Organic synthesis

5(a)(i) M1: (cisplatin) can bond / bind with DNA / (nitrogenous) base A, T, C, G etc. 2 M2: which prevents replication (of the DNA / strand) OR prevents cell division / prevents mitosis OR prevents transcription (and formation of mRNA) 5(a)(ii) square planar 1 5(a)(iii) the distance between two coordinating oxygens is too small to bond trans 1 OR atoms in a bidentate ligand can only bond 90° not 180° 5(a)(iv) +4 1 5(b)(i) 1 L 5(b)(ii) M1: heat / reflux with acidified / alkaline KMnO4 (then acidify) 2 M2: PCl OR SOCl / (heat with) PCl 5 2 3 5(b)(iii) C H O + 2PCl → C H O Cl + 2POCl + 2HCl 1 8 6 4 5 8 4 2 2 3 OR C H O + 2SOCl → C H O Cl + 2SO + 2HCl 8 6 4 2 8 4 2 2 2 OR 3C H O + 2PCl → 3 C H O Cl + 2H PO 8 6 4 3 8 4 2 2 3 3 5(b)(iv) 3 M1: curly arrow from inside hexagon to C of electrophile M2: correct intermediate M3: curly arrow from C—H bond AND formation/loss of H+ © UCLES 2021 Page 9 of 13

9(a) A B 6 Br O O HO C CH 3 HO C CH 3 Br D E C O O CHI 3 HO C OH HO C Cl F O HO C O one mark for each structure © UCLES 2021 Page 17 of 17

5(a) M1: 4 ethylamine phenylamine 4-nitrophenylamine most basic least basic M2 / 3 / 4: explanation two correct for one mark, three correct for two marks, four correct for three marks • (basicity linked to) lone pair / p orbital on N AND being able accept / donate to / coordinate to a proton / H+ • ethyl / alkyl group is electron donating / has a positive inductive effect (so lone pair on N is more able to accept a proton) • (phenylamines are less basic than ethylamine as) p orbital / lone pair on N is delocalised (into the ring so less able to accept a proton) • (4-nitrophenylamine is less basic than phenylamine as) nitro / NO group is electron withdrawing (so lone pair on N is 2 less able to accept a proton) 5(b)(i) 1 5(b)(ii) M1: 2 step 1: HNO , ⩽ 10°C OR NaNO , HCl(aq), ⩽ 10°C 2 2 M2: step 2: NaOH / alkaline AND 1-naphthol / α-naphthol / structure below 5(c)(i) 4-bromo-2-nitrobenzoic acid OR 4-bromo-2-nitro(-1-)benzenecarboxylic acid 1 © UCLES 2021 Page 10 of 16 5(c)(ii) 2 5(c)(iii) M1: step 1 conc. H SO and conc. HNO 3 2 4 3 M2: step 2 Br and AlBr 2 3 M3: step 3 hot (alkaline / acidified) MnO – / KMnO 4 4 © UCLES 2021 Page 11 of 16

9(a) A B 6 Br O O HO C CH 3 HO C CH 3 Br D E C O O CHI 3 HO C OH HO C Cl F O HO C O one mark for each structure © UCLES 2021 Page 17 of 17

7(a) 6 [1] 1 7(b) M1: trend phenylethanoic acid > phenol > ethanol [1] 4 M2: why phenylethanoic acid is the strongest • negative inductive electron withdrawing effect of C=O which weakens O-H bond / stabilises anion [1] M3: why phenol is stronger than ethanol / weaker than phenylethanoic acid • oxygen lone pair is delocalised into the ring system which weakens O-H bond / stabilises anion [1] M4: why ethanol is the weakest • electron donating alkyl / ethyl group which strengthens O-H bond / destabilises anion [1] 7(c)(i) CH Cl + AlCl → +CH + AlCl – [1] 1 3 3 3 4 7(c)(ii) 3 M1: arrow to CH + (arrow must come from inside the hexagon) [1] 3 M2: correct structure of intermediate [1] M3: arrow from C-H bond into the ring AND H+ seen [1] 7(d)(i) Br + UV light [1] 1 2 © UCLES 2021 Page 11 of 16 7(d)(ii) 1 [1] 7(d)(iii) CHECK Q is correct 2 step 2 – KCN in ethanol + heat [1] step 3 – HCl(aq) + heat/reflux/boil [1] 7(d)(iv) 1 [1] ALLOW any viable organic by-product from this radical substitution reaction, e.g. C H CH CH C H 6 5 2 2 6 5 © UCLES 2021 Page 12 of 16

