Cambridge A Level Chemistry 9701 — 2016 Feb/March Paper 4 · Variant 2

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 fluid. DO NOT WRITE IN ANY BARCODES. 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. CHEMISTRY 9701/42 Paper 4 A Level Structured Questions February/March 2016 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level This document consists of 19 printed pages and 1 blank page. [Turn over IB16 03_9701_42/5RP © UCLES 2016 *0516596213*

2 9701/42/F/M/16 © UCLES 2016 Answer all the questions in the spaces provided. 1 (a) Complete the diagrams to show the energies of the electrons in a carbon atom, a C+ ion and a C– ion. increasing energy 1s carbon atom ↑↓ C+ ion ↑↓ C– ion ↑↓ 1s 1s [2] (b) One of the simple molecular allotropes of carbon is buckminsterfullerene, C60. buckminsterfullerene (i) What is the hybridisation of the carbon atoms in C60? … [1] (ii) C60 reacts with an excess of hydrogen to form a single product, C60Hx. Using your answer to (i), suggest a suitable value for x. … [1]

3 9701/42/F/M/16 © UCLES 2016 [Turn over (c) Methylbenzene can undergo different reactions to form the products shown below. CH2Cl Cl CH3 CH3 reaction 1 reaction 2 methylbenzene (i) Give the reagents and conditions for these two reactions. reaction 1 … reaction 2 … [2] (ii) Name the mechanism of reaction 1. … [1] (iii) Draw the structure of the product obtained if reaction 1 is carried out using an excess of chlorine. [1] [Total: 8]

4 9701/42/F/M/16 © UCLES 2016 2 (a) Calcium metal reacts with chlorine gas to form calcium chloride, CaCl 2. (i) Write an equation, including state symbols, to represent the lattice energy of calcium chloride, CaCl 2. … [1] (ii) Complete a fully labelled Born-Haber cycle that could be used to calculate the lattice energy, ,for calcium chloride. Ca(s) + Cl 2(g) CaCl 2(s) [2] (iii) Use your answer to (ii) and the following data, together with relevant data from the Data Booklet, to calculate a value for for calcium chloride. standard enthalpy change of formation of CaCl 2(s), –796 kJ mol–1 standard enthalpy change of atomisation of Ca(s), +178 kJ mol–1 electron affinity of chlorine atoms –349 kJ mol–1 = … kJ mol–1 [3] (b) Entropy is a measure of the disorder of a system. Describe and explain what happens to the entropy of a gas when the temperature is increased. … … … [2]

5 9701/42/F/M/16 © UCLES 2016 [Turn over (c) The table shows four reactions. (i) For each reaction, predict the sign of the entropy change, ∆S o. If you predict no entropy change, write ‘no change’ in the table below. The first one has been done for you. reaction sign of ∆S o CO(g) + O2(g) → CO2(g) negative Mg(s) + 2 1 O2(g) → MgO(s) CuSO4(s) + 5H2O(l) → CuSO4.5H5O(s) NaHCO3(s) + H+(aq) → Na+(aq) + CO2(g) + H2O(l) [2] (ii) Explain why the entropy change for the first process is negative. … … [1] (d) Calculate the standard entropy change, ∆S o, for this reaction. N2(g) + 3H2(g) → 2NH3(g) Standard entropies, S o, in J K–1 mol–1 are given. N2(g) H2(g) NH3(g) +192 +131 +193 ∆S o … J K–1 mol–1 [2] (e) Whether or not a chemical reaction is spontaneous (feasible) can be deduced by calculating the change in free energy, ∆G o, at a given temperature. MgCO3(s) → MgO(s) + CO2(g) ∆H o = +117 kJ mol–1 ∆S o = +175 J K–1 mol–1 (i) Calculate the value of ∆G o at 298 K for the above reaction. [2] (ii) Use your answer to (i) to explain whether or not this reaction is spontaneous at 298 K. … … [1] [Total: 16]

