Higher Chemistry Unit 2

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Higher ChemistryUnit 2

• Natures Chemistry

2
What you should know

• That molecular structure and physical properties
  of hydrocarbons are related.
• The names, molecular and structural formula of
  straight and branched chain alkanes, alkenes and
  cycloalkanes.
• How to identify isomers and draw their
  structural formulae.
• What is meant by saturated and unsaturated
  carbon compounds and how they can be
  distinguished.
• Addition reactions

3

• The name of the functional group present in
  alcohols and the properties of alcohols
• The names, molecular and structural formula of
  straight and branched chain alcohols
• The name of the functional group present in
  carboxylic acids and the properties of
  carboxylic acids
• The names, molecular and structural formula of
  straight and branched chain carboxylic acids

4
Organic Chemistry
Originally, chemical compounds were divided into
2 classes Inorganic or Organic
Organic compounds were derived from living
things. It was believed that they contained a
vital force and could not be made from
inorganic compounds (non-living sources).
Organic chemistry is the study of carbon
compounds
5

• Organic molecules may be as simple as methane,
  CH4
• or as complicated as cholesterol

6
L.I. To learn about esters Section 1 (a) S.C.
By the end of this lesson you should be able
to

• State the name ending present in all esters.
• State the name of the functional group present in
  an ester.
• Draw the functional group present in an ester.
• State the two types of molecules which contribute
  to the name of an ester.
• Draw the structural formula for an ester when
  given its name.
• Draw the structural formula for an ester when
  given the name of the parent alcohol and
  carboxylic acid.
• Describe the characteristic smell of an ester.

7
Fruit Flavours (a) Esters
An ester can be identified by the name endings
-yl oate. Esters contain the carboxylate
functional group (COO-)
Carboxylate group
8
Naming Esters An ester can be named given the
name of the parent alcohol and carboxylic acid or
from shortened and full structural formulae. The
parent alkanol gives the start of the name ending
in -yl. The parent acid gives the second part of
the name ending in -oate. So the ester formed
by reacting ethanol with propanoic acid would be
ethyl propanoate
9

• Workbook activities

10
Name of alcohol Name of carboxylic acid Name of ester Full structural formula of ester
Ethanol Propanoic acid Ethyl propanoate    
Propanol     Methanoic acid Propyl methanoate   
Methanol     Butanoic acid  Methyl butanoate  
Propanol     Ethanoic acid  Propyl ethanoate  
 Butanol     Methanoic acid  Butyl methanoate  
 Ethanol   Butanoic acid  Ethyl butanoate  
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• 1. Methanol Ethanoic acid Methyl
  ethanoate
• 2. Butan-1-ol Ethanoic acid
  Butyl ethanoate

12
L.I. To learn about making esters (b) S.C.
By the end of this lesson you should be able to

• State the name of the reaction in which esters
  are formed from carboxylic acids and alcohols.
• State the definition of a condensation reaction.
• State how an ester linkage can be formed by the
  condensation reaction between a hydroxyl group
  and a carboxyl group, using full structural
  formulae.

13
(b) Making Esters

• In condensation reactions small molecules join
  together to form a bigger molecule by the
  elimination of water.
• Esters are formed by the reaction of a carboxylic
  acid and an alcohol
• Carboxylic acids contain the functional group
  carboxyl (-COOH)
• Alcohols contain the functional group hydroxyl
  (-OH)
• Acids and carboxylic acids can react by
  condensation reactions to form esters.

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Alcohol Acid ? Ester
Water  
The ester link is formed by the reaction of a
hydroxyl group with a carboxyl group. The
reaction is reversible - an equilibrium is set
up. If the water is removed the equilibrium is
shifted to the side of the products .
Concentrated sulphuric acid has a high affinity
for water and is used to remove it and shift the
equilibrium to the right.
Carry out the experiment to make an ester in the
lab (Workcard 1)
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Workbook activity

• What is the purpose of the wet paper towel?
•  
•  
•  
•  
•  
• Why is the reaction mixture heated in a water
  bath?
•  
•  
•  
•  
• Why is the reaction mixture poured onto sodium
  hydrogen carbonate solution?
•  
•  
•  
• Observation

The paper towel acts as a condenser to prevent
volatile gases escaping (it cools them down).
To increase the temperature of the reaction and
make the reaction happen faster. A Bunsen cannot
be used as the reactants are flammable.
It was added to the products to neutralise the
sulphuric acid and any excess carboxylic acid.
The product had a strong smell and formed a layer
over the water since esters are immicible in
water.
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Carboxylic Acid Alcohol Name of ester Shortened structural formula of ester Smell of ester
Ethanoic acid Pentanol Pentyl ethanoate CH3COO(CH2)4CH3      banana
  Ethanoic acid   Ethanol    Ethyl ethanoate   CH3COOC2H5     fruity odour
Methanoic acid Ethanol Ethyl methanoate   HCOOCH2CH3    rum-like odour
Methanoic acid     Propanol Propyl methanoate  HCOOCH2CH2CH3   plum odour
17
L.I. To learn about the uses of esters (c) S.C.
By the end of this lesson you should be able
to

• Describe the three main uses of esters.

18

• (c) Uses of esters
• Esters are oily liquids with generally very
  pleasant fruity smells and have a range of uses.
• Many esters are used as flavourings, in perfumes
  and as solvents.
• Workbook activity

