Download CAPE CHEMISTRY UNIT TWO REVISION PAPER MODULE 1 (a

CAPE CHEMISTRY
UNIT TWO
REVISION PAPER
MODULE 1
1.
(a) On complete combustion, 0.00754 mol of a hydrocarbon E gave 2.654 g of carbon dioxide and 0.543 g of
water.
(i)
Determine the molecular formula of E
Moles CO2 = 2.654 g / 44 gmol-1 = 0.0603 mol = moles of carbon
1 mole of E contains 0.0603 mol / 0.00754 mol = 8 moles of carbon
Moles of H2O = 0.543 g / 18 g mol-1 = 0.03016 mol
Moles of hydrogen = 0.0603 mol
1 mole of E contains 0.0603 mol / 0.00754 mol = 8 moles of hydrogen
Molecular formula is C8H8
(ii)
Construct a balanced equation for the complete combustion of E.
C8H8
+
10 O2

8 CO2
+
(4 marks)
4 H2O
(b) E burns with a smoky flame. 1 mol of E reacts with 1 mol of bromine molecules.
(i) Draw the displayed formula of E
Draw a benzene ring with a CH=CH2 substituent
(ii)
Draw the displayed formula of the product formed between E and bromine
Draw a benzene ring with a CHBr=CH2Br substituent
(2 marks)
(c) What products would you expect if E is heated under reflux with hot concentrated acidified potassium
manganate (VII).
(2 marks)
Benzoic acid / Potassium benzoate, carbon dioxide, (may be methanoic acid)
N.B. potassium manganate (VII) would be decolourized
Draw a benzene ring with a -COO- substituent
(d) Hex-3-ene exists in two isomeric forms but cyclohexene occurs in only one form.
Draw and name the displayed formula of the two isomers of hex-3-ene and explain why such isomerism is not
possible for cyclohexene.
(4 marks)
CH3-CH=CHCH2CH3
The two H are in the cis or trans position
Cyclohexene is a 6-membered ring – only the cis-form is possible (both H outside
the ring. The trans-form would cause distortion of the ring / strain the angles /
and so is not formed
(e) Describe the mechanism of the hydrolysis of bromoethane by aqueous sodium hydroxide. Name the type of
reaction.
(3 marks)
OH- is the nucleophile and it attacks the carbon atom in the C-Br bond where
this carbon carries a partial positive charge due to the neighbouring
electronegative bromine. The C-Br bond begins to elongate as the C-OH bond
starts to form. Eventually, the C-Br bond is broken and ethanol is formed. The
reaction is a nucleophillic substitution.
(TOTAL 15)
2.
(a) Explain how primary, secondary and tertiary alcohols can be distinguished on the basis of their oxidation
Products.
(3 marks)
Primary alcohols are oxidized to carboxylic acid, secondary alcohols are
oxidized to ketones and tertiary alcohols do not undergo oxidation.
To distinguish between the classes of alcohols, acidified potassium
manganate(VII) solution is added to the alcohol and the mixture is warmed.
Decolourisation is observed for primary and secondary alcohols but not for
tertiary alcohols. 2,4-dinitrophenylhydrazine is then added to theresultant
mixture. An orange precipitate of 2,4-dinitrophenylhydrazone is observed if the
alcohol is a secondary alcohol but no orange precipitate is observed if the
alcohol is a primary alcohol.
(b) The ester formed between 3-methylbutan-1-ol and ethanoic acid contributes to the flavour of ripe pears.
(i)
Draw the structural formula of the ester.
O
ll
CH3 C -O- CH2CH2CH(CH3)CH3
(ii)
What conditions and reagents would you use in the laboratory to make the ester from the acid and the
alcohol named above?
3-methylbutan-1-ol is warmed with glacial ethanoic acid using sulphuric
acid as catalyst .
(iii)
Draw the structural formulae of three primary alcohols that are isomers of 3-methylbutan-1-ol, labeling
with an asterisk any chiral carbon atom they contain.
(9 marks)
CH3CH2CH2CH2CH2OH
CH3CH2C*(H)(CH3)CH2OH
CH3CH2C(CH3)(CH3)CH2OH
(c)
Explain why ethanol is the only primary alcohol that undergoes the tri-iodomethane (iodoform) reaction.
(3 marks)
(TOTAL 15)
A primary alcohol has two hydrogen atoms attached to the same carbon to which
the hydroxyl functional group is attached. To undergo tri-iodomethane reaction,
the alcohol has a methyl and a hydrogen atom on the carbon atom to which
hydroxyl functional group is attached. Hence for a primary alcohol to give a
positive tri-iodomethane test, the alkyl group which is attached to the carbon with
the OH group attached has to be a methyl group.
MODULE 2
1.
(a) For each of the following, (a) identify the branch of spectroscopy to which it applies, and (b) explain its use:
 Beer’s law;
Uv / visible spectroscopy
The degree of absorption at a given wavelength of an absorbing compound in a nonabsorbing solvent depends upon the concentration of the compound and upon the path
length of the radiation i.e. the thickness of the cell. This is expressed in the Beer-Lambert
Law, which is generally well obeyed for reasonably dilute solutions.
The more concentrated the absorbing species – the more light will be absorbed
The longer the cell length – the more opportunity for light to be absorbed
Beer-Lambert’s Law: The fundamental equation of spectrophotometry.
A =  cl
A = Absorbance,
[dimensionless]

