2025 DHS H2 Chem Prelim Paper 3 Suggested Solutions

© DHS 2025 9729/03 [Turn over DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 CHEMISTRY Paper 3 Free Response Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/03 25 September 2025 2 hours READ THESE INSTRUCTIONS FIRST Write your centre number, index number, name and class at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. 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. Suggested Solutions For Examiner’s Use Section A 1 20 2 20 3 20 Section B 4 / 5 20 Total 80 This document consists of 22 printed pages.

2 © DHS 2025 9729/03 Section A Answer all the questions in this section. 1 (a) The order of increasing Br Ønsted-Lowry acidity of cyclohexanol, benzoic acid and 2−chlorobenzoic acid is shown. Explain this order. OH cyclohexanol CO2H benzoic acid CO2H Cl 2-chlorobenzoic acid < < [3] Cyclohexanol is the least acidic because its conjugate base, is destabilised by the electron-donating effect of the cyclohexyl/alkyl group, which intensifies the negative charge on the oxygen in the conjugate base. Both 2 -chlorobenzoic acid and benzoic acid are more acidic than cyclohexanol because their conjugate bases are resonance -stabilised by the delocalisation of negative charge over two highly electronegative oxygen atoms. In 2-chlorobenzoic acid, presence of electron withdrawing Cl atom helps to disperse the negative charge, making its conjugate base (2 -chlorobenzoate ion) more stable than benzoate. Hence, 2-chlorobenzoic dissociates to the largest extent and cyclohexanol to the least extent. (b) BF3 acts as a Lewis acid in its reaction with dimethylether, CH 3OCH3, to form product P. (i) Describe how BF3 acts as a Lewis acid in this reaction. [1] The electron deficient B atom in BF3 accepts the (lone) electron pair from O in CH3OCH3. (ii) Draw a structure for P to show the bonding present. [1] F BF F O CH3 CH3 (c) Dimethylether, CH 3OCH3, can be formed from the reaction between carbon monoxide, CO, and hydrogen, H2, as shown by equation 1. equation 1 3CO(g) + 3H2(g) ⇌ CH3OCH3(g) + CO2(g) A mixture of CO and H2 was introduced into a 10 m3 sealed vessel at 500 K. The initial total pressure was 40 atm.

3 © DHS 2025 9729/03 [Turn over After dynamic equilibrium was established at 500 K, the total pressure in the vessel decreased to 28 atm. (i) Write an expression for the equilibrium constant, Kp, for this reaction, and state its units. [1] Kp = PCH3OCH3PCO2 (PCO)3 (PH 2 )3 units: atm−4 OR Pa−4 (ii) Use of the Data Booklet is relevant to this question. The amount of CH 3OCH3 at equilibrium was found to be 732 mol. Show that the equilibrium partial pressure of CH3OCH3 in the sealed vessel is 3 atm. [1] pV = nRT p = nRT/V partial pressure of CH3OCH3 at equilibrium = 732  8.31  500 10 = 304146 Pa = 304146 101325 = 3.0017 atm = 3 atm (shown) OR n = pV/RT total moles of gases at equilibrium = 28  101325  10 8.31 x 500 = 6828.2 mol partial pressure of CH3OCH3 at equilibrium = 732 6828.2 x 28 = 3.0017 atm = 3 atm (shown) (iii) At equilibrium, it was found that 60% of the CO had been converted to the products. Calculate the equilibrium partial pressures of CO and H2 in atm. Hence, determine the value of Kp for the reaction. [3] Let the initial partial pressure of CO be x. 3CO(g) + 3H2(g) ⇌ CH3OCH3(g) CO2(g) initial / atm x 40 – x 0 0 change / atm –0.6x –0.6x +3 +3 equilibrium/ atm 0.4 x 40 – 1.6x 3 3 3CO(g)  1CH3OCH3(g) Change in partial pressure of CO = (3)(3) = 9 atm  0.6x = 9 atm x = 15 atm Equilibrium partial pressure of CO = 0.4x = 0.4 (15)

4 © DHS 2025 9729/03 = 6.0 atm OR (0.4 x) + (40 – 1.6x) + 3 + 3 = 28 x = 15 atm Equilibrium partial pressure of H2 = 28 – 6 – 3 – 3 = 16.0 atm Kp = PCH3OCH3PCO2 (PCO)3 (PH 2 )3 = (3)(3) (6)3(16)3 = 1.02  10−5 atm−4 (iv) The volume of the vessel is reduced from 10 m 3 to 5 m 3 and the system is allowed to reach equilibrium. Explain the effect this will have on the partial pressures of the individual gases and the composition of the reaction mixture. [2] A reduction in volume causes partial pressures of all gas components at equilibrium to double, position of equilibrium shifts to the right, favouring the side with fewer moles of gaseous particles, to reduce the pressure. At new equilibrium, the mixture contains more moles of products, CH 3OCH3 and CO2 and less moles of the reactants, CO and H 2 than that before the reduction in volume. (d) NH3 is commonly used as a nucleophile in organic chemistry. In some instances, it adds to the C=C bond as shown in Fig. 1.1. NH3 + CH2=CH2 CH2(NH2)CH3 NH3 + CH2=CHCN CH2(NH2)CH2CN x Fig. 1.1 (i) Suggest reasons to explain Fig. 1.1. Use concepts of electronegativity and electronic effects in your answer. [2] Ethene is non-polar. There are no electron deficient sites in ethene to attract the NH 3 nucleophile OR NH 3 is a nucleophile and is repelled by the electron -rich C=C bond in ethene. Due to presence of electronegative N / electron -withdrawing -CN group, electron density is withdrawn away from the C=C bond. This causes the carbon of the terminal alkene to be electron deficient, hence susceptible to nucleophilic attack by NH3.

5 © DHS 2025 9729/03 [Turn over Ammonia or primary amines react with aldehydes and ketones to produce imines as shown in Fig. 1.2. The reaction is carried out at carefully controlled pH of between 4 and 5. In Fig. 1.2, R1, R2 and R3 represent alkyl groups or hydrogen atoms. R1 R2 O R1 R2 N R3 + H2O imine + R3NH2 Fig. 1.2 The mechanism for the formation of an imine between a primary amine and propanone is shown in Fig. 1.3. Stage 1: Formation of aminoalcohol O NH2R dipolar intermediate B aminoalcohol N R H OHnucleophilic attack proton transfer Stage 2: Formation of iminium ion intermediate NR H O + H H N + RH + H2O iminium ionintermediate ion C NR H OH H+ removal of water Stage 3: Formation of imine via deprotonation deprotonationN + RH N R + H+ Fig. 1.3

6 © DHS 2025 9729/03 (ii) Complete the mechanism for the formation of B in Fig. 1.3 by • adding curly arrows, a lone pair and a dipole to show how the nucleoph