Download SACE2 Chemistry Workbook Sample Chapter

SACE TWO - AUSTRALIAN CURRICULUM
CHEMISTRY
WORKBOOK
FIRST EDITION
GLEN ARTHUR
RHYS LEWIS
THE AUTHORS
Glen Arthur B.Ed. (Secondary Maths/Science), MRACI
Glen has taught senior Chemistry at Stage 1 and Stage 2 for over 20 years. He has taught in both the public
and independent school sectors over this time.
He has extensive experience in the development of materials, curriculum, and assessment for the SACE
Board of South Australia. He is a member of the SACE Curriculum Leaders Group for Chemistry. He is
an active member of the Royal Australian Chemical Institute, in the Chemical Education Division. He is
also involved in the delivery of Chemistry Curriculum and Methodology in the School of Education, at the
University of Adelaide. He also has over a decade of experience in teaching, moderation, and marking for the
Foundation Studies Program in Chemistry, in conjunction with the South Australian Universities.
Glen is a passionate educator, dedicated to developing the potential of students and Chemistry educators.
Rhys Lewis B.Sc., Grad.Dip.Ed.
Rhys has been a teacher of science for seven years in South Australia. He has taught hundreds of Stage
1 and 2 students in chemistry, physics, and biology. Rhys was Regional winner of the SA Public Teaching
Awards in 2013 and winner of a National excellence in teaching award in 2014. He is also co-author of the
SACE1 Essentials Chemistry Workbook. Rhys excels in his ability to communicate scientific concepts in an
engaging, clear, and concise manner.
PUBLISHING INFORMATION
This Workbook is part of the Essentials series, designed to support the teaching of SACE Stage 1 and 2
subjects in South Australia. It is specially designed to meet the requirements of the SACE Stage 2 Australian
Curriculum Chemistry.
The Essentials Education series is published by
Adelaide Tuition Centre,
21 Fourth Street, Bowden 5007.
TELEPHONE (08) 8241 5568
FACSIMILE (08) 8241 5597
LIBRARY CATALOGUE:
Arthur; Glen - Lewis; Rhys
1. Chemistry, SACE2 Australian Curriculum - 2. Essentials Workbook
ISBN - 978-1-925505-05-4
First edition 2017.
Copyright © Adelaide Tuition Centre 2017.
COPYRIGHT INFORMATION
The copyright of the text of this book remains the property of the authors and the copyright of the diagrams and
cartoons belongs to the publisher. All rights are reserved except under the conditions described in the Copyright
Act 1968 of Australia and subsequent amendments. No part of this publication may be reproduced, stored in a
retrieval system, or transmitted in any form or by any means, without the prior permission of the publisher. While
every care has been taken to trace and acknowledge copyright, the publishers tender apologies for any accidental
infringement where copyright has proved traceable.
TABLE OF CONTENTS
Table of Contents
Preface
Topic One: Monitoring the Environment
1.1 Global Warming and Climate Change
1.2 Photochemical Smog
1.3 Volumetric Analysis
1.4 Chromatography
1.5 Atomic Spectroscopy
Summary Test One: Monitoring the Environment
Topic Two: Managing Chemical Processes
2.1 Rates of Reactions
2.2 Equilibrium and Yield
2.3 Optimising Production
Summary Test Two: Managing Chemical Processes
Topic Three: Organic and Biological Chemistry
3.1 Introduction
3.2 Alcohols
3.3 Aldehydes and Ketones
3.4 Carbohydrates
3.5 Carboxylic Acids
3.6 Amines
3.7 Esters
3.8 Amides
3.9 Triglycerides
3.10 Proteins
Summary Test Three: Organic and Biological Chemistry
Topic Four: Managing Resources
4.1 Energy
4.2 Water
4.3 Soil
4.4 Materials
Summary Test Four: Managing Resources
Solutions
Topic One: Monitoring the Environment
Topic Two: Managing Chemical Processes
Topic Three: Organic and Biological Chemistry
Topic Four: Managing Resources
Appendices
Appendix 1: The Periodic Table of the Elements
Appendix 2: SI prefixes, symbols and values
© Essentials Education
STAGE 2 CHEMISTRY
TOPIC 3: ORGANIC AND BIOLOGICAL CHEMISTRY
3.9 Triglycerides
Draw the structural formula of an edible oil or fat, given the structural formula(e) of the carboxylic acid(s) from
which it is derived.
