards, or supermarket shelves, to see how many foods have these added. .......................................... Here are the structures of ascorbic acid (vitamin C) and α-tocopherol. α-tocopherol Ascorbic acid Q41: Look at the structure of ascorbic acid. Do you think it will be water- or fat-soluble? a) Water-soluble b) Fat-soluble .......................................... Q42: Which one would you use as an antioxidant to add to cream? a) Ascorbic acid b) α-tocopherol .......................................... 5.8.1 The antioxidant mechanism of ascorbic acid 'Reactive oxygen species', such as hydrogen peroxide, the hydroxyl radical (HO)•) and the superoxide anion (O 2 - ), are all produced in organisms as a normal consequence of oxygen metabolism. The hydroxyl radical, a free radical, has an unpaired electron (shown by a •). This makes it a very reactive species, which can 'attack' many organic molecules. It can be formed by homolytic fission of the O-O bond in hydrogen peroxide. H2 O2 → 2HO• When a free radical interacts with nucleic acids, proteins and lipids, it not only changes that molecule into a free radical, but initiates a chain of new free radicals. These can then no longer function correctly. Ascorbic acid and other antioxidants can terminate these chain reactions by forming a radical-ion. RO• + C6 H7 O6 - → ROH + C6 H6 O6 - • The radical-ion then converts into the stable oxidised form of ascorbic acid. © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD Stable oxidised form of ascorbic acid A similar process occurs with α-tocopherol. Stable oxidised form of α-tocopherol This results in a relatively stable radical. This stability is due to the delocalising nature of the ring structures in tocopherols. The radical then reacts with another reactive oxygen radical or tocopherol radical to result in very stable quinones. 5.8.2 Other methods of preventing oxidation If you can prevent oxygen or air reaching the foods, there can be no reaction. This strategy is employed in the manufacture of potato crisps. These are typically fried in a polyunsaturated vegetable oil under an atmosphere of steam to exclude air. They are then packaged under an atmosphere of nitrogen to prolong their shelf-life. Message on cheese packaging Some fruits, particularly, will rapidly turn brown once they are exposed to the air. This is due to copper-containing enzymes, polyphenol oxidases, which catalyse the oxidation of phenols in the fruit to a brown polymer. © H ERIOT-WATT U NIVERSITY 113 114 TOPIC 5. OXIDATION OF FOOD This deterioration is clearly seen in apples and bananas. Apple turning brown As you know, enzymes work best at certain pH values, and polyphenol oxidase is no exception. If you change the pH of the cut fruit, the browning will be considerably reduced. Try it yourself: cut an apple in half; sprinkle lemon juice on one; leave the other alone. Can you see the difference? 5.9 Ion-electron equations Antioxidants are taking part in redox reactions, so it is possible to write ion-electron equations for their reactions. You have looked at these types of equations in the previous unit. This will revise them and introduce some new compounds. Remember the OILRIG mnemonic. Oxidation and reduction concern the transfer of electrons between compounds. Q43: When you put a piece of zinc into a solution of copper(II) sulfate, what do you see, and why? .......................................... Now you can practise yourself. Q44: Write the overall equation and the oxidation and reduction ion-electron equations for the reactions: 1. magnesium metal with copper sulfate solution 2. sodium iodide solution with gaseous chlorine 3. iron(II) chloride solution with gaseous chlorine 4. sulfur dioxide gas and hydrogen sulfide forming sulfur and water © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD 115 .......................................... Now you can look at the equations for ascorbic acid and butylated hydroxyanisole, two antioxidants. The full structures are shown here. Ascorbic acid (C6 H8 O6 ) Butylated hydroxyanisole (C11 H16 O2 ) Q45: Ascorbic acid loses two hydrogen ions when acting as a reducing agent. Using the molecular formula given, write an ion-electron half equations for this oxidation. .......................................... Q46: Butylated hydroxyanisole (BHA) and the similar butylated hydroxytoluene (BHT) are used extensively to reduce the oxidation of hydrocarbons, such as rubbers, plastics, foods and oils. BHA loses the single phenolic hydrogen when acting as a reducing or radical scavenging agent. Using the molecular formula given, write an ion-electron half equations for this oxidation. .......................................... 5.10 Estimating antioxidant levels The vitamin C (ascorbic acid) content before and after cooking This iodine titration simulation is like to the one you have seen for estimating the vitamin in fruit juices. This time there are extracts of uncooked, lightly cooked and 'stewed' broccoli. You can see how the amount of the antioxidant changes with cooking. © H ERIOT-WATT U NIVERSITY 116 TOPIC 5. OXIDATION OF FOOD Estimating antioxidant levels But ... cooking is not always deleterious to antioxidant levels. A study published in The British Journal of Nutrition last year found that a group of 198 subjects who followed a strict raw food diet had normal levels of vitamin A and relatively high levels of beta-carotene (an antioxidant found in dark green and yellow fruits and vegetables), but low levels of the antioxidant lycopene. Lycopene is a red pigment found predominantly in tomatoes and other rosy fruits such as watermelon, pink guava, red bell pepper and papaya. Several studies conducted in recent years (at Harvard Medical School, among others) have linked high intake of lycopene with a lower risk of cancer and heart attacks. Rui Hai Liu, an associate professor of food science at Cornell University who has researched lycopene, says that it may be an even more potent antioxidant than vitamin C. One 2002 study he did (published in the Journal of Agriculture and Food Chemistry) found that cooking actually boosts the amount of lycopene in tomatoes. He tells ScientificAmerican.com that the level of one type of lycopene, cis-lycopene, in tomatoes rose 35 percent after he cooked them for 30 minutes at 190.4 degrees Fahrenheit (88 degrees Celsius). The reason, he says: the heat breaks down the plants' thick cell walls and aids the body's uptake of some nutrients that are bound to those cell walls. © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD 5.11 Summary Summary Oxidation • When applied to carbon compounds, oxidation reactions result in an increase in the oxygen to hydrogen ratio. Oxidation of alcohols • Primary and secondary alcohols can be oxidised by a number of oxidising agents, including copper(II) oxide and acidified potassium dichromate. • Primary alcohols are oxidised first to aldehydes and then to carboxylic acids. Secondary alcohols are oxidised to ketones. • Tertiary alcohols are resistant to oxidation. Aldehydes and ketones • Alkanals and alkanones are homologous series of aldehydes and ketones respectively, identified by the presence of the carbonyl functional group. They are named in a similar way to alkanols. • Aldehydes and ketones can be identified from the ‘-al’ and ‘-one’ name endings respectively. • Straight-chain and branched-chain aldehydes and ketones, with no more than eight carbon atoms in their longest chain, can be named from structural formulae. • Given the names of straight-chain or branched-chain aldehydes and ketones, structural formulae can be drawn and molecular formulae written. Oxidation of food molecules • Aldehydes, but not ketones, can be oxidised to carboxylic acids. • Fehling’s solution, Tollens’ reagent and acidified dichromate solution can be used to differentiate between an aldehyde and a ketone. Antioxidants • Oxygen reacts with edible oils giving the food a rancid flavour. • Antioxidants are molecules which will prevent these oxidation reactions taking place. • Ion-electron equations can be written for the oxidation of many antioxidants. © H ERIOT-WATT U NIVERSITY 117 118 TOPIC 5. OXIDATION OF FOOD 5.12 Resources Texts Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson, ISBN 978-1444167528 © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD 5.13 119 End of topic test End of topic 5 test Q47: Which of the following six structures contains an aldehyde? c) b) a) d) f) e) .......................................... Q48: Which of the six structures in the above example contains an unsaturated ketone? .......................................... Q49: Give the systematic name for the compound below. .......................................... Q50: Choose the structure below whose correct name is 2,2-dimethylpentan-3-one. a) b) c) d) .......................................... © H ERIOT-WATT U NIVERSITY 120 TOPIC 5. OXIDATION OF FOOD Q51: Give the systematic name for the substance shown below. b) a) c) Both structures a) and c) represent the same molecule, and c) is a type of shortened structural formula called a skeletal formula. It is obtained from a) by missing out all the hydrogen atoms and their bonds (this produces b)) and then missing out the carbon symbols. In a skeletal formula, a carbon atom is presumed to be present at the end of a line at a bend in a line and at a junction of lines. .......................................... Q52: The diagram below shows the skeletal formula of dinoprostone, a naturally occuring substance involved in the induction of contractions during labour. From the name ending, suggest what type of functional group is present. .......................................... Q53: Indicate this functional group by highlighting the group on the diagram. .......................................... Q54: Identify the ketone from the six compounds below. c) b) a) e) d) f) .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD 121 Q55: Identify the compound from the six compounds above which could be oxidised to form the following compound. .......................................... Q56: Identify the compound from the six compounds above which could be hydrolysed when warmed with sodium hydroxide solution. .......................................... Q57: Which of the following statements is true of ketones? a) They contain the carboxyl group. b) They contain the group c) They are formed by oxidation of tertiary alcohols. d) They will not reduce Benedict's solution. .......................................... Q58: After heating for several minutes, as shown in the diagram, the pH indicator solution turned red. Liquid Q could be: a) b) c) d) paraffin propan-1-ol propan-2-ol propanone .......................................... Q59: Which of the © H ERIOT-WATT U NIVERSITY following best describes this reaction? 122 TOPIC 5. OXIDATION OF FOOD a) b) c) d) e) f) Addition Condensation Hydration Hydrolysis Oxidation Reduction .......................................... Q60: Which a) b) c) d) e) f) of the following best describes this reaction? Addition Condensation Hydration Hydrolysis Oxidation Reduction .......................................... Q61: Which gas reacts with edible oils to give food a rancid flavour? a) b) c) d) Chlorine Argon Nitrogen Oxygen .......................................... Q62: What name do we give to molecules which will prevent oxidation reactions taking place? a) b) c) d) Reducing Agents Oxidising Agents Antioxidants Oxidisers .......................................... Q63: The reaction of sulfur dioxide gas sulfur and water can be represented and hydrogen sulfide forming using the equations below: Which one shows an oxidation reaction? a) 1 © H ERIOT-WATT U NIVERSITY TOPIC 5. OXIDATION OF FOOD b) 2 c) 3 d) All of the above .......................................... © H ERIOT-WATT U NIVERSITY 123 124 TOPIC 5. OXIDATION OF FOOD © H ERIOT-WATT U NIVERSITY 125 Topic 6 Soaps, detergents and emulsions Contents 6.1 Prior knowledge . . . . . . . . . . . . . . . . . . 6.2 Soap . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Making soap . . . . . . . . . . . . . . . . 6.2.2 The cleansing action of soap . . . . . . . 6.3 Detergents . . . . . . . . . . . . . . . . . . . . . 6.4 Emulsions . . . . . . . . . . . . . . . . . . . . . . 6.4.1 A closer look at emulsions . . . . . . . . . 6.4.2 Synthetic emulsifiers . . . . . . . . . . . . 6.4.3 Emulsifiers in foods and other emulsions 6.5 Summary . . . . . . . . . . . . . . . . . . . . . . 6.6 Resources . . . . . . . . . . . . . . . . . . . . . 6.