6 9701/42/F/M/16 © UCLES 2016 3 (a) Complete the electronic structures of the following species. Co [Ar] … Co2+ [Ar] … [2] (b) In an isolated transition metal atom the five d orbitals have the same energy. When a transition metal ion forms a tetrahedral complex the d orbitals are split into two groups of different energies. Complete an orbital energy diagram to show this, indicating the number of orbitals in each group. increasing energy isolated transition metal ion tetrahedral complex [1] (c) Cobalt(II) forms a six co-ordinate complex containing three water molecules and three chloride ions. (i) Write the formula of this complex showing the overall charge, if appropriate. … [1] (ii) Explain, with the aid of diagrams, how many isomers of the complex in (i) exist. [2] (d) Platinum(II) forms a four co-ordinate complex containing two ammonia molecules and two chloride ions. (i) Write the formula of this complex showing the overall charge, if appropriate. … [1]

7 9701/42/F/M/16 © UCLES 2016 [Turn over (ii) This complex exists as two isomers. Draw the structure of these isomers and give their names. … … [3] (iii) One of the isomers in (ii) is an important anticancer drug. State which isomer this is and explain why this isomer is effective. isomer … reason … … … … [2] (e) Transition metal ions often exist as hexa-aqua complexes in aqueous solution. The reactions which involve ligand exchange are reversible. [Cu(H2O)6]2+ + 4NH3 [Cu(NH3)4(H2O)2]2+ + 4H2O (i) Write an expression for the stability constant, Kstab, for this equilibrium. Give its units. Kstab = units = … [2] (ii) The numerical value for Kstab for this equilibrium at 298 K is 1.20 × 1013. Explain how this value relates to the relative stabilities of the two complexes. … … [1] [Total: 15]

8 9701/42/F/M/16 © UCLES 2016 4 (a) Ethanal, CH3CHO, dimerises in alkaline solution according to the following equation. 2CH3CHO → CH3CH(OH)CH2CHO The initial rate of this reaction was measured, starting with different concentrations of CH3CHO and OH–. The following results were obtained. [CH3CHO] / mol dm–3 [OH–] / mol dm–3 initial rate of reaction (relative values) 0.10 0.015 1 0.20 0.015 2 0.40 0.030 8 (i) Deduce the order of the reaction with respect to CH3CHO. … [1] (ii) Deduce the order of the reaction with respect to OH–. … [1] (iii) State the overall rate equation for this reaction. rate = … [1] (iv) State the units for the rate constant, k. … [1] (v) Calculate the initial rate of reaction (relative value) for a reaction where the [CH3CHO] is 0.30 mol dm–3 and [OH–] is 0.030 mol dm–3. [1]

9 9701/42/F/M/16 © UCLES 2016 [Turn over (b) (i) A three-step mechanism has been proposed for the reaction in (a). step 1 CH3 C + + H O H2O C H O –CH2 –CH2 step 2 CH3 C + + + H O C H O CH3 CH2 C C H H O CH3 CH2 H2O C C H H O OH step 3 CH3 CH2 C C H H O –OH –OH –O –O Using your rate equation in (iii), predict which is the rate-determining step. Explain your answer. rate-determining step … explanation … … [2] (ii) Describe the chemical behaviour of CH3CHO in step 1. … [1] (c) Name the mechanism occurring in steps 2 and 3. … [1] (d) Using the diagram below, show the mechanism for step 2 showing the relevant curly arrows and dipoles. –CH2 C H O CH3 C H O [2] [Total: 11]

10 9701/42/F/M/16 © UCLES 2016 5 Some reactions of chromium ions are shown below. [Cr(H2O)6]2+(aq) [Cr(H2O)6]3+(aq) Cr2O7 2–(aq) CrO4 2–(aq) metal + H2SO4 reaction 1 reaction 2 reaction 3 OH–(aq) H3O+(aq) (a) (i) Use the Data Booklet to suggest a suitable metal to carry out reaction 1. … [1] (ii) Use E o values to explain your answer to (i) by calculating the . … … … [2] (b) A student suggested that reaction 2 could be carried out using acidified hydrogen peroxide solution. Use the Data Booklet to show whether or not this reaction is feasible. … … … [2] (c) Explain using oxidation numbers whether or not reaction 3 is a redox reaction. … … … [2]

11 9701/42/F/M/16 © UCLES 2016 [Turn over (d) The student used an acidified solution of Cr2O7 2–(aq) to electroplate a steel box with chromium metal. Calculate how long it would take for a current of 0.125 A to deposit 0.0312 g of chromium metal. time = … [3] [Total: 10]

