ACJC 2026 Prelim P3 Ans

© ACJC2026 9476/Preliminary Examination/2026 [Turn over Anglo-Chinese Junior College JC2 Chemistry Preliminary Examinations Higher 2 CANDIDATE NAME FORM CLASS INDEX NUMBER SUBJECT TUTOR CHEMISTRY Paper 3 Structured Questions You must answer on the question paper. 9476/03 01 September 2026 2 hours You will need: Data Booklet INSTRUCTIONS • Section A: answer all questions. • Section B: answer one question. • Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. • Write your name, centre number and index number in the boxes at the top of the page. • Write your answer to each question in the space provided. • Do not use an erasable pen. Do not use correction fluid or tape. • Do not write on any bar codes. • You may use an approved calculator. INFORMATION • The total mark for this paper is 75. • The number of marks for each question or part question is shown in brackets [ ]. For Examiner’s Use Question Number Marks 1 2 3 4 5 Presentation of answers Total: 75

© ACJC2026 9476/Preliminary Examination/2026 [Turn over Section A Answer all the questions in this section. 1 (a) A kinetics investigation for the reaction of the halogenoalkanes K, L and M with NaOH(aq) at a particular temperature is carried out. State and explain the relative rate of reaction of K, L and M with NaOH(aq). [2] L > K > M. L is the most reactive while M is the least reactive. Comparing K and L, the C−Br bond is weaker than the C−Cl bond / extent of orbital overlap / bond energy of C−Br bond is smaller than bond energy of C −Cl bond, hence less energy to break. Hence L has a faster reaction rate with OH ‒ than K. Comparing K and M, there are more electron-donating alkyl groups on the tertiary carbon on M, the approach of OH ‒ towards the electron -deficient C in M is more difficult as there is greater steric hindrance / making the tertiary carbon on M less electron deficient, hence M is slower/slowest. Alternative answer: L > M > K. Comparing K and L, the C−Br bond is weaker than the C−Cl bond / extent of orbital overlap / bond energy of C−Br bond is smaller than bond energy of C −Cl bond, hence less energy to break. Hence L has a faster reaction rate with OH ‒ than K. Comparing K and M, there are more electron donating alkyl groups on the tertiary carbon on M. M is a tertiary halogenolkane, hence it undergoes S N1. The carbocation formed by M is stable as the alkyl groups disperse the positive charge, hence M reacts more quickly. Comments A common misconception is using electronegativity of Cl and Br to explain the difference in reactivity of K and L. The rate of reactivity for halogenoalkanes in nucleophilic substitution reactions is dependent on the ease of breaking of the carbon-halogen bond. With respect to M, while it is a tertiary halogenoalkane and undergo a different mechanism (SN1) as the primary halogenoalkane (SN2), it is incorrect to say its rate is slower as there are two steps involved. (b) (i) When K is reacted with sufficient ammonia and heated, a mixture of amine S, C4H11N, and by-product T are obtained. Draw the structure of amine S. [1] S: K L M

© ACJC2026 9476/Preliminary Examination/2026 [Turn over (ii) Some tests were conducted on T. The observations are shown in Table 1.1. Table 1.1 Mr 277.5 addition of aqueous AgNO3 Immediate white precipitate formed Use the information in Table 1.1 to draw the structure of T. [1] T: Comments This question was poorly answered. Many students did not fully utilise the information given in Table 1.1 to deduce the structure of T. (c) K can also undergo similar reaction in (b)(i) with sodium azide, NaN3, to form amine S. The intermediate formed in the process is an alkyl azide. (i) Use VSEPR theory to predict the C -Nx-Ny bond angle in alkyl azide. Explain your answer. [2] Nx has 2 bond pairs, 1 lone pair of electrons on it. Since (repulsion between lone pair-lone pair >) repulsion between lone pair-bond pair > repulsion between bond pair-bond pair, hence to minimise repulsion, the bond angle is 118°.(any suggested value between 110-118°) Comments Many students’ answers were incomplete, or incorrectly identified the number of bond pairs and lone pairs of electrons around Nx. (ii) Draw a labelled diagram to show the orbitals that form the Nx‒Ny bond in the alkyl azide. [2] alkyl azide

© ACJC2026 9476/Preliminary Examination/2026 [Turn over Comments As there are π bonds on Nx and Ny, hence the orbitals involved in bonding between Nx and Ny must include both hybrid orbital and unhybridized p orbital. (iii) The azide ion, N3‒, consists of both σ and π bonds. Draw the ‘dot-and-cross’ diagram of the azide ion, N3 ‒ . [1] Comments Students should ensure the total number of electrons around N atom must not exceed 8 as N is unable to expand octet configuration. (d) (i) The azide ion, N 3− can be converted to the azide radical, •N3(g) under suitable conditions Write an equation to represent the enthalpy change of formation of the azide radical, •N3(g) from nitrogen gas. [1] 3 2 N2(g) →N3(g) Comments The enthalpy change of formation equation refers to the energy required to form one mol of the azide radical. State symbols are required. (ii) The bonding between the nitrogen atoms in the azide radical is similar to the alkyl azide. The enthalpy change of formation of the azide radical, •N3(g) is +452 kJ mol−1. Using data from the Data Booklet, calculate a value for the average bond energy of the nitrogen-nitrogen bond in an N3 radical. [2] ∆Hf(N3) = 3/2 BE(N≡N) – 2 BE(N=N) 2BE(N=N) = 3/2 (+944) ‒ 452 BE(N=N) = +482 kJ mol‒1 Comments Common mistakes were students being unable to correctly apply the formula using bond energy data, as well as not realising there are 2 N=N bonds in an azide radical. (iii) Using your answer to (d)(i) and (ii), state the sign of entropy change of formation of azide radical. Hence, explain why the formation of azide radical from nitrogen does not occur readily at all temperatures. [2]

© ACJC2026 9476/Preliminary Examination/2026 [Turn over The sign of entropy change is negative. In ΔG = ΔH – TΔS, since ΔS is negative, then ΔH is positive / – TΔS is positive. As ΔH is also positive, hence ΔG is always positive at all temperatures. Comments The question already implies that the reaction is not spontaneous at all temperatures , hence students’ explanations have to justify why ΔG is always positive by comparing the signs of ΔH and/or – TΔS. (e) The Claisen-Condensation reaction involves the reaction of 2 esters to form a beta-keto ester, where a ketone group is formed at the second carbon away from the ester. An example of the Claisen -Con