7(a) 6 [1] 1 7(b) M1: trend phenylethanoic acid > phenol > ethanol [1] 4 M2: why phenylethanoic acid is the strongest • negative inductive electron withdrawing effect of C=O which weakens O-H bond / stabilises anion [1] M3: why phenol is stronger than ethanol / weaker than phenylethanoic acid • oxygen lone pair is delocalised into the ring system which weakens O-H bond / stabilises anion [1] M4: why ethanol is the weakest • electron donating alkyl / ethyl group which strengthens O-H bond / destabilises anion [1] 7(c)(i) CH Cl + AlCl → +CH + AlCl – [1] 1 3 3 3 4 7(c)(ii) 3 M1: arrow to CH + (arrow must come from inside the hexagon) [1] 3 M2: correct structure of intermediate [1] M3: arrow from C-H bond into the ring AND H+ seen [1] 7(d)(i) Br + UV light [1] 1 2 © UCLES 2021 Page 11 of 16 7(d)(ii) 1 [1] 7(d)(iii) CHECK Q is correct 2 step 2 – KCN in ethanol + heat [1] step 3 – HCl(aq) + heat/reflux/boil [1] 7(d)(iv) 1 [1] ALLOW any viable organic by-product from this radical substitution reaction, e.g. C H CH CH C H 6 5 2 2 6 5 © UCLES 2021 Page 12 of 16

7(a)(i) phenylamine AND amine AND ester 1 7(a)(ii) sp carbons = 0, sp2 carbons = 7, sp3 carbons = 6 1 7(b) 6 1 7(c) lone pair on the N can accept a proton 1 7(d)(i) 1 7(d)(ii) step 1 M1 concentrated HNO and H SO 2 3 2 4 step 2 M2 hot (alkaline) KMnO (followed by addition of H+) 4 7(e) step 4 M1 HOCH CH N(CH CH ) 3 2 2 2 3 2 step 5 M2 Sn AND HCl M3 concentrated (HCl) AND heat / reflux 7(f)(i) M1 ratio of the concentration of a solute in two solvents 2 M2 at equilibrium © UCLES 2022 Page 14 of 15 7(f)(ii) M1 K = [procaine] / [procaine] 2 pc oct water 1.77 = (x / 50)/(0.5 – x / 75) M2 1.77 = 1.5x / 0.5 – x 0.885 –1.77x = 1.5x x = 0.271 g min 2sf © UCLES 2022 Page 15 of 15

6(a)(i) NO 2 NO 2 3 H +H+ + NO + 2 M1 first curly arrow from within hexagon to N of the NO + 2 M2 correct intermediate M3 second curly arrow from C-H bond into the ring AND H+ formed / lost 6(a)(ii) HSO – + H+  H SO 1 4 2 4 © UCLES 2022 Page 10 of 16 6(b) M1 benzoic acid M2 phenol 2 OR 6(c) M1 phenol > benzene > benzoic acid 4 M2 / M3 phenol:  lone pair / p-orbital on oxygen delocalises into the ring / overlaps with -delocalised ring  accepts / attracts / polarises NO + / electrophiles better 2 benzoic acid:  COOH / C=O is an electron-withdrawing / positive inductive effect two for one mark, three for two marks M4 (phenol-oxygen) increases electron density in the ring (as compared to benzene as a result of the OH group) OR (benzoic acid–COOH) decreases electron density in the ring (as compared to benzene as a result of the COOH group) © UCLES 2022 Page 11 of 16 6(d)(i) 3 6(d)(ii) step 1 M1 Sn and hydrochloric acid / HCl 3 M2 concentrated (HCl) + heat / reflux (followed by NaOH) step 2 M3 HNO and (HCl) ⩽10°C 2 OR NaNO , HCl and ⩽10°C 2 Question Answer Marks

7(a)(i) phenylamine AND amine AND ester 1 7(a)(ii) sp carbons = 0, sp2 carbons = 7, sp3 carbons = 6 1 7(b) 6 1 7(c) lone pair on the N can accept a proton 1 7(d)(i) 1 7(d)(ii) step 1 M1 concentrated HNO and H SO 2 3 2 4 step 2 M2 hot (alkaline) KMnO (followed by addition of H+) 4 7(e) step 4 M1 HOCH CH N(CH CH ) 3 2 2 2 3 2 step 5 M2 Sn AND HCl M3 concentrated (HCl) AND heat / reflux 7(f)(i) M1 ratio of the concentration of a solute in two solvents 2 M2 at equilibrium © UCLES 2022 Page 14 of 15 7(f)(ii) M1 K = [procaine] / [procaine] 2 pc oct water 1.77 = (x / 50)/(0.5 – x / 75) M2 1.77 = 1.5x / 0.5 – x 0.885 –1.77x = 1.5x x = 0.271 g min 2sf © UCLES 2022 Page 15 of 15