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Name Shortened Structural Formula Odour/Flavour  
 Pentyl ethanoate CH3COO(CH2)4CH3   Banana  
 Octyl ethanoate CH3COO(CH2)7CH3   Orange  
Methyl Butanoate    CH3CH2CH2COOCH3   Pineapple  
3-Methylbutyl Butanoate CH3(CH2)2COO(CH2)2CH(CH3)2     Apple
 Propyl ethanoate CH3COOCH2CH2CH3     Pear  
Methyl-1-butyl ethanoate   CH3COOCH(CH3)C4H9       Banana
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2-Methylpropyl methanoate    HCOOCH2CH(CH3)CH3     Raspberry  
 Pentyl butanoate C3H7COOC(CH2)4CH3    Apricot, Strawberry  
Benzyl ethanoate   CH3COOCH2C6H5       Peach, flowers  
Ethyl methanoate    HCOOCH2CH3      Rum
Methyl 2-aminobenzoate   C6H4(NH2)COOCH3     Grapes  
Benzyl butanoate C3H7COOCH2C6H5     Cherry  
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Uses of esters as solvents Some of the smaller
esters are quite volatile and are used as
solvents in adhesives, inks and paints pentyl
ethanoate is used in nail varnish for example.
Ethyl ethanoate is one of a number of solvents
used to extract caffeine from coffee and tea.
De-caffeinated products produced with ethyl
ethanoate are often described on the packaging as
"naturally decaffeinated" because ethyl ethanoate
is a chemical found naturally in many fruits.
Caffeine (C8H10N4O2) is an example of a class of
compounds called alkaloids which are produced by
plants. The name alkaloid means alkali-like,
where alkali is a base and hence refers to these
basic properties.
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Caffeine is more soluble in the organic solvent
ethyl ethanoate than in water, so we will extract
caffeine into the organic solvent to separate it
from glucose, tannins, and other water soluble
compounds using a separating funnel. The ethyl
ethanoate portions can be combined and the ethyl
ethanoate removed by evaporation to leave the
caffeine
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• Workbook activity

VOC stands for Volatile Organic Compounds.
Legislation aims to prevent or reduce the direct
and indirect effects of emissions of volatile
organic compounds (VOCs) on the environment and
human health, by setting emission limits for such
compounds and laying down operating conditions
for installations using organic
solvents. Why? The emissions of volatile
organic compounds (VOCs) in the atmosphere
contribute to the formation of the tropospheric
ozone (ozone in the lower atmosphere). Large
quantities of this ozone may be harmful to
people, vegetation, forests and crops. Sensitive
people may suffer irritation of the throat and
eyes, as well as respiratory difficulties.
Tropospheric ozone is also a greenhouse gas.
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L.I. To learn about the hydrolysis of
esters (d) S.C. By the end of this lesson you
should be able to

• Describe how an ester can be hydrolysed using
  full structural formulae.
• State the definition of a hydrolysis reaction.
• State the name of the products formed by the
  hydrolysis of an ester.

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In hydrolysis reactions large molecules are
broken down into smaller molecules by water.
Esters can be hydrolysed to produce the parent
................... .......... and the parent
.................... . ESTER WATER ?
ACID ALCOHOL An alkali is often used to
catalyse the reaction and the mixture heated
under reflux.
If sodium hydroxide is used a sodium salt of the
acid is formed. If this is reacted with a strong
acid (eg. Hydrochloric acid) the acid can be
displaced from the salt
HCOO- Na HCl
HCOOH NaCl
Sodium hydrochloric
methanoic sodium methanoate
acid
acid chloride
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Carry out the experiment to hydrolyse ethyl
benzoate (Workcard 2)
Mass of filter paper (g)  
Mass of filter paper benzoic acid (g)  
Mass of benzoic acid formed (g)  (actual mass)  
Use the equation below to calculate the
theoretical mass of benzoic acid expected when
0.033 moles (5g) of ethyl benzoate is
hydrolysed.   C6H5COOC2H5 H2O ?
C6H5COOH C2H5OH
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Calculate the Yield using the following
equation yield actual mass x
100 theoretical mass
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L.I. To learn about edible fats and oils Section
2 (a) S.C. By the end of this lesson you
should be able to

• State why fats and oils are an essential part of
  a healthy diet.
• Describe the formation and structure of fats and
  oils.
• Explain using full structural formulae how fats
  and oils are formed.
• State the main difference between the structure
  of saturated and unsaturated fatty acids.

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Natural oils have vegetable and marine (fish)
origins Natural fats have ,
or origins
Workbook Activity
Name of fat or oil Source Animal, vegetable or marine
Sunflower oil    
Linseed oil    
Suet    
Cod liver oil    
Lard    
Castor oil    
Palm oil    
Rapeseed oil    
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Name of fat or oil Source Animal, vegetable or marine
Sunflower oil  Sunflower seeds Vegetable
Linseed oil Flax seeds   Vegetable
Suet  Beef Animal 
Cod liver oil Cod fish  Marine 
Lard Pork  Animal 
Castor oil Castor plant   Vegetable
Palm oil  Fruits of palm oil trees Vegetable 
Rapeseed oil Rapeseed plant  Vegetable 
Fats and oils are an essential part of the diet.
They provide the body with energy. Fats and oils
are a more concentrated source of energy than
.................................
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Carry out the experiment to test the unsaturation
of fats and oils (Workcard 3)

• What does unsaturated mean?
• Which sample is the most saturated and which is
  the most unsaturated?
• This comparison is only approximate. How could
  the method be improved?

Solids fats tend to be saturated Liquid oils
tend to be unsaturated
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Fats and oils are naturally occurring esters of
the alcohol glycerol and long chain carboxylic
acids.
Long chain carboxylic acids are often called
fatty acids
R1, R2, R3 are long carbon chains which can be
the same or different
glycerol
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Propan-1,2,3-triol
Glycerol is a trihydric alcohol it has three
alcohol groups the correct name is
.................................... Fats and
Oils are formed by combination of ..... moles of
fatty acids to .......... mole of glycerol - a
triglyceride. There are ..................
ester linkages in the one molecule.
1
3
3
34
L.I. To learn about the melting points of fats
and oils Section 2 (b) S.C. By the end of
this lesson you should be able to

• State and explain the difference in melting
  points of fats and oils, in terms of structure,
  close packing and strength of intermolecular
  bonds.

35
solid
liquid

• Fats are .................... at room
  temperature while oils are ......................
  .
• The properties of a fat or an oil depend on the
  fatty acids which are combined with the
• glycerol in each triglyceride.
• Most fats and oils in nature are mixtures of
  triglycerides in which the fatty acid
• molecules may or may not be identical.
•  
• Fatty acids are saturated or unsaturated straight
  chain carboxylic acids. They contain an
• even number of carbon atoms and range in size
  from C4 to C24 carbon atoms, but mainly C16
• and C18.

