TJC H2 CHEM P3 Qn & ans Prelim

1 2016 TJC H2 Chemistry Preliminary Exam [Turn Over CHEMISTRY 9647/03 Paper 3 Free Response 13th September 2016 2 hours Candidates answer on separate paper. Additional materials: Answer paper Graph Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, Civics Group, Centre number and Index number in the spaces provided on the cover page and on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, glue or correction fluid. Answer any four questions. 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. This document consists of 15 printed pages.

2 2016 TJC H2 Chemistry Preliminary Exam [Turn Over Answer any four questions 1 Ethanedioic acid is a substance found in many plant foods. Cabbage is among the plant foods with high ethanedioic acid content. However, its anion, C2O42- can bind to iron to form iron (II) ethanedioate, which renders much of the iron in cabbage unusable by the body. (a) 50.0 cm3 sample of iron(II) ethanediaote, FeC2O4 was extracted from 300 g of cabbage, diluted in water and the solution made up to 250 cm 3. A 25.0 cm3 portion of this solution was acidified and required 26.90 cm 3 of 0.0100 mol dm -3 potassium manganate(VII) for oxidation of iron(II) to iron(III) and ethanedioate ions to carbon dioxide. (i) State the change in oxidation number for manganese and carbon in the reaction. [1] Mn: From +7 to +2 [Or decrease by 5] C: From +3 to +4 [Or increase by 1] (ii) Write down all the relevant ion-electron half equations and hence the overall redox equation for the reaction between potassium manganate(VII) and iron(II) ethanedioate. [2] MnO4- + 8H+ + 5e-  Mn2+ + 4H2O  C2O42-  2CO2 + 2e- Fe2+  Fe3+ + e-  3MnO4- + 24H+ + 5C2O42- + 5Fe2+  3Mn2+ + 5Fe3+ + 12H2O + 10CO2 (iii) Calculate the concentration, in mol dm -3, of iron(II) ethanedioate in the original sample [2]  Number of moles of FeC2O4 in 25.0 cm3 = 3 5 x 2.69 x 10-4 = 4.48 x 10-4 mol Number of moles of FeC2O4 in 250 cm3= 4.48 x 10-4 x 250/25.0 = 4.48 x 10-3 mol  Concentration of FeC2O4 in the original sample = 3- -3 10 x 50 10 x 48.4 = 0.0896 mol dm-3 (iv) Calculate the number of moles of iron in each gram of cabbage. [1]  No. of moles of iron per gram of spinach = 300 10 x 48.4 -3 = 1.49 x 10-5 mol/g

3 2016 TJC H2 Chemistry Preliminary Exam [Turn Over (b) Ethanedioic acid dissociates in water according to the following equation. HOOC-COOH + H2O HOOC-COO- + H3O+ The table below compares the Ka values of three organic acids. Formula Ka Ethanoic acid CH3COOH 1.74 x 10-5 Ethanedioic acid HO2CCO2H 6.46 x 10-2 Oxoethanoic acid (CHO)COOH 4.79 x 10-4 With reference to the K a values, comment on the order of acidity of the three organic acids. [2] The bigger the Ka, the stronger the acid. Acid strength is dependent on the stability of the anion formed when the acid ionises. The more stable the anion, the stronger is the acid. Strength of acid: Ethanedioic acid > Oxoethanoic acid > Ethanoic acid  Oxoethanoic acid is a stronger acid than ethanoic acid as -CHO group is electron withdrawing. The negative charge on the anion is dispersed, thereby stabilizing the ion compared to ethanoate ion.  Ethanedioic acid is the strongest acid as stabilisation of the mono anion by intramolecular hydrogen bonding with the unionised –COOH group results in the highest Ka value compared with the other two acids. (c) Compound A, with molecular formula C5H8O, decolourises aqueous bromine and reacts with PC l5 giving off white fumes. Upon refluxing A with acidified potassium manganate(VII), a symmetrical product B, C 5H6O5, is formed. B does not give a red precipitate with Fehling’s solution but a n orange precipitate is observed with 2,4-dinitrophenylhydrazine. 1 mol e of B also reacts with 1 mol e of Na 2CO3 with effervescence observed. B reacts with NaBH 4 to form C. Deduce the structures of compounds A, B and C, explaining the chemistry of the reactions involved. [8]  A decolourises aqueous bromine  C=C present  A reacts with PCl5 giving off white fumes of HCl  -OH present  A undergoes oxidation with KMnO4 where the –OH group is oxidised and C=C in a ring undergoes total bond cleavage to form B OR A undergoes oxidation with KMnO4 to give a symmetrical product B with no loss in C atoms  C=C is in a ring and A is a cyclic compound  B does not give a red precipitate with Fehling’s solution but form an orange precipitate with 2,4-dinitrophenylhydrazine  B is a ketone and A is a 2o alcohol  1 mole of B also reacts with 1 mole of Na2CO3  B has 2 –COOH groups.  B reacts with NaBH 4 to form C  only the carbonyl group in B is reduced but the carboxylic acid groups remain unaffected.

4 2016 TJC H2 Chemistry Preliminary Exam [Turn Over  A: OH  B: C CH2 C CH2 COOH OO HO  C: ( 9 marking points: Maximum 8) (d) (i) Describe what you see when separate samples of sodium and sulfur are burned in excess air. Write equations for the reactions that occur. [2] Sodium burns readily with a brilliant yellow flame in air or oxygen to form white sodium oxide, Na2O. 4Na(s) + O2(g)  2Na2O(s) Sulfur burns slowly with a blue flame on heating in air or oxygen to form colourless sulfur dioxide, SO2. (Note: SO3 is not formed) S(s) + O2(g)  SO2(g) or S8 + 8O2  8SO2  Both equations correct  Both observations correct (ii) The products resulting from the reactions in (d)(i) both react with water. Write equations for these two reactions and describe the effect of the resulting solution