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 128 129 136 137 139 139 140 141 143 144 145 Prerequisite knowledge You should already know that: • an ester can be made by reacting a carboxylic acid and an alcohol (National 5, Unit 2); • some uses of esters are in food flavouring, industrial solvents, fragrances and materials (National 5, Unit 2); Esters (Higher, Unit 2, Topic 1) • an ester can be identified from the ester group and by the name containing the -‘yl-oate’ endings; • an ester can be named given the names of the parent carboxylic acid and alcohol or from structural formulae; • structural formulae for esters can be drawn given the names of the parent alcohol and carboxylic acid or the names of esters; • esters have characteristic smells and are used as flavourings and fragrances; • esters are also used as industrial solvents; 126 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS • esters are formed by the condensation reaction between carboxylic acid and an alcohol; • the ester link is formed by the reaction of a hydroxyl group and the carboxyl group; • in condensation reactions, the molecules join together with the elimination of a small molecule, in this case water; • esters can be hydrolysed to produce a carboxylic acid and alcohol; • given the name of an ester or its structural formula, the hydrolysis products can be named and their structural formulae drawn; • the parent carboxylic acid and the parent alcohol can be obtained by hydrolysis of an ester; • in a hydrolysis reaction, a molecule reacts with water breaking down into smaller molecules; Fats and Oils (Higher, Unit 2, Topic 2) • fats and oils are a concentrated source of energy; • they are essential for the transport and storage of fat-soluble vitamins in the body; • fats and oils are esters formed from the condensation of glycerol (propane-1,2,3triol) and three carboxylic acid molecules; • the carboxylic acids are known as ‘fatty acids’ and are saturated or unsaturated straight-chain carboxylic acids, usually with long chains of carbon atoms; • the lower melting points of oils compared to those of fats is related to the higher degree of unsaturation of oil molecules; • the low melting points of oils are a result of the effect that the shapes of the molecules have on close packing, hence on the strength of van der Waals’ forces of attraction. Learning Objectives At the end of this topic, you should be able to state that: Soaps • production of soaps by the alkaline hydrolysis of fats and oils to form water-soluble ionic salts called soaps; • soap ions have a long covalent tail, readily soluble in covalent compounds (hydrophobic), and an ionic carboxylate head which is negatively charged and water soluble (hydrophilic); • during cleaning using soaps and detergents, the hydrophobic tails dissolve in a droplet of oil or grease, whilst the hydrophilic heads face out into the surrounding water; © H ERIOT-WATT U NIVERSITY TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS • agitation of the mixture results in ball-like structure forming with the hydrophobic tails on the inside and the negative hydrophilic head on the outside; • repulsion between these negative charges results in an emulsion being formed and the dirt released; Detergents • detergents are particularly useful in hard water areas; Emulsions • an emulsion contains small droplets of one liquid dispersed in another liquid; • emulsions in food are mixtures of oil and water; • to prevent oil and water components separating into layers, a soap-like molecule known as an emulsifier is added; • emulsifiers for use in food are commonly made by reacting edible oils with glycerol to form molecules in which either one or two fatty acid groups are linked to a glycerol backbone rather than the three normally found in edible oils; • the one or two hydroxyl groups present in these molecules are hydrophilic whilst the fatty acid chains are hydrophobic. © H ERIOT-WATT U NIVERSITY 127 128 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS 6.1 Prior knowledge Test your prior knowledge Q1: a) b) c) d) Fats and oils can be classified as: soaps. esters. glycerol. fatty acids. .......................................... Q2: a) b) c) d) Identify the true statement about fats and oils. Fats are likely to have low melting points compared to oils. Fats and oils are a less concentrated source of energy than carbohydrates. Fats are likely to have a higher degree of unsaturation than oils. Molecules in fats are packed more closely together than in oils. .......................................... Q3: a) b) c) d) Which of the following is likely to be produced by hydrolysis of a fat? Glycerol Glucose Ethanoic acid Ethanol .......................................... 6.2 Soap Soap is an ancient product, made in Roman times by stirring goat fat or lard with hot "lye" (an alkali) and cooking for hours. This process was sometimes given the name saponification. Nowadays the reaction is understood to involve the hydrolysis of the esters present in fats or oils and uses sodium hydroxide or potassium hydroxide. This yields the sodium or potassium salts of the fatty acids. These are soaps like sodium palmitate. (see below). © H ERIOT-WATT U NIVERSITY TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS Saponification of glyceryl tripalmitate Q4: What name is given to the type of chemical reaction used to produce soaps from oils and fats? .......................................... Q5: How many molecules of soap are produced for every molecule of oil or fat which is completely reacted? .......................................... Key Point Soaps are produced by the hydrolysis of fats and oils. 6.2.1 Making soap Making soap This activity is about making soap at home. Go through the 'Making soap' activity below and answer some questions about the chemistry as you proceed. The starting materials are readily available. You could try it yourself and select your own choice of essential oil. If you do the activity at home take care and wear protective clothing such as safety goggles, rubber gloves and an apron; sodium hydroxide can cause injuries. © H ERIOT-WATT U NIVERSITY 129 130 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS Ingredients for soap ready for use Ingredients: • 450 ml water • 147.5 g caustic soda (pure sodium hydroxide) • 307.5 g coconut oil • 400 ml sunflower oil • 400 ml olive oil Work through the stages of making soap: 1. Prepare the sodium hydroxide solution. © H ERIOT-WATT U NIVERSITY TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS Weighing sodium hydroxide on scales Adding sodium hydroxide to water • Weigh out the sodium hydroxide - take care! • Add the sodium hydroxide to the water - never add water to sodium hydroxide. Mix the solution well. Always do this in a plastic or glass container; metals can be attacked by NaOH. • For the process to make an effective soap, there should be a very slight excess of sodium hydroxide, so that all the oils are converted to soap. Q6: If you started with 400g of an oil with a relative formula mass of 800, what mass of sodium hydroxide (NaOH) would you need to complete the reaction? a) 20g of sodium hydroxide b) 40g of sodium hydroxide c) 60g of sodium hydroxide ............................... 2. Prepare the fats and oils. Melting coconut oil in a pan © H ERIOT-WATT U NIVERSITY Measuring out sunflower oil 131 132 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS • Weigh out the coconut oil in a pan and melt the solid over a gentle heat. • Measure out the sunflower oil into a plastic container and add to the coconut oil. • Measure out the olive oil and add this to the pan with the coconut and sunflower oils. 3. Mix the ingredients together. Mixing the ingredients together Checking the pH of the mixture • Mix together the sodium hydroxide solution, molten coconut, sunflower and olive oils. • Stir the mixture gently for about 40 minutes. As the reaction proceeds, you should see a change in colour and texture. • Check the pH of the mixture. Q7: From the colour of the test paper (below) what is the pH of the product mixture? © H ERIOT-WATT U NIVERSITY TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS a) pH is 10 b) pH is 14 c) pH is 7 ............................... Q8: What does that tell you about the number of moles of sodium hydroxide you added? a) There was an excess of NaOH which makes the mixture alkaline. b) There was insufficient NaOH and the solution was acidic. c) The number of moles of NaOH resulted in a neutral solution. ............................... 4. Add any fragrance oils. In this case Tea Tree oil was added to the mixture. Fragrance oils 5. Pour the warm soap mixture into moulds. © H ERIOT-WATT U NIVERSITY 133 134 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS Warm soap mixture being poured into a mould • Allow the soap to solidify for about 24 hours. 6. Dry out the soaps for several weeks in a cupboard. Drying soaps • Note the pH of the soap once it has dried out. © H ERIOT-WATT U NIVERSITY TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS Q9: What do you estimate the pH of the soap is now? a) pH is 7 b) pH is 9 c) pH is 14 ............................... • The soap is now mildy alkaline and will act as a good detergent without damaging the skin. Q10: Why do you think the pH of the soap changes from 14 to 9 after several weeks in a cupboard? (Hint: Think about what is present in air that could alter the pH.) a) The NaOH is oxidised. b) The NaOH is diluted by reacting with water from air. c) The carbon dioxide in air reacts with NaOH to produce sodium carbonate. ............................... Here (below) is the soap in action. It produces a good, fragrant lather! © H ERIOT-WATT U NIVERSITY 135 136 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS Washing soapy hands This method follows that of Kirstie Allsop on http://www.channel4.com/4homes/how-to/craft/how-to-make-soap. Channel 4, .......................................... 6.2.2 The cleansing action of soap Soap molecules have a cleansing action because their structure contains both a covalent hydrocarbon "tail" and an ionic "head" (see figure below). Soap structure The hydrocarbon tails stick into the greasy dirt and the heads are attracted to the water © H ERIOT-WATT U NIVERSITY TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS molecules (see below and above). Because the charged ionic heads cover the surface of the grease with negative charges, the grease droplets repel each other and can be carried away. water soap molecules globule of grease The cleansing action of soap Q11: The covalent hydrocarbon tail of a soap dissolves easily in greasy dirt. This is an example of: a) b) c) d) like dissolves unlike. like dissolves like. like repels like. like attracts unlike. .......................................... 6.3 Detergents There is an enormous variety of soaps available and until a few decades ago soaps were used for almost all cleaning purposes. But they all have disadvantages: • soaps are not very soluble in water, so they only work well in hot water; • in 'hard' water areas, where there are magnesium and calcium ions in the water, soaps will produce a 'scum'. Not only is this unsightly but it reduces the detergency of the soap. Synthetic detergents In the 1920s and early 1930s long-chain alcohols were sulfonated and sold as solutions of the sodium salt. Long-chain alkyl aryl sulfonates (with a benzene nucleus) were also produced and sold in the USA but did not make significant inroads into the soap market with the exception of Teepol, a secondary olefin sulfate. Produced by Shell from 1938, Teepol is still made in large quantities in Britain and Europe today and is used for dishwashing, floor care, and in the automotive, engineering and marine industries. © H ERIOT-WATT U NIVERSITY 137 138 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS Propylene tetramer (PT) benzene sulfonate (C 18 H29 SO3 Na) very quickly displaced all other basic detergents and for the period 1950-65 considerably more than half the detergents used throughout the world were based on this. All these synthetic detergents are much more soluble in water than soap and the calcium and magnesium salts are also soluble. One reason for this is that sulfonate and sulfate salts are fully ionised, so attracting solvent water molecules more strongly than the weakly ionised carboxylate salts in soaps. Q12: Environmental concerns Dodecylbenzene sulfonate Microbial degradation of hydrocarbons, for example the natural destruction of crude oil after a spill, occurs much more readily for linear hydrocarbons. The long hydrocarbon chains are metabolised by micro-organisms producing carbon dioxide and water. Look at the structure of dodecylbenzene sulfonate above. Do you think it will degrade readily once exposed to environmentally available micro-organisms? .......................................... In response to environmental concerns, from the mid 1960s, these branched alkyl chains were replaced with linear hydrocarbon chains. All alkylbenzene sulfonates used today have linear hydrocarbon chains. An example is shown: Example of a linear hydrocarbon chain © H ERIOT-WATT U NIVERSITY TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS 6.4 139 Emulsions When you use an oil and vinegar salad dressing you need to shake the two ingredients together before you pour it on your food. Soon after you place it back on the table the oil will rise to form a separate layer above the vinegar. These two materials are immiscible and will always separate into two layers. However, mayonnaise consists mainly of vegetable oil and vinegar; it remains a single material because there are egg yolks added. Oil and vinegar Mayonnaise The egg yolk contains lecithin, an emulsifier, and mayonnaise is one of the many foods which contains emulsifiers. Here are some others: Milk 6.4.1 Ice cream Low-fat spread A closer look at emulsions To make an oil-in-water emulsion, like mayonnaise, you start with the vinegar and emulsifier and have to mix small quantities of the oil in very thoroughly to create tiny droplets. The emulsifier acts just like soap in detergents. The hydrophobic, hydrocarbon part will dissolve in the droplets of oil and coat the surface with hydrophilic, water-soluble parts that allow the oil droplets to remain 'dissolved'. This is shown below: © H ERIOT-WATT U NIVERSITY 140 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS Oil droplet surrounded by emusifier In mayonnaise the oil is dispersed in the water - the oil is the dispersed phase and the water the continuous phase. In butter and margarine the oil is the continuous phase surrounding the dispersed aqueous phase; they are water-in-oil emulsions. Phospholipids are a class of lipids that are a major component of all cell membranes. They have important functions in transfer of ions across cell boundaries. Nerve impulse transmission occurs along nerves as sodium and potassium ions are exchanged between the fluids inside and outside the cells. One example of a phospholipid is lecithin. Lecithin Q13: The structure of lecithin is shown below to illustrate the hydrophilic phosphate and the hydrophobic fatty acid chains. In this structure identify the phosphate group and the hydrocarbon chains. .......................................... 