Unit 1 Rings, polymers and analysis
(F324) Module 2 Polymers and synthesis
Topics covered in this module:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
1|Page
Amino acids
Polypeptides and proteins
Optical isomerism
Condensation polymerisation – polyesters
Condensation polymerisation – polyamides
Addition and condensation polymerisation
Breaking down condensation polymers
Organic synthesis of aliphatic compounds
Organic synthesis of aromatic compounds
Chirality in pharmaceutical synthesis
1.
Amino acids
Unit 1 - Rings, polymers and analysis
Polymers and synthesis
Done

State the general formula for an α-amino acid as RCH(NH2)COOH
State that an amino acid exists as a zwitterion at a pH value called
the isoelectric point.
State that different R groups in α-amino acids may result in different
isoelectric points.
Describe the acid-base properties of α-amino acids at different pH
values.
Key areas to concentrate on…

Amino acids feature heavily during your A2 chemistry course and you
must learn all about their important aspects including:
 their structure
 that they can behave as both carboxylic acids and amines, depending
upon whether their environment is acidic or alkaline.
Questions based on the acid-base nature of amino acids are quite common in
exams. You must read the questions carefully – you may be asked for an
equation in which the amino acid is reacting as an acid; in others you may be
asked for an equation of an amino acid reacting with a base. The equation is
the same for both question!
2|Page
Past paper questions
Compound A, shown below, is an amino acid that is being used in the development
of a new anti-inflammatory drug.
H2N
H
H
C
C
O
C
CH3 H
OH
compound A
(a)
(i)
Explain why this molecule is described as an amino acid.
...........................................................................................................
...........................................................................................................
[1]
(ii)
State the general formula of an -amino acid.
Explain whether or not compound A fits this general formula.
...........................................................................................................
...........................................................................................................
[2]
(b)
Compound A exists as a zwitterion in aqueous solution.
(i)
Draw the structure of this zwitterion.
[1]
[Turn over]
3|Page
(ii)
Show how the structure of the zwitterion would change if the
solution was acidified with dilute hydrochloric acid.
[1]
(c)
The anti-inflammatory drug is made by combining compound A with
compound B, shown below. R represents a side chain.
H
R
C
O
C
H
OH
compound B
Show the structure of the anti-inflammatory drug formed from compound
A and compound B.
[2]
[Total 7 marks]
Examiner’s comments
(a)
Most candidates recognised that compound A had amino and carboxylic acid
groups. However many were unable to explain that it didn’t fit the general formula for an αamino acid because these two groups were not attached to the same carbon.
(b)
The unfamiliar amino acid did not prevent the more able candidates drawing the
correctly ionised forms in neutral and acidic conditions.
(c)
Most candidates were also able to combine compound B with compound A with a
peptide bond linking the correct groups. A number of candidates linked the two
compounds with an acid anhydride link, which was also credited.
4|Page
2.
Polypeptides and proteins
Unit 1 - Rings, polymers and analysis
Done
Polymers and synthesis

Explain the formation of a peptide (amide) linkage between α-amino
acids to form polypeptides and proteins.
Describe the acid and alkaline hydrolysis of proteins and peptides
to form α-amino acids or carboxylates.
Key areas to concentrate on…

Amino acids feature heavily during your A2 chemistry course and you
must learn all about their important aspects including:
 their structure
 that they can behave as both carboxylic acids and amines, depending
upon whether their environment is acidic or alkaline.

