Cambridge A Level Chemistry 9701 — 2006 May/June Paper 6 · Variant 1

This document consists of 6 printed pages and 2 blank pages. SP (CW/CGW) T06722/3 © UCLES 2006 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level CHEMISTRY 9701/06 Paper 6 Options May/June 2006 1 hour Additional Materials: Answer Booklet/Paper Data Booklet READ THESE INSTRUCTIONS FIRST If you have been given a Answer Booklet, follow the Instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams, graphs, or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer both questions on two of the Options. A Data Booklet is provided. You may use a calculator. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

2 9701/06/M/J/06 © UCLES 2006 BIOCHEMISTRY Answer both questions on the paper provided. 1 Spider silk is a natural structural protein, which belongs to the same group as collagen and keratin. It contains a high proportion of the amino acids glycine and alanine. NH2 CH2 glycine alanine CO2H NH2 CH CH3 CO2H (a) (i) Draw the structure of the tripeptide Gly-Ala-Gly, showing the displayed structure of the peptide linkages. (ii) What type of reaction takes place during the formation of a peptide? [4] The flexibility of spider silk is due to the glycine-rich regions in the structure. (b) (i) Explain why glycine-rich regions of the protein structure give rise to its flexibility. (ii) Silk is hydrophobic. Explain the meaning of the term hydrophobic. What gives silk this property? [4] (c) Spider silk is considered to be a crystalline structure and has properties similar to nylon and Kevlar. (i) Explain why silk and nylon have similar properties. (ii) If spider silk could be manufactured industrially, suggest a possible use. [2] 2 Deep-sea divers may experience ‘nitrogen narcosis’ during a dive. At sea level, nitrogen is inert and dissolves in the blood. As a diver goes deeper underwater, the partial pressure of nitrogen increases and more nitrogen dissolves in the blood. As a consequence, nitrogen can diffuse into the phospholipid bilayer. (a) Draw a diagram of a phospholipid bilayer. State the nature of the intermolecular forces acting at each end of a phospholipid in a bilayer. [3] (b) Nitrogen collects in the alkyl part of the lipid bilayer. (i) How would the nitrogen interact with this section of the lipid bilayer? (ii) Suggest what effect a high nitrogen concentration might have on the lipid bilayer. [3] (c) The high nitrogen concentration impairs the conduction of nerve impulses. Nerve impulses rely upon movement of Na+ and K+ into and out of the cell. Describe how Na+ and K+ move into and out of cells such as nerve cells, with reference to the parts of the cell and the molecules involved. [4]

3 9701/06/M/J/06 © UCLES 2006 [Turn over ENVIRONMENTAL CHEMISTRY Answer both questions on the paper provided. 3 Two of the characteristics of drinking water are that it must be free from • disease-causing organisms, • harmful or toxic chemicals. (a) Describe, with the aid of an equation, how chlorine helps to remove disease-causing organisms from water. [2] (b) Explain why chlorinated organic compounds may be found in drinking water. [1] (c) River water is often polluted by phosphates that have come from detergents or fertilisers. (i) Explain why phosphates are added to detergents. (ii) Describe the effect of excess phosphates in river water. (iii) River water containing phosphates is a source of drinking water. During the water purification process the phosphates can be removed by reaction with an excess of aluminium ions. A 1000 dm3 sample of river water is mixed with an excess of aluminium ions and a precipitate of mass 488 g is obtained. Assuming all the phosphate exists as PO4 3–(aq) write an equation showing the formation of this precipitate. Calculate the concentration, in mol dm–3, of phosphate ions in the river water sample. [7] 4 Soil is a mixture of decomposing organic material and inorganic compounds such as silicates or aluminosilicates. The diagram shows the silicate repeat unit of a clay. O- O- O- O Si O- Si O- O O O- Si (a) (i) An aluminosilicate has one aluminium atom substituted for a silicon atom in the repeat unit shown above. Give the formula for the repeat unit of the aluminosilicate. (ii) Suggest, with reference to silicates and aluminosilicates, how electrical neutrality is achieved. [3] (b) The pH of soil is controlled by both clays and organic material such as humus. (i) Explain how clays can help to maintain soil acidity. (ii) Explain how humus can help to maintain soil acidity. Assume that humus behaves as a carboxylic acid, RCO2H. [4] (c) Farmers often use lime to reduce the acidity of soil. Explain how lime reduces the acidity of soil and explain the limitations or disadvantages of liming acidic soil. [3]

