Chemistry 9701/43 May/June 2019: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Aqueous solutions of copper(II) salts contain the blue-coloured [Cu(H2O)6]2+ complex ion.
Separate portions of this blue solution react with aqueous sodium hydroxide and with concentrated hydrochloric acid.
Give the following information for each of these reactions.
●● reaction with aqueous sodium hydroxide
ionic equation ……
type of reaction ……
colour and state of the copper-containing product ……
●● reaction with concentrated hydrochloric acid
ionic equation ……
type of reaction ……
colour and state of the copper-containing product ……
[6]
(b) Chloride ions can be identified using aqueous silver nitrate, AgNO3(aq).
Ag+(aq) + Cl –(aq) AgCl (s)
0.303 g of a chloride of sulfur is completely hydrolysed with water. All the chlorine atoms present in the chloride of sulfur are converted into chloride ions. The solution is diluted to 100.0 cm3.
A 25.00 cm3 sample of this solution is titrated with 0.0500 mol dm–3 AgNO3(aq). The titration requires 22.40 cm3 of 0.0500 mol dm–3 AgNO3(aq).
Calculate the empirical formula of the chloride of sulfur. Show all your working.
empirical formula of chloride of sulfur = …… [3]
[Total: 9]
Mark scheme
1(a) M1 [Cu(H2O)6]2+ + 2OH– → Cu(OH)2 + 6H2O 6
M2 precipitation
M3 blue precipitate
M4 [Cu(H2O)6]2+ + 4Cl – → CuCl42– + 6H2O
M5 ligand exchange / displacement / substitution / replacement
M6 yellow solution
1(b) M1 amount of Ag+ = 0.050 × 0.0224 = 1.12 × 10–3 mol (in 25 cm3) 3
amount of Ag+ = 1.12 × 10–3 × 4 = 4.48 × 10–3 mol (in 100 cm3)
M2 amount of Cl – = 4.48 × 10–3 mol (in 100 cm3)
mass of Cl – = 4.48 × 10–3 × 35.5 = 0.159 g (in 100 cm3)
mass of S = 0.303 – 0.159 = 0.144 g (in 100 cm3) ecf
M3 moles of S = 0.144 / 32.1 = 4.49 × 10–3
molar ratio S : Cl 1:1 → SCl ecf
(a) Group 2 nitrates decompose when heated.
Write an equation for the decomposition of strontium nitrate.
. …… [1]
(b) Describe and explain how the thermal stability of Group 2 nitrates changes with increasing atomic number.
. ……
. ……
. ……
. ……
. …… [3]
(c) The variation in the thermal stability of Group 2 amides is similar to that of Group 2 nitrates.
(i) Suggest whether calcium amide, Ca(NH2)2, will decompose more or less readily than barium amide, Ba(NH2)2. Explain your answer.
. ……
. …… [1]
(ii) Ba(NH2)2 decomposes when heated to form barium nitride, Ba3N2, and ammonia as the only products.
Write an equation for this reaction.
. …… [1]
(d) Ba(NH2)2 contains the NH2– ion.
Predict the bond angle of NH2–. Explain your answer using the qualitative model of electron‑pair repulsion.
bond angle ……
explanation ……
. ……
. ……
[3]
[Total: 9]
Mark scheme
2(a) Sr(NO3)2 → SrO + 2NO2 + ½O2 1
2(b) M1 increases 3
M2 cationic radius / ion size increases (down the group)
M3 less polarisation/distortion of anion / nitrate ion / NO3– / nitrate group
2(c)(i) more readily and Ca2+ has a smaller ionic radius or more readily and Ca2+ has a greater charge density 1
2(c)(ii) 3Ba(NH2)2 → Ba3N2 + 4NH3 1
2(d) M1 bond angle 104–105° 3
M2 explanation two lone pairs and two bonding pairs
M3 lone pairs repel more
Chlorate(V) ions are powerful oxidising agents.
