MI H2 CHEM P3 Answer Prelim

1. Lactic acid (2-hydroxypropanoic acid), also known as milk acid, is a chemical compound that plays a role in several biochemical processes. (a) (i) A is an isomer of lactic acid. It rotates plane -polarised light and reacts with Fehling’s solution to give a reddish brown precipitate. 0.01 mol of A reacts with sodium to give 0.24 dm 3 of hydrogen gas under room temperature and pressure conditions. Suggest a structural formula of A, giving reasons for your answer. A is ; A has a chiral centre as it rotates planepolarised light. ; A is an aldehyde since it reacts with Fehling’s solution to give a reddish brown precipitate. ; Amount of hydrogen evolved with sodium per mol of A = 0.24 0.0124  = 1 mol Hence, there are two OH group. ; (ii) Compare the acidity of isomer A and lactic acid. Explain your answer. A is less acidic than lactic acid. ; A is an alcohol. The stability of its anion is lower than the stability of the anion of lactic acid as the anion is destabilised by the electrondonating alkyl group. ; Lactic acid is a much stronger acid as its anion forms two equivalent resonance struc tures with the negative charge on O delocalised over the carbon and two electronegative O atoms. ; [7] (b) Lactic acid can undergo a condensation reaction in the presence of hot concentrated sulfuric acid to form a cyclic diester B, C6H8O4.

The system exists in dynamic equilibrium. (i) Draw the structure of diester B. ; (ii) Explain what is meant by the term dynamic equilibrium. rate of forward reaction equals to rate of reverse reaction ; (iii) Write an expression for the equilibrium constant, K C, for the reaction, stating its units. [ ][ ] [ ] ; units = mol dm-3 ; (iv) In an experiment, 2.0 mol of lactic acid was allowed to reach equilibrium. Given that 95% of lactic acid has reacted at equilibrium and the total volume of the mixture remained constant at 200 cm 3, calculate the K C for this reaction. 2 lactic acid ⇌ B + 2H2O initial mol 2 0 0 change in mol -0.95(2) +0.95 +0.95(2) eqm mol 0.1 0.95 1.9 ICE table ; ( )( ) ( ) mol dm-3 ; (v) In the reverse reaction, i.e. the hydrolysis of the diester B, the rate of reaction gradually increases at first, but subsequently it decreases. Explain why this is so. The rate increases at first when the carboxylic acid is first formed

because it partially dissociates to give H + which catalyses the hydrolysis of ester. ; The rate decreases as the reaction proceeds due to decrease in the concentration of the ester, leading to decrease in frequency of effective collisions. ; [8] (c) Lactic acid, along with ammonium hydrogen carbonate, (NH 4)HCO3, is used in mosquito attractant. Ammonium hydrogen carbonate is a solid which can decompose into ammonia, carbon dioxide and water. (i) Write an equation, with state symbols, for the decomposition of ammonium hydrogen carbonate. (NH4)HCO3 (s)  NH3 (g) + CO2 (g) + H2O (l) ; (ii) Using the enthalpy changes given in the table below, calculate the enthalpy change of reaction for the decomposition of (NH4)HCO3. Compound Enthalpy change of formation / kJ mol-1 (NH4)HCO3 (s) -853 NH3 (g) -46.0 CO2 (g) -394 H2O (l) -286 H = (-46.0 – 394 – 286) – (-853) ; = +127 kJ mol-1 ; (iii) Predict, with a reason, the sign of S for the decomposition of ammonium hydrogen carbonate. S is positive. ; The system becomes more disorderly as there is an increase in number of moles of gases. ; [5] [Total: 20] 2. (a) Compound P is a halogen derivative with molecular formula C 9H10Br2. Reaction of