4 9701/06/M/J/06 © UCLES 2006 PHASE EQUILIBRIA Answer both questions on the paper provided. 5 (a) (i) State Raoult’s Law. (ii) The vapour pressures of hexane and octane at a certain temperature are 48 kPa and 36 kPa respectively. Calculate the vapour pressure at the same temperature of a mixture of the two liquids with a mole fraction of 0.30 of hexane. (iii) State what assumption you have made in this calculation, and explain why the assumption is reasonable in this case. [5] (b) Propanone and trichloromethane form a mixture with a negative deviation from Raoult’s Law, which has a minimum vapour pressure at a mole fraction of 0.50. (i) Explain why this pair of liquids produces a negative deviation. (ii) Suggest why the minimum occurs at a mole fraction of 0.50. [2] (c) A sample of liquid of 0.25 mole fraction trichloromethane and 0.75 mole fraction propanone is distilled using a fractionating column. Describe, with reasons, the sequence of the composition of the distillate obtained when the above mixture is distilled. [3] 6 (a) The triple point of carbon dioxide is 216.6 K and 5.11 atm. Solid carbon dioxide is more dense than liquid carbon dioxide. Use this information to sketch and label the phase diagram of carbon dioxide. [5] (b) Use the phase diagram you have drawn to comment on or explain the following observations about liquid carbon dioxide as present in some fire extinguishers. (i) At 25 °C, the pressure inside an unopened fire extinguisher is over 60 atm. (ii) When the nozzle is opened, the carbon dioxide is seen to form a finely-divided, white, powdery solid. [4] (c) How does the phase diagram for carbon dioxide differ from that of water? [1]

5 9701/06/M/J/06 © UCLES 2006 [Turn over SPECTROSCOPY Answer both questions on the paper provided. 7 (a) Carbon dioxide has the displayed formula O=C=O. Predict the number of absorptions you would expect in an infra-red spectrum of carbon dioxide and identify the reason for each absorption. [3] (b) An organic compound C4H5O2N is known to have two functional groups. These groups are responsible for absorptions at 1710 cm–1, 2260 cm–1 and a broad peak at 2800 cm–1 in the infra-red spectrum of this compound. (i) Identify the bond responsible for each of these absorptions. (ii) Suggest a possible structure for the compound. [4] (c) Another spectroscopic technique could be used to confirm the proposed structure. Name a possible technique and briefly describe the evidence you would expect it to provide. [3] 8 A compound E is known to have either structure I or structure II as shown below. CH3CH2C OCH3 O I II CH3CH2CH2C OH O E was investigated by using infra-red spectroscopy and nuclear magnetic resonance spectroscopy. (a) The infra-red spectrum of E showed strong peaks at 1740 and 1200 cm–1. Indicate which structure this suggests for E, outlining your reasoning. [2] (b) (i) The nuclear magnetic resonance spectrum of E includes a triplet of peaks and a quartet of peaks. Indicate which groups of protons in the molecule are responsible for each of these multiple peaks, and outline how the splitting patterns arise. (ii) The nuclear magnetic resonance spectrum of E shows one other peak. Suggest the chemical shift and the splitting pattern of this peak. (iii) Describe and explain how the addition of deuterium oxide to a sample of E could be used to distinguish between the two structures shown. [6] (c) The mass spectra of compounds I and II are different. Give two differences in these spectra. [2]

6 9701/06/M/J/06 © UCLES 2006 TRANSITION ELEMENTS Answer both questions on the paper provided. 9 (a) (i) Stainless steel is widely used to make cutlery. Suggest a suitable percentage by mass of chromium to use in stainless steel. (ii) Suggest an explanation of how the addition of chromium to iron makes it “stainless”, i.e. reduces its susceptibility to corrosion. [3] (b) Compound A is a red liquid containing the following percentages by mass. Cr, 33.6%; O, 20.6%; Cl, 45.8%. (i) Calculate the empirical formula of A and state the oxidation number of chromium in it. (ii) When dissolved in water, compound A produces an orange, acidic solution that gives a white precipitate when AgNO3(aq) is added. On adding NaOH(aq) to a solution of A in water, the orange solution turns yellow. Explain these observations, giving equations where appropriate. [7] 10 Use of the Data Booklet is relevant to this question. (a) Explain why transition element complexes are often coloured. [3] (b) Suggest explanations for the following colour changes, writing equations where appropriate. (i) The addition of KSCN(aq) to Fe3+(aq) produces a deep blood-red colouration, which disappears when NaF(aq) is added. (ii) The addition of SO2(aq) to an acidified solution of ammonium vanadate(V), NH4VO3, produces a bright blue solution, which turns green when Sn2+(aq) is added. [7]