Amino acids form polypeptides – proteins
 Polypeptides can be converted back to amino acids via hydrolysis.
When you are asked to draw a section of a peptide, remember that one end
of the structure contains a C=O group and the other end an N-H group.
Polypeptides are drawn with their ends open.
Molecular model of human
C-reactive protein (CRP)
showing its pentameric
structure.
The levels of this protein in
the plasma increase greatly up to 1000-fold - in response
to tissue damage and
inflammation.
It is measured in clinical
tests as a marker for
inflammation and infection.
Credit: T. Greenhough & A.
Shrive, Wellcome Images
http://images.wellcome.ac.uk
5|Page
Past paper questions
In this question, one mark is available for the quality of the use and organisation of
scientific terms.
In all living organisms a large variety of polypeptides and proteins are formed
naturally from -amino acids.
State the general formula of an α-amino acid and use it to describe how amino acids
can be combined to give a variety of polypeptides and proteins.
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
[6]
Quality of Written Communication [1]
[Total 7 marks]
Examiner’s comments
This question allowed candidates to demonstrate their knowledge of the
condensation polymerisation of α-amino acids to give a polypeptide. Many knew this
topic well although, for full credit, it is essential to concentrate on the chemistry and
not get distracted by the biological details. A few candidates confused polypeptides
with nylon or Kevlar. For full credit, the peptide linkage had to be identified and the
structures did need to show polymerisation using bonds extending at each end,
rather than just two amino acids combining to give a dipeptide. Candidates also
needed to relate the variety of proteins to the many different sequences possible
with the varying R groups in the amino acids.
6|Page
Past paper questions
The four amino acids shown below are found in proteins and enzymes.
H
H2 N
C
H
COOH
H2 N
H
C
H
COOH
(CH 2 ) 4 NH2
glycine
lysine
H2 N
C
COOH
CH2 SH
cysteine
H
H2 N
C
COOH
CH2CH2 COOH
glutamic acid
Write down the structural formula for a dipeptide formed from one molecule of
glycine and one of lysine.
[Total 2 marks]
Examiner’s comments
Nearly all candidates scored both marks, although some who used an extended
molecular formula ran into trouble. The correct answer for that would be
H2NCH2CONHCH[(CH2)4NH2]COOH. A few did not choose the correct amino acids.
7|Page
Past paper questions
Leucine (2-amino-4-methylpentanoic acid) is a naturally occurring α-amino
acid that is often used in protein supplements.
Leucine has a structural formula of (CH3)2CHCH2CH(NH2)COOH.
(a)
(i)
State the general formula of an α-amino acid.
...........................................................................................................
[1]
(ii)
Draw a displayed formula of leucine.
[1]
(b)
Leucine can exist as a zwitterion.
(i)
State what is meant by the term zwitterion.
...........................................................................................................
...........................................................................................................
[1]
(ii)
Explain with the aid of a diagram how the zwitterion is formed from
the functional groups in leucine.
...........................................................................................................
...........................................................................................................
...........................................................................................................
[2]
[Turn over]
8|Page
(c)
Leucine can be obtained from a source of protein such as meat.
(i)
State suitable reagents and conditions to break down a protein into
amino acids.
...........................................................................................................
...........................................................................................................
[2]
(ii)
State the type of reaction occurring.
...........................................................................................................
[1]
[Total 8 marks]
Examiner’s comments
(a)
Most candidates knew the general formula of an α-amino acid and there were
few who were confusing it with the type of general formula used to represent
an homologous series. Many could also successfully interpret the structural
formula of leucine given.
(b)
A significant number of candidates tried to define a zwitterion in terms of the
proton transfer process (as was required in part (ii) of this question). For the
definition, it needs to be clear that both positive and negative charges are on
the same molecule.
(c)
Conditions for the hydrolysis of a protein were well known, with a suitable
named aqueous acid, base or enzyme being accepted. Dehydrating acids such
as concentrated sulphuric acid are not suitable for hydrolysis.
9|Page
Past paper questions
Compound A is currently being tested as a possible anti-allergic drug.
O
OH
C
CH2
CH2
H
N
H
C
C
H
O
H
CH2
N
C
C
H
O
H
H
N
C
H
OH
C
O
compound A
Compound A can be hydrolysed to form three organic products.
(i)
Name a suitable reagent and conditions for the hydrolysis of compound A.
....................................................................................................................
....................................................................................................................
[2]
(ii)
The three organic products all belong to the same class of compound.
State the general name for this class of organic compound.
....................................................................................................................
[1]
(iii)
Draw the structure of one of the organic products from the hydrolysis of
A using the reagent you have given in (a)(i) above.
[2]
(iv)
Explain what is meant by the term hydrolysis. Use this reaction to
illustrate your answer.
....................................................................................................................
....................................................................................................................
....................................................................................................................
[2]
[Total 7 marks]
10 | P a g e
Examiner’s comments
A suitable aqueous acid or alkali had to be identified for the hydrolysis of a peptide,
but it is also necessary to heat this mixture under reflux, as the peptide bond is
relatively hard to break. Most candidates knew that amino acids are formed and
could draw the structure of one of them. However very few picked up the cue to give
the correct ionic form for the reagent that they had used.
Hydrolysis as the reaction with water to split the compound, in this case by splitting
the peptide bond, was only well described by the better candidates.
Internal structure of an HIV particle showing the capsid surrounding the RNAcontaining core in red and the membrane in blue. The yellow area is electron dense
material, including proteases, between the core and the membrane.
Credit Stephen Fuller, Wellcome Images
11 | P a g e
3.
Optical isomerism
Unit 1 - Rings, polymers and analysis
Polymers and synthesis
Done

Describe optical isomers as non-superimposable mirror images
about an organic chiral centre.
Identify chiral centres in a molecule of given structural formula.
Explain that optical isomerism and E/Z isomerism are types of
stereoisomerism.
Key areas to concentrate on…

You need to know how to spot chiral carbon atoms within structures and
thus optical isomers.
 You must be able to draw both non-superimposable mirror images.
 These diagrams must be in 3D.
You should always draw the isomers as mirror images, with the geometry
around the chiral carbon shown as a tetrahedral shape.
12 | P a g e
Past paper questions
One of the final stages in winemaking involves the fermentation of malic acid
to lactic acid.
An equation for the reaction is shown below.
O
C
HO
H
OH
C
C
H
H
O
C
H
OH
malic acid
H
OH
C
C
H
H
O
C
+
CO2
OH
lactic acid
Both acids contain a chiral centre.
(i)
Identify the chiral centre on the structure of malic acid above using an
asterisk *.
[1]
(ii)
Draw a diagram to show the 3-D arrangement of groups around the
chiral centre in malic acid.
[1]
[Total 2 marks]
Examiner’s comments
Almost all candidates correctly identified the chiral centre in the given malic acid
structure. For the 3-D structure of the molecule, the two structures shown below
were acceptable.
CH2COOH
CH2COOH
H
HO
C
or
COOH
HO
(1)
13 | P a g e
C
H
COOH
(1)
Past paper questions
Glutamic acid and glycine are both -amino acids that occur widely in living
organisms. Their structures are shown below.
H
H2N
C
H
H2N
COOH
CH2
C
COOH
H
CH2
COOH
glutamic acid
(a)
(i)
glycine
State the general formula for an -amino acid.
...........................................................................................................
[1]
(ii)
Explain how glutamic acid and glycine both fit the general formula
given in part (i)
...........................................................................................................
...........................................................................................................
...........................................................................................................
[2]
(b)
Amino acids react with both acids and alkalis.
Draw structures below to show how the glutamic acid molecule is
changed in the presence of excess acid and alkali.
H
H2N
C
COOH
CH2
excess
HCl(aq)
CH2
excess
NaOH(aq)
COOH
glutamic acid
[5]
14 | P a g e
(c)
In this question, one mark is available for the quality of use and
organisation of scientific terms.
Glutamic acid exists as two optical isomers, but glycine does not.
Explain what structural feature causes optical isomerism in organic
molecules. Include appropriate diagrams and use these two amino acids
to illustrate your answer.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[7]
Quality of Written Communication [1]
[Total 16 marks]
15 | P a g e
Examiner’s comments
(a)
Most could state the general formula of an α-amino acid, but only a few could
put into words that the amino and carboxylic acid groups are attached to the
first carbon and that the R group is different in the two examples given.
(b)
The zwitterions were well known by all but the weakest candidates, although a
few who had learnt the general case did not ionise the carboxylic acid group on
the side chain as well.
(c)
This part was well answered by the majority of candidates and it is clear that
most are confident with the explanation of optical isomerism. A few lost marks
by not attempting to draw 3-dimensional structures for glutamic acid, and
some of those that did, were not sure where the two perspective bonds are
relative to the other two bonds. Some candidates would find it useful to
practise drawing the 3-D tetrahedral structure to avoid the more subtle errors
as shown below.
correct 3-d
tetrahedral
16 | P a g e
tetrahedral, but
incorrect 3-d
not tetrahedral
(90° angle)
4.
Condensation polymerisation - polyesters
Unit 1 – Rings, polymers and analysis
Polymers and synthesis
Done

