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 students mistakenly left out the name of the mechanism.  The question stated that A reacted with an optically active sample of 2-bromobutane to form the product which can rotate plane polarised light . Hence, 3D structure must be shown to illustrate the b ackside attack by the nucleophile and the inversion of configuration of the chiral carbon for this SN2 mechanism. (ii) An increase in temperature from T 1 to T2 increases the average kinetic energy of the reactant particles. More reactant particles have energy greater than or equal to the activation energy of the reaction. This results in an increase in effective collision frequency and hence an increase in the rate of the reaction. number of parti cles with a given energy kinetic energy Ea 0 T1 T2 total no. of particles with energy Ea at temperature T1 total no. of particles with energy Ea at temperature T2 T2 > T1

© Raffles Institution 2018 9729/03/S/18 4 Comments  Students need to be familiar with the drawing of the Boltzmann distribution curve s to explain the effect of temperature and/or catalyst on the rate of reaction.  For T2 > T1, the maxima of the curve for T 2 has to be lower than and to the right side of that for T1 (i.e. as the temperature increases, the peak moves to higher kinetic energy and the distribution broadens out). This is because t he t otal area under the curve (which represents the total number of particles) must remain the same.  Students should also note the correct shape o f the curves and that the curves must start from the origin. (iii) The C–Cl bond (BE = 340 kJ mol –1) is stronger than the C –Br bond (BE = 280 kJ mol–1). Hence it is more difficult to break the C –Cl bond, resulting in a slower reaction for 2-chlorobutane. Comments  Most students did well for this question . A handful of students used atomic radii to substantiate their answer. The use of bond energies is preferred here as it will be a more direct way to show which bond is stronger. (iv) Hydroxide ions are less bulky and will experience less steric hindrance when attacking the electron deficient carbon atom. OR In A, the negative charge is delocalised into the adjacent C=O group. Hence, lone pair on A is less available, resulting in a slower reaction. Comments  The electronegativity of an atom in a molecule is a relative measure of its ability to attract bonding electrons. Therefore, it is contradictory to state that the O atom on OH ⁻ is electronegative and hence its lone pair is more available for donation. (c) (i) Comments  Students should note that the final product D has one less carbon than B. Also, D has an ester group which can be formed from the condensation reaction between an alcohol and a dcarboxylic acid. Hence the methyl ketone in B can be converted to a carboxylate (reduction of one carbon) via the iodoform reaction. (ii) Step 1: I2(aq) with NaOH(aq), heat or warm Step 2: (catalytic) conc H2SO4, heat Comments  For step 2, a common mistake was to use H2SO4 (aq), which is dilute H2SO4. Note that excess conc H2SO4 cannot be used as this will cause the alcohol to under go elimination (dehydration) to form an alkene instead.

© Raffles Institution 2018 9729/03/S/18 5 2 (a) (i) PbF2, PbCl2 and PbBr2 have giant ionic structure with strong ionic bonds. PbF2, PbCl2 and PbBr2 have the same cation and the anions have the same charge. Since the ionic radius increases from F– to Br–, the interionic distance increases from PbF2 to PbBr 2. Hence the strength of ionic bond and melting point decreases from PbF2 to PbBr2. Or since |LE|  qq rr   and the ionic radius increases from F – to Br –, the LE become less exothermic , less energy i s required to overcome the ionic bond and melting point decreases from PbF2 to PbBr2. Comments  Many students forgot to state the structure of the compounds (i.e. giant ionic structure) although the question asked to explain the melting points in terms of structure and bonding.  Some students thought that lead(II) halide, PbX 2, is covalent. As PbX 2 is made up of a metal ion, Pb 2+ (which does not have high charge density) and non−metal halide ion, PbX2 is ionic. The high melting point of PbX 2 also shows that it is ionic rather than covalent with simple molecular structure.  Other common mistakes include: o Using decrease in charge density to explain why LE becomes less exothermic from PbF2 to PbBr2. As |LE|  qq rr   , it is the increasing inter -ionic distance (due to increasing anionic radius) that should be used to explain why LE become less exothermic. o Using increase in covalent character to explain the decrease in melting point. This is incorrect as melting involves overcoming electrostatic attraction between Pb 2+ and X− ions, which is dependent on the ionic bond strength and hence LE.