P with ethanolic KOH gives rise to a mixture of two isomeric products, Q and R, with the molecular formula C 9H9Br. Both Q and R decolourise hot acidified KMnO 4 and the reaction produces S (structure given below) and ethanoic acid, CH3COOH. On treatment with excess ethanol ic ammonia, P produces T (C9H12NBr) as the major product. (i) Draw all possible structural isomers of S that have the same chemical properties. ; ; (ii) Deduce the structures of P, Q, R and T. Explain your reasoning. P has C:H  1:1  P is likely to contain a benzene ring P undergoes elimination with ethanolic KOH to give Q and R  Q and R are geometric isomers with only one C=C and one of the Br atoms in P is bonded to the benzene ring Q and R undergo oxidation with hot acidified KMnO 4 to give two carboxylic acids, S and CH3COOH  Q and R are alkenes and each of the C atom of the C=C bond is bonded to one R group and one H atom P undergoes nucleophilic substitution with NH3 to give T (C9H12NBr)  T is a primary amine

C C C H H H H H Br C C H C H Br H H H Q R C C C H H H H H HBr Br P C C C H H H H H HN Br T H H 1m for each deduction (capped at 2m) 1m for each structure [8] (b) Bromocyclohexane (C 6H11Br) and benzoyl bromide (C 6H5COBr) differ in their reactivity with aqueous sodium hydroxide. Describe and explain the differences. Benzoyl bromide reacts with aqueous NaOH at room temperature, while bromocyclohexane reacts with aqueous NaOH only on heating. ; Due to the presence of electronegative O attached to the carbonyl C, the carbonyl C in C 6H5COBr is more partially positive compared to the C to which Br is attached in C6H11Br. OR The carbonyl carbon in C 6H5COBr is sp 2hybridised so that it is less sterically hindered than the sp 3hybridised carbon to which th e Br atom is attached in C6H11Br. ; Thus, C6H5COBr undergoes nucleophilic substitution more readily. [2] (c) The kinetics of the reaction between bromocyclohexane and hot aqueous

sodium hydroxide was investigated. 1.0 mol dm-3 of bromocyclohexane was reacted with 0.01 mol dm -3 of aqueous sodium hydroxide. The pH of the reaction mixture was measured at regular time intervals using a data -logger. The data in the following table was obtained. Time / min pH [OH-] / mol dm-3 0 13.0 2 12.9 4 12.8 6 12.6 8 12.4 10 11.9 (i) Copy the table and fill in the [OH-] for each set of readings. Time / min pH [OH-] / mol dm-3 0 13.0 0.100 2 12.9 0.0794 4 12.8 0.0631 6 12.6 0.0398 8 12.4 0.0251 10 11.9 0.00794 All 6 values calculated correctly 2m At least 3 values calculated correctly 1m Less than 3 values calculated correctly 0m (ii) Using a graphical method, prove that the reaction is zero order with respect to [OH-].

Graph ; Since gradient is constant, the rate is independent of [OH-]. ; (iii) Given that the reaction is first order with respect to [bromocyclohexane], give the rate equation for the reaction. rate = k[bromocyclohexane] ; (iv) The half -life of bromocyclohexane was found to be 75.3 min when 1.0 mol dm -3 of bromocyclohexane was reacted with 0.01 mol dm -3 of aqueous sodium hydroxide. Deduce the half -life of bromocyclohexane when 2.0 mol dm -3 of bromocyclohexane was reacted with 0.02 mol dm -3 of aqueous sodium hydroxide. Explain your reasoning. Since overall order of reaction is one, half life is constant. ; 75.3 min ; (v) Determine the rate constant of the reaction when 1.0 mol dm -3 of bromocyclohexane was reacted with 0.01 mol dm -3 of aqueous sodium hydroxide. k = ln 2 / 75.3 = 9.21 × 10-3 min-1 ; (vi) Using your answer in (c)(iii), propose the reaction mechanism for this reaction. ; 0 0.02 0.04 0.06 0.08 0.1 0.12 0 2 4 6 8 10 12 [OH-] / mol dm-3 Time / min

; [10] [Total: 20] 3. (a) Haemoglobin is a water soluble globular protein which is composed of two α polypeptide chains, two β polypeptide chains and an inorganic prosthetic haem group. Its function is to carry oxygen around in the blood and it is facilitated in doing so by the presence of the haem