1.C [1] 2. B [1] 3. C [1] 4. B [1] 5. (a) boiling points increase (from the first member to the fifth member); increasing size of molecule/area of contact/number of electrons (from the first to the fifth member); strength of intermolecular/van der Waals’/London/dispersion forces increase / more energy required to break the intermolecular bonds (from first member to fifth member); 3 (b) same general formula; successive members differ by CH2; same functional group / similar/same chemical properties; gradual change in physical properties; Accept specific physical property such as melting point, boiling point only once. 2 max [5] IB Questionbank Chemistry 1 6. A: B: C: 1,2-dichloroethane; D: Accept condensed formulas. Penalize missing hydrogens only once. 5 [5] 7. add bromine water/bromine; pentane no change/stays brown and pent-1-ene decolourizes bromine water/bromine; OR add acidified KMnO4; pentane no change/stays purple and pent-1-ene decolourizes acidified KMnO4; Accept any correct colour change. Do not accept “clear” instead of “colourless”. 2 max [2] IB Questionbank Chemistry 2 8. E: primary and F: secondary; G: primary; G / E: only one alkyl group/2 H atoms attached to the carbon atom attached to the Cl / only one carbon atom attached to the carbon atom attached to the Cl; F: two alkyl groups/1 H atom attached to the carbon atom attached to the Cl / two carbon atoms attached to the carbon atom attached to the Cl; 4 [4] 9. Initiation: UV/hf / hv / heat Cl2 → 2Cl•; Reference to UV/hf/hν/heat must be included. Propagation: Cl• + CH4 → CH3• + HCl; CH3• + Cl2 → CH3Cl + Cl•; Termination: Cl• + Cl• → Cl2 / CH3• + Cl• → CH3Cl / CH3• + CH3• → C2H6; Allow representation of radical without • (e.g. Cl, CH3) if consistent throughout mechanism. If representation of radical (i.e. •) is inconsistent, penalize once only. 4 [4] 10. (a) but-1-ene; Allow 1-butene. CH3–CH2–CH=CH2; 2 (b) C2H5–CH(OH)–CH3 → C2H5–CH=CH2 + H2O; heat and (concentrated) phosphoric acid/H3PO4/sulfuric acid/H2SO4; 2 IB Questionbank Chemistry 3 representation of carbocation. 1 IB Questionbank Chemistry 4 . chlorobutane more rapid. 3 max [3] 12. butan-2-ol will give but-2-ene as well as but-1-ene / butan-1-ol will only give but-1-ene. (i) propanone/CH3–CO–CH3 and hydrogen cyanide/HCN/cyanide ion/CN–.(c) curly arrow going from C=C to Br of Br2 and curly arrow showing Br atom leaving Br2 molecule. 3 (d) butan-1-ol gives higher yield / butan-2-ol gives lower yield. C–Cl bond less polar/charge on C–Cl carbon smaller (owing to polarization of delocalized electrons). 2 [2] 13. 2 [9] 11. Any 2 of the above for second and third marks. (CH3)2CH–Mg–Cl. steric hindrance/repulsion by electron cloud from benzene ring in chlorobenzene. Accept opposite statements for chlorobutane. No ECF for second and third marks. C–Cl bond stronger (owing to partial π-bonding by lone pair). curly arrow going from lone pair/negative charge on Br– to C+. anhydrous / absence of water / ether solvent. (a) amine salt. A [1] 18. 2 (b) greater / more alkaline. [(CH3)2NH2]+ + OH– → (CH3)2NH + H2O / [(CH3)2NH2]Br + OH– → (CH3)2NH + Br– + H2O / [(CH3)2NH2]+ + NaOH → (CH3)2NH + Na+ + H2O / [(CH3)2NH2]Br + NaOH → (CH3)2NH + NaBr + H2O.(ii) 1 [2] 14. OR the inductive/e– donating effect of the methyl groups increases the negative charge on the nitrogen of the amine so that it attracts H+ ions more strongly / OWTTE. the inductive/e– donating effect of the methyl groups reduces the charge on the nitrogen atom of the cation / stabilizes the cation / OWTTE. 2 [4] 15. D [1] 16. A [1] IB Questionbank Chemistry 5 . B [1] 17. M2 (for B) and M5 (for E) may also be scored for substitution product if primary chloroalkane used. Award [3 max] if A and D are other way round but both substitution products B and E are correct based on initial choice of A and D. Penalize missing hydrogens once only. Award [2 max] if A and D are other way round but one substitution product B or E is correct based on initial choice of A and D. Accept condensed formulas. Award [1 max] if A and D are other way round (and nothing else correct).19. 5 [5] IB Questionbank Chemistry 6 . 2 [6] 22. Accept curly arrow either going from bond between C and Cl to Cl in 2-chloro-3-methylbutane or in the transition state. formation of organic product 3-methylbutan-2-ol and Cl–. Do not accept just H+ or acid. square brackets and partial bonds. CH3CH2COOH + CH3OH CH3CH2COOCH3 + H2O [1] for reactants and [1] for products. Do not penalize if HO and Cl are not at 180° to each other. Do not allow curly arrow originating on H in HO–. 4 [4] 21. Do not award M3 if OH ---. Allow 1-cyanobutane. CH3CH2CH2CH2CN + 2H2 → CH3CH2CH2CH2CH2NH2. methyl propanoate. greater attraction to the carbon atom (with the partial positive charge) / OWTTE.20. Penalize missing hydrogen once only. representation of transition state showing negative charge. 4 (ii) OH– has a negative charge/higher electron density. Accept CN– for KCN and Cl– for KCl. (concentrated) sulfuric acid/H2SO4. 5 [5] IB Questionbank Chemistry 7 . Do not allow just greater attraction. (i) curly arrow going from lone pair/negative charge on O in HO– to C. curly arrow showing Cl leaving. pentanenitrile. Catalyst: nickel/Ni / palladium/Pd / platinum/Pt. CH3CH2CH2CH2Cl + KCN → CH3CH2CH2CH2CN + KCl. pentan-1-amine / 1-aminopentane / 1-pentylamine / 1-pentanamine.C bond is represented. Allow mechanism with corresponding Kekulé structures. representation of carbocation with correct formula and positive charge on ring. formation of organic product chlorobenzene and HCl and AlCl3. curly arrow going from lone pair on O in H2O to carbonyl C and curly arrow going from C=O bond to O. representation of intermediate anion showing negative charge on O and + on O of H2O. Do not allow curly arrow originating on H in H2O. curly arrow going from lone pair/negative charge on O to C–O to form C=O and curly arrow showing Cl leaving and curly arrow from H–O bond to O+. Lone pair on O not required on representation of intermediate. 4 [4] 24.23. curly arrow going from lone pair/negative charge on Cl in AlCl4– to H and curly arrow going from CH bond to benzene ring. methyl group electron donating and nitro group electron withdrawing. Allow curly arrow going from delocalized electrons in benzene to Cl+ for M1. (a) curly arrow going from delocalized electrons in benzene to Cl in Cl2 and curly arrow going from Cl–Cl bond to AlCl3. 2 [62] IB Questionbank Chemistry 8 . Allow other suitable catalysts such as FeCl3 etc. 4 (b) methylbenzene more reactive / nitrobenzene less reactive. formation of organic product H3CCOOH and Cl– and H+/HCl. B [1] 26. warm / heat / reflux. 2 (ii) acidic solution / H+ / sulfuric acid. (i) C3H 8O + 4 1 O2 → 3CO2 + 4H2O / 2C3H8O + 9O2 → 6CO2 + 8H2O 2 Award [1] for correct products and reactants and [1] for correct balancing. Continuation bonds necessary for mark. D [1] 28. 3 IB Questionbank Chemistry 9 . 2 (ii) addition (polymerization). Do not accept hydrogenation of alkenes. ( CH(CH3)–CH2 ) / –CH(CH3)CH2–. (the solution changes) from orange to green. 2 (iii) hydrogenation (of vegetable oils) / manufacture of margarine / manufacture of ethanol / addition of water. 1 [6] 30. displayed formula or condensed structural formula can be given. Do not accept “clear” instead of “colourless”.25. Accept if more than one repeating unit is shown. Ignore state symbols. (i) addition of bromine/bromine water. D [1] 27. C [1] 29. Accept manufacture of alcohol. the bromine colour remains with propane and propene decolourizes the bromine / solution changes from brown to colourless. 