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A-Level · Chemistry

AC10 Chemistry: Aromatic Chemistry, Carbonyls and Amines

AQA 7405 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Benzene, C6H6, can be represented either by a Kekule model, with alternating single and double carbon-carbon bonds, or by a delocalised model, in which all six ring carbon-carbon bonds are identical and a ring of delocalised π electron density lies above and below the plane of the ring.

(a)State two pieces of experimental evidence that support the delocalised model of benzene over the Kekule model.(2)

(b)Describe the delocalised model of bonding in benzene.(3)

(c)The standard enthalpy of hydrogenation of cyclohexene (one C=C double bond) is -120 kJ/mol. (i) Calculate the enthalpy of hydrogenation that would be predicted for a hypothetical molecule 'cyclohexa-1,3,5-triene', containing three isolated, non-interacting C=C double bonds. (ii) The actual enthalpy of hydrogenation of benzene is -208 kJ/mol. Calculate the extra stabilisation (delocalisation/resonance energy) of benzene compared with this hypothetical molecule.(4)

(Total for Question 1 is 9 marks)

Benzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid to form nitrobenzene.

(a)State the conditions used for this nitration reaction.(2)

(b)Write an equation to show how the electrophile is generated from nitric acid in the presence of sulfuric acid.(2)

(c)Describe the mechanism for the nitration of benzene by the electrophile NO2+, including the structure of the intermediate formed.(4)

(d)Nitration of methylbenzene occurs faster than nitration of benzene, and gives mainly a mixture of 2-nitromethylbenzene and 4-nitromethylbenzene (with very little of the 3-isomer). Explain, in terms of the effect of the methyl group on the ring, (i) why methylbenzene reacts faster than benzene and (ii) why substitution occurs mainly at the 2- and 4-positions.(4)

(Total for Question 2 is 12 marks)

Phenol reacts rapidly with bromine water at room temperature, without the need for a halogen-carrier catalyst, to give an immediate white precipitate. Benzene does not react with bromine water under these conditions.

(a)Explain, in terms of the interaction between the oxygen lone pair and the ring, why phenol is far more reactive than benzene towards electrophilic substitution.(3)

(b)Identify the organic product formed when phenol reacts with excess bromine water, and give its molecular formula.(2)

(c)State the electrophile involved in the reaction between phenol and bromine water, and explain why this electrophile does not need to be generated using a halogen-carrier catalyst (unlike the bromination of benzene).(2)

(Total for Question 3 is 7 marks)

Required practical: a student is asked to prepare a solid ester, compound S, by reacting 3-methylphenol with an excess of benzoyl chloride, C6H5COCl, in the presence of a small amount of pyridine (which acts as a base to remove the HCl by-product), with the reaction mixture cooled in ice. The crude solid product is filtered, washed with cold water, and then purified by recrystallisation. A sample of the purified compound S is then used to determine its melting point.

(a)Describe how the crude solid compound S, once isolated by filtration, would be purified by recrystallisation.(4)

(b)Explain why the solvent chosen for the recrystallisation of compound S must dissolve it well when hot but only sparingly when cold.(2)

(c)The purified compound S is found to have a melting point range of 62-64 degC. The data book (literature) melting point of pure compound S is 65 degC. Evaluate the purity of the sample, and state what you would expect to observe about the melting point range of an impure sample compared with a pure one.(3)

(d)The student reacts 2.16 g of 3-methylphenol (Mr = 108.14) with an excess of benzoyl chloride. After purification, 3.18 g of pure compound S (Mr = 212.24) is obtained. Calculate the percentage yield of compound S.(4)

(e)Suggest one reason, other than an incomplete reaction, why the actual yield of compound S obtained is less than the theoretical yield.(1)

(f)Name the type of reaction occurring between 3-methylphenol and benzoyl chloride, and identify the other product formed alongside compound S.(2)

(g)State one hazard associated with using benzoyl chloride in this preparation, and one precaution the student should take because of it.(1)

(h)The melting point range obtained (62-64 degC) suggests the sample is not yet fully pure. Suggest one modification to the purification procedure that could improve the purity of compound S further, and explain why it would help.(2)

(Total for Question 4 is 19 marks)

Propanal, CH3CH2CHO, and propanone, CH3COCH3, are both reduced by aqueous sodium tetrahydridoborate(III), NaBH4, to alcohols.

