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Chapter 14 From organic molecules to medicines
E1.
List the ailments that can be treated by using acetylsalicylic acid (aspirin).
AE1.
Aspirin is used as
• analgesic or painkiller
• anti-inflammatory, reduces inflammation
• antipyretic, reduces fever
• anticoagulant, reduces blood clotting, prevention heart attack and stroke
Recent research suggests that aspirin can be used to prevent complications arising
from diabetes and dementia and as a preventative to bowel cancer.
E2.
Why should someone with stomach ulcers not take aspirin?
AE2.
In the stomach COX-1 (cyclo-oxygenase-1) enzymes produce prostaglandins that
protect the stomach lining. COX-2 enzymes are present at the site of an inflammation
producing prostaglandins that promote inflammation resulting in pain and fever.
Aspirin inhibits the production of both COX-1 and COX-2 enzymes. So while it can
treat pain, fever and inflammation it can also cause stomach irritation by also
inhibiting those prostaglandins that protect the stomach.
Q1.
a
b
c
Draw a flow chart showing the reaction pathways for the preparation of aspirin
from salicylic acid using:
i ethanoic acid
ii ethanoic anyhdride
Explain why the pathway using ethanoic anhydride is preferred over the one
using ethanoic acid.
Which pathway could be considered greener because it produces the least amount
of waste?
A1.
a
i
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Chapter 14 From organic molecules to medicines
ii
b
c
The rate and yield is greater using ethanoic anhydride.
A green approach to chemical synthesis aims to maximise atom economy, the
number of reactants atoms incorporated into the desired product. The idea of
atom economy was explained in Heinemann Chemistry 1, Chapter 17. Percentage
atom economy is given by:
molar mass of atoms used in the products
percentage atom economy =
× 100
molar mass of all reactants
For the reaction pathway using ethanoic acid, percentage atom economy
180
× 100
=
138 + 60
= 91.1%
For ethanoic anhydride pathway, percentage atom economy
180
× 100
=
138 + 102
= 75%
However, since the rate of reaction and the yield are greater using ethanoic
anhydride, this is the preferred reaction.
Q3.
Why do the instructions on prescription antibiotics tell you to complete the whole
course of tablets?
A3.
At the point when the symptoms of an infection disappear there may still be bacteria
present in the body. If you cease taking antibiotics at this stage the bacteria may start
multiplying again and the infection could return.
Q4.
Why were antibiotics like penicillin V preferred to penicillin G?
A4.
Penicillin V is more stable in the presence of stomach acid so can be given to the
patient in tablet form. Penicillin G is more potent but has to be given in the form of an
injection, requiring a visit to a medical clinic.
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Chapter 14 From organic molecules to medicines
Q5.
What information about the molecular structure of a drug does each of these analytical
techniques provide?
a infrared spectroscopy
b nuclear magnetic resonance spectroscopy
c mass spectroscopy
A5.
IR spectroscopy: Identity of any functional groups that may be present.
Proton NMR: The number of hydrogen environments present in the molecule.
Mass spectroscopy: Molar mass from the molecular ion; structure and identity from
fragmentation pattern.
Q6.
Medicinal proteins such as insulin are administered by injection. Suggest why they
cannot be given orally as tablets or capsules.
A6.
Since insulin is a protein—if it was given orally the substance would be broken down
in the process of digestion before it could be absorbed.
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Chapter 14 From organic molecules to medicines
Chapter review
Q7.
Why is the sodium salt of acetylsalicylic acid more soluble than normal aspirin?
A7.
Sodium acetylsalicylic acid is more soluble in water than acetylsalicylic acid because
it dissociates into ions (Na+ and the acetylsalicylate anion) which then attract water
molecules to form solvation sheaths around them.
Q8.
Briefly explain how aspirin is able to reduce pain and fever.
A8.
Aspirin blocks the action of the enzymes that control the production of prostaglandins
– substances that cause the blood vessels to constrict – resulting in heat building up in
the body to cause fever. Prostaglandins also make it easier for water to pass from
blood capillaries into tissues where it causes swelling and pain.
Q9.
Acetaminophen (paracetamol) is the active agent in some over-the-counter painkillers.