1 (b) addition reactions not favoured energetically since this would involve disruption of cloud of delocalized electrons / stabilization energy would need to be supplied and product would lack delocalized ring of electrons making it less stable / OWTTE. For chlorobenzene: steric hinderance / repulsion by electron cloud in benzene ring/C–Cl less polar. (a) 1. 2 max [2] 33. propan-1-ol gives propanal and propanoic acid and propan-2-ol gives propanone. award [2] for 4 or 5 correct names or structures. 1 IB Questionbank Chemistry 10 . CH3CH2COOH and propanoic acid. CH3COCH3 and propanone/acetone. propan-1-ol has two H atoms bonded to the C containing the –OH whereas propan-2-ol only has one / propan-1-ol is a primary alcohol and propan-2-ol is a secondary alcohol.(iii) CH3CH2CHO and propanal. Award [1] mark for either of the above. Award [1] for 2 or 3 correct names or structures. (a) (i) H3CCHClCH2I . C–Cl bond stronger. For chloromethylbenzene: electron deficient carbon on –CH2Cl group making it susceptible to nucleophilic attack.3-cyclohexadiene. 1 [2] 32. 5 [10] 31. 1 (ii) H3CC(OH)(CN)H. nucleophilic addition. addition-elimination. 1 (c) Reaction (a) (i) (a) (ii) (a) (iii) (a) (iv) Type electrophilic addition.(iii) 1 (iv) (CH3)2CHOH . Grignard. 4 [9] IB Questionbank Chemistry 11 . 1 (b) CH3CHICH2Cl . Catalyst: (concentrated) H3PO4/phosphoric acid/H2SO4/sulfuric acid. Accept ethylene.34. B [1] 37. C [1] IB Questionbank Chemistry 12 . 3 [3] 36. 4 [4] 35. Mechanism: curly arrow going from O of OH to H+ (showing protonation of OH group) and loss of water showing formation of carbocation. Organic product: ethene. Step 2: Br2 /bromine. Step 1: H3PO4 /phosphoric acid / H2SO4 /sulfuric acid. Product from step 1: CH3CH2CH=CH2 /but-1-ene. curly arrow from CH of beta-carbon to CC. If non-optically active isomers given (e. 2-bromo-2-methyl-butane) award [1 max] if name and 3D structure are correct. 3 IB Questionbank Chemistry 13 . and [1 max] for a correct structure of an enantiomer not shown in 3D.g. correct name. (a) correct isomer 3D structure.38. A [1] 39. correct enantiomer 3D structure. B [1] 40. Accept condensed form for alkyl chain throughout. If compound incorrectly named award [2 max] for two correct 3D enantiomers. lone pair/negative charge must be on O). 1-bromobutane award [2 max]. e.(b) (i) curly arrow going from lone pair/negative charge on O in HO– to C bonded to Br. Accept condensed formulas as long as curly arrows can still be shown. originating on negative charge on H.g. transition state showing overall negative charge. Do not allow curly arrow originating on H in HO– (e. IB Questionbank Chemistry 14 . curly arrow from C–Br bond to form Br– (this can also be shown in transition state).e.g. e.g. 3 If wrong formula used for halogenoalkane. i. If wrong formula is used for 2-bromo-2-methylbutane award [2 max]. hence weaker intermolecular forces of attraction/van der Waals’ forces of attraction between molecules of 2-bromo2-methylbutane / OWTTE. 2-bromo-2-methylbutane is more spherical in shape / less surface area in contact between molecules of 2-bromo2-methylbutane than between molecules of 1-bromopentane / OWTTE. If non-bonding pair not shown then arrow must originate from negative sign on O or the minus sign. correct structure of tertiary carbocation. 2-bromo-2-methylbutane forms a tertiary carbocation which is stabilized by the positive inductive effect of the three alkyl groups / OWTTE. Only penalize arrow from H once in (b). the C bonded to the Br in 2-bromo-2-methylbutane has three other (bulky) groups bonded to it so cannot accommodate five groups around it in the transition state / OWTTE. 1-bromopentane would form a primary carbocation (if it went by SN2) which is much less stable as there is only one alkyl group exerting a positive inductive effect / OWTTE. curly arrow going from lone pair/negative charge on O in HO– to C+. CH3–CO–O–(CH2)4CH3 / CH3COO(CH2)4CH3 / CH3COOCH2CH2CH2CH2CH3 / 2 Accept CH3–CO–O–C5H11 IB Questionbank Chemistry 15 . 3 (iii) the C bonded to the Br in 1-bromopentane is also bonded to two H atoms so can accommodate five groups around it in the transition state / OWTTE. 3 (v) esterification / condensation.(ii) curly arrow from C–Br bond to form Br–. 3 max (iv) the boiling point of 1-bromopentane is higher than the boiling point of 2-bromo-2-methylbutane. 1 [4] 42. neither can exist as geometrical isomers as they contain the same two groups/atoms on one of the double bonded carbon atoms / OWTTE. IB Questionbank Chemistry 16 . Award [3] if correct equation given for one molecule of diol reacting with one molecule of dicarboxylic acid. (a) (i) CH3CONHCH2CH3 and HCl. Allow corresponding names also.(c) elimination. i. HO–(CH2)5–OH + HOOC–C6H4–COOH → HO–(CH2)5–O–CO–C6H4–COOH + H2O (ii) formation of polyesters/condensation polymers/synthetic fabrics. 1 (ii) 2-nitromethylbenzene and 4-nitromethylbenzene / 1 Allow use of ortho and para notation. 4 [21] 41.e. (i) nHO–(CH2)5–OH + nHOOC–C6H4–COOH → H–[O–(CH2)5–O–CO–C6H4–CO–]n–OH + (2n – 1)H2O 3 Award [1] for correct reactants. [1] for correct polyester and [1] for balanced water. B [1] IB Questionbank Chemistry 17 . 3 [5] 43. Allow corresponding mechanism involving Kekulé notation. formation of C6H5COCH3 and H+ . 4 [4] 44. curly arrow from CH to + in ring. drawing of intermediate structure with + on ring. Allow corresponding mechanism for formation of para isomer. curly arrow from ring to CH3CO+ electrophile. B [1] 46.(b) curly arrow from ring to +NO2 electrophile. formation of CH3CO+ electrophile. curly arrow from C–H to +. D [1] 45. Allow corresponding mechanism involving Kekulé notation. Do not accept plastics cause litter. 2 IB Questionbank Chemistry 18 . Allow correct condensed structural formula. C2H5Cl. plastics are not biodegradeable / plastics take up large amounts of space in landfill / pollution caused by burning of plastics / OWTTE. Accept any bond angle in the range 113–120°. dots or crosses for electron pairs. Continuation bonds from each carbon are required. Allow abbreviated formulas C2H3Cl. (approximately)120°. (i) Step 1: CH2CHCl + H2 → CH3CH2Cl. Allow NaOH or NaCl etc. Cl atoms can be above or below carbon spine or alternating above and below. Lone pairs required on chlorine. 1 (iii) plastics are cheap/versatile/a large industry / plastics have many uses / OWTTE. 2 (ii) Brackets not required for mark. (i) Accept lines. instead of OH– and Cl–.47. A [1] 48. C2H5OH. Allow plastics don’t decompose quickly / OWTTE. 2 [5] 49. Step 2: CH3CH2Cl + OH– → CH3CH2OH + Cl–. Allow methylene for CH2. Penalize incorrect punctuation. CH3COO–(aq) + H3O+(aq) 2 (iii) CH3COOH(aq) + H2O(l) OR CH3COOH(l) + H2O(l) OR CH3COOH(aq) correct equation. 1 IB Questionbank Chemistry 19 . state symbols and CH3COO–(aq) + H3O+ (aq) CH3COO–(aq) + H+(aq) . D: (2-)methylpropan-1-ol.g. KMnO4 etc.(ii) H2SO4/H+/acidified and Cr2O72–/(potassium/sodium) dichromate.) but only with acid. 50. e. (heat under) reflux. (a) (i) A: butan-1-ol. only once.g. similar chemical properties. 3 [7] BL acid is CH3COOH and cb is CH3COO– / BL acid is H3O+ and cb is H2O. same functional group. same general formula. 4 (ii) C/(2-)methylpropan-2-ol. 1 (iv) B/butan-2-ol. successive members differ by CH2. 