Describe condensation polymerisation forming polyesters, such as
Terylene and poly(lactic acid).
State the use of polyesters as fibres in clothing.
Key areas to concentrate on…

You must be able to show condensation polymerisation from given
monomers.
 It is often a good idea to draw more than one complete unit of a
polymer and to circle one complete unit.
 Check that all the atoms circled are equivalent to both the monomers
minus two hydrogen atoms and one oxygen atom.

Polyesters have an ester link between the monomers – e.g. Terylene,
Polylactic acid.
Polyamides have an amide (peptide) link between the monomers – e.g.
Nylon, Kevlar.
 Both types of polymer can be converted back to monomers by
hydrolysis.

17 | P a g e
5.
Condensation polymerisation - polyamides
Unit 1 - Rings, polymers and analysis
Polymers and synthesis
Done

Describe condensation polymerisation to form the polyamides
nylon-6,6 and Kevlar.
State the use of polyamides as fibres in clothing.
Key areas to concentrate on…

You must be able to show condensation polymerisation from given
monomers.
 It is often a good idea to draw more than one complete unit of a
polymer and to circle one complete unit.
 Check that all the atoms circled are equivalent to both the monomers
minus two hydrogen atoms and one oxygen atom.

Polyesters have an ester link between the monomers – e.g. Terylene,
Polylactic acid.
Polyamides have an amide (peptide) link between the monomers – e.g.
Nylon, Kevlar.
 Both types of polymer can be converted back to monomers by
hydrolysis.

Iran hosting its first fashion and costume-design festival, December 2008
18 | P a g e
Past paper questions
Kevlar is a very tough polymer made from 1,4-diaminobenzene and benzene1,4-dicarboxylic acid.
(i)
State a use for Kevlar.
....................................................................................................................
[1]
(ii)
Describe the polymerisation reaction that forms Kevlar. Include in your
answer:
•
an explanation of the type of polymerisation involved
•
an equation for the reaction
•
a repeat unit to show the structure of Kevlar.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[5]
[Total 6 marks]
Examiner’s comments
Most candidates had a good understanding of condensation polymerisation and knew
the formation of Kevlar from its monomers. Uses of Kevlar were accepted as long as
they required a particularly strong fibre.
19 | P a g e
6.
Addition and condensation polymerisation
Unit 1 - Rings, polymers and analysis
Polymers and synthesis
Done

Compare condensation polymerisation with addition polymerisation.
Suggest the type of polymerisation from a given monomer or pair of
monomers, or a given section of a polymer molecule.
Identify the monomer(s) required to form a given section of a
polymer (and vice versa).
Key areas to concentrate on…

You must be able to show condensation polymerisation from given
monomers.
 It is often a good idea to draw more than one complete unit of a
polymer and to circle one complete unit.
 Check that all the atoms circled are equivalent to both the monomers
minus two hydrogen atoms and one oxygen atom.

Polyesters have an ester link between the monomers – e.g. Terylene,
Polylactic acid.
Polyamides have an amide (peptide) link between the monomers – e.g.
Nylon, Kevlar.
 Both types of polymer can be converted back to monomers by
hydrolysis.


