2019 EJC Prelims H2 Chemistry Paper 1 Worked Solution

2019 JC2 Preliminary Examination H2 Chemistry 9729 Paper 1 Worked Solution 1 From Group 1318, the number of unpaired electrons es from 1 (Group 13) to 3 (Group 15), then es to 0 (Group 18). Hence, W, X, Y and Z cannot be main group elements, i.e. they are TM. W must have been 4d8 5s2 and not 4d2 5s2 as the number of unpaired electrons es with more electrons added.  B 2 Angle of deflection, z m For a proton, 1 1H , 1 11 z m    For particle X,  51 115 3 z m     A : 3 1 A ; B : 5 2 B ; C : 6 3 C ; D : 93 4 D  A 3 A : trans-isomer is non -polar (C–Cl bond dipole cancels), while cis-isomer is polar. Hence cis-isomer is more soluble in polar H2O. B : C=O is more polar than C –Cl as O is more electronegative. Also, oxygen of C=O can form H -bond with water via its lone pair.  CH3COCH3 is more soluble. C : CH3CH2CH2CH2CH(OH)CO2H can form intermolecular H -bond with H2O, while CH 3CH2CH2CH2CH(NH3 +)CO2 – is zwitterionic and forms stronger ion - dipole interaction with H 2O, hence more soluble. D : 2-nitrophenol can form intramolecular H-bond, while 4-nitrophenol forms intermolecular H -bond with H 2O, hence more soluble.  B 4 A : Like graphite, each C is sp2 hybridised with one unpaired e– in the remaining p-orbital, which is delocalised, allowing CNT to conduct electricity. B : Like graphite, the C –C bonds within each molecule is stronger than that in diamond as they result from sp 2-sp2 overlap. However, between two molecules, there is only weak intermolecular id -id attraction, hence softer than diamond. C : Like graphite, the high melting point is due to energy required to break the strong intramolecular C –C bond and not to overcome the weak intermolecular id-id attraction. D : Like graphite, the C –C bonds within each molecule is stronger than that in diamond as they result from sp 2-sp2 overlap.  A 5 I : H2SO4 oxidises Br– to Br2, itself reduced to SO2 II : H2SO4 catalyses the condensation reaction between an acid and an alcohol to give an ester (and water), but making the C=O more electrophilic. III : Acid-base reaction where H2SO4 is the acid and NaOH the base, to give salt (Na2SO4) and water.  D 6 1 : A more exothermic lattice energy just implies that it takes more energy to vaporise the ionic lattice into gasous ions. No bearing on the decomposition. 2,4: CO3 2– is more easily polarised as planar CO3 2– has a larger surface area than the near spherical tetrahedral SO4 2– ion. Hence the C–O bond is more readily weakened, causing CaCO3 to decomposing at a lower temperature. 3 : Due to smaller size of the CO3 2– ion compared to SO4 2–, the charge density is higher. However, higher anionic charge density has no bearing on the decomposition.  C 7 Oxidation state of S in SF6 is +6, SCl2 is +2 and S2Br2 is +1, which correlates with the ability of the halogen to oxidise S.  D 8 13 0.13Number of C atoms 0.0113 LL   Each 13 6C atom has 13 – 6 = 7 neutrons Number of neutrons 0.01 7 0.07LL    B 9  B 10 I2(s) + Cl2(g)  2ICl(s) H1 --------- (1) ICl(s) + Cl2(g)  ICl3(s) H2 --------- (2)  2 2 3 1 1 3(1) (2) : (s) C (g) C s2 2 2   llII    f 3 1 2 1 11C (s) 14 8822 81 kJ mol H H H          lI  C 11 Combustion of fuel (hydrocarbon) is an exothermic process, so H is –ve. Combustion leads to the production of more gaseous CO2 and H2O molecules than O2(g) consumed. Hence, entropy increases, i.e. S is +ve. G = H – TS is –ve for all T.  