= molar absobptivity, c = concentration of sample or standard, l = path length
[mol-1dm3 cm-1]
[mol dm-3]
[cm]
 Nujol mull;
Infra-red spectroscopy
Solid samples can be prepared in four major ways. One method is to crush the sample with
a mulling agent, (usually nujol / paraffin oil) in a marble or agate mortar, with a
pestle. A thin film of the mull is applied onto salt plates and measured.
 Ionization chamber;
Mass spectroscopy (not really spectroscopy)
The first process in mass spectroscopy is to bombard the atoms or molecules with a stream
of high-energy electrons producing positive ions which are accelerated in an electric
field. The process occurs in the ionization chamber of the mass spectromenter.
Complexing agent.
[7 marks]
A ligand (neutral molecule, cation or anion) that forms a complex ion with a simple
ion (cation or anion).
Example: Vanadate- molybdate reagent + aqueous phosphate  yellow phosphovanadomolybdate complex
(b) Scientists use several spectroscopic methods for qualitative and quantitative analysis.
(i)
Draw and label a diagram of the electromagnetic spectrum. Show the approximate wavelength range of each
region.
[4 marks]
This is on page UWI spectroscopy booklet
(ii)
Select any two regions of the electromagnetic spectrum and explain how the differences in their relative
energies determine their analytical application.
[4 marks]
(TOTAL 15)
There are many different types of electromagnetic radiation but the only difference
between them is the amount of energy found in the photons.
Radiation
Frequency,
Hz
1016
Energy, kJmol-1
Practical application
Chemical interacrions
Uv
Wavelength,
, nm
10
12,000
Sun lamps
Visible
103
1014
310
Light bulb
Infrared
105
1012
150
Heat lamps
Quantitative analysis
Redistribution of outer electrons in molecular orbitals
Quantitative analysis
Redistribution of outer electrons in molecular orbitals
Qualitative analysis
Vibration of chemical bonds
Ultra violet [190 to 400 nm] / visible region [400 to 800 nm]
Energy can cause electronic transitions
Radiation is absorbed by atoms and molecules when the energy of the photons exactly
matches the energy difference between the lower energy state (ground state) and one
of the higher energy state of the atoms or molecules.
Ultra-violet and visible spectroscopy deals with the absorption of energy bringing
about an increase in the energy of electrons. These energy changes are quantized,
that is, only certain transitions are permitted. The energy absorbed or emitted during
d to d transitions within transition metal complexes are measured by uv-visible
spectrophotometers.
The absorbance is used to determine the quantity of materials
Infrared region [2500 to 25000 nm or 400 to 4000 cm-1]
Energy can cause bond vibration and rotation (not sufficient for electronic
transitions)
Infrared spectra originate from the different modes of vibration and rotation of a
molecule. IR radiation generally is not sufficient to cause electronic transitions but can
induce transitions in the vibrational and rotational states associated with the ground
electronic states.
The absorbance is used to determine the chemical bonds present but is not very reliable to
determine the quantity of material present.
2.
(a) Define the following chromatographic terms:
 Retention time
In GLC chromatography retention time for each component is the time needed after
injection of the mixture on to the column until that component reaches the detector. The
time taken for each component to pass through the column is found by measuring the
distance on the chromatogram between the injection of the mixture (time = 0) and the
centre of the peak for that component.

