2024 Prelims DHS H2 Chem P2 (Ans)

© DHS 2024 9729/02 [Turn over Suggested Solutions DUNMAN HIGH SCHOOL Preliminary Examination Year 6 H2 CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 11 September 2024 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. The use of an approved scientific calculator is expected, where appropriate. You may lose marks if you do not show your working or if you do not use appropriate units. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 11 2 12 3 11 4 16 5 25 Total 75 This document consists of 19 printed pages.

2 © DHS 2024 9729/02 1 (a) The concentration of aluminium ions in a water sample can be determined accurately by titrating it with a solution of EDTA4–. The representative balanced equation for the reaction between aluminium ions and EDTA4– is shown below. Al3+(aq) + EDTA4–(aq) → [Al(EDTA)]–(aq) A 10.0 cm3 water sample is transferred to a 250 cm3 volumetric flask and made up to the mark with deionised water. An aliquot of 25.0 cm3 is pipetted and titrated with 0.0200 mol dm –3 EDTA4–. 35.45 cm3 of this EDTA4– solution is required for complete reaction with the aluminium ions. (i) Calculate the amount of EDTA4– that reacted with the aluminium ions. [1] No. of moles of EDTA4– = 0.0200 × 35.45 1000 = 7.09 × 10–4 mol (ii) Calculate the amount of aluminium ions in the water sample. [1] No. of moles of Al3+ in 25.0cm3 = 7.09 × 10–4 mol No. of moles of Al3+ in 250cm3 = 7.09 × 10–4 × 250 25 = 7.09 × 10–3 mol (b) The process of anodising aluminium increases its resistance to wear and corrosion. (i) State the two half -equations in the anodising of aluminium and the overall equation to form the protective layer. [2] At the cathode: 2H+(aq) + 2e– → H2(g) At the anode: 2H2O(l) → O2(g) + 4H+(aq) + 4e– The O2(g) evolved at the anode then reacts with Al metal to form Al2O3(s): 4Al(s) + 3O2(g) → 2Al2O3(s) (ii) Draw a labelled diagram to show the cell set-up used to anodise aluminium. Include details of the cathode, anode and electrolyte. [1]

3 © DHS 2024 9729/02 [Turn over (c) Aluminium chloride is used extensively as a Lewis acid catalyst in organic reactions with the notable example of Friedel–Crafts alkylation of arenes. (i) State the type of reaction occurring in Friedel–Crafts alkylation. [1] Electrophilic substitution (ii) In the presence of aluminium chloride, phenylamine does not form 2–methylphenylamine but forms neutral compound B instead. Suggest the structure of B and hence, explain why 2–methylphenylamine is not formed as expected. NH2 + AlCl3 + CH3Cl NH2 CH3 B 2-methylphenylamine [3] B is Cl Al - N + Cl Cl H H Phenylamine will form an adduct with aluminium chloride as there exists vacant low-lying orbitals in aluminium that can accept the lone pair of electrons from N atom of phenylamine. Hence, aluminium chloride cannot function as a catalyst to generate the methyl carbocation eletrophile to form 2– methylphenylamine. (d) Using relevant data from the Data Booklet, explain whether fluorine or chlorine will have greater reactivity with aluminium. [2] From the Data Booklet, F2 + 2e– ⇌ 2F– E = +2.87 V Cl2 + 2e– ⇌ 2Cl– E = +1.36 V Since fluorine has a more positive E value, fluorine is a stronger oxidising agent than chlorine. Hence, fluorine shows greater reactivity towards aluminium than chlorine would to aluminium. [Total: 11]

4 © DHS 2024 9729/02 2 Carbon, nitrogen and oxygen are pivotal elements that form a vast array of organic compounds. (a) Explain the following observations. • Nitrogen has a smaller atomic radius than carbon. • Oxygen has a lower first ionisation energy than nitrogen. [3] Atomic radius of N vs C: Nitrogen has a higher nuclear charge but comparable screening effect to carbon. Hence, nitrogen has a higher effective nuclear charge. There is a greater electrostatic force of attraction between the nucleus and valence electrons, hence the electrons pulled closer to the nucleus. 1st I.E. of N vs O: The paired 2p electrons in oxygen experiences inter-electronic repulsion. Hence, less energy is required to remove the valence electron from oxygen as compared to the unpaired 2p electron in nitrogen. Carboxylic acids, esters and ketones are examples of organic compounds containing carbon and oxygen atoms. (b) Table 2.1 shows a list of organic compounds and the p Ka values of their carboxylic acid group. Table 2.1 name structure pKa value benzoic acid O OH 4.2 4–hydroxybenzoic acid O OH OH 4.6 2–hydroxybenzoic acid O OH OH 4.1 (i) With reference to the carboxylate anion, explain why carboxylic acids are generally stronger acids than alcohols. [1] The negative charge on the carboxylate anion is delocalised over 2 electronegative oxygen atoms. This reduces the intensity of the negative charge and makes the carboxylate anion more stable than the alkoxide anion. Hence, carboxylic acids will dissociate to a greater extent to give H +, making them stronger acids than alcohols.

5 © DHS 2024 9729/02 [Turn over (ii) Suggest a reason why • 4–hydroxybenzoic acid has a higher pKa value than benzoic acid, • 2–hydroxybenzoic acid has a lower pKa value than 4–hydroxybenzoic acid. [3] 4–hydroxybenzoic acid vs benzoic acid The lone pair of electrons on the oxygen of the phenol group delocalises into the benzene ring, intensifying the negative charge of the conjugate base of 4– hydroxybenzoic acid. This makes it less stable than the benzoate anion. Hence, 4–hydroxybenzoic acid dissociates to a lesser extent and is a weaker acid than benzoic acid. 2–hydroxybenzoic acid vs 4–hydroxybenzoic acid Intramolecular hydrogen bonding exists between the negatively charged oxygen of the carboxylate group and the adjacent phenol group in the conjugate base of 2 –hydroxybenzoic acid. This reduces the intensity of the negative charge and makes the conjugate base more stable. O O- O H hydrogen bonding +− Hence, 2 –hydroxybenzoic acid dissociates to a greater extent and is a stronger acid. (iii) Carboxylic acids react with Group 2 elements to give an effervescence of hydrogen gas. State the role of the Group 2 elements in this reaction and describe how the reactivity of the Group 2 elements in this reaction will vary down the group. [1] Role: Reducing agent Reactivity: Reactivity increases down the group (c) Compound W contains an ester functional group. O O W W was heated in an aqueous solution containing KMnO 4 and H2SO4. Two organic products, compounds X and Y, were isolated. Table 2.2 contains information about these two products.

6 © DHS 2024 9729/02 Table 2.2 X Has poor solubility in dilute NaOH. Y Dissolves readily in dilute NaOH to give a crystalline solid, compound Z, after removing the solvent. Draw the structures of X, Y and Z. [2] O O OH O O- Na+ X Y Z (d) State a reagent that can be used to distinguish cyclohexanone from ethanoic acid. This reagent should give a positive observation for cyclohexanone. O cyclohexanone Write the equation for the reaction occurring in this chemical test. [2] Reagent 2,4–dinitrophenylhydrazine Observations (not required