NYJC H2 Chem 2012 Prelim P3 Soln

2 H2 Chemistry 9647/03 NYJC J2/12 PX [Turn Over Answer any four questions 1 Aldehydes and ketones react with primary amines in slightly acidic solution to form imines which have the C=N functional group. For example, propanal reacts with methylamine to give N-methyl-1-propanimine. CH3CH2 C H O CH3NH2 H+ N CH3 H CH3CH2 C propanal N-methyl-1-propanimine (a) Aldehydes and ketones also react with 2,4-dinitrophenylhydrazine (2,4-DNPH) to give compounds that are closely related to imines. (i) Write a balanced equation for the reaction between benzaldehyde and 2,4- DNPH. NO2 O2N H2N N H +C H O H2O+NO2 O2N N N H C H (ii) State what you would observe in the reaction with 2,4-DNPH. What is the type of reaction? orange precipitate condensation [ 4]

3 H2 Chemistry 9647/03 NYJC J2/12 PX [Turn Over (b) The mechanism for the reaction between benzaldehyde and methylamine is shown below. + H2O – C O H + CH3NH2 C O H N H H CH3 + C OH H N H CH3 carbinolamine H+ C OH 2 H N H CH3 + G + step I step II step III step IV −− −− H+ −− −−imine iminium ion G step V (i) What is the type of reaction in step I? nucleophilic addition (ii) The positive charge on the iminium ion, G+, does not reside on a carbon atom. Suggest the structure of G+. CH3C H N H +

4 H2 Chemistry 9647/03 NYJC J2/12 PX [Turn Over (iii) Write equations for step I and step IV to show the movement of electrons, using curved arrow notation. Show the lone pairs of electrons, if any, that are involved in each step. (iv) The maximum rate of formation of imines occurs at a pH of about 4.5. Explain why the rate is slow under very acidic conditions and under alkaline conditions. acidic: amine protonated; no nucleophile (Step I) alkaline: very low H+ conc; no catalyst (Step III) [ 6] (c) Acrylamide is a carcinogen. It is formed when potato chips are heated to above 120 °C. Its formation has been linked to the presence of glucose (C 6H12O6) and an α -amino acid L found in relatively high amounts in potato. C C H CONH2 H H H C C OH H C OH H C OH H C OH H C OH H H O acrylamide glucose Glucose reacts with L to form an imine M. At high cooking temperatures, M decomposes to produce only three compounds in equimolar amounts: acrylamide, carbon dioxide and compound N (C6H13NO5). (i) Name the functional groups in acrylamide. alkene , amide

5 H2 Chemistry 9647/03 NYJC J2/12 PX [Turn Over (ii) Deduce the structure of L. Show clearly how you obtained your answer. L : H2N C CH2CONH2 H CO2H (asparagine). Accept other possible alternatives. [5] (d) Suggest the reagents and conditions required to synthesise the following imine derivative. Identify all the intermediate compounds. O N N O OH aq NaBH4 xs conc. H2SO4 heat Br OH NH2 OH aq Br2 xs ethanolic NH3 heat in sealed tube NH2acidified K2Cr2O7 warm O H+ N N [5] [Total: 20]

6 H2 Chemistry 9647/03 NYJC J2/12 PX [Turn Over 2 (a) A container holds a gaseous mixture of nitrogen and propane. The pressure in the container at 200 ° C is 4.5 atm. At − 40 ° C, the propane completely condenses and the pressure drops to 1.5 atm. Calculate the mole fraction of propane in the original gaseous mixture. [3] Let the amount of nitrogen and propane be n 1 and n2 respectively. Since the gas constant, R, and volume of container, V, are constants, the gas equation becomes: P = nR = kn where k = R T V V ∴ 4.5 = k (n1 + n2) = 9.513 x 10−− −− 3 473 and 1.5 = kn 1 = 6.437 x 10−− −− 3 233 ∴ kn2 = 9.513 x 10−− −− 3 − 6.437 x 10−− −− 3 = 3.076 x 10−− −− 3 mole fraction of propane = 3.076 x 10−− −− 3 = 0.323 9.513 x 10−− −− 3 ( b) The graph below shows the variation in electromotive force (e.m.f.) of the following electrochemical cell with lg [Ag+(aq)] at 298 K. Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s) 0 0.1 0.2 0.3 0.4 0.5 0.6 -8.0 -7.0 -6.0 -5.0 -4.0 -3.0 -2.0 -1.0 0.0 (i) Using the information from the graph, calculate the standard electrode potential of the half-cell, Ag+(aq) | Ag(s), at 298 K. lg [Ag+(aq)] e.m.f / V

