2021 HCI SAJC TJC H3 Chemistry Answers

RESTRICTED Temasek Junior College Advanced Level Higher 3 Preliminary Examination 9813 Chemistry September 2021 Answer Key 1 HCI SAJC TJC JC2 Preliminary Examination CHEMISTRY Paper 9813/01 Paper 1 This answer key is published as an aid to students, to indicate the requirements of the examination. It shows the basis on which the markers award marks. It does not indicate the details of the discussions that took place before marking began, which would have considered the acceptability of alternative answers. Note: 1. If the candidate’s answer has the same “meaning” or can be clearly inter preted as being of equivalent significance, detail and validity as that in the markscheme then award the mark. Where this point is considered to be particularly relevant in a question it is emphasized by OWTTE (or words to that effect). 2. Occasionally, a part of a question may require an answer that is required for subsequent marking points. If an error is made in the first marking point then it should be penalized. However, if the incorrect answer is used correctly in subsequent marking points then follow through marks should be awarded. When marking, indicate this by adding ECF (error carried forward) on the script. 3. Do not penalize candidates for errors in units or significant figures, unless it is specifically referred to in the answer key. 4. If a question specifically asks for the name of a substance, do not award a mark for a correct formula unless directed otherwise in the answer key. Similarly, if the formula is specifically asked for, unless directed otherwise in the answer key, do not award a mark for a correct name. 5. If a question asks for an equation for a reaction, a balanced symbol equation is usually expected, do not award a mark for a word equation or an unbalanced equation unless directed otherwise in the answer key. 6. Ignore missing or incorrect state symbols in an equation unless directed otherwise in the answer key.

RESTRICTED Temasek Junior College Advanced Level Higher 3 Preliminary Examination 9813 Chemistry September 2021 Answer Key 2 HCI SAJC TJC JC2 Preliminary Examination Question Answer Marks 1(a) Chromium- and selenium-based reagents are toxic; large scale production will generate a lot of wasteful by-products. Getting these by-products removed will add to the cost of production, which is not sustainable. 1 Palladium- and rhodium-based catalyst are expensive and will add to the cost of production. 1 Removing the above -mentioned elements entirely from the product cannot be guaranteed – this is a problem especially when the products are used in medicines. The syntheses thus have no industrial application. (Allow other valid reasons) 1(b) Electronic effect: Use the above and highlight that the radical is resonance stabilised (the lone electron on the 2o carbon is delocalised into the adjacent  system). 1 Hammond postulate: Homolytic cleavage of C–H is endothermic and hence has a late transition state (TS resembles the radical intermediate). The activation barrier leading to the allylic radical is likely to be lower than that leading to the 1o radical. 1 1(c)(i) 1 1 1(c)(ii) The allylic substrate does not get oxidised at the anode directly; instead it is being oxidised to the allylic radical by R2N–O● which is produced when the conjugate base of R 2N-OH gets oxidised. Hence the role of the mediator is to function like a ‘middleman’ / electron shuttle between the anode and the substrate. (OWTTE) 1 tBuOOH works in synergy with R2N–OH, helping to complete the oxidation by playing the role of an oxidant after the allylic radical is being formed. 1

RESTRICTED Temasek Junior College Advanced Level Higher 3 Preliminary Examination 9813 Chemistry September 2021 Answer Key 3 HCI SAJC TJC JC2 Preliminary Examination Question Answer Marks 1(c)(iii) Working (not required to score the mark): 1 1(d)(i) 1st step correct (accept SET to the meta C followed by full curly arrows towards the ortho C to give the negative charge) 1 2nd & 4th steps correct 1 3rd step correct 1 1(d)(ii) Mg → Mg2+ + 2e– 1 1(d)(iii) The reduced product will be re-oxidise back to TBS-cresol (accept bromide will be oxidised to bromine) 1

RESTRICTED Temasek Junior College Advanced Level Higher 3 Preliminary Examination 9813 Chemistry September 2021 Answer Key 4 HCI SAJC TJC JC2 Preliminary Examination Question Answer Marks 1(e) RDS of Pathway A: D1 to D2; Ea = 52 kJ mol–1 1 RDS of Pathway B: D3ads to Eads; Ea = 41 kJ mol–1 (the correct RDS must be identified and Ea corrected computed to score the credit) The reduction of D to E undergoes the Pathway B. (ECF if students chose the wrong RDS but conclude the pathway with a lower Ea) 1 1(f) (equipment)  Specialised equipment is required for maintaining the temperature of the reaction at –35 oC and this is especially so for large scale production. Expensive to build, maintain and operate.  The systematic removal of NH3 and H2 is challenging after the reaction is completed. 1 1 (risk)  If the equipment breaks down in the process, it will result in the vapourisation of 5 dm 3 of liquid NH3 and this will translate to an enormous volume of gaseous NH 3. This might risk explosion/ compromise the health of scientist. (Allow other valid reasons) 1 1(g)(i) Assuming the rate of reduction is constant, (TBS-cresol) reduced after 16 hours = 7.1 / 222.1 = 3.1967  10–2 mol (TBS-cresol) reduced per second = 3.1967  10-2 / (16  60  60) = 5.5499  10–7 mol A = 10.0  4.0 = 40.0 cm3 [R]surface = 5.5499  10–7 / 40.0 = 1.39  10–8 mol cm–2 1 1(g)(ii) kred = 0.75 / (96500  40.0 x 1.39  10–8) = 13.9 s–1 (units not required) 1 Question Answer Marks 2(a) 1 – 2 correct structures [1] 3 – 4 correct structures [2] All 5 correct structures [3] (Award [2 max] if Na+ is not included in the structure) 3

RESTRICTED Temasek Junior College Advanced Level Higher 3 Preliminary Examination 9813 Chemistry September 2021 Answer Key 5 HCI SAJC TJC JC2 Preliminary Examination 2(b)(i) OR Correct reagents and conditions for the 1st step + intermediate 1 Correct reagents and conditions for the 2nd step + intermediate 1 Correct reagents and conditions for the 3rd step 1 Question Answer Marks 2(b)(ii) With the fourth priority group (H atom) pointing into the plane, the top three priority groups are arranged clockwise, hence configuration is R. working is needed for the mark (Working is required for the credit) 1 2(c)(i) 1) K2Cr2O7, dilute H2SO4 warm with immediate distillation 2) NaOH(aq) 1) K2Cr2O7, dilute H2SO4 warm with immediate distillation 2) NaOH(aq)

RESTRICTED Temasek Junior College Advanced Level Higher 3 Preliminary Examination 9813 Chemistry September 2021 Answer Key 6 HCI SAJC TJC JC2 Preliminary Examination Largest substituent (-NBn2) perpendicular to the plane of C=O (marks awarded regardless of configuration) 1 Two correct Newman projections drawn (check configuration before awarding marks) 1 2(c)(ii) Correct explanations for the two paths of the first conformation (2 x [0.5]) 1 Correct explanations for the two paths of the second conformation (2 x [0.5]) 1 Show via a Newman projection (or otherwise) that the least hindered path gives K 1 Question Answer Marks 2(c)(iii) Hence, the configuration of chiral centre x matches that in dolaproline. 1 1

RESTRICTED Temasek Junior College Advanced Level Higher 3 Preliminary Examination 9813 Chemistry September 2021 Answer Key 7 HCI SAJC TJC JC2 Preliminary Examination Hence, the configuration of chiral centre y does not match that in dolaproline. Question Answer Marks 3(a)(i) When radiation in the IR region is being absorbed, it is associated with the vibrational energy levels within the m