(a)Describe the mechanism for the reduction of propanal by NaBH4, including the role of the hydride ion.(4)

(b)State a reagent, and the conditions needed, that would distinguish propanal from propanone using a simple test-tube test, and state what would be observed with propanal.(3)

(c)2,4-dinitrophenylhydrazine (Brady's reagent) reacts with propanal (and with propanone) to give an orange precipitate, confirming the presence of a carbonyl group but not distinguishing between the two compounds. Explain how this orange precipitate could then be used to identify which specific carbonyl compound (e.g. propanal or propanone) was originally present.(2)

(d)Calculate the atom economy for the conversion of propanone, C3H6O (Mr = 58), into propan-2-ol, C3H8O (Mr = 60), by reduction, treating the overall reaction as simple addition of H2 (Mr = 2) across the C=O bond, with propan-2-ol as the only product.(2)

(Total for Question 5 is 11 marks)

Ethanoic acid reacts with phosphorus(V) chloride, PCl5, to form ethanoyl chloride, CH3COCl.

(a)Write a balanced equation for this reaction and state what would be observed.(3)

(b)Ethanoyl chloride reacts with water far more readily (violently) than an ester such as ethyl ethanoate does. Explain, in terms of the bonding and leaving group involved, why acyl chlorides are much more reactive towards nucleophiles such as water than esters.(3)

(c)A 2.50 g sample of ethyl butanoate, C6H12O2 (Mr = 116), is completely hydrolysed by refluxing with an excess of water in the presence of a dilute acid catalyst. Calculate the maximum mass of butanoic acid, C4H8O2 (Mr = 88), that could be formed.(3)

(d)State the other organic product formed in this hydrolysis.(1)

(Total for Question 6 is 10 marks)

Ethylamine, ammonia and phenylamine differ considerably in their base strength.

(a)Explain, in terms of the availability of the nitrogen lone pair, why ethylamine is a stronger base than ammonia, and why phenylamine is a weaker base than ammonia.(4)

(b)State the type of mechanism occurring when bromoethane reacts with excess ethanolic ammonia to form ethylamine.(1)

(c)Explain why an excess of ammonia (rather than an excess of bromoethane) is used in this reaction.(2)

(d)Chloromethane can be converted into ethylamine by a two-step route. In step 1, chloromethane reacts with potassium cyanide dissolved in ethanol, under reflux. State the type of mechanism for this step and name the organic product formed.(2)

(e)State a suitable reagent and condition for step 2, the reduction of this nitrile to ethylamine.(1)

(Total for Question 7 is 10 marks)

Triethylamine, (C2H5)3N, reacts with excess chloromethane, CH3Cl, to form a quaternary ammonium salt.

(a)Write an equation for this reaction, and state the type of bond formed between the nitrogen atom and the new methyl group.(3)

(b)Quaternary ammonium salts with one long hydrocarbon chain and three short alkyl/methyl groups are used industrially as cationic surfactants (e.g. in fabric conditioners). Suggest, in terms of its structure, why such a compound can act as a surfactant.(2)

(c)Nitrobenzene can be converted into phenylamine (aniline) in the laboratory. Outline the two steps of a suitable synthetic route, including the reagents and conditions for the reduction step.(2)

(d)State one safety hazard of this reduction and one precaution that should be taken.(2)

(Total for Question 8 is 9 marks)

Glycine (aminoethanoic acid), H2NCH2COOH, is the simplest naturally occurring amino acid.

(a)Define the term 'zwitterion' and state the type of ion glycine forms at its isoelectric point.(2)

(b)Explain why glycine exists mainly as a cation in strongly acidic solution, but mainly as an anion in strongly alkaline solution.(3)

(c)Glycine and alanine can react together to form a dipeptide. State the type of reaction that occurs and name the type of linkage formed between the two amino acid residues.(2)

(Total for Question 9 is 7 marks)

Poly(ethylene terephthalate), PET, is a polyester formed from benzene-1,4-dicarboxylic acid (terephthalic acid) and ethane-1,2-diol. Nylon-6,6 is a polyamide formed from hexanedioic acid and 1,6-diaminohexane.

