2022 DHS Y6 H2 Prelim Paper 2 Suggested Solutions

© DHS 2022 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 13 September 2022 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 12 2 12 3 14 4 16 5 21 Total 75 This document consists of 18 printed pages.

2 © DHS 2022 9729/02 Answer all the questions in the spaces provided. 1 (a) Describe the thermal decomposition of the hydrogen halides HCl, HBr and H I and explain any variation in their thermal stabilities. [3] Hydrogen halides thermally decompose to give its constituent elements / show equation: 2HX → H2 + X2. HCl does not decompose even on strong heating. HBr decomposes on strong heating while HI decomposes readily in the presence of a hot rod. Thermal stability decreases in the order HC l > HBr > H I as bond energy / bond strength of the H –X bond decreases from HC l to HBr to H I. As such, decreasing amount of energy is required to decompose the hydrogen halide from HCl to HI. (b) Tert-butyl alcohol reacts with hydrogen chloride according to the equation shown. (CH3)3COH + HCl → (CH3)3CCl + H2O This reaction occurs in three steps. step 1 protonation of –OH group in (CH3)3COH to produce (CH3)3COH2+ cation step 2 loss of H2O molecule from (CH3)3COH2+ to produce a carbocation step 3 chloride ion reacts with carbocation to produce (CH3)3CCl (i) Describe the mechanisms which occur in steps 2 and 3. Use curly arrows to show the movement of electrons and label the slow step. [3] CH3 C O + CH3 CH3 H H CH3 C + CH3 CH3 + H2O CH3 C + CH3 CH3 + Cl slow CH3 C CH3 CH3 Cl (ii) An enantiomerically pure alcohol, where the carbon atom bonded to the –OH group is chiral, was used for the reaction in (b). Use your answer in (b)(i) to deduce the stereochemical outcome of this reaction. Explain your reasoning. [2] The reaction will produce a racemic mixture / the product will be formed as a 50 : 50 mixture of enantiomers.

3 © DHS 2022 9729/02 [Turn over The carbocation intermediate is (trigonal) planar around the positively charged C atom. The chloride ion can attack the positively charged C atom from the top and bottom of the plane with equal probability. Tert-butyl alcohol also reacts with solid phosphorus pentachloride, PCl5, to produce (CH3)3CCl. (iii) With the aid of a suitable equation, explain why the reaction is not carried out in aqueous medium. [2] PCl5 + 4H2O → H3PO4 + 5HCl PCl5 hydrolyses in water and will not be available to un dergo nucleophilic substitution with tert-butyl alcohol. (c) (CH3)3CCl is one of the two monochlorinated products of the reaction between an alkane, X, and chlorine gas in the presence of UV light. (i) Draw the structure of the alkane, X, and state the IUPAC name of the other monochlorinated product. [1] Structure of Alkane, X: CH3 C CH3 CH3 H Name of the other monochlorinated product: 1-chloro-2-methylpropane (ii) The rate of formation of (CH3)3CCl is faster than that of the other monochlorinated product. Suggest an explanation for the different rates of reaction. [1] (CH3)3CCl is formed from a tertiary radical intermediate, (CH 3)3C•, which is more stable than the primary radical intermediate, (CH 3)2CHCH2•, that forms the other monochlorinated product, (CH3)2CHCH2Cl. [Total: 12]

4 © DHS 2022 9729/02 2 (a) The following equilibrium occurs when ethanal is mixed with water. G = −0.282 kJ mol–1 Use relevant data from the Data Booklet to calculate the equilibrium constant, K, for the reaction. [2] G = –RT ln K (–0.282 x 103) = –(8.31)(298)ln K ln K = 0.11388 K = 𝑒0.11388 = 1.12 (b) The aldol reaction is a useful reaction that forms a carbon–carbon bond between two carbonyl compounds. For example, two ethanal molecules can be combined using the aldol reaction. The carbonyl carbon, a, of one ethanal molecule forms a covalent bond with a carbon atom, b, of another ethanal molecule. Carbon atom, b, must be adjacent to carbonyl carbon, c. (i) When different carbonyl compounds are used in an aldol reaction, a mixture of structural isomers is formed. Suggest two possible structural isomers that can be formed if propanone, CH3COCH3, and propanal, CH3CH2CHO, are mixed. [2] Any two of the following: (ii) Both propanal and propanoic acid can be formed from propan –1–ol in the same reaction. Describe the reagents and conditions needed to ensure that the reaction yields propanal as the major product. [2]

5 © DHS 2022 9729/02 [Turn over Heat propan–1–ol with K2Cr2O7, H2SO4(aq) with immediate distillation to collect mainly propanal as distillate. (c) Some tin reagents are useful in organic chemistry. Tin forms two chlorides, SnCl2 and SnCl4. (i) A mixture of these chlorides was found to contain 50.0% by mass of tin. Calculate the percentage by mass of SnCl2 in the mixture. [3] Relative formula mass of SnCl2 = 189.7 Relative formula mass of SnCl4 = 260.7 Percentage by mass of Sn in SnCl2 = 118.7 189.7 × 100 = 62.572% Percentage by mass of Sn in SnCl4 = 118.7 260.7 × 100 = 45.531% Let y be the fraction of mass of SnCl2 in the sample. y (62.572 100 )+ (1–y) 45.531 100 = 50 100 y = 0.26229 percentage by mass of SnCl2 in the sample = 0.26229 x 100 = 26.2% (ii) Tin exists in +2 or +4 oxidation states in many of its compounds. Great care must be taken to ensure the correct oxidation state of tin is formed. A student proposed the following preparation methods to prepare the two chlorides, SnCl2 and SnCl4. Preparation method for SnCl2 Heating tin with hydrochloric acid produces hydrogen gas. Careful evaporation of the water and dehydration produces white solid SnCl2. Preparation method for SnCl4 Passing chlorine gas over heated tin produces colourless liquid SnC l4 as the only product. Explain if the preparation methods proposed above are feasible without reference to any calculation. Use relevant standard electrode potentials from the Data Booklet. [3] Sn2+(aq) + 2e– ⇌ Sn(s) E = –0.14V 2H+(aq) + 2e– ⇌ H2(g) E = 0.00V Sn4+(aq) + 2e– ⇌ Sn2+(aq) E = +0.15V The preparation method of SnCl2 proposed is feasible.

6 © DHS 2022 9729/02 EH+/H2 O is more positive than ESn2+/Sn O but less positive than for ESn4+/Sn2+ O . Hence H+ can oxidise Sn to Sn2+ to form SnCl2 but cannot oxidise Sn2+ to Sn4+. Cl2(g) + 2e– ⇌ 2Cl–(aq) E = +1.36V The preparation method of SnCl4 proposed is feasible. ECl2/Cl-O is more positive than both ESn2+/Sn O and ESn4+/Sn2+ O . Hence C l2 can oxidise Sn to Sn4+ to form SnCl4. [Total: 12]

7 © DHS 2022 9729/02 [Turn over 3 (a) Compound N, C3H4O3, liberates a gas when treated with aqueous sodium carbonate. (i) Identify the gas and state the functional group that is present in compound N. [1] Gas is carbon dioxide. Functional gro