2021 RVHS Prelim Paper 3 (Solutions)

River Valley High School 9729/03/PRELIMS/21 2021 Preliminary Examination [Turn over RIVER VALLEY HIGH SCHOOL JC 2 PRELIMINARY EXAMINATION CANDIDATE NAME CLASS 2 0 J CENTRE NUMBER S INDEX NUMBER H2 CHEMISTRY 9729/03 Paper 3 Free Response 21 September 2021 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet 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. 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 the questions. Section B Answer one question. Circle the question number of the question you attempted. 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. For Examiner’s Use Question Number 1 2 3 4 5 s.f. units Total Marks 20 22 18 20 20 80 This document consists of 32 printed pages and 0 blank pages.

2 River Valley High School 9729/03/PRELIMS/21 2021 Preliminary Examination Section A Answer all the questions in this section. 1 In a 2014 paper published in the Journal of Agricultural and Food Chemistry, Hendon and Colonna-Dashwood discovered the effect of water hardness on coffee flavour. Compounds in hard water tend to attach to the flavo urful elements in roasted coffee beans during brewing. Water with higher levels of magnesium will likely extract more flavour from a coffee bean. Water described as "hard" is high in concentration of Total Dissolved Solids (TDS) , specifically calcium and magnesium. The hardness of water may be reported in parts per million (ppm). The solute concentration of a dilute aqueous solution in units of mg dm3 is called parts per million, or ppm. Classification ppm Soft 0  17.1 Slightly hard 17.1  60.0 Moderately hard 60.0  120 Hard 120  180 Very hard > 180 (a) In a sample of Singapore’s tap water, the concentration of magnesium and calcium ions present are found to be 5.97  105 mol dm 3 and 5.49  104 mol dm3 respectively. These two ions can be separated by selective precipitation with potassium hydroxide. The numerical values of solubility product of magnesium hydroxi de and calcium hydroxide at 25 ° C are 1.50  1011 and 5.50  106 respectively. (i) Calculate the total concentration of magnesium and calcium ions in ppm, and hence classify the hardness of water in this sample of tap water. [2] [Mg2+] in ppm = 5.97  105  24.3  1000 = 1.45 ppm [Ca2+] in ppm = 5.49  104 × 40.1 × 1000 = 22.0 ppm Total [Mg2+] and [Ca2+] = 1.45 + 22.0 = 23.5 ppm The sample of water is slightly hard.

3 River Valley High School 9729/03/PRELIMS/21 2021 Preliminary Examination [Turn over (ii) Calculate the minimum pH of the solution at which the magnesium ion precipitates as magnesium hydroxide. [2] Ppt is formed when IP (Mg(OH)2) ≥ Ksp (Mg(OH)2), (5.97 × 105)[OH]2 ≥ 1.5 x 1011 mol3 dm9 [OH] ≥ 5.01 x 104 mol dm3 Minimum pH = 14 – pOH = 14 – 3.30 = 10.7 (iii) The magnesium hydroxide continues to precipitate out of the solution as potassium hydroxide is being added continuously. Eventually, the concentration of the hydroxide becomes high enough to precipitate the calcium ions as well. What is the concentration of magnesium ions when calcium ions begin to precipitate? [2] When IP (Ca(OH)2) = Ksp (Ca(OH)2), (5.49 × 104)[OH]2 = 5.5 x 106 mol3 dm9 [OH] = 0.100 mol dm3 When IP (Mg(OH)2) = Ksp (Mg(OH)2), [Mg2+](0.100)2 = 1.5 x 1011 mol3 dm9 [Mg2+] = 1.50 x 109 mol dm3 (b) A balanced extraction is a well -brewed cup of coffee that is aromatic and rich in flavours. Eugenol is a flavour note with a “woodsy” taste found in coffee, wine and whisky. Eugenol Like other alkenes, it undergoes hydrohalogenation when treated with hydrogen halides. (i) Draw a label led diagram showing the orbital overlap between the carbon atoms C1 and C2 and state the hybridisation involved. Do not include other atoms. [2]

4 River Valley High School 9729/03/PRELIMS/21 2021 Preliminary Examination (ii) Hydrohalogenation of unsymmetrical alkenes results in a mixture of products. In such cases, the major product can be predicted using Markovnikov’s rule. Describe the mechanism of the reaction between eugenol and hydrogen chloride. You may represent eugenol using . [2] (iii) With reference to your mechanism in (b)(ii), explain why the major product is formed. [2] In step 1, the more stable secondary carbocation intermediate is formed instead of a primary carbocation. (More) a lkyl groups exert an electron–donating effect , helping to reduce/ disperse the positive charge on the carbocation, stabilising it. (c) Hydration of alkenes via hydroboration favours formation of the anti –Markovnikov product. The hydroboration reaction involves 2 stages; first with limited borane, BH 3, followed by treatment with alkaline hydrogen peroxide. R = alkyl/aryl group

5 River Valley High School 9729/03/PRELIMS/21 2021 Preliminary Examination [Turn over It is suggested that the mechanism goes through the formation of the intermediate below. C C BH3 (i) The initial reaction between the alkene and borane can be considered to occur in a similar fashion as the hydrohalogenation reaction in (b)(ii). Suggest the mechanism showing the formation of the intermediate given above when eugenol reacts with borane. Indicate clearly the polarity of the B–H bond in borane by drawing + and – on the appropriate atoms. [2] (ii) By determining the change in the oxidation number of the reactive carbon , suggest the role of hydrogen peroxide in step 2 in the reaction. [1] Oxidation number of C bonded to –BH2 changes from 3 to 1. H2O2 acts as an oxidising agent. (d) 3 bottles of halogens are labelled as X2, Y2 and Z2. They are chlo rine, bromine and iodine. The table shows the results of experiments in which the halogens X2, Y2 and Z2 were added to separate solutions containing X–, Y– and Z– ions. X–(aq) Y–(aq) Z–(aq) X2 no reaction no reaction no reaction Y2 X2 formed no reaction Z2 formed Z2 X2 formed no reaction no reaction With reference to the table above, identify the halogens X, Y and Z. Explain your reasoning. [2]

6 River Valley High School 9729/03/PRELIMS/21 2021 Preliminary Examination Y2 displaces (oxidises) both X– and Z– from solution, hence it must be the strongest oxidising agent, Cl2 (or Y– is the weakest reducing agent). X2 cannot displace (oxidise) Y– and Z– from solution, hence it must be the weakest oxidising agent, I2 (or X– is the strongest reducing agent). Therefore, Z2 is Br2. (e) A glass rod was heated in a Bunsen burner flame and placed into a jar of hydrogen chloride gas. The experiment was repeated using a jar of hydrogen iodide gas. A colour change was observed in one of the samples. Using relevant data from Data Booklet, explain these observations. [3] The sample with HI will give a colour change. Bond energy of HCl = 431 kJ mol1 Bond energy of HI = 299 kJ mol1 The thermal stability of hydrogen halides decreases down the Group due to decreasing H –X bond energy. The I atom is larger than the C l atom and its valence orbitals are more diffuse. This results in less effective overlap between the small H atom and the larger I atom and less energy is required to break the weaker H–I bond, forming H2 and (violet fumes)