(a)Identify the type of polymerisation that forms PET, and state the small molecule released during its formation.(3)

(b)Write an equation to show the formation of one amide (peptide-type) linkage between one molecule of hexanedioic acid and one molecule of 1,6-diaminohexane, identifying the small molecule by-product.(3)

(c)Evaluate the environmental impact of using condensation polymers such as PET and nylon in packaging and clothing, compared with using naturally occurring materials, and discuss strategies that could be used to reduce this impact.(6)

(Total for Question 10 is 12 marks)

Compound Q is an amine used as a precursor in a pharmaceutical synthesis route. Compound Q contains only carbon, hydrogen and nitrogen. Complete combustion of 12.1 g of Q produces 35.2 g of carbon dioxide and 9.9 g of water. On combustion analysis, all of the nitrogen in Q is converted to gaseous nitrogen, N2, and 1.40 g of N2 is also collected.

(a)Calculate the empirical formula of compound Q.(4)

(b)Given that the relative molecular mass, Mr, of compound Q is 121, determine its molecular formula.(1)

(c)Compound Q reacts with dilute hydrochloric acid to form a soluble salt. Suggest the class of organic compound to which Q belongs.(1)

(Total for Question 11 is 6 marks)

A four-step synthetic route converts benzene into 1-phenylethylamine, C6H5CH(NH2)CH3, as follows. Step 1: benzene -> phenylethanone (C6H5COCH3), using ethanoyl chloride and a catalyst. Step 2: phenylethanone -> 1-phenylethanol, C6H5CH(OH)CH3, using NaBH4. Step 3: 1-phenylethanol -> (1-chloroethyl)benzene, C6H5CHClCH3, using a suitable chlorinating reagent. Step 4: (1-chloroethyl)benzene -> 1-phenylethylamine, using excess ethanolic ammonia.

(a)Name the catalyst used in Step 1, state the type of mechanism, and describe the catalyst's role in generating the electrophile.(3)

(b)State the type of reaction occurring in Step 2, and identify the nucleophile involved in its mechanism.(2)

(c)Name a suitable reagent for Step 3, and state one observation that might be made during this reaction.(2)

(d)Explain why, in Step 4, the ammonia used must be (i) in excess and (ii) dissolved in ethanol rather than water.(2)

(e)Calculate the atom economy of Step 4, given Mr[(1-chloroethyl)benzene] = 140.5, Mr(NH3) = 17.0, Mr(1-phenylethylamine) = 121, and Mr(HCl) = 36.5, if only 1-phenylethylamine is considered the desired product.(3)

(Total for Question 12 is 12 marks)

Mark scheme · AC10 Chemistry: Aromatic Chemistry, Carbonyls and Amines

Question 1

(a) B1 X-ray diffraction shows all six C-C bond lengths in benzene are equal (about 0.139-0.140 nm), a value between a C-C single bond (0.154 nm) and a C=C double bond (0.134 nm) oe
(a) B1 the enthalpy of hydrogenation of benzene is much less exothermic than the value predicted for a Kekule structure with three localised double bonds oe; accept: benzene does not readily decolourise bromine water/undergo the addition reactions typical of alkenes
(a) Answer: Equal C-C bond lengths (X-ray data) and a less exothermic enthalpy of hydrogenation than predicted for three localised double bonds.
(b) B1 each carbon atom is sp2 hybridised and forms σ bonds to two other carbon atoms and one hydrogen atom, all in the same plane oe
(b) B1 each carbon atom has one electron in a p-orbital, perpendicular to the plane of the ring
(b) B1 the p-orbitals overlap sideways (above and below the plane) to form a ring of delocalised π electron density spread over all six carbon atoms
(b) Answer: sp2 carbons each contribute one p-electron; sideways overlap of the six p-orbitals forms a delocalised π system of electron density above and below the ring plane.
(c) M1 predicted enthalpy = 3 x (-120) = -360 kJ/mol
(c) A1 -360 kJ/mol for the hypothetical triene
(c) dM1 stabilisation = predicted - actual = (-360) - (-208), ft their predicted value from (i)
(c) A1 152 kJ/mol (allow awrt 150-152); benzene is 152 kJ/mol more stable (less exothermic hydrogenation) than the hypothetical Kekule triene
(c) Answer: Predicted enthalpy of hydrogenation = -360 kJ/mol; delocalisation (resonance) stabilisation = 152 kJ/mol

