Cambridge A Level Chemistry 9701 — 2015 May/June Paper 4 · Variant 2

CHEMISTRY 9701/42 Paper 4 Structured Questions May/June 2015 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet Cambridge International Examinations Cambridge International Advanced Level This document consists of 17 printed pages and 3 blank pages. [Turn over IB15 06_9701_42/FP © UCLES 2015 *9412850145* READ THESE INSTRUCTIONS FIRST 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 an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fl uid. DO NOT WRITE IN ANY BARCODES. Section A Answer all questions. Section B Answer all questions. Electronic calculators may be used. You may lose marks if you do not show your working or if you do not use appropriate units. A Data Booklet is provided. 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. For Examiner’s Use 1 2 3 4 5 6 7 8 9 Total

2 9701/42/M/J/15 © UCLES 2015 Section A Answer all the questions in the spaces provided. 1 (a) Complete the electronic confi gurations of the following atoms. fl uorine: 1s2… sulfur: 1s2… [1] (b) (i) Write an equation to show the thermal decomposition of HCl. … [1] (ii) Using all relevant bond energy values from the Data Booklet, explain why the thermal stability of HF is much more than that of HCl. … … … [1] (c) Explain what is meant by the term electronegativity, and how it relates to the concept of bond polarity. … … … … [2]

3 9701/42/M/J/15 © UCLES 2015 [Turn over (d) Sulfur and fl uorine react together to give the covalent compound SF4. (i) Draw a ‘dot-and-cross’ diagram to show the bonding in SF4. Include all outer shell electrons in your diagram. [2] (ii) State whether a molecule of SF4 has a dipole moment. Explain your answer. … … [1] (e) Suggest a reason why sulfur can form both SF4 and SF6 whereas oxygen can only form OF2. … … [1] (f) (i) State a major source of atmospheric sulfur dioxide. … [1] (ii) State one environmental consequence of atmospheric sulfur dioxide. … [1] [Total: 11]

4 9701/42/M/J/15 © UCLES 2015 2 (a) A sample of lead consists of the following isotopes in the percentage abundances stated. isotope % abundance 204Pb 1.9 206Pb 24.8 207Pb 21.4 208Pb 51.9 Use these data to calculate the relative atomic mass of the sample of lead to two decimal places. Ar (Pb) = … [2] (b) Tin and lead both form oxides in oxidation states (II) and (IV). (i) How does the acid-base nature of tin(II) oxide compare to that of tin(IV) oxide? … [1] (ii) Illustrate your answer to (i) with equations, showing the reaction of each oxide with a suitable acid or base, as appropriate. SnO … SnO2 … [2] (iii) Describe the reactions, if any, that occur when separate samples of tin(IV) oxide and lead(IV) oxide are heated in air. Include any relevant observations and write equations for any reactions that occur. … … … … … … [3] [Total: 8]

5 9701/42/M/J/15 © UCLES 2015 [Turn over 3 (a) Complete the table with the symbol of the ion that contains the number of protons, electrons and neutrons stated in the following table. The fi rst line has been completed as an example. protons electrons neutrons symbol 3 2 4 7Li+ 15 16 18 [2] (b) Describe and explain the trend in the solubilities of the sulfates of the Group II elements down the group. … … … … … [4] (c) Calcium sulfate is sparingly soluble in water. Describe and explain what you would see when a few cm3 of concentrated Na2SO4(aq) were added to a saturated solution of CaSO4(aq). … … … … [2] (d) When a solution of a chromium salt X is electrolysed, chromium metal is deposited on the cathode, according to the following equation. Crn+(aq) + ne– → Cr(s) When a current of 1.8 A was passed for 40 minutes through a solution of salt X, it was found that 0.776 g of chromium had been deposited. Calculate the value of n in the above equation. Show your working. n = … [4] [Total: 12]

