RI 2018 Prelim P3 Ans

© Raffles Institution 2018 9729/03/S/18 1 2018 Y6 H2 Chemistry Preliminary Exams Paper 3 (Suggested Solutions) Section A 1 (a) (i) Comments  Students should follow the pattern given in reaction 1. The OH ⁻ removes a H + on an  carbon (C next to the carbonyl carbon) to form a nucleophile , OHC-CH2⁻, which can take part in nucleophilic addition with an ethanal molecule. In the same way, when ethanal or propanone are used, OH ⁻ can remove H + from the  carbon in ethanal or propanone to form a nucleophile that rea cts with the other or the same carbonyl compound. (ii) Since OH – is a catalyst, [OH –] is constant during each experiment and does not affect the reaction rate. Using Experiment I, First t½ = 4.7 min. Second t½ = 9.5 – 4.7 = 4.8 min. Since both t½ are approximately the same, the reaction is first order w.r.t. ethanal. Comments  Students should show the construction lines on the graph and determine the values of at least two half-lives. It is insufficient to find only one half-life.  Some students read the graph wrongly and obtained a half-life of 4.4 min or 4.8 s.  Even though either graph can be used to show that reaction is first order with respect to ethanal, the question specified for students to use the graph for expt I.

© Raffles Institution 2018 9729/03/S/18 2 (iii) By drawing tangents at t = 0 min, For Experiment I, initial rate = | 1.20 - 0 0 - 6.8 | = 0.176 mol dm–3 min–1 For Experiment II, initial rate = | 1.20 - 0 0 - 3.4 | = 0.353 mol dm–3 min–1 When [OH–] doubles, rate doubles. Hence, reaction is first order w.r.t. OH–. Comments  The tangent drawn at t=0 should follow the curve for about 2 to 3 small squares . This will better allow students to get a tangent with a gradient within the acceptable range.  By convention, rate is always positive. Hence, initial rate = − d[rxt] dt . (iv) rate = k[OH–][ethanal] Comments  Very well done.  Only a few students repeated the mistake of writing ‘rate equation = k[OH⁻][ethanal]’. (v) Initial rate method Using Experiment I, 0.1765 = k(1.0)(1.2)  k = 0.147 mol–1 dm3 min–1 OR Half-life method t½ = ln 2 k [OH-] Using Experiment I, 4.75 = ln 2 k (1.0)  k = 0.146 mol–1 dm3 min–1

© Raffles Institution 2018 9729/03/S/18 3 Comments  Many students did not realise that the unit for time in this experiment was min. Some converted min to seconds, which is fine, but not necessary.  Using the half -life method, many students left out [OH ⁻] but still obtained the correct answer because in this case, [OH⁻] = 1.0. (b) (i) Mechanism: SN2 (or bimolecular nucleophilic substitution) Let RCH2− represent A. The mechanism should show the following:  curly arrows to show electron movement  lone pair on C of nucleophile  dipoles on C and Br  inversion of configuration  all partial charges / charges Comments  Many stud