It can be synthesised from p-hydroxyaniline (4-hydroxyaminobenzene) and acetic
anhydride. An incomplete equation for this reaction is shown in Figure 14.25.
a What by-product is formed in this reaction?
b Determine the atom economy of this reaction.
A9.
a
b
ethanoic acid, CH3COOH. Note the similarity between this reaction and the
reaction for the synthesis of aspirin.
molar mass of atoms used in the products
Percentage atom economy =
×100
molar mass of all reactants
123
=
× 100
109 + 102
= 58%
Q10.
a
b
The drugs celecoxib and rofecoxib were specifically designed to be COX-2
inhibitors. They do not affect the production of COX-1. Aspirin inhibits both
COX-1 and COX-2. What are the benefits of these COX-2 inhibiting drugs over
aspirin?
Find out why rofecoxib, marketed as Vioxx, was withdrawn from use by the
pharmaceutical company Merck.
A10.
a
The COX-2 inhibitors only works on the enzymes that are present at the site of
inflammation and not on those present in the stomach. Hence it does not cause
stomach irritation or bleeding.
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Chapter 14 From organic molecules to medicines
b
Vioxx was withdrawn in 2004 because of concern that it has adverse side effects
and may lead to the development of heart disease or stroke.
Q11.
From the description of the use of the Australian native plant with the Aboriginal
name of bunaangu, what active ingredient might be present to give it these medicinal
properties?
‘You only eat the sap of the bunaangu. It has a bitter taste. You boil it up to drink. It
helps toothache and a bad chest.’
A11.
From the description of the bitter taste of bunaangu and the symptoms that it helps to
relieve, it seems as though the plant might be a source of salicin or possibly salicylic
acid.
Q12.
As well as aspirin, analgesics include acetaminophen, ibuprofen, naproxen sodium,
ketoprofen. Use the web to find out:
a the names under which they are sold
b their side effects
c their chemical structure
d how they are synthesised.
A12.
The website ‘Aspirin Adventures’ provides an easy to read summary of various pain
killers. http://www.chemheritage.org/educationalservices/pharm/asp/asp00.htm
Q13.
Codeine, morphine and heroin are drugs derived from the opium poppy. The
molecular structure of these drugs is shown in Figure 14.26. Identify the part of each
molecular structure that is common to all three drugs. How does the structure of these
molecules differ?
Figure 14.26
Molecular structures of codeine, morphine and heroin. Caution—it is not possible to convert
codeine into morphine or heroin by any simple laboratory process.
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Chapter 14 From organic molecules to medicines
A13.
Different functional groups are attached to the benzene rings in each structure. Small
changes to the molecular structure can result in significant changes to chemical or
biochemical activity.
Q14.
Suggest some advantages and disadvantages of using genetically modified plants to
produce therapeutic proteins.
A14.
Advantages: convenient source of the medicinal protein that can be grown rather than
produced via chemical synthesis. Could be included in the diet rather than having to
be taken as a medication or could be used to supplement inadequate food supplies.
Disadvantages: the effects of adding a medicinal protein to food has not been
subjected to long-term studies. Also the modified gene may be transferred to other
plants and result in unwanted harmful effects.
Q15.
Use the web to find out about the diseases that are treated by the following medicinal
proteins:
a growth hormone
b interferon
c monoclonal antibodies
A15.
a
b
c
Growth hormone is used to treat children who fail to grow at a normal rate due to
growth hormone deficiency.
Interferon is used to treat hepatitis C.
Specific monoclonal antibodies can be used to treat certain types of cancer of the
kidney, leukaemia or lymphoma. Cytotoxic (cancer killing) substances are joined
to the antibody and the resulting complex is given to the patient. The antibody,
with the cytotoxic agent attached, then binds only to the cancer cells and
eliminates them. Sometimes just the antibody itself is sufficient to trigger the
body’s normal defence mechanism and stimulate it to eliminate the cancer cells.
Q16.
Illustrate, using aspirin as an example, how biochemists modify the structure of drugs
to eliminate undesirable side effects.
A16.