Question 2

(a) B1 concentrated nitric acid and concentrated sulfuric acid (as a catalyst) used together
(a) B1 temperature kept at or below about 50 degC (reflux); higher temperatures/prolonged heating cause further substitution to give dinitrobenzene products oe
(a) Answer: Conc. HNO3 + conc. H2SO4 (catalyst), maintained at around 50 degC.
(b) M1 HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4- (state symbols not required)
(b) A1 correctly balanced with NO2+ identified as the electrophile (nitronium ion)
(b) Answer: HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4-
(c) B1 NO2+ is attracted to/attacks the delocalised π electron ring (electrophilic attack)
(c) B1 a pair of electrons from the ring forms a new C-N bond to the nitrogen of NO2+
(c) B1 this generates a positively charged intermediate (arenium/Wheland ion) in which the delocalised ring system now extends over only five carbon atoms (the ring is no longer fully aromatic; positive charge delocalised over the remaining five carbons)
(c) B1 a proton, H+, is lost from the carbon bearing both the H and the NO2 group, which restores the full delocalised aromatic ring, regenerating a catalytic H+ that reforms H2SO4 with HSO4-
(c) Answer: NO2+ attacks the ring; a C-N bond forms, breaking full delocalisation and giving a positively charged arenium intermediate (delocalised over 5 C); loss of H+ restores the aromatic ring, regenerating the H2SO4 catalyst.
(d) B1 the methyl group is electron-donating (positive inductive/hyperconjugative effect), which increases the electron density of the delocalised ring π system
(d) B1 this makes the ring more strongly attract (more strongly polarise) the electrophile NO2+, so the reaction is faster than with benzene
(d) B1 the increase in electron density is greatest at the carbon atoms ortho (2-) and para (4-) to the methyl group oe
(d) B1 so the arenium (Wheland) intermediates formed by attack at the 2- and 4- positions are more stabilised/lower in energy than attack at the 3-position, so 2- and 4- substitution predominate
(d) Answer: The methyl group donates electron density into the ring (activating it, so reaction is faster than benzene) and increases electron density most at the 2- and 4- positions, favouring attack (and forming more stable intermediates) there rather than at the 3-position.

Question 3

(a) B1 one of the lone pairs on the oxygen atom of the -OH group overlaps with (is delocalised into) the ring π system
(a) B1 this increases the electron density of the delocalised ring, particularly at the 2-, 4- and 6- positions
(a) B1 so the ring more strongly polarises/attracts electrophiles such as Br2, so no halogen-carrier catalyst is needed and reaction occurs readily at room temperature
(a) Answer: The oxygen lone pair delocalises into the ring, increasing electron density (especially at 2/4/6 positions), so phenol reacts with electrophiles such as Br2 far more readily than benzene, without a catalyst.
(b) B1 2,4,6-tribromophenol
(b) B1 C6H3Br3O
(b) Answer: 2,4,6-tribromophenol, C6H3Br3O
(c) B1 electrophile is Br+ (a bromine atom carrying a partial/full positive charge)
(c) B1 the electron-rich ring of phenol polarises the Br-Br bond sufficiently on its own (induces a dipole in Br2) for heterolytic fission to occur, so a catalyst (e.g. FeBr3/AlBr3) is not required
(c) Answer: Br+ (from polarisation of Br2 by the electron-rich ring); phenol's ring is electron-rich enough to induce the necessary polarisation of Br-Br itself, without a halogen-carrier catalyst.