7 H2 Chemistry 9647/03 NYJC J2/12 PX [Turn Over Let the standard electrode potential of Ag+(aq) || || Ag(s) half-cell be x. When [Ag+(aq)] = 1.00 mol dm−− −− 3 , lg [Ag+(aq)] = 0 E/eshshortrev/eshshortrev /eshshortrev/eshshortrevcell = +0.46 = x −− −− (+0.34) x = +0.46 + (+0.34) = +0.80 V (ii) If the Ag +(aq) solution of the electrochemical cell is replaced by a saturated solution of silver bromate(V), AgBrO 3, in 0.1 mol dm − 3 potassium bromate(V) and the e.m.f. of the cell measured at 298 K is +0.27 V, determine (I) the concentration of Ag+(aq) ions in the saturated solution, and F rom the graph, when e.m.f. of cell is +0.27 V, lg [Ag+(aq)] = −− −− 3.2. [Ag+(aq)] = 6.309 x 10−− −− 4 mol dm−− −− 3 (II) the solubility product of silver bromate(V) at 298 K. Ksp = 6.309 x 10−− −− 4 x 0.1 = 6.309 x 10−− −− 5 mol2 dm−− −− 6 [5] (c) Solid silver nitrate was slowly dissolved in a solution Q containing ethanedioate, C2O4 2− , and chromate(VI), CrO 4 2− , ions of concentrations 2.50 x 10 − 2 mol dm − 3 and 1.44 x 10− 5 mol dm− 3 respectively. (i) When a permanent precipitate of silver ethanedioate first appeared, the concentration of silver ions in the solution was 2.10 x 10 − 5 mol dm − 3. Calculate the solubility product of silver ethanedioate. Ksp(Ag2C2O4) = (2.10 x 10−− −− 5)2 x 2.50 x 10−− −− 2 = 1.10 x 10 −− −− 11 mol3 dm−− −− 9 (ii) The dissolving of solid silver nitrate in Q was continued until a permanent red precipitate of silver chromate(VI) first appeared. Calculate the concentrations of silver ions and ethanedioate ions at that instant. (Ksp of silver chromate(VI) is 1.2 x 10− 12 mol3 dm− 9.) Given: Ksp(Ag2CrO4) is 1.2 x 10−− −− 12 mol3 dm−− −− 9 . 1.2 x 10−− −− 12 = [Ag+]2 x 1.44 x 10−− −− 5 [Ag+]2 = 8.333 x 10−− −− 8 [Ag+] = 2.89 x 10−− −− 4 mol dm−− −− 3 1.102 x 10−− −− 11 = (2.886 x 10−− −− 4 )2 x [C2O4 2−− −− ] [C2O4 2−− −− ] = 1.32 x 10−− −− 4 mol dm−− −− 3 (iii) What is the amount of silver ethanedioate precipitated from 1.00 dm 3 of the solution?

8 H2 Chemistry 9647/03 NYJC J2/12 PX [Turn Over n(C2O4 2−− −− ) removed from solution = 2.50 x 10−− −− 2 −− −− 1.322 x 10−− −− 4 = 0.02486 mol ∴∴ ∴∴ amount of silver ethanedioate ppted = 0.0249 mol. [7] (d) An electrochemical cell containing an oxygen cathode and a hydrogen anode is shown below. The pistons above the gas chambers are frictionless. (i) Write balanced equations for the half reactions and for the overall reaction in the cell. Half reactions: H 2 →→ →→ 2H+ + 2e−− −− (anode) O2 + 4H+ + 4e−− −− →→ →→ 2H2O (cathode) Overall reaction: 2H 2 + O2 →→ →→ 2H2O (ii) How does the concentration of sulfuric acid affect the equilibria of the half reactions? 2H+ + 2e−− −− ÝÝ ÝÝ H2 0.00V (anode) O2 + 4H+ + 4e−− −− ÝÝ ÝÝ 2H2O +1.23V (cathode) Increase in concentration of sulfuric acid will increase [H+] and by Le Chatelier’s Principle, cause the position of equilibrium of anode and cathode reactions to shift right. high resistance voltmeter pistons 1 atm 25 ° C H2 O2 1 mol dm− 3 H2SO4 Pt Pt

9 H2 Chemistry 9647/03 NYJC J2/12 PX [Turn Over (iii) If weights are added to the pistons of both chambers, how would the reading of the voltmeter change? Explain your answer. If weights are added, the pressure will increase, res