7(a) C > A > B [1] 3 chlorine and C=O are electronegative / withdraw charge and this causes greatest weakening of O–H bond or greatest stabilisation of the anion [1] 2nd oxygen / C=O is electronegative / withdraws charge and this weakens O–H bond or stabilises anion [1] 7(b)(i) chloromethane 1 aluminium chloride 7(b)(ii) CH Cl + AlCl → AlCl – + CH + 1 3 3 4 3 7(b)(iii) curly arrow from within benzene to CH + [1] 3 3 positively charged intermediate [1] curly arrow from C–H bond into ring, C H CH , H+ [1] 6 5 3 7(b)(iv) hot alkaline KMnO [1] 2 4 C H CH + 3[O] → C H CO H + H O OR 6 5 3 6 5 2 2 C H CH + 3[O] + OH– → C H CO – + 2H O [1] 6 5 3 6 5 2 2 7(b)(v) 2 or 4-nitromethylbenzene and 3-nitrobenzoic acid 1 7(c)(i) HNO , T between 0° and 10°C 1 2 7(c)(ii) warm / T>10°C and H O 1 2 © UCLES 2022 Page 9 of 11 7(c)(iii) CH C H –N=N–C H (CH )OH [1] 2 3 6 4 6 3 3 T between 0° and 10°C and NaOH(aq) [1] Question Answer Marks

4(a) M1 bond angle = 120° AND carbons are sp2 3 M2  bonds are formed by end-on-end / head on / head to head / linear overlap of orbitals M3  bonds are formed by sideways / lateral overlap of p orbitals 4(b)(i) 2 [1] [1] 4(b)(ii) M1 step 1 (CH ) CHBr and FeBr / Al Br 3 3 2 3 3 M2 step 2 conc HNO and conc H SO 3 2 4 M3 step 3 Sn and conc HCl 4(c) 3,4,5-trimethylphenylamine 1 4(d) 1 compound number of peaks observed W 6 Z 6 Both correct for one mark Question Answer Marks

4(a) M1 bond angle = 120° AND carbons are sp2 3 M2  bonds are formed by end-on-end / head on / head to head / linear overlap of orbitals M3  bonds are formed by sideways / lateral overlap of p orbitals 4(b)(i) 2 [1] [1] 4(b)(ii) M1 step 1 (CH ) CHBr and FeBr / Al Br 3 3 2 3 3 M2 step 2 conc HNO and conc H SO 3 2 4 M3 step 3 Sn and conc HCl 4(c) 3,4,5-trimethylphenylamine 1 4(d) 1 compound number of peaks observed W 6 Z 6 Both correct for one mark Question Answer Marks

8(a) phenol, amide AND alkene / C=C all three needed [1] 1 8(b) 3 M1: HBr [1] u / c M2: structure of the organic product[1] M3: correct balanced equation [1] ecf 8(c) 2 M1: ring reduced [1] M2: alkene reduced [1] 8(d)(i) hot AND concentrated acidified AND MnO – / KMnO all [1] 1 4 4 © UCLES 2023 Page 14 of 18 8(d)(ii) 3 [1] for  values [1] for splitting [1] for explanations 8(e)(i) 2 [1] [1] 8(e)(ii) hydrolysis AND neutralisation / acid-base[1] 1 © UCLES 2023 Page 15 of 18 8(f) 1 [1] Question Answer Marks

8(a) phenol, amide AND alkene / C=C all three needed [1] 1 8(b) 3 M1: HBr [1] u / c M2: structure of the organic product[1] M3: correct balanced equation [1] ecf 8(c) 2 M1: ring reduced [1] M2: alkene reduced [1] 8(d)(i) hot AND concentrated acidified AND MnO – / KMnO all [1] 1 4 4 © UCLES 2023 Page 14 of 18 8(d)(ii) 3 [1] for  values [1] for splitting [1] for explanations 8(e)(i) 2 [1] [1] 8(e)(ii) hydrolysis AND neutralisation / acid-base[1] 1 © UCLES 2023 Page 15 of 18 8(f) 1 [1] Question Answer Marks