Sometimes you will be asked to predict the monomers which made up a
polymer.
 This could be applied to the addition polymers you covered in AS Unit
F322: Module 2 - Alcohols, halogenoalkanes and analysis.
20 | P a g e
Past paper questions
Poly(phenylethene) is one of the most versatile and successful polymers.
The 3-D skeletal formula of a section of atactic poly(phenylethene) is shown in
the diagram below.
(i)
State the type of polymerisation used to make poly(phenylethene).
....................................................................................................................
[1]
(ii)
Draw a skeletal or displayed formula to show the monomer used to make
poly(phenylethene).
[1]
(iii)
Outline how the polymer is formed from the monomer molecules. (You
do not need to give any details of the catalyst or conditions involved.)
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[2]
[Total 4 marks]
Examiner’s comments
Addition polymerisation and the structure of phenylethene were well known. However
fewer candidates pointed out that this process involves the π-bond breaking and that many
molecules join in order to give a polymer. Many just described addition generally.
21 | P a g e
Past paper questions
The fibres used in carpets are made from synthetic or natural polymers such
as nylon-6,6, OrlonTM and wool.
(a)
Complete the table below.
OrlonTM
nylon-6,6
O
monomer(s)
HO
O
C
(CH2)4
C
H2N
(CH2)6
NH2
OH
repeat unit of
the polymer
H
CN
C
C
H
H
type of
polymerisation
[4]
(b)
Nylon-6,6 can be made from its monomers in the laboratory in two
stages as shown below.
O
HO
C
O
(CH2)4
C
OH
stage 1
Not in the new
specification, but
PCl5 or SOCl2 is
an appropriate
reagent to convert
carboxylic acids to
acyl chlorides.
O
Cl
C
O
(CH2)4
C
Cl
H2N
(CH2)6
NH2
stage 2
nylon-6,6
(i)
State a suitable reagent to carry out stage 1.
...........................................................................................................
[1]
(ii)
Deduce the inorganic product that is also formed in stage 2.
...........................................................................................................
[1]
22 | P a g e
(c)
Industrially, nylon-6,6 is not manufactured by the method in (b). Instead,
the two monomers are mixed directly at room temperature to give a salt.
This salt is then heated to convert it to nylon-6.6.
Suggest the structures of the two ions present in this salt.
[2]
(d)
Wool is a protein. It is a natural polymer made by the same type of
polymerisation as nylon-6,6.
A section of the polymer chain in a protein is shown below.
H
O
N
C
C
H
CH2
H2C
O
N
C
C
H
H
H
O
N
C
C
H
H
H
O
N
C
C
H
CH3
OH
(i)
How many monomer units does this section contain?
...........
[1]
(ii)
Draw the structure of one of the monomer molecules that was
used to form this section.
[1]
[Turn over]
23 | P a g e
(iii)
State three ways in which the monomer units of a protein differ
from those of nylon-6,6.
...........................................................................................................
...........................................................................................................
...........................................................................................................
...........................................................................................................
...........................................................................................................
[3]
[Total 13 marks]
Examiner’s comments
(a)
Many candidates scored full marks on this part and demonstrated a good
understanding of addition and condensation polymerisation. The most
common error was to not show a correct repeat for nylon-6,6.
(b)
Once again this reaction is new to the specification, and relatively few
candidates could identify PCl5 or SOCl2 as an appropriate reagent to convert a
carboxylic acid into an acyl chloride. However, a greater number did deduce
that HCl would be made when the amide was formed.
(c)
A pleasing number of candidates correctly identified the diamino and
dicarboxylate ions that would form by the acid/base reaction between the
nylon monomers. A few interpreted the question as just asking for the two
ionic functional groups and did not give the complete structures. Partial credit
was given for this.
(d)
Nearly all candidates identified that the section of protein was formed from
four monomers, and most successfully drew the structure of one of the amino
acids. The last part elicited a wide range of answers and candidates were
credited for all as long as they didn’t try to give the same point twice or restate
what was given in the question. Good suggestions included that protein
monomers were bifunctional, that they were chiral, that they could form
zwitterions, that there were a greater variety of monomers or specific
references to their atomic composition compared to that of nylon-6,6.
24 | P a g e
Past paper questions
Nylon is sometimes used for electrical insulation. However, if there is a risk of
high temperatures then a polymer such as Nomex®, with a higher melting
point, is used.
The repeat unit of Nomex® is shown below.
(i)
O
O
C
C
N
N
H
H
n
Draw the structures of two monomers that could be used to form Nomex®.
[2]
(ii)
Suggest a reason why the melting point of Nomex® is higher than that of
nylon.
....................................................................................................................
....................................................................................................................
....................................................................................................................
[1]
[Total 3 marks]
Examiner’s comments
Most candidates could deduce the structures of the monomers that could be used to
form the given polymer, although a common error was to draw 1,4-substituted
benzene rings instead of the 1,3-substituted rings needed in this case. For the last
part, any sensible suggestion that could explain greater intermolecular forces
between the polymer chains was accepted.
25 | P a g e
7.
Breaking down condensation polymers
Unit 1 - Rings, polymers and analysis
Polymers and synthesis
Done

Describe the acid and base hydrolysis of polyesters and
polyamides.
Outline the role of chemists in the development of degradable
polymers.
Explain that condensation polymers may be photodegradable and
may be hydrolysed.
Key areas to concentrate on…


