2023 ACJC H2 Chem Prelim Paper 2 - Suggested Solutions

1 © ACJC2023 9729/02/Prelim/2023 [Turn over Anglo-Chinese Junior College JC2 Preliminary Examinations Higher 2 CANDIDATE NAME SUGGESTED SOLUTIONS FORM CLASS TUTORIAL CLASS INDEX NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 22 August 2023 2 hours READ THESE INSTRUCTIONS FIRST Write your index number and name in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams, graphs or rough working. 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. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiners’ use only 1 / 15 2 / 9 3 / 21 4 / 14 5 / 16 Total / 75 This document consists of 19 printed pages and 1 blank page.

2 © ACJC2023 9729/02/Prelim/2023 [Turn over 1 This question examines the chemistry of Group 17 elements and their halides. (a) A2 and B2 are halogens, and they are known to be more soluble in organic solvents. In an experiment, excess A2 was mixed with Na2S2O3(aq). When cyclohexane was added, two immiscible layers were observed which were later separated using a separatory funnel. An orange-red organic layer was obtained, and the aqueous layer was divided into two portions. To one portion of the aqueous layer, a solution of Ba(NO3)2 was added and a white precipitate formed which is insoluble in excess dilute nitric acid. To another portion of the aqueous layer, B2(aq) was added and shaken. When CHCl3 was added, a purple organic layer was obtained. (i) State the identities of A2 and B2. A2: ………………………… B2: ………………………… [1] (ii) Write a balanced ionic equation for the reaction between Na2S2O3 and A2. …………………………………………………………………………………………. [1] (iii) Explain why halogens are more soluble in organic solvents than in water, in terms of the energy changes involved. ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ………………………………………………………………………………………….. [2] (b) In another reaction, halogens can react with alkanes to form mono-substituted products. An example is the reaction of chlorine with 2-methylpentane. (i) Alkanes are generally unreactive. Explain why this is so. ………………………………………………………………………………………….. [1] Bromine Iodine 4Br2(l) + S2O32–(aq) + 5H2O(l) → 2SO42–(aq) + 8Br–(aq) + 10H+(aq) The energy evolved from forming id-id interactions btw halogen and organic solvent molecules is able to compensate the energy that is needed to overocome the id-id interactions btw halogen molecules and id-id interactions btw organic solvent molecules. Hence, it is energetically favourable. Water is polar, energy released from forming id -id interactions btw water molecules and halogens is insufficient to overcome the energy needed to break the stronger hydrogen bonds btw water molecules. The C‒H bonds are non-polar and very strong.

3 © ACJC2023 9729/02/Prelim/2023 [Turn over (ii) This reaction is seldom used for synthesis as there are many associated problems. Firstly, several isomeric products are formed. The relative ratio of the isomeric products may be more accurately determined if relative rates of abstraction of H atoms are considered. The relative rates of abstraction of H atoms are shown in Table 1.1. Table 1.1 type of H atoms relative rate of abstraction primary 1 secondary 4 tertiary 6 By examining the difference in stability of the intermediates formed when different types of H atom are abstracted, explain the trend in the relative rate of abstraction. ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ………………………………………………………………………………………….. [2] (iii) Predict the ratio of the following two products X and Y, from the reaction of chlorine with 2-methylpentane, taking into account the relative rates of abstraction given in Table 1.1. Explain your reasoning. [2] Tertiary radical is the most stable followed by secondary radical followed by primary radical. The tertiary radical is the most stable as it contains the most electron donating alkyl groups which help to stabilise the electron -deficient radical centre. Hence, it is abstracted more easily/faster. 6 possible (primary) hydrogens (Ha) can be substituted to form X 1 possible (tertiary) hydrogen (Hb) can be substituted to form Y Assuming equal probability of abstraction, Ratio of X:Y = 6:1 However, since a tertiary H is abstracted 6 times faster than a primary H, Ratio of X:Y = 6:1 x 6 = 1:1

4 © ACJC2023 9729/02/Prelim/2023 [Turn over (iv) Describe how you can distinguish between compounds X and Y in (b)(iii) by chemical means. ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ……………………………………………………………………………………………... ………………………………………………………………………………………….. [2] (v) Another problem of the reaction is poly-substitution. Suggest the condition that will give rise to formation of poly-substituted products. ………………………………………………………………………………………….. [1] Add NaOH(aq) to both compounds and heat (to form alcohols) followed by KMnO4, NaOH(aq) and heat Observation for X: Purple KMnO 4 is decolourised and brown ppt of MnO 2 is formed. Observation for Y: Purple KMnO4 remains Excess chlorine or limited alkane

5 © ACJC2023 9729/02/Prelim/2023 [Turn over (c) Hydrogen halides are dissociated at high temperatures according to the following equation: 2HX(g) H2(g) + X2(g) The approximate Kc values for the above equilibrium at various temperatures for the respective hydrogen halides are shown in Table 1.2. Table 1.2 temperature / oC Kc values for dissociation of HX HCl HBr HI 800 10−13 10−9 10−5 1000 10−10 10−7 10−4 1200 10−9 10−5 10−3 1400 10−7 10−4 10−2 Using the information in Table 1.2 and relevant data from the Data Booklet, describe and explain the relative thermal stability of the hydrogen halides. ……………………………………………………………………………………………………… ……………………………………………………………………………………………………... ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ………………………………………………………………………………………………….. [3] [Total: 15] From the table, it is observed that at each temperature, decreasing order of Kc value is HI > HBr > HCl. Since the larger the Kc values, the higher the degree of dissociation of HX, the order of degree of dissociation is HI > HBr > HCl. From the Data Booklet, based on decreasing bond strength HCl > HBr > HI as shown: H – C