NJC 2022 H2 Chemistry Prelim P3 Ans

1 NJC/H2 Chem Preliminary Examination/03/2022 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 3 Free Response Candidates answer on Question Paper. Additional Materials: Data Booklet 9729/03 13 September 2022 2 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 /18 2 /19 3 /23 Section B 4 /20 5 /20 Paper 3 Total /80 This document consists of 28 printed pages.

2 NJC/H2 Chem Preliminary Examination/03/2022 Section A Answer all the questions in this section. 1 (a) Amines can be synthesised from methylbenzene, shown by the following steps. Fig. 1.1 (i) Suggest the structure for compounds H. [1] (ii) Suggest reagents and conditions for steps 1, 2, 3 and 4 in Fig. 1.1. [4] (iii) Suggest why the yield for step 2 is not high. [1] The Kb values of three bases, at 25oC, are shown in table 1.2. Table 1.2 base Kb/moldm−3 ammonia 1.8 x 10−5 compound K 4.5 x 10−4 compound J 7.4 x 10−10 (iv) Explain the relative magnitudes of the Kb values in table 1.2. [2] …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. CH3 CH2Cl Compound H CH2NH2 CH2Cl H2N Step 1 Step 2Step 3 Compound J Compound K (i) 1 or 0 (ii) Step1: UV light, limited Cl2. 1 or 0 Step 2: ethanolic NH3, heat in sealed tube. Step 3: conc H 2SO4, Conc HNO 3, 60oC (−CH2Cl has negligible effect on benzene) -accept T lower than 60oC. Step 4: conc HCl, Sn, heat followed by NaOH(aq) or (step 1 then 2) 1 or 0 for each step Step 4

3 NJC/H2 Chem Preliminary Examination/03/2022 [Turn over ……………………………………………………………………………………….. (b) The rate of reaction for step 2 in Fig. 1.1 can be followed by measuring the change in concentration of C7H7Cl with time. The reaction was carried out with the other reactant in large excess. Fig. 1.2 shows the concentration of C7H7Cl monitored against time for step 2. Showing all your working and drawing clearly any construction lines, use Fig.1.2 to determine: (i) The order with respect to C7H7Cl. Explain your reasoning. [2] (ii) The initial rate, in mol dm−3 min−1 [1] Given that the half -life magnitude of C7H7Cl is not affected by the change in concentration of the other reactant. (iii) Write the rate equation for step 2 in Fig.1.1, and calculate a value for the rate constant, stating its units. [2] 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 0 100 200 300 400 500 600 700 800 Time/min [C7H7Cl ] / moldm-3 330 660 (iii) The primary amine produced, due to the electron donating alkyl group ½ , is a stronger nucleophile than NH 3, ½ hence can compete/further react with NH 3 for C7H7Cl to form a secondary amine instead of the compound K. (½ for stating polysubstitution) (iv) Smaller Kb value means a weaker base. Lone pair on N in J is less available ½ compared to that of K to accept H+ as the electron density at N is reduced½ via resonance with the adjacent benzene ½ (lone pair on N is delocalised into the benzene ring ) while the electron density at N in K is increased by the electron donating effect of the alkyl group½.

4 NJC/H2 Chem Preliminary Examination/03/2022 (iv) Hence, outline a mechanism for step 2 in Fig.1.1 to form all the products. Show all charges and relevant lone pairs and show the movement of electron pairs by using curly arrows. [3] (v) Explain why step 2 proceeds via the mechanism you describe in b(iv). [1] (vi) Suggest and explain the difference in reactivity when 4-bromomethylbenzene undergoes the same reaction. [1] [Total : 18] …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. (b)(i) 1m for showing how the half-life is obtained. Since half-life of to C7H7Cl is constant at 330 min, it is 1st order wrt to C7H7Cl. 1m (ii) Initial rate = − (gradient at t = 0) = − (−0.0200) 400 = 5 × 10−5moldm−3s−1 1m (iii) Rate = k [C7H7Cl] 1m Two methods: Making use of the initial rate calculated in (ii) k = 5 × 10−5  0.02 = 2.5 × 10−3 s−1 Making use of the half-life magnitude: k = ln 2 𝑡1 2⁄ = 2.1 × 10−3 s−1 ½ m each for answer and units

5 NJC/H2 Chem Preliminary Examination/03/2022 [Turn over …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. …………………………………………………………………………………………….. ………… (iv) Accept show formation of HCl in separate step. Draw SN2 based on SN1 rate equation on (iii) – 1m for name of mechanism Draw SN2 based on SN2 rate equation in part (iii) – full credit 1m for name of mechanism ½ for slow 1 m for curly arrows and charges 1m for balanced equations (v) Explain why step 2 proceeds via the mechanism you describe in b(iv). [1] The carbocation intermediate can be resonance stabilised i.e. the positive charge of the cation can be dispersed via resonance with the adjacent benzene ring. 1m Or The bulky benzene ring poses steric hindrance to the incoming nucleophile if it occurs via SN2 or one step mechanism. 1m Cannot accept sn2 explanation (vi) C-Br in 4 -bromomethylbenzene is resistant to substitution [ ½ ] , as it has double bond character due to the continuous p orbital overlap [ ½ ] involving Br and benzene ring.

6 NJC/H2 Chem Preliminary Examination/03/2022 2 Hydrogen cyanide, HCN, is extremely toxic and with sufficient concentrations it leads to rapid death. During the Second World War, a form of hydrogen cyanide known as Zyklon B was used in the Nazi gas chambers. (a) (i) Draw a dot–and–cross diagram to illustrate the bonding in HCN. [1] HCN can be oxidised to cyanogen, C2N2. C2N2(g