AJC N2014 H2 P1 solns

©2020ASRJC/CHEM 1 H2 Chemistry 9647 2014 A level Paper 1 Suggested Solutions 1 n(Be) present = 9 09.0 = 0.01 mol no. of neutrons present per 9 4 Be atom = 5  no. of neutrons present in 0.09 g of 9Be = 5(0.01)L = 0.05L B 2 ClO2 + e– ClO2– [R] Since no. of electrons transferred must be equal  1 mol of ClO2 gained 1 mol of e  another 1 mol of ClO2 lost 1 mol of e to form Q  O.N. of chlorine in Q changes from +4 (original) to +5 (final) FYI ClO2 + 2OH– ClO3– + H2O + e– [O] balanced equation: 2ClO2 + 2OH– ClO3– + ClO2– + H2O C 3 molecular formula of G = C7H12O3 C7H12O3 + 2 17 O2 7CO2 + 6H2O A 4 Presence of H covalently bonded to O and available lone pairs of e lectrons on O in both water and methanol, hence hydrogen bonds will be the strongest IMF. Methoxymethane does not have H covalently bonded to O , hence only permanent dipoles. A 5 no. of bp no. of lp shape should be A,B 2 2 bent (non–linear) C,D 3 1 trigonal pyramidal A 6 The behaviour of real gas deviates from ideality because volume of gas particles is not insignificant compared to the overall volume occupied by the gas (option D). There are also significant intermolecular forces of attraction between the gas particles (option C), resulting in inelastic collisions (option A). B 7 Lattice energy is proportional to )rr( qq −+ −+ + . Thus the bigger the charge, and the smaller the inter–ionic distance, the more exothermic the lattice energy is. Size of cation: Cs+ > Na+ ; size of anion: Cl– > F– Since Cs+ and Cl– are the larger cation and anion respectively, CsCl will have the least exothermic lattice energy (–661 kJ mol–1). A 8 Hr = Hf (products) – Hf (reactants) = [–1273 + 6(0)] – [6(–394) + 6(–286)] = +2807 kJ mol–1 > 0 As the forward reaction results in a formation of solid (with no change in no. of gaseous molecules), there will be less ways of arranging the particles and S is < 0. C

©2020ASRJC/CHEM 2 9 Cl2 + 2e 2Cl– +1.36 V Br2 + 2e 2Br– +1.07 V I2 + 2e 2I– +0.54 V (SCN)2 + 2e 2SCN– x V Since both aq. C l2 and aq. Br 2 are able to oxidise SCN – ion to (SCN) 2 while aq. I2 is not able to, the Eo(SCN)2/SCN– must be less positive than +1.07 but more positive than +0.54. C 10 rate = k [P] [Q]2 If [P] is doubled while [Q] is halved, rate is halved. Hence only half of the volume collected in first experiment will be produced in the first minute of the 2nd experiment. B 11 To determine the order of reaction wrt H + ions, we have to monitor how the rate of reaction changes as the concentration of H + ions is changed , while keeping all other factors constant. C 12 [H+] = [HNO3] after dilution = )1090( 01.0x10 + = 0.001 mol dm–3 pH = 3 C 13 2H2(g) + CO(g) CH3OH(g) initial amount 2.0 1.0 0 change in amount –x – 1 2 x + 1 2 x amount at eqm 2.0 – x 1.0 – 𝟏 𝟐x 𝟏 𝟐x concentration at eqm 5.0 )x0.2( − 5.0 )x2 10.1( − 5.0 )x2 1( B 14 A bluish green – copper B green – barium C lilac – potassium D orange/red/brick red - calcium Group II metal colour of flame in oxygen Mg intense / brilliant white Ca intense / brilliant white with tinge of red (brick–red) Sr almost white with tinge of red Ba white with pale green tinges B 15 deduction likely identity of Z A Z is a metal. could be sodium, magnesium or aluminium B Z is probably in Group V. is phosphorus C Z is a covalent chloride. could be aluminium or phosphorus based on the formula (ZCl3) D Z forms an amphoteric hydroxide with NaOH. is aluminium Alternatively Option A and B are mutually exclusive, while the conclusion for C is correct for both A and B. Hence D is the key to answering this question. Using your knowledge of A lCl3 and PCl3 with water, you can then conclude that Z is aluminium. B 16 Dissolving CuCl2 in water will give the blue [Cu(H2O)6]2+(aq) and Cl–(aq). D