© SACE 2016
Fatty acids are carboxylic acids that have long (4 to 28 carbon atoms) hydrocarbon chains (Figure 3.89). The
hydrocarbon chain of a fatty acid is either saturated (single carbon‑carbon bonds only) or unsaturated (at least one
carbon‑carbon double or triple bond).
Figure 3.89: Saturated and unsaturated fatty acid molecules.
Fatty acids with one double or triple bond are termed monounsaturated. Fatty acids with more than one double
or triple bond are termed polyunsaturated. Glycerol (propane‑1,2,3‑triol) is a small alcohol with three hydroxyl
groups (Figure 3.90).
Figure 3.90: Glycerol (propane‑1,2,3‑triol).
Glycerol (alcohol) reacts with three fatty acid (carboxylic acid) molecules to form an ester called a triglyceride.
Three molecules of water are formed in each reaction (Figure 3.91).
Figure 3.91: Formation of a triglyceride.
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In many cases, the three fatty acid molecules that combine and form the triglyceride are identical. However, three
different fatty acid molecules can form one triglyceride molecule. Figure 3.92 shows a triglyceride formed in the
reaction of glycerol with stearic acid, linoleic acid, and palmitic acid.
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Figure 3.92: Triglyceride formed from three different fatty acids.
Triglyceride molecules are often represented using a partially condensed structural formula. Figure 3.93 shows the
partially condensed structural formula of triglyceride shown in Figure 3.92.
Figure 3.93: Partially condensed structural formula of the triglyceride molecule shown in Figure 3.92.
Question
50. Draw the structural formula of the triglyceride formed from the fatty acids below.
CH3(CH2)16COOHCH3(CH2)5CH=CH(CH2)7COOHCH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH
(2 marks) KA4
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TOPIC 3: ORGANIC AND BIOLOGICAL CHEMISTRY
Identify and draw the structural formulae of the alcohol and acid(s) from which a triglyceride is derived, given its
structural formula.
© SACE 2016
Triglycerides are esters that undergo hydrolysis in acidic or alkaline solution. Three fatty acids and glycerol are
formed when a triglyceride is hydrolysed in an acidic solution (Figure 3.94).
Figure 3.94: Hydrolysis of a triglyceride molecule in an acidic solution.
Three carboxylate anions are formed when the triglyceride is hydrolysed in an alkaline solution (Figure 3.95).
Figure 3.95: Hydrolysis of a triglyceride molecule in an alkaline solution.
The hydrolysis of a triglyceride occurs inside the cells of living organisms including animals and plants. The
hydrolysis reaction is catalysed by a class of enzymes called lipases. Lipases are present in the saliva and other
digestive secretions inside the human body. The role of lipases is to rapidly break down triglycerides into fatty
acids that can be used for a variety of functions from energy production to the maintenance of the immune,
circulatory and nervous systems.
Question
51. Draw the structural formula of the three fatty acids formed in the hydrolysis of the triglyceride below.
(6 marks) KA4
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TRIGLYCERIDES
Describe and explain the use of a solution of bromine or iodine to determine the degree of unsaturation of a
compound. Draw the structural formula of the reaction product.
Explain how the degree of unsaturation causes differences in the melting points of edible oils and fats.
© SACE 2016
Edible fats and oils are mixtures of organic compounds that contain a high proportion of triglycerides. Edible fats
are distinguished from edible oils by their melting points. Edible fats are solid at standard laboratory temperature
(25°C) whereas edible oils are liquid.
Edible fats
Edible fats are solid mixtures containing a higher concentration of saturated fats when compared with edible oils.
Saturated fats are triglyceride molecules containing only saturated fatty acids (Figure 3.96).
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Figure 3.96: Structural formulae of a saturated fat.
Saturated fats are derived from animals and plants. Land animals contain a higher percentage of saturated fats
than marine animals. Some vegetable products, including coconut oil and palm kernel oil, are higher in saturated
fat than fat derived from any land animal.