Polyesters have an ester link between the monomers – e.g. Terylene,
Polylactic acid.
Polyamides have an amide (peptide) link between the monomers – e.g.
Nylon, Kevlar.
 Both types of polymer can be converted back to monomers by
hydrolysis.
The conditions for hydrolysis are always aqueous acid or aqueous base in the
presence of heat. Aqueous implies water and hydrolysis means the breaking
of a bond with a water molecule. The means that water must be present!
Fish swim along a coral reef near a water bottle label and a plastic bag off the coast of a Red Sea resort
Photograph: Mike Nelson/EPA
26 | P a g e
Biodegradable plastic bags carry
more ecological harm than good
Decomposing bags sound environmentally friendly but they require a lot of energy to make,
won't degrade in landfills and may leave toxic leftovers.
Biodegradable plastic bags – as handed out by Tesco, the Co-op and once even sold by the
Soil Association – must be good, surely? They have a magic ingredient that means they
self-destruct after a few months, breaking up into tiny pieces made of simple molecules
that bugs and fungi can happily munch up. Dozens of major corporations use them,
including Pizza Hut, KFC, News international, Walmart and Marriott hotels.
But last week, the European Plastics Recyclers Association warned that they "have the
potential to do more harm to the environment than good."
Technically what we are talking about here is "oxo-degradable" plastics. These are plastics
made to degrade in the presence of oxygen and sunlight, thanks to the addition of tiny
amounts of metals like cobalt, iron or manganese.
British manufacturers – headed by Symphony Technologies of Borehamwood – are at the
sharp end of a revolution that could banish bag-strewn beauty spots and back alleys alike.
But the criticisms are twofold. First, some research suggests that the bags don't degrade as
well as claimed. And second, priming plastic bags for destruction is itself an ecological
crime.
So, do they really biodegrade away to nothing? Symphony, which supplies the Co-op and
Tesco, says its bags are "able to degrade completely within about three years, compared to
standard bags which take 100 years or longer". Tesco reckons they all decompose within
18 months "without leaving anything that could harm the environment".
But whether it actually happens seems to depend a lot on where the "biodegradable" plastic
ends up. If it gets buried in a landfill it probably won't degrade at all because there is no
light or oxygen. But what about elsewhere?
Studies of one brand in the US, commissioned by the Biodegradable Products Institute,
found that breakdown is very dependent on temperature and humidity. It goes slow in cold
27 | P a g e
weather. And high humidity virtually stops the process, making long, wet winters sound like
bad news.
You might think a compost heap full of biodegrading bugs would be ideal. But a recent
Swedish study found that polyethylene containing manganese additive stops breaking down
when put in compost, probably due to the influence of ammonia or other gases generated
by microorganisms in the compost.
And, while most manufacturers say that to put only tiny amounts of metals into the plastic,
the US study found that one brand contained "very high levels of lead and cobalt", raising
questions about the toxicity of the leftovers. Neither of these studies relates specifically to
Symphony's products. But they raise questions.
The European Plastics Recyclers Association last week argued that biodegradable bags are
not the right environmental option anyway. Plastic bags take a lot of energy and oil to make
so why waste them by creating bags that self-destruct? "It is an economic and
environmental nonsense to destroy this value," the recyclers' trade association concluded.
Of course, we consumers can reuse or recycle biodegradable bags as easily as any other
kind. Symphony and other manufacturers stress making bags biodegradable is just an
insurance policy for those that don't get recycled or reused. But surely we are less likely to
bother if we are told the bags are eco-bags that biodegrade.
This European backlash against oxo-biodegradable plastics follows similar rumblings in the
US. In March, the New York Times announced it would not be wrapping its paper in bags
made of the stuff because claims that the plastic was "100% biodegradable" did not stand
up. This followed a ruling last December by an advertising industry watchdog, part of the
US Council of Better Business Bureaus, that makers should stop calling the bags "ecofriendly".
(In marked contrast, the UK Periodical Publishers Association two years ago recommended
that all its members use oxo-biodegradable film to wrap their magazines)
Industry websites, including Symphony's, do proudly proclaim one green endorsement –
that the organic trade body the Soil Association buys their bags. But Clio Turton at the Soil
Association told me: "We've had problems with people making these claims. We have asked
for them to be removed. It's very frustrating."
Plastic bags are not the biggest environmental issue on the planet, as George Monbiot
explained in a blog here recently.
But most of us probably make "bag choices" several times a day. Brits get through 8bn
plastic bags a year. For that reason, they are one of the choices that tend to show if we
care about the environment or not. And we should be clear. Re-using bags is best.
Recycling is second best. Throwing them away in the hope that a magic formula will
guarantee their rapid disappearance is laziness, not environmental care. And anybody who
tries to persuade us otherwise is guilty of Greenwash.
28 | P a g e
• This article was amended on Friday 19 June 2009. We should have made clear that the
Soil Association no longer sells the biodegradable plastic bags referred to in this article.
This has been corrected.
guardian.co.uk © Guardian News and Media Limited 2009
Thursday 18 June 2009 11.00 BST
Past paper questions
Short sections of the molecular structures of two polymers are shown below.
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
polymer C
O
O
C
C
O
H
H
C
C
H
H
O
O
O
C
C
O
H
H
C
C
H
H
O
polymer D
(a)
(i)
Circle, on the diagrams above, the simplest repeat unit in each polymer.
[2]
(ii)
In the boxes below, draw the displayed formulae of the two
monomers that could be used to prepare polymer D.
[2]
(b)
Chemists have developed degradable polymers to reduce the quantity of
plastic waste being disposed of in landfill sites. Polymer D is more likely
to be a ‘degradable polymer’ than polymer C.
Suggest two reasons why.
....................................................................................................................
29 | P a g e
....................................................................................................................
....................................................................................................................
[2]
[Total 6 marks]
30 | P a g e
8.
Organic synthesis of aliphatic compounds
Unit 1 - Rings, polymers and analysis
Polymers and synthesis
Done

Identify the functional group in an aliphatic molecule containing
several functional groups.
Predict properties and reactions of aliphatic molecules containing
several functional groups.
Devise multi-stage synthetic routes for preparing aliphatic organic
compounds.
Key areas to concentrate on…

Synthesis can involve any functional group from within the specification.
 It relies upon the candidate’s ability to convert from one functional
group to another, so you must have a good knowledge of all aspects
of the organic chemistry covered during your A level course.
31 | P a g e
9.
Organic synthesis of aromatic compounds
Unit 1 - Rings, polymers and analysis
Polymers and synthesis
Done

Identify the functional groups in an aromatic molecule containing
several functional groups.
Predict the properties and reaction of an aromatic molecule
containing several functional groups.
Devise multi-stage synthetic routes for preparing aromatic organic
compounds.
Key areas to concentrate on…

Synthesis can involve any functional group from within the specification.
 It relies upon the candidate’s ability to convert from one functional
group to another, so you must have a good knowledge of all aspects
of the organic chemistry covered during your A level course.
32 | P a g e
10.
Chirality in pharmaceutical synthesis
Unit 1 - Rings, polymers and analysis
Polymers and synthesis
Done

