RI 2018 Prelim P2 Ans

© Raffles Institution 2018 9729/02/S/18 1 2018 Y6 H2 Chemistry Preliminary Exams Paper 2 (Suggested Solutions) 1(a) X is likely to be in Group 14 as there is a large jump from the 4th to 5th ionisation energies. This shows that the 5th electron is removed from an inner electron shell and requires more energy for removal as it is more strongly attracted to the nucleus. Hence, there are 4 valence electrons in X. Since X has at least 10 electrons but fewer than 20 electrons, X is Si. Comments • Students need to be familiar with the Group number in the Periodic Table given in the Data Booklet. Silicon is a Group 14 (NOT Group 4 or Group IV) element. • Since X has at least 10 electrons, carbon is not a possible answer for X. • Answers need to be specific when referring to the 5th electron being removed from the inner electron shell. Avoid using vague phrases like ‘different electron shell’ or ‘another electron shell’. 1(b) Mn and Fe are first-row transition metals. Although Fe has a higher nuclear charge (greater proton number) than Mn, Fe has more 3d electrons which provide more shielding between the nucleus and the outer 4s shell of electrons. This increase in shielding effect offsets the increase in nuclear charge and hence, the increase in effective nuclear charge from Mn to Fe is minimal, resulting in the first ionisation energies of Mn and Fe to be similar. Comments • Most students were able to identify that Fe has both higher nuclear charge and shielding effect than Mn. However, students need to be clear er in their explanation on why the inc rease in effective nuclear charge from Mn to Fe is minimal. 1(c) Ca2+ has a larger cationic radius than Cu2+, thus Ca2+ has a lower charge density and weaker polarising power. Consequently, there is lower extent of distortion of the electron cloud of the carbonate anion and hence covalent bonds within the carbonate anion are weakened to a lesser extent . Hence, CaCO3 decomposes at a higher temperature. Comments • Most students were able to link the larger cationic radius of Ca 2+ to its lower charge density and weaker polarising power. • Students are required to make the link between the stronger covalent bonds within the carbonate anion of CaCO3 to its higher decomposition temperature. 2(a)(i) Anthracene is a cyclic molecule. All carbon atoms of anthracene are sp2 hybridised and are trigonal planar. The unhybridised p -orbitals of the carbon atoms in anthracene can form a continuous p-orbital overlap, thus the  electrons are delocalised throughout the molecule. Anthracene has 4(3) + 2 = 14  electrons where or n = 3.

© Raffles Institution 2018 9729/02/S/18 2 Comments • Most students correctly identified the hybridisation of the carbon atoms as sp 2 and also stated that the unhybridised p-orbital of the carbon atoms form a continuous overlap allowing the elect rons in the p-orbitals to be delocalised forming the -electron cloud. • Most students also correctly calculated the number of -electrons or the value of n. • Answers which simply restated the bulleted points , i.e. criteria for aromaticity (e.g. the -electrons are delocalised throughout the molecule), without stating that this is possible through the continuous p-orbital overlap were not given any credit. 2(a)(ii) Enthalpy change of hydrogenation = 7(−118) = −826 kJ mol−1 Comments • Most students correctly calculated this value. 2(a)(iii) Enthalpy change of hydrogenation = −356 −129 = − 485 kJmol−1 Comments • Most students correctly calculated the enthalpy change of hydrogenation. • For the energy level diagram, some incorrect or missing state symbols were commonly observed, particularly for anthracene and tetradodecahydroanthracene. • Incorrect coefficient for H2(g) was also commonly observed. • Some students missed out the arrow for ∆H2. 0 14 C(s) + 12 H2(g) (s) + 7 H2(g) (s) +129 − 356 ∆H2 energy / kJ mol−1

© Raffles Institution 2018 9729/02/S/18 3 2(a)(iv) Resonance energy = 826 − 485 = 341 kJ mol−1 (energy level diagram – not required by question) Comments • Most students correctly calculated the value of the resonance energy. 2(b)(i) TEQ = (0.029 x 0.01) + (0.135 x 0.01) + (0.204 x 1.00) = 0.20564 = 0.206 Comments • Most students correctly calculated the value of the TEQ. 2(b)(ii) ECR = 0.20564 70 x 1.37 x 10−3 = 4.00 x 10−6 Since the ECR from the consumption of smoked fish exceeds the value set by USEPA, it will be a cancer risk for the 70 kg person. Comments • Most students correctly calculated the value of the ECR and came to the correct conclusion. • Based on the question, answers need to conclude if there will be a cancer risk. 2(c)(i)

© Raffles Institution 2018 9729/02/S/18 4 Comments • Generally well done. • Answers need to show the c orrect positions of the methyl and isopropyl (-CH(CH3)2) groups on thymolphthalein. • The phenol -OH group is a 2,4 -directing group. Hence students should not give the minor product where the substitution occurred at position 3 with respect to the phenol group of thymol. 2(c)(ii) Thymol reacts more readily as the alkyl groups of thymol are electron donating and increase the electron density on the benzene ring making it more susceptible to electrophilic substitution. Comments • This part was very well done by most students. • Answers which concluded that phenols react more readily as there is less steric hindrance due to absence of alkyl groups were accepted for this question. • It is incorrect to state that the alkyl groups increased the intensity of negative charge on thymol as the thymol molecule has no charges. • Students need to avoid contradiction in their answers. 2(d) The carboxylate anion in A is more stable than the phenoxide anion in B as the negative charge on oxygen is dispersed over the two highly electronegative oxygen atoms resulting in two equivalent resonance structures in A as compared to B. Comments • This part was very well done by most students. • It is incorrect to mention steric hindrance/strain in B as the single bonds connecting the phenol rings to phthalic anhydride can rotate to minimise steric hindrance. • In general, 5-membered and 6-membered rings are stable. Hence answers should not discuss about ring strain in the 5-membered ring in B. 3(a)(i) The stronger the base, the more available is its lone pair of electrons for donation to acid. Hence, a strong Lewis base will readily donate its lone pair to the electron deficient carboxyl C (Lewis acid) to reform the starting acid derivative . Thus, strong Lewis bases are poor leaving groups. or Strong Lewis bases donate electron pair more easily and are less likely to accept an electron pair. Thus, it is more difficult to break the C-Y bond to form Y–. Comments • Many students concluded that strong Lewis bases are poor leaving groups as they form stronger bonds with the carbonyl carbon. This may not be correct as C−N bond is weaker than C−O bond but NH2− is a stronger base than OH−. • There are also students who suggested that strong Lewis bases repel the bond pair of electrons in the C−Y bond. This is incorrect because when Y is covalently bonded to C, it is not a Lewis base. Only when the C−Y bond breaks and Y leaves as :Y−, then Y− is considered a Lewis base as it has an electron−pair that can be donated.

© Raffles Institution 2018 9729/02/S/18 5 3(a)(ii) Since pKb of CH3COO− is more positive than NH2−, CH3COO− is a weaker base and is hence a better leaving group so that the reactivity of (CH3CO)2O is higher than that of CH3CONH2. Reaction 1 will proceed with a faster rate than reaction 2. or The pKb of CH3O− is less positive than that of CH3COO− but more positive than that of NH2−. Hence the product CH 3COOCH3 is less re