6.4.2 Synthetic emulsifiers Many foods have synthetic emulsifiers added made by partial hydrolysis of fats and oils. © H ERIOT-WATT U NIVERSITY TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS When one or two ester bonds are hydrolysed in a triglyceride molecule, a di- or monoglyceride is formed (see figure below). These are common food emulsifiers. Make sure you can identify the hydrophilic and hydrophobic parts of these emulsifiers. Hydrolysis of a triglyceride .......................................... 6.4.3 Emulsifiers in foods and other emulsions Emulsifiers in foods Have a look to see how many foods in your kitchen or the supermarket have emulsifiers added. Emulsifiers derived from fatty acids have E numbers 470 to 489. Mono and diglycerides are E471. Here is an illustration of what you might see. List of ingredients in ice cream .......................................... © H ERIOT-WATT U NIVERSITY 141 142 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS Other emulsions Foods are not the only places where emulsions are found. You will have heard of 'emulsion paints'. They are water-based paints where the paint polymers are produced in a water continuous phase which stabilises the paint. They are solid-liquid emulsion unlike the liquid-liquid ones in many foods. The polymers are not soluble in water, so will produce a water resistant coating when the water evaporates. Other important emulsions are the silver halide emulsions used in photography. The minuscule crystals of silver chloride and bromide are held in a gelatin continuous phase, instead of falling to the bottom under gravity. Silver chloride and bromide crystals Emulsions are also important in pharmacy. Many creams, ointments, and liniments are oil-in-water emulsions prepared so that they can be applied easily to skin (topically). Emulsions containing a high proportion of water are used in oral preparations where it is important to deliver a fixed dose of the active ingredient. Of course, there are situations where emulsions are not wanted. For example, when crude oil is extracted from wells it usually comes to the surface with emulsified 'produced water'. This has to be separated using a 'demulsifier', so that the oil can be sent for processing and the water sent for disposal. © H ERIOT-WATT U NIVERSITY TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS 6.5 Summary Summary You should be able to state that: Soaps • production of soaps by the alkaline hydrolysis of fats and oils to form watersoluble ionic salts called soaps; • soap ions have a long covalent tail, readily soluble in covalent compounds (hydrophobic), and an ionic carboxylate head which is negatively charged and water soluble (hydrophilic); • during cleaning using soaps and detergents, the hydrophobic tails dissolve in a droplet of oil or grease, whilst the hydrophilic heads face out into the surrounding water; • agitation of the mixture results in ball-like structure forming with the hydrophobic tails on the inside and the negative hydrophilic head on the outside; • repulsion between these negative charges results in an emulsion being formed and the dirt released; Detergents • detergents are particularly useful in hard water areas; Emulsions • an emulsion contains small droplets of one liquid dispersed in another liquid. Emulsions in food are mixtures of oil and water; • to prevent oil and water components separating into layers, a soap-like molecule known as an emulsifier is added; • emulsifiers for use in food are commonly made by reacting edible oils with glycerol to form molecules in which either one or two fatty acid groups are linked to a glycerol backbone rather than the three normally found in edible oils; • the one or two hydroxyl groups present in these molecules are hydrophilic whilst the fatty acid chains are hydrophobic; • when applied to carbon compounds, reduction reactions result in a decrease in the oxygen to hydrogen ratio. © H ERIOT-WATT U NIVERSITY 143 144 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS 6.6 Resources Texts Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson, ISBN 978-1444167528 © H ERIOT-WATT U NIVERSITY TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS 6.7 End of topic test End of topic 6 test Q14: A soap can be made from palm oil by: a) b) c) d) dehydrogenation. acidic hydration. catalytic hydrogenation. alkaline hydrolysis. .......................................... Q15: When a soap is added to grease, which part of the molecule is attracted to the grease? a) b) c) d) The ester group The hydrocarbon tail The carboxyl group The sodium ion .......................................... Q16: Which of the following substances acts as a soap? a) b) c) d) Glyceryl tristearate Sodium stearate Glycerol Ethyl stearate .......................................... Q17: All detergent molecules consist of two distinct regions: a) b) c) d) a water-soluble hydrophobic periphery and a fat-soluble hydrophilic centre. a water-soluble hydrophobic centre and a fat-soluble hydrophilic periphery. a water-soluble hydrophilic part and a fat-soluble hydrophobic part. a water-soluble hydrophobic part and a fat-soluble hydrophilic part. .......................................... Q18: Soap ions have: a) b) c) d) a long non-polar hydrocarbon 'tail' and an ionic carboxylate 'head'. a long non-polar carboxylate 'tail' and an ionic hydrocarbon 'head'. an ionic carboxylate 'tail' and a long non-polar hydrocarbon 'head'. a long polar hydrocarbon 'tail' and a non-polar ionic carboxylate 'head'. .......................................... Q19: Which of the following are reasons for the growth in the use of synthetic detergents? You may choose more than one option. a) Soaps can be harmful to skin. b) Soaps can produce a 'scum' in hard water. c) Soaps are difficult to degrade in the environment. © H ERIOT-WATT U NIVERSITY 145 146 TOPIC 6. SOAPS, DETERGENTS AND EMULSIONS d) Soaps are not very effective in cold water. .......................................... Q20: Using your knowledge of Chemistry, describe how soaps are effective at removing grease. .......................................... Q21: A typical food emulsifier, a monoglyceride, is shown below. hydrophobic part. H 2C O HC OH H 2C OH CO Identify the (CH2)14CH3 .......................................... Q22: Identify the hydrophilic structures in the above monoglyceride. .......................................... Q23: You are presented with some unlabelled bottles containing samples of possible emulsifiers. Describe a simple experiment you could carry out to see if any were effective. .......................................... © H ERIOT-WATT U NIVERSITY 147 Topic 7 Fragrances Contents 7.1 Prior knowledge . . . . . . . 7.2 Essential oils . . . . . . . . . 7.2.1 Steam distillation . . . 7.2.2 Solvent extraction . . 7.2.3 Uses of essential oils 7.3 Terpenes . . . . . . . . . . . 7.3.1 Isoprene . . . . . . . . 7.3.2 Classes of terpenes . 7.3.3 Household examples . 7.4 Summary . . . . . . . . . . . 7.5 Resources . . . . . . . . . . 7.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 150 150 151 152 153 154 155 158 160 160 161 Prerequisite knowledge You should already know: • about the functional groups ester link, carbonyl, and carboxyl (Higher, Unit 2); • some uses of esters are in food flavouring, industrial solvents, fragrances and materials (National 5, Unit 2); • what is meant by the terms hydrocarbon, homologous series, saturated and unsaturated (National 5, Unit 2); • the name and draw structural formulae for straight chain alkanes (C1 - C8), alkenes (C2 - C6) and cycloalkanes (C3 - C6) (National 5, Unit 2); • how to use, appropriately, molecular formulae and full and shortened structural formulae (National 5, Unit 2); • how to identify and draw structural formulae for isomers (National 5, Unit 2); • an ester can be made by reacting a carboxylic acid and an alcohol (National 5, Unit 2); • carboxylic acids can be identified by the carboxyl ending, the COOH functional group and the ‘-oic’ name ending (National 5, Unit 2); 148 TOPIC 7. FRAGRANCES • an alcohol is identified from the -OH group and the ending ‘-ol’ (National 5, Unit 2); • the following can be identified and named from structural formulae: – straight-chained carboxylic acids (National 5, Unit 2); – straight chain alcohols (National 5, Unit 2); – branched-chain carboxylic acids, with no more than eight carbon atoms in their longest chain; – straight-chain and branched-chain aldehydes and ketones, with no more than eight carbon atoms in their longest chain. • structural formulae can be drawn and molecular formulae written when given the names of: – straight-chain alcohols (National 5, Unit 2); – branched-chain carboxylic acids (Higher 5, Unit 2); – straight-chain or branched-chain aldehydes and ketones (Higher 5, Unit 2); • structural formulae can be drawn when given the name of straight chained carboxylic acids (National 5, Unit 2); • aldehydes and ketones can be identified from the ‘-al’ and ‘-one’ name endings respectively (Higher, Unit 2); • aldehydes, but not ketones, can be oxidised to carboxylic acids (Higher, Unit 2). Learning Objectives At the end of this topic, you should be able to state that: Essential oils • essential oils are concentrated extracts of the volatile, non-water soluble aroma compounds from plants; • essential oils are widely used in perfumes, cosmetic products, cleaning products and as flavourings in foods; • essential oils are mixtures of organic compounds; • terpenes are key components in most essential oils; Terpenes • terpenes are unsaturated compounds formed by joining together isoprene (2methylbuta-1,3-diene) units; • terpenes are components in a wide variety of fruit and floral flavours and aromas; • terpenes can be oxidised within plants to produce some of the compounds responsible for the distinctive aroma of spices. © H ERIOT-WATT U NIVERSITY TOPIC 7. FRAGRANCES 7.1 Prior knowledge Test your prior knowledge Q1: What is the name of the following ester? a) b) c) d) Ethyl ethanoate Ethyl propanoate Propyl ethanoate Propyl Propanoate .......................................... Q2: What is the name of the following carboxylic acid? a) b) c) d) Methanoic acid Ethanoic acid Propanoic acid Butanoic acid .......................................... Q3: Which of the following best describes the following molecule? a) b) c) d) Alkane, saturated Alkane, unsaturated Alkene, unsaturated Alkene, saturated .......................................... Q4: Which of the following statements is true of ketones? a) b) c) d) They contain the C=O group. They are formed by oxidation of tertiary alcohols. They contain the carboxyl group. They will reduce Benedict's solution. .......................................... © H ERIOT-WATT U NIVERSITY 149 150 TOPIC 7. FRAGRANCES 7.2 Essential oils The word "essential" is applied to natural products in two very distinct ways. You have already met "essential" amino acids. These amino acids are "necessary" to humans because our own metabolism cannot make them. Without them in our diet we become ill. "Essential" in the case of 'essential oils' derives from the noun "essence" - the oils embody the plant's odour, its essence. They are not necessary for our well-being. They are concentrated extracts of the volatile, generally oily, aroma compounds from plants. Like many natural products they consist of several different compounds. Extraction of essential oils from plant materials Some plants that yield common oils are shown. Citrus Lavender Eucalyptus Essential oils are derived from various sections of plants. There are two different procedures used to extract the oils from the bulk plant material: • Steam distillation • Solvent extraction 7.2.1 Steam distillation In steam distillation, steam from a boiler is passed through the plant material. At first condensation occurs, but once the plant is hot, steam will continue to travel through carrying with it the volatile components of the plant. The steam and oils are then condensed and the aromatic oil will form a concentrated layer above the water. © H ERIOT-WATT U NIVERSITY TOPIC 7. FRAGRANCES 151 Condenser Steam Boiling Water Plant material Hot plate Essential oil Water Steam distillation process The process works because when immiscible liquids are heated the total vapour pressure is the sum of the vapour pressures of the individual components. Thus the volatile oils, having a significant vapour pressure at 100 ◦ C, will be carried over with the steam. Non-volatile materials will be left. Q5: If these oils are 'volatile', why not just heat the plant and drive off the volatile oil, which could then be condensed to a concentrated product? .......................................... Steam distillation Try setting up an experiment to carry out a simple steam distillation of an essential oil from a fragrant plant. The apparatus pictured below and any fragrant plant material can be used. Simple steam distillation process .......................................... 