Download SQA CfE Higher Chemistry Unit 2

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 . . .
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1
3
4
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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 . . . . . . .
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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 . . . .
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41
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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 . . . . .
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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 . . . . . . . . . . .
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91
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98
100
103
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109
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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 . . . . . . . .
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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
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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 . . . .
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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
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202
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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 . . . . . . . . . . . .
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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);
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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.
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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
..........................................
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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.
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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).
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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.
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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.
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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.
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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.
..........................................
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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.
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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.
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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.
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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
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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?
..........................................
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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.
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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?
..........................................
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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.
..........................................
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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.
..........................................
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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
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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
..........................................
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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.
..........................................
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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
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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.
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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.
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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.
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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.
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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.
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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.
..........................................
..........................................
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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.
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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
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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.
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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
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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.
..........................................
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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
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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.
..........................................
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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
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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
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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.
..........................................
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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
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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 . . . . . . . . . . . .
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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.
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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
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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.
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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
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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
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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
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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.
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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.
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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?
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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
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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.
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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
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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.
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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
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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
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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)
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59
b)
c)
d)
e)
f)
..........................................
Q40: Which of the following is a β - amino acid?
a)
b)
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TOPIC 3. PROTEINS
c)
d)
e)
f)
..........................................
Q41: Identify the molecules which could be produced when a protein is hydrolysed.
a)
b)
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61
c)
d)
e)
f)
..........................................
Q42:
Identify the molecule which would be present in the hydrolysed sample of the section of
protein above.
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TOPIC 3. PROTEINS
a)
b)
c)
d)
e)
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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.
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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
..........................................
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65
Q48: Which of these protein sections could be P?
1.
2.
3.
4.
..........................................
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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
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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 . . . . . . . . . . . . . .
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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
..........................................
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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)
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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.
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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.
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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
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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?
..........................................
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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
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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:
..........................................
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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?
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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
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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?
..........................................
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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
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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
..........................................
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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
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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?
..........................................
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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?
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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.
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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.
..........................................
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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
..........................................
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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
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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
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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.
..........................................
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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 . . . . . . . . . . . . . . . . . . . .
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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;
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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.
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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
..........................................
..........................................
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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.
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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?
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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
..........................................
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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.
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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
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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.
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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).
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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.
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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.
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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.
..........................................
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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
..........................................
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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
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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).
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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
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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.
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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.
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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
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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.
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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.
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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
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..........................................
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.
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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.
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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.
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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
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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)
..........................................
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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)
..........................................
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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
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following
best
describes
this
reaction?
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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
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TOPIC 5. OXIDATION OF FOOD
b) 2
c) 3
d) All of the above
..........................................
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TOPIC 5. OXIDATION OF FOOD
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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 . . . . . . . . . . . . . . . . . .
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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;
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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.
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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).
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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.
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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.
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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
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Measuring out sunflower oil
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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?
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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.
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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.
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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!
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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
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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.
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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
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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:
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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.
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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
..........................................
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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.
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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.
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6.6
Resources
Texts
Higher Chemistry for CfE, J Anderson, E Allan and J Harris, Hodder Gibson, ISBN
978-1444167528
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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.
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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.
..........................................
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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 . . . . . . .
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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.
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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.
..........................................
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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.
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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.
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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.
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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.
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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
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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.
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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
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α-Selinene
The isoprene units in selinene
Q11: Can
you
identify
the
three
5-carbon
isoprene
..........................................
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units
in
selinene?
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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.
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Vulcanised rubber structure
Latex resin from trees is processed to produce tyres.
Latex resin taken from trees
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Tyres on a truck
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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
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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
..........................................
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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);
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• 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.
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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
..........................................
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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.
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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.
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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.
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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.
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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?
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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.
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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.
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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
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+
+
Y
-
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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.
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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.
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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.
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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.
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Anti-ageing lotion
..........................................
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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
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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.
..........................................
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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
..........................................
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TOPIC 8. SKIN CARE
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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
..........................................
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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
..........................................
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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.
..........................................
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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?
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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.
..........................................
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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.
..........................................
..........................................
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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
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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
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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 .
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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.
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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
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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.
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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
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ANSWERS: TOPIC 2
Q25:
c)
a)
Q26: Hydrogenation
Q27: 5
Q28: c) Palmitoleic: C15 H29 COOH and d) Linoleic: C 17 H31 COOH
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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.
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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
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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
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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
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ANSWERS: TOPIC 4
Q9: volatile; soluble in oil
Q10: volatile; soluble in water
Q11: not volatile; soluble in oil
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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!
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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
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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
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ANSWERS: TOPIC 5
Naming alkanals (page 103)
Q17:
Answers from page 104.
Q18: 3
Q19:
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ANSWERS: TOPIC 5
Q20:
From name to structure (page 105)
Q21:
Q22:
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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?
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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.
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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
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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.
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ANSWERS: TOPIC 5
Q58: b) propan-1-ol
Q59: e) Oxidation
Q60: d) Hydrolysis
Q61: d) Oxygen
Q62: c) Antioxidants
Q63: c) 3
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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.
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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
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ANSWERS: TOPIC 6
mark). The best emulsifiers will prevent the oil and water separating as fast (1 mark) as
the control sample (1 mark).
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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.
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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
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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
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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
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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
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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:
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ANSWERS: TOPIC 9
Q12:
Q13: Essential amino acids
Q14: a)
Q15: Esters
Q16: The alcohol is glycerol (propane-1,2,3-triol).
Q17: Soap
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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.
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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.
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