DHS H1 CHEM P2 Suggested Solutions

This document consists of 18 printed pages and 0 blank page. © DHS 2018 [Turn over Suggested Solutions DUNMAN HIGH SCHOOL Preliminary Examination 2018 Year 6 H1 CHEMISTRY 8873/02 Paper 2 Structured Questions 13 September 2018 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your index number and name on all the work you hand in. 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. Section A Answer all the questions. Section B Answer one question. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use P1 P2 Section A P2 Section B Total % 30 Q1 Q2 Q3 Q4 Q5 Q6 110 13 9 20 18 20 20

2 © DHS 2018 8873/02 Section A Answer all the questions from this section, in the spaces provided. 1 (a) Group 1 elements are strong reducing agents. State and explain the trend of reducing power of Group 1 elements down the Group. [3] Reducing power of Group 1 elements increases down the group. Number of filled principle quantum shell increases and valence electrons are increasingly further away from the nucleus Valence electrons are less strongly attracted to the nucleus and smaller amount of energy is needed to remove the valence electron so it is more easily oxidised. (b) Group 1 elements and its compounds have m any uses. Sodium hydrogencarbonate (NaHCO3) is often used to treat ant stings which contains methanoic acid (HCOOH), a weak acid. Write an equation for the dissociation of methanoic acid in water. Indicate which species are the acid, the base and their conjugate pairs in the reaction. [2] HCOOH + H2O ⇌ HCOO– + H3O+ Acid: HCOOH Conjugate base: HCOO– Base: H2O Conjugate acid: H3O+ (c) When the ant bites, it injects a solution containing 50 % by volume of unionised methanoic acid. A typical ant may inject around 6.0 × 10–3 cm3 of this solution. (i) Given that the density of methanoic acid is 1.2 g cm –3, what is the amount (in moles) of methanoic acid that a typical ant injects? [1] volume of HCOOH = 6.0 x 10–3 x 0.5 = 3.0 x 10–3 cm3 mass of HCOOH = 3.0 x 10–3 x 1.2 = 0.00360 g no of moles of HCOOH = 0.00360 46.0 = 7.83 x 10–5 mol (3 s.f.) As soon as the methanoic acid is injected, it dissolves in water in the body to produce a solution of methanoic acid with pH 2.43. (ii) Assuming that it dissolves fully in 1.0 cm 3 of water in the body, calculate the concentration of the methanoic acid solution that is formed initially. You may ignore the volume of methanoic acid injected in this calculation. [1]

3 © DHS 2018 8873/02 [Turn Over [HCOOH] = 7.826 x 10-5 0.001 = 7.83 x 10–2 mol dm–3 (3 s.f.) (iii) Calculate the percentage of methanoic acid molecules which have dissociated in 1.0 cm3 of water. [2] [H+] = 10–2.43 = 3.7154 x 10–3 mol dm–3 (5 s.f.) percentage = [H+] ሾHCOOHሿ x 100% = 3.72 x 10-3 7.826 x 10-2 x 100% = 4.75 % (3 s.f.) (d) A student was given an unlabelled bottle and was told that it contained pure sample of one of the following three compounds. • methanol • methanal • propanone (i) State the reagent used for a chemical test that could show that the sample can be either methanal and propanone but not methanol. Describe what would be observed. [2] Reagent: 2,4–DNPH Observations: orange precipitate observed. (ii) Describe one chemical test that could distinguish between methanal and propanone. [2] Add Tollens’ reagent and warm to both samples. Only methanal will form black/ grey solid or silver mirror OR Add Fehling’s solution and warm to both samples. Only methanal will form reddish– brown precipitate. OR Add alkaline aqueous iodine and warm to both samples. Only propanone will form yellow precipitate. [Total: 13]

4 © DHS 2018 8873/02 2 A sequence of reactions starting from compound A is shown below. (a) Draw the structures of compounds A, B, C, D, E and F in the boxes below. [6] O CN OH CN Cl COOH Cl COOH ClOH OH COO - Na + ClNa + O - Na + O - OH OH COOHOH OH C O O CH2CH2CH3 OH OH Br Br COOH HCN with trace amount of NaOH (aq) cold (10 °C to 20 °C) I PCl5 II HCl (aq) heat under refluxIII IV Na VI V CH3CH2CH2OH conc H2SO4 heat under reflux VII Br2 (l)VIII AB C D E F (b) For the reaction scheme shown above, state (i) the type of reaction occurring in reaction I. [1] (Nucleophilic) Addition (ii) the reagents and conditions for reaction V. [1] KOH in alcohol, heat under reflux

5 © DHS 2018 8873/02 [Turn Over (c) The alcohol CH 3CH2CH2OH, used as the reagent in reaction VII above, can be converted into CH3CH=CH2. How may this conversion be achieved in a laboratory? [1] Add excess concentrated H2SO4 to a sample of CH3CH2CH2OH and heat. [Total: 9] 3 (a) Pure hydrogen peroxide, H 2O2, was long believed to be unstable. Its decomposition follows a first order reaction. H2O2 → H2O + ½O2 (i) Sketch a graph of [H 2O2] against time to show that the reaction is first order with respect to H2O2. [2] (ii) As H 2O2 decomposes slowly at room temperature, catalysts such as platinum metal are often added to lower the activation energy to increase the rate of reaction. With reference to the information provided below, sketch the energy profile diagram showing the catalysed reaction only. enthalpy change of the decomposition –196 kJ mol –1 activation energy (without catalyst) +75 kJ mol–1 activation energy (with catalyst) +49 kJ mol–1 [2] [H2O2] / mol dm−3 time/ s x ½ x ¼ x t1/2 t1/2’ t1/2 = t1/2’ 0

6 © DHS 2018 8873/02 (b) In the presence of UV light, H2O2 decomposes to form hydroxyl free radicals, •OH. (i) Draw the ‘dot-and-cross’ diagram for H2O2. [1] (ii) Using relevant bond energy values from the Data Booklet, suggest the relative rate of the formation of •C l from chlorine gas as compared that of •OH from hydrogen peroxide. [2] From the Data Booklet, Bond energy of O–O = 150 kJ mol–1 Bond energy of Cl–Cl = 244 kJ mol–1 Ease of cleavage of bond: O–O > Cl–Cl Rate of formation of •OH radicals is faster than •Cl. (iii) Propylamine, CH 3CH2CH2NH2, may be formed via the following reaction pathway involving free radical substitution in the first step. CH 3CH2CH2Cl CH3CH2CH3 CH 3CH2CH2NH2 Name the type of reaction occurring in Step II. State the reagent and conditions required for this step. [2] Type of reaction: (Nucleophilic) Substitution Reagent: NH3 in excess Condition: in ethanol, heat in sealed tube Step I Step II Cl2, uv Energy / kJ mol–1 Progress of reaction H2O2 H2 + ½O2 +49 kJ mol–1 –196 kJ mol–1

7 © DHS 2018 8873/02 [Turn Over (iv) As the yield from the reaction in (b)(iii) is low, propose a 3-step synthetic route to produce propylamine from ethene instead. Show clearly the reagents and conditions as well as the intermediates involved. [3] CH2CH2 CH3CH2Cl CH3CH2CN CH3CH2CH2NH2 HCl(g) alcoholic KCN