12 9701/42/F/M/16 © UCLES 2016 6 Two elements, V and W, are in adjacent groups in the Periodic Table. V reacts with oxygen to form an acidic gas, X. V forms an anion with formula VOm –. W reacts with oxygen to form an acidic gas, Y. W forms an anion with formula WOn 2–. A solution of WOn 2– forms a white precipitate with Ba2+(aq) but shows no visible reaction with Mg2+(aq). (a) Complete the table below. identity or value V X m W Y n [3] (b) By referring to enthalpy changes, explain why WOn 2– forms a white precipitate with Ba2+(aq) but shows no visible reaction with Mg2+. … … … [3] [Total: 6]

13 9701/42/F/M/16 © UCLES 2016 [Turn over Question 7 starts on the next page.

14 9701/42/F/M/16 © UCLES 2016 7 (a) (i) State and explain the relative acidities of ethanol and phenol. … … … … [2] (ii) In the table below, give the reaction conditions for the formation of organic products D and E and draw their structures. OH phenol HNO3 Br2 D E reagent conditions structure HNO3 dilute, 5 °C D Br2 E [3] (iii) Name the mechanism of the reaction forming compound E. … [1] (b) (i) Phenylamine reacts with aqueous bromine to give compound F. Describe the appearance of compound F. … [1] (ii) Phenylamine reacts with nitrous acid to form a diazonium salt. State the conditions for this reaction. … [1]

15 9701/42/F/M/16 © UCLES 2016 [Turn over (iii) The diazonium salt in (ii) reacts with an alkaline solution of phenol to produce a coloured compound, G. Draw the structure of G. [2] (c) Compound H is a primary amine which has three peaks in its carbon-13 NMR spectrum. H3C CH3 CH2 NH2 C H CH2 compound H (i) An isomer of H is another primary amine J which also has three peaks in its carbon-13 NMR spectrum. Use this information to suggest the structure of J. [1] (ii) Another isomer of H is the tertiary amine K. It has three peaks in its proton NMR spectrum. One of the peaks is a doublet. CH3 H3C CH3 CH3 N CH K Circle the protons responsible for the doublet. [1] [Total: 12]

16 9701/42/F/M/16 © UCLES 2016 8 Acebutolol is a drug that can be used to lower blood pressure. O O OH O N H H N CH3 CH3 CH3 H3C P Q R acebutolol (a) Give the full name of the circled functional groups labelled P, Q and R in acebutolol. P … Q … R … [3] (b) On the diagram of acebutolol below, draw a circle around any chiral carbon atoms. O O OH O N H H N CH3 CH3 CH3 H3C [1] (c) On warming with dilute hydrochloric acid, acebutolol splits to form two molecules. (i) Draw a line through the bond broken by heating with dilute hydrochloric acid. [1] (ii) Draw the structure of the smaller molecule produced by this reaction. [1]

17 9701/42/F/M/16 © UCLES 2016 [Turn over (d) Suggest what would be observed when acebutolol reacts with the following reagents. If no reaction would take place, write ‘none’ in the table below. reagent observation alkaline iodine solution universal indicator solution 2,4-dinitrophenylhydrazine Tollens’ reagent [3] (e) Butanoic acid can be reduced to form compound N. Compound N reacts with sodium. (i) Suggest a suitable reducing agent for this reaction. … [1] (ii) Draw the skeletal formula of the isomer of N that exists as a pair of optical isomers. [1] (iii) Another isomer of N does not react with acidified dichromate(VI) solution but does react with sodium. Draw the structure of this isomer. [1] [Total: 12]

18 9701/42/F/M/16 © UCLES 2016 9 (a) (i) Name an example of a synthetic polyester and a synthetic polyamide. polyester … polyamide … [1] (ii) Polyesters and polyamides are formed by condensation reactions. Name a molecule which is commonly eliminated in such reactions. … [1] (b) (i) The table shows the repeat units of a number of polymers. Place a tick () against the ones which are biodegradable. polymer repeat unit biodegradable A O O n CH3 B O O O O n C O CH3O n D N H O n O N H [2] (ii) Draw the structures of two monomers used to form polymer B. [2]

