NJC Prelims Paper 3 Answers Final

1 NJC/H2 Chem Preliminary Examination/03/2021 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 3 Free Response Candidates answer on Question Paper. Additional Materials: Data Booklet 9729/03 27 August 2021 2 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 /20 2 /16 3 /24 Section B 4 /20 5 /20 Paper 3 Total /80 This document consists of 24 printed pages.

2 NJC/H2 Chem Preliminary Examination/03/2021 Section A Answer all the questions in this section. 1 (a) Nicotine (C10H14N2) is a drug present in tobacco. In aqueous solution, nicotine ionises as shown. C10H14N2 + H2O C10H15N2+ + OH− pKb (C10H14N2) = 6.0 (i) Calculate the pH of a 0.100 mol dm−3 C10H14N2 solution. [2] C10H14N2 is a weak base, [OH−] = √10ି଺×0.100 = 3.162 × 10−4 mol dm−3 pOH = −lg[OH−] = 3.50 pH = 14 – pOH = 10.5 (ii) Suggest a suitable indicator for the titration of aqueous nicotine with HNO3(aq). Explain your answer. [2] Methyl orange [1]. The equivalence point pH would be less than 7 [1/2], the working range of methyl orange (pH 3-5) would overlap with the region of rapid pH change at equivalence point. [1/2] Note: The salt, C10H15N2+, (with Ka value) is the conjugate acid of weak base C10H14N2 (Kb). Hence the salt is acidic and has a pH of < 7 at equivalence point. (iii) Calculate the pH of a 5.00 dm3 solution consisting of 0.100 mol dm‒3 C10H14N2 and 0.200 mol dm‒3 C10H15N2+. [1] The mixture contains a weak base and its conjugate acid, it is a buffer solution. pOH = pKb + lg[௖௢௡௝௨௚௔௧௘ ௔௖௜ௗ] [௪௘௔௞ ௕௔௦௘] = 6.0 + lg଴.ଶ ଴.ଵ = 6.30 [1/2] pH = 14−6.30 = 7.70 [1/2]

3 NJC/H2 Chem Preliminary Examination/03/2021 [Turn over (iv) Calculate the number of moles of HNO3 that needs to be added to the solution in (a)(iii) to obtain a buffer solution of pH 7.40. [2] Weak base C10H14N2 of the buffer will react with the HNO3 added, use ICF table (in mol) to determine limiting reagent. Initial amount of C10H14N2 = 5 × 0.100 = 0.5 mol Initial amount of C10H15N2+ = 5 × 0.200 = 1.0 mol Let amount of HNO3 added = x mol C10H14N2 + H+ o C10H15N2+ I / mol 0.5 x 1.0 C / mol x x +x F / mol 0.5 – x 0 1 + x pH = 7.40, pOH = 6.60 pOH = pKb + lg[௖௢௡௝௨௚௔௧௘௔௖௜ௗ] [௪௘௔௞௕௔௦௘] 6.60 = 6.0 + lg(ଵ ା ௫ ଴.ହ ି ௫) lg(ଵ ା ௫ ଴.ହ ି ௫) = 0.60 (ଵ ା ௫ ଴.ହ ି ௫) = 100.60 [1] or similar idea ଵ ା ௫ ଴.ହ ି ௫ = 3.981 1 + x = 1.991 – 3.981x 4.981 x = 0.991 x = 0.199 Amount of HNO3 needed = 0.199 mol

4 NJC/H2 Chem Preliminary Examination/03/2021 When a cigarette is smoked, nicotine-rich blood stimulates the release of many chemical messengers including dopamine and epinephrine. OH OH OH N CH3 H OH OH NH2 epinephrinedopamine (b) (i) Name the type of isomerism exhibited by epinephrine and draw the isomers. [2] Optical isomerism / enantiomerism [1] C CH2NH(CH3) H OH OH OH C CH2NH(CH3) HOH OH OH [1] Note : must show 3D tetrahedral drawing around the chiral carbon. (ii) Give the structure of the product when dopamine is reacted with excess concentrated HNO3. [2] OH OH NH3 + O2N NO2 NO2 1m for 3 × E.sub of −NO2 on phenol (0m if monosub of −NO2) 1m for acid base reaction with −NH2 to give −NH3+ Note: conc HNO3 lead to tri sub of −NO2 at 2,4,6 position of EACH phenol.