Salicylic acid was used as a pain killer but found was to found to cause bleeding of
the stomach walls. Its structure was modified by reaction salicylic acid (aspirin) with
ethanoic acid to form the ester, acetylsalicylic acid which while an effective pain
killer did not have this side effect. A soluble form of aspirin was developed by
converting the carboxylic acid functional group to a sodium salt.
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Chapter 14 From organic molecules to medicines
Q17.
Explain why pharmaceutical companies manufacture drugs such as aspirin using
chemical reactions, while biotechnological processes are used to produce drugs such
as insulin.
A17.
The active ingredients in drugs such as aspirin consist of organic molecules that can
be made via a chemical pathway from some fairly simple feedstock. However,
substances such as insulin are proteins that are made by the body. It would be very
difficult to manufacture such a large molecule via a chemical pathway. Instead, they
are made by adapting the structure of a related protein obtained from a biological
source.
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Chapter 14 From organic molecules to medicines
Unit 3 Area of Study 2 review
Multiple-choice questions
Q1.
The smallest molecule in which homologous series must contain two carbon atoms?
A an alkane
B an alkanol
C an alkene
D an alkanoic acid
A1.
C. There is a double bond between two carbon atoms in alkenes. The first members of
the other homologous series, methane, methanol and methanoic acid all contain
1 carbon atom.
Q2.
The systematic name for CH3CH2CH2CH(CH3)2 is:
A 1,1-dimethylbutane
B 2-methylpentane
C 2-methylpentene
D propyldimethylmethane
A2.
B. It may assist you to draw a full structural formula. This molecule is an alkane as
there are no multiple bonds, therefore the name ends in ane There are 5 carbon atoms
in the longest chain name must end in pentane. There is a methyl group on the second
carbon atom from the end. This is represented as 2-methyl. The full name is
2-methylpentane.
Q3.
Which compound would be expected to have the highest boiling point?
A CH3CH2CH2OH
B CH3CH2CH2CH3
C CH3CH2CH2Cl
D CH3CH2CH3
A3.
A. The alkanols have higher boiling points as the strength of the intermolecular forces
are greater. The strength of hydrogen bonding in alkanols is greater than those in
chloroalkanes. Only dispersion forces act between alkane molecules.
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Chapter 14 From organic molecules to medicines
Q4.
Which of the following statements are true of the homologous series of primary
alkanols?
I The members differ by one CH2 unit.
II They are all strong bases.
III They can be oxidised to form carboxylic acids.
A I and II
B II and III
C I and III
D I, II and III
A4.
C. Members of a homologous series differ by CH2. Alkanols can be oxidised to
carboxylic acids by strong oxidants such as acidified permanganate or dichromate.
Q5.
What is the product formed by the reaction of CH2CH2 with Br2?
A CH2BrCH2Br
B CH2CHBr
C CH3CH2Br
D CHBrCHBr
A5.
A. Ethene, CH2CH2, will undergo an addition reaction with bromine. The double bond
is broken and a bromine atom bonds to each carbon atom to form 1,2-dibromethane.
Q6.
When ethanol is heated under reflux with an acidified solution of potassium
dichromate (VI), the final product has the formula:
A CH3CH2CH2OH
B CH3COOCH3
C CH3COOH
D CH3CH2CH3
A6.
C. Ethanol is oxidised to ethanoic acid by an acidified potassium dichromate solution.
Q7.
Consider the following reaction pathway:
I
II
III
ethene ⎯⎯
→ chloroethane ⎯⎯
→ ethanol ⎯⎯
→ ethanoic acid
The reactions that occur in steps I, II and III of the pathway are:
A cracking, addition, hydrolysis
B chlorination, substitution, addition
C addition, hydrolysis, oxidation
D addition, reduction, hydrolysis
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Chapter 14 From organic molecules to medicines
A7.
C. Ethene undergoes an addition reaction with HCl to form chloroethane.
Chloroethane undergoes a hydrolysis reaction with water or NaOH to produce
ethanol. Ethanol is oxidised to ethanoic acid by an acidified potassium dichromate
solution.
Q8.
Acetylsalicylic acid could be formed from salicylic acid by warming it with:
(Chemically inactive hydrogen atoms
have been omitted for clarity.)
A
B
C
D
ethanol
methanol
ethanoic acid
methanoic acid
A8.
