RI 2021 Prelim P3 Answers

26 2021 H2 Chemistry Prelim Paper 3 Suggested Solutions C1 (a)(i) Cl2(aq) is added to solution of KI, followed by hexane. The mixture is shaken and the hexane layer turns purple, indicating the presence of I2. Cl2 + 2I– ⎯→ I2 + 2Cl– Examiner Comments • This question was poorly attempted as many students were unable to properly describe a redox reaction involving the two halogens. In order to show the relative oxidizing abilities of Cl2 and I2, you must choose a redox reaction. • It is also necessary to describe the use of hexane for a visual observation of the reaction. (a)(ii) Cl2 + 2e – ⇌ 2Cl– EӨ = 1.36 V I2 + 2e– ⇌ 2I– EӨ = 0.54 V EӨcell = 1.36 – 0.54 = +0.82 V Since the EӨcell value is positive, the reaction is spontaneous. OR Since the EӨ value for C l2 is higher than that of I2, chlorine has a higher tendency for reduction and is hence, a stronger oxidising agent. Examiner Comments • Many students were able to quote the correct EӨ to use and/or calculate the correct EӨcell value. However, do remember to compare them and state a conclusion. • Unfortunately, many students did not apply this knowledge into answer part (1). (a)(iii) Half-equation for iodine: I2(aq) + 2e– ⇌ 2I–(aq) Hence, ascorbic acid and iodine react in a 1:1 mole ratio. Amount of iodine titrated = 0.005  0.02205 = 1.1025  10-4 mol Amount of ascorbic acid present in 10.0 cm3 = 1.1025  10-4 mol Amount of ascorbic acid present in 100 cm3 = 10  1.1025  10-4 = 1.1025  10-3 mol Mass of ascorbic acid present = 1.1025  10-3  176.0 = 0.194 g Mass of ascorbic acid present in 100.0 g of candy = 0.194 g % by mass of ascorbic acid = 0.194 100.0 × 100% = 0.194% Examiner Comments • Generally well done. • Students are reminded to include units in the intermediate steps of your working. (b) Add equal amounts of bromopropane, chloropropane and iodopropane in separate test tubes and warm the test tubes in a water bath maintained at 50 °C. Then add 5.0 cm3 of silver nitrate solution in ethanol to each compound and note the time taken for the precipitates to first appear. The mixture containing iodopropane will form a yellow ppt of AgI first, followed by the mixture containing bromopropane with a pale cream ppt of AgBr and in the mixture containing chloropropane with a white ppt of AgCl.

27 Examiner Comments • Many students who attempted this question gave reasonable answers but left out important details, and hence, did not earn full credit. While this is not a full planning question as you would see in Paper 4, your experiment outline must still include steps to ensure the aim of the experiment can be achieved. • For instance, many students left out the heating step so the reaction did not even take place. Note that heating in a water bath is recommended for this experiment as temperature should be kept constant for all three reactions (since this experiment is used to determine rate of reaction). • Most answers used equal volumes of the RX compounds, but in order to observe the rate of the reaction, you must start with equal amounts of the compound. Since the three compounds have different densities, using the same volume or mass will not equal the same amount. • A number of students did not measure the time taken for the p recipitate to first appear. Since there are three reactions, it’s impossible to start the three reactions and a stopwatch simultaneously. Hence, you must use a stopwatch to record time (instead of a visual observation of which is faster). • Many students did not include the expected sequence of formation of ppt as part of the expected observations. • Many incorrect answers used NaOH(aq) as the reagent with heat, followed by neutralization with HNO3 before adding AgNO3. This will not work because the hydrolysis reaction would be complete by the time the acid is added. Adding AgNO3 would then give an immediate ppt for all three compounds. • Some answers suggested carrying out the hydrolysis for a fixed interval of time and then weighing the mass of ppt form. This will not work because the molar mass of each ppt is different! (c)(i) Examiner Comments • Generally well done. (c)(ii) NaBH4 / H2, Ni, heat / LiAlH4 in dry ether Reduction reaction Examiner Comments • Generally well done, although some students read the question wrongly and gave for reagents and conditions for the oxidation of alcohol instead. • Students are reminded to use the correct conditions for the reaction.

28 (c)(iii) There are two functional groups on rhododendrol: phenol and aliphatic alcohol. When added to sodium hydroxide, only phenol will react to form phenoxide. This is because phenol is a stronger acid than the aliphatic alcohol. In the phenoxide ion, the p–orbital of the O atom overlaps with the –electron cloud of the benzene ring and the negative charge/lone pair on the O atom is delocalised into the ring. Hence, the phenoxide ion is resonance-stabilised and more stable than the alkoxide ion. OR This is because the aliphatic alcohol is a weaker acid than phenol. The O atom on the alkoxide ion is bonded to two electron donating alkyl groups which intensifies the negative charge on the O atom. The alkoxide ion is less stable than the phenoxide ion. Examiner Comments • Most students who attempted this question gave good answers which correctly explained the stability of the two conjugate bases. However, students wrote too much or were too detailed in their explanations, which would lead to insufficient time for the rest of the paper. • Students need to be clear in their answers whether they are referring to the acid or the conjugate base. (c)(iv) Add 2,4-dinitrophenylhydrazine/2,4-DNPH to each compound in a test tube. • For RK, orange ppt is formed • For rhododendrol, no ppt is formed Add K2Cr2O7(aq), H2SO4(aq) to each compound in a test tube and heat in a hot water bath. • For rhododendrol, orange acidified K2Cr2O7(aq) turns green • For RK, the solution remains orange Examiner Comments • Generally well done. • Oxidation using acidified KMnO 4 will not distinguish the two compounds as both will undergo side-chain oxidation. • The iodoform test also will not distinguish the two compounds as both will give positive tests (RK has a –COCH3 group, and rhododendrol has a –CH(OH)CH3 group at the end of the chain). • Students should give the observations for both compounds. (d)(i) The reaction is acidic hydrolysis. • Gardenol is an ester • B is a carboxylic acid • C is an alcohol/phenol Examiner Comments • Generally well done. • Many students did not name the functional group in gardenol, but were not penalized.

29 (d)(ii) Gardenol has formula C10H12O2 C:H ratio ≈ 1:1  benzene ring present Gardenol reacts with acidified KMnO4 to give B and D Gardenol undergoes oxidation and acidic hydrolysis. • C undergoes oxidation to form D • C is not phenol OR C is an alcohol • D is a ketone / carboxylic acid C reacts with acidified KMnO4 to give D C undergoes side-chain oxidation • D is benzoic acid CH3COOH B C D gardenol Examiner Comments • Although many students were able to identify B from its chemical formula, the rest of the structures proved challenging. • Students should use the information already deduced in part (i) to help in the elucidation without having to repeat those deductions ( i.e. the hydrolysis reaction or the acid - carbonate reaction). • Since gardenol is C 10H12O2 and the hydrolysis reaction involves H2O (opposite of condensation), given that B is C 2H4O2, the chemical formula of C is C 8H10O. Understanding that C is an alcohol w ith a chiral centre, there is only one possible structure for C. • Many students forgot that C will undergo side-chain oxidation to form benzoic a