(a) The reduction of chlorate(V) ions, Cl O3–, with SO2 forms chlorine dioxide, Cl O2, and sulfate ions,
SO42–, as the only products.
Construct an equation for this reaction.
. …… [1]
(b) (i) Chlorine dioxide, Cl O2, disproportionates with hydroxide ions, OH–(aq), to form a mixture of Cl O2– and Cl O3– ions.
2Cl O2 + 2OH– Cl O2– + Cl O3– + H2O
Explain, using this reaction as an example, what is meant by disproportionation.
. ……
. …… [1]
(ii) Deduce the ionic half-equations for the reaction in (b)(i).
. ……
. …… [2]
(c) A lithium-iodine electrochemical cell can be used to generate electricity for a heart pacemaker.
The cell consists of a lithium electrode and an inert electrode immersed in body fluids. When current flows lithium is oxidised and iodine is reduced.
(i) Use the Data Booklet to write half-equations for the reactions taking place at the two electrodes. Hence write the overall equation for when a current flows.
● . ……
● . ……
overall equation ……
[2]
(ii) Use the Data Booklet to calculate the for this cell.
= …… V [1]
(iii) A current of 2.5 × 10–5 A is drawn from this cell.
Calculate the time taken for 0.10 g of lithium electrode to be used up. Assume the current remains constant throughout this period.
time = …… s [3]
[Total: 10]
Mark scheme
3(a) 2ClO3– + SO2 → 2ClO2 + SO42– 1
3(b)(i) Cl in ClO2 gets both oxidised and reduced or Cl goes from +4 → +5 and +4 → +3 1
3(b)(ii) M1 ClO2 + 2OH– → ClO3– + H2O + e– 2
M2 ClO2 + e– → ClO2–
3(c)(i) M1 Li → Li+ + e– and I2 + 2e– → 2I– 2
M2 2Li + I2 → 2Li+ + 2I–
3(c)(ii) Eocell = 0.54 –(–3.04) = +3.58 V [1] 1
3(c)(iii) M1 amount of Li = 0.10 / 6.9 = 1.45 × 10–2 mol [1] 3
M2 Q needed = 96500 × 1.45 × 10–2 = 1399 (1398.55) C [1] ecf
M3 t = 1399 / (2.5 × 10–5) = 5.6 × 107 s [1] ecf 2sf min
(a) Sketches of the shapes of some atomic orbitals are shown.
Identify the type of orbital, s, p, or d.
y z y z y z
shape of orbital x x x
type of orbital
[1]
(b) Cadmium forms the two ions, Cd22+ and Cd2+. The electronic configuration of cadmium in these ions is shown.
●● [Kr] 4d105s1
●● [Kr] 4d10
Use this information to explain why cadmium is not a transition element.
. ……
. …… [1]
(c) Methylamine, CH3NH2, is a monodentate ligand.
(i) State what is meant by the term monodentate in this context.
. ……
. …… [1]
In the presence of aqueous methylamine, [Cd(H2O)6]2+ reacts to form a mixture of two isomeric octahedral complexes.
equilibrium 1 [Cd(H2O)6]2+ + 4CH3NH2 [Cd(CH3NH2)4(H2O)2]2+ + 4H2O = –57 kJ mol–1
(ii) Complete the three-dimensional diagrams to show the isomers of [Cd(CH3NH2)4(H2O)2]2+.
Use L to represent CH3NH2 in your diagrams.
Cd Cd
[2]
(d) (i) State what is meant by the term stability constant.
. ……
. …… [1]
(ii)
Complete the table by placing one tick () in each row to suggest how increasing temperature will affect Kstab and the equilibrium concentration of the cadmium complex,
[[Cd(CH3NH2)4(H2O)2]2+], for equilibrium 1. Explain your answer.
decreases no change increases
Kstab
[[Cd(CH3NH2)4(H2O)2]2+]
explanation ……
. ……
[2]
EDTA4– is a polydentate ligand. When a solution of EDTA4– is added to [Cd(H2O)6]2+ a new complex [CdEDTA]2– is formed.