Retention factor
The chromatogram: Solutes eluted from a chromatography column are observed with
various detectors. Simple systems may be measured directly.
Rf measured: Rf = distance traveled by component (from origin)
distance traveled by solvent front (from origin)
 Visualizing agent
Many compounds are not visible to the eye when dissolved in a solvent or adsorbed on a
adsorbent. Visualization processes make these substances visible. The visualizing agents
used for this purpose include UV lights that cause fluorescence or phosphorescence and
chemical reactions that give colored compounds [e.g. ninhydrin, iodine etc].

Solvent front
[4 marks]
The farthest distance moved by the solvent or mobile phase along the stationary phase.
(b) Explain the principle of partition chromatography in relation to paper chromatography.
[2 marks]
In partition chromatography the solutes are partitioned between the mobile phase and a
thin film of liquid (commonly water) firmly absorbed on the surface of the stationary
phase.
Stationary phase: paper - cellulose fibre composed of glucose molecules, which have a
large number of hydroxyl groups. Water molecules hydrogen-bond to these groups, so that
a sheet of ‘dry’ paper contains around 10% by mass of water. This water acts as the
stationary phase.
Mobile phase: a solvent consisting of an aqueous solution or an organic solvent such as
ethanol or ethanoic acid
The mixture to be separated is dissolved in the mobile phase, which moves along the
paper. The movement comes about by capillary action, which results from the forces
between the solvent and the solid fibres of the paper.
The paper may be referred to as the support.
The separation is stopped when the solvent has traveled nearly to the top of the paper.
The solutes are transferred from the mobile phase to the stationary phase by partition
between the two liquids (the water on the paper and the moving solvent). Solutes in the
mobile phase move along with it. Solutes in the stationary phase move more slowly are
remain stationary.
(c) The fragrance for the perfume CHANEL No. 5, is extracted from the flower of the Ylang Ylang tree.
Draw a diagram of the apparatus that could be used to extract the fragrance oils from a batch of crushed flowers.
[4 marks]
The steam distillation apparatus
(d) Scientists use standard materials and statistical techniques to ensure the reliability of analytical data.
(i)
What purpose do standard materials serve?
[2 marks]
Analytical standards are stable materials of high purity and known composition. They
are used as reference materials in quantitative analysis.
(ii)
An analyst measured the iron content of each of twenty iron tablets. The mean value was 0.0224 mg/g and
the standard deviation was 0.0010 mg/g. The following day the analyst measured another iron tablet and
obtained a value of 0.0223 mg/g. Comment on the reliability of the last measurement.
[3 marks]
The value obtained on the second day (0.0223 mg/g) differed from the mean
by – 0.0001 mg/g which is less than one standard deviation. Therefore the measurement is
precise. However the accuracy is unknown because the true value in not given.
The true value may be determined by a number of methods. One method is ‘spiking’. A set
of tablets are used to prepare at least two test solutions. One solution is spiked with a
known quantity of iron standard before being made up to volume. The measurement for
the spiked sample should be equal to the measurement for the unspiked sample plus the
known quantity of iron added (the calculated spike).
YOU GOT PAGES 1 TO 5 BEFORE
HERE IS THE REST
MODULE 3
1. (a) The ozone layer in the stratosphere offers a vital protection to life on the Earth’s surface. It is maintained by the
equilibrium that exists between ozone and oxygen.
(i)
Define the term ‘photodissociation’
(Also photolysis or photodecomposition)
A chemical reaction in which a chemical is broken down by photons.
(ii)
Explain, in detail, how the equilibrium between ozone and oxygen is maintained.
A dynamic photochemical equilibrium develops between the formation and
decomposition of ozone.
In the stratosphere , under the influence of UV radiation with wavelength below 242 nm,
diatomic molecular oxygen is split (cleaved) into two oxygen atoms, O(g) and excited high
energy O*(g)..
O2(g) 
O(g)
+
O*(g)
Ozone, O3, is then formed due to the reaction between the excited oxygen atom and
diatomic oxygen in the presence of an inert 3rd body (M) which is usually N2 or O2
(also NO).
O*(g)
+
O2(g) + M(g) 
O3(g) +
M*(g)
The ozone reforms diatomic oxygen and oxygen atoms by absorbing ultraviolet radiation
of nwavelengths between 290 and 330 nm (decomposed by photodissociation).
O3(g) 
O2(g) + O(g)
Ozone reacts with with O to form two molecules of oxygen
O3(g)
(iii)
+
O(g)