Question 4

(a) B1 dissolve the crude solid in a minimum volume of a hot, suitable solvent (one in which the solid is soluble hot but only sparingly soluble cold)
(a) B1 filter the hot solution (e.g. through a fluted/hot filter funnel, or add a little decolourising charcoal and filter) to remove any insoluble impurities
(a) B1 allow the filtrate to cool slowly (e.g. to room temperature, then in ice) so that crystals of compound S form, leaving soluble impurities in solution
(a) B1 filter the crystals under reduced pressure (Buchner filtration), wash with a small volume of ice-cold solvent, and allow the crystals to dry
(a) Answer: Dissolve in minimum hot solvent, hot-filter, cool slowly to crystallise, then filter (Buchner) under reduced pressure, wash with cold solvent, and dry.
(b) B1 dissolving well when hot allows a small (minimum) volume of solvent to be used, and allows the solid to be hot-filtered to remove insoluble impurities
(b) B1 being only sparingly soluble when cold means the compound crystallises out (while soluble impurities, present in low concentration, remain dissolved in the solvent) on cooling, giving a purer, higher-yield crop of crystals
(b) Answer: Good hot solubility allows use of a small volume and hot filtration; poor cold solubility forces the pure compound to crystallise out while soluble impurities stay in solution, improving purity and recovery.
(c) B1 the sample is not completely pure, since its melting point (62-64 degC) is lower than, and does not match, the literature value (65 degC)
(c) B1 impurities lower the melting point (freezing point depression) and broaden the melting point range
(c) B1 a pure sample would be expected to melt sharply over a narrow range (typically within about 1-2 degC) at (or very close to) the literature value
(c) Answer: The sample is impure (melts below and over a wider range than the pure literature value); impurities depress and broaden the melting point range, whereas a pure sample melts sharply, close to 65 degC.
(d) M1 moles of 3-methylphenol = 2.16 / 108.14 = 0.01997 mol (0.0200 mol to 3 sf)
(d) M1 since 3-methylphenol : compound S is a 1:1 ratio, theoretical (maximum) mass of S = 0.01997 x 212.24 = 4.24 g
(d) dM1 percentage yield = (actual mass / theoretical mass) x 100 = (3.18 / 4.24) x 100, ft their theoretical mass
(d) A1 75.0% (accept 74.9-75.1%; awrt 75%)
(d) Answer: Percentage yield = 75.0%
(e) B1 any one: product lost during transfers between apparatus / some product remains dissolved in the mother liquor (recrystallisation solvent) and is not recovered / product lost during filtration or washing / a competing side reaction occurs oe, any sensible practical loss
(e) Answer: For example, some of the product is lost in solution in the recrystallisation mother liquor (or lost during transfers/filtration).
(f) B1 nucleophilic addition-elimination (condensation)
(f) B1 hydrogen chloride, HCl (removed as it forms by reaction with the pyridine present)
(f) Answer: Nucleophilic addition-elimination (condensation); HCl is the other product (mopped up by the pyridine).
(g) B1 hazard: corrosive/lachrymator (irritant, causes eyes to water) and reacts with moisture to release HCl fumes; precaution: carry out the reaction in a fume cupboard and wear eye protection/gloves (either a matching hazard/precaution pair for the mark)
(g) Answer: Benzoyl chloride is corrosive/a lachrymator (irritant); the reaction should be carried out in a fume cupboard, wearing eye protection.
(h) B1 modification: e.g. repeat (a second) recrystallisation of the crystals obtained / use a smaller volume of solvent / add decolourising charcoal during the hot filtration stage, any one sensible suggestion
(h) B1 explanation: a further recrystallisation removes more of the remaining soluble impurities (which stay in solution) at each stage, giving a narrower melting point range closer to the literature value oe, matched to the suggestion given
(h) Answer: For example, carry out a second recrystallisation: each recrystallisation leaves more impurities behind in solution, so repeating it removes more impurity and gives a sharper, higher melting point range closer to 65 degC.

Question 5

(a) B1 NaBH4 provides a hydride ion, H-, which acts as a nucleophile
(a) B1 the carbon atom of the C=O group is delta-positive (polarised, due to oxygen being more electronegative), so it is attacked by the hydride ion
(a) B1 a new C-H bond forms as the hydride ion's electron pair moves to carbon, while the electron pair in the C=O π bond moves onto the oxygen, forming an alkoxide ion intermediate
(a) B1 the alkoxide ion is then protonated (by water or on addition of dilute acid) to give the alcohol product, propan-1-ol
(a) Answer: H- (nucleophile) attacks the delta+ carbonyl carbon; the C=O π electrons move onto O forming an alkoxide intermediate, which is then protonated to give propan-1-ol.
(b) B1 Tollens' reagent (ammoniacal silver nitrate)
(b) B1 warmed (in a water bath)
(b) B1 propanal gives a silver mirror (Ag+ reduced to Ag), while propanone gives no visible change/no silver mirror
(b) Answer: Tollens' reagent, warmed: propanal (an aldehyde) forms a silver mirror; propanone (a ketone) gives no reaction.
(c) B1 the orange precipitate (2,4-dinitrophenylhydrazone derivative) is recrystallised to purify it, and its melting point is measured
(c) B1 the measured melting point is compared with data book (literature) values of melting points of known 2,4-dinitrophenylhydrazone derivatives, to identify which specific aldehyde/ketone was present
(c) Answer: Purify (recrystallise) the derivative, measure its melting point, and match this against data book values for known derivatives to identify the specific carbonyl compound.
(d) M1 atom economy = (Mr of desired product / sum of Mr of all reactants) x 100 = 60 / (58 + 2) x 100
(d) A1 = 100%
(d) Answer: Atom economy = 100%