7.2.2 Solvent extraction Mixing the plant material with a suitable solvent and filtering off the resulting solution will extract the oils. On account of the hydrophobic, water-insoluble nature of oils, the solvents must be of similar nature. © H ERIOT-WATT U NIVERSITY 152 TOPIC 7. FRAGRANCES In commercial solvent extraction, supercritical carbon dioxide can be used. This is carbon dioxide maintained at a pressure where it turns into a liquid with very good solvent properties. Q6: a) b) c) d) Which of these do you think might be used as a solvent? Water Methanol Ethyl ethanoate Dilute hydrochloric acid .......................................... Q7: Compared with a normal solvent, what do you think the advantages of supercritical carbon dioxide might be? .......................................... Q8: There is a problem with solvent extraction compared with steam distillation. Can you think what this might be? Think about what you are trying to extract from the plants. .......................................... 7.2.3 Uses of essential oils The mind map below illustrates some the ways that essential oils are used. Uses of essential oils Look at a wide range of the consumer products in your home and see if you can identify any materials that could be classed as essential plant products. Many commercial products, of course, will just state 'fragrance', so you may not find this task so easy. Here are some examples of consumer products you might find. © H ERIOT-WATT U NIVERSITY TOPIC 7. FRAGRANCES Lypsyl contains: citronellol, geraniol, and limonene 153 Sudocrem contains: lavender fragrance Flash wipes contain: citronellol and Dettol wipes contain: hexyl cinnamal geraniol and limonene .......................................... Key Point 'Essential oils' can be extracted from suitable plant sources by steam distillation or solvent extraction. 7.3 Terpenes Many natural fragrance and aroma compounds are in a large and varied class of organic compounds called terpenes. Many are found in the resins from conifer trees (see picture below) that make turpentine, hence the name. They also have important roles in biochemistry; steroid hormones, for example, are derived from terpenes. © H ERIOT-WATT U NIVERSITY 154 TOPIC 7. FRAGRANCES Conifer tree (Scots pine, Fritham Plain by Jim Champion is licensed under CC BY SA 2.0) They were the first compounds to be extracted from natural sources in ancient times for use as flavourings and fragrances. For example, both frankincense and myrrh, used by ancient Egyptians and mentioned in The Bible, are terpenes. • Frankincense, an oil containing about 75% terpenes or terpene derivatives, is obtained by steam distillation of olibanum resin. • Myrrh is a resinous material containing two terpenes with analgesic (pain relieving) activities. 7.3.1 Isoprene Terpenes are derivatives of isoprene. The simplest terpenes can be formed from a head-to-tail union of two isoprene units. Look carefully at the hydrocarbon below with five carbon atoms. All terpenes can be formed by combinations of this compound. Isoprene structure Q9: Isoprene is a 'common' name. From the structure above what is the IUPAC name for this compound? a) 2-Methyl-buta-1,3-diene b) 2-Methyl-butene c) Buta-1,3-diene © H ERIOT-WATT U NIVERSITY TOPIC 7. FRAGRANCES d) Methyl-buta-1,3-diene .......................................... Formation of myrcene from isoprene The image below shows the formation of myrcene from isoprene. Myrcene is one of the main terpenes found in hops, used to add a bitter flavour to beer. .......................................... 7.3.2 Classes of terpenes Myrcene is described as a monoterpene as it is made from joining two isoprene units together. The addition of further isoprene units, each with five carbon atoms, gives rise to a classification of terpenes, as shown in the table below. This is further information and you are not required to learn the information in this table. © H ERIOT-WATT U NIVERSITY 155 156 TOPIC 7. FRAGRANCES Class Number of carbon atoms monoterpene 10 sesquiterpene 15 diterpene 20 sesterterpene 25 triterpene 30 tetraterpene 40 Terpene classes Once the basic carbon skeleton of the terpene is formed, it can undergo two important changes which give rise to the large number of different compounds. These are: • the formation of derivatives, particularly those with oxygen-containing functional groups; • cyclisation (ring formation). Q10: Which of the following families of organic compounds contain oxygen? a) Alcohols b) Aldehydes c) Ketones d) Carboxylic acids e) Esters f) Alkanes g) Alkenes .......................................... Some examples of terpenes can be seen below. Farnesol, a linear terpene, is an important intermediate in the biosynthesis of steroids. α-Selinene is a sesquiterpene found in celery. Farnesol © H ERIOT-WATT U NIVERSITY TOPIC 7. FRAGRANCES 157 α-Selinene The isoprene units in selinene Q11: Can you identify the three 5-carbon isoprene .......................................... © H ERIOT-WATT U NIVERSITY units in selinene? 158 TOPIC 7. FRAGRANCES 7.3.3 Household examples Monoterpenes Q12: Looking at the names of the following five monoterpenes, which functional groups do you think might be present? 1. α-Phellandrene (eucalyptus) 2. Menthol (peppermint) 3. Citral (lemon grass) 4. Limonene (citrus fruit) 5. Terpineol (lilac) .......................................... These smaller terpenes are all used as fragrances: • Diterpenes - The diterpenes cembrene (pine) and retinol (vitamin A) are used as antioxidants and some have antimicrobial and anti-inflammatory properties. • Triterpenes - Squalene is found in large amounts in shark liver. In humans squalene is one of the terpene precursors on the pathway to steroids. • Tetraterpenes - As you have seen before, in Topic 5 Antioxidants, carotenes and lycopene (plant colourings) are used as antioxidants. • Polyterpenes Natural rubber is a polyisoprene. It has a molecular mass of 100,000 to 1,000,000, and has traces of other materials. It is very elastic and, because there are many double bonds, it is subject to oxidation (perishing). Chemical structure of natural rubber Charles Goodyear, in 1839, developed vulcanisation. The natural rubber is heated with sulfur. This forms cross-links with the linear chains of polyisoprene. These have the effect of strengthening the polymeric material. The greater the amount of sulfur crosslinking, the harder will be the material. Finely-divided carbon (carbon black) is also added. This has the effect of improving its strength and durability, particularly for rubber tyres. © H ERIOT-WATT U NIVERSITY TOPIC 7. FRAGRANCES 159 Vulcanised rubber structure Latex resin from trees is processed to produce tyres. Latex resin taken from trees © H ERIOT-WATT U NIVERSITY Tyres on a truck 160 TOPIC 7. FRAGRANCES 7.4 Summary Summary You should now be able to state that: Essential oils • essential oils are concentrated extracts of the volatile, non-water soluble aroma compounds from plants; • essential oils can be extracted from suitable plant sources by steam distillation or solvent extraction; • essential oils are widely used in perfumes, cosmetic products, cleaning products and as flavourings in foods; • essential oils are mixtures of organic compounds; • terpenes are key components in most essential oils; Terpenes • terpenes are unsaturated compounds formed by joining together isoprene (2-methylbuta-1,3-diene) units; • terpenes are components in a wide variety of fruit and floral flavours and aromas; • terpenes can be oxidised within plants to produce some of the compounds responsible for the distinctive aroma of spices. 7.5 Resources Texts Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson, ISBN 978-1444167528 © H ERIOT-WATT U NIVERSITY TOPIC 7. FRAGRANCES 7.6 End of topic test End of topic 7 test Q13: Essential oils are "essential" because: a) b) c) d) they are major components of the plant extract. our bodies cannot be healthy without them. plants need them for their metabolism. they cannot be replaced by other substances. .......................................... Q14: A common technique for extracting essential oils from plant material is: a) b) c) d) steam reforming. fractional distillation. esterification. steam distillation. .......................................... Q15: Many essential oils contain terpenes formed by joining: a) b) c) d) isoprene units. ester units. isopentane units. alcohols. .......................................... Q16: Which of the following are uses for essential oils? a) Cosmetics b) Cleaning products c) Perfumes d) Flavourings e) Textiles f) Fuels .......................................... © H ERIOT-WATT U NIVERSITY 161 162 TOPIC 7. FRAGRANCES © H ERIOT-WATT U NIVERSITY 163 Topic 8 Skin care Contents 8.1 Prior knowledge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 166 8.3 Ultraviolet (UV) radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Classification of UV radiation . . . . . . . . . . . . . . . . . . . . . . . . 166 168 8.3.2 Reducing the effects of UV radiation . . . . . . . . . . . . . . . . . . . . 8.3.3 Sun protection factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 171 8.4 Free radical reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Free radical chain reactions . . . . . . . . . . . . . . . . . . . . . . . . . 173 174 8.4.2 Free radical scavengers . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 179 8.6 Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 180 Prerequisite knowledge You should already know that: • many plants are used by chemist in the design and manufacture of many everyday products such as pharmaceuticals soaps, cosmetics, dyes, medicines, foods or food colourings (National 4, Unit 2); • plants are a source of carbohydrates and oils which can be used for food or fuel (National 4, Unit 2); • some uses of esters are in food flavouring, industrial solvents, fragrances and materials (National 5, Unit 2). You should already be able to identify, and understand the chemistry of: • alcohols (Higher, Unit 2, Topic 4); • carboxylic acids (National 5, Unit 2); • esters (Higher, Unit 2, Topic 1); • flavor molecules (Higher, Unit 2, Topic 3); 164 TOPIC 8. SKIN CARE • proteins during cooking (Higher, Unit 2, Topic 3); • oxidation reactions in carbon compounds (Higher, Unit 2, Topic 4); • soaps (Higher, Unit 2, Topic 5); • detergents (Higher, Unit 2, Topic 5); • emulsions (Higher, Unit 2, Topic 5); • fragrance molecules (Higher, Unit 2, Topic 6). Learning Objectives At the end of this topic, you should be able to state that: UV radiation • ultraviolet radiation (UV) is a high-energy form of light, present in sunlight; • exposure to UV light can result in molecules gaining sufficient energy for bonds to be broken; • this is the process responsible for sunburn and also contributes to aging of the skin; • sun-block products prevent UV light reaching the skin; Free radicals • when UV light breaks bonds, free radicals are formed; • free radicals have unpaired electrons and, as a result, are highly reactive; • free radical chain reactions include the following steps: initiation, propagation and termination; Free radical scavengers • many cosmetic products contain free radical scavengers; • free radical scavengers are also added to food products and to plastics. © H ERIOT-WATT U NIVERSITY TOPIC 8. SKIN CARE 8.1 Prior