2O2(g)
What effect does the ozone layer have on the sunlight passing through it?
The ozone layer an altitude between 15 and 50 km above the Earth’s surface (in the
Stratosphere) is of existential significance for life on Earth because it absorbs the shortwavelength, high energy UV portion of the solar radiation
The ozone layer absorbs a significant amount of ultraviolet radiation from the sunlight
that passes through it.
(iv)
What type of chemical reaction at the Earth’s surface would be inhibited by the presence of the ozone layer?
[8 marks]
Photochemical reactions that occur by the absorption of ultraviolet radiation.
(b)
Explain what is meant by the term ‘residence time’ for gases in the atmosphere. Illustrate your answer by
reference to two important sources (excluding the combustion of fossil fuels) and two important sinks for carbon
dioxide.
[4 marks]
Residence time refers to the average time that a molecule of a particular gas spends
in the atmosphere from its introduction by a source to its removal by a sink.
Residence time is affected by the concentration of the gas that is naturally present
and the rate by which it is removed by the sinks.
Two important sources for carbon dioxide are the aerobic respiration of living
organisms and the manufacture of cement.
Two important sinks (reservoir) (for carbon dioxide are photosynthesis of green
plants and dissolution of carbon dioxide in water bodies such as the oceans.
(c) Both carbon monoxide and carbon dioxide are present in exhaust gases of cars. The proportion of each gas is partly
determined by the following equilibrium:
2CO2(g) 
2CO(g)
+ O2(g)
This reaction is endothermic in the forward direction.
Explain what are the optimum conditions of temperature, pressure and airflow which favour the production of carbon
dioxide.
[3 marks]
(TOTAL 15)
Production of carbon dioxide is favoured by low temperature (since its production is
exothermic), high pressure (since less gaseous molecules are present after its formation)
and a faster air flow so that more oxygen is available for complete combustion.
2. (a) (i) State three different substances in everyday or industrial use which contains chlorine.
(ii) For each substance in (i) above, state one use and explain how the use depends on a relevant property of the
substance.
[5 marks]

(chloroethene) – make pipes – relatively inert, durable, strong, can be easily
molded
 Sodium chlorate (I) – make bleach – oxidizes coloured stains to colourless
products
 Tetrachloromethane – solvent – inert, not flammable
 CCl2F2 – dichlorodifluoromethane – refrigerant fluid – inert – low melting /
boiling point.
(iii)
Comment on any problems which arise from the disposal of materials which contain chlorine.
[2 marks]
If materials containing chlorine are buried , the material may be non-biodegradable so
that they remain around for a long time occupying precious land space. If they are
biodegradable , the chlorine may be converted by microorganisms into organochlorine
compounds which may be harmful.
Materials containing chlorine on incineration produces toxic gases such as hydrogen
chloride and may also form the very toxic dioxins which are suspected of causing cancer
in humans.
(b) With the aid of a labeled diagram, outline the process by which chlorine is manufactured from brine in the diaphragm
cell. Include all relevant equations
[5 marks]
This is in your text book
(c ) Outline three major impacts of the aluminium industry on the terrestrial environment.
[3 marks]
(TOTAL 15)
Mining:
 Deforestation – vegetation cover removed – may lead to soil erosion
 Destruction of ecosystems – top soil removed followed by soil to a depth of 4 to 6
metres
 Oil spill from mining vehicles kill organisms
 Original ecosystems may not be fully restored after land reclamation.
[Mining also produce fossil fuel fumes, noise and red dust].
Digestion:
 Waste digestion liquids – seeps into soil and rocks – alter pH - kills organisms
[Sodium hydroxide fumes diffuse into the air]
Smeltering:
 Spent lining from electrolysis pot – contains cyanide and fluorides – occupy
landfills
Products
 Discarded cars, planes, pots and other products occupy valuable land
 Degrade slowly – protected by aluminium oxide