The values for the stability constants for two Cd2+ complexes are shown.
Kstab
[Cd(CH3NH2)4(H2O)2]2+ 4.0 × 106
[CdEDTA]2– 4.0 × 1016
(iii) A solution containing equal numbers of moles of CH3NH2 and EDTA is added to [Cd(H2O)6]2+.
Predict which complex is formed in the larger amount. Explain your answer.
. ……
. …… [1]
(e) Methylamine is a Brønsted-Lowry base.
Write an equation showing how methylamine dissolves in water to give an alkaline solution.
. …… [1]
(f) Methylamine is a useful reagent in organic chemistry.
(i) Write an equation for the reaction of ethanoyl chloride with methylamine.
. …… [2]
(ii) Methylamine also reacts with propanone to form compound P as shown.
P
H 3C H3C
C O + H 2N CH3 C N + H 2O
H 3C H 3C CH3
Deduce the type of reaction shown here.
. …… [1]
[Total: 13]
Mark scheme
4(a) 1
All shapes
required for
mark
p s d
4(b) both cadmium ions have full d subshells 1
4(c)(i) donates one lone pair to the central metal ion 1
4(c)(ii) M1 one 3D diagram of [Cd(CH3NH2)4(H2O)2]2+ 2
M2 cis and trans structures
4(d)(i) equilibrium constant for the formation of a complex ion in solution / solvent [1] 1
4(d)(ii) 2
decreases no change increases
Kstab 9
[[Cd(CH3NH2)4(H2O)2]2+] 9
M1 both ticks correct [1]
M2 equilibrium moves to the left as the (forward) reaction is exothermic [1]
4(d)(iii) [CdEDTA]2– and larger Kstab value 1
4(e) CH3NH2 + H2O ⇌ CH3NH3+ + OH– 1
4(f)(i) CH3COCl + CH3NH2 → CH3CONHCH3 + HCl 2
M1 Correct formulae of CH3COCl or CH3CONHCH3
M2 rest of the equation
4(f)(ii) condensation or addition-elimination 1
(a) Chlorate(I) ions undergo the following reaction under aqueous conditions.
2NH3 + Cl O– N2H4 + Cl – + H2O
A series of experiments was carried out at different concentrations of Cl O– and NH3.
The table shows the results obtained.
[Cl O–] [NH3] initial rate experiment
/ mol dm–3 / mol dm–3 / mol dm–3 s–1
1 0.200 0.100 0.256
2 0.400 0.200 2.05
3 0.400 0.400 8.20
(i) Use the data in the table to determine the order with respect to each reactant, Cl O– and
NH3.
Show your reasoning.
. ……
. ……
. ……
. ……
. ……
. …… [2]
(ii) Write the rate equation for this reaction.
rate = …… [1]
(iii) Use the results of experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k.
k = ……
units = ……
[2]
(iv) On the axes sketch a graph to show how the value of k changes as temperature is increased.
rate constant, k
temperature
[1]
(b) In another experiment, the reaction between chlorate(I) ions and iodide ions in aqueous alkali was investigated.
A solution of iodide ions in aqueous alkali was added to a large excess of chlorate(I) ions and
[I–] was measured at regular intervals.
(i) Describe how the results of this experiment can be used to confirm that the reaction is first-order with respect to [I–].
. ……
. ……
. …… [2]
A three-step mechanism for this reaction is shown.
step 1 Cl O– + H2O HCl O + OH–
step 2 I– + HCl O HIO + Cl –
step 3 HIO + OH– H2O + IO–
(ii) Use this mechanism to deduce the overall equation for this reaction.
. …… [1]
(iii) Identify a step that involves a redox reaction. Explain your answer.
