ACJC H2 CHEM P3 ANS Prelim

PAPER III 1 Penicillin can be made in the laboratory by reacting 6-aminopenicilanic acid with a suitable acyl chloride. For a particular penicillin, the acyl chloride is the CH3CH2CH=CHCH2COCl. Since the acyl chloride used can be easily made from the corresponding carboxylic acid, the chemist has decided to work with CH3CH2CH=CHCH2CO2H because the acyl chloride is more reactive. H3CH2CH=CHCH2CO2H + PCl5  CH3CH2CH=CHCH2COCl + POCl3 + HCl (a) (i) The acyl chloride, CH 3CH2CH=CHCH2COCl, can be obtained by reacting CH3CH2CH=CHCH2CO2H with PCl5. Write a balanced equation for the reaction. [1] C (ii) ergie given alculate the enthalpy change for the above reaction. [2] Bond Energies /kJ mol-1 Using relevant data from the Data Booklet and the bond en s below, c P-Cl P=O 423 322 rokenBonds B Bonds formed 2 P-Cl 2(+322) P=O -423 C-O +360 C-C l -340 O-H +460 H-C l -431 H 60-423-340-431 = +270 kJ mol-1  = 2(+322)+360+4 (b) (i) e reaction. [1] team; heat in the presence of H PO catalyst 300 oC and 65 atm lternative: conc H2SO4 at 0 oC, followed by boiling with H2O The carboxylic acid, CH 3CH2CH=CHCH2COOH, can be converted into the hydroxyacid, CH 3CH2CH(OH)CHCH2COOH. Give the reagent and condition for th S 3 4 A (ii) romine. Name and describe the mechanism involved. [4] lectrophilic Addition The hydroxyacid obtained can be distinguished from its starting material by using b E C C H CH2COOH H CH2CH3 + Br Br   Slow C C H CH2COOH H CH2CH3 electrophile Br Br © ACJC 2009 9746/03/Prelim/09 [Turn over

Br Fast C C CH2COOH H H CH2CH3 Br Br nucleophile C C H CH2COOH H CH2CH3 Br (iii) Suggest another reagent and condition in which the hydroxyacid obtained can be distinguished from its starting material. [2] Reagent: acidified K2Cr2O7 and heat (c) (i) In the course of preparing the hydroxyacid, an internal chiral ester was also detected. Draw the structure of this ester. Circle the chiral centre(s). [2] CH CH2 CH2 C O CH2CH3 O (ii) State the number of sp2 and sp3 hybridised carbons in the ester. [2] sp2 - 1 sp3 - 5 (iii) This ester may be made into a diol. Give the reagent and condition for the conversion and draw the structure of this diol. [2] Reagent and condition: LiAlH4 in dry ether; reflux OH CHCH2CH2CH2OHCH3CH2 (d) (i) An experiment was carried out to determine the enthalpy change of combustion of this diol. Write a balanced equation including state symbols for the combustion of the diol under standard condition. [1] CH3CH2CH(OH)CH2CH2CH2OH(l) + 17/2O2(g)  6CO2(g) + 7H2O(l) (ii) A large beaker of water was placed on the stove and heated. The temperature rise was recorded. The cylinder was weighed before and after the experiment to determine the mass of diol used. The following results were obtained. © ACJC 2009 9746/03/Prelim/09 [Turn over

Mass of diol used = 3.4 g Mass of water heated = 500 g Temperature rise = 44 oC Using relevant data from the Data Booklet and results given above, calculate the enthalpy change for the combustion of the diol. [3] Q = mcT = (500)(4.18)(44) = -91960 J Mr of diol = 6(12.0) + 14(1.0) + 2(16.0) = 118.0 H = Q/ n = -91.96/ (3.4/118.0) = -3191 kJ mol -1 = -3190 kJ mol-1 [Total: 20] © ACJC 2009 9746/03/Prelim/09 [Turn over

