HCI 2021 Prelim Paper 2 Questions

This document consists of 18 printed pages. HWA CHONG INSTITUTION C2 Preliminary Examinations Higher 2 CANDIDATE NAME CT GROUP 20S CENTRE NUMBER INDEX NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 25 August 2021 2 hours READ THESE INSTRUCTIONS FIRST Write your name, Centre number, index number and CT group clearly in the spaces at the top of the page. Write in dark blue or black pen. You may use a HB pencil for any diagrams or graphs. 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 Examiner’s Use 1 / 16 2 / 18 3 / 16 4 / 25 Deductions s.f. units Total / 75

2 © Hwa Chong Institution 2021 9729/02/C2Prelim 2021 [Turn over Answer all the questions in the spaces provided. 1 Hypoglycin A occurs naturally in fruits such as lychees and longans. In 2017, scientists advised against over-consuming lychees on an empty stomach after discovering its role in a mysterious recurring outbreak of acute neurological illness since 1995 in Bihar, the largest lychee cultivation area in India. hypoglycin A, C7H11O2N For Examiner's use (a) (i) Hypoglycin A contains a carboxylic acid functional group. Explain the acidity of the carboxylic acid functional group in terms of its structure. ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………[2] (ii) Identify as fully as you can the other functional groups present in hypoglycin A. ……………………………………………………………………………………………… ……………………………………………………………………………………………[2] (iii) Explain whether hypoglycin A can show cis-trans isomerism. ……………………………………………………………………………………………… ……………………………………………………………………………………………[1]

3 © Hwa Chong Institution 2021 9729/02/C2Prelim 2021 [Turn over (b) Draw the structures of the major organic products, P, Q and R, when hypoglycin A reacts with each of the following reagents. For Examiner's use [3] (c) Hypoglycin A can be synthesised from a 3-stage process. (i) In the journal article where this synthesis was first published, stage I was described as a “condensation” reaction. Identify the small molecule that was produced. ……………………………………………………………………………………………[1] (ii) Suggest the type of reaction that occurs in stage II. ……………………………………………………………………………………………[1] (iii) In stage III, one of the two –CO2C2H5 groups in the intermediate compound is lost. State the type of reaction the other –CO2C2H5 group undergoes to obtain hypoglycin A. ……………………………………………………………………………………………[1]

4 © Hwa Chong Institution 2021 9729/02/C2Prelim 2021 [Turn over (iv) A by-product of stage III is methanoic acid, HCO2H. Draw a dot-and-cross diagram of a molecule of methanoic acid. [1] For Examiner's use (d) The dipeptide, hypoglycin B, is also found in lychees and possesses the same toxic effects as hypoglycin A. Suggest reagents and conditions to hydrolyse hypoglycin B in the laboratory and draw the organic products formed. reagents and conditions: ……………………………………………………………………… organic products formed: [3] (e) The toxicity of hypoglycin A and B arises when they metabolise in the body to form a compound, MCPA, which causes acute hypoglycemia leading to coma and fatality in the outbreaks. x MCPA retains the group and has a molecular formula of C6H8O2. x It has only one chiral centre and gives effervescence with NaHCO3(aq). x No carbon atom in MCPA is bonded to four other carbon atoms. Suggest a structure for MCPA. [1] [Total: 16]

5 © Hwa Chong Institution 2021 9729/02/C2Prelim 2021 [Turn over 2 (a) Among the elements of Group 14, those towards the top, carbon to germanium, have very different properties from those at the bottom, tin and lead. For example, the melting points show a marked change after germanium. element C Si Ge Sn Pb mp / °C >3550 1410 937 232 327 Suggest, in terms of structure and bonding, why carbon has a higher melting point than tin. For Examiner's use ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. …………………………………………………………………………………………………[3] Benzene can be converted into phenylamine in 3 steps. (b) (i) State the role of H2SO4 in the formation of NO2+ during step 1. ……………………………………………………………………………………………[1] (ii) Identify the type of mechanism involved in step 1. ……………………………………………………………………………………………[1] As with most organic reactions, the yield of steps 1 and 2 is less than 100%, resulting in a mixture of reactants and products after each step. Purification is carried out to separate the products from the reactants before each subsequent step can be carried out. Table 2.1 gives the melting and boiling points of benzene and nitrobenzene. Table 2.1 compound melting point / °C boiling point / °C benzene 5.5 80.1 nitrobenzene 5.7 211 (iii) Using the data in Table 2.1, state a physical method that can be used to separate nitrobenzene from benzene after step 1. ……………………………………………………………………………………………[1]

6 © Hwa Chong Institution 2021 9729/02/C2Prelim 2021 [Turn over (c) Explain why nitrobenzene has a higher boiling point than water. For Examiner's use ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. …………………………………………………………………………………………………[2] (d) [SnCl6]2− ions are formed together with phenylammonium ions in step 2. By considering the changes in the oxidation number of tin, construct a half-equation to illustrate the formation of [SnCl6]2− in step 2. …………………………………………………………………………………………………[1] (e) (i) Explain, in terms of their structures, why phenylamine is a weaker base than ethylamine. ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………[2] (ii) Explain why the phenylammonium salt is soluble in water. ……………………………………………………………………………………………… ……………………………………………………………………………………………[1]

7 © Hwa Chong Institution 2021 9729/02/C2Prelim 2021 [Turn over (iii) After step 2, the reaction mixture containing phenylammonium ion, [SnCl6]2− and nitrobenzene is shaken with water in a separating funnel and the two layers are allowed to separate. The layer containing phenylammonium ion and [SnCl6]2− is then transferred into another separating funnel and shaken with NaOH(aq) and organic solvent ethyl acetate, in step 3, to obtain phenylamine. The two layers are then allowed to separate. Phenylamine is more soluble in ethyl acetate than in water. compound density / g cm3 solubility in water nitrobenzene 1.20 insoluble water 1.00 - ethyl acetate 0.90 insoluble In the boxes below, indicate the locations of x phenylammonium ion and nitrobenzene after step 2, x phenylamine and [SnCl6]2− after step 3. For Examiner's use After step 2: