Cambridge A Level Chemistry 9701 — 2011 Oct/Nov Paper 2 · Variant 3

This document consists of 11 printed pages and 1 blank page. DC (CB (SE)) 32712/5 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level * 8 4 6 9 9 0 6 1 6 1 * CHEMISTRY 9701/23 Paper 2 Structured Questions AS Core October/November 2011 1 hour 15 minutes Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, Centre number and candidate number 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. DO NOT WRITE ON ANY BARCODES. Answer all questions. You may lose marks if you do not show your working or if you do not use appropriate units. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together. For Examiner’s Use 1 2 3 4 5 Total

2 9701/23/O/N/11 © UCLES 2011 For Examiner’s Use Answer all the questions in the space provided. 1 Sulfur, S, and polonium, Po, are both elements in Group VI of the Periodic Table. Sulfur has three isotopes. (a) Explain the meaning of the term isotope. … … …[2] (b) A sample of sulfur has the following isotopic composition by mass. isotope mass 32 33 34 % by mass 95.00 0.77 4.23 Calculate the relative atomic mass, Ar, of sulfur to two decimal places. Ar = … [2] (c) Isotopes of polonium, proton number 84, are produced by the radioactive decay of several elements including thorium, Th, proton number 90. The isotope 213Po is produced from the thorium isotope 232Th. Complete the table below to show the atomic structures of the isotopes 213Po and 232Th. number of isotope protons neutrons electrons 213Po 232Th [3]

3 9701/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use Radiochemical reactions, such as nuclear fission and radioactive decay of isotopes, can be represented by equations in which the nucleon (mass) numbers must balance and the proton numbers must also balance. For example, the nuclear fission of uranium-235, 235 92U, by collision with a neutron, 1 0n, produces strontium-90, xenon-143 and three neutrons. 235 92U + 1 0n 90 38Sr + 143 54Xe + 3 1 0n In this equation, the nucleon (mass) numbers balance because: 235 + 1 = 90 + 143 + (3x1). The proton numbers also balance because: 92 + 0 = 38 + 54 + (3x0). (d) In the first stage of the radioactive decay of 232 90 Th, the products are an isotope of element E and an alpha-particle, 4 2 He. (i) By considering nucleon and proton numbers only, construct a balanced equation for the formation of the isotope of E in this reaction. 232 90Th … + 4 2 He Show clearly the nucleon number and proton number of the isotope of E. nucleon number of the isotope of E … proton number of the isotope of E … (ii) Hence state the symbol of the element E. … …[3] [Total: 10]

4 9701/23/O/N/11 © UCLES 2011 For Examiner’s Use 2 When 0.42 g of a gaseous hydrocarbon A is slowly passed over a large quantity of heated copper(II) oxide, CuO, A is completely oxidised. The products are collected and it is found that 1.32 g of CO2 and 0.54 g of H2O are formed. Copper is the only other product of the reaction. (a) (i) Calculate the mass of carbon present in 1.32 g of CO2. Use this value to calculate the amount, in moles, of carbon atoms present in 0.42 g of A. (ii) Calculate the mass of hydrogen present in 0.54 g of H2O. Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.42 g of A. (iii) It is thought that A is an alkene rather than an alkane. Use your answers to (i) and (ii) to deduce whether this is correct. Explain your answer. … … [5]

5 9701/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (b) Analysis of another organic compound, B, gave the following composition by mass: C, 64.86%; H, 13.50%, O, 21.64%. (i) Use these values to calculate the empirical formula of B. (ii) The empirical and molecular formulae of B are the same. B is found to be chiral. Draw displayed formulae of the two optical isomers of this compound, indicating with an asterisk (*) the chiral carbon atom. (iii) There are three other structural isomers of B which are not chiral but which contain the same functional group as B. In the boxes below, draw the structural formulae of these isomers. [7] [Total: 12]

6 9701/23/O/N/11 © UCLES 2011 For Examiner’s Use 3 The Periodic Table we currently use is derived directly from that proposed in 1869 by Mendeleev who had noticed patterns in the physical and chemical properties of the elements he had studied. The diagram below shows the first ionisation energies of the first 18 elements of the Periodic Table. 2500 2000 1500 first ionisation energy / kJ mol–1 1000 500 0 0 1 2 3 4 5 6 7 8 9 proton number 10 11 12 13 14 15 16 17 18 H He Ne Na Ar Li (a) Give the equation, including state symbols, for the first ionisation energy of carbon. … [2] (b) (i) Explain why sodium has a lower first ionisation energy than magnesium. … … (ii) Explain why magnesium has a higher first ionisation energy than aluminium. … … (iii) Explain why helium, He, and neon, Ne, occupy the two highest positions on the diagram. … … (iv) Explain why the first ionisation energy of argon, Ar, is lower than that of neon, which is lower than that of helium. … … … [8]

