ACJC 2026 Prelim P2 Ans

2 ACJC2026 9476/Preliminary Examinations/2026 [Turn over Answer all the questions in the spaces provided. 1 Butenedioic acid, HOOCCH=CHCOOH, is a weak diprotic Brønsted-Lowry acid that exists as two stereoisomers: cis-butenedioic acid (maleic acid) and trans-butenedioic acid (fumaric acid). It has two acidic hydrogen atoms, and this is illustrated by the following equations 1.1 and 1.2. equation 1.1 HOOCCH=CHCOOH + H2O HOOCCH=CHCOO− + H3O+ equation 1.2 HOOCCH=CHCOO− + H2O −OOCCH=CHCOO− + H3O+ (a) (i) Identify a conjugate acid-base pair from the equations above. acid: ………………………………….. conjugate base: ………………………………… [1] HOOCCH=CHCOOH and HOOCCH=CHCOO− HOOCCH=CHCOO− and −OOCCH=CHCOO− H2O and H3O+ Comments Conjugate acid-base pairs are species that differ by one H+ ion. (ii) Draw and label the structures of the cis and trans isomers of butenedioic acid. [2] cis isomer trans isomer Comments Correct structures that show the trigonal planar shape about C=C is expected. Students must also label cis and trans isomer correctly. (b) Table 1.1 shows the acid dissociation constants for both isomers at 298 K. isomer pKa1 pKa2 cis-butenedioic acid 1.93 6.23 trans-butenedioic acid 3.02 4.23 Table 1.1

3 ACJC2026 9476/Preliminary Examinations/2026 [Turn over (i) Draw the displayed formula of the conjugate base of cis-butenedioic acid. [1] all bonds to be shown Comments Many students fail to show the full displayed formula especially forgetting to show the O – H bond clearly. (ii) With reference to your answer in (b)(i), suggest why the pKa1 of cis-butenedioic acid is significantly lower than that of trans-butenedioic acid. ………………………………………………………………………………………………... [2] monoanion of cis-butenedioic acid is more stable due to the intramolecular hydrogen bond formed between -COOH and -COO– groups as they are near each other , hence stronger cis-butenedioic acid and therefore lower pKa1. monoanion of trans-butenedioic acid monoanion is less stable due to the carboxylic acid and carboxylate ion far apart, making intramolecular hydrogen bond unlikely. Hence, less acidic trans-butenedioic acid and therefore higher pKa1. Comments The lower pKa1 of acid implies a stronger acid and thus a more stable conjugate base, monoanion of the acid. Students must state clearly the groups that cause the intramolecular hydrogen bonds.

4 ACJC2026 9476/Preliminary Examinations/2026 [Turn over (c) A student adds a small portion of 0.100 mol dm -3 sodium hydroxide, NaOH(aq) to 20 cm 3 of 0.100 mol dm-3 cis-butenedioic acid. The student uses a pH meter to measure the pH of the mixture. (i) Using pKa1 value, calculate the initial pH of the cis-butenedioic acid solution before any NaOH(aq) is added. [2] 𝑲𝒂 = [𝑯+ ]𝟐 𝟎. 𝟏 [H+] = √10−1.93 (0.1) [H+] = 0.0342768 mol dm-3 pH = -lg (0.0342768) = 1.47 Comments Students will calculate the initial pH of this weak acid as if it is a monobasic acid using only pKa1. As usual, students will assume that the extent of ionisation is small and that self-ionisation of water is ignored. (ii) State the pH of the mixture after 10 cm 3 of NaOH(aq) has been added to 20 cm 3 of 0.100 mol dm-3 cis-butenedioic acid. ………………………………………………………………………………………………... [1] pH = pKa1 = 1.93 Comments 10 cm 3 of 0.100 mol dm -3 NaOH(aq) neutralises exactly half of the first proton of the 20.0 cm3 acid. This creates a maximum capacity acidic buffer where [H2A] = [HA-]

5 ACJC2026 9476/Preliminary Examinations/2026 [Turn over (iii) Calculate the pH of the mixture at the second equivalence point , when 40 cm 3 of NaOH(aq) has been added to 20 cm3 of 0.100 mol dm-3 cis-butenedioic acid. [3] At the 2nd equivalence point, all H2A is converted to A2- (salt Na2A is formed) Total volume = (20 + 40) = 60 cm3 H2A + 2NaOH → Na2A + 2H2O Concentration of salt A2- formed = (0.100 x 20.0)/60.0 = 0.0333 mol dm-3 A2- + H2O ⇌ HA– + OH– Kb2 = 1.0 × 10−14 10−6.23 = 1.698 x 10-8 mol dm-3 [OH-] = √(1.698 x 10-8)(0.0333) = 2.378 x 10-5 mol dm-3 pOH = -lg (2.378 x 10-5) = 4.623 pH = 14 – 4.623 = 9.38 Comments Students must first recognise that the salt formed at second equivalence point is Na2A. Students are to determine the concentration of the salt Na2A formed and then use the salt hydrolysis equation and Kb2 to determine the concentration of OH– and then pH. (iv) Table 1.2 provides information about some acid-base indicators. Table 1.2 pH indicator working range pKa methyl orange 3.1 – 4.4 3.5 bromothymol blue 6.0 – 7.6 7.1 phenolphthalein 8.2 – 10.0 9.3 thymolphthalein 9.3 – 10.5 9.8 Identify the most suitable indicator for the detection of the second equivalence point. Explain your answer. ……………………………………………………………………………………………...… [2] Phenolphthalein the end point of the titration pH = 9.38 is closest to the pKa = 9.3 OR in the middle of the range of pH for colour change (8.2–10.0). Note: rapid pH change – not allowed. Students are to use the information from the table. Comments Both phenolphthalein and thymolphthalein can be used to detect the second equivalence point since the final pH is in the working range of both indicators. However, phenolphthalein is the most suitable indicator as the titration pH = 9.38 is closest to the pKa value.

6 ACJC2026 9476/Preliminary Examinations/2026 [Turn over (d) trans-butenedioic acid reacts with steam in the presence of a concentrated H3PO4 catalyst at high temperature and pressure to form product Y via the electrophilic addition mechanism. (i) Draw the structure of product Y. [1] Comments H2O is added to the alkene to form an alcohol in Y. The alkene is symmetrical so there is only one structure of product Y. (ii) State the number of chiral centres in a molecule of product Y. ………………………………………………………………………………...……………… [1] 1 chiral centre Comments *chiral centre [Total: 16]

7 ACJC2026 9476/Preliminary Examinations/2026 [Turn over 2 (a) The halogens Cl2, Br2, and I2 are often used in the synthesis of halogenoalkanes. Describe and explain the trend in volatility of the halogens down Group 17. ……………………………………………………………………………………………………….. [2] Volatility decreases down Group 17 from chlorine to iodine. Down the group, the size of the electron cloud increases, leading to greater polarisability. This results in stronger instantaneous dipole-induced dipole (id-id) interactions between molecules, requiring more energy to overcome. √volatility decreases (i.e. boiling point increases) √larger and more polarisable electron cloud √stronger instantaneous dipole-induced dipole (id-id) between molecules √require more energy to overcome Comments Students must recognise that the higher the boiling point, the lower the volatility. Students generally attribute the cause of the higher boiling point to stronger id -id but failed to explain the correct factor explanation that leads to stronger id-id. There were some students who misinterpreted this question and explained the difference in terms of bond energies, while other answers referred to shielding effect and nuclear charge. (b)