6 9701/42/M/J/15 © UCLES 2015 4 (a) (i) What is meant by the term buffer solution? … … … [2] (ii) Write equations to show how the hydrogencarbonate ion, HCO3 –, controls the pH of blood. … … [2] (iii) A solution containing both Na2HPO4 and NaH2PO4 is commonly used as a buffer solution. The following equilibrium is present in the solution. H2PO4 –(aq) HPO4 2–(aq) + H+(aq) Ka = 6.2 × 10–8 mol dm–3 Calculate the pH of a buffer solution made by mixing 100 cm3 of 0.5 mol dm–3 Na2HPO4 and 100 cm3 of 0.3 mol dm–3 NaH2PO4. pH = … [2] (b) Silver phosphate, Ag3PO4, is sparingly soluble in water. (i) Write an expression for the solubility product, Ksp, of Ag3PO4, and state its units. Ksp = units: … [1] (ii) The numerical value of Ksp is 1.25 × 10–20 at 298 K. Use this value to calculate [Ag+(aq)] in a saturated solution of Ag3PO4. [Ag+(aq)] = … mol dm–3 [3]

7 9701/42/M/J/15 © UCLES 2015 [Turn over (c) The half-equation for the redox reaction between phosphoric(III) acid and phosphoric(V) acid is shown. H3PO4(aq) + 2H+(aq) + 2e– H3PO3(aq) + H2O(l) E o = –0.28 V Find suitable data from the Data Booklet to write an equation for the reaction between H3PO3 and Fe3+(aq) ions, and calculate the for the reaction. equation: … = … V [2] [Total: 12]

8 9701/42/M/J/15 © UCLES 2015 5 (a) Compound B is a component of several perfumes and fl avourings. It can be obtained by the hydrogenation of compound A. During the reaction, the hydrogen atoms all add onto the same side of the benzene ring. A OH B OH + 6[H] (i) Suggest reagents and conditions for this reaction. … [1] (ii) Circle all the chiral atoms on the structure of B above. [1] (iii) How many possible optical isomers are there with the same structural formula as B? … [1] (iv) Complete the following part-structure to show the structure of one of the isomers of B that would be formed during the above reaction. [1] (b) Compound A can be obtained from propan-2-ylbenzene by the following route. propan-2-ylbenzene A step 1 step 2 step 4 heat in water step 3 C NO2 OH NH2 (i) Suggest the structure of the intermediate cation C and draw it in the box above. [1]

9 9701/42/M/J/15 © UCLES 2015 [Turn over (ii) Suggest reagents and conditions for the following steps. step 1 … step 2 … step 3 … [4] (c) Suggest the structures of the organic products of the reactions between each of the compounds A and B and the following reagents. If no reaction occurs write ‘no reaction’ in the relevant box. reagent OH product with A, OH product with B, HBr Na NaOH(aq) [5] [Total: 14]

10 9701/42/M/J/15 © UCLES 2015 6 (a) Carboxylic acids can be converted into primary amines by the following sequence of reactions. step 1 RCO2H RCONH2 RCH2NH2 step 2 step 3 D (i) Suggest the identity of intermediate D and write its structure in the box above. [1] (ii) Suggest the reagents for step 1 … step 2 … step 3 … [2] (b) Four compounds, E, F, G and H, are isomers of each other. Each compound contains an aromatic ring and two functional groups from the following list. • alcohol • amide • amine • carboxylic acid • ester • phenol (i) Which of these functional groups react readily with cold HCl (aq)? … [1] (ii) Which of these functional groups react readily with cold NaOH(aq)? … [1] The molecular formula of the four isomers, E, F, G and H, is C8H9NO2. All four compounds are insoluble in water. Table 1 shows their solubilities in acid or alkali. compound solubility in HCl (aq) solubility in NaOH(aq) E insoluble insoluble F soluble soluble G soluble insoluble H insoluble soluble Table 1

11 9701/42/M/J/15 © UCLES 2015 [Turn over (iii) Use this information to suggest the two functional groups, taken from the list on page 10, that each compound contains. compound fi rst functional group second functional group E F G H [4] (iv) Suggest a structure for each compound. E F G H [4] [Total: 13]

12 9701/42/M/J/15 © UCLES 2015 Section B Answer all the questions in the spaces provided. 7 This question is about the structures and roles of DNA and RNA in protein synthesis. (a) Study the structures of the three molecules below. One of the molecules could be a building block for a protein while the other two could be building blocks for other biological polymers. L K J OH O HO NH2 H OH H HO OH H O CH2OH H OH H OH HO H H OH H H H O Which of the three could be a building block for a protein? Explain your answer. … … [1] (b) Outline the different roles played by mRNA and tRNA in producing a protein with a specifi c primary structure. mRNA … … … tRNA … … … [4]