Edible oils
Edible oils are liquid mixtures containing a higher concentration of unsaturated fats when compared with edible
fats. Unsaturated fats are triglyceride molecules containing one or more unsaturated fatty acids (Figure 3.97).
Figure 3.97: Structural formulae of an unsaturated fat.
Unsaturated fats are derived from animals and plants. Fish contain many different unsaturated (omega‑3) fats
which are an essential components of the human brain. Most edible oils are derived from plants.
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Material
Source
Saturated fat
(% of total fat)
Unsaturated fat
(% of total fat)
Coconut oil
Plant
92
8
Palm kernel oil
Plant
86
14
Butter fat
Animal
66
34
Beef fat
Animal
52
48
Palm oil
Plant
51
49
Cottonseed oil
Plant
27
73
Olive oil
Plant
15
85
Sunflower oil
Plant
10
90
Melting points
The melting points of edible fats and oils are dependent on the strengths of the intermolecular forces between
triglyceride molecules. Edible fats are solid at standard laboratory temperature (25°C) as the melting point is
greater than 25°C. The higher melting points of edible fats are attributed to the greater proportion of triglyceride
molecules with saturated carbon chains. Shorter distances separate triglycerides with saturated carbon chains
which increases the strength of the dispersion forces between molecules (Figure 3.98).
Figure 3.98: Separation of two triglyceride molecules with saturated carbon chains.
Edible oils are liquids at standard laboratory temperature as the melting point is lower than 25°C. The lower
melting points of edible oils are attributed to the greater proportion of triglyceride molecules with unsaturated
carbon chains. Unsaturated carbon chains often prevent triglycerides from moving close enough together to form
a solid at standard laboratory temperature. Figure 3.99 shows an unsaturated edible oil derived from fish.
Figure 3.99: Separation of two triglyceride molecules with unsaturated carbon chains.
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Degree of unsaturation
Diatomic halogen molecules (X2) undergo addition reactions with carbon‑carbon double bonds. One mole of
diatomic halogen molecules is required to react completely for every mole of carbon‑carbon double bonds
(Figure 3.100).
Figure 3.100: Addition reaction of unsaturated triglyceride with diatomic halogen molecules.
The degree of unsaturation of a triglyceride or fatty acid can be approximated by reacting the material with a
known quantity of a halogen such as bromine (Br2) or iodine (I2). The mass of iodine or bromine that reacts with
exactly 100 g of a triglyceride or fatty acid is called the iodine value or bromine number.
The iodine values of some unsaturated fatty acids and their triglycerides are shown in the table below. Note that
the iodine value increases with the number of double bonds present.
Material
Number of carbon
atoms
Number of C=C
double bonds
Iodine value of
isolated fatty acid
Iodine value of
triglyceride
Myristic acid
14
1
110
105
Linoleic acid
18
2
180
175
Linolenic acid
18
3
275
260
Arachidonic acid
20
4
370
350
In practice, iodine monochloride (ICl) is used instead of iodine (I2) or bromine (Br2) but dilute solutions of bromine
and iodine will provide satisfactory results for a comparison between different fats and oils. It should also be noted
that unwanted side reactions (such as halogen substitution) also occur between unsaturated fatty acids and
halogen molecules which limits the accuracy of any test for unsaturation using bromine or iodine.
Questions
52. Students were given samples of coconut oil, olive oil, cottonseed oil and sunflower oil.
Bromine water was supplied and the students worked in a fume cupboard.
The students placed 1.0 g of each oil into four separate test tubes.
Bromine water was transferred dropwise to each of the four test tubes. Each test tube was then stoppered
and shaken gently.
More bromine water was added if the colour of the bromine water changed from orange to colourless.
If the bromine water remained orange, no more bromine was added.
The students’ results are shown in the table.
Material
Drops of bromine transferred before no colour change was observed.
Coconut oil
1
Cottonseed oil
4
Olive oil
6
Sunflower oil
7
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(a) State and explain which material contains the most saturated fatty acids and triglycerides.
Material: (3 marks) IAE3
(b) Coconut oil is a mixture of saturated and unsaturated fatty acids and triglycerides.
