CJC 2022 JC2 H2 Chem Prelim P2 MS Examiners Comments

[Turn over 9729/2 CJC JC2 Preliminary Examination 2022 CANDIDATE NAME CLASS 2T CHEMISTRY 9729/02 Paper 2 Structured Questions 26 August 2022 2 hours Candidates answer on the Question Paper Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen. You may use an HB 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 calculation 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. This document consists of 20 printed pages. For Examiner’s Use Paper 1 30 Paper 2 Q1 /13 Q2 /11 Q3 /15 Q4 /12 Q5 /11 Q6 /13 75 Paper 3 80 Paper 4 55 OVERALL (100%) GRADE Catholic Junior College JC2 Preliminary Examination Higher 2 MARK SCHEME WITH EXAMINER’S COMMENTS

2 9729/2 CJC JC2 Preliminary Examination 2022 1 (a) Fehling’s solution is used in Qualitative Analysis of organic compounds. It is prepared fresh by mixing a solution of CuSO4 (Fehling’s A solution) with a mixture of tartrate ions and NaOH (Fehling’s B solution). When the two solutions are mixed, the tartrate ions act as ligands to form a complex ion with the Cu2+ ions. (i) Another reagent with similar action to Fehling’s solution is Benedict’s solution, which is also an alkaline solution of Cu2+, but the ligand forming the complex ion is citrate ion instead of tartrate ion. Suggest why the presence of the ligand is necessary in Fehling’s (and Benedict’s) solution. …………………………………………………………………………………………… …………………………………………………………………………………………… ….………………………………………………………………………………………[1] Fehling’s solution was used to distinguish between three unknown compounds, A, B and C: Compound Observation with Fehling’s solution A blue solution B red ppt C blue solution It is known that compounds A, B, and C are the following (not necessarily in order): C6H5COCH3 C6H5CHO C6H5CH2CHO (ii) Based on the above observations, state the identity of compound B. ….………………………………………………………………………………………[1] Without the ligand to form the complex ion, Cu2+ ions would precipitate out as Cu(OH)2 due to the NaOH present. B is C6H5CH2CHO EXAMINER’S COMMENTS Qualitative Analysis – Precipitation of insoluble metal hydroxides in NaOH(aq) In this question, students were expected to note that Cu2+ ions were being mixed with a solution of NaOH (and tartrate or citrate ions), and use their QA knowledge to predict that a blue precipitate of Cu(OH)2 would result – which does not occur, and this must be due to the formation of the copper(II) complex with tartrate or citrate, which brings the Cu2+ down such that ionic product of Cu(OH)2 does not exceed its Ksp. The question was poorly answered. Most students instead resorted to a template explanation of the origin of colour in transition metal complexes or stated that the ligand is necessary for the observation of colour in the distinguishing test (which does not make sense as Cu2+ itself is a coloured ion – due to the [Cu(H2O)6)2+ complex which already exists in CuSO4(aq), i.e., without the need for tartrate or citrate ligands! EXAMINER’S COMMENTS Fehling’s Solution – test for aliphatic aldehydes As Fehling’s solution gives a red ppt only for aliphatic aldehydes, only C6H5CH2CHO will give a positive result. The aromatic aldehyde C6H5CHO will not give a red ppt, neither will the ketone C6H5COCH3. The observation is blue solution for these two compounds, since Fehling’s solution is blue in colour. A handful of students responded with Cu2O, which showed poor comprehension ability.

3 9729/02/CJC JC2 Preliminary Examination 2022 [Turn over (iii) Give a simple chemical test to distinguish compounds A and C. In your answer, the observations for each compound must be stated. …………………………………………………………………………………………… ..………………………………………………………………………………………….. ……………………………………………………………………………………………. ….………………………………………………………………………………………[2] 1 (a) Compounds A, B and C are all able to form hydroxynitriles when subject to the same reagents and conditions. (iv) State the reagents and conditions required for the above transformation. ….………………………………………………………………………………………[1] To a sample of each compound (A and C) in separate test tubes, add Tollens’ reagent and heat. C6H5CHO would give a silver mirror while there would be no silver mirror for C6H5COCH3. HCN, trace NaOH (or trace NaCN), 10-20 oC EXAMINER’S COMMENTS A significant number of students had the incorrect idea that the compounds were presented in order of A, B, C, in spite of the question stating that the compounds were not necessarily in this order. These students referred to the observations for each compound as “A would give [observation] while C would give [observation]”. A number of students even referred to compound B instead of C for this question! Students who gave the ketone in (ii) would not be able to distinguish the remaining aldehydes (since the only reagent that can do so, Fehling’s solution, was already used in the question. Some students missed out the “heat” condition (or gave “heat under reflux” when reflux is not suitable for distinguishing test) or did not specify the acid used with K2Cr2O7. Others, in spite of clear instructions (which need not be stated), did not properly give the observation for the compound with the negative test. Students are to note that “KMnO4(aq)/H2SO4(aq), heat” was generally not accepted because the ketone given in this question, phenylethanone, is a special ketone that can be oxidized by KMnO4 (see Carbonyl Compounds lecture notes, pg 15-18). EXAMINER’S COMMENTS Carbonyl Compounds – Nucleophilic Addition This was a basic question with a very generic compound, testing on whether students were able to correctly recall the reagents and conditions for nucleophilic addition of HCN to carbonyl compounds (aldehydes & ketones). Finer details that were missed out include the use of trace amounts of either NaOH or NaCN to generate the CN- nucleophile (only small amounts needed as it is regenerated – a catalyst), and the temperature conditions (a handful failed to mention which temperature scale – Celsius or Fahrenheit?, while others wrote “heat”).

4 9729/2 CJC JC2 Preliminary Examination 2022 2sp 2p (v) The N atom of the nitrile group is sp hybridi sed. Complete the energy level diagram to show how the electrons are arranged in the second quantum shell of the N atom so that the bonding in the nitrile can occur. [1] (vi) Hence, suggest why nitriles are weaker bases than amines, given that N atom in amine is sp3 hybridised. …………………………………………………………………………………………… ..………………………………………………………………………………………….. …………………………………………………………………………………………… ….………………………………………………………………………………………[1] The lone pair of electrons on N of the nitrile is less available to accept H+, because it is located in an sp hybrid orbital, which is held closer to the nucleus than the corresponding lone pair of electrons on N of the amine, which is located in an sp3 hybrid orbital (or accept p orbital). . Energy EXAMINER’S COMMENTS Chemical Bonding – hybridization, overlap of atomic orbitals in bonding The N atom in the nitrile forms 1  b