DHS H2 CHEM P3 ANS

Answer all questions 1(a) Pyruvic acid, CH 3COCOOH, occurs naturally in th e body and is an end product of the metabolism of sugar. When 10.0 cm 3 of a solution of pyruvic acid were titrated against 0.01 mol dm –3 sodium hydroxide, with a data logger, the following curve was obtained. pH Volume of NaOH / cm3 7 Equivalence point 2.5 30 (i) Suggest a suitable indicator, if the titr ation were to be repeated without the use of a data logger. Explain your reasoning. Phenolphthalein. The pH trans ition range of the indicator lies within the sharp pH change over its equivalence point. (ii) Calculate the value of Ka for pyruvic acid. No. of mol of NaOH = 01 01000 30 . = 3 x 10–4 No. of mol of pyruvic acid = 3 x 10–4 [pyruvic acid] = 1000 10 10 34 = 0.03 mol dm–3 From graph, pH = 2.5 Hence [H +] = 10–2.5 = 3.162 x 10–3 Ka = ] _ [ ] [ acidpyruvic H 2 = 3 2 3 10 162 3 03 0 10 162 3      . . ). ( = 3.73 x 10–4 mol dm–3 2 [Turn Over

(iii) Explain, with the aid of an appropriate equation, why the pH at equivalenc e point is greater than 7. [7] CH3COCOO– (aq) + H2O (l) CH3COCOOH (aq) + OH– (aq) CH3COCOO– undergoes salt hydrolysis. [OH] > [H+]. (b) Pyruvic acid is converted to the amino acid , alanine, in just one single step in the body by the enzyme alanine transaminas e. However, to achieve the same conversion in a laboratory, multiple steps are required. Given the chemical structure of alanine below, show how pyruvic acid can be converted to alanine in not more than 4 steps. Indicate the r eagents, conditions and all intermediates in your answer. The alanine formed from yo ur suggested synthesis steps may be in its cationic, anionic or zwitterionic forms. C C H 3 H NH2 C O OH Alanine [5] C C C H H H O O OH CCH H H OH O OHH CCC Cl O ClH H H H I : H 2 with Ni, heat OR NaBH 4 in ethanol / methanol (room temp) (LiA lH4 not acceptable) II: PCl 5, room temperature III: NH3 in sealed tube, heat. IV: HCl / H2SO4, heat OR NaOH, heat CCCH H H NH2 O OHH CCCH H H NH2 O NH2H C I II III IV 3 [Turn Over

Alternative answer: I : H2 with Ni, heat OR NaBH 4 in ethanol / methanol (room temp) (LiA lH4 not acceptable) II: concentrated H2SO4, 170 oC III: HCl (g) IV: NH3 in sealed tube, heat. (c) Alanine, together with the 19 other standard amino acids are the building blocks of proteins. State the two factors that cause protein denaturation and explain how the two stated factors denature the protein. [4] Any 2 of the following 1. Addition of heavy metal ions. Heavy metal form salt or complex ions, thereby breaking / interfering with the ionic interactions heavy metals break disulphide links. 2. Extremes of temperature / High temperature / Heating Heat breaks the weak van der waals’s forces and hydrogen bonds holding together the quart enary, tertiary and seconda ry structure of the protein, resulting in a change in the original conformation of the protein. 3. Extreme pH changes Adding H+ or OH– protonate or deprotonate the ionic R groups, disrupting the ionic or hydrogen bonds holding together the quarternary and tertiary structure of the protein. C C C H H H O O OH CCCH H O OHH CCCH H H OH O OHH I II III CCCH H H NH2 O OHH CCCH H O OHH H ClIV 4 [Turn Over

(d) When 0.200g of alanine is subjected to co mplete combustion, carbon dioxide, nitrogen dioxide and water is formed. These gases are absorbed by 20 cm 3 of 1.5 mol dm –3 NaOH. The resultant solution is then titrated with 0.3 mol dm –3 H SO . Calculate the volume of H SO required to reach equivalence point. [4]2 4 2 4 H int: 2NO2 + 2NaOH  NaNO3 + NaNO2 + H2O C3H7NO2 + 19/4O2  3CO2 + 7/2H2O + NO2 No. of mol of alanine = 2 16 14 7 12 3 2 0      . = 0.002247 No. of mol of CO2 = 3 x 0.002247 = 0.006741 No. of mol of NO 2 = 0.002247 CO 2 + 2NaOH  Na2CO3 + H2O No. of mol of NaOH that reacted with CO 2 = 2 x 0.006741 = 0.01348 NO2  NaOH No. of mol of NaOH that reacted with NO 2 = 0.002247 No. of mol of original NaOH = 5 11000 20 . = 0.03 No. of mol of NaOH remaining = 0.03 – 0.01348 – 0.002247 = 0.01427 2NaOH  H2SO4 No. of mol of H2SO4 required = ½ x 0.01427 = 7.136 x 10–3 Volume of H2SO4 required = 3 0 10 136 73 . .  = 0.0238 dm3 or 23.8 cm3 [Total: 20] 5 [Turn Over

2(a) Hydrogen chloride gas and ammonia gas react to form ammonium chloride. NH3(g) + HCl(g) NH4Cl(s) (i) By constructing a Hess’ Law Cycle, calculate the standard enthalpy change of the reaction, given the following information:  ) NH ( f3 H = –46.1 kJ mol–1  ) HCl ( fH = –92.3 kJ mol–1  ) Cl NH ( f4 H = –314.4 kJ mol–1 NH3(g) + HCl(g) NH4Cl(s) ½ N 2(g) + 2H2(g) + ½ Cl2(g) + =  rxnH+  ) NH ( f3 H  ) HCl ( fH  ) Cl NH ( f4 H – –  rxnH=  ) Cl NH ( f4 H  ) NH ( f3 H  ) HCl ( fH = –314.4 – (–46.1) – (–92.3) = –176 kJ mol–1 (ii) The standard entropy change of the reaction is –284 J K –1mol–1. Explain why this value is negative. There is a decrease in entropy due to the decrease in total number of mol of gases from 2 to 0. (iii) State and explain whether the reaction is s pontaneous at high or lo w temperature. Reaction is spontaneous at low temperature. For , since    S T H G   ) HCl ( fH ) NH ( f3 H  ) Cl NH ( f4 H S and H are both negative , the value of T needs to be low enough such that   H S T , in order that G is negative. 6 [Turn Over

(iv) Calculate the temperature for which the reaction is spontaneous. [6] For reaction to be spontaneous, Gө < 0 Hence Hө – TSө < 0 (–176) – T(–284 x 10-3) < 0 T(–284 x 10-3) > –176 T < 618K (b) When hydrogen chloride gas is added to organic compound P, C 6H12, two products Q and R, which are isomer s of each other, C 6H13Cl, are formed in unequal proportion. When Q and R are separately boiled with aqueous sodium hydroxide, S is formed from Q, while T is formed from R. S and T are also isomers of each other, with the same molecular formula, C 6H14O. When S and T are separately boiled with ac idified potassium manganate( VII), no decolourisation is observed in the test tube containing S; while decolourisation is observed in the test tube containing T which results in the formation of U, C6H12O2. When some sodium carbonate powder is added to U, effervescence of carbon dioxide gas is observed. When P is boiled with acidified po tassium manganate (VII), compound V, C5H10O and CO2 are formed. V gives an orange precipitate when warmed with 2,4-dinitrophenylhydrazine. When V is heated with lithium aluminium hydride in dry ether, W, C5H12O, which does not exhibit optical activity, is formed. Deduce the structures fo r each lettered compound, P to W. Explain the chemistry of the reactions involved. Balanced equations a