Explain that the synthesis of pharmaceuticals often requires the
production of a single optical isomer.
Explain that synthetic molecules often contain a mixture of optical
isomers, whereas natural molecules often have only one optical
isomer.
Explain that the synthesis of a pharmaceutical that is a single
optical isomer increases costs, reduces side effects and improves
pharmacological activity.
Describe strategies for the synthesis of a pharmaceutical with a
single optical isomer.
Key areas to concentrate on…

You must understand the problems of chirality when synthesising novel
pharmaceuticals and how pharmaceutical companies go about removing
the unwanted isomer.
33 | P a g e
Past paper questions
Compound A, shown below, contributes to the smell and taste of black tea and
is a component in jasmine oil.
O
O
O
(i)
A
Deduce the molecular formula of compound A.
...........................................................
[1]
(ii)
Compound A contains several functional groups.
Identify, by name, the functional groups in compound A.
....................................................................................................................
....................................................................................................................
....................................................................................................................
[3]
(iii)
Compound A is a stereoisomer.
On the structure above,
•
mark each feature responsible for stereoisomerism with an
asterisk, *,
•
label each feature with the type of stereoisomerism.
[2]
(iv)
Outline two important factors that pharmaceutical companies need to
consider when manufacturing chiral compounds for use as medicines.
....................................................................................................................
....................................................................................................................
....................................................................................................................
[2]
[Total 8 marks]
34 | P a g e
Past paper questions
Noradrenaline is produced naturally by nerve cells in the brain. Compound P is
a synthetic compound that has been widely used as an appetite suppressant.
It is thought to be effective because it is similar to noradrenaline.
OH
OH
HO
NH2
CH3
HO
noradrenaline
(a)
(i)
NH2
compound P
Suggest a reagent that could be used in a test to distinguish
compound P from noradrenaline.
...........................................................................................................
[1]
(ii)
Draw a ring round the functional group responsible for the positive
result in the test you have chosen.
[1]
(iii)
State the expected observation for the positive result in the test
you have chosen.
...........................................................................................................
[1]
(b)
Both these compounds have stereoisomers due to the presence of chiral
centres.
(i)
Identify the chiral centres in each molecule by labelling them
clearly with asterisks (*) on the structures above.
[2]
(ii)
State the type of stereoisomerism caused by the presence of chiral
centres.
...........................................................................................................
[1]
(iii)
Using noradrenaline as an example, explain how a chiral centre
gives rise to the stereoisomers. Illustrate your answer with a
suitable diagram of the stereoisomers.
...........................................................................................................
...........................................................................................................
...........................................................................................................
[3]
35 | P a g e
(c)
Explain why a pharmaceutical company may produce an appetite
suppressant containing only one stereoisomer of compound P, rather
than a mixture of the stereoisomers.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[3]
[Total 12 marks]
Examiner’s comments
(a)
For this part a reagent that reacted with phenols and not with alcohols was
needed. Decolourising bromine water was the most common correct answer
although other valid responses with correct observations, such as NaOH(aq)
and neutral FeCl3 were also accepted. Candidates generally found this harder
than the chemical tests earlier in the paper.
(b)
Nearly all candidates identified the chiral centre in noradrenaline, but only a
few spotted that compound P would have two chiral centres. However, many
candidates had a good attempt at drawing the optical isomers of
noradrenaline. To obtain full credit, valid use of the 3-D ‘wedge’ and ‘dotty’
bonds was needed and the groups had to be connected to the chiral carbon by
the correct atom. See the mark scheme for examples of correct
representations. Candidates do need to practise drawing these to become
confident at 3-D representations.
(c)
Many candidates knew that a drug containing a mixture of stereoisomers could
result in harmful side-effects and the need for higher doses. However, for full
credit candidates also needed to explain that this was because only the correct
isomer would have the right 3-D shape to be pharmacologically active.
36 | P a g e
Past paper questions
Compound A is currently being tested as a possible anti-allergic drug.
O
OH
C
CH2
CH2
H
N
H
C
C
H
O
H
CH2
N
C
C
H
O
H
H
N
C
H
OH
C
O
compound A
Compound A can exist as a number of stereoisomers, but only one of them is
pharmacologically active as the anti-allergic drug.
(i)