Question 6

(a) M1 CH3COOH + PCl5 -> CH3COCl + POCl3 + HCl
(a) A1 correctly balanced (all species; state symbols not required)
(a) B1 steamy white fumes (of HCl gas) are observed
(a) Answer: CH3COOH + PCl5 -> CH3COCl + POCl3 + HCl; steamy white fumes of HCl are seen.
(b) B1 in both compounds, water (a nucleophile) attacks the delta-positive carbonyl carbon (nucleophilic addition-elimination)
(b) B1 chlorine is more electronegative than oxygen, so it withdraws electron density more strongly from the carbonyl carbon in the acyl chloride, making that carbon more delta-positive/more strongly attacked by the nucleophile than in an ester
(b) B1 chloride ion, Cl-, is also a much better/more stable leaving group than the alkoxide/RO- group that would have to leave in the corresponding step for an ester, so elimination occurs much more readily for the acyl chloride
(b) Answer: The more electronegative Cl makes the carbonyl carbon of an acyl chloride more strongly delta+ than in an ester, and Cl- is a far better leaving group than RO-, so acyl chlorides react with nucleophiles like water much faster than esters do.
(c) M1 moles of ethyl butanoate = 2.50 / 116 = 0.02155 mol
(c) M1 since the ratio of ester : acid is 1:1, mass of butanoic acid = 0.02155 x 88
(c) A1 = 1.90 g (3 sf)
(c) Answer: Maximum mass of butanoic acid = 1.90 g
(d) B1 ethanol
(d) Answer: Ethanol

Question 7

(a) B1 ethylamine: the ethyl group is electron-donating (positive inductive effect), pushing electron density onto the nitrogen atom
(a) B1 this increases the electron density (availability) of the nitrogen lone pair in ethylamine, making it more able to accept (form a dative bond with) a proton, H+, than the lone pair in ammonia
(a) B1 phenylamine: the nitrogen lone pair overlaps with (is partially delocalised into) the aromatic ring's π system
(a) B1 this reduces the electron density/availability of the lone pair on nitrogen in phenylamine, making it much less able to accept a proton than the lone pair in ammonia
(a) Answer: Ethylamine's alkyl group donates electron density onto N, making its lone pair more available (stronger base than ammonia); phenylamine's N lone pair delocalises into the ring, making it far less available (weaker base than ammonia).
(b) B1 nucleophilic substitution
(b) Answer: Nucleophilic substitution
(c) B1 ethylamine formed is itself a nucleophile and could go on to react further with remaining bromoethane, giving secondary/tertiary amines and eventually a quaternary ammonium salt
(c) B1 using excess ammonia makes it far more likely that a molecule of bromoethane reacts with an ammonia molecule than with an ethylamine molecule, so further substitution is minimised and the primary amine is favoured as the major product
(c) Answer: Ethylamine can itself act as a nucleophile towards unreacted bromoethane; a large excess of ammonia makes it much more likely a bromoethane molecule reacts with ammonia rather than with product amine, minimising further substitution to secondary/tertiary amines or quaternary salts.
(d) B1 nucleophilic substitution
(d) B1 ethanenitrile, CH3CN (accept: a nitrile)
(d) Answer: Nucleophilic substitution; the product is ethanenitrile, CH3CN.
(e) B1 LiAlH4 (lithium tetrahydridoaluminate) in dry ether (or: H2 gas with a Ni catalyst, heat/pressure); either accepted
(e) Answer: LiAlH4 in dry ether (or H2/Ni catalyst).