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• Workbook activity
• Complete the table below

Fatty acid Formula Saturated or unsaturated
palmitic acid    
stearic acid    
linoleic acid    
oleic acid    
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Fatty acid Formula Saturated or unsaturated
palmitic acid  C16H32O2  Saturated
stearic acid  C18H36O2  Saturated
linoleic acid  C18H32O2  Unsaturated
oleic acid  C18H34O2  Unsaturated
38
Fat and oil molecules are roughly tunning fork
in shape with three limbs being hydrocarbon
chains. If the chains are saturated, the
molecules pack neatly together even at quite high
temperature
Fat higher melting point
Oil lower melting point

• If the chains contain one or more double bonds
  the zig-zag chains become more distorted and the
• close packing of the molecules is less easy.
• How will the close packing of fat molecules
  affect the strength of the Van der Waals
  (intermolecular) bonds?
•  
• What effect will this have on the melting point?

The more closely the molecules can pack together,
the stronger the van der Waals forces.
The stronger the van der Waals forces, the more
energy is needed to break them and the melting
point will be higher.
39
L.I. To learn about the function of
proteins Section 3 (a) S.C. By the end of
this lesson you should be able to

• Describe the role of proteins in the body.

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growth
repair
We need protein for ................. and
............... . Foods such as meat and fish are
rich in protein. Proteins are the major
structural materials of animal tissues. Proteins
are also involved in the maintenance and
regulation of life processes. Proteins can be
classified as fibrous or globular.
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Fibrous proteins are long and thin and are the
major structural materials of animal tissue.
Workbook activity
Give examples of fibrous proteins
Keratin found in hair and nails Collagen and
elastin found in connective tissue in the body.
Globular proteins have the spiral chains folded
into compact units. Globular proteins are
involved in the maintenance and regulation of
life processes and include enzymes and many
hormones.
Workbook activity
Give examples of globular proteins
Albumin found in blood plasma Haemoglobin found
in blood which carries oxygen around the body
42
Proteins are chemicals containing the element
nitrogen.
Carry out the experiment to heat proteins with
soda lime (Workcard 4)
Protein Effect on moist pH paper
   
   
   
   
When a protein is heated strongly in the presence
of soda lime (a mixture of calcium and sodium
hydroxide) an unpleasant smelling alkaline gas
is produced which turns moist pH paper
blue. These gases are amines or ammonia -
chemicals which contain nitrogen.
43
Enzymes are proteins, which act as biological
catalysts. They are specific to particular
chemical reactions e.g. the enzyme pepsin only
catalyses the hydrolysis of proteins and not any
other chemical reaction Certain sequences of
amino acids form a region known as the active
site.   The shape of the active site allows
specific reactants known as substrates to attach,
like a lock and key. Incorrect substrates are
unable to fit the shape of the active site and
are not changed. Enzyme function is therefore
related to the molecular shapes of proteins
44
Carry out the experiment to find out if pH
affects enzyme activity (Workcard 5)
45
             
  normal reaction   Enzyme reaction   Enzyme reaction  
                 
Rate     Rate   Rate    
                 
  Temperature   Temperature     pH
1.
2.
3.
If the temperature rises above a critical level
the shape of the enzyme molecule becomes
irreversibly altered although the peptide links
remain intact. This means that the enzyme is
unable to function. The enzyme is said to be
.............................. . All proteins,
not just enzymes, can be denatured by temperature
and pH.
denatured
46
L.I. To learn about amino acids Section 3
(b) S.C. By the end of this lesson you should
be able to

• State the name of the building blocks from which
  protein molecules are formed.
• Describe the structure of amino acid molecules
  and how they differ to produce 20 common
  naturally occurring amino acids.
• Explain why some amino acids are known as
  essential amino acids.

47
Amino acids contain an amino group (-NH2) at one
end and an acid group (-COOH) at the other end of
the molecule. The majority of amino acids found
in proteins are of the type
R represents a carbon side chain which may even
contain the elements nitrogen and sulphur.
Name R (side chain) Name R (side chain)
Glycine -H Aspartic Acid -CH2 COOH
Alanine -CH3 Methionine -CH2 CH2 S CH3
Valine -CH CH3 CH3 Phenylalanine -CH2 C6H5
48
There are 20 or so amino acids which go into
making up protein. The body cannot make all the
amino acids required for body proteins and is
dependent on dietary   protein for the supply of
certain ............................. amino acids.
essential
49
L.I. To learn about amide links Section 3
(c) S.C. By the end of this lesson you should
be able to

• State the type of reaction in which many amino
  acid molecules can link together to form a
  protein molecule.
• Explain using full structural formulae how an
  amide (peptide) linkage can be formed by a
  condensation reaction between amino acids.
• State how the diverse range of proteins needed to
  fill the different roles in the body is produced
  from just 20 amino acids.

50
Proteins specific to bodys needs are built up
within the body. Proteins are ...................
........ polymers formed by combining amino acids
to form long chain molecules of maybe several
thousand amino acid units long.
condensation
Workbook activity The following amino acids
react to form a tripeptide. Draw the structure
of the product molecule below.
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When amino acids join together an amide link (or
peptide link) is formed and water is eliminated.
The amide link is formed by the reaction of an
amino group with a carboxyl group. The
structure of a section of protein is based on the
constituent amino acids. The sequence by which
the amino acids are joined differs in different
protein molecules. The sequences of amino acids
are controlled from information in the nucleus of
the cell. Less complex than the proteins are the
peptides. A tripeptide has 3 amino acid
units. A shorter chain of amino acids, up to 100
units or so is often referred to as a polypeptide.
52
L.I. To learn about the hydrolysis of
proteins Section 3 (d) S.C. By the end of
this lesson you should be able to

• State and explain how proteins can be hydrolysed
  (digested) to produce amino acids.
• Draw the full structural formulae of the amino
  acids obtained from the hydrolysis of a given
  section of a protein.
• Explain how the amino acids present in a sample
  of hydrolysed protein can be analysed and
  identified.