5 NJC/H2 Chem Preliminary Examination/03/2021 [Turn over (iii) A reaction between dopamine and chloromethane, CH3Cl, forms a compound with formula C11H18NO2Cl. Suggest a structure for this compound and how the yield of this compound can be maximised. [2] OH OH N+ CH3 CH3 CH3 Cl React Dopamine with excess CH3Cl. [1m] Note: dopamine, C8H11NO2 reacts with 3 mol of CH3Cl to obtain 11 C. Amine group function as the nucleophile to react with C−Cl. (iv) Suggest a chemical test to distinguish the two chemical messengers, dopamine and epinephrine. [2] 1m chemical test for 2q alcohol in epinephrine. 1m observations Anhydrous PCl5 or SOCl2 White fumes of HCl observed for epinephrine but no white fumes for dopamine. K2Cr2O7,H2SO4(aq), heat Orange K2Cr2O7 turns green for epinephrine but K2Cr2O7 remains orange for dopamine. Do not accept hot KMnO4 as both compounds would undergo benzene side chain oxidation. Cigarette smoke contains many harmful chemicals such as Period 4 elements, chromium, nickel and arsenic. (c) Give the full electronic configuration of chromium and arsenic. [2] Cr : 1s22s22p63s23p63d54s1 As : 1s22s22p63s23p63d104s24p3 (Group 15 configuration) (d) Chromium and nickel are transition elements. (i) State what is meant by the term transition elements. [1] Transition elements concept (ii) Suggest why the first ionisation energies of chromium and nickel are similar. [2] Transition elements concept [Total : 20]

6 NJC/H2 Chem Preliminary Examination/03/2021 2 Baeyer-Villiger reaction is an organic reaction that forms an ester from a ketone. Pentan-2-one can be converted into propyl ethanoate using a peroxyacid, RCO3H. equation 2.1 O RCO3H O O Ester can be reduced by LiA lH4 to give alcohols. An example of the reduction of propyl ethanoate is shown below. equation 2.2 O O OHOH +LiAlH4 (a) (i) Suggest the type of reaction shown in equation 2.1. [1] Oxidation [gain O atom] (ii) Suggest the products formed when ethyl benzoate is reacted with LiAlH4. [2] O O OH +LiAlH4 OH (b) Fig 2.1 shows a reaction scheme involving a cyclic ester, compound C. A (C6H10) step 1 B (C5H8O) RCO3H O O C LiAlH4 D Fig 2.1 (i) State the reagents and conditions required for step 1 and suggest structures for the organic compounds A, B and D. [4] KMnO4, H2SO4(aq), heat RCO3H O O C LiAlH4 D CH2 O OH OH A B 1m for step 1 B to C follows the reaction stated in equation 2.1. B must be a carbonyl compound. C to D follows the reaction stated in equation 2.2. D contains 2 −OH groups after the ester reacts with LiAlH4. A must be an alkene with =CH2 such that it loses one C atom as CO2 after vigorous oxidation with hot acidified KmnO4

7 NJC/H2 Chem Preliminary Examination/03/2021 [Turn over (ii) Compound C can also be synthesised from HOOC(CH2)3CH2OH. Suggest the reagents and conditions required for this synthesis. [1] Concentrated H2SO4, heat Note: The −COOH and −OH group undergoes intramolecular condensation to form cyclic ester. (c) Peroxyacid, RCO3H, also converts alkene into epoxide, a cyclic ether with three-atom ring that approximates an equilateral triangle. Epoxide reacts with water readily to give a diol. RCO3H O H2O OH OH Use your knowledge of VSEPR theory to explain the high reactivity of epoxide. [2] Based on VSEPR theory, the C (or O atoms) in epoxide should have a bond