2 max [2] 51. Accept suitable oxidizing agents (e. gradually changing physical properties. Ignore missing or incorrect oxidation states in reagents. 1 (iii) A/butan-1-ol. Second mark can be scored even if reagent is incorrect. C: (2-)methylpropan-2-ol. B: butan-2-ol. Accept answers in the form of 1-butanol and 2-methyl-2-propanol etc. commas for hyphens. Do not award third mark if OH----C bond is represented. 1 (ii) curly arrow going from lone pair/negative charge on O in OH– to C. (a) C2H5MgBr + H2O → C2H6 + Mg(OH)Br Award [1] for C2H6 and [1] for correct equation. Accept curly arrow either going from bond between C and Br to Br in 1-bromobutane or in the transition state. curly arrow showing Br leaving. square brackets and partial bonds.(v) 1 (b) (i) SN2. representation of transition state showing negative charge. 3 [12] 52. Do not penalize if HO and Br are not at 180° to each other. Do not allow curly arrow originating on H in OH–. 2 IB Questionbank Chemistry 20 . 1 (ii) curly arrow going from lone pair on O to H+. 2 [6] 53.(b) (i) butan-2-ol/2-butanol. 3 [4] 54. (i) but-1-ene/1-butene. (a) as the bromine approaches the alkene an induced dipole is formed / OWTTE. No mark awarded if C+ is not represented. 2 (ii) 2-methylbutan-2-ol. 1 IB Questionbank Chemistry 21 . curly arrow going from lone pair on O of H2O/H2PO4– to H and curly arrow going from CH bond to C–C+ to form C=C.3-dibromobutane. CH3CH2CH(OH)CH3. 1 (b) (i) 2. representation of positively charged O intermediate and curly arrow showing H2O leaving. C2H5C(CH3)2OH. 2-disubstituted benzene compounds. Do not accept average bond enthalpy in benzene is between that of C–C and C=C. stating that the secondary carbocation will be formed in preference to the primary carbocation.(ii) 2-bromobutane. only one isomer exists for 1. IR absorption of C–C bonds in benzene is different to that of both C–C single bonds and C=C double bonds. 2 max (b) substitution rather than addition occurs / bromine is not decolourized (without a catalyst)/OWTTE. 1 [3] IB Questionbank Chemistry 22 . showing curly arrow from double bond to H (in H–Br) and curly arrow from bond in H–Br to Br. the two positive/electron releasing inductive effects due to the two R– groups on the secondary carbocation make it more stable. 4 [4] 56. (a) the C–C bond lengths are all the same. showing the curly arrow from the lone pair/negative charge on Br– to the secondary carbocation and 2-bromobutane as correct product. chemical shift of protons in benzene is different to that of protons in alkenes. 1 [3] 55. only once. Accept 2-bromomethylpropane and 1-bromomethylpropane for C and D respectively. B: 2-bromobutane. Allow use of 2-bromo-2-methylpropane instead of RBr. D [1] 60. commas for hyphens. RBr → R+ + Br–. Penalize incorrect punctuation.g. C: 2-bromo-2-methylpropane. (a) A: l-bromobutane. A [1] 59. 4 (b) (i) C/2-bromo-2-methylpropane. unimolecular nucleophilic substitution.57. e. C [1] 61. D [1] 58. D: 1-bromo-2-methylpropane. 2 (ii) 1 IB Questionbank Chemistry 23 . square brackets and partial bonds. Accept curly arrow either going from bond between C and Br to Br in 1-bromobutane or in the transition state. representation of transition state showing negative charge. (plane-) polarized light shone through. Do not allow curly arrow originating on H in OH–. 1 IB Questionbank Chemistry 24 . (b) (iii) rate doubles as the rate is proportional to [OH–] / OH– appears in the rate-determining/slow step / first order with respect to OH–. 2 (e) 2-bromobutane/B. 5 (f) (i) elimination. curly arrow going from lone pair/negative charge on O in OH– to C. C–Br bond is weaker/breaks more easily than C–Cl bond. Do not award fourth mark if OH----C bond is represented. chemical properties are identical (except with other chiral compounds). 4 (c) 2 (d) rate of 1-bromobutane is faster. physical properties identical (apart from effect on plane-polarized light).(iii) A/1-bromobutane/D/1-bromo-2-methylpropane. Do not penalize if HO and Br are not at 180° to each other. Accept “turn” instead of “rotate” but not “bend/reflect”. Award [1] if correctly predicts no rate change for SN1 and doubling of rate for SN2 of without suitable explanation. Do not accept “similar” in place of “identical”. (b)(i) no change as [OH–] does not appear in rate equation/in the rate determining step. curly arrow showing Br leaving. enantiomers rotate plane of plane-polarized light to left or right/ opposite directions (by same amount). (ii) curly arrow going from lone pair/negative charge on O in OH– to H on β-C. two products formed: but-1-ene / but-2-ene/(cis) but-2-ene/ (trans) but-2-ene. curly arrow going from CH bond to form C=C bond. curly arrow going from lone pair/negative charge on O in OH– to H on C adjacent to C+ and curly arrow going from CH bond to form C=C bond. 4 max [25] IB Questionbank Chemistry 25 . curly arrow showing Br leaving. Accept the following for first 3 marks. representation of carbocation. Allow C2H5O– instead of OH–. curly arrow showing Br leaving. Do not allow curly arrow originating on H in OH–. Award [1] for two correct answers. Accept AlCl3 / AlI3 / FeCl3. 2 [2] 63. aluminium chloride / iodide / iron chloride. Accept 2-nitromethylbenzene and 4-nitromethylbenzene. Accept mechanism with corresponding Kekulé structures. 1 IB Questionbank Chemistry 26 . correctly showing intermediate. HLPE: use of a Ziegler-Natta catalyst / ionic mechanism / coordination polymerization. Accept prior formation of electrophile to give CH3+ and AlICl3–/AlI4–. Accept o-methylnitrobenzene and p-methylnitrobenzene. the H2NO3+ formed loses water / H2NO3+ → H2O + NO2+. curly arrow from –H into ring and curly arrow from catalyst anion to H. Accept HNO3 + H2SO4 → NO2+ + H2O + HSO4– for the second and third points. (i) 1-methyl-2-nitrobenzene and 1-methyl-4-nitrobenzene. 1 (ii) 3 [4] 65.62. B [1] and p-methylnitrobenzene. the (concentrated) sulfuric acid protonates the nitric acid / HNO3 + H2SO4 → H2NO3+ + HSO4–. 4 [4] 64. NO2+ / nitronium ion. LDPE: free radical mechanism. showing curly arrows from benzene ring to CH3– and curly arrow from bond in CH3–I to catalyst. Accept 2-methylnitrobenzene and 4-methylnitrobenzene. (synthesis of) ethylene glycol/1. (hydrogenation of unsaturated oils in the manufacture of) margarine. (synthesis of) drugs/pesticides. 3 [4] 66.2-ethanediol/ethane-1. or for decolourized with no reference to original colour. A [1] 67. Poly(chloroethene): . dots or crosses for e– pairs. 2 (iii) (hydration of ethene for the manufacture of) ethanol/ C2H4 + H2O → C2H5OH. Accept HNO3 + H2SO4 → NO2+ + H2O + HSO4– for the second and third points. (synthesis of) CH3COOH /ethanoic/acetic acid. Accept other commercial applications. the (concentrated) sulfuric acid protonates the nitric acid / HNO3 + H2SO4 → H2NO3+ + HSO4–. 1 (ii) Chloroethene: No mark if the lone pairs missing on Cl. the H2NO3+ formed loses water / H2NO3+ → H2O + NO2+. Continuation bonds must be shown.2-diol. n and square brackets are not required. Accept lines. No mark for change to clear. (i) colour change from yellow/orange/rust colour/red/brown to colourless.(ii) NO2+ / nitronium ion. 2 max [5] IB Questionbank Chemistry 27 . B [1] 68. Award [1] if CH3CH2CHO and CH3CH2COOH identified but conditions not given/incorrect. Accept C2H5 as CH2CH3. concentrated sulphuric acid/ H2SO4 and heat/steam / phosphoric acid/H3PO4 (catalyst) and heat/steam. dichromate(VI) (ion)/Cr2O72– and acidic/H+. 