IJC H2 Chem 2012 Prelim P3 Soln

2012 H2 Prelim 2 Essay solutions_13 aug   2    Step 2 (ii) The bromine atom in the product from b(i) is very reactive towards nucleophilic reagents. The bromine atom in 1–bromopropene is unreactive towards nucleophilic reagents. Suggest an explanation for the unreactivity of the bromine atom in 1–bromopropene. [2] In 1-bromopropene, the p orbital of Br overlaps with the π orbital of adjacent C atom, strengthening the C – Br bond. Thus, the C – Br bond is too strong to be broken and hence unreactive towards nucleophilic reagents. OR In 1-bromopropene, the C atom is adjacent to the double bond and hence less electron deficient and thus less susceptible to nucleophilic attacks. (c) para-Methoxyamphetamine first came into circulation in the early 1970s and went by the street names of "Chicken Powder" and "Chicken Yellow" and was found to be the cause of a number of drug overdose deaths in the United States and Canada at that time. It can be synthesised by the following route. (i) Suggest reagents and conditions for steps II and III. II: CH3Br / CH3I/ CH3Cl, heat with reflux

2012 H2 Prelim 2 Essay solutions_13 aug   3    III: NH3 in ethanol, heat in a sealed tube (ii) What type of reaction is step II? Nucleophilic substitution (iii) Suggest why compound J is converted into K before step II is carried out. To generate a stronger nucleophile, phenoxide ion. (iv) What type of stereoisomerism does para-methoxyamphetamine exhibits? Draw the structures of the stereoisomers of para-methoxyamphetamine. Optical isomerism C CH2 H NH2 CH2N CH2 H mirror CH3 OO CH3 [6] (d) Compound P is a neutral, sweet-smelling liquid with molecular formula C 5H8O2. It reacts with hot sulfuric acid to give a single compound Q, C 5H10O3. Q has two stereoisomers and gives a pale yellow precipitate with alkaline iodine and compound R. Deduce the structures of P, Q and R. P undergoes acidic hydrolysis with hot sulfuric acid to give Q. P is an ester. Q contains CH3CH(OH) since it undergoes mild oxidation with alkaline iodine. Q only contains 1 chiral carbon since it has two stereoisomers.

2012 H2 Prelim 2 Essay solutions_13 aug   4    [5] [Total:20]

2012 H2 Prelim 2 Essay solutions_13 aug   5      2 (a) When a precipitate is formed, ∆Gθ ppt, in kJ mol -1, is given by the following expression. ∆Gθ ppt = 1000 K log 2.303RTsp (i) Given that the Ksp value of BaF 2 is 1.70x10 -6 at 298K, calculate ∆Gθ ppt, in kJ mol-1, for BaF2. ∆Gθ ppt = [2.303 x 8.31 x 298 x log (1.70x 10-6)] ÷ 1000 = -32.9 kJ mol-1 (ii) The standard enthalpy change of formation of BaF 2 is −858 kJ mol-1. Use your answer in (a)(i) to calculate ∆Sθ ppt, in J mol -1 K-1 for the formation of the precipitate at 298K. ∆G = ∆H –T∆S -32.9 = -858 – (298) ∆S ∆S = -2.77 kJ mol-1K-1 = -2770 J mol-1K-1 (iii) Explain the significance of the sign of your answer in (ii). The sign is negative which means that entropy decreases as the system is more ordered/less disordered when there is a phase change from aqueous to solid state. (iv) Predict and explain whether the precipitation will be feasible at high or low temperature. Since ∆S is negative, (-T ∆S) is always positive. At high temperature, the magnitude of (-T ∆S) would be greater than ∆H. Hence ∆G would be positive and the reaction would not be feasible. Therefore the reaction is feasible at low temperature. (v) Suggest how the magnitude of the lattice energy of BaF2 might compare to that of BaCl2. Explain your answer.  Although fluoride ion and chloride ion have the same charge,  fluoride ion is smaller than chloride ion.  |L.E| α | qq rr   | or Since lattice energy is inversely proportional to the