Explain what causes stereoisomerism in compounds such as A.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[3]
(ii)
Explain why there are four different stereoisomers of compound A.
....................................................................................................................
....................................................................................................................
....................................................................................................................
[2]
(iii)
Suggest how a drug company could synthesise compound A so that the
drug contains only the pharmacologically active stereoisomer.
....................................................................................................................
....................................................................................................................
[1]
[Turn over]
37 | P a g e
(iv)
Sometimes it is difficult to manufacture a drug containing only the one
pharmacologically active stereoisomer.
Describe two possible disadvantages of producing a drug containing a
mixture of several stereoisomers.
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
[2]
[Total 8 marks]
Examiner’s comments
Optical isomerism and its implications for drug synthesis were well described overall.
There is a range of possible ways to synthesise only the correct isomer, including
starting with chiral compounds using a stereospecific catalyst (such as an enzyme) or
separating the mixture by a suitable method such as chromatography. Some
candidates thought that production of a mixture might be disadvantageous because it
would be expensive to separate. However this is still sometimes the cheapest
approach if a suitable catalyst is not available.
38 | P a g e
Sample student answers
Module 2: Polymers and synthesis
Question 1
(a)
Total marks: 15
Glycine H2N–CH2–COOH and alanine H2N–CH(CH3)–COOH are two amino acids.
(i)
Give the systematic name of alanine.
(ii)
Explain why glycine does not have optical isomers but alanine does.
(iii)
Draw a diagram to show the two optical isomers of alanine.
(iv)
Write the equation to show the reaction between glycine and dilute hydrochloric
acid.
(v)
Write the equation to show the reaction between glycine and aqueous sodium
hydroxide.
Marks available:
(i) 1 (ii) 1 (iii) 2
(iv) 1 (v) 1
Student answer:
(a) (i) 2-aminopropanoic acid
(ii) 2-aminopropanoic acid (alanine) has a carbon (at position number two) with four
different groups bonded to it. Aminoethanoic acid (glycine) does not have any such
carbon atoms.
(iii)
(iv) H2N–CH2–COOH + HCl
+
 H3N–CH2–COOH + Cl
–
–
+
(v) H2N–CH2–COOH + NaOH  H2N–CH2–COO Na + H2O
Examiner comments:
(a) (i) The 2 is very important here as it denotes that the amine group (NH2) is present on the
second carbon in the chain.
(ii) It would be acceptable to use the term chiral without explaining what this term means.
The glycine molecule does not possess a carbon atom that has four different groups
attached to it, having: one –NH2 group; one –COOH group; but two –H groups.
(iii) The diagram must be drawn to represent a 3D drawing. The bonds from the central
carbon atom must be attached to the correct atom on the group – i.e. the N of the NH2
group and the C of the COOH group.
(iv) It is also acceptable to write the equation in structural form.
(v) You must remember to include the water molecule as a product of this reaction.
39 | P a g e
(b)
Amino acids have isoelectric points.
(i)
What is meant by the term isoelectric point?
(ii)
The isoelectric point of alanine is pH = 6.0. Draw the structure of alanine at pH = 7.0.
Marks available:
(i) 1 (ii) 1
Student answer:
(b) (i) The pH at which an amino acid exists as a zwitterion
(ii)
Examiner comments:
(b) (i) This point varies according to the amino acid.
(ii) Despite being neutral, the pH of the solution is more alkaline than the isoelectric point
for alanine, so the alkaline form of the zwitterion exists.
(c)
Draw the displayed formula of a dipeptide formed when alanine and glycine react. Circle the
functional group and give its name.
Marks available: 3
Student answer:
(c)
Examiner comments:
(c)
There is an alternative form in which the amino acids swap places, but the peptide link
remains in the same position.
40 | P a g e
(d)
Alanine can be produced in the laboratory by the following reaction scheme.
CH3COCOOH  CH3CH(OH)COOH  CH3CHClCOOH  CH3CH(NH2)COOH
(i)
However, this manufactured alanine could not readily be used for making proteins
within a living organism. Explain why not.
(ii)
There are 20 or so naturally-occurring amino acids. These can combine and link
together to form chains consisting of varying orders of amino acids. Amino acid
chains can be formed from just the following three amino acids: A; B; and C.
Determine how many different chains of three amino acids can be formed using just
these three amino acids.
Marks available:
(i) 3 (ii) 1
Student answer:
(d)
(i) Alanine is a chiral compound. Any chiral molecules which are found within living
organisms consist of the one optical isomer only. The manufactured sample of the
amino acid will contain a 50:50 mixture of both optical isomers. Because when 2hydroxypropanoic acid, CH3CH(OH)COOH, is made in stage 1, there is a 50% chance
that either optical isomer would form.
(ii) 27.
Examiner comments:
(d) (i) Any chiral product produced in a laboratory will be a 50:50 mixture unless certain
measures are taken during the manufacturing process.
(ii) One amino acid can only be arranged in one way, two amino acids can be arranged in
2 × 2 i.e. four different ways, and three amino acids can be arranged in 3 × 3 ×3
different ways.
41 | P a g e