Lauric acid and oleic acid (shown below) are both present in coconut oil.
(i)
Draw the structural formula of oleic acid after sufficient bromine water has been added.
(2 marks) KA4
(ii) The concentration of lauric acid is 52% w/w.
Determine the mass of lauric acid in 1.0 g of coconut oil.
(1 mark) KA4
Explain the role of pressure, temperature, and a catalyst in the hydrogenation of liquid triglycerides to form
triglycerides of higher melting point.
© SACE 2016
Hydrogen gas (H2) reacts with unsaturated fatty acids and triglycerides in the same way as bromine or iodine.
The addition of hydrogen to an unsaturated fatty acid results in the carbon‑carbon double bonds being converted
to into single bonds (Figure 3.101). The process is called hydrogenation.
Figure 3.101: Hydrogenation of linoleic acid and its triglyceride.
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Science and human endeavour:
Hydrogenation is a process that changes unsaturated fatty acids and triglycerides
into saturated compounds. The process was first used in the 1890s by American
chemist James F Boyce Senior. Boyce mixed powdered nickel with a heated
sample of cottonseed oil before bubbling hydrogen gas into the mixture. The
cottonseed oil was converted into a solid fat which was a more useful material for
producing food products and soap. Boyce’s process was developed by French
chemist Paul Sabatier (right). Sabatier was awarded the 1912 Nobel Prize for
his method of hydrogenating organic compounds in the presence of powdered
metals. The process of hydrogenation is used in the food industry to convert plant
oils into more saturated fats which are used in cooking as well as the petrochemical
industry to convert alkenes into alkanes.
The process of hydrogenation is carried out using a finely powdered catalyst such as nickel, platinum or palladium.
Nickel is used most commonly as it is inexpensive relative to platinum and palladium. Metals act as a catalyst by
providing an alternate reaction pathway that has a lower activation energy.
Hydrogenation reactions proceed very slowly at standard pressure and temperature. Increasing the pressure and
temperature results in an increase in the rate of the hydrogenation reaction. Increasing the pressure results in an
increase in the number of collisions between hydrogen and the unsaturated fatty acid or triglyceride molecules.
Increasing the temperature increases the number of collisions as well as the energy of the colliding particles which
results in more successful collisions in each unit of time.
Hydrogenation converts lower melting point oils into higher melting point fats by increasing the proportion of
saturated fatty acids and triglycerides with saturated carbon chains (Figure 3.98).
Question
53. Cottonseed oil is a mixture of fatty acids and triglycerides.
Linoleic acid (55%) and oleic acid (20%) are two of the largest components of cottonseed oil.
The condensed structural formulae of linoleic acid and oleic acid are shown below.
Linoleic acid
Oleic acid
CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH
CH3(CH2)7CH=CH(CH2)7COOH
(a) Draw the structural formula of a triglyceride composed of two linoleic acid molecules and one oleic acid
molecule.
(2 marks) KA4
(b) The triglyceride above can be partially hydrogenated.
During partial hydrogenation, one double bond from each of the three fatty acid chains reacts with
hydrogen.
Draw the structural formula of the triglyceride after partial hydrogenation.
(2 marks) KA4
(c) State the change to the melting point of the triglyceride after the hydrogenation reaction is complete.
(1 mark) KA1
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(d) Partial hydrogenation is carried out at a temperature and pressure that is three times the laboratory
standard (25°C, 1 atm).
Explain the benefit of increasing the pressure and temperature in a hydrogenation reaction.
(3 marks) KA2
Alkaline hydrolysis of triglycerides produces carboxylate ions, which have both hydrophilic and hydrophobic
regions. Explain how such particles form micelles in solutions.
Explain how micelles can dissolve and move non‑polar substances through an aqueous medium or vice versa.
© SACE 2016
Three carboxylate anions are formed when a triglyceride undergoes hydrolysis in an alkaline solution (Figure 3.102).
Figure 3.102: Alkaline hydrolysis of a triglyceride.