knowledge Test your prior knowledge Q1: What is the name of the following ester? a) b) c) d) Methyl ethanoate Methyl propanoate Propyl ethanoate Propyl propanoate .......................................... Q2: The following alcohol could be classified as: a) b) c) d) primary. secondary. tertiary. quaternary. .......................................... Q3: What will be the final product of oxidation of the following molecule? a) b) c) d) Aldehyde Ketone Carboxylic acid Alcohol .......................................... © H ERIOT-WATT U NIVERSITY 165 166 TOPIC 8. SKIN CARE 8.2 Introduction This topic looks at skin care products, mainly from the effects of the sun's radiation and products which offer protection. It is divided into sections on ultraviolet radiation, free-radical reactions and free-radical scavengers. Sunlight can have a profound effect on skin causing cancer and a range of skin changes. Many skin changes that were previously thought to be due to ageing can now be attributed to ultraviolet radiation damage. One of the most aggressive forms of skin cancer is melanoma. Melanoma on skin 8.3 Ultraviolet (UV) radiation Ultraviolet (UV) radiation is part of the spectrum of electromagnetic radiation that spans the extremely short wavelength gamma radiation (from atomic radioactive decay) to the long wavelength radio waves. You can see that the wavelength of radio waves is similar to a football pitch; in contrast, the wavelength of gamma radiation is much smaller than an atom. The whole spectrum is shown in the following image. © H ERIOT-WATT U NIVERSITY TOPIC 8. SKIN CARE 167 Energy Wavelength (λ) in metres 10-12 atom 10-10 Gamma rays virus germs pollen 10-8 10-6 10-4 X-rays Ultraviolet Infrared bee child football pitch 10-2 1 102 Microwaves Radio waves visible Electromagnetic spectrum The spectrum of visible light (all the colours of the rainbow) forms only a small part of the whole. Visible light has wavelengths from about 700 nm* for red light to 400 nm for violet. [* nanometres: 1 nm = 10 -9 m] The part of the spectrum immediately to the shorter wavelengths of violet is called the 'ultraviolet'. Humans cannot see this radiation, although it is visible to some insects which use it to 'see' the nectar-bearing parts of flowers, and to navigate. Q4: Look at the electromagnetic spectrum again. Comparing the wavelengths/frequencies of visible light and UV radiation, which has the greater energy? a) Visible light b) Ultraviolet radiation .......................................... The Sun is our own star. It is ultimately responsible for providing all the energy we use on Earth. Without the Sun's radiation the Earth would soon become a lifeless, intensely cold place. Throughout antiquity man has always realised the importance of the Sun, and even today we continue to study its structure and are greatly influenced by its changes. © H ERIOT-WATT U NIVERSITY 168 TOPIC 8. SKIN CARE Sun rising over Stonehenge The Sun radiates energy in most regions of the electromagnetic spectrum. In fact the study of the universe in regions of the spectrum other than the traditional visible has enhanced our knowledge of the life cycle of stars and galaxies, and opened up our understanding of its development since the 'Big Bang'. Click ESA site to see several images of a near neighbour galaxy, the Andromeda galaxy, viewed at several different wavelengths. The Andromeda galaxy is about 2.5 million light years away! Let's return to our Sun, only about 8.5 light minutes away! Remember a light year is the distance light travels in one year; about 10 13 km. 8.3.1 Classification of UV radiation Radiation can be absorbed and interact with matter in several ways. For example, when you sit in the sun on a warm, sunny day your body is physically warmed by the infrared radiation. This is absorbed by the molecules in your body and causes an increase in the vibrations of the bonds in these molecules. Since 'heat' is a measure of the vibration energy in a substance, our body is warmed up. Absorption of ultraviolet and visible radiation in organic molecules is restricted to certain functional groups (chromophores) that contain valence electrons of low excitation energy. This means that many molecules with strong absorptions in the visible and ultraviolet frequencies have conjugated double bonds, that is alternating double and single bonds. The structure of indigo, the blue dye used to colour jeans, is shown. You can see how many double bonds there are. © H ERIOT-WATT U NIVERSITY TOPIC 8. SKIN CARE 169 Structure of indigo Ultraviolet light from the Sun is divided into several bands depending on wavelength or frequency. You will see these types mentioned on sunscreen creams for example. Names Abbreviation Wavelength Ultraviolet A, long wave, near UVA 400 - 315 nm Ultraviolet B, medium wave, middle UVB 315 - 280 nm Ultraviolet C, short wave, far UVC 280 - 100 nm low, super, extreme LUV, SUV, EUV 100 - 10 nm The sun emits ultraviolet radiation in the UVA, UVB, and UVC bands. Most of this is in the UVA band. The smaller proportion of UVB, however, is largely responsible for the sunburn and other damage caused by exposure to the Sun. 8.3.2 Reducing the effects of UV radiation UV radiation from the sun is dangerous to living organisms. Fortunately, the Earth has developed a natural defence to this radiation. As the Earth's atmosphere became richer in oxygen and more and more species became terrestrial the Earth developed a layer of ozone (O3 ) in the upper atmosphere. This ozone layer blocks 97-99% of the Sun's UV radiation from penetrating through the atmosphere. Sun-protection creams Some molecules can absorb UV light and so prevent the damage it causes to molecules in our skin. These usually act by absorbing UVB radiation, the wavelength that causes skin damage. This can range from mild sun burn to severe skin cancers. © H ERIOT-WATT U NIVERSITY 170 TOPIC 8. SKIN CARE Sun protection creams contain one or more of the following ingredients: • Organic chemicals that absorb UV radiation • Inorganic particulate materials that reflect, scatter or absorb UV, for example, titanium dioxide (TiO2 ) or zinc oxide (ZnO) • Organic particulates that act by a combination of absorption and reflection Typical molecules that absorb UV radiation without harming the skin are octyl methoxycinnamate, avobenzone and benzophenone-3. Octyl methoxycinnamate Avobenzone Benzophenone-3 Q5: Look at the structures of these three molecules. Which functional groups do you think are used to absorb UV radiation? © H ERIOT-WATT U NIVERSITY TOPIC 8. SKIN CARE a) b) c) d) C=C bonds Carbonyl groups Methoxy (CH3 -O-) groups All of the above .......................................... Q6: Why do you think that more than one UV-absorbing groups are present in these molecules? .......................................... Sunscreen ingredients Have a look at some sunscreen products and see if any of the compounds named above are present. You will see a long list of ingredients. Many of these are concerned with producing a cream that is pleasant to use. The compounds present to absorb a wide range of UV wavelengths are degraded by this action, so there are also stabilisers added. These have a narrower wavelength range of UV absorption, but save the sunscreen from becoming ineffective. Q7: Many creams also contain α-tocopheryl acetate and BHT. Why do you think these are added? .......................................... 8.3.3 Sun protection factor The sun protection factor (SPF) of a sunscreen is a laboratory measure of the effectiveness of a sunscreen. The higher the SPF, the more protection a sunscreen offers against UVB (the ultraviolet radiation that causes sunburn). The SPF is the amount of UV radiation required to cause sunburn on skin with the sunscreen on, relative to the amount required without the sunscreen. It is not directly related to time: an SPF of 15 does not allow you to remain exposed for 15 x longer than without the cream, since the actual protection may be lost due to swimming, for example, and the amount of radiation will change dramatically depending on then time of day. Two sunscreen creams Two sunscreen creams, with SPFs of 8 and 35 are shown. © H ERIOT-WATT U NIVERSITY 171 172 TOPIC 8. SKIN CARE The ingredients for the creams, including their place in the ingredient list, are listed below. Remember, ingredients are listed in order of concentration - highest first. Cream A contains: • benzophenone-3 (third); • octyl methoxycinnamate (fifth); • avobenzone (sixth). Cream B contains: • avobenzone (fifth); • octyl methoxycinnamate (eighth). Q8: Which cream do you think has SPF 35? a) A b) B .......................................... Q9: Explain the reason for your answer to the question above. .......................................... None of the sun creams is as effective as the body's own defence mechanism - the production of a sun tan. The brown pigment, melanin, is able to absorb a wide range of UV frequencies and dissipate the energy effectively as harmless heat. © H ERIOT-WATT U NIVERSITY TOPIC 8. SKIN CARE 8.4 173 Free radical reactions All chemical reactions involve the breaking and making of bonds. The products of a reaction and the actual way in which it occurs will be strongly influenced by the way in which the bonds break. In covalent bonding, electrons are generally shared in pairs between two atoms, e.g. in the hydrogen bromide molecule (see below). H Br X Y or a general bond Covalent bond During bond breaking (also called bond fission), the electrons are redistributed between the two atoms. There are two ways in which this can occur. Homolytic and heterolytic bond breaking In homolytic fission, the two shared electrons separate equally, one going to each atom, as shown below. X Y X + Y Homolytic fission The single dot (·) beside each atom represents the unpaired electron that has been retained by each atom from the shared pair. However, the atoms are highly reactive because unpaired electrons tend to attack other species. Highly reactive atoms or groups of atoms containing unpaired electrons are called free radicals. Free radicals are most likely to be formed when the bond being broken is essentially non-polar, i.e. the electrons are more or less equally shared. This type of bond fission occurs typically when the energy of ultraviolet radiation is used to break bonds. In heterolytic fission, both of the shared electrons go to only one of the two atoms producing ions, as shown below. X Y X Heterolytic fission © H ERIOT-WATT U NIVERSITY + + Y - 174 TOPIC 8. SKIN CARE Heterolytic fission is more likely when a bond is already polar. For example, the carbon to bromine bond in bromomethane is polar and can break heterolytically, the pair of electrons going to the more electronegative bromine atom (see below). δ+ H 3C δ- Br + H 3C Br - C-Br bond fission This type of bond fission results in the production of positive and negative ions. You will study these types of bond breaking in more detail when you consider the ways in which reactions actually take place. Free radicals and skin care There is a considerable amount of material in the media linking changes in people's skin, for example the reduction in its elasticity as we age, to the damage that occurs to structural components of skin as a consequence of free radicals. You should use the Internet and look at a range of commercial products which purport to reduce or even reverse these changes. Whilst there undoubtedly is evidence that free radical reactions can damage biological structure, you should look behind the advertised claims for any evidence that would stand scientific scrutiny. .......................................... 8.4.1 Free radical chain reactions One of the interesting features of some free radical reactions is that they can become extremely fast. The reaction of methane with chlorine does not occur in the dark but occurs explosively on exposure to sunlight to produce chloromethane and hydrogen chloride. A hydrogen atom has been replaced by chlorine on each methane molecule. A substitution reaction has taken place. This is an example of a chain reaction. Chain reactions take place in three stages initiation, propagation and termination : 1. Initiation The reaction is started when a chlorine molecule absorbs a photon of light of the correct wavelength. Homolytic fission of the Cl-Cl bond occurs, producing some chlorine atoms. Note the dot representing an unpaired electron. © H ERIOT-WATT U NIVERSITY TOPIC 8. SKIN CARE Since these contain unpaired electrons they are free radicals and highly reactive. 2. Propagation The chlorine radicals then react with methane molecules. The chlorine radical removes a hydrogen atom from methane to form hydrogen chloride at the same time producing a methyl radical. (CH3 · is a methyl radical) The reactive methyl radical attacks a chlorine molecule to form the other product, chloromethane. This produces another chlorine radical which continues the reaction and propagates the chain, which is aself-sustaining cycle called a free radical chain reaction. 