YIJC 2025 Prelim P2 Suggested Solutions (for exchange) H2 Chem

YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME Suggested Solutions CG INDEX NO CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 1 September 2025 2 hours READ THESE INSTRUCTIONS FIRST Write your name, class and index number in the spaces 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. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 28 printed pages. For Examiner’s Use 1 / 18 2 / 24 3 / 17 4 / 16 Penalty units significant figures Overall / 75

2 ©YIJC 9729/02/JC2/PE/2025 [Turn over Answer all the questions in the spaces provided. 1 (a) Compounds A, B and C are shown in order of increasing basicity. Explain this order. C is most basic as the N is attached to an electron donating alkyl group which increases the electron density on the N atom and hence increasing the availability of lone pair of electrons to accept H+. In compound B, there is delocalisation of lone pair of electrons on N atom into the benzene ring. Hence this decreases the availability of lone pair of electrons on the N atom to accept H+, making compound X less basic than C. A is an amide and is neutral because the lone pair of electrons on the N atom is delocalised into the electron -withdrawing C=O group , hence unavailable to accept a H + ion. Hence compound A is the least basic amongst the three compounds. [3] (b) Amides can be found in many drugs such as paracetamol and procainamide. Procainamide can be used for the treatment of cardiac arrhythmias. procainamide Predict the products obtained when procainamide undergoes reaction with hot, dilute H2SO4. and [2]

3 ©YIJC 9729/02/JC2/PE/2025 (c) Compound J can be synthesised by the following route in Fig. 1.1, with all the carbon atoms coming from compound E. Fig. 1.1 • Compound E does not react with NaOH(aq) but reacts with Na to give a gas that extinguishes a lighted splint with a ‘pop’ sound. • Compound H is soluble in dilute HCl and can also be obtained from the reaction of compound K with LiAlH4. K • Compound J is neutral and is a cyclic molecule. (i) Draw the structure of compounds E to H, and J. E F G H J [5] (ii) State the reagents and conditions for steps 2 and 4. step 2 (anhydrous) PCl5 / PCl3 / SOCl2 step 4 excess NH3 in ethanol, heat in sealed tube [2]

4 ©YIJC 9729/02/JC2/PE/2025 [Turn over (d) The compounds responsible for the umami flavour of soy sauce are salts of glutamic acid. glutamic acid Glutamic acid has pKa values of 2.1, 4.1 and 9.5. Draw the structure of the zwitterion. Suggest a pH at which the predominant species of glutamic acid is a zwitterion. pH between 2.1 and 4.1 is accepted (students are to give a pH, not a pH range) [2] (e) A polypeptide contains 9 amino acid residues. It was partially hydrolysed to give a mixture of tripeptides. asp-gly-tyr glu-tyr-lys gly-glu-tyr met-asp-gly tyr-ala-gly Determine the sequence of amino acids that make up the primary structure of the polypeptide. met-asp-gly-tyr-ala-gly-glu-tyr-lys [1]

5 ©YIJC 9729/02/JC2/PE/2025 (f) Halogenoalkanes can react with NH2– to produce amines. A sample that contains only one enantiomer of 2 -bromobutane reacts completely with NH 2– to produce a mixture that does not rotate plane-polarised light. Draw a mechanism for the reaction between NH2– and 2-bromobutane. Include all relevant lone pairs, dipoles, curly arrows and charges. Nucleophilic substitution, SN1 [3] [Total: 18]

6 ©YIJC 9729/02/JC2/PE/2025 [Turn over 2 Citric acid, C6H8O7, is a naturally occurring weak organic acid found in citrus fruits. It has a wide range of applications in the food, cleaning products and healthcare industries. It is triprotic and has the following structure. (a) (i) Citric acid is a Brønsted-Lowry acid. Explain what is meant by this statement. Citric acid is a proton/H⁺ donor. [1] (ii) The dissociation of citric acid in water occurs in three steps. Using H3A as a simplified representation of citric acid, the first dissociation step is as shown: H3A + H2O ⇌ H2A– + H3O+ Write the balanced equation for the second dissociation step of citric acid in water. H2A– + H2O ⇌ HA2– + H3O+ [1] (iii) Identify the two conjugate acid-base pairs in the dissociation step you have written in (a)(ii). acid H2A– conjugate base HA2– base H2O conjugate acid H3O+ [1] (iv) Explain why the carboxylic acid group on citric acid is a stronger Brønsted -Lowry acid than the hydroxyl group. The carboxylate ion is more stable than the alkoxide ion as the negative charge on the O atom of the carboxylate ion is more effectively delocalised between the two electronegative oxygen atoms. [1]

7 ©YIJC 9729/02/JC2/PE/2025 (b) The pKa values for citric acid are shown in Table 2.1. Table 2.1 pK1 pK2 pK3 citric acid 3.1 4.8 6.4 (i) Calculate the pH of 0.10 mol dm−3 citric acid at 298 K (ignore the effect of pK2 and pK3 on the pH). Show your working. K1 = 10−3.1 = 0.00079433 [H3O+] = √K1 × c = √0.00079433 × 0.1 = 0.0089125 mol dm−3 pH = − log 0.0089125 = 2.05 [2] (ii) A buffer solution with a pH of 3.40 is made by adding 50 cm 3 of solution L containing monosodium citrate to 100 cm3 of 0.0200 mol dm−3 citric acid. Calculate the concentration of monosodium citrate in solution L. You may use NaH2A to represent monosodium citrate, and H3A to represent citric acid. Let the conc of monosodium citrate in solution L be x mol dm-3. Amount of monosodium citrate used = 0.05x mol Amount of citric acid used = 0.1 x 0.0200 = 0.002 mol pH = pKa + lg [salt]/[acid] 3.40 = 3.1 + lg [(0.05x ÷ 0.150) / (0.002 ÷ 0.150)] lg (0.05x / 0.002) = 0.3 x = 0.0798 mol dm-3 [3] (iii) Using an equation, explain how the citric acid/monosodium citrate buffer solution in (b)(ii) resists pH changes when a small amount of acid is added to it. C6H7O7– + H3O+ → C6H8O7 + H2O or H2A– + H3O+ → H3A + H2O When a small amount of H3O+ is added, most of the H3O+ ions are removed by C6H7O7– / H2A–. Hence, the [H3O+] in the solution does not increase much / does not change significantly and the pH of the solution is kept relatively constant. [2]

8 ©YIJC 9729/02/JC2/PE/2025 [Turn over (iv) 10 cm3 of 0.100 mol dm−3 citric acid was titrated against 0.100 mol dm−3 sodium hydroxide. The titration curve is shown in Fig. 2.1. Fig. 2.1 Fill in the boxes above with the correct pH values and NaOH volumes. [2] (c) A sample of citric acid is heated with excess ethanol in the presence of a small amount of concentrated sulfuric acid. (i) In the box above, draw the skeletal structure of the organic product formed. [1] (ii) State the type of reaction that has occurred. Condensation [1]