2 (ii) CH3CH=CHCH3 + H2O → CH3CH(OH)CH2CH3. (i) Penalize missing H atoms once only. 3CH3CH(OH)CH2CH3 + Cr2O72– + 8H+ → 3CH3COCH2CH3 + 2Cr3+ + 7H2O. from propan-1-ol: CH3CH2CHO(propanal) obtained by distillation (as product is formed). Accept either condensed or full structural formulas. CH3CH2CH2OH: primary and CH3CH(OH)CH3: secondary. propan-2-ol gives CH3COCH3 by heat / reflux. Need both formula and name for mark. CH3COCH3. 5 [7] 70. propan-1-ol gives CH3CH2COOH (propanoic acid) by (heating under) reflux. propan-1-ol/1-propanol. heat/reflux. Accept CH3CH(OH)CH2CH3 + [O] → CH3COCH2CH3 + H2O . propan-2-ol/2-propanol. CH3CH2COOH. 5 max [8] IB Questionbank Chemistry 28 . Accept either condensed or full structural formulas.69. (i) CH3CH2CH2OH. 3 (ii) CH3CH2CHO. CH3CH(OH)CH . Accept MnO4– in place of Cr2O72– in third and fourth marks. 4 [4] IB Questionbank Chemistry 29 .71. Allow ECF if major and minor products are interchanged. Allow more detailed formulas throughout the option. 72. curly arrow showing movement of electron pair from the double bond to H+. correct formula of the organic product. correct formulas of the reactants and the inorganic product. curly arrow showing either C–Br bond formation / mechanism for either product. Award [3 max] for mechanism. 2 [2] IB Questionbank Chemistry 30 . 4 max [4] 73. (CH3)2CH+ is more stable / CH3CH2CH2+ is less stable. OR equation for HBr dissociation. correct structures of both carbocations. formation of Br–. curly arrow showing movement of electron pair from the double bond to hydrogen in HBr. (a) increases acidity / OWTTE. 1 [5] 76. B [1] IB Questionbank Chemistry 31 . Reaction type for Step 1: elimination/E / dehydration. A [1] 77. A [1] 78. (acceptors) increase O–H bond polarity / increase δ+ on H / decrease O–H bond strength / favour dissociation of O–H bond / OWTTE.74.98. Step 2: CH3CH2CH=CH2 + Br2 → CH3CH2CHBrCH2Br. Step 1: H . The actual pKa value is 3. 3 max (b) chloroethanoic acid > 3-chloropropanoic acid > propanoic acid / OWTTE. increases the stability of the conjugate ion.87– 4. heat CH3CH2CH2CH2OH → CH3CH2CH=CH2 + H2O. 1 (c) any pKa value or range of values within the range 2. concentrated H3PO4 or concentrated H2SO4 as catalyst. Reaction type for Step 2: electrophilic addition/ AE. halogens are electron acceptors / halogens withdraw/pull electrons / halogens are more electronegative than carbon. Allow more detailed formulas. 5 [5] 75. + Accept H+.86. Penalize missing H atoms. cis (more) polar / trans non-polar/less polar. cis experiences stronger (permanent) dipole-dipole interaction / trans experiences no/(much) less dipole-dipole interaction. 3 [7] 81. 1 IB Questionbank Chemistry 32 . Award [1] for 2 correct structures without names.79. (a) compounds with same structural formula.3-dichlorocyclobutane) Need clear cis/trans structure and name for each mark. 2 (ii) cis (higher boiling point).3-dichlorocyclobutane) Trans(-1. but different arrangement of atoms in space/spatial arrangement. Do not allow “same molecular or chemical formula without the same structural formula”. (i) Accept CH3CH2CH2Br. Accept CH3CH2CH2CH2NH2. Do not accept just strong forces without reference to dipole-dipole interaction. 2 (b) (i) Cis(-1. B [1] 80. (ii) CH3CH2CH2Br + KCN → CH3CH2CH2CN + KBr. Accept ionic equation with C2H5O– or OH–. C2H5Br + C2H5ONa → C2H4 + NaBr + C2H5OH / C2H5Br + NaOH → C2H4 + NaBr + H2O. Accept ionic equation. Equation must be balanced for mark. CH3CH2CH2CN + 2H2 → CH3CH2CH2CH2NH2. heat/150 °C. 3 IB Questionbank Chemistry 33 . (i) hot. alcoholic OH– /NaOH/KOH. 4 [5] 82. For the second equation: Ni (as catalyst). Accept LiAlH4 in place of reaction with hydrogen. Accept arrow origin from OH– but do not allow curly arrow originating on H in OH–. FeCl3.g. 1 IB Questionbank Chemistry 34 . 5 [8] 83. curly arrow showing Br leaving. structural formula of organic product CH2=CH2. Award [4 max] for E1 mechanism (unstable primary carbocation) curly arrow showing Br leaving. representation of primary carbocation. Accept other workable Lewis acids e. curly arrow going from lone pair on O in H2O to H on C adjacent to C+ and curly arrow going from CH bond to form C=C bond. (a) Accept C6H6 and C6H5CH3 instead of their structural formulas. curly arrow going from CH bond to form C=C bond. structural formula of organic product CH2=CH2. Equilibrium sign and Lewis acid not required for mark. Accept OH– in place of C2H5O–(to form H2O). curly arrow going from lone pair/negative charge on O in C2H5O– /CH3CH2O– to H on β–C.(ii) OH– reacts with ethanol to form ethoxide ion/C2H5OH + OH– → C2H5O– + H2O. representation of carbocation with correct formula and positive charge on ring. condensed or displayed. 4 [5] 84. 2 max [4] IB Questionbank Chemistry 35 . of (4-methylphenyl)ethanone. (a) correct structural formulas of reactants.positions / OWTTE. Accept 2-methyl isomer. correct structural formula. curly arrow going from lone pair/negative charge on Cl in AlCl4– to H and curly arrow going from CH bond to benzene ring.and 6-positions / stabilizes σ-complexes with sp3 hybrid carbon in ortho. CH3CO is an electron-withdrawing/deactivating/Type II substituent (which) prevents/decreases the rate of further substitution / OWTTE. formation of organic product methylbenzene and HCl and AlCl3. 2 (b) sterical hindrance of the methyl group / CH3 is bulky enough to prevent substitution at 2.(b) curly arrow going from delocalized electrons in benzene to +CH3. condensed or displayed.and 6-positions / OWTTE. 4. Do not penalize if CH3+ is written. Allow mechanism with corresponding Kekulé structures.and para. CH3 is a (weakly) electron-releasing/Type I substituent / directs substitution to 2-. B [1] 90. addition-elimination / condensation / esterification / nucleophilic substitution. (a) methylpropene. 3 [3] 86. 1 (b) (i) brown/orange/yellow to colourless / bromine is decolorized. correct structural formula of ethanoic acid. 1 IB Questionbank Chemistry 36 .85. B [1] 87. C [1] 89. Accept 2-methylpropene. A [1] 88. correct structural formula of acetylsalicylic acid. Accept condensed formula for ethanoic acid. (electrophilic) addition.2-dibromo-2-methylpropane / 1. cheap. with weaker intermolecular/Van der Waals’/London/dispersion forces. chloroethane. Award [1] if structure and correct name are given for 2-bromo-2methylpropan-1-ol. Accept monomers have lower molecular mass. chemically unreactive materials produced. 2 max (ii) addition. hyphens or added spaces. Award [2] for any two. wide range of uses/physical properties / versatile. 1 (iv) monomers are smaller molecules / have smaller surface area than polymers. 2 [2] IB Questionbank Chemistry 37 . Must have 6 carbons joined to each other along chain. Do not penalize missing commas.2-dibromomethylpropane / 1-bromo-2methylpropan-2-ol / 1-bromomethylpropan-2-ol.(ii) 1. 1 (iii) Must show continuation bonds. large industry. Do not accept free radical/nucleophilic addition. Accept opposite argument for polymers. Ignore bracket around the 6 carbons. uses a limited natural resource. (i) synthesis of materials not naturally available/plastics. 2 [6] 92. 2 [4] 91. Award [1] for any termination step. 2 (ii) 2 (d) Initiation: Br2 → 2Br•.01 1. Propagation: Br• + RCH3 → HBr + RCH2•. Termination: [1 max] Br• + Br• → Br2. C3H8. 3 (b) C 3H 8. Do not penalize the use of an incorrect alkane in the mechanism.01 ratio of 1: 2. 