3. Termination The chain reaction stops whenever two free radicals collide and combine. Two chlorine atoms form a chlorine molecule. Two methyl radicals combine to form an ethane molecule. A methyl radical and a chlorine atom combine to form chloromethane. All the stages in this reaction occur very rapidly which explains the explosive nature of the nature. Hydrogen and bromine Hydrogen and bromine react together to form hydrogen bromide. The following four statements describe aspects of the reaction mechanism. © H ERIOT-WATT U NIVERSITY 175 176 TOPIC 8. SKIN CARE 1. No reaction occurs in darkness. 2. If light of an appropriate wavelength is supplied, there is a violent reaction. 3. Thousands of HBr molecules are produced for each photon absorbed. 4. The presence of substances with unpaired electrons slows down the reaction. Answer the following questions. Q10: Why is light needed? .......................................... Q11: What type of step is this? .......................................... Q12: Write an equation for this step. .......................................... Q13: Write the number of the statement which gives the best evidence that a chain reaction is involved. a) b) c) d) 1 2 3 4 .......................................... Q14: Explain your answer to the previous question. .......................................... Q15: Write equations for the propagation steps. .......................................... Q16: Why should substances with unpaired electrons slow down the reaction? .......................................... 8.4.2 Free radical scavengers In order to mitigate the effects of free radicals on organisms, natural free radical scavengers are produced. These act as terminators, reacting with available free radicals and halting the chain reactions. Two of these are melatonin and vitamin E. Melatonin Melatonin is found in animals, plants and microbes. In mammals it is produced during darkness by the pineal gland in the brain from where it enters the circulation. The blood concentration of this hormone controls many of our daily (circadian) rhythms, for example, causing our bodies to prepare for sleep. Melatonin has been found to be produced within certain cells where it acts as a powerful free radical scavenger, particularly protecting the cell's DNA. It has been shown to react directly with some carcinogens. © H ERIOT-WATT U NIVERSITY TOPIC 8. SKIN CARE 177 Melatonin structure Vitamin E Vitamin E (α-tocopherol) is a lipid-soluble antioxidant and free radical scavenger which protects cell membranes form lipid peroxidation by reactive oxygen free radicals. It was discussed earlier with regard to food spoilage (Topic 5.8 - Oxygen and food). Vitamin E is important for the prevention of anaemia, but its precise role in metabolism is not clear. Since α-tocopherol is a fat-soluble vitamin it can accumulate in body tissues and may cause unwanted effects. The recommended daily amount (RDA) in the UK (EU) is currently 12 mg, with some studies indicating that taking more than 200 mg for an extended period can have deleterious effects on health. Vitamin E structure Free radical scavengers in consumer products Anti-oxidant free radical scavengers are added to many food products to prevent oxidation to undesirable products which cause rancid flavours. They are also added as preservatives to pharmaceuticals. The omega 3 fish oils capsules mentioned in the section on polyunsaturated fats also contain DL-alpha-tocopherol. Many cosmetics contain scavengers as important components in their composition. Whether the addition of these compounds into skin care products can justify the claims made for the products is another matter. Boots 'Protect and Perfect' products Use the internet, or other source material, to investigate the recent research by a dermatologist on a range of anti-ageing products. Look particularly at the results found for the 'Protect and Perfect' range from Boots. © H ERIOT-WATT U NIVERSITY 178 TOPIC 8. SKIN CARE Anti-ageing lotion .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 8. SKIN CARE 8.5 179 Summary Summary You should now be able to state that: UV radiation • ultraviolet radiation (UV) is a high-energy form of light, present in sunlight; • exposure to UV light can result in molecules gaining sufficient energy for bonds to be broken; • this is the process responsible for sunburn and also contributes to aging of the skin; • sun-block products prevent UV light reaching the skin; Free radicals • when UV light breaks bonds, free radicals are formed; • free radicals have unpaired electrons and, as a result, are highly reactive; • free radical chain reactions include the following steps: propagation and termination; initiation, Free radical scavengers • many cosmetic products contain free radical scavengers; • free radical scavengers are also added to food products and to plastics. 8.6 Resources Text • Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson, ISBN 978-1444167528 © H ERIOT-WATT U NIVERSITY 180 TOPIC 8. SKIN CARE 8.7 End of topic test End of topic 8 test Q17: Ultraviolet radiation is part of the spectrum of a) b) c) d) radiation. electromotive electromagnetic electronic electrical .......................................... Q18: The frequency of ultraviolet radiation is visible light. a) lower than b) the same as c) higher than .......................................... Q19: The energy of ultraviolet radiation is similar to: a) b) c) d) the energy of chemical bonds. gamma radiation. atmospheric energy. the vibration energy of molecules. .......................................... Q20: On Earth, what provides us with the protection from much of the Sun's UV radiation? .......................................... Q21: UV radiation breaks chemical bonds by: a) b) c) d) ion formation. heterolytic fission. ionisation. homolytic fission. .......................................... Q22: UV radiation breaks chemical bonds resulting in the formation of free radicals, which are characterised by: a) not being attached to anything else. b) moving rapidly. c) having unpaired electrons. d) being very reactive. .......................................... Q23: Free radical chain reactions have three steps; enter these below in the order in which they occur. .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 8. SKIN CARE Q24: In a chain reaction initiated by the energy of a single photon of UV radiation: a) the photon is regenerated at the end of the reaction. b) the reaction is unlikely to occur if the radiation is absent. c) there is a large number of product molecules produced. d) only one product molecule is very rapidly produced. .......................................... Q25: Sunscreen products contain 'free radical scavengers'. Using your knowledge of chemistry, comment on how these work. .......................................... Q26: The reactions below all occur in a region of the Earth's atmosphere called the stratosphere. Which two reactions help to maintain a constant level of ozone in the stratosphere? a) O3 → O2 + O (energy from UV radiation) b) O2 + O → O3 c) O2 → O + O (energy from UV radiation) d) O3 + Cl → O2 + ClO e) O3 + ClO → 2O2 + Cl .......................................... © H ERIOT-WATT U NIVERSITY 181 182 TOPIC 8. SKIN CARE © H ERIOT-WATT U NIVERSITY 183 Topic 9 End of unit test Contents 184 TOPIC 9. END OF UNIT TEST End of unit 2 test Q1: a) b) c) d) e) f) g) h) Esters can be produced by the reaction of alcohols with . increase hydrolysis hydrogenation decrease hydration esters carboxylic acids alcohols .......................................... Q2: a) b) c) d) e) f) g) h) . Oils can be converted into fats by the process called increase hydrolysis hydrogenation decrease hydration esters carboxylic acids alcohols .......................................... Q3: a) b) c) d) e) f) g) h) Many fruit flavours are due to the presence of . increase hydrolysis hydrogenation decrease hydration esters carboxylic acids alcohols .......................................... Q4: When applied to carbon compounds, reduction results in a oxygen to hydrogen ratio. a) b) c) d) e) f) g) h) in the increase hydrolysis hydrogenation decrease hydration esters carboxylic acids alcohols .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 9. END OF UNIT TEST Q5: The product of the oxidation of the above compound is: a) b) c) d) 2-methylpentan-4-one. 2-methylpentanal. 4-methylpentan-2-one. 4-methylpentanal. .......................................... Q6: Rum flavouring is based on the compound with the formula shown above. It can be made from: a) b) c) d) butanol and methanoic acid. ethanol and propanoic acid. ethanol and butanoic acid. propanol and ethanoic acid. .......................................... Q7: What product(s) would be expected upon dehydration of the following alcohol? a) b) c) d) 2-methylbut-2-ene only 2-methylbut-1-ene only 3-methylbut-1-ene and 2-methylbut-2-ene 2-methylbut-2-ene and 2-methylbut-1-ene .......................................... © H ERIOT-WATT U NIVERSITY 185 186 TOPIC 9. END OF UNIT TEST Q8: a) b) c) d) Fats have higher melting points than oils because: fats are more loosely packed. fat molecules are more saturated. fats have more cross links between molecules. fats have more hydrogen bonds. .......................................... Q9: Aspirin is one of the most widely used painkillers in the world. It has the following structure. Which two functional groups are present in an aspirin molecule? a) b) c) d) Aldehyde and ketone Hydroxyl and carboxyl Carboxyl and ester Ester and aldehyde .......................................... Q10: Proteins can be denatured under acid conditions. During this denaturing, the protein molecule: a) b) c) d) changes shape. is neutralised. is hydrolysed. is dehydrated. .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 9. END OF UNIT TEST Q11: Highlight the area in the above diagram to identify a peptide link. .......................................... Q12: Highlight the area in the above diagram to identify the hydroxyl group. .......................................... Q13: On hydrolysis, this section of protein produces three amino acids, none of which can be made by the human body. What name is given to such amino acids? .......................................... Q14: Which of the following amino acids could not be produced on hydrolysis of this section of protein? © H ERIOT-WATT U NIVERSITY 187 188 TOPIC 9. END OF UNIT TEST b) a) c) d) .......................................... The following triglyceride is found in fats and oils Q15: Which type of compound do the triglycerides in fats and oils belong to? .......................................... Q16: The hydrolysis of the triglyceride above produces an alcohol and long chain fatty acids. Name the alcohol produced. .......................................... © H ERIOT-WATT U NIVERSITY TOPIC 9. END OF UNIT TEST Q17: If the hydrolysis is carried out in alkaline conditions, give the general name of the other product. .......................................... Q18: The structural formula of a sunscreen product, octyl methoxycinnamate is shown below. Indicate the ester bond on this structure. .......................................... Here is a list of the ingredients listed on a bottle of sunscreen: aqua, glycerol, silicone oil, long chain glycol monoesters, parfum, octyl methoxycinnamate, alpha-tocopherol. Q19: Comment on the effectiveness of this as a skin care product offering protection against UV damage. .......................................... Q20: Why do you think that alpha-tocopherol has been included in the formulation? .......................................... Q21: Aspirin is a widely used medicine - see the structure of aspirin below. It is advised that it is stored in dry, cool conditions. Using your knowledge of chemistry, comment on the reasons why aspirin should be stored under these conditions. .......................................... Q22: A student makes the following statement: "Sugar can be used to produce alcohol, a carboxylic acid and the ester ethyl ethanoate". Using your knowledge of chemistry, comment on the accuracy of the student's statement. .......................................... .......................................... © H ERIOT-WATT U NIVERSITY 189 190 GLOSSARY Glossary Aldehyde an organic compound with a carbonyl functional group (C=O) at the end of the molecule Alkanals a homologous series of aldehydes based on the corresponding alkanes by changing one of the terminal carbon atoms into a carbonyl group Alkanones a homologous series of ketones based on the corresponding alkanes by changing one of the middle chain carbon atoms into a carbonyl group Amide links a group of atoms formed by condensation polymerisation of amino acids during the formation of proteins. The amide link can be identified as -CO-NH- and occurs where each pair of amino acids has joined together Condensation a condensation reaction is one in which two molecules combine to form a larger molecule at the same time eliminating a small molecule