Module 2: Polymers and synthesis
Question 2
(a)
Total marks: 15
Kevlar is a polyamide used in the manufacture of bulletproof material. It is made from
benzene-1,4-diamine and benzene-1,4-dicarboxylic acid. Draw a section of this polymer
showing two repeat units.
Marks available: 2
Student answer:
(a)
Examiner comments:
(a) Ensure the end bonds go through the surrounding brackets.
(b)
Nylon-6,6 has the following structure: - (CO–(CH2)4–CONH–(CH2)6–NH) n.
(i)
Draw the structures of the molecules from which nylon-6,6 is synthesised.
(ii)
A number of other nylons also exist, for example, nylon-6,8 and nylon-5,10. What do
the numbers 6,8 and 5,10 represent?
(iii)
Suggest the structure of two repeat units of nylon-6.
Marks available:
(i) 2 (ii) 1 (iii) 2
Student answer:
(b) (i) HOOC–CH2–CH2–CH2–CH2–COOH
H2N–CH2–CH2–CH2–CH2–CH2–CH2–NH2
(ii) The number of carbon atoms in the molecules from which the polymer is synthesised
from
(iii) -(CO– (CH2)5–NHCO–(CH2)5–NH) peptide (amide) link two units.
Examiner comments:
(b) (i) The structures can be abbreviated to HOOC–(CH2)4–COOH and H2N–(CH2)6–NH2.
(ii) In these examples nylon-6,8 would be synthesised from one molecule containing six
carbon atoms and another containing eight carbon atoms; while nylon-5,10 would be
formed from a molecule containing five carbon atoms and another containing 10 carbon
atoms.
(iii) Again, ensure the end bonds go through the surrounding brackets.
42 | P a g e
(c)
Terylene is a polyester made from benzene-1,4-dicarboxylic acid and ethane-1,2-diol. Draw a
section of this polymer showing two repeat units.
Marks available: 2
Student answer:
(c)
Examiner comments:
(c) Note the end bonds go through the surrounding brackets.
(d)
Lactic acid (2-hydroxypropanoic acid) can undergo polymerisation to form poly(lactic acid).
(i)
Write the equation for the production of this polymer.
(ii)
Poly(lactic acid) is a biodegradable polymer. Give an example of one use for
poly(lactic acid) which depends upon its biodegradability.
(iii)
Why should society be encouraging further development of biodegradable
polymers?
Marks available:
(i) 3 (ii) 1 (iii) 2
Student answer:
(d) (i) nHO–CH(CH3)–COOH  ( O–CH(CH3)–CO) n + nH2O
monomer
polymer
(ii) Internal medical stitches.
(iii) Ease of disposal. Unless recycled, current non-biodegradable polymers need to be either
buried in landfill sites which are filling up or combusted releasing polluting gases.
Examiner comments:
(d) (i) Condensation polymers often eliminate H2O molecules when the monomers combine.
(ii) Any use which depends upon biodegradability of material would do – e.g. bioplastics,
food packaging, nappies, compost bags and disposable tableware, PLA.
(iii) There is no requirement here to list the toxic gases – e.g. CO, HCl, HCN, etc.
43 | P a g e
2
9.0
Be
beryllium
4
24.3
Mg
magnesium
12
40.1
Ca
calcium
20
87.6
Sr
strontium
38
137.3
Ba
barium
56
[226]
Ra
radium
88
1
44 | P a g e
6.9
Li
lithium
3
23.0
Na
sodium
11
39.1
K
potassium
19
85.5
Rb
rubidium
37
132.9
Cs
caesium
55
[223]
Fr
francium
87
89
actinium
[227]
Ac*
57
lanthanum
39
138.9
La*
yttrium
21
88.9
Y
scandium
45.0
Sc
•
•
•
•
•
•
73
72
neodymium
60
uranium
92
140.9
Pr
praseodymium
59
[231]
Pa
protactinium
91
140.1
Ce
cerium
58
232.0
Th
thorium
90
238.1
U
144.2
Nd
106
105
104
seaborgium
dubnium
[266]
Sg
74
93
neptunium
[237]
Np
61
promethium
144.9
Pm
107
bohrium
[264]
Bh
75
rhenium
43
186.2
Re
tungsten
technetium
42
183.8
W
25
[98]
Tc
manganese
54.9
Mn
molybdenum
24
95.9
Mo
chromium
52.0
Cr
rutherfordium
[262]
Db
tantalum
hafnium
[261]
Rf
41
180.9
Ta
niobium
23
92.9
Nb
vanadium
50.9
V
40
178.5
Hf
zirconium
22
91.2
Zr
titanium
47.9
Ti
atomic (proton) number
name
Key
relative atomic mass
atomic symbol
94
plutonium
[242]
Pu
62
samarium
150.4
Sm
108
hassium
[277]
Hs
76
osmium
44
190.2
Os
ruthenium
26
101.1
Ru
iron
55.8
Fe
58.9
Co
58.7
Ni
63.5
Cu
65.4
Zn
95
americium
[243]
Am
63
europium
152.0
Eu
109
meitnerium
[268]
Mt
77
iridium
45
192.2
Ir
rhodium
27
102.9
Rh
cobalt
96
curium
[247]
Cm
64
gadolinium
157.2
Gd
110
darmstadtium
[271]
Ds
78
platinum
46
195.1
Pt
palladium
28
106.4
Pd
nickel
97
berkelium
[245]
Bk
65
terbium
158.9
Tb
111
roentgenium
[272]
Rg
79
gold
47
197.0
Au
silver
29
107.9
Ag
copper
zinc
81
thallium
49
204.4
Tl
indium
31
114.8
In
gallium
82
lead
50
207.2
Pb
tin
32
118.7
Sn
germanium
14
72.6
Ge
silicon
6
28.1
Si
carbon
83
bismuth
51
209.0
Bi
antimony
33
121.8
Sb
arsenic
15
74.9
As
phosphorus
7
31.0
P
nitrogen
14.0
N
84
polonium
52
[209]
Po
tellurium
34
127.6
Te
selenium
16
79.0
Se
sulfur
8
32.1
S
oxygen
16.0
O
85
astatine
53
[210]
At
iodine
35
126.9
I
bromine
17
79.9
Br
chlorine
9
35.5
Cl
fluorine
19.0
F
86
radon
54
[222]
Rn
xenon
36
131.3
Xe
krypton
18
83.8
Kr
argon
10
39.9
Ar
neon
20.2
Ne
98
californium
[251]
Cf
66
dysprosium
162.5
Dy
99
einsteinium
[254]
Es
67
holmium
164.9
Ho
100
fermium
[253]
Fm
68
erbium
167.3
Er
101
mendelevium
[256]
Md
69
thulium
168.9
Tm
102
nobelium
[254]
No
70
ytterbium
173.0
Yb
103
lawrencium
[257]
Lr
71
lutetium
175.0
Lu
Elements with atomic numbers 112–116 have been reported but not fully
authenticated
80
mercury
48
200.6
Hg
cadmium
30
112.4
Cd
13
69.7
Ga
aluminium
5
27.0
Al
boron
12.0
C
2
10.8
B
0
helium
7
1
6
hydrogen
5
4.0
He
4
1.0
H
3
The Periodic Table of the Elements
The Periodic Table of the Elements
Characteristic infrared absorptions in organic molecules
bond
location
wavenumber/cm
–1
C–O
alcohols, esters, carboxylic acids
1000–1300
C=O
aldehydes, ketones, carboxylic acids, esters,
amides
1640–1750
C–H
organic compound with a C–H bond
2850–3100
O–H
carboxylic acids
2500–3300 (very broad)
N–H
amines, amides
3200–3500
O–H
alcohols, phenols
45 | P a g e
3200–3550 (broad)