RI 2021 Prelim P2 Answers

9 © Raffles Institution 2021 9729/02/S/21 2021 H2 Chemistry Prelim Paper 2 – Suggested Solutions General comments • There is a common theme in the examiner comments – it is apparent that many students do not read the question carefully to answer the question. Many students do not give the necessary details dictated by the question. Please read carefully! • The handwriting of some scripts left much to be desired. No marks are awarded for answers which cannot be read. Please write clearly for your own sake. • Please space out the answers. Do not squeeze the drawing of structures into the space of one line. Structures which cannot be clearly seen were not awarded marks. There is sufficient space given for each part. • In the drawing of mechanisms, please show the lone pair of electrons clearly – make them bigger and darker. • Please use ink throughout. Do not use pencil as pencil markings are often unclear and may be deemed as rough work. 1(a)(i) Heat gained by water = 4.18 x 10.0 x 300 = 12540 J Heat gained by copper can = 0.384 x 10 x 250 = 960 J Total heat gained = 12540 + 960 = 13500 J = 13.5 kJ Examiner Comments • This was generally well done. • Some students forgot to account for the heat gained by copper can even though the question clearly stated that consideration. 1(a)(ii) Amt of ester = 0.980 / 74.0 = 0.0132 mol Theoretical heat energy released = 0.0132 x 1592 = 21.1 kJ Examiner Comments • This was generally well done. • Note that there is no need for put negative sign here as the heat energy calculated is for the heat released. • A sign would be necessary for heat change as it could be a positive or negative change. 1(a)(iii) Method 1 - Find % heat loss to surroundings 13.5 kJ (expt) / 21.1 kJ (theoretical) = 64% of heat energy transmitted (36% heat loss) Amt of ethyl ethanoate = 0.948 / 88.0 = 0.010773 mol Heat transferred = 4.18 x 11.5 x 300 + 0.384 x 11.5 x 250 = 15525 J = 15.5 kJ Theoretical heat released = 15525 / 0.64 = 24257 J = 24.3 kJ H = − 24.3 kJ / 0.01077 = −2252 = −2250 kJ mol−1 Method 2 - Find thermal capacity of apparatus OR by proportion Thermal capacity = theoretical energy released / observed temperature change Thermal capacity = 21.1 kJ / 10 K = 2.11 kJ K–1 Theoretical heat produced from combustion = 2.11 kJ K–1 x 11.5 K = 24.3 kJ Amt of ethyl ethanoate = 0.948 / 88.0 = 0.010773 mol H = − 24.3 kJ / 0.01077 = −2252 = −2250 kJ mol−1 Examiner Comments • Similar to part (i), students must account for the heat transfer to both water and the copper can. • It is stated clearly in the question that the percentage heat loss (not absolute heat loss) is the same ac ross both experiments. A number of students failed to consider this important information resulting in loss of marks. • Many students struggled with the calculation of percentage heat loss to determine the theoretical heat transferred during reaction. Firstly, students need to realize that the heat that

10 © Raffles Institution 2021 9729/02/S/21 was transmitted (100% – heat loss%) is that which made it to the calorimeter and observed to raise the temperature. So the calculated q (from mcT) represents 64% of the total heat given out by the reaction. Secondly, to scale any quantity from x% to 100%, we need to divide by x to get 1% and then x 100 for 100%, OR to use simple ratios and take q total/qtransmitted = 100/64, and therefore q total = q transmitted x100/64). Students need to master these basic mathematical manipulation. • As there is no such thing as %loss or %transmission of H, it is not acceptable to calculate those terms. The question specifically hinted at the use of % loss of heat, which is not the same as H. • Quite a few students also made careless mistakes in calculations or did not read the question and make use of the Mr data that has been given . 1(b)(i) Examiner Comments • This question proved to be challenging. • Students should make use of the hints given in the questions as well as the molecular formulae of A and B given to deduce the answers. o The approach should be to first deduce the structure of carboxylic acid from the reaction. It can be deduced from the structure of the product ester that P is . o Since A decolourises aqueous Br2, this implies the presence of C=C double bond. The use of conc. H2SO4 can result in elimination of H2O in P, resulting in the formation of C=C.

11 © Raffles Institution 2021 9729/02/S/21 o Since B is a neutral compound and the number of C has doubled (six carbons up from three carbons in P), B is likely to be an ester and formed through condensation reaction between two molecules of P, catalysed by conc. H2SO4. 1(b)(ii) pKa for P will be smaller. The conjugate base for P CH3CH(OH)COO− is more stable than CH3CH2COO− . The –OH group is electron-withdrawing and disperses the negative charge on O of –COO– to a large extent . Hence, P is a stronger acid than propanoic acid. Examiner Comments • This question allowed incorrect understanding of the effect of –OH group to be surfaced. • A good number of students incorrectly thought that –OH is electron-donating, resulting in the wrong conclusion drawn on relative acidicity. Note the –OH is only “electron-donating” when it is attached to a benzene ring. • –OH in phenol contributes to resonance (which outweighs inductive effect) because the lone pair of electrons on O can be delocalised into the pi electron cloud of benzene ring, resulting in increased electron density of the ring . This is only possible because, there is  overlap between the orbital containing the lone pair electrons of O, and the p -orbital of the neighbouring C of the benzene ring. No such thing happens in P. • Students should note that resonance effect is possible only if resonance structures are allowed, which usually require some sort of  overlap. Students should not overgeneralize –OH groups as “electron donating ”. In fact, –OH groups are electron withdrawing by default (due to electronegative O). Sharing of lone -pair electron density by resonance is a separate phenomenon that is expected only in structures that allow resonance. • Any answer that implies resonance is present in P is not accepted (such as “electrons are spread over 3 electronegative O atoms in P”) • Students should also avoid wasting time on irrelevant points, such as similarities between P and propanoic acid. Both compounds have the COOH group, and gives the –COO- in their conjugate base s which is resonance stabilized. There is thus no need to write about –COO– resonance stabilization at all. 1(b)(iii) [H+] = 10−2.43 = 3.72 x 10−3 mol dm−3 Ka = (3.72 x 10−3)2 / (0.10 – 3.72 x 10−3) = 1.43 x 10–4 mol dm−3 pKa = –log 1.43 x 10−4 = 3.84 Since [H+] is known, there isno need to make assumption here. With assumption made, Ka is 1.38 x 10 −4 mol dm −3 and p Ka is 3.86. This answer was also accepted. Examiner Comments • This calculation was straightforward and well done.

12 © Raffles Institution 2021 9729/02/S/21 1(c)(i) Examiner Comments • The first reaction involving hot KOH(aq) is a hydrolysis reaction, and the ester is hydrolysed to an alcohol (linalool) and a carboxylic acid. However, due to the alkaline conditions, the carboxylic acid reacts with KOH to form the carboxylate salt. • Students should remember to include the correct counter ion (K +) in C, since the identity of the reagent (which provided the counter ion) was known. • When skeletal structures are drawn, the usual practice is to leave out the H atoms, but in the case of compound C, including the H adds clarity to the structure. A small number of students drew a bond without H, which implies ethanoate instead of methanoate. • On the other hand, if non -skeletal structures are drawn (e.g. labelling the central C atom in C), you must