C. The –OH group on salicylic acid undergoes an esterification reaction with ethanoic
acid to form acetylsalicylic acid (aspirin).
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Chapter 14 From organic molecules to medicines
Q9.
Consider the following molecular structures of substances that all function as drugs.
(Chemically inactive hydrogen atoms have been omitted for clarity.)
The substances that are expected to have similar pharmaceutical effects are:
A I and II
B II and III
C I and III
D I, II and III
A9.
C. Compounds I and III have similar molecular structures and would be expected to
have similar pharmacological effects.
Q10
The ester methyl ethanoate could be made by reacting together:
A CH3CH2OH and CH3COOH
B CH3CH2OH and HCOOH
C CH3OH and CH3CH2COOH
D CH3OH and CH3COOH
A10.
D. Methyl ethanoate is formed by an esterification reaction between methanol,
CH3OH and ethanoic acid, CH3COOH.
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Chapter 14 From organic molecules to medicines
Q11.
The correct systematic name for this substance is
A
B
C
D
2,4-dimethylhexane
2-methyl-4-ethylpentane
2-ethyl-4-methylpentane
1,1,3-trimethylpentane
A11.
A. The substance is an alkane. The longest chain contains 6 carbon atoms. There are
methyl side chains on the 2nd and 4th carbon atoms.
Q12.
A carboxylic acid with five carbon atoms per molecule could have the formula:
A C5H11COOH
B CH3(CH2)4COOH
C CH3CH2CH2CH2COOH
D C3H7COOCH3
A12.
C. The carboxylic acid having five carbon atoms is pentanoic acid.
Q13.
Which statement about compounds I and II is not true?
A
B
C
D
The 13C NMR spectrum of compound I has three peaks, and that of compound II
has two peaks.
The low resolution 1H NMR spectrum of I has four peaks, and that of II has three
peaks.
The mass spectrum of both will show a peak at a mass-to-charge ratio of 60.
The fingerprint region of the IR spectra will be identical for both compounds.
A13.
D. The compounds shown are propan-1-ol and propan-2-ol. They are isomers. The
fingerprint region of the IR spectra of these compounds would be different as their
structures are different.
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Chapter 14 From organic molecules to medicines
Questions 14 and 15 refer to the following information.
The following list gives the formulas of a number of important biochemical
molecules.
C6H12O6
I
C17H33COOH
V
CO(NH2)2
II
C3H8O3
VI
C2H5NO2
III
CO2
VII
C3H7NO2
IV
C17H35COOH
VIII
Q14.
From the list, identify two molecules that would react together to form a dipeptide.
A II and III
B III and IV
C IV and VI
D IV and II
A14.
B. Amino acids undergo condensation reactions to form dipeptides; condensation
polymerisation of amino acids produces proteins.
Q15.
When a saturated fat undergoes hydrolysis, two products that might be formed are:
A VI and VIII
B I and III
C V and VI
D II and VIII
A15.
A. Fats hydrolyse to form fatty acids (V and VIII) and glycerol (VI); molecule VIII
has only single C-to-C bonds in the hydrocarbon chain (CH3(CH2)16COOH) and is a
saturated fatty acid, formed from a saturated fat.
Q16.
When starch is formed from glucose, the condensation reaction that occurs involves:
A a carboxyl functional group and an amino functional group
B two hydroxyl functional groups
C a hydroxyl functional group and a carboxyl functional group
D a hydroxyl functional group and an amino functional group.
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Chapter 14 From organic molecules to medicines
A16.
B. Carbohydrate molecules combine with one another because of condensation
reactions between hydroxyl functional groups, –OH. For example:
Q17.
Olive oil contains a high proportion of oils made from a fatty acid called oleic acid.
The formula of oleic acid is C18H34O2. The number of carbon-to-carbon double
covalent bonds in oleic acid is:
A 0
B 1
C 2
D 3
A17.
B. Oleic acid is CH3(CH2)7CH=CH(CH2)7COOH, i.e. there is one C-to-C double
bond. When written in this format (C18H34O2) the saturated fatty acid has (2 × 18) =
36 H atoms, and this molecule has two fewer hydrogens, thus one double bond.
Q18.