19 9701/42/F/M/16 © UCLES 2016 [Turn over (c) A section of polypeptide was hydrolysed and the following amino acids identified. amino acid formula T CH3CH(NH2)CO2H U C6H5CH2CH(NH2)CO2H V H2N(CH2)4CH(NH2)CO2H (i) Which of the amino acids T, U or V has the highest pH in aqueous solution? Explain why. amino acid … … … [1] (ii) State how many different dipeptides could be formed from a reaction mixture consisting of amino acids T and U. … [1] (iii) Polypeptides contain a high proportion of carbon and hydrogen in their structures, yet many are soluble in water. By referring to the structure of a polypeptide, explain why. … … … … [2] [Total: 10]

20 9701/42/F/M/16 © UCLES 2016 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 Subsidiary and Advanced Level MARK SCHEME for the March 2016 series 9701 CHEMISTRY 9701/42 Paper 4 (A Level 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 March 2016 series for most Cambridge IGCSE® and Cambridge International A and AS Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – March 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Mark 1 (a) 2 (b) (i) sp2 1 (ii) x = 60 / C60H60 1 (c) (i) reaction 1: Cl 2 and UV light; reaction 2: Al Cl 3, Cl 2 (NOT aqueous); 1 1 (ii) (free) radical substitution 1 (iii) CCl3 Cl Cl Cl Cl Cl CCl3 or 1

Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – March 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Mark 2 (a) (i) Ca2+(g) + 2Cl –(g) → CaCl 2(s) (state symbols required) 1 (ii) 2 (iii) ∆Hlatt o = –796 – 242 – 178 – 590 – 1150 + (2 × 349) = – 2258 kJ mol–1 3 (b) (higher temperature means that) particles have more energy; entropy (of the gas / system) increases because of an increase in the amount of disorder / randomness; 2 (c) (i) reaction sign of ∆So CO(g) + O2(g) → CO2(g) negative Mg(s) + ½O2(g) → MgO(s) negative CuSO4(s) + 5H2O(l) → CuSO4.5HsO(s) negative NaHCO3(s) + H+(aq) → Na+(aq) + CO2(g) + H2O(l) positive 2 (ii) there is a reduction in the overall number of gaseous molecules 1 (d) ∆Sf o = 386 – (192 + (3 × 131)) = –199 (J K–1 mol–1) 2 (e) (i) ∆Go = ∆Ho – T∆So = 117 – ((298 × 175) / 1000) = (+) 64.85 (kJ mol–1) 2 (ii) ∆Go is positive and so the reaction is not spontaneous (at 298 K) 1 Ca2+(g) + 2Cl (g) (+ 2e–) 2nd I.E of Ca 1st I.E of Ca EA of Cl × 2 Atomisation/∆Hat of Ca E(Cl-Cl)/2∆Hat of Cl ∆Hlatt o ∆Hf o CaCl2(s)

P Q 3 Pag Qu ge es (a) (b) (c) (d) (e) 4 tio ) ) ) ( (i ) ( (i (ii ) ( (i on (i) ii) (i) ii) ii) (i) ii) Co Co [C Cl C [P M1 M3 (ci thi wh M1 M2 Ks (la En C o o2+ Co(C l Cl t(C 1, M 3: n c is i is c hic 1: f 2: u stab arge ner Cam Cl ) C H Cl )2 M2 nam cis so can h p for uni = e v rgy mb [A [A )3(H Co Cl H2O 2(N 2: d me -di me n re pre mu its [C [C valu y brid r] 3 r] 3 H2O l O NH3 diag es am er) eac eve ula (ec Cu Cu ue dge 3d7 3d7 O)3 O O 3)2] gra mm ct / nts cf f (N (H of ― e In © 74s 7 3]– OH OH am ci ine bo s re fro H3) 2O Ks ― ― nte Ca s2 H2 H2 s is-p edic ond epl m f )4(H O)6 2 stab ― ― is ern amb pla chl d / b ica for H2O +][N sh ― ― sol ; M nati brid atin loro bind atio rmu O)2 NH how ― late ; Mar ion dge H n / op d w on o ula 2 2+] H3]4 ws ― ed k S nal e In H2O Cl lati with of t ) ] 4 tha ― io Sch AS nter O l inu h D the mo at) on hem S/A rnat C um( DNA e st ol–4 the me A L tion A Co Cl Cl (II) A; tra 4 dm e te e Lev nal Ans l ) nd m12 etra vel Ex swe O O / p 2 am te – xam er OH2 OH2 tra rev mm ― etr Ma mina 2 2 ans ven ine ― ― ― rah arc atio s-d nts e co ― ― ― ― hed ch ons diam ce om ― ― dra 20 s 20 t mm ell d mple al c 16 016 tra min div ex com 6 6 ns- ned visio is mp -pl dich on mo plex atin hlo ; ore x S n / rop e st Syl 9 pla tab llab 970 atin ble bu 01 um s m(II P I) Pa 4 pe 42 Ma er ark 1 1 1 1 2 1 2 1 1 1 1 1 1 k 1 1 1 1 2 1 2 1 1 1 1 1 1