13 9701/42/M/J/15 © UCLES 2015 [Turn over (c) Sickle cell anaemia is a genetic-based disease in which one of the glutamic acid residues is replaced by a valine residue. OH HO NH2 O glutamic acid O OH NH2 valine O Suggest and explain how this change in the primary structure of the protein would affect the overall structure and function of the protein. … … … … … [3] [Total: 8]

14 9701/42/M/J/15 © UCLES 2015 8 (a) NMR spectroscopy and X-ray crystallography can both be used to examine the structure of organic compounds. NMR is very useful at examining hydrogen atoms in compounds, but hydrogen atoms are invisible to X-rays. (i) Explain why NMR spectroscopy can detect hydrogen atoms in molecules. … … [1] (ii) Explain why hydrogen atoms are invisible to X-rays. … … [1] (iii) The molecular formula of the amino acid cysteine is C3H7O2NS. Explain which of the atoms present would show the greatest absorption on exposure to X-rays. … … [1] (b) The NMR spectrum below was obtained from an organic liquid, P, which contains fi ve carbon atoms per molecule. 10 9 8 7 6 5 4 3 2 1 0 2 1 9 δ / ppm (i) How many protons are present in one molecule of P? Explain your answer. number of protons … … … [1]

15 9701/42/M/J/15 © UCLES 2015 [Turn over (ii) When a little D2O is added to P, the absorption at δ 2.0 disappears. Explain what this tells you about the group responsible for this absorption and why. … … … [2] (iii) What does the absorption at δ 0.9 tell you about the adjacent carbon atom? … … [1] (iv) What group(s) is/are responsible for the absorption at δ 0.9? … [1] (v) Suggest a structure for P. [1] (c) When an isomer of P is heated with concentrated H2SO4 it forms a new compound, Q. This new compound Q reacts with bromine to give a dibromide, R. (i) A mass spectrum was obtained of R. The ratio of the heights of the M : M+1 peaks was 9.3 : 0.5. Show that there are fi ve carbon atoms present in one molecule of R. [1] (ii) Predict the ratio of the heights of the M : M+2 : M+4 peaks as a result of the two bromine atoms in the dibromide R. Show your working. ratio … [1] (iii) What is the molecular formula of R? … [1] [Total: 12]

16 9701/42/M/J/15 © UCLES 2015 9 Polymers consist of monomers joined either by addition or condensation reactions. (a) Complete the table by placing a tick () in the correct column to indicate the type of reaction that would polymerise each of the monomers. monomer addition condensation both N C C H H O OH H H C C OH H H H C C CH3 H H H3C [3] (b) Poly(ethene) bags pollute the environment for a long time because they are non-biodegradable. Suggest why. … … … [2] (c) There has been considerable research into making biodegradable plastic bags. The repeat unit for one of the polymers used, polylactic acid (PLA), is shown. O O CH3 n (i) Draw the structure of the monomer for PLA. [1]

17 9701/42/M/J/15 © UCLES 2015 [Turn over (ii) Suggest why PLA breaks down more easily in the environment than poly(ethene). … … … [1] (d) The table shows the melting points of three polymers. polymer melting point / °C polyethene 137 polychloroethene (PVC) 212 nylon 6,6 265 Explain the differences in melting point of these three polymers in terms of the intermolecular forces between the chains. … … … … … … … … [3] [Total: 10]

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20 9701/42/M/J/15 © UCLES 2015 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Level MARK SCHEME for the May/June 2015 series 9701 CHEMISTRY 9701/42 Paper 4 (Structured Questions), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2015 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9701 42 © Cambridge International Examinations 2015 1 (a) fluorine: 1s22s22p5 [1] sulfur: 1s22s22p63s23p4 (b) (i) 2HCl → H2 + Cl2 [1] (ii) bond energies: HF (562) is stronger than HCl (431) [1] or F2 (158) is weaker than Cl2 (244) (c) electronegativity: [2] The attraction by an atom / nucleus / element of the electrons in a bond or a shared pair or a molecule bond polarity: ..is due to atoms / elements of different electronegativities at each end of a bond (d) (i) F S F F F (ii) Yes, it will have a dipole moment, [3] either because it has an uneven distribution of electrons or because it contains a lone pair or the S–F dipoles don't cancel or molecule is not symmetrical or diagram of see-saw shape. (allow an ecf for "no dipole" if their structure in (d)(i) has no lone pair) (e) Sulfur can use its d-orbitals or has low-lying / accessible / available d-orbitals or can [1] expand its octet. (allow reverse argument for oxygen; do NOT allow just "sulfur has d-orbitals") (f) (i) Burning of fossil fuels or coal / oil / petrol / natural gas (NOT methane or hydrocarbons) or volcanoes or roasting / burning sulfide ores (ii) Acid rain [2] [Total: 11]