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level and GCE Advanced Subsidiary Level MARK SCHEME for the May/June 2006 question paper 9701 CHEMISTRY 9701/06 Paper 6 Maximum raw mark 40 This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the Report on the Examination. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2006 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 1 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9701 06 © University of Cambridge International Examinations 2006 Biochemistry 1 (a) (i) O CH3 O
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NH2 – CH2 – C- NH -CH – C-NH – CH2 – COOH [1] Displayed structure [1] One peptide linkage shown [1] (ii) Condensation [1] [4] (b) (i) Weak intermolecular forces of attraction (1) Van der Waals (1) [2] (ii) No attraction/ffinity for water [1] Non-polar structure [1] [4] (c) (i) Both contain the polyamide structure/-CONH- [1] (ii) Bullet proof vests ; body armour; ropes; airbags; kayaks; gloves; skis; run-flat tyres; shields for jet engines; helmets; racquets; clothing any one [1] [2] 2 (a) Diagram: [1] R end – van der Waals forces [1] Phosphate end - ionic/polar [1] [3] (b) (i) van der Waals interactions/dipole –dipole interactions/temporary dipoles/hydrophobic interactions [1] with the hydrocarbon part of the bilayer [1] (ii) Disrupt it/distort it/weakens the interactions between the bilayers [1] [3] (c) K+ moves into cell, Na+ moves out of cell [1] This occurs by active transport [1] ATP/adenosine triphosphate provides the energy [1] Integral proteins in the membrane are used [1] [4]

Page 2 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9701 06 © University of Cambridge International Examinations 2006 Environmental Chemistry 3 (a) Cl2 + H2O
HOCl + HCl / Cl2 + H2O == ClO– + 2H+ + Cl– [1] HOCl is an oxidising agent which kills bacteria/ClO– is an oxidising agent which kills bacteria [1] [2] (b) Dissolved chlorine will react with organic pollutants in water [1] (c) (i) Water softener/removes magnesium or calcium ions from water (as insoluble magnesium or calcium phosphate) [1] (ii) Phosphate encourages growth of algae and bacteria to form an algal bloom [1] Algal bloom prevents sunlight from reaching plants lower down in water so they stop photosynthesising and die [1] Bacteria feeding on dead plants multiply and their respiration uses up all the available dissolved oxygen [1] (iii) Al3+ + PO4 3–
AlPO4 [1] Moles of AlPO4 = 4.00/ecf from wrong formula or molar ratio [1] Concentration = 0.004 mol dm–3/ecf from wrong number of moles [1] [7] 4 (a) (i) Si2AlO9 7– [1] (ii) Cations can be adsorbed onto surface of clay/attraction between the negative clay and the cation [1] Plants need certain cations such as potassium and by attraction to clay they cannot be washed out of soil easily [1] [3] (b) (i) Any two from Hydrogen ions adsorbed onto surface of clay / attraction between hydrogen ion and negative clay [1] Some cations attached to clay can raise pH because cation is replaced by proton from water [1] Other cations such as aluminium can lower pH when replaced by hydrogen ion from water [1] (ii) If H+ is low RCO2H dissociates to produce H+/dissociation equilibrium moves to the right [1] If [H+] is high RCO2H forms RCO2H2 + [1] [4] (c) Any three from Liming involves adding calcium hydroxide or calcium carbonate to soil [1] OH– + H+
H2O/CO3 2– + H+
HCO3 –/more complex equations involving clay [1] Can reduce nitrogen content of the soil/NH4 + + OH–
NH3 + H2O [1] Efficiency of liming reduced by acid surges caused by melting of ice containing acidic water [1] [3]