Carboxylate anion derivatives of triglycerides have a large non‑polar carbon chain and a small, highly polar carboxyl
group. The carboxyl group interacts strongly with water through the formation of ion‑dipole forces. The carboxyl
group is described as hydrophilic given the strong attraction for water. The hydrocarbon chain is described as
hydrophobic given its limited interaction with water (Figure 3.103).
Figure 3.103: Structural formula of a carboxylate anion showing the hydrophilic and hydrophobic regions.
Carboxylate ions form nanoscopic (between 1 and 100 nm) structures called micelles when placed in a volume
of liquid water or an aqueous solution. The hydrophilic heads are in contact with the molecules of water in
the surrounding solution and the hydrophobic hydrocarbon chains directed towards the centre of the micelle
(Figure 3.104).
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Figure 3.104: Chemical structure of a micelle particle in water.
Micelles and drug delivery
Polymer molecules containing both hydrophobic and hydrophilic regions are used to create large micelle particles
(nanoparticles) that transport drugs around the body. Single polymer molecules called unimers containing
hydrophobic and hydrophilic regions self‑assemble in water or an aqueous solution (such as blood plasma) and
form micelle polymer nanoparticles (Figure 3.105). The unimers are in equilibrium with the micelle polymers.
Increasing the concentration of unimers shifts the equilibrium to the right which increases the size and stability of
the polymer micelles.
Figure 3.105: Formation of a micelle polymer (nanoparticle) in water.
Micelle polymers are nanomaterials that can deliver drugs, genes and other materials to the target site within the
body where the drug has a therapeutic effect. Micelle polymers are amphiphilic as they have both hydrophilic
and hydrophobic components. The spherical core is hydrophobic and this region stores the drug, gene or other
molecule to be transported around the body. The hydrophilic shell allows the micelle to dissolve in an aqueous
medium such as blood plasma and tissue fluid where the drug can be transported around the body. The unimers
are biodegradable and are excreted from the body without causing harmful side effects.
Science and human endeavour:
Micelle polymers were developed in the 1980s and 1990s and have been used to deliver a range of
therapeutic agents to target sites around the body. Micelle polymers used in medical therapy are between
10 and 100 nm in diameter and they can transport very large molecules including DNA and proteins. Micelle
polymers are effective at transporting anti‑cancer drugs to the site of tumours, as well as transporting
drugs to the brain to treat neurodegenerative diseases and delivering radioactive isotopes to a target area
in the body for medical imaging and diagnostics.
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3.10 Proteins
Write the general formula of amino acids and recognise their structural formulae.
Draw the structural formula of the product formed when an amino acid self‑ionises, given its structural formula.
© SACE 2016
Amino acids are organic compounds which contain an amino group (NH2), a carboxyl group (COOH) and an
R group (Figure 3.106).
Figure 3.106: Structural formula of an amino acid (left) and phenylalanine (right).
The R group is an organic side chain that is unique to the twenty (20) different amino acids (Figure 3.107).
Figure 3.107: Amino acid side chains (R‑groups)
Amino acids are amphiprotic compounds that act as proton donors (acid) and proton acceptors (base) depending
on the pH of the solution. In an acidic solution, the amino group accepts a proton forming a protonated amino
acid (Figure 3.108).
Figure 3.108: Formation of a protonated amino acid in acidic solution.
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PROTEINS
In an alkaline solution (pH greater than 7), the carboxyl group donates a proton forming a deprotonated amino
acid (Figure 3.109).
Figure 3.109: Deprotonation of an amino acid in an alkaline solution.
In a pH‑neutral solution, the carboxyl group can donate its proton to the amino group forming an ionised amino
acid called a zwitterion (Figure 3.110).
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Figure 3.110: Formation of a zwitterion.
Zwitterions are electrically‑neutral but are highly polar given the separation of positive and negative charge.
Question
54. Cysteine and serine are amino acids with similar chemical structures.
(a) State which compound has the greater molar mass (no calculation is required).
(1 mark) KA1
(b) Draw the structural formula of cysteine when it is placed in an alkaline solution.
(2 marks) KA4
(c) Draw the structural formula of serine when it is placed in an acidic solution.
(2 marks) KA4
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(d) Draw the structural formula of cysteine when it is placed in a pH‑neutral solution.