7 9701/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (c) (i) The first ionisation energies of the elements Na to Ar show a variation. Some physical properties show similar variations. The atomic radius of the elements decreases from Na to Cl. Give a brief explanation of this variation. … … (ii) The cations formed by the elements Na to Al are smaller than the corresponding atoms. Give a brief explanation of this change. … … [3] (d) The oxides of the elements of the third Period behave differently with NaOH(aq) and HCl (aq). In some cases, no reaction occurs. Complete the table below by writing a balanced equation for any reaction that occurs, with heating if necessary. If you think no reaction takes place write ‘no reaction’. You do not need to include state symbols in your answers. …MgO(s) + … NaOH (aq) …MgO(s) + … HCl (aq) …Al2O3(s) + … NaOH (aq) + …H2O (l) …Al2O3(s) + … HCl (aq) …SO2(g) + … NaOH (aq) …SO2(g) + … HCl (aq) [6] [Total: 19]

8 9701/23/O/N/11 © UCLES 2011 For Examiner’s Use 4 The structural formulae of six different compounds, P – U, are given below. CH3CH=CHCH2CH3 CH3CH2COCH2CH3 CH2=CHCH2CH2CH3 P Q R CH3CH2CH2CH2CH2OH HOCH2CH2CH(OH)CH3 CH3CH2CH2OCH2CH3 S T U (a) (i) What is the empirical formula of compound T? … … (ii) Draw the skeletal formula of compound S. [2] (b) (i) Compounds S and U are isomers. What type of isomerism do they show? … … (ii) Two of the six formulae P – U can each be drawn in two forms which are known as stereoisomers. Which two compounds have formulae that can be drawn in two forms? What type of stereoisomerism does each show? Identify each compound by its letter. compound type of stereoisomerism [3]

9 9701/23/O/N/11 © UCLES 2011 [Turn over For Examiner’s Use (c) Compound S can be converted into compound R. (i) What type of reaction is this? … … (ii) What reagent would you use for this reaction? … … (iii) Write the structural formula of the compound formed when T undergoes the same reaction using an excess of the reagent you have used in (c)(ii). … …[3] (d) Compound P may be converted into compound Q in a two-step reaction. CH3CH=CHCH2CH3 step 1 intermediate step 2 CH3CH2COCH2CH3 P Q (i) What is the structural formula of the intermediate compound formed in this sequence? (ii) Outline how step 1 may be carried out to give this intermediate compound. … … … (iii) What reagent would be used for step 2? … …[4] [Total: 12]

10 9701/23/O/N/11 © UCLES 2011 For Examiner’s Use 5 Each of the three organic compounds, V, W, and X, has the empirical formula CH2O. The number of carbon atoms in each of their molecules is shown in the table. compound number of C atoms V 1 W 2 X 3 V gives a brick red precipitate when warmed with Fehling’s reagent; W and X do not. W is a fruity smelling liquid. In X, the carbon atoms are bonded directly to one another. X gives an effervescence when shaken with Na2CO3(aq); V and W do not. (a) Give the structural formula of V. [1] (b) (i) What functional group is present in W? … … (ii) Give the structural formula of W. [2] (c) When X is heated under reflux with acidified K2Cr2O7, the product, Y, gives no reaction with 2,4-dinitrophenylhydrazine reagent. (i) Give the structural formula of X. (ii) Give the structural formula of Y, the compound formed from X. [2]

11 9701/23/O/N/11 © UCLES 2011 For Examiner’s Use (d) When X is warmed with a little concentrated sulfuric acid, a small amount of a cyclic compound, Z, is formed. Z has the molecular formula C6H8O4. (i) Suggest a displayed formula for Z. (ii) What type of reaction occurs when Z is formed from X? … …[2] [Total: 7]

12 9701/23/O/N/11 © UCLES 2011 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. University of 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. BLANK PAGE