Question 8

(a) M1 (C2H5)3N + CH3Cl -> (C2H5)3(CH3)N+ Cl-
(a) A1 correctly balanced, with the ionic product shown as a salt (cation and Cl- anion)
(a) B1 dative covalent (coordinate) bond
(a) Answer: (C2H5)3N + CH3Cl -> (C2H5)3(CH3)N+ Cl-; the N-CH3 bond formed is a dative covalent (coordinate) bond.
(b) B1 the molecule has an ionic, charged (hydrophilic) 'head' at the nitrogen, which can interact with/dissolve in water
(b) B1 and a long, non-polar hydrocarbon (hydrophobic) 'tail', which can interact with/dissolve in oils, grease or the surface of fabric fibres, allowing the molecule to bridge between polar and non-polar environments
(b) Answer: It has a charged, hydrophilic 'head' (the N+ centre) that associates with water, and a long, hydrophobic hydrocarbon 'tail' that associates with grease/fabric, so the molecule can bridge water and non-polar surfaces like a surfactant.
(c) B1 Step 1: nitrobenzene is heated under reflux with tin (or another suitable reducing metal) and excess concentrated hydrochloric acid, reducing the nitro group to give a salt of phenylamine (the anilinium/phenylammonium salt)
(c) B1 Step 2: excess concentrated sodium hydroxide solution is then added, to liberate the free phenylamine base (which may be separated by steam distillation)
(c) Answer: Reflux nitrobenzene with tin and excess conc. HCl (reduction, forming the anilinium salt); then add excess NaOH(aq) to release free phenylamine.
(d) B1 hazard: concentrated hydrochloric acid is corrosive (heated under reflux, so hot corrosive vapours/liquid are present); nitrobenzene and phenylamine are both toxic/harmful
(d) B1 precaution: carry out the procedure in a fume cupboard and wear eye protection/gloves, any sensible matching precaution
(d) Answer: Concentrated HCl is corrosive and nitrobenzene/phenylamine are toxic; the reduction should be carried out in a fume cupboard with eye protection and gloves.

Question 9

(a) B1 a zwitterion is a species (molecule) that carries both a positive and a negative charge (on different atoms within the same species), with an overall net charge of zero
(a) B1 at its isoelectric point, glycine exists predominantly as the zwitterion, +H3NCH2COO-
(a) Answer: A zwitterion carries both a positive and a negative charge but has zero overall net charge; at its isoelectric point glycine exists mainly as the zwitterion +H3NCH2COO-.
(b) B1 in strongly acidic solution, there is a high concentration of H+ ions, which protonates the -NH2 (or -COO-) group, giving the fully protonated cation, +H3NCH2COOH
(b) B1 in strongly alkaline solution, there is a high concentration of OH- ions, which removes/accepts a proton from the -COOH (or -NH3+) group, giving the fully deprotonated anion, H2NCH2COO-
(b) B1 the position of these acid-base equilibria shifts depending on the pH (H+ concentration) of the solution, so the dominant species present changes with pH
(b) Answer: In strongly acidic solution, excess H+ protonates the amine/carboxylate group to give the cation +H3NCH2COOH; in strongly alkaline solution, OH- removes a proton from the ammonium/acid group to give the anion H2NCH2COO-.
(c) B1 condensation reaction (with loss of a molecule of water)
(c) B1 peptide bond (amide linkage)
(c) Answer: Condensation reaction (loss of H2O); a peptide (amide) bond is formed.

Question 10

(a) B1 condensation polymerisation
(a) B1 water, H2O, is released (at each ester linkage formed)
(a) B1 the polymer chain contains repeating ester linkages (-C(=O)-O-) formed between the -COOH groups of the diacid and the -OH groups of the diol
(a) Answer: Condensation polymerisation; water is released at each ester linkage formed between the diacid and the diol.
(b) M1 HOOC-(CH2)4-COOH + H2N-(CH2)6-NH2 -> HOOC-(CH2)4-CONH-(CH2)6-NH2 + H2O
(b) A1 structure correctly shows one new amide (-CONH-) linkage formed, with one -COOH and one -NH2 group remaining unreacted at the ends
(b) B1 water, H2O, identified as the by-product
(b) Answer: HOOC-(CH2)4-COOH + H2N-(CH2)6-NH2 -> HOOC-(CH2)4-CONH-(CH2)6-NH2 + H2O
(c) L3 (5-6): A well-developed, balanced evaluation is given, covering both environmental drawbacks and possible benefits/counterpoints, together with at least two distinct, relevant strategies to reduce impact, supported by chemical reasoning (e.g. reference to bond types/structure). The answer is coherent and logically structured throughout.
(c) L2 (3-4): A reasonably clear discussion is given, covering some environmental drawbacks and at least one relevant strategy to reduce impact, with some supporting reasoning. The answer shows some structure but may lack balance or full development.
(c) L1 (1-2): Basic, isolated relevant points are made (e.g. a simple statement that polymers do not biodegrade, or one disposal method named), with little or no supporting reasoning or structure.
(c) L0 (0): No relevant content, or the response does not address the question.
(c) Answer: Awarded via levels of response (0-6) against the indicative content: a full-mark answer balances drawbacks (non-biodegradability, fossil-fuel origin) against benefits (durability, reduced waste elsewhere), and gives at least two well-explained reduction strategies (e.g. mechanical recycling, chemical hydrolysis/re-polymerisation, biodegradable alternatives, energy recovery).