5(a)(i) AlCl3 + CH 3 Cl → AlCl4 – + CH 3 + 1 © Cambridge University Press & Assessment 2024 Page 10 of 13 5(a)(ii) 1 + H+ curly arrow from inside hexagon to electrophile (CH +) 3 intermediate 1 curly arrow from C—H bond into the hexagon AND H+ formed 1 5(a)(iii) C 6 H 4 (Cl)CH 3 + 2[O] → C 6 H 4 (Cl)CHO + H 2 O 1 5(a)(iv) step 3 HCN AND KCN (cat) OR KCN AND H 2 SO 4 /HCl 1 step 4 H 2 SO 4 (aq) OR HCl (aq) 1 5(a)(v) addition–elimination / condensation 1 5(a)(vi) (a strong base) it would hydrolyse the ester 1 5(a)(vii) (a substance able to) rotate the plane of plane-polarised light 1 5(a)(viii) any two of: 2 • reduced / different biological activity of ‘other’ enantiomer ORA • avoids need to separate the optical isomers to form the pure active isomer • lower dosage required OR (drug is) more potent • higher yield (of biologically-active molecule) • no / less (harmful) side effects OR other isomer can have side effects 5(b)(i) M1 [HA] = (75  10–3 / 180) ÷ 0.100 = 4.17  10–3 OR 1 / 240 (mol dm–3) 1 M2 [H+] = (K  [HA])½= (10–3.49  4.17  10–3)½ = 1.16  10–3 (mol dm–3) 1 a M3 pH = –log [H+] = 2.93 to 2.94 min 2sf 1 © Cambridge University Press & Assessment 2024 Page 11 of 13 5(b)(ii) 1 1 CH COOH 3 Question Answer Marks

7(a) methyl pentanoate 1 7(b)(i) time between injection and detection 1 7(b)(ii) 29.5% 1 © Cambridge University Press & Assessment 2024 Page 12 of 18 7(c)(i) 2 ester number of peaks seen number of peaks seen in proton (1H) NMR in carbon-13 NMR B 5 6 C 4 5 M1 row 1 correct M2 row 2 correct 7(c)(ii) A and B 1 © Cambridge University Press & Assessment 2024 Page 13 of 18 7(d) 6 one mark for each structure: F, G, H, J, K and L © Cambridge University Press & Assessment 2024 Page 14 of 18 7(d) six alternatives for F © Cambridge University Press & Assessment 2024 Page 15 of 18

7(a)(i) 2-nitrobenzoic acid OR 2-nitrobenzenecarboxylic acid 1 7(a)(ii) hot / reflux / heat AND (alkaline / acidified / neutral) MnO – / KMnO 1 4 4 7(a)(iii) COOH / carboxyl group is electron-withdrawing / electronegative AND 3- and 5- / meta- directing 1 7(b)(i) 3 7(b)(ii) M1 step 1: Fe / Sn, conc. HCl 3 M2 step 2: HNO OR NaNO AND HCl 2 2 M3 step 1: heat / reflux / hot AND step 2: ⩽10 °C Question Answer Marks

7(a) methyl pentanoate 1 7(b)(i) time between injection and detection 1 7(b)(ii) 29.5% 1 © Cambridge University Press & Assessment 2024 Page 12 of 18 7(c)(i) 2 ester number of peaks seen number of peaks seen in proton (1H) NMR in carbon-13 NMR B 5 6 C 4 5 M1 row 1 correct M2 row 2 correct 7(c)(ii) A and B 1 © Cambridge University Press & Assessment 2024 Page 13 of 18 7(d) 6 one mark for each structure: F, G, H, J, K and L © Cambridge University Press & Assessment 2024 Page 14 of 18 7(d) six alternatives for F © Cambridge University Press & Assessment 2024 Page 15 of 18

7(a) CH CN / ethanenitrile [1] 1 3 7(b) dilute acid / HCl (aq) 1 AND hot / heat [1] three components required (acid / aq / heat) 7(c) CH COCl / ethanoyl chloride [1] 1 3 7(d) CH COOH + SOCl → CH COCl + HCl + SO [1] 1 3 2 3 2 7(e) C H Br / C H Cl / bromoethane / chloroethane [1] 1 2 5 2 5 7(f) heat in ethanol AND 1 under pressure / in sealed tube [1] 7(g) LiAlH / lithium aluminium hydride [1] 1 4 © Cambridge University Press & Assessment 2024 Page 12 of 14 7(h) Examiner checks first that each of 1–6 is only used once and 4 IGNOREs any repeated unless a CON hydrolysis 1 addition 2 reduction 2, 6 substitution 3, 4, 5 two correct [1] four correct [2] five correct [3] six correct [4] 7(i) M1 C H NHC H ammonia C H NHCOCH [1] 4 2 5 2 5 2 5 3 M2 (basicity linked to) lone pair/ p orbital on N AND being able accept / donate to a proton/H+ [1] M3 (amine stronger because of) +ve inductive effect / electron donating alkyl / R / ethyl group [1] M4 (amide weaker because) lone pair / p orbital on N is delocalised into C=O group [1] Question Answer Marks