Page 3 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9701 06 © University of Cambridge International Examinations 2006 Phase Equilibria 5 (a) (i) V.P. of A = vapour pressure of A on own x mol fr. of A [1] OR PA = PA xA (ii) 0.3 x 48 = 14.4 [1] 0.7 x 36 = 25.2 [1] 39.6 k Pa (iii) Raoult‘s law obeyed [1] components are similar/ideal mixture/components not interact [1] [5] (b) (i) Molecules attract each other OR dipoles align [1] Stronger intermolecular forces than components (ii) OR Interact in 1:1 ratio [1] [2] (c) pure propanone [1] since this has lowest b.p. OR highest VP OR is most volatile (allow discussion of b.p./composition curves) [1] Then azeotrope or 0.50 composition [1] [3]

Page 4 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9701 06 © University of Cambridge International Examinations 2006 6 (a) both axes [1] solid/liquid slope [1] areas [1] T.P. (1) values or label [1] shape [1] [5] (b) (i) line drawn at 298K or indicated [1] value of 60 atm indicated [1] [Explanation without ref to diagram only scores [1]] (ii) CO2 expands from over 60 atm to 1 atm cools [1] to below triplet point, explains solid [1] [4] (c) The solid/liquid line has a positive slope for CO2 rather than the negative slope of water [1] [1]

Page 5 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9701 06 © University of Cambridge International Examinations 2006 Spectroscopy 7 (a) Two absorptions [1] Asymmetric bend (or diagram) [1] Asymmetric stretch (or diagram) [1] [3] (b) (i) 1710 cm–1– C=O [1] 2260 cm–1 – C≡N [1] 3200 cm–1 – O–H [1] (ii) NC-CH2-CH2-CO2H [1] [4] (c) Nmr [1] + indication of absorptions (CH2 ~ 1.3 δ, -O-H ~ 4.5 δ) [2 x 1] OR Mass spectrometry [1] + two examples of likely fragmentations e.g. M-28 (loss of CN) and M-17 (loss of –OH) [2 x 1] [3] 8 (a) Structure II [1] A peak at 3450 cm–1 is characteristic of -OH would be seen for structure II [1] [2] (b) (i) Triplet-quartet is characteristic of a CH3 next to CH2 group [1] Standard 1,3,3,1 and 1,2,1 diagrams [1] (ii) Singlet (1) at δ 2.0 – 3.8 (1) [2] (iii) Deuterium oxide will exchange protons with -OH group in structure II [1] Since deuterium does not absorb in this part of the spectrum the –OH peak would disappear [1] [6] (c) Structure II will show (M-17)+ - loss of OH Structure I will show (M-31)+ loss of CH3O Structure II will show (M-43)+ loss of C3H7 any two [2]

Page 6 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9701 06 © University of Cambridge International Examinations 2006 Transitions Elements 9 (a) (i) somewhere between 4% and 20% chromium [1] (ii) Cr forms its oxide/Cr2O3 on the steel’s surface [1] which is impermeable to oxygen/hard [1] [3] (b) (i) Cr = 33.6/52 = 0.646 O = 20.6/16 = 1.288 Cl = 45.8/35.5 = 1.290 [1] thus A is CrO2Cl2 [1] O.N. of chromium = +6 [1] (ii) orange solution contains Cr2O7 2– [1] 2CrO2Cl2 + 3H2O → Cr2O7 2– + 6H+ + 4Cl– [1] white ppte is AgCl or Ag+ + Cl– → AgCl(s) [1] yellow solution contains CrO4 2– [1] Cr2O7 2– + 2OH– → 2CrO4 2– + H2O [1] [8 max 7] [7] 10 (a) colour dues to absorption of visible light [1] d-orbitals are split into two sets at different energies [1] photon is absorbed when an electron is promoted to higher orbital [1] [3] (b) (i) [Fe(SCN]2+ is formed - this is red [1] F– is a stronger ligand than SCN– or ligand exchange occurs [1] [FeF6]3– is colourless or energy gap between d-orbitals is large [1] (ii) reduction occurs [1] to VO2+ (which is blue) [1] SO2 + 4H+ + 2VO3 – → SO4 2– + 2VO2+ + 2H2O [1] (further reduction to) V3+ (which is green) [1] Sn2+ + 4H+ + 2VO2+ → Sn4+ + 2V3+ + 2H2O [1] [8 max 7] [7]