(2 marks) KA4
Draw the structural formula of a section of a protein chain that could be formed from amino acids, given their
structural formulae or vice versa.
© SACE 2016
Two amino acids undergo a condensation reaction forming a dipeptide (Figure 3.111).
Figure 3.111: Formation of a dipeptide from two amino acids.
An amide group is formed in the reaction. Amide groups formed in the reaction of two amino acids are often called
peptide groups. The covalent bond between the carbon and nitrogen atoms of the peptide group is called the
peptide bond (Figure 3.112).
Figure 3.112: Structure of a dipeptide showing the peptide group and peptide bond.
Many amino acids combine forming a large (polymer) molecule called a polypeptide (Figure 3.113).
Figure 3.113: Chemical structure of a polypeptide showing peptide groups (in red).
Large polypeptide molecules are called proteins. Proteins perform a large number of essential tasks inside the
cells of biological organisms.
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Question
55. Insulin is a protein that consists of two polypeptide molecules (named chain A and chain B).
(a) A section of chain B is shown below.
Draw the structural formula of the four amino acids shown in this section of the polypeptide.
3
(8 marks) KA4
(b) Chain A consists of 21 amino acids.
The structural formula of four amino acids present in chain A are shown below.
(i)
Draw a section of chain A using the following sequence: cysteine, threonine, serine, isoleucine.
(2 marks) KA4
(ii) Circle one peptide group in the polypeptide you have drawn.
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(1 mark) KA1
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Identify where secondary interactions can occur, given the structural formula of a section of a protein chain.
© SACE 2016
Proteins are large polypeptides containing hundreds to tens of thousands of amino acid subunits. Proteins are
essential to the structure and function of all biological organisms. There are thousands of different types of proteins
in every cell of the body. Proteins carry out a variety of vital functions in the cell including the synthesis of DNA,
sending and responding to chemical signals, providing structural support and catalysing chemical reactions
(carrying out metabolism). The biological function of a protein is dependent on its three dimensional structure and
folding. The structure of proteins is described at four levels.
Primary structure
The primary structure of a protein is in reference to the sequence of the amino acid subunits in the molecule.
Figure 3.114 shows a section of the primary structure of the insulin protein.
Figure 3.114: Section of the primary structure of insulin protein (chain A).
Secondary structure
Large protein molecules are able to fold into three‑dimensional structures that have a unique shape. The protein
molecule rotates such that two peptide groups are close enough for hydrogen bonding to occur. Hydrogen
bonding between peptide groups on different amino acid subunits facilitates the formation of the secondary
structure of the protein. Two examples of secondary structure in proteins are the alpha helix and the beta sheet
(Figure 3.115).
Figure 3.115: Alpha helix (left) and beta sheet (right)
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Figure 3.116 shows hydrogen bonding between the peptide groups of the amino acid subunits in the alpha helix
and beta sheet.
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Figure 3.116: Hydrogen bonding between peptide groups in an alpha helix (left) and beta sheet (right).
Tertiary structure
The alpha helices and beta sheet structures in a protein molecule fold and rotate into a unique three dimensional
shape called the tertiary structure of the protein (Figure 3.117).
Figure 3.117: Tertiary structure of a protein showing the location of two alpha helices.
The tertiary structure of a protein is determined by the interactions between the R groups (organic side chains)
of the amino acid subunits at different locations in the protein molecule. The type of interaction depends on the
polarity of the interacting R groups. Figure 3.118 shows four common interactions between the R groups of
different amino acid subunits at different locations in the protein molecule.
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Figure 3.118: Interactions between amino acid side chains (R groups)
that give rise to the tertiary structure of a protein molecule.
Quaternary structure
Protein molecules combine to form larger protein complexes that have a unique function. The protein collagen is
the main component of skin and tendons and this protein complex consists of three protein molecules. Haemoglobin
is a protein that transports oxygen in the blood and this protein complex consists of four protein molecules. Protein
complexes are stabilised by attractive forces between the R groups on the different protein molecules.
Explain why the biological function of a protein (e.g. an enzyme) may be affected by changes in pH and
temperature.