53
During digestion enzymes hydrolyse the proteins
in food to amino acids. These can then pass
through the gut wall into the blood stream. In
the stomach, the enzyme pepsin starts the
digestion of protein by hydrolysing to smaller
polypeptides. Further enzymes in the gut
continue the hydrolysis to the constituent amino
acids. In hydrolysis large molecules are broken
down into smaller molecules by water. PROTEIN
WATER ? AMINO ACIDS
Workbook activity
Draw the product molecules as this tripeptide
molecule is hydrolysed.
54
In the lab a protein can be hydrolysed back to
its constituent amino acids by refluxing with
concentrated hydrochloric acid for several
hours. Amino acids can be identified by the use
of paper (or thin layer) chromatography. A piece
of chromatography paper is spotted with some
amino acids suspected as being present and also
with the hydrolysed protein.
By comparing the position of the spots of the
known amino acids with that of the hydrolysed
protein, the amino acids in the protein can be
identified.    
55
L.I. To learn about flavour in food Section 4
(a) S.C. By the end of this lesson you should
be able to

• Explain why volatile molecules are important in
  flavour.
• Identify the functional groups present in flavour
  molecules and explain whether they are likely to
  be water soluble or oil soluble.
• Make predictions about the boiling point (and
  volatility) of flavour molecules based on
  molecular size and the functional groups present.

56
Many of the flavours in foods are due to the
presence of volatile molecules. Normally these
molecules are trapped in the cell. However,
during cooking the cell walls can be broken by
moisture within the cell evaporating and
rupturing the walls. Also chemicals can damage
the walls which are made of a structural
carbohydrate called cellulose.
Volatile flavour molecules can then leave the
food and enter the air. Also some may dissolve in
the cooking liquid and be lost. So the method
and time of cooking is important in preserving
flavours. Flavour molecules can be
water-soluble or oil/fat-soluble. You can
preserve the flavour by using water for cooking
foods that contain oil/fat-soluble flavours such
as green beans and broccoli, and using oil/fat
to cook foods that are richer in water-soluble
flavours such as asparagus. Overcooking food
using either method will cause more flavour to
be lost.
57
To recognise flavours we use our sense of smell
and sense of taste. Try eating flavoured food
without looking at it, while pinching your
nose. It is not as simple are you might think.
There are at least five basic taste qualities
sweet, sour, bitter, salty, and umami. (Umami, or
savoury, is the taste we comes from glutamate,
found in chicken broth, meat extracts, and some
cheeses.) Our nose can detect more than 10,000
different smells. If there is hydrogen bonding
between the molecules responsible to smell,
this will reduce the volatility of the
molecule.
58

• To complicate matters some chemical reactions
  that take place during cooking can result in
  new,
• desirable flavours. The chemistry here is
  complicated but a simple example to illustrate
  that more
• desirable flavours appear happens when bread is
  toasted. Bread contains a carbohydrate and its
• carbonyl group (gtCO) reacts with amino groups
  (-NH2) in protein to make this desirable smell.

The functional groups present in flavour
molecules will give an indication whether they
are likely to be water or oil soluble. The size
and functional groups present can be taken into
account in predicting their relative boiling
point and hence probable volatility. The
important feature here is whether the
functional groups present can hydrogen bond with
water, making them water soluble, or whether
they are unable to hydrogen bond with water in
which case they will be fat- or oil-soluble.
59
Some functional groups are included in the table
below for reference. However, the presence of a
particular functional group is no guarantee of
the substance being water- or oil-soluble. e.g. a
long hydrocarbon chain (which is non-polar)
with a single polar group would be unlikely to be
able to hydrogen bond with water and would make
the molecule more likely to be oil-soluble.
Functional group name Functional group structure Can it form hydrogen bonds? Water- or oil-soluble
Alkane C-C No oil-soluble
Alkene CC No oil-soluble
Alkyne CC No oil-soluble
Hydroxyl -O-H Yes water-soluble
Carbonyl gtCO Yes water-soluble
Carboxylic acid -COOH or Yes water-soluble
Aldehyde -CHO or Yes water-soluble
Amino -NH2 or Yes water-soluble
Phenyl C6H5- No oil-soluble
Amide -CONH- or Yes water-soluble
Ester -COO- or Yes water-soluble
60
As a general rule, if the molecule contains only
C and H atoms is will be non-polar, while if it
also contains O or and N atoms it is more
likely to be polar. However, the non-polarity of
long hydrocarbon chains/rings can cancel the
effect of a single polar group and make the
molecule behave as non-polar.
Workbook activity
The molecular structures below represent the
formulae of some common flavours. Examine them
and complete the table to show if they are likely
to be water or oil soluble.  
61
Molecular Name Flavour Water soluble or Oil soluble Functional groups present in molecule
Nona-2,6-dienal   Cucumber    
4-hydroxy-2,5-dimethylfuran-3-one Strawberry    
Propyl pentanoate   Pineapple    
Limonene   Orange    
1-methyl-1-butylethanoate   Banana    
3,7-dimethyloct-2,6-dienal   Lemons    
2-methoxy-3-isobutyl-pyrazine Green peppers    
Vanillin   Vanilla    
Methyl anthranilate   Pineapple    
62
Molecular Name Flavour Water soluble or Oil soluble Functional groups present in molecule
Nona-2,6-dienal   Cucumber    
4-hydroxy-2,5-dimethylfuran-3-one Strawberry    
Propyl pentanoate   Pineapple    
Limonene   Orange    
1-methyl-1-butylethanoate   Banana    
3,7-dimethyloct-2,6-dienal   Lemons    
2-methoxy-3-isobutyl-pyrazine Green peppers    
Vanillin   Vanilla    
Methyl anthranilate   Pineapple    
63
In general, molecules with a molecular mass of
under 300 are likely to be volatile whereas much
larger molecules are not likely to be volatile.
The type of bonding present will also have a
significant influence on how volatile the
molecule is likely to be. If hydrogen bonding is
not present between molecules then these
molecules will be more volatile due to the weak
intermolecular forces present.
Watch Hestons cooking video clips
64
L.I. To learn about changes in protein structure
upon heating Section 4 (b) S.C. By the end of
this lesson you should be able to

• Explain the importance of intermolecular bonding
  in protein structure.
• State the difference between fibrous and globular
  proteins.
• State the changes that take place on heating
  proteins and relate this to the texture of foods
  such as eggs and meat.