When a protein, such as insulin, is formed, the condensation reaction that occurs
involves:
A a carboxyl functional group and an amino functional group
B two hydroxyl functional groups
C a hydroxyl functional group and a carboxyl functional group
D a hydroxyl functional group and an amino functional group.
A18.
A. The amino group, –NH2, of one amino acid reacts with the carboxyl functional
group, –COOH, of the other to form a peptide link, –NHCO–.
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Chapter 14 From organic molecules to medicines
Q19.
Protein molecules are formed by condensation reactions between amino acids. The
secondary structure of many proteins is a helical shape. The type of chemical bond
responsible for maintaining this shape is:
A hydrogen bonds
B ion–dipole bonds
C ionic bonds
D covalent bonds.
A19.
A. Hydrogen bonds form between the positive dipole of an –N–Hδ+ group in one
peptide link of the protein and the negative dipole of a –C=Oδ– group in another
peptide link of the same protein molecule as shown in the diagram.
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Chapter 14 From organic molecules to medicines
Short-answer questions
Q20.
a
b
c
d
What feature of the bonding of alkenes makes them so useful?
The presence of an alkene can be detected by shaking a sample with a dilute
aqueous solution of bromine.
i Describe what changes you would observe if an alkene was present.
ii Write a balanced equation for the reaction of aqueous bromine solution with
ethene.
i Give the name of an instrumental method that could be used to confirm that
the compound was an alkene.
ii Outline briefly what observation would be expected in the output of the
method if the substance was an alkene.
Under suitable conditions chloroethene can be converted into a polymer.
i What type of polymerisation process would you expect chloroethene to
undergo?
ii Write an unbalanced equation to represent the polymerisation reaction.
iii Give the systematic name of the product.
A20.
a
b
c
d
Alkenes have a double covalent bond that allows them to make a range of useful
addition compounds.
i Brown bromine solution becomes colourless.
ii C2H4(g) + Br2(aq) → CH2BrCH2Br(l)
i Infrared spectroscopy
ii Characteristic absorption of C=C bond at 2070 to 2250 cm–1
i addition polymerisation
ii n CH2CH2 → ~[CH2CH2]n~
iii polyethene
Q21.
a
b
c
What is meant by the terms:
i renewable
ii sustainable?
Assume that petrol is made entirely of octane.
i Write a balanced equation to represent the combustion of octane in an
abundant supply of air.
ii Explain why the use of petrol as a fuel for vehicles is of environmental
concern.
The use of ethanol as an alternative biochemical fuel to petrol is increasing.
i Write a balanced equation for the combustion of ethanol in a good air supply.
ii Why is the use of biofuel ethanol seen as being beneficial for the
environment?
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Chapter 14 From organic molecules to medicines
A21.
a
b
c
i
ii
i
ii
i
ii
Resource is capable of being regenerated.
Resource is capable of meeting the demands of the user.
2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)
The main product of the combustion of petrol in an internal combustion
engine is carbon dioxide. Raised levels of carbon dioxide in the atmosphere
are believed to be causing global warming, with all the undesirable climatic
changes that could have severe impact on the entire world.
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
Whilst combustion of ethanol produces carbon dioxide, the ethanol is made
from plant sources (biomass) and the plants have taken in atmospheric
carbon dioxide by photosynthesis. Hence the use of ethanol as a fuel is not
adding to the total amount of carbon dioxide in the atmosphere.
An additional advantage is that the fuel burns more cleanly, causing less air
pollution by particulate matter and oxides of nitrogen.
Q22.
a
b
c
Crude oil consists mainly of alkanes of varying molecular sizes. What feature of
the intermolecular forces of alkanes allows the alkanes to be separated into
fractions at an oil refinery?
Ethanol is an alternative biochemical fuel to petrol.
i Draw the structural formula of ethanol.
ii Why does ethanol have an affinity for water whilst the alkanes in petrol
do not?
E10 petrol contains 10% ethanol. Outline the environmental advantages using
E10 petrol.