4 max (e) (i) substitution and nucleophilic and bimolecular/two species in rate-determining step. Accept radical with or without • throughout. Allow second order in place of bimolecular.93. Accept hf/hv/sunlight. RCH2• + Br2 → RCH2Br + Br•.1. Cr2O72– /MnO4– and acidified/ H+ /H3O+.7. heat / reflux. UV/ultraviolet light. 1 IB Questionbank Chemistry 38 . Accept names. No penalty for using 12 and 1. 12. RCH2• + RCH2• → RCH2CH2R.80 and nH = = 18. 1 (c) (i) Br2 /bromine.3 = 6.67 /1: 2.7 18. (a) nC = 81. RCH2• + Br• → RCH2Br. planar. delocalization / resonance. all C–C–C bond angles 120°. all carbon-carbon bond lengths equivalent / all carbon-carbon bond lengths intermediate between single and double bonds / carbon-carbon bond order of 1.(ii) curly arrow going from lone pair/negative charge on O in OH– to C. Do not allow curly arrow originating on H in OH– curly arrow showing Br leaving. Allow ecf if wrong structures suggested. representation of transition state showing negative charge. Do not penalize if HO and Br are not at 180° to each other. (i) CH3OCH2CH. Allow more detailed structural formulas. 2 max [4] 95. 3 max IB Questionbank Chemistry 39 . CH3CHOHCH. 2 (ii) CH3CHOHCH3 has higher boiling point due to hydrogen bonding. CH3OCH2CH3 has lower boiling point due to Van der Waals’/London/ dispersion/dipole-dipole forces. Accept curly arrow either going from bond between C and Br to Br in bromoethane or in the transition state. Do not award M3 if OH ---. Do not penalize the use of an incorrect alkyl chain in the mechanism. 3 [16] 94.5. square brackets and partial bonds. Allow sp2 (hybridization for Cs).C bond is represented unless already penalized in M1. hydrogen bonds in CH3CHOHCH are stronger. (a) hexagonal / ring of six carbon atoms (each with one hydrogen). (addition of) CO2/carbon dioxide and H2O/water. undergoes (electrophilic) substitution reactions / does not undergo addition reactions / does not decolorize bromine water. negative charge/lone pair on oxygen can interact with delocalized electrons of benzene/aromatic ring so spreads out charge more / OWTTE. (CH3)CH=CH2. (a) phenol. 2 max [5] 96. (a) A: CH3CH2CH(CH3)Br. only one isomer exists for 1. 2 IB Questionbank Chemistry 40 . Allow diffraction pattern or contour map for electron density map. Br2 /bromine. CH3CH2CH2CO2H / CH3CH2CH2COOH. concentrated phosphoric acid/ H3PO4 /sulfuric acid/ H2SO4.2-disubstituted benzene compounds / only three disubstituted benzene compounds (rather than four). BrCH2CHBrCH3. (addition of) Mg/magnesium. 4 [4] 98. CH3CH2CH2MgBr. 4 [4] 99. electron density map (of benzene) showing equal electron density/all carbon-carbon bond lengths equivalent / OWTTE. 1 (b) 1 [2] 97.(b) enthalpy change of hydrogenation not equal to three times enthalpy change of hydrogenation of cyclohexene. making it break more easily to release H+ ions / electron-withdrawing nature of chlorine / greater electronegativity of chlorine / OWTTE. rate = k[X].(b) chloroethanoic acid. (a) First and second structures should be mirror images. 3 [5] 100. ClCH2CO2– more stable than CH3CO2– because negative charge spread more / OWTTE. order with respect to OH– = 0. chlorine (atoms) withdraws electrons from OH bond. C [1] 104. D [1] 101. 2 (b) (i) 3 IB Questionbank Chemistry 41 . B [1] 103. B [1] 102. order with respect to X = 1. Award [3] for final correct answer. Tetrahedral arrangement around carbon must be shown. 2 (iv) 2 [11] 105.(ii) 0. Allow other suitable catalysts such as FeCl3 etc. formation of organic product chlorobenzene and HCl and AlCl3. 4 [4] IB Questionbank Chemistry 42 . Do not penalize if curly arrow originates inside circle. 2 (iii) 2-bromo-2-methyl-propane. min–1. Do not penalize missing hyphens or added spaces. tertiary (structure).2(0). curly arrow going from delocalized electrons in benzene to Cl in Cl2 and curly arrow going from Cl–Cl bond to AlCl3. No penalty if primary structure is shown. No credit for SN2 mechanism. Allow H+ + AlCl4– → HCl + AlCl3 as alternative to curly arrow going from lone pair/negative charge on Cl in AlCl4– to H. except by ECF. curly arrow going from lone pair/negative charge on Cl in AlCl4– to H and curly arrow going from CH bond to benzene ring. C4H9+ + OH– → C4H9OH / in equation with curly arrows and fast. Allow curly arrow going from delocalized electrons in benzene to Cl+ for M1. Accept 2-bromomethylpropane. Allow mechanism with corresponding Kekulé structures. representation of carbocation with correct formula and positive charge on ring. C4H9Br → C4H9+ + Br– / in equation with curly arrows and slow. 1 (c) addition-elimination. C [1] 111. 1 [4] 108. A [1] 109. CH3CH2NH3+Cl– /CH3CH2NH3Cl. A [1] IB Questionbank Chemistry 43 . (addition of) Br2 / bromine and UV light/hf/hv. Order of E and F does not matter. 2 (b) CH3CH2NH2 / ethylamine. C [1] 110. Accept acylation. Order of steps does not matter (so first product could form 2-bromomethylbenzene). Allow other suitable catalysts such as FeBr3 etc.106. structure of first product i. 3 [3] 107. (addition of) Br2 / bromine and AlBr3 / aluminium bromide / AlCl3 aluminium chloride. (a) CH3CONHCH2CH3.e. butane has van der Waals’/London/dispersion forces. propanal has dipole-dipole attractive forces. propanone / acetone. 4 (ii) butane is least soluble. it cannot form hydrogen bonds/attractive forces with water molecules. 2 (iii) propanal and propanoic acid. (i) hydrogen bromide / hydrobromic acid. 1 (ii) sodium hydroxide / hydroxide ions (name required).112. dilute and aqueous / dilute and warm / aqueous and warm. 2 [12] 113. Do not accept HBr. as name is asked for. Treat references to bond breaking as contradictions if the imfs are correct. imf marks are independent of the order. 2 IB Questionbank Chemistry 44 . (i) butane < propanal < propan-1-ol. 3 (iv) 1 (v) secondary (alcohol). propan-1-ol has hydrogen bonding. 134 nm) / has delocalized electrons / drawing showing two resonance forms e. 2 [8] 114. Accept from lone pair or minus sign or O. transition state including negative charge and partial bonds.154 nm) and double bonds (0. (a) contains a six-membered carbon ring. all bond angles are 120°/the same.139 nm. ( HO =) –360 (kJ mol–1).(iii) curly arrow from OH– to C atom.g. C–C bond lengths are intermediate between single (0. Do not award marking point if arrow originates from the H of OH–. Accept all carbons are sp2 hybridization. curly arrow from bond between C and Br to bromine atom on bromoethane or the transition state. planar. 3 (iv) hydration of ethene / steam + ethene. 4 max (b) 1 [5] IB Questionbank Chemistry 45 . Allow equation (ethanol used as) solvent/fuel/antiseptic/intermediate to form other compounds. all C–C bond distances are equal/0. CH3CH2COOH etc. 2 (ii) 2 [4] 116. 1 (ii) Allow correct condensed structural formula in each case e. C–Br bond is stronger/harder to break because lone pair of electrons on Br interact with delocalized electrons / OWTTE. (i) SN(2)/nucleophilic substitution (bimolecular). 2 [3] IB Questionbank Chemistry 46 .115. (i) CH3MgBr/methyl magnesium bromide. OH– ions repelled by the delocalized electrons in aromatic ring / OWTTE.g. Allow correct condensed structural formula in each case. also students do not have to show double and triple bonds. 