2026 H2 Timed Practice Paper 2 Solutions + Examiner Comments (updated 17 July)

NJC SH2 Timed Practice 9476/2/26 [Turn over NATIONAL JUNIOR COLLEGE SH2 MID YEAR TIMED PRACTICE Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on Question Paper. Additional Materials: Data Booklet 9476/02 7 July 2026 2 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 /7 2 /19 3 /15 4 /18 5 /12 6 /9 Paper 2 Total /80 This document consists of 24 printed pages.

2 NJC SH2 Timed Practice 9476/2/26 1 (a) Two organic compounds, A and B, can co -polymerise to produce a polyester. On treatment with hot acidified sodium dichromate( VI), compound A, C4H10O2, is converted into compound B. Compound B can also be formed from fumaric acid, HO2CCH=CHCO2H, upon treatment with hydrogen over a nickel catalyst (i) Draw the structures for compounds A and B. A B HOCH2CH2CH2CH2OH HO2CCH2CH2CO2H Examiner's Comment: Many students cannot recall that H2 with Ni only reduces C=C and COOH group can only be reduced by LiAlH4 in dry ether. To solve for A, students need to draw structure of B first. [2] (ii) Draw the structural formula of one repeat unit of the polymer formed from compounds A and B. [1] Examiner's Comment: Students should check if their drawn repeat unit to give the polymer when repeated. Poly(A-B) is a biodegradable plastic and can be used for food packaging like disposable tablewares and paper cups. (iii) Suggest why poly(A-B) is biodegradable. [1] Poly(A-B) has ester linkages that can be hydrolysed. Examiner's Comment: Many students did not highlight it is the presence of ester linkages which can be hydrolysed.

3 NJC SH2 Timed Practice 9476/2/26 [Turn over (b) Ethane-1,2-diol, HOCH2CH2OH, has a wide variety of uses including in antifreeze in cars. Ethane-1,2-diol undergo a series of steps to form NH2CH2CH2NH2. HOCH2CH2OH Step 1 Y ethanolic NH3 heat in sealed tube H2NCH2CH2NH2 (i) Suggest the structure of Y and state the reagents and conditions for step 1. [2] Y: BrCH2CH2Br / ClCH2CH2Cl Step 1: anhydrous PBr3 / PCl3 / PCl5 / SOCl2 or HBr (g) or HCl(g) (ii) Deduce which compound, HOCH 2CH2OH or NH 2CH2CH2NH2 has a higher boiling point. ………….………………………………………………………………………... [1] HOCH2CH2OH can form stronger H-bonds between molecules since O-H bond is more polar than N-H bond. Therefore, boiling point of HOCH2CH2OH will be higher. Examiner's Comment: Many students mistaken N to be more electronegative than O, hence N-H bond is more polar, giving rise to stronger hydrogen bonds. Many did not realise that both NH2CH2CH2NH2 and HOCH2CH2OH have the same extensiveness in hydrogen bonding (consider the no. of lone pairs on O/N and no. of protonic H). A few highlighted the higher Mr of HOCH2CH2OH, hence larger electron cloud and stronger idid. While this is correct, the difference in 2 in no. of electrons is not significant. [Total: 7]

4 NJC SH2 Timed Practice 9476/2/26 2 Bromine can form several compounds under suitable conditions as shown in Table 2.1. Table 2.1 Bromine containing species Oxidation number of Br Br − −1 BrO− +1 BrF4− +3 BrO3− +5 BrO4− +7 (a) (i) Draw the dot-and-cross diagram for BrO3−. [1] (i) Examiner's Comment: O being more electronegative, will attract the extra electron to form the anion. Br is able to expand octet, hence double bonds to O atoms. (ii) State and explain the bond angle in BrO3−. [2] There are 3 bond pairs (BP) and 1 lone pair (LP) around Br, electron geometry is tetrahedral. There is stronger repulsion between LP -BP than BP-BP. thus lone pair presses the bond pairs towards each other resulting in bond angle to be 107o < 109.5o and the shape is trigonal pyramidal around Br. Examiner's Comment: Many students did not highlight the stronger repulsion of lone pair-bond pairs.

5 NJC SH2 Timed Practice 9476/2/26 [Turn over (b) When a sample of 9.909 g of Br2(l) was reacted with an excess of aqueous NaOH, two different bromine containing ions, X and Y, are formed. The identities of X and Y can be found in Table 2.1. The resulting mixture was then made up to 250 cm 3 with deionised water to form solution Q. The following procedures were carried out on solution Q. 1. When 25.0 cm3 of solution Q was acidified with an excess of acid, followed by the addition of AgNO3(aq), a cream precipitate was formed. 2. This precipitate was insoluble in excess NH3(aq), and its mass was found to be 1.942 g. 3. To a separate 25.0 cm3 portion of solution Q, excess KI solution was added. The I2 liberated required 24.80 cm 3 of 0.500 mol dm -3 Na2S2O3 for complete reaction as shown. 2S2O32– + I2 → S4O62– + 2 I – (i) Given that ion X is responsible for the formation of the cream precipitate, state the identity of X. Ion X: ………………………… [1] Br− Examiner’s comment: • Students struggled with extracting the important key words in the question for answering the question. • Some students gave “AgBr” as answer ignoring the fact that question asks for ion. • A few students suggest “Ag+ / I−” even though question stated that X is bromine containing ions. • A few students suggest BrO− as the ion, forgetting that AgNO3 typically test for halides. (ii) Calculate the amount of X in 25.0 cm3 of solution Q. Amount of Br– in 25.0 cm3 of solution Q = 1.942 79.9+107.9 = 0.01034 mol Examiner’s comment: • Students struggled with this calculation when they failed to recognise that the Br- ion is precipitated as AgBr. • Some failed to realise that the mass collected in step 2 is that of AgBr not just Br− ion and definitely not NaBr. [1]

6 NJC SH2 Timed Practice 9476/2/26 (iii) Calculate the amount of bromine atoms in 9.909 g of Br2(l). Hence, show that there are 0.00206 mol of Y in 25.0 cm3 of solution Q. Amount of Br2 in 250.0 cm3 of solution Q = 9.909 2×79.9 = 0.06201 mol Total moles of Br atoms in Br− and Y = (2)(0.06201) = 0.12402 Amount of Y in 250.0 cm3 of solution Q = 0.12402 – ( 250 25.0) (0.01034) = 0.02062 mol Amount of Y in 25.0 cm3 of solution Q = ( 25.0 250) (0.02062) = 0.002062 (shown) Examiner’s comment: • Students struggled with this calculation when they failed to recognise Br2 is the reactant that leads to formation of bromine containing ions X and Y. • Some are confused whether they should calculate the number of Br atoms or moles of bromine atoms. Students are reminded to associate “amount” to moles than the number. • Some struggled with whether they are calculating moles in 25.0 cm3 or 250 cm3. • Some manage to get the numerical value by working with math instead of using chemistry principles. [2] (iv) In step 3, Y reacts with KI to form X. Deduce the oxidation state of Br in Y.