2020 MI Prelim Paper 3 mark scheme

Class Adm No Candidate Name: This question paper consists of 28 printed pages and 2 blank page. 2020 End-of-Year Examinations Pre-University 3 H2 CHEMISTRY 9729/03 Paper 3 Free Response 18th Sep 2020 2 hours Additional materials: Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A: Answer all questions Section B: Answer any 1 question A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for good English and clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Section A Section B Total Question 1 2 3 4 5 Marks 25 20 15 20 20 80

2 Section A Answer all questions from this section. 1 Phenylamine, C6H5NH2, otherwise known as aniline, is commonly used in the manufacture of industrial chemicals. The family of amines tend to exhibit strong odours. The more volatile the amine, the stronger the odour. Trimethylamine, (CH 3)3N, is the compound responsibl e for the fishy odour associated with rotting fish. (a) (i) Suggest, with explanation, whether phenylamine or trimethylamine is likely to exhibit a stronger fishy odour. [2] Trimethylamine has a smaller electron cloud size than phenylamine, and thus has weaker instantaneous dipole -induced dipole forces of attraction between its molecules. Less energy is required to overcome them, and hence ; trimethylamine is likely to be more volatile / have a lower boiling point , exhibiting a stronger odour. ; (ii) When phenylamine and trimethylamine are burnt, only phenylamine burns with a smoky flame. Explain the above observation. [1] Phenylamine has a larger number of carbon atoms per molecule than trimethylamine, hence it undergoes incomplete combustion to produce soot / C(s). ; (iii) The reaction between phenylamine and ethanoyl chloride, CH3COCl, takes place rapidly at room temperature, releasing a gas in the process. Write a balanced chemical equation for the reaction between phenylamine and ethanoyl chloride, and identify the functional group formed in the product. [2] C6H5NH2 + CH3COCl → CH3CONHC6H5 + HCl ; Amide ; (iv) A laboratory assistant wants to synthesise the product in (iii). However, ethanoyl chloride is unavailable, so he opts to use ethanoic acid in place of ethanoyl chloride. Explain why the desired product is not obtained. [1] Acid-base reaction takes place between the basic phenylamine and carboxylic acid instead ; or

3 [Turn over The carboxyl carbon in ethanoic acid is not sufficiently electron -deficient / electrophilic enough. ; (v) State the reagents and conditions needed to produce ethanoyl chloride from ethanoic acid. [1] PCl5, r.t. / PCl3, r.t. / SOCl2, r.t. / NaCl(s), conc H2SO4, heat under reflux; (b) Phenylamine can be synthesised from nitrobenzene via many different reactions. One such reaction involves using palladium on carbon (Pd/C) catalyst, at 200–300 °C: (i) Identify the type of reaction carried out above. [1] Reduction. ; (ii) Palladium on carbon (Pd/C) is a fine powder made by depositing palladium metal in the porous cavities of activated charcoal. State the type of catalysis involved, and suggest why palladium metal needs to be in the form of a fine powder. [2] Heterogeneous catalysis ; As a fine powder, Pd/C will have a larger surface area, increasing frequency of effective collisions and hence a faster rate of reaction. ; accept any reasonable answer (iii) A chemist carries out this reaction and obtains a percentage yield of 96%. Suggest why it is difficult to achieve 100% yield. [1] Unwanted side-reactions could have occurred in the synthesis / some reagents or product was lost during the transfer across apparatus / some product was lost during the purification process. accept any reasonable answer (c) The –CH=CH2 group of phenylethene directs incoming substituents to the 3 -position of the benzene ring, while the –NH2 group of phenylamine directs incoming substituents to the 2-position of the benzene ring. This is known as the orienting effect.

4 A student claims that this orienting effect is due to “–CH=CH2 and –NH2 having orbital overlaps with the benzene ring”. Under the right set of reagents and conditions used, the following products are obtained from phenylethene and phenylamine respectively: (i) State the corresponding reagents and conditions for steps I and II. [2] I: conc HNO3, conc H2SO4, heat II: Br2 in CCl4, room temperature (ii) Describe the mechanism of Reaction 1. In your answer, show all relevant lone pairs and curly arrows involved in the mechanism. [4] Electrophilic Substitution ; HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4– ; ;;

5 [Turn over (iii) State the hybridisation for each of the two carbon atoms in the –CH=CH2 substituent of phenylethene. [1] Both are sp2. ; (iv) With the aid of an appropriate orbital diagram, illustrate in phenylethene, the orbital overlaps of –CH=CH2 with the benzene ring. You need only show the relevant orbitals of the carbon atoms involved. [2] p orbitals of benzene 6C ; overlapped with p orbitals of 2C in –CH=CH2 ; denoted by dotted lines and labelled (v) Briefly explain why it would be incorrect to say that “ –CH=CH2 is an electron-withdrawing group”. [1] Phenylethene is a hydrocarbon, and all the C and H atoms have similar electronegativities / –CH=CH2 does not have a lone-pair that can delocalise into the ring. OWTTE (vi) Suggest a chemical test to distinguish between phenylethene and phenylamine. [2] R&C: Br2(aq), r.t. Obsv: Orange Br2(aq) decolourises for both phenylethene and phenylamine, but a white precipitate forms for only phenylamine. or R&C: KMnO4, dil H2SO4, heat Obsv: Purple KMnO 4 decolourises for phenylethene with effervescence of a gas that forms a white precipitate in limewater / Ca(OH)2. Purple KMnO4 remains for phenylamine. (d) Given that –NO2 is an electron-withdrawing group, explain the following observation. [2] The electron-withdrawing –NO2 group reduces the intensity of the negative charge on the carboxylate group, stabilising its conjugate base more. ; Hence is a stronger acid and has a lower pKa. ;

6 [Total: 25] 2 (a) (i) The Period 3 elements vary in their physical properties, such as melting point and electrical conductivities. Sketch the melting point trend of the elements in Period 3 (from Na to Cl). [1] (ii) The Period 3 metals Na and Al undergo vigorous reactions with chlorine gas upon heating to form their respective metal chlorides. When these chlorides are dissolved into water separately, solutions of different pH values are obtained. Write balanced equations to explain the differences in pH, stating the pH value of the solutions formed for each metal chloride. [3] Polarising power of Na+ < Al3+ as it is proportional to charge density. ;  NaCl (s) dissolves in water to form Na+(aq) and Cl–(aq) with no hydrolysis. NaCl(s) → Na+(aq) + Cl–(aq) pH ≈ 7  AlCl3 dissolves in water with hydrolysis, polarising water molecules and weakening the O–H bond to give H+ ions. AlCl3 (s) + 6 H2O(l) [Al(H2O)6]3+ (aq) + 3Cl- (aq) [Al(H2O)6]3+ (aq) [Al(H2O)5(OH)]2+ (aq) + H+ (aq)