RI 2019 Prelim P3 Ans

© Raffles Institution 2019 9729/03/S/19 1 2019 Y6 H2 Chemistry Preliminary Exams Paper 3 Suggested Solutions Section A 1 (a) (i) Examiner’s Comments • Most students were able to correctly draw the structure of the amine. However, many did not balance the equation with HBr. (ii) The p orbital of the Cl atom overlaps with the  electron cloud of the benzene ring. This allows the lone pair of electrons in the p orbital of the chlorine atom to delocalise into the benzene ring and results in partial double bond character in the C–Cl bond. This bond is strengthened and hence not easily broken. Sterically, nucleophilic attack at the electron -deficient carbon atom from the side directly opposite the Cl atom is blocked by the benzene ring . In addition, the pi - electron cloud of the benzene ring will repel the lone pair of electrons of an incoming nucleophile, rendering attack of the nucleophile difficult. Examiner’s Comments • Many students gave incomplete answers, either omitting the C l p-orbital overlapping with the  electron cloud of benzene or omi tting the delocalisation of electrons. • Students should phrase their answers clearly. For example, it is the C -Cl bond that has partial double bond character and not the Cl atom. • Most students did not discuss the steric hindrance effect of the benzene ring. (iii) Add NaOH(aq) to each sample in a test -tube and heat each mixture in a hot water bath. Cool each mixture and acidify each one with dilute HNO3. Then add AgNO3(aq) to each mixture. For compound H, a pale cream precipitate of AgBr will be observed. For bupropion, no cream precipitate is observed. Alternative answer: Br2(aq) – compound H will decolourise orange Br2(aq) Examiner’s Comments • Most students were able to correctly state the distinguishing test and accompanying observations. Do remember to refer to the Data Booklet for the correct colour of the silver halides. • When describing observations, avoid stating “no observable change” and instead, state the opposite of the positive test. • Students should also take note to specify the solvent used. For example, NaOH(aq), Br2(aq), HNO3(aq). (b) (i) To quench the reaction / to stop the reaction at one minute / to neutralise the unreacted amine from the reaction mixture. Examiner’s Comments • Statements such as “slow down” the reaction, no matter how “drastically” or “significantly”, were not accepted. + HBr

© Raffles Institution 2019 9729/03/S/19 2 (ii) The rate of the reaction is dependent on temperature. Having the two solutions at the same temperature before the start of the reaction will minimise any temperature changes during the mixing of solutions. Examiner’s Comments • This question was poorly done. Many answers did not address the question. • For example, “…so that temperature will not affect the rate of the reaction.” Nothing you can do experimentally will stop temperature from affecting the rate of a reaction. • Another example, “…so that temperature is the only independent variable in the experiment.” This would be correct if the question asked, “Why was the reaction mixture kept in a temperature-controlled water bath?” (iii) At 60 C, the average kinetic energy of the reactant particles increases. As such, significantly 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. Examiner’s Comments • Students must be careful to replicate the shape of the distribution curve exactly. • The T2 curve must peak at a lower maximum than the T1 curve and be broader. • The area under both curves must be about the same, and they should pass through the origin, without intersecting the x axis. • The point that more reactant particles have energy  Ea must be clearly stated or shown in the diagram. (c) Initial rate = − (0.05−0) (0−180) = 2.78 × 10-4 mol dm-3 min-1 Also accepted: 4.63 × 10-6 mol dm-3 s-1 Examiner’s Comments • A range of initial rate values were accepted due to a certain level of subjectivity in drawing a tangent to the curve at t = 0. Values outside of the range were not accepted. • Many students neglected the units or stated incorrect units ( such as s-1 instead of min-1). • Although the gradient is negative, students should note that initial rate is a positive value. Note: T2 > T1 Maxwell-Boltzmann distribution of kinetic energies at two different temperatures. 0 kinetic energy T2 K T1 K Ea (activation energy) total number of particles with energy > Ea at T2 K total number of particles with energy > Ea at T1 K number of particles with a given energy

© Raffles Institution 2019 9729/03/S/19 3 (d) Using experiment 1 and 2: When [triethylamine] is doubled (0.10/0.05), the relative initial rate is also doubled (1/0.5). Hence order of reaction w.r.t. triethylamine is 1. Using experiment 2 and 3: When [PhCOCH2Br] is doubled (0.10/0.05), the relative initial rate is also doubled (1/0.5). Hence order of reaction w.r.t. PhCOCH2Br is 1. The reaction is overall second order. Rate = k[PhCOCH2Br][triethylamine] units of rate constant: mol-1 dm3 min-1 or mol-1 dm3 s-1 Examiner’s Comments • This question was generally well done. However, reasoning must be shown in determining the orders of the reaction. • Also, students should take note of the correct form of the rate equation. (e) Mechanism: SN2 Examiner’s Comments • The name of the mechanism must be stated, as required in the question. Please also note the correct format of writing SN2 – note which are terms are in subscript. • All partial charges and arrows were required for full credit. • Many students omitted the positive charge on the N in the product. • Students should also note that in a SN2 mechanism, the nucleophile should attack the R-X from the “backside” (see above) and not the “front side” (see below). (f) (i) Using PhCOCH2I will result in a higher rate than Experiment 1 because the C-I bond is weaker than the C-Br bond. Examiner’s Comments • Generally well done. However, many students went on to explain why the bond was weaker. This was not necessary since the question was only given 1 mark.

© Raffles Institution 2019 9729/03/S/19 4 (ii) Using compound G, which is a secondary RX, will result in a slower rate than Experiment 1. During the SN2 reaction, the nucleophile attacks the electron-deficient carbon atom from the side directly opposite the halogen atom (i.e. a backside attack). For compound G, this approach and attack is sterically hindered by the methyl group on the C-Br carbon. Examiner’s Comments • Many students were able to identify the difference in Compound G being that this was a secondary RX with a methyl group on the C-Br carbon. • However, a large number of students stated that the “electron -donating methyl group decreased the + partial charge on the C -Br carbon”. This was not accepted. In the SN2 mechanism, the nucleophile must be able to attack the + carbon, i.e., the steric effect is the key factor here. • Note that t he electron donating effect is important in the S N1 mechanism to stabilise the carbocation. • Marks were not awarded for answers which were v ague in describing the position of the methyl group, e.g. “additional methyl group in G” or “methyl group on the carbon atom”. (iii) Using PhCH2CH2Br will result in a slower rate than Experiment 1. Compared to PhCOCH 2Br, (2-bromoethyl)benzene does not have an additional C=O electron withdrawing group on the C -Br carbon. This decreases the intensity of the + charge on the carbon atom, making it less susceptible to nucleophilic attack. -or- The C-Br carbon in PhCOCH2Br is bo