2 (a) The triodide ion, I3 -, is known to exist as a stable species in aqueous solution. Draw a dot and cross diagram for I3 - and hence deduce its shape. [2] I Ix I xx xx xx Linear (b) tionalize the relative stability of CsI3 and LiI3. [3] o Cs+ ent by Li+ sI3 is more stable than LiI3. Solid CsI3 is stable with respect to CsI and I 2 but LiI 3 is not stable with respect to LiI and I 2. Based on your knowledge of the relative stabilities of group II carbonates, ra Li+ is much smaller than Cs+ Charge density of Li+ is higher than that f I3 - is polarized to a greater ext C (c) (i) [2] Ca(s) + Ca2+(g)  2Ca+(g) a+(g) → Ca2+(g) + e 2 nd IE = +1150 kJ mol-1 Ca(s) + Ca2+(g) → 2Ca+(g) Ca(g) + Ca2+(g) H = +176.4 - 1150 + 590 = -383. 6 kJ mol-1 Given that the enthalpy of atomization of Ca is +176.4 kJ mol -1, and using relevant ionization energy values from the Data Booklet, calculate the enthalpy change for the reaction: Ca(g) → Ca +(g) + e 1 st IE = +590 kJ mol-1 C Δ (ii) u onclud about the relative stability of Ca +(g th r spect nd Ca2+? [1] a+(g) is more stable with respect to Ca(s) and Ca2+(g) What can yo c e ) wi e to Ca a C (d) tions between the aqueous ions of Ca and the solid Ca metal is the light o f your answer to (c)(ii), account for the differences (if any) noted. [2] 2+(aq) The energetic rela given as follows: Ca(s) + Ca2+(aq)  2Ca+(aq) ΔH = +63 kJ mol-1 In Ca+(aq) is less stable with respect to Ca(s) and Ca Hy dration of Ca2+ ion is more extensive than Ca+ © ACJC 2009 9746/03/Prelim/09 [Turn over

HE of Ca2+ >> HE of Ca+ (e) (i) t the enthalpy changes associated with the reactions vary considerably. bly hat factor is responsible for the dissimilarity of the set ΔH values? [1] Based on data, the standard electrode potentials of Ca, Sr and Ba are nearly the same, bu . Although there is found to be a good correlation between the ΔH for M2+(aq) + 2e → M(s) and the corresponding E θ (M2+/M), the degree of similarity between the E θ (M2+/M) is greater whilst that for the Δ H is noticea lower. W Mg Ca Sr Ba Eθ (M2+ / M) / V -2.38 -2.87 -2.89 -2.90 ΔH / kJ mol-1 M2+(aq) + 2e → M(s) -403.2 -256.2 -289.8 -365.4 a ood measure of the spontaneity / tendency of the reaction. Entropy changes during hydration are neglected and hence ΔH is not g © ACJC 2009 9746/03/Prelim/09 [Turn over

(ii) For Mg, ΔH deviates so greatly, that no good correlation can be observed between ΔH and E θ. Suggest a reason for this inconsistency between the two v a l u e s f o r M g . [2] Mg2+ is relatively smaller with a high charge density compared with the other ions which are much larger. Hence a greater degree of hydration of Mg2+ leads to greater orderliness in the solution thus affecting the entropy change significantly. (iii) In the light of your answers to (e)(i) and (e)(ii), what energy term will give a more exact correlation with the Eθ? Why? [2] ΔG for M2+(aq) + 2e  M ( s ) ΔGθ accounts for entropy changes in the reaction (f) HF(aq) is the weakest acid compared with the other aqueous hydrogen halides. One factor relevant to its weakness is that the entropy change that accompanies its dissociation is the most negative. What other factor is relevant in explaining the weakness of HF(aq) as an acid? [1] The dissociation of aqueous hydrogen halides as represented by the equation: HX(aq) → H +(aq) + X -(aq), is least favourable energetically for HF due to high B E . (g) (i) The enthalpy of vaporisation of the hydrogen halides is markedly different for one of the HX as compared with the others. Identify the hydrogen halide and suggest what phenomenon is responsible for this deviation. [2] HF Hydrogen bonds between HF molecules (ii) The values of entropies of vaporization ( ΔSb), that is the molar latent enthalpy of evapor