ACSI Paper 3 Prelim

YEAR 6 PRELIMINARY EXAMINATION 2023 INTERNATIONAL BACCALAUREATE DIPLOMA PROGRAMME CHEMISTRY HIGHER LEVEL PAPER 3 Tuesday 12 th September 2023 1 hour 15 minutes INSTRUCTIONS TO CANDIDATES  Do not open this examination paper until instructed to do so.  Write your candidate session number in the box above.  A calculator is required for this paper.  A copy of the Chemistry Data Booklet is required for this paper.  Write your answers in the boxes provided.  If you use additional sheets of paper for your answer, attach them to the booklet. Indicate the question number clearly on these sheets.  All drawings must be in ink. For examiner’s use Qn 1 /6 Qn 2 /9 Qn 3 /4 Qn 4 /6 Qn 5 /5 Qn 6 /6 Qn 7 /5 Qn 8 /4 Wrong s.f. /units Total /45 ________________________________________________________________________ This question paper consists of 15 printed pages including this cover page. Name Candidate Number

2 Section A Answer all questions. Answers must be written in the answer boxes provided. 1. Molecules that have a different arrangement as a result of bond rotation are called conformers. Conformers can be represented by the Newman projection. A Newman projection views the carbon-carbon bond directly end- on and represents the two carbon atoms by a circle. Bonds attached to the front carbon are represented by lines to the centre of the circle, and bonds attached to the rear carbon are represented by lines to the edge of the circle. Ethane can have two conformers as a result of the bond rotation: staggered and eclipsed. Using the Newman projection, the angle between the C-H bonds of the front carbon and the C-H bonds on the back carbon is known as the dihedral angle. Butane can have four types of conformers as a result of bond rotation. Table 1 shows the four conformers of butane. The gauche conformer occurs when the dihedral angle between the methyl groups are 60o apart and the anti conformer occurs when the dihedral angle is 180o apart. (This question continues on the following page)

3 (Question 1 continued) Table 1 An equilibrium between the gauche 1 and the staggered (anti) conformer can be achieved: ⇌ The equilibrium constant, Kc, between the gauche 1 and the staggered (anti) conformer at 298 K can be calculated by the equation given below:  3630 = RT ln Kc R = molar gas constant in J mol1 K1 T = temperature in Kelvins (a) Which of the two conformers of butane is the most stable? Explain your answer. [1] .................................................................................................................................................. .................................................................................................................................................. (This question continues on the following page)

4 (Question 1 continued) (b) (i) There are two gauche conformers of butane. With reference to Gauche 1, draw the other gauche conformer of butane. [1] (ii) At equilibrium, butane will consist the staggered (anti) conformer and the two gauche conformers. Calculate the equilibrium constant Kc for the equilibrium between the gauche 1 and the staggered (anti) conformer of butane at 298 K. Hence, calculate the percentage of the staggered (anti) conformer of butane at 298 K by using section 2 from the Data Booklet. [3] .................................................................................................................................................. .................................................................................................................................................. .................................................................................................................................................. .................................................................................................................................................. .................................................................................................................................................. .................................................................................................................................................. (This question continues on the following page)

5 (Question 1 continued) (c) A Newman projection for Compound X is shown below. State the IUPAC name for Compound X. [1] .................................................................................................................................................. .................................................................................................................................................. 2. The Finkelstein reaction is a nucleophilic substitution reaction and is carried out using dry propanone as a solvent. One example of the Finkelstein reaction is given. CH3CH2CH2Br + NaI ⇌ CH3CH2CH2I + NaBr (a) (i) Explain why it is important for propanone to be dry. [1] .................................................................................................................................................. .................................................................................................................................................. (This question continues on the following page)

6 (Question 2 continued) (a) (ii) The solubilities of NaBr and NaI in propanone are shown. compound solubility at 25 ᴼC in g / 100g of propanone NaBr 0.00841 NaI 39.9 Use this information to explain why the reaction produces a very high yield despite being a reversible reaction. [1] .................................................................................................................................................. .................................................................................................................................................. (b) State and explain one precaution, other than using protective equipment such as hand gloves, a lab coat or eye protection, that should be taken when carrying out this experiment. [1] .................................................................................................................................................. .................................................................................................................................................. (This question continues on the following page)

7 (Question 2 continued) (c) A student plans an experiment to show that the rate of the reaction is proportional to the concentration of NaI. Propanone is used as the solvent in this reaction. CH3CH2CH2Br (pr) + NaI (pr) ⇌ CH3CH2CH2I (pr) + NaBr (s) (pr) = substance is dissolved in propanone The student plans to record the time it takes for the solid formed to obscure a cross on a piece of paper below the conical flask, as shown. To carry out this experiment, the following materials are available.  CH 3CH2CH2Br (l)  Na I (s)  dry propanone, CH3COCH3 (l)  usual laboratory apparatus (i) The student recorded data in the table below. Complete the table with appropriate volumes that the student could have used in four further experiments. [2] volume of 0.5 mol dm−3 NaI (pr) / cm3 volume of CH3CH2CH2Br (l) / cm3 volume of CH3COCH3 (l) / cm3 total volume / cm3 time / s 10.0 2.0 30.0 42.0 (This question continues on the following page)

8 (Question 2 continued) (c) (ii) Write an expression to show how the student could calculate the rate of the reaction.