Question 11

(a) M1 moles CO2 = 35.2 / 44 = 0.800 mol, so moles C = 0.800 mol
(a) M1 moles H2O = 9.9 / 18 = 0.550 mol, so moles H = 2 x 0.550 = 1.10 mol
(a) M1 moles N2 = 1.40 / 28 = 0.0500 mol, so moles N = 2 x 0.0500 = 0.100 mol
(a) A1 ratio C : H : N = 0.800 : 1.10 : 0.100 = 8 : 11 : 1, so empirical formula = C8H11N
(a) Answer: Empirical formula = C8H11N
(b) B1 empirical formula mass of C8H11N = (8x12)+(11x1)+14 = 121, which equals the given Mr, so the molecular formula is the same as the empirical formula, C8H11N; ft their empirical formula from (a) if the mass matches
(b) Answer: Molecular formula = C8H11N
(c) B1 an amine (the nitrogen lone pair accepts a proton from HCl to form an ammonium/aminium salt, characteristic of amines)
(c) Answer: Q is an amine.

Question 12

(a) B1 aluminium chloride, AlCl3 (a halogen carrier)
(a) B1 electrophilic substitution (Friedel-Crafts acylation)
(a) B1 AlCl3 accepts a lone pair of electrons from the chlorine atom of ethanoyl chloride (CH3COCl), polarising and breaking the C-Cl bond to generate the acylium ion, CH3CO+, as the electrophile (and [AlCl4]-)
(a) Answer: AlCl3 (halogen carrier catalyst); Friedel-Crafts acylation (electrophilic substitution); AlCl3 accepts a lone pair from Cl in CH3COCl, generating the electrophile CH3CO+ and [AlCl4]-.
(b) B1 reduction (nucleophilic addition)
(b) B1 hydride ion, H- (from NaBH4)
(b) Answer: Reduction, by nucleophilic addition; the nucleophile is the hydride ion, H-.
(c) B1 phosphorus(V) chloride, PCl5 (accept SOCl2 or PCl3)
(c) B1 steamy white fumes (of HCl) are seen (if using PCl5 or SOCl2)
(c) Answer: PCl5 (or SOCl2); steamy white fumes of HCl are observed.
(d) B1 (i) excess ammonia favours reaction of the haloalkane with NH3 rather than with the primary amine product formed, minimising further substitution to secondary/tertiary amines or a quaternary ammonium salt
(d) B1 (ii) an aqueous solvent would allow water (a nucleophile) to compete with ammonia in a hydrolysis side reaction with the haloalkane (substitution by OH-, giving the alcohol back); ethanol dissolves both the organic haloalkane and ammonia without this competing reaction
(d) Answer: (i) Excess NH3 minimises further nucleophilic substitution to secondary/tertiary amines or a quaternary salt. (ii) Ethanol (rather than water) avoids competing hydrolysis of the haloalkane by water and dissolves both reactants.
(e) M1 sum of Mr of reactants = 140.5 + 17.0 = 157.5
(e) M1 atom economy = (Mr desired product / sum Mr reactants) x 100 = (121 / 157.5) x 100
(e) A1 = 76.8% (awrt 76.8-76.9%)
(e) Answer: Atom economy = 76.8%

Chemistry: Aromatic Chemistry, Carbonyls and Amines - Worksheets, Questions and Revision