65
Proteins are complex molecules made from long,
amino acid chains which are also branched. The
chains are held together by intermolecular
bonding between the side chains of the
constituent amino acids. Hydrogen bonds occur
between the amide links and between other groups
present in the molecule. Proteins
form three-dimensional structures which are
either sheets, spirals or coils.
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.  
For example, egg whites contain many molecules of
a globular protein called albumen. When an egg is
boiled or fried, the protein structure is
irreversibly changed and a solid is made. During
cooking, the protein is denatured and the
protein chains unwind and, as they can now form
intermolecular bonds with neighbouring albumen
molecules, a network of interconnected proteins
forms causing the egg white to solidify.
66
Collagen molecule
The structure of meat changes when cooked.
Different temperatures are required for cooking
meats with different levels of connective
tissue. Joints containing a lot of connective
tissue become tender if cooked at over 60oC as
the collagen forming the tough connective tissue,
denatures. The tender lean meat found in cuts
such as fillet steaks have less connective tissue
and should not be cooked at too high a
temperature, because in this case the protein
molecules start to bunch together resulting in
the meat becoming tougher.
67
Workbook activity

• Carry out research and write a short note on the
  chemistry of browning foods. Use the ipads to
  help you
• answer the following questions...
•  
• What are Maillard reactions?
•  
•  
•  
• What reaction takes place during caramelisation?
•  
•  
•  
•  
• Why do apples brown?

68
L.I. To learn about the oxidation of
alcohols Section 5 (a) S.C. By the end of
this lesson you should be able to

• Draw the structural formula of branched chain
  alcohols when given the name.
• State the name of branched chain alcohols when
  given the structural formula.
• State the molecular formula of branched chain
  alcohols when given the name or structural
  formula.
• State how to classify alcohols as primary,
  secondary or tertiary.
• State the product formed when primary alcohols
  are oxidised.
• State the product formed when secondary alcohols
  are oxidised.
• State what happens when attempting to oxidise a
  tertiary alcohol.
• State what is meant by oxidation in terms of the
  oxygen to hydrogen ratio.
• State the name of common oxidising agents used in
  the lab and describe the colour changes which
  occur.
• State the ion equations associated with the
  oxidation of primary and secondary alcohols.

69

• State what is meant by the terms dihydric and
  diol.
• State what is meant by the terms trihydric and
  triol.
• Draw the structural formulae of dihydric and
  trihydric alcohols when given the name.
• State the effect of hydrogen bonding on the
  properties of diols and triols.

70
Alcohols can be classified according to the
position of the OH group.
71

• In a primary (1) alcohol, the carbon which
  carries the -OH group is only attached to one
  alkyl group (chain of carbon atoms). Methanol,
  CH3OH, is counted as a primary alcohol even
  though there are no alkyl groups attached to the
  carbon with the -OH group on it.
• In a secondary (2) alcohol, the carbon with the
  -OH group attached is joined directly to two
  alkyl groups, which may be the same or different.
• In a tertiary (3) alcohol, the carbon atom
  holding the -OH group is attached directly to
  three alkyl groups, which may be any combination
  of same or different.

72
Structural formula Name 1o 2o 3o
        ethanol  
   
   
        2-methylpropan-2-ol  
73
Alcohols burn in oxygen and air to produce carbon
dioxide and water. This is complete oxidation
(combustion).
C2H5OH 3O2 2CO2 3H2O
Less vigorous oxidising conditions can be
provided by reacting an alcohol with an oxidising
agent.
74
Your teacher will demonstrate the oxidation of
alcohols using hot copper oxide (Workcard
6a) Carry out the experiment using potassium
dichromate (Workcard 6b)
Oxidising alcohol using hot copper oxide
The alcohol vapour is passed over hot CuO. The
copper oxide is reduced to copper metal and the
alcohol is oxidised.
Ethanol copper oxide ethanoic acid
copper water C2H5OH CuO
CH3COOH Cu H2O
75
Oxidising alcohols using potassium dichromate
The dichromate ion is reduced to the Cr3 ion and
the alcohol is oxidised.
Cr2O72- 14H 6e- ? 2Cr3 7H2O
76
Primary alcohols are oxidised, first to aldehydes
and then to carboxylic acids. Secondary alcohols
oxidise to ketones. Tertiary alcohols are
resistant to oxidation
77
Oxidation of ethanol (C2H5OH) to ethanal (CH3CHO)
has increased the OH ratio. Ethanal can also be
reduced back to ethanol and this would be a
decrease of the ratio.
78
Workbook activity
Name Formula OxygenHydrogen Ratio
    C2H5OH  
   Ethanal CH3CHO  
   Ethanoic acid CH3COOH  
79
Name Formula OxygenHydrogen ratio
   Ethanol C2H5OH   16
   Ethanal CH3CHO   14
   Ethanoic acid CH3COOH   24 12
Now complete the workbook activity by calculating
the OH ratio for each of the compounds
given. Remember to draw the structural formulae
of each compound too.
80

• More than one hydroxyl group
• There are some alcohols that have more than one
  hydroxyl group in the molecule. For example,
  glycol, which is used as the antifreeze in car
  radiators, and glycerol, which you may have eaten
  as an ingredient in soft ice cream.
• The alcohol used in ordinary car antifreeze has
  two hydroxyl groups and for this reason is known
  as a dihydric alcohol or diol. Its common name is
  glycol, but this says little about its structure.
  The systematic name is ethane-1,2-diol...

Two numbers are needed in the name to describe
the positions of the two hydroxyl groups. Notice
that di now appears infront of ol (di 2,
ol hydroxyl group).
81
With two carbon atoms in the molecule, the
systematic name is based on ethane. In this case
the final e of ethane is not dropped in the
sysetmatic name because it is not followed by a
vowel.   The alcohol commonly known as glycerol
is found in a variety of foods. It has the
systematic name propane-1,2,3-triol.
This molecule has three hydroxyl groups and as a
result is known as a trihydric alcohol or triol.
82

• As the number of hydroxyl groups in a molecule
  increases, the number of hydrogen bonds between
  molecules also increases. This will mean that a
  greater amount of energy is needed to overcome
  these intermolecular forces of attraction.
• This explains why propane-1,2,3-triol has a
  higher boiling point than ethane-1,2-diol.