4 [4] IB Questionbank Chemistry 47 . Do not penalize students if they draw a structure that attaches NO2 to benzene ring via O and not N.117. D [1] 121. mechanism showing: curly arrow going from (lone pair of electrons on) O to H+. Award [3] for a concerted mechanism. B [1] IB Questionbank Chemistry 48 . C [1] 120. correct identification of products as (CH3)2C=CH2/methylpropene. Correct geometry is not required for structure of carbocation. curly arrow from (lone pair on) oxygen of water to H shown. structure of carbocation. 4 [4] 119.118. B [1] 123. E2 curly arrow showing OH– acting as base on H bonded to C. reflux / heat. D [1] 124. (a) (i) CH3CH2CH2Br → CH3CH2CH=CH2 + HBr / CH3CH2CH2CH2Br + OH– → CH3CH2CH=CH2 + H2O + Br–. concerted curly arrows showing Br leaving C–Br. alcoholic NaOH/OH–. OH– acting as base on the intermediate carbocation.122. 3 IB Questionbank Chemistry 49 . Then accept either E1 or E2 mechanism. E1 curly arrow showing bromine leaving the halogenoalkane. Penalize missing Hs once only throughout the question 3 (ii) elimination reaction. warm / excess ammonia (to prevent secondary amines etc. Accept a second molecule of NH3 removing H+ from the transition state to give NH4+ and Br– as products. HOOC(CH2)8COOH / ClOC(CH2)8COCl. 2-bromo-2-methylpropane goes by SN1 / 1-bromobutane by SN2.(iii) CH3CH2CH2CH2Br + NH3 → CH3CH2CH2CH2NH2 + HBr. 1 (ii) H2N(CH2)6NH2. no space around tertiary carbon for five groups (in SN2 transition state). formation of HBr and organic product.). 2 (iii) 2-bromo-2-methylpropane is tertiary / 1-bromobutane is primary. 2 (ii) polarimeter (to measure angle of rotation). 2 IB Questionbank Chemistry 50 . 3 max [20] 125. correct transition state. the plane of plane-polarized light rotates in opposite directions (by the different enantiomers). 3 (iv) curly arrow from ammonia (to form transition state). (i) amide / peptide. 4 (b) (i) Award [1] for correct structure and [1] for correct 3-D representation of both enantiomers. intermediate carbocation more stable for tertiary. curly arrow from bond to Br atom in either the first or second step. ammonia/NH3. (i) HNO3 + 2H2SO4 → NO2+ + 2HSO4– + H3O+ / HNO3 + H2SO4 → NO2+ + H2O + HSO4– / HNO3 + H2SO4 → H2NO3+ + HSO4– and H2NO3+ → H2O + NO2+ Award [1] for correct reactions and products and [1] for balancing. Award [1] for correct organic product and [1] for (2n – 1)H2O. Accept --(--OCC6H4COOCH2CH2O--)n— for the organic product. Also accept two step equations or curly arrow equations.(iii) nHOOCC6H4COOH + nHOCH2CH2OH → HO--(--OCC6H4COOCH2CH2O--)n–H + (2n – 1)H2O. 2 IB Questionbank Chemistry 51 . 2 [5] 126. Award [1 max] for the equation: 3 [5] IB Questionbank Chemistry 52 .(ii) equation showing the formation of the CH3CO+ ion. curly arrow going from benzene to electrophile and subsequent formation of intermediate correctly represented in mechanism. curly arrow showing removal of proton and second curly arrow showing the reformation of the aromatic ring to form the new product and hydrogen chloride. C [1] 131. 2 (ii) (nucleophilic) substitution only on alkyl group/nucleophilic substitution cannot occur on benzene ring / OWTTE. 2 [4] 129.127. A [1] IB Questionbank Chemistry 53 . Br2 and AlBr3 / AlCl3 (to substitute in benzene ring). first product. correct reagents and first step. Accept in either order. Accept correct names or condensed formulas in place of structures. (i) reaction with Br2 and UV light (to react with methyl group). reaction with Br2 and UV light in second step. 3 [3] 128. Accept other reasonable suggestions. Chlorobenzene is not necessary for mark. D [1] 130. 5–23 and 47.88 kJ. 1 IB Questionbank Chemistry 54 . 1 g ethanol produces 22. B: reflux. Award [2] for +1040 kJ. average values obtained from a number of similar bonds/ compounds / OWTTE. Bonds formed (2 × 2)(C=O) + (3 × 2)(O–H) = (4)(746) + (6)(464) = 5768 (kJ). 4 (v) ethanol/CH3CH2OH.132. (i) (2-)methylbutane / (2.1 and Mr(C8H18) = 114. B: CH3COOH/CH3CO2H. Accept values ranges of 22.3. (i) energy required to break (1 mol of) a bond in a gaseous molecule/state. dyes / drugs / cosmetics / solvent / (used to make) esters / (used in) esterification/disinfectant. Award [3] for final correct answer. 3 (iii) Mr(C2H5OH) = 46. hydrogen bonding (in ethanol). 2 (ii) Bonds broken (1)(C–C) + (1)(O–H) + (5)(C–H) + (1)(C–O) + (3)(O=O) = (1)(347) + (1)(464) + (5)(413) + (1)(358) + (3)(498) = 4728(kJ). H = 4728 – 5768 = –1040 kJ mol–1 / –1040 kJ. C [1] 133. Accept energy released when (1 mol of) a bond is formed in a gaseous molecule/state / enthalpy change when (1 mol of) bonds are made or broken in the gaseous molecule/state. 2 (iv) A: CH3CHO.8–48 kJ respectively. 2 (vi) (concentrated) H3PO4 /(concentrated) phosphoric acid / H2SO4/sulfuric acid. Units needed for last mark. 2 [16] 134. A: distillation.26/114. Award second point only if the first is obtained.57 kJ and 1 g octane produces 47. Accept either full or condensed structural formulas but not the names or molecular formulas. No penalty for use of Mr = 46 and Mr = 114.08 / 46.2-)dimethylpropane. leaving of Br. correct products CH3(CH2)4OH and Br–. 1 IB Questionbank Chemistry 55 . (a) bonding electron pair spread over three (or more) nuclei or atoms/ not restricted/confined between two nuclei or atoms / OWTTE. 4 [5] 135. transition state representation with both Br and OH attached to C–1.(ii) curly arrow going from O/lone pair of OH–(but not H) to the C attached to Br. C–OH and C–Br bonds in TS must be represented by dashed lines. Charge must be shown for TS. 3. benzene produces less (due to delocalization). OR electron density maps. 4 max [5] IB Questionbank Chemistry 56 . show even electron density over ring. OR not (structural) isomers / same compounds. chemical evidence [2 marks]: hydrogenation of C6H6 (1. as it is more stable (due to delocalization). instead of longer single bonds and shorter double bonds / intermediate between single and double. not 4. OR 3 isomers of C6H4X2.5-cyclohexatriene) expected to produce three times as much energy as cyclohexene.(b) physical evidence [2 marks]: same carbon–carbon bond lengths in benzene / all carbon–carbon bonds are equal in length / forms regular hexagon. OR (1H/proton) NMR spectrum shows only one peak / all the Hs in the same chemical environment. not 2 peaks / not 2 different chemical environments. different bond energies. OR enthalpy of combustion of C6H6 less than expected. OR benzene undergoes substitution rather than addition reactions. reaction is faster with C6H5CH2Cl. minor product: (CH3)2CHCHBrCH3/2-bromo-3-methylbutane. attacking OH– nucleophile not repelled by the delocalized electrons. C–Cl bond stronger/harder to break. C–Cl bond weaker/easier to break. because of overlap between a lone electron pair of Cl with ring delocalized electrons / delocalization increases electron density on ring / delocalization reduces δ+ charge on C (attached to Cl atom). more electron releasing/pushing R/alkyl/C groups in tertiary. attacking OH– nucleophile repelled by delocalized electrons / attraction of nucleophile decreases / less polar C does not attract OH– as much. Accept condensed structural formulas 2 [2] IB Questionbank Chemistry 57 . OR reaction is slower with C6H5Cl. tertiary more stable than secondary carbocation / more alkyl groups with positive inductive effect. 3 max [3] 137. major product: (CH3)2CBrCH2CH3/2-bromo-2-methylbutane. because no overlap between a lone electron pair of Cl with ring delocalized electrons.136. 5 [5] 138. mention of carbocation/(CH3)2C+CH2CH3/(CH3)2CHCH+CH3. 