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