Page 3 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9701 42 © Cambridge International Examinations 2015 2 (a) Ar = 204 × 0.019 + 206 × 0.248 + 207 [2] = 207.21 (correct ans = [2]) The last answer written by the candidate needs to be written with 2 d.p. to get the last mark. (b) (i) Tin(II) oxide is more basic than tin(IV) oxide [1] or tin(II) oxide is less acidic than tin (IV) oxide (ii) e.g. SnO + 2HCl → SnCl2 + H2O(or ionic or with H2SO4) [2] SnO2 + 2NaOH → Na2SnO3 + H2O (or ionic or with KOH etc.) (iii) SnO2 stays the same (white) or is stable or no reaction [3] PbO2 changes colour (from brown/black to yellow/orange/red) PbO2 → PbO + 2 1 O2 or 3PbO2 → Pb3O4 + O2 [Total: 8]

Page 4 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9701 42 © Cambridge International Examinations 2015 3 (a) 33P– [2] (b) Solubility decreases (from Mg to Ba or down the group) [4] Both lattice energy / ∆Hlatt and enthalpy change of hydration / ∆Hhyd are involved enthalpy change of hydration decreases more than lattice energy So enthalpy change of solution / ∆Hsol becomes more endothermic or more positive or less exothermic or less negative (NOT ∆Hsol decreases, or increases) (c) precipitate / solid CaSO4 would form [2] due to the common ion effect or Ksp is exceeded or the following equilibrium shifted over to the right Ca2+(aq) + SO42–(aq) ⇌ CaSO4(s) (d) charge passed = 1.8 × 40 x 60 (= 4320 C) [4] n(e–) = 4320 / 96500 (= 4.477 × 10–2 mol) ecf n(Cr) = 0.776 / 52 (= 1.492 × 10–2 mol) ecf n = 4.477 × 10–2 / 1.492 × 10–2 = 3.00 (=3) [Total: 12]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9701 42 © Cambridge International Examinations 2015 4 (a) (i) a solution that resists / minimises a change in its pH or helps maintain its pH… [2] (NOT any of: "maintains pH"; "keeps pH constant"; "no change in pH") …when small amounts of acid / H+ or base / OH– are added (both acid and base are needed) (ii) HCO3 – reacts with H+ ions as follows: [2] HCO3– + H+ → H2CO3 (or H2O + CO2) and with OH– ions thus: HCO3 – + OH– → CO3 2– + H2O (the equation arrows can be equilibrium arrows, as long as HCO3 – is on the left) (iii) (pKa = –log(Ka) = 7.21) [2] pH = pKa + log([base] / [acid] = 7.21 + log(0.5 / 0.3) = 7.43 (7.4) (b) (i) Ksp = [Ag+]3[PO4 3–] and units: mol4dm–12 [1] (ii) call [PO4 3–] = x, then [Ag+] = 3x, and Ksp = 27x4 [3] x = (Ksp / 27)1/4 = (1.25 × 10–20 / 27)1/4 = 4.64 × 10–6 mol dm–3 [Ag+] = 3x = 1.39 × 10–5 (mol dm–3) (allow 1.4 × 10–5) (c) H3PO3 + 2Fe3+ + H2O → H3PO4 + 2Fe2+ + 2H+ [2] Eocell = 0.77 –(–0.28) = (+)1.05 V or 3H3PO3 + 3H2O + 2Fe3+ → 3H3PO4 + 6H+ + 2Fe Eocell = –0.04 –(–0.28) = (+)0.24 V [Total: 12]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9701 42 © Cambridge International Examinations 2015 5 (a) (i) H2 + Pt or H2 + Ni / Pd + heat / warm or 50o < T < 500 oC [1] (ii) OH [1] (iii) 22 = 4 [1] (iv) CH(CH3)2 OH H H or H H OH CH(CH3)2 2 Hs have to be on the same side of the ring. Allow –C3H7 or –R for –CH(CH3)2 [1] (b) (i) [1] N N C or N2 or N N (ignore additional Cl or Cl-, but don't allow covalent -N=N-Cl) (ii) step 1: conc HNO3 + H2SO4 (@ 25 oC < T < 60 oC – see below) ("aq" negates) [4] step 2: Sn / Fe + HCl step 3: HNO2 or NaNO2 + HCl (@ T< 10 °C – see below) both temperatures correct for steps 1 + 3 (temperature not required for step 2) (inclusion of the word "heat" or "reflux" in step 3 negates the temperature mark) (c) [5] HBr no reaction Br Na ONa ONa NaOH(aq) ONa no reaction [Total: 14]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9701 42 © Cambridge International Examinations 2015 6 (a) There are three acceptable alternatives – follow each column down vertically: (i) D is RCOCl RCOOCH2CH3 RCO2 – NH4 + (ii) step 1 SOCl2 (or PCl3 or PCl5) ethanol (e.g.) + conc H2SO4 NH3 (ii) step 2 NH3 (NaOH negates this mark) heat (ii) step 3 LiAl H4 (aq) negates(NOT NaBH4; Sn + HCl etc.) (b) (i) amine (other groups negate) [1] (ii) phenol and carboxylic acid (both needed) [1] (iii) [4] compound first functional group second functional group E amide alcohol F amine carboxylic acid G amine ester H amide phenol (iv) Mark this in the following way. For each structure of E, F, G and H: [4] • check whether the structure fits the molecular formula C8H9NO2, i.e. that it has: one nitrogen, two oxygens and eight carbons. • check that it contains the two groups that the candidate's answers to part (ii) says it contains. [Total: 13]