© SACE 2016
The biological function of a protein is related to its chemical structure. For example, the catalytic function of enzymes
(proteins) is determined by the tertiary structure of the protein molecule. Enzymes are folded in a way that creates
regions called active sites that bind target molecules called substrates. For example, the enzyme zymase has
an active site that bonds to glucose (substrate). While bonded to the active site, glucose is broken down into
ethanol and carbon dioxide (chemical reaction). The shape of the active site is created by the interactions between
the R groups of the amino acid subunits in the enzyme molecule (refer to Figure 3.118 on the previous page).
Changes in the physical environment can alter the interactions between the R groups resulting in a destabilising
of the tertiary structure of the enzyme.
Heat
Heat is a flow of energy. When heat flows into a protein molecule, the atoms vibrate more rapidly. The rapid
vibrations destabilise the interactions between the R groups which maintain the tertiary structure of the protein
molecule. As more heat flows into a protein molecule, the vibrations become large enough to destabilise all
secondary interactions within the protein. The protein unfolds but covalent bonds between the amino acid subunits
preserve the primary structure of the protein. The protein is no longer able to carry about its biological function and
is said to be denatured (Figure 3.119).
Figure 3.119: Denaturation of a folded and functional protein.
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pH
Many amino acid side chains (R groups) contain functional groups that either donate or accept protons. R‑groups
with acidic side chains become deprotonated when the pH of the environment is above 7 (Figure 3.120).
Figure 3.120: Deprotonation of amino acid side chain (R group) in an alkaline (high pH) environment.
R groups with basic side chains become protonated when the pH of the environment is below 7 (Figure 3.121).
Figure 3.121: Protonation of amino acid side chain (R group) in an acidic (low pH) environment.
Secondary interactions between R groups are altered when R groups become protonated or deprotonated. In
most cases, attractive forces are replaced by repulsive forces which leads to the unfolding and denaturation of the
protein molecule. Figure 3.122 shows some theorised changes to the interactions between R groups in protein
molecules in a low pH and high pH environment.
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Figure 3.122: Theorised changes to the interactions between R groups
in protein molecules in a low pH (H+) and high pH (OH–) environment.
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SUMMARY TEST 3: ORGANIC AND BIOLOGICAL CHEMISTRY
Summary Test 3: Organic and Biological Chemistry
1. Whisky is an alcoholic beverage containing hundreds of polar and non-polar organic compounds.
Isoamyl acetate is a compound present in some whiskies that provides an aroma of bananas.
The structural formula of isoamyl acetate is given below.
(a) Write the systematic name of isoamyl acetate.
(2 marks) KA4
(b) Name the functional group present in isoamyl acetate.
(1 mark) KA1
(c) Isoamyl acetate is prepared in the laboratory using ethanoic acid and an alcohol.
Draw the structural formula and provide the systematic name of the alcohol used to produce isoamyl
acetate.
Space for diagram
Systematic name: (4 marks) KA4
(d) A pair of students prepared isoamyl acetate in the laboratory.
The students heated a mixture of the alcohol and ethanoic acid under reflux for several hours.
(i)
State the purpose of heating under reflux when preparing isoamyl acetate.
(2 marks) KA1
(ii) Sulfuric acid is added to the reaction mixture before it is heated under reflux.
Sulfuric acid protonates ethanoic acid forming a reactive carboxylic acid intermediate.
Use this information to state how sulfuric acid increases the rate of reaction between the alcohol
and ethanoic acid.
(1 mark) KA1
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(e) One student spilled a drop of sulfuric acid onto their polyester shirt during the experiment.
A hole developed in the students’ shirt several days later.
Explain how a hole developed in the students’ shirt.
(2 marks) KA2
(f) Vanillin is a flavour compound found in whisky.
Vanillin is extracted from oak (wood) barrels into the whisky during the ageing process.
(i)
Explain how whisky removes vanillin from the wooden barrels.
(2 marks) KA2
(ii) Draw the structural formula of vanillin after it is reacted with Tollens’ reagent.
(2 marks) KA2
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SUMMARY TEST 3: ORGANIC AND BIOLOGICAL CHEMISTRY
Credit will be given for answers to part (g) which correctly use appropriate chemical terms and effectively
communicate knowledge and understanding of chemistry.
