NYJC H2 Chem 2013 Prelim P3 Soln

1 H2 Chemistry 9647/03 NYJC J2/2013 Prelim [Turn over Answer for Paper 3 1 The elements of Group VII of the Periodic Table are called halogens, which mean "salt formers". They lack only one electron to fo rm a complete shell, and are extremely active chemically. This question examines the chemistry of two reactions of halogens. (a) In the first reaction, a 1:2 mixture of H 2 and I 2 gases was mixed at a constant temperature of 430 oC and a constant pressure of 2 atm. The following equilibrium was set up. H2 (g) + I2 (g) 2HI (g) ∆H = −11 kJ mol1 (i) Given that degree of dissociation of H 2 is 86 %, calculate the equilibrium constant, Kp for the above reaction under these conditions. H 2 (g) + I 2 (g) 2 HI (g) Initial amt / mol X 2x 0 Change / mol 0.86x 0.86x +1.72x Equilbrium / mol 0.14x 1.14x 1.72x . n(total) = 1.14x + 0.14x + 1.72x = 3x K p = 22 2 2 1.72(2 )() 3 0.14 0.14() ( ) (2 ) (2 )33 HI HI x P x x xPP xx  = 18.5 (ii) Suggest, with reasoning, a way to increase degree of dissociation of H 2 for the above reaction. Decrease temperature. By Le Chatelier’s Principle, when temperature is decreased, The forward exothermic reaction is favoured. Hence POE shift to the right, and degree of dissociation of H 2 increases. 66

(iii) Deduce how Kp for the following system would be compared to your answer in (i). H2 (g) + Br2 (g) 2HBr (g) It will be larger since a stronger HBr bond is fo rmed. Extent of equilibrium lies more to the right. [7] In another reaction, halogens can react with unreactive alkanes to form mono- substituted product. An example is the reaction of chlorine with 2-methylpentane shown below. 2-methylpentane (b) (i) Explain why alkanes are generally unreactive. The C-H bonds are non-polar and very strong (ii) This reaction is seldom used for synthesis as there are many associated problems. Firstly, several isomeric pr oducts are formed. The relative ratio of the isomeric products may be more a ccurately determined if relative rates of abstraction of H atoms are taken into account. The relative rates of abstraction of H atoms are Type of H atoms Relative rate of abstraction Primary 1 Secondary 4 Tertiary 6 By examining the difference in stabilit y of the intermediates formed when different types of H atom are abstracted, explain the trend in relative rate of abstraction. The tertiary radical is the most stable as it contains the most electron- donating alkyl groups which help to stab ilise the electron-deficient radical centre. Hence, it is abstracted more easily/faster. (iii) Predict the ratio of the following two products A and B, taking into account the relative rates of abstraction given in (ii). Explain your reasoning. Cl and A B Cl 67

3 H2 Chemistry 9647/03 NYJC J2/2013 Prelim [Turn over CC CHa3 CCCHa Ha Ha Hb H H H H H H H 6 possible (primary) hydrogens (Ha) can be substituted to form A 1 possible (tertiary) hydrogens (Hb) can be substituted to form B Assuming equal probability of abstraction, Ratio of A : B = 6:1 However since a tertiary H is abstracted 6 times faster than a primary H, Ratio of A : B = 6:1x6 = 1:1 (iv) Describe suitable chemical test(s) to distinguish compounds A and B in (iii). Add NaOH(aq) to both unknowns, heat (to form alcohols) [1] followed by KMnO 4, NaOH(aq), heat. Observation for A: Purple KMnO 4 decolourised, Brown ppt of MnO 2 formed. (as 1o alcohol formed is oxidised). Observation for B: Purple KMnO 4 remains (as 3 o alcohol formed cannot be oxidised). [7] (c) Another problem of the reaction is multi-substitution. (i) Suggest the conditions that will give rise to formation of multi-substituted products. Excess Cl 2 or limited alkane. [1] (ii) A di-substituted product C with molecular formula C 6H12Cl2 was obtained in the reaction of chlorine with 2-methylpentane. C does not exhibit optical activity. On warming with alcoholic KOH, a major product D is formed. When D is treated with hot acidified potassium manganate(VII) solution, effervescence was observed and the only organic product remaining gives a yellow precipitate with alkaline aqueous iodine. Use this information to deduce the structure of C and D , explaining your reasoning. 68