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9701 42 © Cambridge International Examinations 2015 7 (a) L – it is the only compound that is an amino acid or can form (NOT contain) [1] –NH–CO– / amide / peptide linkages / bonds or it contains an N atom / NH2 group / CO2H group (b) mark both parts of this together – max [4] from the following six points [4] M1 mRNA is complementary to or a copy of (a portion of) DNA M2 mRNA encodes the sequence of amino acids in proteins or each of its codons (base triplets) codes for one amino acid M3 mRNA binds to / associates with the ribosome M4 tRNAs are specific to their amino acids M5 tRNA contains an anticodon or bonds to the codon / mRNA through base pairing or translates the RNA code into the amino acid sequence M6 tRNA carries the amino acid to the ribosome/mRNA (c) max [3] from the following six points. [3] M1 the pH of that area of the protein would change M2 protein becomes less hydrophilic / soluble or more hydrophobic M3 fewer hydrogen bonds or more van der Waals’ (id–id) forces M4 fewer ionic bonds form M5 the tertiary structure / folding / (3D) shape (of the protein) would change M6 the active site would be different / less efficient [Total: 8]

Page 9 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9701 42 © Cambridge International Examinations 2015 8 (a) (i) The nucleus / proton of a hydrogen atom has spin [1] (ii) Hydrogen doesn't have enough electrons / electron density [1] (iii) S / sulfur – it has the greatest number of electrons or highest electron density [1] (b) (i) 12 protons (=9+2+1) [1] (ii) The group responsible for this peak is –OH (allow NH) [2] The D in D2O exchanges with the H in –OH or H is replaced by D or "–OH → –OD", (iii) The adjacent carbon atom has no hydrogen atoms bonded to it [1] (iv) Methyl / CH3 group [1] (v) P is (CH3)3C–CH2OH [1] (c) (i) 50/10.23 9.3 1.1 0.5 100 M 1.1 1) (M 100 n = × × = × + × = [1] = 4.89 hence 5 carbons (ii) (Ratio of 79Br:81Br is 1 : 1), [1] hence ratio of M : M+2 : M+4 is 1 : 2 : 1 (iii) Molecular formula of R is C5H10Br2 [1] [Total: 12]

Page 10 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2015 9701 42 © Cambridge International Examinations 2015 9 (a) [3] monomer addition condensation both ✓ ✓ ✓ (b) polythene is non-polar or its bonds are non-polar [2] so not (easily) hydrolysed (c) (i) [1] OH O OH Cl O OH OH O Cl or NOT (Allow displayed, skeletal, part-skeletal, structural etc.) (ii) The ester (or –COO–) linkage / bond is hydrolysed or reacts with water [1] (d) Polythene has (weak) van der Waals’ (or id–id) forces [3] PVC has stronger van der Waals’ forces or additional dipole forces Nylon has (strong) hydrogen bonding [Total: 10]