©2020ASRJC/CHEM 3 When concentrated HCl is added to an aqueous solution of Cu 2+, ligand exchange occurs where Cl– displaces the H2O ligands, forming the yellow [CuCl4]2–. [Cu(H2O)6]2+ + 4Cl– [CuCl4]2– + 6H2O 17 Cold condition Cl2(g) + 2OH–(aq) Cl–(aq) + ClO–(aq) + H2O(l) Hot condition The above reaction will take place 1 st before the C lO– formed undergoes further disproportionation 3ClO–(aq) 2Cl–(aq) + ClO3–(aq) Combining the 2 equations, 3Cl2(g) + 6OH–(aq) 5Cl–(aq) + ClO3–(aq) + 3H2O(l) C 18 A Down the group, the charge density of M2+ decreases since ionic radius increases. B Reactivity of Group II metals increases down the group. C Hsolution becomes more negative down the group and solubility increases. D From option A and since polarising power  charge density, extent of weakening of covalent bond in nitrate ion decreases and thermal stability increases. A 19 Given that the unknown is dibasic [2 x –CO2H], you can deduce the molecular mass of the alkyl chain = 146 – 2(45) = 56. Hence, the no. of C atoms = 4 [56  14 (each >CH2)] + 2 (from –CO2H) = 6. C 20 C CCH3 CH3 H C H H CH3 H H CH3 C C H C CH3 CH3 H H H H P Q Note: There is a plane of symmetry within a molecule of Q. B 21 The reaction pathway involving bromoalkane shows a 2–step mechanism . Hence, you can conclude that it occurs via unimolecular nucleophilic substitution (SN1). point X Y Z species present C Br lengthening and subsequent breaking of C–Br bond C trigonal planar carbocation intermediate formed C OH attack on C+ by OH– nucleophile and subsequent forming of C–OH bond A a b c d e f g h i i h

©2020ASRJC/CHEM 4 22 H2N CH2OH O NH CH2CH2CONH2 O N OH O N O OH CH2CH2CO2H hydrolysis acid-base H H RCO2H + NaOH RCO2Na + H2O C6H5OH + NaOH C6H5ONa + H2O RCONHR’ + + NaOH RCO2Na + R’NH2 Since both hydrolysis and acid –base involves a 1:1 reacting mole ratio with NaOH , 0.7 mol of NaOH (0.4 mol for hydrolysis and 0.3 mol for acid–base) will react with 0.1 mol of T. Note: NaOH has no reaction with the alcohol present (RCH2OH). D 23 CH2HO CH C OH H O 2,3-dihydropropanal HO CH2 C CH2 OH O 1,3-dihydropropanone A only 2,3–dihydropropanal contains a chiral carbon. B only 2,3–dihydropropanal gives silver precipitate. C absence of CH3CH(OH)R and CH3COR  both do not give yellow precipitate. D they have the same molecular (C3H6O3) and empirical (CH2O) formulae. FYI (mechanism of the iodoform test) C CR O H H H OH- C CR O H H I I C CR O H H I C CR O I I I OH- C OHR O + CI3 - C O-R O + CHI3 Step 1: acid–base reaction Step 4: nucleophilic (acyl) substitution Step 4 and 5 can be combined and is known as base–catalysed hydrolysis Hence it is not possible for both to give yellow precipitate with I2/OH–. However, RCOCH2I (RCOCHI2, RCOCI3) gives yellow precipitate with I2/OH–. D * aldehyde ketone

©2020ASRJC/CHEM 5 24 CH3 OH CH3 Br (CH3)3P CH3 (CH3)3P substitution Br substitution nucleophilenucleophile Br D 25 A nucleophilic (acyl) substitution occurs readily and an immediate white precipitate would be observed. B nucleophilic substitution does not occur due to the delocalisation of the lone pair of electrons on the C l atom into the benzene ring, creating a partial double bond character in the C–Cl bond. C same as A D nucleophilic substitution occurs upon boiling with NaOH(aq) and the resulting mixture contains C l– ions. upon cooling, acidification and addition of AgNO 3(aq), white precipitate would be observed. D 26 ala–gly lys–ser ser–gly–ala met–ala gly–lys met–ala–gly–lys–ser–gly–ala Peptide P contains 7 amino acid residues. Alternatively A not possible to get gly-lys and ser-gly-ala fragments B not possible to get gly-lys fragment D not possible to get lys-ser fragment C 27 CH3 N R O N