7(a) CH CN / ethanenitrile [1] 1 3 7(b) dilute acid / HCl (aq) 1 AND hot / heat [1] three components required (acid / aq / heat) 7(c) CH COCl / ethanoyl chloride [1] 1 3 7(d) CH COOH + SOCl → CH COCl + HCl + SO [1] 1 3 2 3 2 7(e) C H Br / C H Cl / bromoethane / chloroethane [1] 1 2 5 2 5 7(f) heat in ethanol AND 1 under pressure / in sealed tube [1] 7(g) LiAlH / lithium aluminium hydride [1] 1 4 © Cambridge University Press & Assessment 2024 Page 12 of 14 7(h) Examiner checks first that each of 1–6 is only used once and 4 IGNOREs any repeated unless a CON hydrolysis 1 addition 2 reduction 2, 6 substitution 3, 4, 5 two correct [1] four correct [2] five correct [3] six correct [4] 7(i) M1 C H NHC H ammonia C H NHCOCH [1] 4 2 5 2 5 2 5 3 M2 (basicity linked to) lone pair/ p orbital on N AND being able accept / donate to a proton/H+ [1] M3 (amine stronger because of) +ve inductive effect / electron donating alkyl / R / ethyl group [1] M4 (amide weaker because) lone pair / p orbital on N is delocalised into C=O group [1] Question Answer Marks

6(a) 4 test I2(aq) warm with sodium metal Na 2 CO 3 (aq) 2,4-DNPH + Fehling’s Br 2 (aq) OH–(aq) solution alcohol / carboxylic carbonyl / (methyl) aldehyde alkene carboxylic acid ketone / ketone / ( acid aldehyde methyl) alcohol Any two [1] any three [2] any five [3] all six [4] 6(b)(i) (primary) alcohol 1 6(b)(ii) M1: A → B oxidation/dehydrogenation 3 M2: C → E reduction/(nucleophilic) addition M3: E → F elimination/dehydration © Cambridge University Press & Assessment 2025 Page 13 of 17 6(b)(iii) 2 each correct structure [1] 6(b)(iv) 5 D F E C A any one of the following each correct structure [1] © Cambridge University Press & Assessment 2025 Page 14 of 17

9(a) • all C atoms are sp2 hybridised 3 • bonds between C atoms are  and  • C–C  bonds are formed by overlap of sp2 hybrid orbitals • C–C  bonds are formed by overlap of p orbitals • bonds between C and H atoms are  • C–H bonds are formed by overlap of sp2 hybrid orbitals and s orbitals Any two [1], any four [2], all six [3] © Cambridge University Press & Assessment 2025 Page 14 of 15 9(b) CH 3 Cl and AlCl3 1 9(c) 3 M1: curly arrow from within hexagon towards +C M2: intermediate, usual rules for horseshoe and + charge M3: curly arrow from C–H bond into ring AND H+ product 9(d) M1: HNO AND H SO 2 3 2 4 M2: concentrated AND 25 °C ⩽ T ⩽ 60 °C 9(e) alkaline KMnO 1 4 9(f) 4-nitrobenzoic acid 1 9(g) (hot) concentrated HCl and Sn 1 9(h) M1: perform step 3 before step 2 OR perform step 2 before step 1 2 M2: COOH is 3-directing OR NO is 3-directing 2 © Cambridge University Press & Assessment 2025 Page 15 of 15

6(a) sp = 1 AND sp2 = 1 AND sp3 = 3 1 6(b)(i) acid-base / neutralisation AND hydrolysis 1 6(b)(ii) * 4 * – – * reaction 7 * – * – – * – reaction 8 reaction 9 * * – – any two * [1], any three * [2], any five * [3], all six * [4] 6(c)(i) five / 5 1 6(c)(ii) 2 proton environment a b c name of splitting pattern doublet multiplet doublet chemical shift range, 2.2–3.0 3.2–4.0 2.0–3.0  / ppm any three [1], all six [2] © Cambridge University Press & Assessment 2025 Page 13 of 16