D 12 A : Since HI, a strong acid is produced, although H+ is a catalyst, the [H+] es. B : The rate constant, aE RTk Ae   , is independent of concentration. C : Given  33rate H CH COCHk   . So rate es by 4  when concentrations of all reactants are doubled. D : pH 1  [H+] = 10–1 = 0.1 mol dm–3 pH 2  [H+] = 10–2 = 0.01 mol dm–3 [H+] es by 10 , rate es by 10   A 13 A : Adding Ar at constant volume will not cause a spike in the concentration of all three gases. B : Decrease in temperature of a large body of gas cannot be achieved instantaneously. C : As amountconc volume , when volume of container is decreased, the total pressure and conc of all three gases will increase suddenly. By LCP, eqm will shift to the side with fewer gaseous particles (right side), in attempt to bring down the pressure, leading to consumption of N 2 and H2 to form more NH3 at the new eqm. D : Removal of NH 3 will not result in a spike in concentration of all 3 gases.  C 14   14 w 6.22 7 3 pOH p pH lg 2.4 10 7.4 6.22 OH 10 6.03 10 mol dm K                  B 15 Since there are two chiral centers and one cis-trans double bond, Number of stereoisomers = 23 = 8  D 16 There are 8 different types of H on the molecule:  C 17  C 18 A : B : C : D :  C

19 1 : 2 : 3 : There is no intermediate in SN2 4 :  A 20 From the rate of formation of ppt (solvolysis), P is C6H5CH2Br, Q is X–C6H4CH3, R is C6H5CH2I, S is C6H5CH2Cl A : R should have the longest C –X bond since it is the weakest and most reactive. B : Free radical substitution does not work with I2. C : Q can be BrC 6H4CH3 formed from C6H5CH3 + Br2 in the presence of AlBr3. D : Only P, R and S gives benzoic acid upon vigorous oxidation.  C 21 A : hot acidified K2Cr2O7 can hydrolyse the ester to release ethanol which can be oxidised into ethanoic acid, reducing orange K2Cr2O7 into green Cr3+. B : 2RCO2H + Na 2CO3  2RCO2 –Na+ + CO2 + H 2O. 1 mole of benazepril only contains 1 mole of –CO2H. C : R2NH + HC l  RNH3 +Cl–. 1 mole of benazepril contains 1 mole of 2º amine. D : Benazepril hydrolyses in NaOH to give  B 22 1 : 2 : 5 Ar–H relative (2 -, 4-, 6-) to phenol or phenylamine can be substituted by Br: Ar–H + Br2  Ar–Br + HBr 3 : There is one –CH(OH)CH3 and one –COCH3 group in pactamycin.  A 23 A : B : C : D :  D 24 The ester, amide and halogenoalkane moieties will be hydrolysed by NaOH(aq), giving 2 -amino acids and 1 -amino acid:  B 25 1 : As HCl is a much stronger acid than H2O, C l– are less basic than OH –, hence they are better leaving group, making RC(=O)–Cl better acylating agents than RC(=O)–OH. 2 : Due to the similarity in size of O and C, l one pair of electrons on O in RCO2H is much more effectively delocalised into the C=O than that on much bigger Cl in RCOCl, rendering the C=O carbon in RCO 2H less electrophilic. 3 : While O is more electronegative then Cl, the C=O carbon in RCO2H is less electrophilic than that in RCOC l as the lone pair of electrons on O is more effectively delocalised into the C=O of RCO2H.  C 26 In the gas phase, diisopropylamine is the stronger base (more spontaneous protonation, i.e. G is more negative). In aqueous phase, azepane is the stronger base (smaller pKb, i.e. larger Kb). This is primarily due to steric hindrance from the two bulky –CH(CH3)2 groups which hinders hydration of the R 2NH2 + ion through H-bonding.  D 27 Extraction is dependent on the oxidation of Au to Au(I) by O2 in alkaline medium: [R] O2 + 2H2O + 4e–  4OH– E 0.40 V For the oxidation to be feasible  cellE 0.40 E     Au Au 0 I  E   Au Au 0.40 VI  A 28 The left-hand-side cell is a electrochemical cell, i.e. battery source of +1.52 V: [O]: Mn  Mn2+ + 2e– E 1.18 V [R]: Cu2+ + 2e–  Cu E 0.34 V cellE  0.34 1.18 1.52 V    Electrolysis occurs on the right-hand-side: Cathode (Reduction) : E  Ag Ag 0.80 V  E  Na Na 2.71 V  E  2H H 0.00 V  E   2Zn Zn 0.76 V  E  22H O H 0.83 V Ag is deposited for A, while H+ or H2O is reduced to H2(g) for the res