P Q 4 Pag Qu ge es (a) (b) (c) (d) 5 tio ) ( (i (ii (iv (v ) ( (i ) ) on (i) ii) ii) v) v) (i) ii) 1st 1st rat mo ca rat ex ac nu M1 M2 C t or t or te ol–1 alcu te- xpla cid / ucle 1: b 2: d Cam rde rde = 1 dm ulat de ana / pr eop bot dip mb er er k[ m3 tio ter atio roto phi th c pole brid CH s–1 n f rmi on: on lic cur e c δ+ dge H3C 1 (o rom nin bo do ad rly corr + e In © CHO or p m c ng oth ono ddit arr rec δ nte Ca O][ per can ste h re or / tion row ctly δ– ern amb [OH r an ndi ep: eac ac n ws y sh M nati brid H–] ny da st ctan idic how Mar ion dge ] su te’s ep nt s c b wn k S nal e In uita s a 1 spe beh Sch AS nter able ans eci hav hem S/A rnat e ti swe es viou me A L tion A me er t ar ur e Lev nal Ans e u to ( re i vel Ex swe nit (iii) n s – xam er ) ) (e ste Ma mina exp p 1 arc atio pec 1 / r ch ons cte rate 20 s 20 d a e-d 16 016 ans dete 6 6 swe erm er min = 6 ning S 6) g s Syl 9 step llab 970 p bu 01 s Pa 4 pe 42 Ma er ark 1 1 1 1 1 1 1 1 1 1 1 k 1 1 1 1 1 1 1 1 1 1 1

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – March 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Mark 5 (a) (i) any metal with an Eo value more negative than –0.41 V, e.g. Fe, Mn, Zn, Mg, Cr, Al R: Li / Na / K / Ca / Ba 1 (ii) M1: value of Ecell correctly calculated (with correct sign) for metal named in (i) M2: Eo cell is positive and so reaction is feasible 1 1 (b) M1: (Cr2O7 2– + 14H+ + 6e– ⇌ 2Cr3+ + 7H2O) Eo = +1.33 V (H2O2 + 2H+ + 2e– ⇌ 2H2O) Eo = +1.77 V Eo cell = 0.44 (V) M2: Eo cell (0.44 V) is positive (so the reaction is feasible) / Eo(Cr2O7 2– / Cr3+) is less positive than Eo(H2O2 / H2O) 1 1 (c) M1: Cr2O7 2–: ox.no Cr = +6 because –2 = 2 × ox.no(Cr) + (7 × –2) CrO4 2–: ox.no Cr = +6 because –2 = ox.no(Cr) + (4 × –2) M2: no change in oxidation number, so reaction is not redox 1 1 (d) M1: no. moles Cr deposited = 0.0312 / 52 = 6.0 × 10–4 moles M2: deduction that 6 moles of e– needed per mole of Cr / reaction is Cr2O7 2- + 14H+ + 12e– → 2Cr + 7H2O M3: no. moles of e– = 6 × 6.0 × 10–4 = (0.125 × t) / 96 500 so t = (6 × 6.0 × 10–4 × 96 500) / (0.125 × 60) = 46.3 min / 0.772 h / 2780 s 1 1 1

Page 7 Mark Scheme Syllabus Paper Cambridge International AS/A Level – March 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Mark 6 (a) identity or value V nitrogen or chlorine X NO / NO2 Cl O2 / Cl O3 m 2, 3 1,2,3, or 4 W sulfur Y SO2 or SO3 n 4, 3 3 (b) M1: (white precipitate is BaSO4) descending the group ∆Hsol becomes more endothermic / positive; M2, M3 any two from: ∆Hlatt decreases / becomes more endothermic / becomes less exothermic ∆Hhyd decreases / becomes more endothermic / becomes less exothermic ∆Hhyd decreases more than ∆Hlatt 1 2