83
L.I. To learn about aldehydes and
ketones Section 5 (b) S.C. By the end of
this lesson you should be able to

• State the name of the functional group present in
  aldehydes and ketones.
• State the name ending present in aldehydes.
• State the name ending present in ketones.
• State the name of straight chain and branched
  chain aldehydes when given the
• structural formula.
• State the name of straight chain and branched
  chain ketones when given the structural formula.
• Draw the structural formula of aldehydes and
  ketones when given the names.
• State the molecular formula of aldehydes and
  ketones when given the names.
• State and explain the results of a chemical test
  used to distinguish between
• aldehydes and ketones.

84

• State the name of the products formed when
  aldehydes (and ketones) are oxidised.

• State the effect of oxidising aldehydes using
  chemicals such as Benedicts solution, Tollens
  reagent and acidified potassium dichromate
  solution.

85
Both aldehydes and ketones contain the carbonyl
group. In aldehydes a hydrogen atom is bonded
to the carbonyl group but in ketones the carbonyl
group is always flanked by carbon atoms
This structural difference accounts for the fact
that aldehydes can undergo mild oxidation to
form carboxylic acids but ketones resist
oxidation. Oxidising agents can therefore be
used to distinguish between aldehydes and
ketones.
86
Carry out the experiment to oxidise aldehydes and
ketones (Workcard 7)
Aldehydes but not ketones can be oxidised by a
number of oxidising agents, including Benedict's
solution, Tollens reagent and acidified
dichromate to carboxylic acids.
  Benedicts solution Tollens reagent Acidified dichromate (H/Cr2O72-)
with compound X   Colour change from blue to orange-red  A solid silver precipitate forms (the silver mirror effect) Colour change from orange to green
with compound Y  No reaction   No reaction No reaction
Compound X must be Compound Y must
be ..
87
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89
Aldehydes have a hydrogen atom joined to the
carbonyl group whereas ketones have two carbon
atoms joined to the carbonyl group. Naming of
alkanones is almost exactly the same as for
alkanols. e.g. butanone. For longer chain ketones
the position of the carbonyl group must be given.
Workbook activities

1. Draw the full structural formulae and shortened
   structural formulae for the compounds listed.
2. Draw the full structural formulae and name each
   of the compounds listed.

90
L.I. To learn about antioxidants Section 5
(c) S.C. By the end of this lesson you should
be able to

• State how food containing edible oils turns
  rancid.
• State how to prevent food from turning rancid.
• State the ion-electron equations for the
  oxidation of antioxidants.
• State the name of carboxylic acids when given the
  structural formula.
• Draw the structural formula of branched chain
  carboxylic acids when given the
• name.
• State the molecular formula of branched chain
  carboxylic acids when given the names or
  structural formula.
• State the products formed when carboxylic acids
  undergo reduction reactions.
• State the name of the product formed when
  carboxylic acids react with bases.
•  

91
Oxygen reacts with edible fats and oils in food
giving the food a rancid flavour.   For example,
butter, a mainly saturated fat, can undergo
hydrolysis releasing small quantities of
volatile, unpleasant smelling butanoic acid which
is responsible for a rancid smell.   The
reaction with oils is more complex and appears to
affect unsaturated fatty acids in the oil. The
point of attack is at the CC double bonds and
volatile, smaller molecules such as aldehydes and
ketones are released which cause the rancid
smell.  
Oxidation reactions can produce free radicals. A
free radical is a highly reactive species
containing an unpaired electron. Free radicals
can damage food by removal of an
electron. Antioxidant molecules mop up free
radicals to protect the foodstuff.
92

• The antioxidant molecule donates an electron to
  the potentially damaging free radical.
•  
• A stable electron pair is formed, stabilising the
  free radical.
• The antioxidant itself becomes oxidised (loses an
  electron).
• Antioxidants are molecules which will prevent
  oxidation reactions taking place, antioxidants
  are reducing agents. By undergoing oxidation, the
  antioxidant provides electrons to prevent the
  oxidation of fats/oils in food.
• In crisp manufacture, potatoes are typically
  fried under an atmosphere of steam and packaged
  under nitrogen to help reduce oxidation.
• Ascorbic acid (or vitamin C), found in many
  fruits is an antioxidant.

93
When it behaves as an antioxidant, two of the
hydroxyl groups become changed into carbonyl
groups as found in ketones. The product has two
hydrogen atoms less than the ascorbic acid and is
know as dehydroascorbic acid. It can be seen as
an oxidation because there is a increase in the
OH ratio and also that electrons are released in
the reaction.
C6H8O6 C6H6O6 2H 2e-  
94
When apples are cut, they go brown because of
oxidation by the air. A few drops of lemon
juice will prevent this oxidation, though the
lemon juice is not an antioxidant.
HOC6H4OH
OC6H4O 2H 2e-
Citric acid and benzoic acid is used in many
foods such as jam to prevent oxidation by helping
other antioxidants do their work.
95
Antioxidants in action
The apple is protected when dipped in orange
juice containing the antioxidant vitamin C
Oxidation occurs when the apple is left exposed
to air
96
The table shows some typical antioxidants
Antioxidant E-number Typical foods
Ascorbic acid (vitamin C) E300 Beers, cut fruits, jams, dried potato. Helps to prevent cut and pulped foods from going brown by preventing oxidation reactions that cause the discolouration. Can be added to foods, such as potato, to replace vitamin C lost in processing.
Tocopherols E306 Oils, meat pies. Obtained from soya beans and maize. Reduces oxidation of fatty acids and some vitamins.
Butylated hydroxyanisole (BHA) E320 Oils, margarine, cheese, crisps. Helps to prevent the reactions that break down fats and cause the food to go rancid .
Citric acid E330 Jam, tinned fruit, biscuits, alcoholic drinks, cheese, dried soup. Naturally-occuring in citrus fruits like lemons. Helps to increase the anti-oxidant effects of other substances. Helps to reduce the reactions that can discolour fruits. May also be used to regulate pH in jams and jellies.
http//www.understandingfoodadditives.org
97
Carry out the SSERC experiments on antioxidants
(Workcard 8)
Now carry out the active talk activity with
your group
Carboxylic acids   Vinegar is a dilute solution
of ethanoic acid which is commonly used in food
preservation. Ethanoic acid is a member of the
carboxylic acid family. As a homologous series,
carboxylic acids all share the same functional
group and fit the same general formula
(CnH2n1COOH).
Carboxyl functional group
It is necessary to be able to name and draw
structural formulae and molecular formulae for
carboxylic acid molecules. This can be done by
using the principles we have used before...
98
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99

• Carry out the workbook activities on naming
  carboxylic acids and drawing full and shortened
  structural formulae.

