2024 MI PU3 MYE P3 (Answers) with marker's comments

2024 Mid-Year Examination Pre-University 3 H2 CHEMISTRY 9729/03 Paper 3 Free Response 5 July 2024 2 hours Section A 1 (a) (i) 1s2 2s2 2p6 3s2 3p6 3d5 4s2 Marker’s comment Many candidates did well for this question. (ii) 3𝑑𝑥𝑦 3𝑑𝑥2−𝑦2 3𝑑𝑧2 Marker’s comment Some candidates did not draw the 3dxy orbital between the x and y axis. (iii) forms ions of variable oxidation states / variable oxidation states in compounds forms coloured compounds high density Marker’s comment Candidates understand what the question wants. Weaker candidates wrote ‘more oxidation number’ instead of forms ions of ‘variable oxidation state’.

2 (b) MnO has giant ionic structure electrostatic forces of attraction between oppositely-charged ions Mn2O7 has simple molecular structure instantaneous dipole-induced dipole forces of attraction between molecules the stronger forces in MnO require more energy to overcome points: (1) identify structure and bonding in MnO (2) identify structure and bonding in Mn2O7 (3) identify MnO has stronger forces, compare energy [2] all 3 points [1] either (1) or (2) [0] just (3) Marker’s comment This question proved to be challenging to candidates. Many who attempted were unable to identify the correct structure and bonding in MnO and Mn 2O7. Some candidates explained why Mn 2O7 has a simple molecular structure, but did not continue to compare with the structure and bonding in MnO. (c) (i) Mn2O7 + H2O ⟶ 2HMnO4 Marker’s comment Many candidates did well for this question. (ii) hydrolysis Marker’s comment Some candidates did not attempt this question. Those who did, some wrote ‘hydration’ instead of hydrolysis. (iii) Marker’s comment Majority of the candidates were unable to do this question. (d) from Ti to Cu, relatively similar atomic radii nuclear charge increases shielding increases but due to electrons added to 3d 4s electrons experience similar effective nuclear charge [2] 5 / 5 [1] 3 / 5 Marker’s comment Many candidates confused this explanation with the trend of atomic radii across a period for the main group elements.

3 [Turn over 2 (a) (i) 2,4-dinitrophenylhydrazine / if orange precipitate forms, sample is O if no precipitate forms, sample is P Marker’s comment Some candidates gave tri-iodomethane test as the distinguishing test. For candidates who gave the correct distinguishing test, some gave ‘no visible reaction for P’, ‘P remains the same’ and ‘solution remains colourless for P’. (ii) information reaction inference L was heated with concentrated H2SO4 dehydration / elimination of water (1) L was an alcohol (3) M and N were alkenes (4) M and N were heated with KMnO4 / H+ oxidative cleavage / vigorous oxidation (2) for O, the carboxyl carbon and ketone carbon were once joined with C=C in M for P, the two carboxyl carbon atoms were once joined with C=C in N hence M N L [1] for every 2 reactions / inferences Marker’s comment This question proved to be challenging to candidates. Some candidates wrote ‘M and N undergo oxidation’ instead of ‘vigorous oxidation’. (iii) M alkene functional group is bonded to more substituents / alkene functional group is bonded to 3 substituents in M versus 2 in N Saytzeff’s rule (iv) number of enantiomers = 23 = 8

4 (b) (i) (ii) generate nucleophile: CH3CH2MgBr ⟶ CH3CH2– + MgBr+ (1) Mg2+ and Br– okay nucleophilic attack: protonation: [3] 8 / 8 [2] 5 / 8 [1] 2 / 8 (iii) fewer substituents on carbonyl carbon unhindered nucleophile approach to carbonyl carbon less electron-donating effect to carbonyl carbon carbonyl carbon is more electron-deficient attracts nucleophile more strongly [2] 4 / 4 [1] 2 / 4 (2) (3) (4) (5) (6) (7) (8)