5(a)(i) HNO + H SO → +NO + H O + HSO – 1 3 2 4 2 2 4 OR HNO + 2H SO → +NO + H O+ + 2HSO – 3 2 4 2 3 4 © Cambridge University Press & Assessment 2026 Page 15 of 18 5(a)(ii) 3 + H+ M1: curly arrow 1 from touching / inside the hexagon to nitrogen of NO + 2 M2: correct intermediate M3: curly arrow 2 from C-H bond into hexagon (touching / inside) AND H+ 5(a)(iii) CH C H NO + 6[H] → CH C H NH + 2H O 1 3 6 4 2 3 6 4 2 2 5(a)(iv) Stage 1 M1: HNO OR NaNO with (dilute strong) acid AND ⩽ 10 °C 2 2 2 Stage 2 M2: warming with H O / aqueous 2 5(a)(v) CH Cl AND Al Cl 2 3 3 NO directs to the meta / 3(,5)-position (in the ring so 4-methylation unlikely) 2 5(b)(i) it is inert / unreactive 1 5(b)(ii) a high-boiling point / non-volatile AND (non-polar) liquid / solvent 1 5(b)(iii) it is the least polar 1 OR it is more attracted to the stationary phase OR forms a strong IMF to the stationary phase OR attracted to the stationary phase for a longer time OR it is the least volatile / highest bp © Cambridge University Press & Assessment 2026 Page 16 of 18 5(b)(iv) % = 100%  22 / (29 + 34 + 22) = 25.88 1 % = 25.9 OR 26 min 2sf 5(c) M1: weakest = ethanol … water … phenol = strongest 4 M2: why acids are stronger or weaker strongest acid has weakest O—H OR H+ more easily lost OR anion stabilised / conjugate base stabilised M3: for phenol: as p-orbital / lone pair on oxygen overlaps / delocalised into the ring /  system / delocalised system M4: for ethanol: positive inductive effect/electron donating of alkyl / R group Question Answer Marks

7(a)(i) 120° and 1 Only answers. sp2 7(a)(ii) • sp2 (and sp2) 2 CON s orbital unless clearly referring to C- • head on / direct / end-on / linear / acceptable description H, where you ignore s orbital. • p (and p) A orbital type written in overlap space if • sideways / lateral / acceptable description not written in ‘orbitals involved’ space. Otherwise consider only orbitals on the ✓ ✓ ‘orbitals involved lines’ and only overlap ⚫ ⚫ on the ‘orbital overlap’ lines. Read ‘6p orbitals’ as the number of p orbitals. 7(b)(i) conc HNO + conc H SO , T = 25–60 °C 1 Conc, con or c seen once. 3 2 4 CON aq. A any T in range. I heat / warm / reflux. 7(b)(ii) C H NH 1 A C H NH + or C H NH Cl. 6 5 2 6 5 3 6 5 3 Trace ID: PC-0F078E40FE Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(b)(iii) M1 HNO / HCl + NaNO , T = 0–10 °C 2 A any T in range. 2 2 I HNO / NaNO . 3 3 M2 H O / (aq) and warm / heat / T > 10 °C A [1] for all correct reagents or both 2 correct temperatures. M2 CON addition of NaOH but ignore acids. 7(c)(i) M1 curly arrow from within / touching hexagon towards C+ 3 CON additional curly arrows in M1. I additional curly arrows in M3. CON curly arrow from H rather than middle of bond for M3 intermediate – normal rules apply. ‘Horseshoe’ must include C3, C4 and C5, may include C2 and / or C6, CON including C1. + charge must be above a line joining C2 and C6. M2 intermediate M3 curly arrow from C–H bond goes within hexagon and H+ I connectivity of side group if drawn as OCCH for M2. 3 A Kekulé for M2 if correct. A HCl for M3 if Cl – introduced earlier. 7(c)(ii) C H COCH + 4[O] → C H COOH + CO + H O 1 6 5 3 6 5 2 2 © Cambridge University Press & Assessment 2026 Page 17 of 21 Trace ID: PC-0F078E40FE Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(d) M1 NaOH(aq) / OH–(aq) / alkaline and iodine / I 2 A NaOH and I (aq). 2 2 A Cl / OH–. 2 M2 any acid / H+ unc A HCl, H SO . 2 4 (pass alternative (correct) method to TL for advice, e.g. LiAlH then alkaline I CON additional reagents. 4 2 followed by acid) Question Answer Marks Guidance