RI_Prelim_P3 ANS

© RI(JC) 200 9 1 Raffles Institution (Junior College) 2009 Preliminary Examination H2 Chemistry Suggested Solutions Question 1 (a)(i) Amount of lactic acid = Amount of NaOH needed for neutralisation = 120 . 0x1000 00 .20 = 0.0024 mol Concentration of lactic acid = 0.0024 ÷ 1000 0 .25 = 0.0960 mol dm − 3 (ii) pK a = pH at half-neutralisation = 3.85 (when V NaOH = 10 cm 3) K a = 10 − 3.85 = 1.41 x 10 − 4 mol dm − 3 Alternatively, K a = ]HA [ ]A][ H [ −+ Initially, [H +] = [A − ] = 10 − 2.45 = 3.548 x 10 − 3 mol dm − 3 K a = 3 33 10 x 3.548 0960 . 0 10 x 3.548 x10 x 3.548 − −− − = 1.36 x 10 − 4 mol dm − 3 where HA = lactic acid (iii) At equivalence point, the resultant mixture is a solution of sodium lactate, H 3CH(OH)COO − Na +. Being the conjugate base of a weak acid, lactate i on hydrolyses in water to give OH − : CH 3CH(OH)COO − (aq) + H 2O(l) ⇋ CH 3CH(OH)COOH(aq) + OH − (aq) The formation of OH − causes the pH at equivalence point to be greater t han 7. (b)(i) Samples A and B contain lactic acid but the lactic acid in B is the enantiomer/ optical isomer of the lactic acid in A. Hence A and B behave differently towards plane − polarised light. (ii) Sample C is a racemic mixture of the two optical isomers of lactic acid (contain s equal amounts of the two optical isomers of lactic acid) and hence does not exhibit optical activity. Due to the mixture of the optical isomers, the packing of lactic acid molecules in C is not as regular as the packing of lactic acid molecules in A where only one optical isomer of lactic acid is present. Hence melting point of sample C is lower than that of A. (c) Compound D is D does not contain −− −− OH nor −− −− COOH since it does not react with sodium. It is a di ester formed when − OH and − COOH groups of one lactic acid molecule reacts with the − COOH and − OH groups respectively of another lactic acid molec ule. COOH C H HO CH 3 COOH C H OH CH 3 mirror plane O C C O C C O O CH 3 CH 3 H H

© RI(JC) 200 9 2 (d)(i) E has a chiral centre as it rotates plane − polarised light. E 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 E = 0.24 0.01 24 ÷ = 1 mol Hence two displaceable H are present ⇒ there are two −− −− OH group . E is (ii) E 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 − CH 2CH(OH)CHO group. Lactic acid is a much stronger acid as its anion f orms two equivalent resonance structures with the negative charge on O delocalised over two electronegative O atoms . (e)(i) maleic acid fumaric acid (ii) Intramolecular h