5 [Turn over 3 (a) C Br CH3 H C2H5 slow C + CH3 H C2H5 Br – + CN -C + CH3 H C2H5 C CN CH3 H C2H5 [3] 7 / 7 [2] 5 / 7 [1] 2 / 7 Marker’s comments: Many are confused between SN1 and SN2, some gave mechanisms with fragments of both mechanisms. Most of those who gave SN1 mechanism, wrote the mechanism in one continuous manner, often missing out the Br- in the first step as a result. (b) (i) make CN– concentration remain constant with time only decrease of bromoalkane concentration will affect rate Marker’s comments Many gave incomplete answers, which include not mentioning that [CN -] becomes almost constant. Some also gave vague answers such as ‘so that the reaction can be pseudo’. (ii) 1 molecule of bromoalkane is present in the rate-determining step of both SN1 and SN2 mechanisms Marker’s comments Many candidates do not understand the question and went on to state that since the half life is constant, it is first order. Most answers were irrelevant. (iii) from the graph, 𝑡1/2 = 21 min with two interpolations shown [1] rate = k [1-bromo-1-methylpropane] [CN–]0.5 = k’ [1-bromo-1-methylpropane] where k’ = k [CN–]0.5 𝑡1/2 = ln 2 𝑘′ hence 𝑡1/2 = ln 2 𝑘[𝐶𝑁−]0.5 sub in values 21 𝑚𝑖𝑛 = ln 2 𝑘(2.00)0.5 [1] correct substitutions hence rate constant 𝑘 = 0.0233 mol–0.5 dm1.5 min–1 Marker’s comments Many just made use of the data given in the table and the order of reaction given to solve for k. They were not given full credit. (1) (2) (3) (4) (5) (6) (7)

6 (iv) [CN–] / mol dm–3 initial rate / mol dm–3 min–1 1.5 0.00285 or 0.00286 2.0 0.00330 3.0 0.00404 5.0 0.00521 or 0.00522 Marker’s comments A large number of students did not attempt the question. Some who attempted did not take into account information given in (iii) assumed that both reactants are first order. (v) SN2 because unhindered nucleophile approach only secondary bromoalkane / only 2 hydrocarbon groups are present SN1 because stable carbocation intermediate electron-donating hydrocarbon groups disperse positive charge on carbon atom Marker’s comments Majority did not attempt this question. (vi) SN2 mechanism forms 1 part (–)-enantiomer (complete stereoinversion) SN1 mechanism forms 0.5 part (+)-enantiomer and 0.5 part (–)-enantiomer hence (+) : (–) enantiomers = 1 : 3 or 75% of enantiomers are (–) working needs to be shown Marker’s comments Majority did not attempt this question. Those who did, did not show working and the answers were incorrect. (c) R does not hydrolyse correct graph drawn [1] completely flat okay 𝜋 overlap between p orbitals of iodine atom and benzene ring / C-I bond has partial double bond character / electron delocalise to C-I bond + C-I bond is strong and will not break [1] P hydrolyses slower than Q correct graph drawn [1] C-Cl bond is stronger than C-Br bond + due to smaller Cl atomic radius / better orbital overlap / shorter bond length [1] 0.00 0.20 0.40 0.60 0.80 1.00 0.0 10.020.030.040.050.060.070.080.090.0100.0time concentration / mol dm–3 0 Q P R

7 [Turn over Marker’s comments Mark scheme was loosened a little. Explanations given by candidates were mostly incomplete. Some quoted bond energy from data booklet even though question did not mention the use of DB. Many knew that R doesn’t undergo hydrolysis but still drew a curve downwards for R, albeit with a more gradual gradient. Students must show understanding that R doesn’t undergo hydrolysis to gain credit. This question shows that some students still do not understand that the trend of ease of hydrolysis is due to the bond strength. Some compared the electronegativity instead. (d) (i) Ksp = [Ag+]2 [CO32–] let the solubility of Ag2CO3 = 𝑥 mol dm–3