Your answer should be confined to the space provided.
(g) The boiling point of vanillin is twice that of isoamyl acetate.
Compound
Boiling point (°C)
isoamyl acetate
142
vanillin
285
•
Explain the difference in the boiling points of vanillin and isoamyl acetate using your knowledge of
intermolecular (secondary) forces.
•
Explain why whisky drinkers can smell isoamyl acetate but not vanillin in their drink.
3
(8 marks) KA2
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77
STAGE 2 CHEMISTRY
TOPIC 3: ORGANIC AND BIOLOGICAL CHEMISTRY
(h) Whisky is produced from the fermentation of sugars extracted from barley (plant).
During the first stage of whisky production, raw barley grains are soaked in water for several days.
During this time the grains produce enzymes called amylases and peptidases.
(i)
State the function of enzymes in biochemical reactions.
(1 mark) KA1
(ii) Peptidases break down the protein matrix that protects the starch granules forming amino acids.
A section of the primary structure of one protein found in the protein matrix is shown below.
Draw the extended structural formula of leucine after the protease breaks down the protein.
(2 marks) KA4
(iii) Name the type of chemical reaction that converts proteins into amino acids.
(1 mark) KA1
(iv) Amylases break down starch (amylose) into maltose.
Write the molecular formula of maltose.
(2 marks) KA4
(v) Bracket the repeating unit in amylose (starch). 78
© Essentials Education
(1 mark) KA4
TOPIC 3
SUMMARY TEST 3: ORGANIC AND BIOLOGICAL CHEMISTRY
(vi) Some amylose is converted to glucose (C6H12O6) in the process.
Write a fully balanced chemical equation to show the conversion of starch to glucose.
(2 marks) KA4
(vii) Glucose is used by the barley plant for growth.
To prevent growth of the plant, the barley grains are transferred to a kiln and heated to a temperature
between 70°C and 90°C for 2-4 hours.
Explain how heating the barley grains prevents growth of the barley plant.
(3 marks) KA2
(viii) Some Scotch whisky distilleries use peat fires to dry the barley grains.
Peat is a solid mixture of organic compounds that is similar to coal.
Organic compounds from peat smoke are transferred to the barley grains which influence the flavour
of the final spirit.
The chemical structure of three compounds transferred from peat smoke are shown below.
Name a functional group present in the three compounds above.
(1 mark) KA1
(ix) Draw the structural formula of eugenol after its reaction with hydrogen gas.
(2 marks) KA4
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79
3
STAGE 2 CHEMISTRY
(i)
TOPIC 3: ORGANIC AND BIOLOGICAL CHEMISTRY
The dried barley mixture is called malt.
In the second stage of the whisky production process, malt is ground into a powder and mixed with
warm water.
Under these conditions, maltose undergoes hydrolysis forming glucose.
Use the diagrams above to explain why both maltose and glucose are highly soluble in water.
(2 marks) KA2
(j)
A microorganism called yeast is added to the mixture of soluble sugars in the third stage of the process.
Yeast contains an enzyme that converts sugars into ethanol and carbon dioxide.
The table below shows the mass of carbon dioxide collected from the fermentation of a number of
carbohydrates in the production of whisky from barley.
(i)
Carbohydrate
Class of carbohydrate
Carbon dioxide produced (g)
glucose
monosaccharide
88
fructose
monosaccharide
90
sucrose
disaccharide
85
maltose
disaccharide
80
lactose
disaccharide
0
starch
polysaccharide
0
In fermentation, one mole of glucose (C6H12O6) is converted to two moles of ethanol (C2H5OH) and
two moles of carbon dioxide (CO2).
Write a fully balanced chemical equation to show the process of fermentation.
(1 mark) KA4
(ii) Determine the minimum mass of glucose that underwent fermentation to produce 88 g of carbon
dioxide.
(4 marks) IAE3
(iii) State the most likely reason why no carbon dioxide was produced when yeast was mixed with
lactose or starch.
(1 mark) KA2
80
© Essentials Education
New Science Workbooks
SACE Stage 2 Australian Curriculum
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