A22.
a
b
The dispersion forces between the alkane molecules increase as the size of the
molecules increase. This leads to an increase in the boiling temperature. Hence
the molecules can be separated in fractions (molecules of similar size) according
to the temperature at which they liquefy.
i
The hydroxyl functional group in ethanol is capable of hydrogen bonding to
water. The alkane molecules in petrol do not possess this functional group
(only held together by dispersion forces) so will not mix with water.
E10 petrol contains 10% ethanol. The use of E10 reduces the amount of carbon
dioxide put into the atmosphere from fossil fuel sources as the carbon atoms in
the ethanol have come from plant sources. The plants in turn have obtained their
carbon from current atmospheric carbon dioxide.
ii
c
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Chapter 14 From organic molecules to medicines
Q23.
A student attempted to prepare a sample of acetylsalicylic acid (aspirin) starting from
salicylic acid.
a
b
c
d
Apart from salicylic acid (2-hydroxybenzoic acid), what other reagent is needed?
Write a balanced equation for the reaction. You may use either structural or
molecular formulas.
Salicylic acid and acetylsalicylic acid are both white crystalline solids. The
student collected and purified the product of the reaction before sending it off for
analysis by proton NMR spectroscopy. How would the spectrum produced by this
technique allow the student to identify whether or not any acetylsalicylic acid had
been formed in the reaction?
A sample of the product was also sent off for analysis by mass spectroscopy. In
the spectrum returned there was a peak at m/e = 138 and a peak at m/e = 180.
What does this tell you about the sample?
A23.
a
b
c
d
Ethanoic anhydride
Alternative molecular formula equation:
2C7H6O3 + C4H6O3 → 2C9H8O4 + H2O
1
H NMR spectrum would show a large peak at 2.2 ppm if the CH3 group was
present, showing that acetylation had occurred.
Peak at m/e = 138 is molecular ion peak of salicylic acid.
Peak at m/e = 180 is molecular ion peak of acetylsalicylic acid.
If both are present, the sample contains both reactant and product.
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Chapter 14 From organic molecules to medicines
Q24.
Traditional medicine used infusions of willow bark to help reduce pain and fever.
Analysis has shown that the infusion contains the substance salicin. It is now known
that digestive enzymes are able to convert salicin into salicylic acid.
A student proposes to synthesise a soluble form of aspirin from salicin via the reaction
pathway shown below.
a
b
Name the type of reactions represented by the letters: A, B, C
Copy the two structures (salicin and salicylic acid) shown below and circle their
common features.
c
Acetylsalicylic acid (aspirin) has the structure shown below:
i
ii
Circle the part of this molecular structure that would not contribute to its
pharmaceutical activity.
Acetylsalicylic acid is hydrolysed to salicylic acid in the body. Explain why
it is preferable to take acetylsalicylic acid rather than salicylic acid.
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Chapter 14 From organic molecules to medicines
A24.
a
A. Oxidation; the alkanol functional group is oxidised to the carboxylic acid
functional group
ii B. Esterification reaction between the hydroxy functional group on salicylic
acid and ethanoic acid
iii C. Acid base reaction between the COOH group and Na2CO3 or NaOH
i
b
c
i
ii
Acetylsalicylic acid does not have the acidic properties or unpleasant taste of
salicylic acid so can be taken much more easily and is much less irritating to
the stomach. It does not hydrolyse back to salicylic acid until it reaches the
small intestine.
Q25.
Aspirin, acetylsalicylic acid, C9H8O4, is made by reacting salicylic acid, C7H6O3
(Mr = 138.0), with ethanoic anhydride, C4H6O3 (Mr = 102.0), according to the
equation:
2C7H6O3 + C4H6O3 → 2C9H8O4 + H2O
a Determine which is the limiting reagent if 10.0 g salicylic acid is reacted with
10.0 g ethanoic anhydride.
b Find the maximum mass of acetylsalicylic acid that could be produced by this
reaction.
c Explain why sufferers of pain and fever would prefer to take acetylsalicylic acid
(aspirin) rather than salicylic acid. Are there any situations where aspirin should
not be given?
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Chapter 14 From organic molecules to medicines
A25.
a
b
c
10.0
= 0.07246 mol
138.0
10.0
n(ethanoic anhydride) =
= 0.09804 mol
102.0
From the equation, 2 mol salicylic acid reacts with 1 mol ethanoic anhydride.