such as water Denaturing denaturing of a protein involves physical alteration of the molecular shape as a result of temperature or pH changes Electronegative electronegativity is a measure of the ability of an atom to attract a bonded pair of electrons - the more electronegative, the stronger the attraction Enzymes protein molecules which act as catalysts in biological processes Essential (in the sense of an amino acid) is a necessary material required by living organisms for normal growth Free radicals atoms or molecule containing unpaired electrons Free radical scavengers molecules which can react with free radicals to form stable molecules and prevent chain reactions Heterolytic fission both of the shared electrons go to only one of the two atoms producing ions Homolytic fission the two shared electrons separate equally, one going to each atom © H ERIOT-WATT U NIVERSITY GLOSSARY Hydrogenation the addition of hydrogen to a carbon to carbon multiple bond Hydrolysis the breakdown of a molecule by reaction with water Ketone an organic compound with a carbonyl functional group (C=O) within the carbon chain (ie. not on one of the end carbons) Oxidation when applied to carbon compounds, oxidation reactions result in an increase in the oxygen to hydrogen ratio Peptide links an amide link which is found in a living organism Polyunsaturated a polyunsaturated molecule has more than one carbon to carbon unsaturated bond Proteins biological polymers of small molecules called amino acids Redox reaction a reaction in which one reactant gains electrons and another reactant loses electrons Reduction when applied to carbon compounds, reduction reactions result in a decrease in the oxygen to hydrogen ratio Saponification the process by which soaps are made from fats and oils in a hydrolysis reaction Triglycerides molecules formed through the condensation of one glycerol molecule with three fatty acid molecules Unsaturated an unsaturated molecule has at least one carbon to carbon double bond. An unsaturated hydrocarbon does not contain the maximum number of hydrogen atoms for a given carbon atom framework Volatile a volatile substance evaporates very easily to form a gas Volatility a measure of how easily a molecule will evaporate © H ERIOT-WATT U NIVERSITY 191 192 ANSWERS: TOPIC 1 Answers to questions and activities 1 Esters Test your prior knowledge (page 3) Q1: b) Ethanol Q2: d) Butanoic acid Q3: c) Flavourings Esters in fruit smells (page 4) Q4: Methyl butanoate Q5: Propyl ethanoate Answers from page 8. Q6: Methyl Q7: Butyl Answers from page 8. Q8: Ethanoate Q9: Methanoate Answers from page 9. Q10: Ethanoic acid Answers from page 9. Q11: Methanol Answers from page 9. Q12: Methyl ethanoate Naming and structure of esters (page 9) Q13: Ethyl butanoate Q14: Propyl propanoate © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 1 Q15: Propyl ethanoate Q16: Methyl methanoate Q17: Pentyl ethanoate Q18: Ethyl methanoate Drawing ester structures (page 12) Q19: Q20: Q21: There are several possible answers: CH3 CH2 CH2 CH2 OOCH; HCOOCH 2CH2 CH2 CH3 ; C4 H9 OOCH; HCOOC4 H9 . Q22: There are several possible answers: CH3 CH2 CH2 OOCCH3 ; CH3 COOCH2 CH2 CH3 ; C3 H7 OOCCH3; CH3 COOC3 H7 . © H ERIOT-WATT U NIVERSITY 193 194 ANSWERS: TOPIC 1 Making an ester (page 13) Q23: Some of the substances involved are volatile, notably the alcohol and ester. The vapours will condense back to a liquid on the cold surface and drip back into the mixture. In this way, no material is lost. Q24: c) Basic Q25: The basic sodium hydrogencarbonate solution will neutralise the sulfuric acid and any unreacted carboxylic acid, preventing the smell of the carboxylic acid from interfering with the smell of the ester. Q26: carbon dioxide Esterification (page 14) Q27: water Identifying the parent acid and alcohol (page 16) Q28: ethanoic Q29: methanol Identifying esters (page 17) Q30: Propanoic acid and ethanol Q31: Ethanoic acid and methanol Q32: Ethanoic acid and ethanol Q33: Methanoic acid and ethanol Q34: Ethanoic acid and methanol Q35: Methanoic acid and methanol Q36: Ethanoic acid and butanol Q37: Ethanoic acid and butanol Q38: Propanoic acid and ethanol Q39: Butanoic acid and propanol End of topic 1 test (page 20) Q40: a) ethanol and butanoic acid. © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 1 Q41: b) Q42: c) Ethyl ethanoate Q43: b) Q44: Any one from: flavouring, cosmetics, solvent, nail varnish, or lipstick. Q45: Methyl butanoate Q46: A condensation reaction. Q47: Ethanoic acid Q48: Butan-2-ol © H ERIOT-WATT U NIVERSITY 195 196 ANSWERS: TOPIC 2 2 Fats and oils Answers from page 27. Q1: sunflower Q2: sunflower Answers from page 28. Q3: b) Packing in unsaturates Testing with bromine (page 28) Q4: sunflower Q5: sunflower Q6: sunflower Q7: b) The lower melting point of oils compared to fats is caused by the higher unsaturation of oil molecules. Healthy diet exercise (page 32) Q8: Hydrolysis of fats and oils (page 34) Q9: ester Q10: 3 Q11: b) even. © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 2 197 Q12: c) may or may not be identical. Summary exercise (page 36) Q13: End of topic 2 test (page 38) Q14: a) esters. Q15: d) fat molecules are more saturated. Q16: b) hydrogenation. Q17: c) Mutton fat Q18: b) Glycerol Q19: d) Double bonds C C Q20: d) a lower melting point and more alkene double bonds. Q21: c) Hydrolysis Q22: b) Addition and e) Hydrogenation Q23: a) Fats and oils in the diet can supply the body with energy and d) Molecules in fats are packed more closely together than in oils. O Q24: b) CH3 C OC2H5 © H ERIOT-WATT U NIVERSITY 198 ANSWERS: TOPIC 2 Q25: c) a) Q26: Hydrogenation Q27: 5 Q28: c) Palmitoleic: C15 H29 COOH and d) Linoleic: C 17 H31 COOH © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 3 3 Proteins Test your prior knowledge (page 43) Q1: b) Ethanol Q2: d) Butanoic acid Q3: a) Methyl Ethanoate The structure of amino acids (page 45) Q4: glycine Q5: cysteine Q6: cysteine Condensation of amino acids (page 46) Q7: water Q8: amide link Q9: dipeptide Q10: The three letter words made with these letters don't have to make sense. There are SIX in total. The combinations are: B-A-D, B-D-A, A-B-D, A-D-B, D-B-A, D-A-B. Note that B-A-D and D-A-B are not the same molecule. Answers from page 48. Q11: water Q12: The ester still has a carboxylic acid group, and a hydroxyl group. Q13: Since there are still free carboxyl and hydroxyl groups, these will continue to react to form esters, producing a large molecule with ester links. Polyester formation (page 49) Q14: 3 Q15: 3 Answers from page 50. Q16: phenylalanine Q17: No. The wheat protein is lacking in lysine, which is an essential amino acid. Extra glycine and alanine would not make up for this. © H ERIOT-WATT U NIVERSITY 199 200 ANSWERS: TOPIC 3 Q18: c) Beef Hydrolysis of protein (page 51) Q19: water Q20: • Peptide • Peptide link • Amide • Amide link Q21: 3 Q22: alanine Chromatography (page 52) Q23: 3 Q24: cysteine Q25: d) A, B and D Q26: Solvent A B C D Start Q27: Hydrolysed protein Y contains two amino acids. (1 mark). It contains cysteine because it has a spot which travelled the same as spot "C". (1 mark). It contains an amino acid which cannot be named but it is not A, B or D. (1 mark). Q28: 3 Q29: phenylalanine Q30: alanine © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 3 Answers from page 55. Q31: 3 Q32: 4 Q33: serine Q34: aspartic acid End of topic 3 test (page 58) Q35: B: Enzymes Q36: D: humans must acquire through their diet. Q37: A: digested plant proteins. Q38: C: Condensation Q39: B: F: Q40: A: Q41: B: E: Q42: E: Q43: Q44: Amino acid Q45: Glycine and alanine Q46: There are three spots and so three amino acids. Q47: 1. A and B Q48: 4. Q49: A and D © H ERIOT-WATT U NIVERSITY 201 202 ANSWERS: TOPIC 4 4 Chemistry of cooking Test your prior knowledge (page 69) Q1: b) Ethyl propanoate Q2: b) Ethanoic acid Q3: c) Alkene, unsaturated Answers from page 74. Q4: All except limonene have carbonyl (C=O) groups, as aldehydes or ketones. They have fairly low molecular masses compared with the high molecular mass polymers which constitute much of our food. Q5: a) Volatile Q6: fm6 Volatile or not? (page 75) Q7: 1. Casein - not volatile 2. Anisole - volatile 3. Hex-3-ene-1-ol - volatile 4. Cellulose - not volatile 5. Lignin - not volatile 6. Menthol - volatile Which cooking method? (page 76) Q8: not volatile; soluble in water © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 4 Q9: volatile; soluble in oil Q10: volatile; soluble in water Q11: not volatile; soluble in oil © H ERIOT-WATT U NIVERSITY 203 204 ANSWERS: TOPIC 4 Answers from page 77. Q12: a) Yes Q13: a) Yes Q14: b) No Answers from page 79. Q15: hydrogen Answers from page 81. Q16: c) sequence of amino acids in the polypeptide chain. Q17: d) folding of the backbone of the chain into helices and pleated sheets. Q18: b) interaction of amino acid side chains to give an overall folding to the protein. Q19: a) aggregation of several polypeptide chains into a functional unit. Q20: a) Primary Answers from page 83. Q21: c) protein. Q22: d) Quaternary Other advantages of cooking (page 83) Q23: Cooking will caused denaturing and possibly some hydrolysis of food, both of which will help the body to digest them. Q24: Pineapple contains a protease - an enzyme that breaks down proteins - so eating the two together will enhance digestion! © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 4 Q25: Cooking will caused destruction of many micro-organisms. Some of these could cause stomach upsets. In fact, this process can probably still be thought of as denaturing and hydrolysis, this time of pathogens. Answers from page 84. Q26: b) Poor cut of steak cooked Q27: b) Poor cut of steak cooked Summary questions (page 85) Q28: A - alcohol (or more correctly, because it is attached to a benzene ring - a phenol group); B - amide link; C - carbon-carbon double bond; [D is also a functional group, a methoxy or ether group that you will meet later in your chemistry course]. Q29: b) Non-volatile - will remain in the cooked product Q30: Capsaicin has a fairly high molecular mass, so increasing the London dispersion forces of attraction. It also has alcohol and amide groups which can hydrogen bond together, again increasing the intermolecular attractions and raising the boiling point. Q31: oxygen and nitrogen (magnesium is also present, but that is a metal) Q32: Chlorophyll contains electronegative elements (oxygen and nitrogen) which will produce polar bonds that can form hydrogen bonds with water making it soluble. βcarotene is a hydrocarbon and cannot form hydrogen bonds with water so it does not dissolve. Q33: Vegetable oil (which is often coloured pale yellow owing to traces of dissolved carotenes). Q34: It has a lot of connective tissue. This is not surprising when you think that there is no skeleton to support the musculature. Q35: c) 3 Q36: c) Quaternary, tertiary and secondary Q37: b) Hydrogen bonds © H ERIOT-WATT U NIVERSITY 205 206 ANSWERS: TOPIC 4 Q38: c) Peptide links Q39: d) Hydrolysis End of topic test (page 89) Q40: b) be volatile Q41: c) Beta-carotene Q42: 2. Molecular mass and 4. Type of intermolecular bonding Q43: d) small with weak bonds between molecules Q44: Ethane (C 2 H6 ), (CH3 COOH) Ethanal (CH3 CHO), Ethanol (C2 H5 OH), Ethanoic Acid Q45: b) Hydrogen bonding Q46: a) short time in boiling water Q47: d) The sequence of amino acids in the chain. Q48: The structure of the protein is disrupted (1 mark), causing the shape to change (1 mark). Q49: d) Intermolecular Q50: d) Secondary, tertiary and quaternary Q51: Stewing steak contains a much greater amount of the connective protein, collagen, (1 mark) than prime steak. This makes the meat much more difficult to chew (1 mark), so a cooking method which destroys the collagen (1 mark) is required. When heated above 60 ◦ C, collagen denatures (1 mark), the fibres unwind (1 mark) eventually becoming the soft, gelatinous gelatin (1 mark). Stewing for a long period, converts the tough collagen to soft gelatin, making the meat palatable (1 mark). © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 5 5 Oxidation of food Test your prior knowledge (page 94) Q1: b) Boiling point Q2: b) volatile and insoluble in water. Q3: a) The sequence of amino acids in the chain. Q4: d) Quaternary Oxidation and reduction (page 96) Q5: 0.50 Q6: 0.25 Q7: b) Reduction Q8: c) Neither Q9: a) Oxidation Q10: b) Reduction Q11: c) Neither Q12: a) Oxidation Answers from page 101. Q13: ethanal Q14: ethanoic acid Answers from page 101. Q15: carbonyl Answers from page 102. Q16: propanone © H ERIOT-WATT U NIVERSITY 207 208 ANSWERS: TOPIC 5 Naming alkanals (page 103) Q17: Answers from page 104. Q18: 3 Q19: © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 5 Q20: From name to structure (page 105) Q21: Q22: © H ERIOT-WATT U NIVERSITY 209 210 ANSWERS: TOPIC 5 Q23: Answers from page 106. Q24: c) Tertiary Q25: c) 0 Q26: b) Ketones What are the compounds? (page 107) Q27: e) aldehyde Q28: CH3 CH2 CHO Q29: b) alcohol Q30: (a) CH3 CH2 CH2 OH and (b) (CH3 )2 CHOH Q31: Yes. Dichromate oxidation produced