2021 YIJC Prelim P2 Answer

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME TEACHERS’ COPY CG INDEX NO CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper Additional Materials: Data Booklet 9729/02 30 August 2021 2 hours This document consists of 19 printed pages and 3 blank pages. For Examiner’s Use Paper 1 /30 Paper 2 1 /7 2 /5 3 /7 4 /17 5 /11 6 /22 7 /6 Penalty /75 Paper 3 /80 Paper 4 /55 Overall Percentage (%) READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an 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. At the end of the exam ination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

2 ©YIJC Answer all the questions in this section in the spaces provided. 1 Use of the Data Booklet is relevant to this question. This question is about period 3 elements. (a) (i) State the electronic configuration of argon. [1] 1s22s22p63s23p6 (ii) Using the Cartesian axes shown in Fig. 1.1, draw a fully labelled diagram of the valence orbitals of argon. Fig 1.1 [1]

3 ©YIJC [Turn over (iii) Two glass vessels B and C are connected by a closed valve as shown in Fig 1.2. Fig. 1.2 B contains argon at 20 C at a pressure of 1 x 10 5 Pa. C has been evacuated and has three times the volume of B. In an experiment, the valve was opened and the temperature of the whole apparatus was raised to 100 C. Calculate the final pressure in the system. Since amount of argon is the same throughout, 𝑃1𝑉1 𝑇1 = 𝑃2𝑉2 𝑇2 1 𝑥 105𝑥 𝑉 20+273 = 𝑃2𝑥 4𝑉 100+273 P2(1172V) = 3.73 x 107V P2 = 31825.9 = 31800 Pa (3 sf) [2] (b) Fig 1. 3 shows the fifth, sixth, seventh, eighth and ninth ionisation energies of another element T in Period 3. Fig.1.3 (i) State and explain which period 3 element has these ionisation energy values. [2] 1000 6000 11000 16000 21000 26000 31000 36000 41000 5 6 7 8 9 ionisation energy/ kJ mol1 electron removed

4 ©YIJC Chlorine as it has a large increase in ionisation energy from the removal of 7th to 8 th electron, indicating that there are 7 valence electron in the outermost shell (OR and the 8th electron is removed from an inner quantum shell). (ii) Element U is directly below element T in the periodic table. Explain how the 7th ionisation energy of element U is compared to element T. [1] Element U would have a lower 7th ionisation energy compared to element T as it has one more inner quantum shell of electrons. (OR increase in shielding effect outweighs increase in nuclear charge hence effective nuclear charge decreases and less energy is required to remove the 7th electron). [Total: 7] 2 Diamine can ionise in stages. (a) Table 2.1 compares the pKb values of ethylamine and 1,2-ethanediamine at 25 C. Table 2.1 Base Formula pK1 pK2 ethylamine CH3CH2NH2 3.19 - 1,2ethanediamine H2NCH2CH2NH2 4.11 7.39 (i) Suggest a reason why the p K1 value of ethylamine is less than the p K1 value of 1,2ethanediamine. [1] Ethylamine is a stronger base (smaller pK1 value) as the lone pair of electron is more available to accept a proton as compare d to 1,2ethanediamine, where there is electron withdrawing nitrogen atom OR there is intramolecular hydrogen bonding between the amine groups. This makes the lone pair of electron on another N atom less available to accept a proton , hence weaker base (larger p K1 value).

5 ©YIJC [Turn over (ii) Suggest why the pK2 value of 1,2ethanediamine is higher than its pK1 value. [1] The second ionisation of 1,2ethanediamine is less favourable than the first ionisation because the second ionisation of 1,2ethanediamine involved accepting H + ion to a positively charged CH3CH2NH3+(repulsion), while the first ionisation involves accepting H + to a uncharged 1,2-ethanediamine (less repulsion). (iii) 0.10 mol dm3 of HCl (aq) was added to ethylamine solution at 25 C. CH3CH2NH2 (aq) + H2O (l) ⇌ CH3CH2NH3+ (aq) + OH(aq) Deduce without calculation, what happens to the position of equilibrium and the value of pK1. [2] Addition of acid will cause the concentration of OH− to decrease. The equilibrium position will shift to the right so as to increase the concentration of OH− OR to remove excess H+. Value of pK1 remains constant at 3.19 as temperature is constant or it is temperature dependent. (b) Adding equimolar of 1,2ethanediamine and phosgene, COCl2 produced a cyclic compound V with molecular formula C3H6ON2. Suggest the structure of compound V. [1] Compound V [Total: 5]

6 ©YIJC 3 Lysine is an essential amino acid that cannot be synthesised by the human body and must be obtained from the diet. It is found in legumes such as peas, and animal products such as beef and fish. The structure of lysine is given below. (a) A solution was prepared by reacting 1 mole of lysine with 2 moles of hydrochloric acid. This solution was titrated with aqueous sodium hydroxide to obtain the titration curve below. (i) Draw the structure of the species present at points W, X, Y and Z on the titration curve. [4] Amount of OH− /mol pH W X Y Z

7 ©YIJC [Turn over W X Y Z [4] (ii) Explain why the melting point of lysine is high, in terms of structure and bonding. [2] Lysine exists as a zwitterion and has a giant ionic structure with strong electrostatic forces of attraction between the oppositely charged –COO– and –NH3+ groups . Since a lot of energy is required to overcome the (OR) strong ionic bonds, lysine has high melting point. (b) Leucine is another essential amino acid that is mainly found in legumes. Write an equation to explain how the zwitterionic form of leucine behaves as a buffer when a small amount of base is added. [1] [Total: 7]

8 ©YIJC 4 Benzene is a natural constituent of crude oil and is mainly used as an intermediate to produce other chemicals. (a) In the presence of anhydrous AlCl3, benzene can undergo Friedel-Crafts alkylation with CH3Cl to form methylbenzene. The alkyl side chain can further react to form compound A, C14H14, along with side products like hydrogen chloride gas. (i) Suggest the structure of compound A. [1] (ii) Unlike benzene, phenylamine does not undergo Friedel -Crafts alkylation with halogenoalkanes in the presence of AlCl3. This is because the amine reacts with AlCl3 to form a neutral compound B. Draw the displayed formula of B. [1] (b) Friedel-Crafts alkylation can also be achieved using alkenes, such as the example below. (i) In the first step, (CH3)3C+ is produced in the presence of AlCl3 and HCl. Write an equation to show the formation of (CH3)3C+. [1]

9 ©YIJC [Turn over (ii) Name and outline the mecha