Hence, 0.07246 mol salicylic acid reacts with ½ × 0.07246 mol ethanoic
anhydride
= 0.03623 mol ethanoic anhydride
As there is 0.09804 mol ethanoic anhydride available, salicylic acid is the
limiting reagent.
From the equation, 2 mol salicylic acid forms 2 mol acetylsalicylic acid.
Hence, 0.07246 mol salicylic acid forms 0.07246 mol acetylsalicylic acid.
M(C9H8O4) = 180.0 g mol–1
Maximum mass of acetylsalicylic acid = 0.07246 × 180.0
= 13.0 g (three significant figures)
Aspirin is preferred to salicylic acid as it does not have an unpleasant taste and
does not irritate the lining of the mouth and stomach.
However, aspirin should not be given to children or adolescents suffering viral
infections such as influenza or chickenpox as they can develop Reye’s Syndrome
(can be fatal). Also people who are taking prescribed amounts of aspirin as part of
their treatment for other conditions should not take additional aspirin to
counteract pain or fever. They should consult a pharmacist who will advise on
appropriate action.
n(salicylic acid) =
Q26.
a
b
c
Draw the structural formula for:
i hexanoic acid
ii butan-2-ol
iii pent-2-ene
Draw the structural formula and provide the systematic name of four isomers of
C4H9Cl.
Give systematic names for:
i CH3CH2CH2OH
ii CH3CH2CH2COOH
iii CH3CH2CH2COOCH2CH3
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Chapter 14 From organic molecules to medicines
A26.
a
i
ii
iii
b
b
i propan-1-ol
ii butanoic acid
iii ethylbutanoate
Q27.
Ethyl hexanoate is an ester commonly used as artificial pineapple essence.
a What starting materials are required to prepare a sample of this ester by one
reaction?
b Describe the reaction conditions that you would use in this preparation. Explain
the purpose of any additional reagents.
c Write a balanced equation to represent the reaction.
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Chapter 14 From organic molecules to medicines
A27.
a
b
c
ethanol, hexanoic acid
Concentrated sulfuric acid is added as a catalyst.
CH3CH2OH(l) + CH3(CH2)4COOH)(l) → CH3(CH2)4COOCH2CH3 (l)+ H2O(l)
Q28.
The boxes below represent structures that make up a polymer of biological
importance.
a
b
c
d
e
f
What is the name of this biologically important polymer?
Draw a simple diagram using boxes to show the arrangement of the components
of a nucleotide containing cytosine.
Nucleotides can bond together to form a long chain. What name is given to this
type of reaction? What molecule is also produced in this reaction?
Draw a diagram to show the arrangement of nucleotide monomers in a single
strand of DNA.
Draw a simple diagram to show how two single polynucleotide strands are
aligned in DNA. Show any bonds that are formed between two single strands of
DNA
The structure of DNA is often described as a double helix. Explain what is meant
by this term.
A28.
a
b
deoxyribonucleic acid, DNA
c
Condensation polymerisation, water
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Chapter 14 From organic molecules to medicines
d
e
f
Two strands of polynucleotides coiled around each other can be thought of as
similar in shape to a spiral staircase where the polynucleotides represent the
handrails and the complementary base pairs are the steps.
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Chapter 14 From organic molecules to medicines
Q29.
Explain how a DNA profile can be obtained.
How is a DNA profile used in forensic analysis?
a
b
A29.
a
•
•
•
DNA is extracted from a sample of saliva, blood or tissue.
Restriction enzymes are used to cut the DNA into fragments.
The number of fragments are duplicated using polymerase chain reaction
(PCR).
• Fragments are sorted by length using gel electrophoresis.
• Radioactive or fluorescent probes are attached to the fragments.
• Bands are detected using X-ray film of light of various wavelengths.
b The profile obtained from a DNA sample is compared with the DNA profile
obtained from a source whose identity is known. In criminal investigations, DNA
profiles may be used to identify a suspect or remove an individual from
suspicion. DNA profiling is also used to determine paternity and to identify
victims of natural disasters.
Q30.