2 (b) Accept either of the two following alternatives for the second and third mark. D [1] IB Questionbank Chemistry 58 . Accept equation with condition specified.139. C2H5MgBr. OR C2H5COCH3. A [1] 142. CH3COCH3. in an anhydrous/dry solvent / (C2H5)2O /diethyl ether. 3 max [5] 140. C [1] 143. If CH3COCH3 and CH3MgBr or C2H5COCH3 and C2H5MgBr combination given. CH3MgBr. B [1] 141. then award only [1]. (a) reaction of Mg/magnesium with halogenoalkane/named compound. 144. dyes / drugs / cosmetics / solvent / (used to make) esters / (used in) esterification / disinfectant. A: distillation. (i) CH 3CH 2 OH 2 Cr2 O 7 → CH3COOH → CH3CH2CH2CH3 + H2O CH3CH2OH K H+ H2SO4 Structural formulas of reactants and products CH3CH2OH and CH3COOH/CH3CO2H and CH3CO2CH2CH3 (+ H2O). B: CH3COOH/CH3CO2H. Award [1] for structural formulas of reactants and products and [1] for the correct conditions/reagents used. B: reflux. A: CH3CHO. Accept H+/H2SO4 instead of sulfuric acid and acidified. Suitable oxidizing agents are potassium dichromate/K2Cr2O7 / sodium dichromate/Na2Cr2O7 / dichromate/Cr2O72– / potassium manganate(VII)/potassium permanganate/KMnO4 / permanganate/ manganate (VII)/MnO4–. (concentrated) H2PO4/(concentrated) phosphoric acid / H2SO4/sulfuric acid. 2 [2] 146. 2 IB Questionbank Chemistry 59 . Accept either full or condensed structural formulas but not the names or molecular formulas. 4 [4] 145. Conditions/reagents used reflux with named suitable acidified oxidizing agent and then heat with alcohol and sulfuric acid. formation of (CH3)2C=CH2 and Br–. curly arrow going from the C–H bond on the β carbon to the bond joining the α carbon to the β carbon and curly arrow showing Br acting as leaving group. Accept alternative E1 type mechanism curly arrow showing Br acting as leaving group to form carbocation. Use of NaOCH2CH3 with curly arrow originating on O of NaOCH2CH3 is penalized already in the first marking point. Allow formation of NaBr for third marking point. and everything else correct. curly arrow going from O of –OCH2CH3 attacking hydrogen.(ii) 2 Cr2 O 7 2O H2C=CH(CH3) H → CH3CH(OH)CH3 K → (CH3)2CO H2SO4(conc. Conditions/reagents used water/H2O and sulfuric acid/H2SO4 / dilute acid medium and heat/reflux with suitable acidified oxidizing agent. Accept either full or condensed structural formulas throughout the question.) H+ Structural formulas of reactants and products H2C=CH(CH3) and CH3CH(OH)CH3 and (CH3)2CO. 2 [4] 147. No marks awarded if a substitution mechanism is given. Note: If primary alcohol is given as product of first step. Suitable oxidising agents are potassium dichromate/K2Cr2O7 / sodium dichromate/Na2Cr2O7 / dichromate/Cr2O72– / potassium manganate(VII)/ potassium permanganate/KMnO4 / permanganate/manganate (VII)/MnO4–. 3 [3] IB Questionbank Chemistry 60 . Do not award first mark if curly arrow originates from O of NaOCH2CH3. award [1 max]. if was used (incorrectly) in the mechanism. formation of (CH3)2C=CH2 and Br–. Accept H+/H2SO4 instead of acidified. curly arrow going from O of –OCH2CH3 attacking hydrogen. Allow the curly arrow to originate from either the lone pair or O of –OCH2CH3 but not from H of –OCH2CH3. 1-chloro-3-nitrobenzene / meta-chloronitrobenzene / m-nitrochlorobenzene / 3-chloro-1-nitrobenzene / 3-chloronitrobenzene / 1-nitro-3-chlorobenzene / 3-nitrophenylchloride. (iii) Allow [1 max] if the structures are correct but it is not clear that they are mirror images.and 4-positions. 1 (ii) Allow [1 max] if structures are correct but arrangement of groups in space does not clearly show the cis/ trans isomerism. 2 [3] 149.148. 5 [5] IB Questionbank Chemistry 61 . (i) compounds with the same (molecular formula and) structural formula but different arrangements of atoms in space / OWTTE. deactivates ring / reduces electron density on ring. NO2 electron withdrawing/attracting. slower rate compared to benzene. greater charge distribution in the 3-position / lesser charge in the 2. D [1] ¥ ì154. intermediate. Partial polarity on C=O and C–Cl not required for mark. addition–elimination mechanism. C [1] 152. Curly arrows not required for mark. movement of electrons from OH– to C. A [1] 153. 5 [5] 151. originating from negative charge or oxygen atom. C [1] IB Questionbank Chemistry 62 . movement of electrons in the C=O. CH3COCl + NaOH → CH3COOH + NaCl / CH3COCl + OH– → CH3COOH + Cl–.150. 2 (iii) 1-chlorobutane: SN2. Do not accept 2-dimethylpropane. or 2. 2 (b) C5H12. the total bond strength in the pentanol molecule is higher than the total bond strength in pentane. 2 IB Questionbank Chemistry 63 . pentanol requires more energy to break intermolecular forces/ hydrogen bonding / OWTTE.2-dimethyl propane/dimethyl propane. hyphens or added spaces. Do not penalize missing commas. the total amount of energy produced in bond formation of the products per mole is the same. weaker intermolecular force/van der Waals’/London/dispersion forces. unimolecular / OWTTE. (a) (i) Isomer Boiling point A 36 °C B 28 °C C 10 °C Award [1] if correct boiling points are assigned to 3 isomers. C: 2. Accept any two of the following explanations. fewer moles of pentanol in 1 g. 2-chloro-2-methylpropane: SN1. analogous compounds such as butane and butan-1-ol show a lower value for the alcohol per mole in the data book / OWTTE. increase in branching / more side chains / more spherical shape / reduced surface contact / less closely packed. (i) C4H9Cl + KOH → C4H9OH + KCl. C5H11OH has greater molar mass / produces less grams of CO2 and H2O per gram of the compound / suitable calculations to show this. 1 (ii) (substitution) nucleophilic.155. C5H11OH contains an O atom which contributes nothing to the energy released / partially oxidized / OWTTE. Accept the opposite arguments 3 (ii) B: 2-methylbutane/methylbutane.2-methylpropane. 3 max [8] 156. SN1 2-chloro-2-methylpropane–allow ECF from (iii) curly arrow showing Cl leaving. formation of carbocation. with negative charge and dotted lines to represent bonds. curly arrow from lone pair or negative charge on O in OH– to C+. curly arrow for Cl leaving. formation of the transition state in bracket. curly arrow going from lone pair or negative charge on O in OH– to C. 6 [11] IB Questionbank Chemistry 64 . Can be shown in transition state.(iv) SN2 1-chlorobutane–allow ECF from (iii). 1 (iv) orange to green. 4 (ii) CH3–CH2–CH2–CHO / (CH3)2CHCHO.157. (i) Penalize missing H atoms once only. 2 (iii) CH3–CH2–CO–CH3. CH3–CH2–CH2–COOH / (CH3)2CHCOOH. 1 IB Questionbank Chemistry 65 . Accept correct condensed structural formulas. (v) 1 [9] 158. (a) dimethylamine / (CH3)2NH. No ECF from (a). electronegative/electron-withdrawing chlorine draws electrons away from carboxylate/ COO–/CO2– (group) / attracts electrons in the OH bond closer to oxygen. 2 max [3] 159. more likely to attract/accept proton from water molecule. 2 max [2] 161. 1 IB Questionbank Chemistry 66 . CH3CN + H2O + H3O+ → CH3COOH + NH4+. Allow any other reasonable pathway. making conjugate base weaker (and hence making the acid stronger) / reduces electron density on oxygen / so making it easier for a proton to leave. stabilization of positive ion / so dimethylamine contains an N-atom that is more electron-rich. 1 (b) methyl groups electron-donating/electron-releasing/ involve positive inductive effect. (a) (i) A: CH3CH2C(CH3)2OH. CH3MgCl + CO2 H2O required for mark OR CH3Cl + NaCN → CH3CN + NaCl. CH3Cl + Mg → CH3MgCl. H 2O → CH3COOH + Mg(OH)Cl. 2 [2] 160. Award [3 max] for a correct mechanism involving the formation of the 1-iodo product. secondary/ 2° carbocation more stable than primary/1° carbocation. curly arrow from C=C to H of HI and curly arrow showing iodide leaving. Allow CH3C+HCH2CH3. structure of carbocation and iodide attacking carbocation from either lone pair or negative charge. 