Reactions of carboxylic acids Carboxylic acids
can be reduced in stages back to the original
parent aldehyde and alcohol using a reducing
agent such as lithium aluminium hydride. For
example...


(Carboxylic acid) (Aldehyde)
(Primary Alcohol)
Ethanoic acid Ethanal Ethanol
100
Workbook activity
Like ordinary acids carboxylic acids can react
with bases to form salts. For example...
Ethanoic acid Sodium hydroxide Sodium
ethanoate Water
NaOH
H2O
Workbook activity
Vitamin C tablets experiment
101
L.I. To learn about soaps and emulsions Section
6 (a) making soap S.C. By the end of this
lesson you should be able to

• Explain how soaps are produced by alkaline
  hydrolysis of fats and oils. 

102
Soaps are produced by the alkaline hydrolysis of
fats and oils. Fats and oils are esters. The
hydrolysis of fats and oils produces fatty acids
and glycerol (propane-1,2,3-triol) in the ratio
of three moles of fatty acid to one mole of
glycerol. In the presence of an alkali, a water
soluble ionic salt of the carboxylic acid is
formed.
These salts are the important ingredients of soap
- the ones that do the cleaning.
Watch the animation of making a soap on
http//www.educationscotland.gov.uk/higherscienc
es/chemistry/animations/soapformation.asp
103
L.I. To learn about soaps and emulsions Section
6 (b) cleansing action of soaps S.C. By the
end of this lesson you should be able to

• State what is meant by a hydrophobic tail in a
  soap molecule.
• State what is meant by a hydrophilic head in a
  soap molecule.
• Describe the cleansing action of soaps in terms
  of the structure of the soap molecule.

104
Watch the animation to investigate the cleansing
action of soaps... http//www.educationscotland.g
ov.uk/highersciences/chemistry/animations/cleansin
gsoap.asp Cleaning with water alone has little
effect if the stains consist of non-polar
substances, such as grease and sweat. Soaps and
detergents are emulsifying reagents. These are
simply chemicals which can make oil and water
become permanently mixed to produce a stable
emulsion. Soaps and detergents do this job due
to the structure of the molecules Soap
molecules have a long non polar hydrocarbon chain
tail which is readily soluble in non polar
(hydrophobic) compounds and a polar ionic
carboxylate head which is water soluble.
During cleaning, the hydrophobic tails dissolve
in the droplet of grease, whilst the hydrophilic
heads face out into the surrounding water,
resulting in ball like structures. The following
diagrams show how this works.
105
The dirt or grease is held inside the ball and
suspended in the water.
106
L.I. To learn about soaps and emulsions Section
6 (c) emulsions in food S.C. By the end of
this lesson you should be able to

• State the meaning of the term emulsion.
• Explain why emulsifiers are added to food.
• State how emulsifiers are made.
• Describe how emulsifiers work.

107
An emulsion contains small droplets of one liquid
dispersed in an 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.
The presence of this emulsifier is shown on
packaging by E-numbers, E471 and is one of the
most common on food packaging.
108
Many foods contain emulsions. Mayonnaise is
mainly a mixture of vegetable oil and
water/vinegar/lemon juice. Egg yolk or a
synthetic emulsifier (xanthan gum) can be used to
keep the normally immiscible liquid evenly mixed.
Without the emulsifier the two liquids would
separate and would not appear appetising.
Emulsifiers are added to a very large range of
different foods including sauces, bread,
biscuits, ice cream, low fat spread and even
dried pastas where they help to prevent pasta
pieces sticking to each other during cooking.
Watch the animation on emulsions and
emulsifiers http//www.educationscotland.gov.uk/h
ighersciences/chemistry/animations/emulsions.asp
109
Emulsifiers
Mayonnaise contains oil and water. The emulsifier
keeps these mixed and without it the oil and
water separate.
110
L.I. To learn about fragrances Section 7 (a)
essential oils S.C. By the end of this lesson
you should be able to

• State what is meant by an essential oil.
• State the uses, properties and products of
  essential oils.
• Describe how essential oils can be extracted from
  plant material.
• State the key component in many essential oils.

111
Essential oils are hydrophobic liquids containing
mixtures of volatile aroma compounds that
evaporate easily in air giving distinctive
fragrances. The oils have the aroma of the
plant from which they are extracted. They include
lavender, peppermint, orange, lemon, and
eucalyptus oils. Every essential oil contains a
mixture of organic compounds rather than just one
pure compound, but it is often the aromatic
molecules that provide the distinct
fragrances. Essential oils are extracted from
plant material, the difficulty is obtaining the
liquid before it evaporates.
112
Steam Distillation In this process the volatility
of the oil is important so it is carried by the
steam. More water can be added from the funnel.
The essential oil collects as a mixture of oil
and water, but as the two do not mix, they are
easily separated. In the laboratory extraction
can be achieved with the apparatus shown....
113
Essential oils are used in perfumes, cosmetics,
soaps and other products, for flavoring food and
drink, and for adding scents to incense and
household cleaning products.
Essential oil can contain esters, alcohols,
aldehydes and ketones. An important component in
essential oils are known as terpenes.
114
Modern uses
Cosmetics
Flavours
Cleaning
Essential oils
Dentistry
Adhesives
Perfumes
Medical
Insect repellents
115
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116
Terpenes are unsaturated compounds formed by
joining together isoprene (2-methylbuta-1,3-diene)
units.
5 carbon molecule.
(Looks like a horse.)
The basic molecular formulae of terpenes are
multiples of that, (C5H8)n where n is the number
of linked isoprene units. This is called the
isoprene rule. The isoprene units may be linked
together "head to tail" to form linear chains or
they may be arranged to form rings. One can
consider the isoprene unit as one of nature's
common building blocks
Complete the workbook activity to highlight the
isoprene units in each molecule.
117
Terpenes are components in a wide variety of
fruit and floral flavours and aromas. Terpenes
can be oxidised within plants producing some of
the compounds responsible for the distinctive
aroma of spices.
118
Limonene a cyclic terpene
119
Menthol a cyclic terpenoid
This terpene has been oxidised to a terpenoid
120
a-Selinene a cyclic terpene
3 isoprene units 15 carbon atoms
121
ß-carotene a linear terpene
8 isoprene units 40 carbon atoms
122
Find the horse
123
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124
L.I. To learn about skin care products Section
8 (a) effect of UV light S.C. By the end of
this lesson you should be able to

• State the effect of ultraviolet radiation on
  molecules.
• State the effect of ultraviolet radiation on the
  human body.
• Describe how sunblock works.