Cysteine is an amino acid that forms part of the polypeptide chain in albumin, a
protein that occurs in eggs. The ‘Z-group’ of cysteine has the structure –CH2SH.
a Draw the structure of the amino acid cysteine.
b On your diagram, label and name the two functional groups that are common to
all amino acids.
c Draw the structure of cysteine at low pH.
d Draw the structure of the two possible dipeptides that can form when a molecule
of cysteine reacts with a molecule of alanine (alanine has a ‘Z-group’ of –CH3).
A30.
a, b
c
+
NH3CH(CH2SH)COOH
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Chapter 14 From organic molecules to medicines
d
Q31.
Condensation reactions are responsible for the formation of many important
biomolecules, e.g. cellulose, proteins, starch, lipids.
a Draw the structure of the products of the condensation reactions between the
following pairs of compounds:
i
ii
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Chapter 14 From organic molecules to medicines
iii
glycine
b
serine
Name the functional group that has been produced as a result of:
i the reaction between glucose and fructose
ii the reaction between glycerol and palmitic acid
iii the reaction between glycine and serine
A31.
a
i
ii
iii
b
i glycoside link
ii ester link
iii peptide bond
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Chapter 14 From organic molecules to medicines
Q32.
Compound X contains hydrogen, carbon and oxygen. 4.50 g of X was burnt in excess
air. 2.70 g of water was produced. 3.67 L of carbon dioxide measured at 25°C and
1.00 atm pressure was also produced.
a Determine the percentage composition of the compound.
b What is the empirical formula of the compound?
c The mass spectrum of compound X is shown below. Deduce the molecular
formula of X.
Mass spectrum of X.
d
e
The molecular formula of compound X suggests that it is either a carboxylic acid
or an ester. Write the structural formula and give the name of the two isomers that
correspond to the molecular formula of X.
Use the IR spectrum of X below to explain why the compound is not an acid.
IR spectrum of X.
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Chapter 14 From organic molecules to medicines
f
Explain why there are only two peaks in the 1H NMR spectrum of compound X
shown below.
1
H NMR spectrum of compund X.
g
Identify compound X.
A32.
a
b
c
2 × 2.70
= 0.3mol
18
m(H) = 0.3 × 1 = 0.3 g
0.3 × 100
%(H)=
= 6.67%
4.50
pV
1 × 101.3 × 3.67
n(C) =
=
= 0.15 mol
RT
8.31 × 298
0.15 × 12 × 100
%C =
= 40.0%
4.50
%O = 100 – (%H + %C) = 100 – (6.67 + 40.0) = 53.33%
C
: H
: O
% 40.0 : 6.67 : 53.33
40.0
6.67
53.33
n
:
:
12
1.0
16
3.33 : 6.67 : 3.33
empirical formula = CH2O
From the mass spectrum, the molecular ion has a m/e of 60, i.e. molecular mass
is 60.
The molar mass of CH2O unit is 30. There must be two CH2O units in the
molecule. The molecular formula is C2H4O2.
n(H) =
d
e
Refer to a table of IR absorptions. The IR spectrum of ethanoic would have a
broad absorbance between 2500 cm–1 and 3200 cm–1 due the OH bond in
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Chapter 14 From organic molecules to medicines
f
g
carboxylic acid functional group. Since this absorption is not present, the
compound is not ethanoic acid.
The two peaks represent two different hydrogen ‘environments’. The peak at
chemical shift 8 is due to H–COO hydrogen and the peak at chemical shift of 3.7
is due to the three hydrogens in –O–CH3.
methyl methanoate
Q33.
The DNA profiles below were used as evidence in two separate court cases in which
women (M) claimed particular men (F) were the fathers of their children (C).
a
b
What do you think should have been the verdict in each case?
Each band in these profiles represents a repeating base sequence region. Why are
multiple repeating base sequence regions compared in DNA profiling?
A33.
a
b
The DNA profile of the child in case 1 has a number of bands in common with
that of the male. In this case, the male would be the father. In case 2, there are no
common bands in the DNA profile of the child and the male. The male is not the
child’s father.
The number of repeating base sequences in the non coding region of DNA vary
between individual. In DNA profiling, the base sequences at 10 or more locations
on the chromosomes are compared. Each individual (except identical twins) will
have their own unique DNA profile.
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