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2011 question paper for the guidance of teachers 9701 CHEMISTRY 9701/23 Paper 2 (AS Structured Questions), maximum raw mark 60 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 must be read in conjunction with the question papers and the report on the examination. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the October/November 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9701 23 © University of Cambridge International Examinations 2011 1 (a) same proton number/atomic number (1) different mass number/nucleon number (1) [2] (b) Ar = (32 × 95.00) + (33 × 0.77) + (34 × 4.23) (1) 100 = 3040 + 25.41 + 143.82 = 3209.23 100 100 which gives Ar = 32.09 (1) [2] (c) number of isotopes protons neutrons electrons 213Po 84 129 84 232Th 90 142 90 allow one mark for each correct column if there are no ‘column’ marks, allow maximum one mark for a correct row (3 × 1) [3] (d) (i) nucleon no. is 228 (1) proton no. is 88 (1) (ii) Ra not radium (1) [3] [Total: 10] 2 (a) (i) mass of C = 12 × 1.32 = 0.36g (1) 44 n(C) = 0.36 = 0.03 (1) 12 (ii) mass of H = 2 × 0.54 = 0.06 g (1) 18 n(H) = 0.06 = 0.06 (1) 1 (iii) yes because 0.03 mol of C are combined with 0.06 mol of H or C : H ratio is 1 : 2 or empirical formula is CH2 (1) [5]

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9701 23 © University of Cambridge International Examinations 2011 (b) (i) C : H : O = 64.86 : 13.50 : 21.64 (1) 12 1 16 = 5.41: 13.50 : 1.35 = 4 : 10 : 1 gives C4H10O (1) (ii) correct compound and correct chiral C* (1) correct mirror object/ mirror image relationship in 3D (1) (iii) (1) (1) (1) [7] [Total: 12] 3 (a) C(g) → C+(g) + e– correct equation (1) correct state symbols (1) [2] (b) (i) Na and Mg Mg has greater nuclear charge/more protons than Na (1) in both atoms, the 3s electrons are in the same orbital/ same energy level/same shell (1) (ii) Mg and Al in Al outermost electron is in 3p rather than 3s (1) 3p electron is at higher energy or is further away/is more shielded from nucleus (1)

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9701 23 © University of Cambridge International Examinations 2011 (iii) He and Ne both He and Ne have the highest nuclear charges in their Period (1) (iv) He, Ne, and Ar going down the group, valence/outer shell electrons are farther from the nucleus (1) there is greater shielding (1) attraction between valence electrons and nucleus is less or effective nuclear charge is less (1) [8] (c) (i) from Na to Cl increased nuclear charge/nuclear attraction (1) (ii) cation has fewer electrons than atom or cation has lost outer electrons or cation has fewer shells (1) but cation has same nuclear charge as atom or proton number is the same (1) [3] 3 (d) ignore any state symbols MgO(s) + NaOH(aq) → NO REACTION (1) MgO(s) + 2HCl(aq) → MgCl2 + H2O (1) Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4 or Al2O3(s) + 2NaOH(aq) + H2O(l) → 2NaAlO2 + 2H2O or Al2O3(s) + 6NaOH(aq) + 3H2O(l) → 2Na3Al(OH)6 (1) Al2O3(s) + 6HCl(aq) → 2AlCl3 + 3H2O or Al2O3(s) + 6HCl(aq) → Al2Cl6 + 3H2O (1) SO2(g) + NaOH(aq) → NaHSO3 or SO2(g) + 2NaOH(aq) → Na2SO3 + H2O (1) SO2(g) + HCl(aq) → NO REACTION (1) [6] [Total: 19] 4 (a) (i) C2H5O (1) (ii) (1) [2]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9701 23 © University of Cambridge International Examinations 2011 (b) (i) functional group isomerism or structural isomerism (1) do not allow 'functional isomerism' or positional isomerism (ii) compound type of isomerism P cis-trans or geometrical T optical (1 + 1) [3] (c) (i) dehydration/elimination (1) (ii) conc. H2SO4 / P4O10 / Al2O3 / H3PO4 / pumice (1) (iii) CH2=CHCH=CH2 allow CH2=C=CHCH3 (1) [3] (d) (i) CH3CH2CH(OH)CH2CH3 (1) (ii) steam with H3PO4 catalyst or conc. H2SO4 then water (1 + 1) only allow condition mark if reagent mark has been given (iii) Cr2O7 2– /H+ or MnO4 –/H+ (1) [4] [Total: 12] 5 (a) V is HCHO (1) [1] (b) (i) ester (1) (ii) W is HCO2CH3 (1) [2] (c) (i) X is HOCH2CH2CO2H (1) (ii) Y is HO2CCH2CO2H (1) [2]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9701 23 © University of Cambridge International Examinations 2011 (d) (i) Z is (1) (ii) esterification or dehydration or elimination or condensation (1) [2] [Total: 7]