an aldehyde. Structure (a), is a primary alcohol which will oxidise to an aldehyde: structure (b) is a secondary alcohol producing a ketone. So B is structure (a). Q32: c) carboxylic acid Q33: Formula must be (CH 2 O)n . Carboxylic acid have the COOH group with 2 Os so n = 1 is not possible. n = 2; formula C2 H4 O2 ; structure CH3 COOH. n = 3; formula C3 H6 O3 ; structures HOCH2 CH2 COOH and CH 3 CH(OH)COOH. n = 4; formula C4 H8 O4 ; structure HOOCCH2CH2 COOH. etc. Q34: The acid is distilled off, so must be volatile. (CH2 COOH)2 is a strongly hydrogen bonded solid acid (in fact succinic acid). The second formulae are likely to be nonvolatile, with both hydroxyl and carboxylic groups. But what about compound A? Could that help? © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 5 Q35: d) ester Q36: The ester CH 3 COOCH2CH2 CH3 . So the acid is ethanoic, CH3 COOH. Q37: Carboxylic acids and carbonates react to form carboxylate salts with the evolution of gas. So copper(II) carbonate and ethanoic acid will form copper(II) ethanoate and carbon dioxide. Q38: A is propyl ethanoate CH 3 COOCH2CH2 CH3 B is propanol CH3 CH2 CH2 OH C is propanal CH 3 CH2 CHO D is ethanoic acid CH3 COOH E is copper(II) ethanoate (CH 3 COO)2 Cu Answers from page 111. Q39: The initial attack by oxygen on the hydrocarbon chain in fats is on the carboncarbon double bonds. In saturated fats there are fewer of these than in unsaturated and polyunsaturated ones. Answers from page 111. Q40: You will probably be surprised by the number. But remember, when you cook at home, you eat your food soon after it is prepared with little opportunity for it to oxidise. Commercial foods are different. Answers from page 112. Q41: a) Water-soluble Q42: b) α-tocopherol Answers from page 114. Q43: The zinc becomes covered with metallic copper and the solution turns pale. The zinc atoms lose electrons, becoming colourless zinc ions; the copper(II) ions gain electrons becoming copper atoms. In other words the zinc metal oxidises the copper ions. © H ERIOT-WATT U NIVERSITY 211 212 ANSWERS: TOPIC 5 The overall equation (omitting spectator ions) is: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) This can then be split into two ion-electron equations for the oxidation and reduction half-equations. Zn(s) → Zn2+ (aq) + 2e- (oxidation) Cu2+ (aq) + 2e- → Cu(s) (reduction) Remember in these types of equations not only do you have to balance the number of atoms of each element on both sides of the equation, but you have to balance the charge as well. Answers from page 114. Q44: 1. magnesium metal with copper sulfate solution to form magnesium sulfate solution and copper metal © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 5 Mg(s) + Cu2+ (aq) → Mg2+ (aq) + Cu(s) Mg(s) → Mg2+ (aq) + 2e- (oxidation) Cu2+ (aq) + 2e- → Cu(s) (reduction) 2. sodium iodide solution with gaseous chlorine to form sodium chloride and iodine 2I- (aq) + Cl2 (g) → I2 (aq) + 2Cl- (aq) 2I- (aq) → I2 (aq) + 2eCl2 (g) + 2e- → 2Cl- (aq) 3. iron(II) chloride solution with gaseous chlorine to form iron(III) chloride solution 2Fe2+ (aq) + Cl2 (g) → 2Fe3+ (aq) + 2Cl- (aq) Fe2+ (aq) → Fe3+ (aq) + eCl2 (g) + 2e- → 2Cl- (aq) 4. sulfur dioxide gas and hydrogen sulfide forming sulfur and water SO2 (g) + 2H2 S(g) → 4S(s) + 2H2 O(l) SO2 (g) + 4e- → S + 2O2H2 S(g) → 2H+ + S + 2e- Answers from page 115. Q45: C6 H8 O6 → C6 H6 O6 + 2H+ + 2eQ46: C11 H16 O2 → C11 H15 O2 • + H+ + e- End of topic 5 test (page 119) Q47: b) and e) Q48: c) Q49: 4-methylpentan-2-one Q50: d) Q51: 4-methylpent-2-ene Q52: The carbonyl group Q53: Q54: b) Q55: a) Q56: c) Q57: d) They will not reduce Benedict's solution. © H ERIOT-WATT U NIVERSITY 213 214 ANSWERS: TOPIC 5 Q58: b) propan-1-ol Q59: e) Oxidation Q60: d) Hydrolysis Q61: d) Oxygen Q62: c) Antioxidants Q63: c) 3 © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 6 6 Soaps, detergents and emulsions Test your prior knowledge (page 128) Q1: b) esters. Q2: d) Molecules in fats are packed more closely together than in oils. Q3: a) Glycerol Answers from page 129. Q4: hydrolysis Q5: 3 Making soap (page 129) Q6: c) 60g of sodium hydroxide Q7: b) pH is 14 Q8: a) There was an excess of NaOH which makes the mixture alkaline. Q9: b) pH is 9 Q10: c) The carbon dioxide in air reacts with NaOH to produce sodium carbonate. Answers from page 137. Q11: b) like dissolves like. Answers from page 138. Q12: No. The hydrocarbon chain has several branches which will stop the degradative enzymes. © H ERIOT-WATT U NIVERSITY 215 216 ANSWERS: TOPIC 6 Lecithin (page 140) Q13: End of topic 6 test (page 145) Q14: d) alkaline hydrolysis. Q15: b) The hydrocarbon tail Q16: b) Sodium stearate Q17: c) a water-soluble hydrophilic part and a fat-soluble hydrophobic part. Q18: a) a long non-polar hydrocarbon 'tail' and an ionic carboxylate 'head'. Q19: b) Soaps can produce a 'scum' in hard water and d) Soaps are not very effective in cold water. Q20: The hydrophobic hydrocarbon (fat-loving) part of soap (1 mark) can dissolve in the fat droplets (1 mark) and surround them with a coating of hydrophilic (water-loving) carboxylate ions (1 mark). These will allow the droplet to dissolve in the water (1 mark) and be removed (1 mark). Q21: Q22: H 2C O HC OH H 2C OH H 2C O HC OH H 2C OH CO (CH2)14CH3 CO (CH2)14CH3 Q23: Take a mixture of water and vegetable oil and add a small amount of the sample material (1 mark). Shake vigorously to disperse the oil in the water (1 mark). Have a control sample of just oil and water, shaken the same as the one with added sample (1 © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 6 mark). The best emulsifiers will prevent the oil and water separating as fast (1 mark) as the control sample (1 mark). © H ERIOT-WATT U NIVERSITY 217 218 ANSWERS: TOPIC 7 7 Fragrances Test your prior knowledge (page 149) Q1: b) Ethyl propanoate Q2: b) Ethanoic acid Q3: c) Alkene, unsaturated Q4: a) They contain the C=O group. Answers from page 151. Q5: 1. These are oils with quite high boiling points. Heating to the temperature required to distil these could result in some degradation of the oils and possibly the plant materials. Steam distillation does not allow the temperature to rise above 100 ◦ C. 2. As the temperature rises oxidation of the oils could occur. Keeping the distillation in an atmosphere of steam will exclude oxygen and reduce oxidation. Answers from page 152. Q6: c) Ethyl ethanoate Q7: It requires considerable energy input to separate the essential oil and a liquid solvent, often by fractional distillation. Once the pressure is reduced the carbon dioxide will change to a gas and be easily removed. It is also less toxic than some solvents. Q8: The non-polar solvents used to dissolve the essential oil you want will also dissolve other hydrophobic plant components such as waxes and non-volatile oil that you don't want. Answers from page 154. Q9: a) 2-Methyl-buta-1,3-diene Answers from page 156. Q10: Alcohols (-OH), aldehydes (-CHO), ketones (-CO-), carboxylic acids (-COOH) and esters (-COO-). There are others such as phenols (aromatic OH) that you may also know. You should be able to recognise which functional group is present from the name. © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 7 The isoprene units in selinene (page 157) Q11: Monoterpenes (page 158) Q12: 1. α-Phellandrene - carbon-carbon double bond -C=C- 2. Menthol - alcohol -OH © H ERIOT-WATT U NIVERSITY 219 220 ANSWERS: TOPIC 7 3. Citral - aldehyde -CHO 4. Limonene - carbon-carbon double bond -C=C- 5. Terpineol - alcohol -OH End of topic 7 test (page 161) Q13: a) they are major components of the plant extract. Q14: d) steam distillation. Q15: a) isoprene units. Q16: a) Cosmetics, b) Cleaning products, c) Perfumes, and d) Flavourings © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 8 8 Skin care Test your prior knowledge (page 165) Q1: a) Methyl ethanoate Q2: b) secondary. Q3: c) Carboxylic acid Answers from page 167. Q4: b) Ultraviolet radiation Answers from page 170. Q5: d) All of the above Q6: Each functional group will absorb only a relatively narrow band of wavelengths (frequencies) at a particular wavelength. Having more than one group will make the compound absorb over a much wider range, so blocking out a greater amount of the Sun's radiation. Sunscreen ingredients (page 171) Q7: Both of these are antioxidant free radical scavengers which will help to reduce the skin damage caused by UV radiation which escapes absorption. Two sunscreen creams (page 171) Q8: a) A Q9: Cream A has three 'active' compounds; cream B has only two, giving A a better range of UV absorbing power. Because of the order in the ingredient list, cream A probably has more of the active components, offering better protection. Hydrogen and bromine (page 175) Q10: To supply energy to break bonds to start the reaction. Q11: initiation Q12: Br2 + light → 2Br· Q13: c) 3 Q14: Absorption of one photon produces two bromine atoms each of which then reacts to produce a hydrogen bromide molecule and a free radical. This free radical reacts © H ERIOT-WATT U NIVERSITY 221 222 ANSWERS: TOPIC 8 further, propagating the chain and producing more and more hydrogen molecules until the chain is terminated. If a chain reaction was not involved each photon absorbed would produce one product molecule. Q15: H2 + Br· → H· + HBr H· + Br2 → HBr + Br· Q16: These substances would be able to react with the free radicals and so terminate the chain. End of topic 8 test (page 180) Q17: b) electromagnetic Q18: c) higher than Q19: a) the energy of chemical bonds. Q20: The ozone layer Q21: d) homolytic fission. Q22: c) having unpaired electrons, and d) being very reactive. Q23: Initiation, propagation and termination. Q24: b) the reaction is unlikely to occur if the radiation is absent, and c) there is a large number of product molecules produced. Q25: These are molecules which react with free radicals (1 mark) forming stable molecules (1 mark) and terminate or prevent chain reactions (1 mark). Q26: a) O3 → O2 + O (energy from UV radiation), and O 2 + O → O3 © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 9 9 End of unit test End of unit 2 test (page 184) Q1: g) carboxylic acids Q2: c) hydrogenation Q3: f) esters Q4: d) decrease Q5: c) 4-methylpentan-2-one. Q6: c) ethanol and butanoic acid. Q7: d) 2-methylbut-2-ene and 2-methylbut-1-ene Q8: b) fat molecules are more saturated. Q9: c) Carboxyl and ester Q10: a) changes shape. Q11: © H ERIOT-WATT U NIVERSITY 223 224 ANSWERS: TOPIC 9 Q12: Q13: Essential amino acids Q14: a) Q15: Esters Q16: The alcohol is glycerol (propane-1,2,3-triol). Q17: Soap © H ERIOT-WATT U NIVERSITY ANSWERS: TOPIC 9 Q18: Q19: Ingredients are in order of concentration (1 mark), so there is probably very little of the active material (octyl methoxycinnamate) (1 mark) in this product. It might not be good at protecting against uv radiation (1 mark). Q20: Alpha-tocopherol is a free radical scavenger (1 mark), so will act by limiting any skin damage (1 mark) that might occur and reducing degradation of the cream's components due to oxidation (1 mark). Q21: As a general rule, award yourself a half mark (up to a maximum of three marks in total) for each point you made. [NB this is a general rule only, there are no half marks awarded at Higher] • Aspirin should be stored under cool dry conditions as it will eventually decompose (as will all drugs) and heat or damp will speed up this process. • An increase in temperature will increase the rate at which the aspirin decomposes. Aspirin contains an ester link and if this comes into contact with water then this group will start to hydrolyse into the parent alcohol and acid. • The parent acid is ethanoic acid and so a sign that the aspirin has gone off would be a vinegar smell. • Conversely, the carboxyl group on the aspirin could react with any alcohol present to undergo a condensation reaction and form and ester and water. • The formation of the water would further hasten the breaking down of the ester link. Q22: As a general rule, award yourself a half mark (up to a maximum of three marks in total) for each point you made. [NB this is a general rule only, there are no half marks awarded at Higher.] • Not all sugars can be used to produce alcohol. • Glucose can be fermented in the presence of yeast to produce a mixture of carbon dioxide and ethanol. • This alcohol is a primary alcohol and so can be oxidised completely to produce ethanoic acid (a carboxylic acid). • This could be done using Tollen's reagent, Fehlings solution or acidified potassium dichromate. • However, if the alcohol is only partially oxidised then ethanal (an aldehyde) will be produced instead. © H ERIOT-WATT U NIVERSITY 225 226 ANSWERS: TOPIC 9 • If the products of each of these reactions were reacted together in a water bath with the addition of a few drops of concentrated sulphuric acid then the ester ethyl ethanoate would be produced. • However, this reaction is an equilibrium so you will not get 100% conversion into the ester unless you remove the product as it is being formed. © H ERIOT-WATT U NIVERSITY