P Q 7 Pag Qu ge es (a) (b) (c) 8 tio ) ( (i (ii ) ( (i (ii ) ( (i on (i) ii) ii) (i) ii) ii) (i) ii) M1 we gro M2 rin ele wh be M1 M2 CH CH C eit C 1: p eak ou 2: p ng ect hite etw 1: d 2: r H3 H3 CH3 the Cam phe ken p p o trop e p wee dou res 3 er o mb eno ned orb phi rec en 0 ub st o CH C CH N one brid ol i d / t ita rea H ilic cip 0° C le b of m H3 C H2N e o dge is m the l / lo age NO Br2 su itat C a bon mo N CH NH2 r b e In © mo e ph one ent O3 2 ubs te / and nd lec N N H3 2 CH oth nte Ca ore hen e p t stitu / so d 1 be cule N H h C ern amb e ac nox pair utio olid 0 ° etw e CH C CH3 M nati brid cid xid r of on d C wee H3 CH3 3 g Mar ion dge dic e a f e ( en n rou k S nal e In tha ani lec d (I: t nitr ups Sch AS nter an on ctro co dilu aq tem rog s c hem S/A rnat et is ons nd ute que mp gen O ircl me A L tion A ha sta s o itio e, 5 eou era n a OH led e Lev nal Ans no ab n O ons 5 °C us atu tom H d vel Ex swe l be ilis O c s C ure) ms – xam er eca ed can ) Ma mina aus / e n be arc atio se etha e d ch ons the ano delo 20 s 20 e O ol h oca 16 016 O–H has alis S 6 6 H b s a sed Str bon an e d o uct nd ele ve tur S in ctr r / o re Syl 9 ph ron ove llab 970 hen do erla bu 01 nol ona aps s is atin s w P ng with Pa 4 h pe 42 Ma er ark 1 1 3 1 1 1 1 1 1 1 k 1 1 3 1 1 1 1 1 1 1

P Q 8 Pag Qu ge es (a) (b) (c) (d) (e) 9 tio ) ) ) ( (i ) ) ( (i (ii on (i) ii) (i) ii) ii) P Q R Q se LiA (m C CH C a k s = c ee l a 2,4 Al H mus CH H3 Cam am ket sec car ine alk u 4-d H4 st b H3 mb mide ton co rbo e o kali uni din To be s C CH brid e ne nd ony on d ne ive itro olle ske H3 dge dary yl a dia rea e io ersa oph ens ele O e In © y a and gra age din al i hen s’ r etal OH nte Ca alco d R am ent ne nd nyl rea ) ern amb oho R = m in t so ica hyd ge M nati brid ol alc n (b luti ator dra ent Mar ion dge coh b) ion r azin k S nal e In hol n ne Sch AS nter sc hem S/A rnat core me A L tion A es bl ye e Lev nal Ans [1] lue ello vel Ex swe ] ye e / p ow – xam er ob llow pur w / o n Ma mina bse w p ple ran o r arc atio erv ppt e co nge rea ch ons vati t. fo olo e p acti 20 s 20 ion orm our ppt on 16 016 n med fo for 6 6 d rm rm ed ed Syl 9 llab 970 bu 01 s Pa 4 pe 42 Ma er ark 1 1 1 1 1 1 3 1 1 1 k 1 1 1 1 1 1 3 1 1 1

Page 10 Mark Scheme Syllabus Paper Cambridge International AS/A Level – March 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Mark 9 (a) (i) polyester : Terylene / polylactic acid (PLA) / polyamide : nylon / Kevlar / Nomex 1 (ii) water or hydrochloric acid / hydrogen chloride 1 (b) (i) polymer biodegradable A yes B yes C no D yes 2 (ii) HOCH2CH2OH and or equivalent 1,4-diacyl chloride or equivalent 1,4-diester 2 (c) (i) V: it has two amine / NH2 groups (which can be protonated) or it has an amine / NH2 group on its side chain / R group 1 (ii) four (TT, TU, UT, UU) 1 (iii) hydrogen bonds; between the O / N atoms or named group (in the polypeptide) and water; or ion-dipole attractions; between NH3 + / CO2 – and water; 2