ACJC 2022 Prelim P3 Answers

© ACJC2022 9729/03/Prelim/2022 ACJC solutions for H2 Chemistry Prelim Paper 3 2022 Section A 1 (a) Explain what is meant by the term Lewis acid. [1] Lewis acid is a substance that accepts an electron pair from a base via a dative covalent bond (i.e. an electron-pair acceptor) Examiner’s comments:  Do not be confused with the Bronsted-Lowry or Arrhenius’ definition of acids. (b) The water molecule can react in various ways: as an acid, as a base, as a nucleophile, as an oxidising agent and as a reducing agent. Study the following reactions and decide in which way water is reacting in each case. Explain your answers fully. [5] (i) 2F2 + 2H2O  4HF + O2 Reducing agent as itself is being oxidised. The oxidation number of O increases from −2 in H2O to 0 in O2. Or water reduces F2 to HF as the oxidation number of F decreases from 0 in F2 to –1 in HF. Examiner’s comments:  Change in oxidation number and relevant substance must be clearly stated as question stated to explain your answers fully. (ii) CH3COCl + H2O  CH3COOH + HCl Nucleophile. The carbon of the acyl chloride functional group is electron deficient and is susceptible to nucleophilic attack by water, which has a lone pair of electrons on its oxygen atom. Examiner’s comments:  Your answer should clearly reflect the definition of a nucleophile.  Base is not acceptable as a base involve donation of elec tron pair but without the expulsion of any groups. In this case, Cl is expelled and replaced with OH hence the reaction is a nucleophilic substitution rather than acid base. (iii) H2PO4− + H2O HPO42− + H3O+ Base. H2O accepts a proton from H2PO4− to form H3O+ and HPO42−. Examiner’s comments:  Your answer should clearly reflect the definition of a base, i.e. mention of the accepting of proton.

© ACJC2022 9729/03/Prelim/2022 (iv) Li + H2O  LiOH + ½ H2 Oxidising agent. Water oxidises Li to Li + as the oxidation number of Li increases from 0 in Li to +1 in LiOH Examiner’s comments:  Change in oxidation number and relevant substance must be clearly stated as question stated to explain your answers fully. (v) NO2− + H2O HNO2 + OH− Acid. H2O donates a proton to NO2− to form OH− and HNO2. Examiner’s comments:  Your answer should reflect the definition of an acid, i.e. proton donor. (c) The pKa values of three acids are listed in the Table 1.1 below: Table 1.1 acid formula pKa 1 CH3CH2COOH 4.9 2 CH3CHClCOOH 2.8 3 CH2ClCH2COOH z (i) Explain the difference in pKa values between acid 1 and acid 2. [2] The presence of the electron withdrawing chlorine group in acid 2 disperses the negative charge on the conjugate base, CH3CHClCOO−, hence making it a more stable conjugate base than CH3CH2COO−. Therefore, acid 2 has a lower pKa as this increases the acid strength of CH3CHClCOOH. Examiner’s comments:  The difference between acids 1 and 2 lie in the presence of the electron withdrawing Cl group hence this must be clearly stated.  Dispersal of negative charge in a conjugate base contributes to its stability so this should also be clearly stated. (ii) Suggest a value for z and explain your answer. [1] 4.0 (accept any value between 2.8 and 4.9) As the distance between the –COOH group and the C l atom increases, the negative charge on the conjugate base is less dispersed. The conjugate base becomes less stable and hence acid strength decreases. However, acid 3 should still be more acidic than acid 1 due to the presence of the electron withdrawing Cl group.

© ACJC2022 9729/03/Prelim/2022 Examiner’s comments:  The main difference between acid 2 and 3 is the proximity of the Cl group.  Dispersal of negative charge in a conjugate base contributes to its stability so this should also be clearly stated. (iii) Peroxyacids are weak acids. One way to prepare peroxypropanoic acid is to treat the corresponding carboxylic acid with hydrogen peroxide. acid 1 peroxypropanoic acid Suggest why the pKa of peroxypropanoic acid is higher than that of acid 1. [1] CH3CH2CO3H is less acidic than CH 3COOH as CH3CH2CO3− is less stable than CH3CH2CO2− because the negative charge on CH3CH2CO3− cannot be delocalised over the C=O group due to the additional oxygen atom , hence the resonance stabilisation of the conjugate base is lost. Examiner’s comments:  The main difference between acid 1 and peroxypropanoic acid is the fact that the negative charge on the conjugate base of peroxypropanoic acid , CH3CH2CO3− cannot be delocalised as the p orbitals of O with the negative charge cannot overlap with the pi electron cloud of the C=O.  As shown below, the oxygen atom preventing the continuous p orbital overlap CH3CH2CO3− is sp3 hybridised as it has 2 lone pairs and 2 bond pairs. Conjugate base of acid 1 Conjugate base of peroxypropanoic acid (d) A, B and C are isomers with the molecular formula C 5H6O2. All three compounds decolourise bromine water in the dark. A produces effervescence in the presence of Na 2CO3(aq) whereas B and C do not. A also reacts with hot acidified KMnO4 to form D, C3H2O5. B forms a brick-red precipitate when heated with Fehling’s solution. C reacts with hot aqueous sodium hydroxide. Upon acidification, it forms C5H8O3.

© ACJC2022 9729/03/Prelim/2022 C5H8O3 When B and C are separately reacted with hot acidified KMnO4, they form the same mixture of E C3H4O4 and F C2H2O4. F undergoes further oxidation to give effervescence. (i) Draw the structures of D, E and F. [3] D is E is F is HOOC–COOH Examiner’s comments:  Since D, E and F are formed from oxidation with acidified KMnO 4, they contain functional groups that do not undergo further oxidation. Thus D, E and F have carboxylic acid, ketone and / or tertiary alcohol functional groups, but not primary alcohol, secondary alcohol or aldehyde. (ii) Deduce the structures of A, B and C with reasoning. [7] A, B and C have 3 degrees of unsaturation. A, B and C have alkene functional group that undergoes electrophilic addition with bromine water. A, B and C undergoes vigorous oxidation / oxidative cleavage of the alkene functional group. A has the carboxylic acid functional group that undergoes acid-carbonate / acid- base reaction with carbonate. B has the aldehyde functional group that undergoes oxidation in the presence of Fehling’s solution. C undergoes alkaline hydrolysis, so it has an ester functional group. A B C Examiner’s comments:  Good answers made deductions based on the types of reactions and functional groups present  “Double bond” is not an acceptable description of the alkene functional group  “Carboxylic group” or “acidic group” is not an acceptable des