125
Image of the sun taken with ultraviolet imaging
telescope
126
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 ageing of the skin. Sun-block
products prevent UV light reaching the skin. UV
light is divided into UVA, UVB and UVC light. UVA
is the highest energy and causes most damage. UVC
light does not penetrate the atmosphere and so
causes no problems. UVA and UVB cause wrinkles
by breaking down collagen, creating substances
called free radicals that inhibit the natural the
repair of the skin. UVB light, but not UVA
light is stopped by glass. You may notice labels
on sunglasses to indicate their effectiveness.
127
UV Photography reveals the effects of
"photoaging", or ageing of skin caused by
light. Photoageing refers to the damage that is
done to the skin from prolonged exposure to UV
radiation, over a person's lifetime. Most of
the skin changes that occur as we get older are
accelerated by sun exposure. Examples of skin
changes from photoaging include dark spots,
wrinkles, leathery skin and skin cancer
(malignant melanoma).
Theres no such thing as a healthy tan
128
Skin cancer malignant melanoma
In the UK, 2,000 people a year die from malignant
melanoma, and the number is increasing.
129
Photoageing caused by UVA
Taxi driver
Exposed to sun Not exposed to
sun (through car window)
130
The effects of UV - ageing of skin
131
How to protect yourself-Sunscreen

• Sunscreen works by combining organic and
  inorganic active ingredients.
• Inorganic ingredients like zinc oxide or
  titanium oxide reflect or scatter ultraviolet
  (UV) radiation.
• Organic ingredients absorb UV radiation,
  dissipating it as heat.

132
L.I. To learn about skin care products Section 8
(b) Free Radical Reactions S.C. By the end of
this lesson you should be able to

• State what is meant by a free radical.
• Describe the reactivity of a free radical.
• Describe how free radicals are formed.
• State what is meant by the terms initiation,
  propagation and termination.

133
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.
  Hydrogen and chlorine video clip   Hydrogen
reacts with explosively with chlorine in the
presence of U.V. light. The reaction can be shown
as follows   H2(g) Cl2(g) ? 2HCl(g)   The
presence of acid HCl in the product can be shown
with moist pH paper.   The reaction follows a
free radical chain reaction, initiated by U.V.
light. For convenience, the reaction can be split
into three stages.  
134
1) Initiation U.V. light provides the energy
for the homolytic fission of halogen into
reactive halogen atoms or free radicals (atoms
with an unpaired electron). Cl2(g) ? Cl.(g)
.Cl(g)
2) Propagation In this stage, free radicals
collide with other species but the number of free
radicals is maintained (hence the term
propagation). H2(g) .Cl ? H.(g)
HCl(g) H.(g) Cl2(g) ? HCl(g) Cl. (g) These
reactions continue until reactants are used up,
or until free radicals are used up by collision
with each other.
135
3) Termination In this stage, free radicals are
used up by collision with each other. H.(g)
.Cl(g) ? HCl(g) H.(g) .H(g) ? H2(g) Cl.(g)
.Cl(g) ? Cl2(g)
136
Free radical Substitution Methane Another free
radical reaction takes place when a halogen is
substituted into an alkane in the presence of UV
light. This reaction is not explosive and results
in the decolourisation of bromine.  Alkanes
react with bromine in the presence of U.V. light,
though the reaction with bromine is slow. The
reaction can be shown as follows CH4(g)
Br2(g) ? CH3Br(g) HBr(g)   The presence of
acid HBr in the product can be shown with moist
pH paper. However, the reaction does not end
here and further substitution can occur with
hydrogen atoms progressively replaced by halogen
atoms. The slow substitution reaction follows a
free radical chain reaction, initiated by U.V.
light. Again, the reaction can be split into
three stages
137

• Initiation
• U.V. light provides the energy for the homolytic
  fission of halogen into reactive halogen atoms or
  free radicals (atoms or molecular fragments with
  an unpaired electron).
• Br2(g) ? Br.(g) .Br(g)
• 2) Propagation
• In this stage, free radicals collide with other
  species but the number of free radicals is
  maintained (hence the term propagation).
• CH3-H(g) .Br ? CH3.(g) HBr(g)
• CH3.(g) Br2(g) ? CH3-Br(g) Br. (g)
• These reactions continue until reactants are
  used up, or until free radicals are used up by
  collision with each other.

138

• 3) Termination
• In this stage, free radicals are used up by
  collision with each other.
• Br.(g) .Br(g) ? Br2(g) CH3.(g) .Br(g) ?
  CH3-Br(g) CH3.(g) .CH3(g) ? CH3-CH3(g)
•  
• The product of the last equation is ethane.
  However, to minimise the range of possible
  products, an excess of the original alkane is
  used and the products separated from the excess
  alkane by distillation.  Evidence to support
  this mechanism
• The reaction is initiated by U.V. light and, once
  started, can continue in the dark.
• Other substitution products are made such as
  CH2Br2, CHBr3, CBr4 together with longer alkanes
  (and smaller amounts of substitution products of
  these alkanes.
• However, these other substitution products can be
  minimised by using an excess of the original
  alkane to try to ensure collision of the
  relatively small number of free radicals produced
  by sunlight quickly uses up the bromine.

139
L.I. To learn about skin care products Section 8
(c) Free Radical Scavengers S.C. By the end
of this lesson you should be able to

• State the effect that free radicals have on the
  body.
• Describe how free radical scavengers work.
• State examples of natural free radical
  scavengers.

140
Many cosmetic products contain free radical
scavengers. These are molecules which can react
with free radicals to form stable molecules and
prevent chain reactions. Melatonin and Vitamin
E are examples of natural free radical
scavengers.
Melato