1 [3] 162. because it is stabilized by a greater number of electron-releasing alkyl groups.(ii) 1 (b) addition-elimination / condensation. structure of CH3CHI(CH2CH3) as major organic product (C). 5 [5] IB Questionbank Chemistry 67 . curly arrow showing lone pair on oxygen attacking H+. allow other alternatives such as H2PO42– ion removing H+ (which shows its action as a catalyst) or simple loss of H+. Only penalize once for missing lone pair on O. Allow condensed formula if + charge on correct C atom curly arrow showing lone pair of oxygen on water attacking hydrogen / curly arrow from C–H bond to form C=C / curly arrow showing H2PO42– removing hydrogen. Product D: CH2=CH2. For M4. e. Do not allow departure of OH–.g. 5 [5] IB Questionbank Chemistry 68 . curly arrow showing departure of water. Allow no charge on O. formula/structure of carbocation.163. 164. (i) 2 (ii) no rotation possible due to double bond/pi bond. 2 [5] IB Questionbank Chemistry 69 . Accept hindered or restricted rotation. chiral carbon atom identified. A [1] 165. B [1] 168. C [1] 166. 1 (iii) correct structural formula. D [1] 167. cis isomer has (predominantly) intramolecular hydrogen bonding. curly arrow showing Br leaving. (i) CH3CH2Br + OH– → CH3CH2OH + Br–. Accept KOH and KBr in the balanced equations dilute KOH compared to concentrated KOH. Ni / Pt / Pd. Accept any two reaction conditions. representation of transition state. 4 max IB Questionbank Chemistry 70 . 3 (ii) cis isomer readily releases water (vapour forming a cyclic anhydride). 4 (iii) CH3CH2CN + 2H2 → CH3CH2CH2NH2. warm/40–50 °C compared to hot/80-100 °C. CH3CH2Br + OH– → CH2=CH2 + H2O + Br–. (i) SN2. Curly arrow may be represented on transition state. 1 (ii) curly arrow going from CN– to C. products. 2 [6] 171.169. aqueous KOH compared to ethanolic KOH. 1 [4] 170. (i) trans has the higher melting point. trans isomer has (predominantly) intermolecular hydrogen bonding. showing negative charge and dotted lines. Accept opposite arguments for trans isomer. G: CH3Cl and AlCl3.4 / ortho/o. 3 [3] 173.(ii) curly arrow from O to H. 4 (iii) addition. products CH2=CH2 + Br– + C2H5OH/H2O. curly arrow showing Br leaving. (i) activating. C [1] IB Questionbank Chemistry 71 . 1 [9] 172. curly arrow from C–H to C–C. para/p directing. 2 [4] 174. B [1] 175. 1 (iii) electron-withdrawing nature of –NO2 group. For M1 and M3 award [1 max] if CH3COCl is given for E and CH3Cl is given for G only without AlCl3. E: CH3COCl and AlCl3. F: HNO3 and H2SO4. 1 (ii) 2. so makes benzene ring less susceptible to attack by electrophiles. (a) one general formula / same general formula. similar chemical properties. due to stronger hydrogen bonding/2 hydrogen bonds per molecule. 1 (b) ethanol lower / ethanoic acid higher. gradual change in physical properties.176. [1] for any two correct isomers. due to larger mass of ethanoic acid/stronger van der Waals’/ London/dispersion forces. Award [1] for any two of the above characteristics. [5] IB Questionbank Chemistry 72 . Award [2] for all three correct isomers. differ by CH2. B [1] 178. A [1] 177. 2 (c) 2 Allow condensed structural formulas such as CH3CH2CH2CH2OH. A [1] 179. Accept either answer for second mark. 180. (i) (Empirical formula =) C8H8O3; H H O O H C C H O H ; H H H 2 Allow double bonds on arene in alternate positions, or allow delocalized representation (of pi electrons). (ii) the bond at 0.1373 nm is a double bond and the bond at 0.1424 nm is a single bond; in CO2(g) both bonds are double bonds and would have a value around 0.137 nm; 2 (iii) Ester; Arene/benzene ring; Alcohol; Award [2] for any three correct, award [1] for any two correct. Do not accept alkane as a type of functional group in this molecule. 2 [6] 181. (i) boiling point increases as the number of carbons increases / OWTTE; Greater Mr and hence greater van der Waals’/London/dispersion forces present; 2 IB Questionbank Chemistry 73 (ii) / UV light CH4 + Cl2 hv → CH3Cl + HCl; Do not award mark if hv/uv light is not given. Initiation step: / UV light Cl2 hv → 2Cl•; Do not award mark if hv/uv light is not given. Penalize once only. Propagation step: CH4 + Cl• → CH3• + HCl; CH3• + Cl2 → CH3Cl + Cl•; Termination step: Cl• + Cl• → Cl2 or Cl• + CH3• → CH3Cl or CH3• + CH3• → CH3CH3; Allow fish-hook half-arrow representations i.e. use of Penalize use of full curly arrows once only. Penalize missing dots on radicals once only. . 5 [7] 182. (i) A. = CH3(CH2)7CHO; B. = CH3(CH2)7COOH/CH3(CH2)7CO2H; C. = (CH3)3COH; D. = (CH3)2CO; E. = BrCH2CH2Br; Allow correct structural formulas. 5 (ii) addition; /-(CH2-CH2)3-/-(CH2)6-; 2 [7] 183. D [1] 184. B [1] IB Questionbank Chemistry 74 185. D [1] 186. C [1] 187. (a) Cl Cl and 1,1 dichlorocyclopropane; Cl Cl Cl Cl Cl Cl Cl Cl (cis- or trans-) 1,2 dichlorocyclopropane; Award point for the correct name corresponding to the related isomer. Accept diagrams that do not display 3 dimensional structure. Award [1 max] for correct structures only, without the corresponding names. 2 (b) Cl CH 3 and Cl Cl H Cl Cl CH 2 Cl and H Cl CH 3 H H H H CH 2 Cl 2 [4] IB Questionbank Chemistry 75 curly arrow showing formation of double bond. award [2 max]. Award [1] each for any three. curly arrow showing C–Br bond fission. (ii) plane polarized light. 2 (iii) CH 3 C H C H CH 3 . rotation in opposite/different directions. CH3OH + HCOOH → HCOOCH3 + H2O Award [1] for both reactants and [1] for both products (accept C2H4O2). If but-2-ene formed.188. CH 3 2 (iv) curly arrow showing attack by –OH on end H. (i) OH C CH3 CH2 H CH3 H CH3 OH C CH 2 CH 3 2 Award [2] for both tetrahedral structures. 3 [3] IB Questionbank Chemistry 76 . CH 3 C H C H . methyl methanoate. 3 max [9] 189. or [1] if tetrahedral structure is not clear. H2O and Br– shown as products. curly arrow showing pi bond breaking.154 and 0. (CH3)2C(OH)CN + H+ + 2H2O → (CH3)2C(OH)COOH + NH4+. Accept hydroxy(l) instead of alcohol. curly arrow from :O to H+. award [1]. (a) all C—C bonds in benzene or structure B are 0.190. 150 (kJ mol–1)/difference between –360 and –210 represents greater stability of benzene/structure B. 1 :CN – CN δ+ δ– C O H3C C CH 3 H+ H3C CN O – : H3C C CH 3 OH H3C Suitable diagram with curly arrow showing attack by :CN– on carbonyl Cδ+. 4 (c) 2 [7] 191.134 (nm)/ different bond lengths. (a) addition-elimination/condensation. Accept more detailed formula. structure A would have C—C bond lengths of 0. 1 [5] 192.139 (nm) (long)/the same length. If no reference to carbon-carbon bonds. 1 IB Questionbank Chemistry 77 . structure of product (CH3)2C(OH)CN. structure A would have value of (about) –360 (kJ mol–1). (a) (b) (CH3)2CO + HCN → (CH3)2C(OH)CN.134 (nm)/ benzene does not have C—C bond lengths of 0. 2 (b) (i) 2 (ii) delocalized electrons. carboxylic acid and alcohol.154 or 0. (b) H N N NO2 + H 2 O. due to its overall electron withdrawing capacity. 4 [4] IB Questionbank Chemistry 78 . 1 [4] 193. NO2 Award [1] for correct structural formula of the organic product and [1] for water. –NO2 is deactivating. and causes it to form more slowly. which destabilizes the carbocation intermediate. 2 (c) the (crystalline) solid has a characteristic melting point.