2012 H2 Prelim 2 Essay solutions_13 aug   6    ionic radius  Therefore the magnitude of the lattice energy of BaF 2 would be greater than that of BaCl2. [7] (b) In the past, chemical analysis was carried out by chemists using traditional laboratory apparatus. Many qualitative tests used depended on an application of the principles of solubility product. (i) Write an expression for the Ksp of barium fluoride. Ksp = [Ba2+][F- ]2 (ii) Predict whether precipitation occurs if 50.0 cm 3 of 0.150 mol dm –3 of Ba(OH) 2 solution is mixed with 50.0 cm3 of 0.100 mol dm–3 of KF solution in the laboratory. The Ksp of BaF2 is 1.70 x 10-6 mol3 dm-9. [Ba2+]new = 0.150 2 = 0.075 mol dm-3 [F- ]new = 0.100 2 = 0.05 mol dm-3 IP = [Ba2+][F- ]2 = 0.075 x (0.05)2 = 1.875 x 10-4 mol3 dm-9 1.875 x 10-4 mol3 dm-9 > Ksp Precipitation occurs. [3] (c) Myrcene is a naturally occurring compound found in the leaves of bay trees. It is known to be a polyunsaturated hydrocarbon. It can react with hydrogen to produce a saturated hydrocarbon. In a laboratory investigation, a 1.00 g sample of pure myrcene fully reacted with exactly 510 cm 3 of hydrogen gas measured at 20.0°C and 105.0 kPa. In this reaction, myrcene was converted to a saturated alkane with a molecular formula C 10H22. (i) What type of reaction has occurred between the myrcene and hydrogen? Addition of hydrogen /hydrogenation/redox/ reduction (ii) Calculate the amount, in moles, of hydrogen reacting. no of moles of H2 = pV / RT = 105 x 103 x 510 x 10-6 / (8.31 x [20.0 +273]) = 0.0220 mol (iii) Calculate the mass of C10H22 produced in the reaction. mass of C10H22 = mass of myrcene + mass of H2

2012 H2 Prelim 2 Essay solutions_13 aug   7    mass of H2 = 0.0220 x 2.0 = 0.044 g mass of C10H22 = 1.00 + 0.0440 = 1.04 g (iv) Determine the number of double bonds in each molecule of myrcene. no of moles of C10H22 = 1.044 / 142 = 7.35x10-3 mol = n(myrcene) equation for reaction is myrcene + xH2 → C10H22 n(H2) / n(myrcene) = x / 1 0.0220 / 7.35x10-3 = x 3 = x 3 molecules of H 2 added to each myrcene molecule. Hence there are 3 C=C double bonds. The following is an alternate approach. no of moles of C 10H22 = 7.35x10-3 mol no of moles of myrcene = 7.35x10-3 Molar mass of myrcene = 1.0 / 7.35x10-3= 136 g mol-1 molar mass of myrcene is 6 less than molar mass C10H22. → 3 H2 molecules added to each myrcene molecule → 3 C=C double bonds. One mark was awarded if 3 double bonds was stated, but to obtain the second mark a logical explanation of how the number of double bonds was determined was required. [6] (d) In September 2009, the wholesale of weedkiller containing chlorate(V) ions was banned in various European countries. Chlorate(V) ions can act as a strong oxidising agent in acid solution according to the following half equation: ClO 3 − (aq) +6H+ (aq) + 6e  Cl− (aq) + 3H2O(l) In an experiment, 25.0 cm 3 of a sample of sodium chlorate(V) solution reacted with an excess of sodium iodide, Na I. The iodine produced required 25.00 cm 3 of 1 mol dm-3 of sodium thiosulfate, Na2S2O3, for complete reaction. (i) Write a balanced equation between chlorate(V) ions and iodide ions in acidic medium. [R] ClO3 - +6H+ + 6e  Cl- + 3H2O [O] 2I-  I2 + 2e Overall: ClO3 - +6H+ + 6I-  Cl- + 3H2O + 3I2 (ii) Calculate the number of moles of iodine liberated by the chlorate(V) solution. [R] : I2 + 2e   2I  [O] : 2S2O3 2   S4O6 2 + 2e

2012 H2 Prelim 2 Essay solutions_13 aug   8    I2 + 2S2O3 2  2I  + S4O6 2 No of moles of thiosulfate = 1000 25 x 1 = 0.025 mol No of moles of iodine = 0.025 ÷ 2 = 0.0125 mol (iii) Calculate the concentration of sodium chlorate(V) in the solution.   Mole Ratio ClO- : I2 1:3 No of moles of chlorate = 0.0125 ÷3 = 4.167 x 10-3 mol Concentration of sodium chlorate(V) = 1000 25 10 167 . 43x = 0.167 mol dm-3 [4] [Total: 20]

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