C: CC CH3 CCCH H H Cl H H H H Cl H H D: CC CH3 CCCH H H H HH H  C does not exhibit optical activity as it doesn’t contain any chiral centre (no carbon with 4 different groups).  Alkyl halide in C undergoes elimination with alcoholic KOH to form alkene in D.  Alkene in D undergoes oxidative cleavag e to form the following products: CC CH3 H H H O CC OHO H OO CO2  CC CH3 H H H O undergoes oxidation to form yellow precipitate with alkaline aq iodine.  The terminal alkene is oxidised to CO 2.  Ethanedioic acid formed is also oxidised spontaneously to CO2. [6] [Total: 20] 2 Precipitation plays an import ant function in growth of bo nes and teeth in our bodies. 69

5 H2 Chemistry 9647/03 NYJC J2/2013 Prelim [Turn over Teeth and bones are largely compos ed of calcium phosphate salt Ca 3(PO4)2. In order for this precipitation to occur, the concentrations of the ions in blood must exceed the solubility product in the immediate region of deposition. (a) Write an ionic equation with st ate symbols to show the formation of teeth and bones at the deposition region and hence an expression for the solubility product of the Ca3(PO4)2 deposits. [2] 3Ca2+(aq) + 2PO4 3- (aq) Ca3(PO4)2(s) Ksp of Ca3(PO4)2 = 3223 4Ca PO     (b) (i) At a temperature of 37 oC, the solubility product for calcium phosphate Ca3(PO4)2 has a numerical value of 4 × 10 -27. In a sample of blood at a pH of 7.4, the concentrations of calcium and phosphate ions are found to be 1.2 × 10-3 mol dm-3 and 1.6 × 10-8 mol dm-3 respectively. State and explain if the above composition in blood will resu lt in the growth or degeneration of teeth and bones. ionic product = 3223 4Ca PO     = (1.2 × 10-3)3 × (1.6 × 10-8)2 = 4.42 × 10-25 mol5 dm-15 Since IP > Ksp (4 × 10-27), precipitation occurs resulting in growth. (ii) Hence, state and explain if t he pH environment of 7.4 in (i) is that for a growing child, a fully grown man or an old person. At pH = 7.4, since there is depositio n of calcium phosphate, it is the environment for a growing child. [3] (c) The pH at the area of growth may change due to metabolic processes in the body. The hydrogen ions that are present combine phos phate ions as hydrogen phosphate ions. This prevents growth of strong teeth and bones. PO 4 3- (aq) + H+ (aq) HPO4 2- (aq) Assuming that the [HPO4 2-] is maintained at 1.7 × 10-4 mol dm-3 by cell functions, (i) calculate the pH that will result in neither growth nor degeneration at the teeth and bones deposition region when the concentration of calcium ions in blood is maintained at 1.2 × 10-3 mol dm-3. [Ka of HPO4 2- = 4.4 × 10-13 mol dm-3] To prevent neither growth nor degeneration, the [PO 4 3-(aq)] must be at saturation (or equilibrium with the Ca3(PO4)2(s), 70

[PO4 3-(aq)] = 27 33 41 0 (1.2 10 )     =1.52 × 10-9 moldm-3 pH = -log[H +] =-log( 13 4 9 4.4 10 1.7 10 1.52 10       ) = 7.3 (ii) Suggest another acid–conjugate base e quation that accounts for the constant supply of HPO4 2- ions from cell functions