ACSI Paper 3 Prelim (Student Version) Answer

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) 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 conformer of butane is the most stable? Explain your answer. [1] Anti conformer is most stable as the two bulky groups are furthest from each other / it has the lowest potential energy / Kc > 1 / G < 0 so forward reaction is more spontaneous. Marker comments: Generally well done. A good handful of candidates interpreted the question wrongly, not recognising that the question is asking for one out of the two conformers given in the equilibrium. Although not penalised, candidates are advised to always read the context given to address the question correctly. Some candidates seemed to think that the different conformers are structural isomers. They should take note conformers are different spatial arrangements of the same compound due to the free rotation of sigma bonds.

4 (b) (i) There are two gauche conformers of butane. With reference to Gauche 1, draw the other gauche conformer of butane. [1] Marker comments: Generally well done. (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] 3630 = RT ln Kc 3630 = 8.31  298 ln Kc ln Kc = 1.466 Kc = 4.331 = [௔௡௧௜] [௚௔௨௖௛௘ ଵ] [anti] = 4.331 [gauche 1] Let [gauche 1] = x = [gauche 2] Total concentration = [anti] + [gauche 1] + [gauche 2] = 4.331 x + 2x = 6.331 x Percentage of anti conformer = ସ.ଷଷଵ ௫ ଺.ଷଷଵ ௫  100 % = 68.4% Alternative mark scheme: anti ⇌ gauche 1 + gauche 1a Kc = 4.331 = [௔௡௧௜] [௚௔௨௖௛௘]మ Let the percentage of anti = x %

5 Kc = 4.331 = x ((100-x 2 ))2 x = 90.8 % Marker comments: Not well done. Majority of the candidates calculated the KC value but unable to interpret the information about the equilibrium between anti and the two gauche conformers to calculate the percentage of anti conformer (within the 3 conformers). (d) A Newman projection for Compound X is shown below. State the IUPAC name for Compound X. [1] Cyclohexane There are 6 carbon atoms connected in a ring. Marker comments: Very poorly done. Most candidates either could not recall that the intersection of two lines represents a carbon atom in skeletal structure or understand from the question that the circle represents a carbon atom in the Newman projection structure. 1 2 3 4 5 6

6 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] H2O is a nucleophile which can interfere/take part in the reaction. To prevent hydrolysis of 1–bromopropane Marker comments: Candidates’ answers were rather varied. Common answers that were not accepted: - H 2O is a polar protic solvent, which reduces the rate of SN2 Although this is a true statement, it is not relevant in this question as the concern with non-dry propanone is not about the minute amount of water replacing the bulk of propanone as the solvent. - H 2O reacting with propanone Nucleophilic substitution / hydrolysis / reaction of propanone with water is not feasible as propanone is relatively much more stable than the gem-diol product. (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] Solubility of NaBr in propanone is very low, hence will precipitate out as solid, its concentration decreases and shifts the position of equilibrium to the right, favouring the formation of products. Marker comments: Generally well done.

7 (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] Organic compounds / solvents are flammable AND should not be used near naked flame / direct heat source OR Organic compounds / solvents are volatile / toxic / irritant to respiratory system / may cause dizziness / drowsiness AND should be used in a fume cupboard / fumehood / be properly disposed into organic waste bin OR NaI used in powder form to prepare standard solution can reach harmful concentration of airborne particles, which is irritating to the eyes, skin and respiratory tract AND should be used in a fume cupboard / fumehood / handled with the use of facial mask OR Iodide ions / 1–iodopropane can be oxidised by air / oxygen in air / to form I2, which is volatile / toxic / irritant to respiratory system AND should be used in a fume cupboard / fumehood OR Iodide ions / 1–iodopropane is photosensitive and can form I2, which is volatile / toxic / irritant to respiratory system AND should be used in a fume cupboard / stored in a opaque bottle Marker comments: Candidates’ answers were rather varied but a good number were able to giv