. ……
. …… [1]
[Total: 10]
Mark scheme
5(a)(i) M1: using expt 2 and 3, [NH3] × 2, rate × 4 so order with respect to [NH3] = 2 2
M2: using expt 1 and 2, [ClO–] × 2 and [NH3] × 2, as rate × 8 (=22 * x) so order with respect to [ClO–] = 1
5(a)(ii) rate = k[NH3]2[ClO–] 1
5(a)(iii) M1: k = 0.256 / (0.200 × 0.1002) k = 128 2
M2: Units dm6 mol–2 s–1
5(a)(iv) curve / line showing k increasing as temperature increases 1
5(b)(i) M1: plot a graph of [I–] against time 2
M2: constant half-lives
5(b)(ii) ClO– + I– → IO– + Cl – 1
5(b)(iii) step 2 and Cl is reduced / oxid no. decreases / oxid no. +1 → –1 1
or
step 2 and I is oxidised / oxid no. increases / oxid no. –1 → +1
(a) Complete the table by placing one tick () in each row to indicate the sign of each type of energy change under standard conditions.
always always either negative energy change positive negative or positive
bond energy
enthalpy change of formation
[1]
(b) Explain what is meant by the term enthalpy change of atomisation.
. ……
. …… [1]
(c) The overall reaction for the atomisation of liquid bromine molecules, Br2(l), is shown.
Br2(l) 2Br(g)
This happens via a two-step process.
●● Construct a labelled energy cycle to represent this atomisation process, including state symbols.
●● Use your cycle and relevant data from the Data Booklet to calculate the enthalpy change of vaporisation of Br2(l), .
The enthalpy change of atomisation of bromine, ∆Hat, = +112 kJ mol–1.
= …… kJ mol–1 [3]
(d) Suggest how the of iodine, I2(l), would compare to that of bromine, Br2(l). Explain your answer.
. ……
. …… [1]
(e) (i) Explain what is meant by the term enthalpy change of hydration.
. ……
. …… [1]
(ii) Suggest why the enthalpy change of hydration of Br –(g) is more exothermic than that of
I–(g).
. ……
. ……
. …… [2]
[Total: 9]
Mark scheme
6(a) 1
always always either negative
energy change
positive negative or positive
bond energy 9
enthalpy of formation 9
both ticks correct
6(b) (energy change) when 1 mole of gaseous atoms are formed (from an element in its standard state) 1
6(c) 3
M1: correct cycle: formulae and state symbols
M2: use of 1 × 193 and 2 × (112)
M3: for the correct sum and answer ecf from M2
∆Hovap (= (2 × 112) – (193) ) = +31 kJ mol–1 [scores M2 and M3]
6(d) more endothermic and greater Van der Waals / London / induced dipole-dipole forces both 1
6(e)(i) (energy change) when 1 mole of gaseous ions is dissolved in (an excess of) water 1
6(e)(ii) M1: Br- has a smaller ionic radii 2
M2: stronger (ion-dipole) attractions with water molecules
(a) Benzene can be converted into cyclohexane.
(i) For this reaction name the type of reaction and identify the reagent and conditions needed.
type of reaction ……
reagent and conditions ……
[2]
(ii) State the bond angles in benzene and cyclohexane.
bond angle in benzene …… bond angle in cyclohexane ……
Explain your answers.
. ……
. ……
. ……
[2]
(b) When benzene reacts with SO3, benzenesulfonic acid is produced.
benzenesulfonic acid
+ SO3
SO3H
The mechanism of this reaction is similar to that of the nitration of benzene. Concentrated
H2SO4 is used in an initial step to generate the SO3H+ electrophile as shown.
SO3 + H2SO4 SO3H+ + HSO4–
(i) Draw a mechanism for the reaction of benzene with SO3H+ ions. Include all necessary curly arrows and charges.
benzenesulfonic acid
SO3H+
SO3H
[3]
(ii) Write an equation to show how the H2SO4 catalyst is reformed.
. …… [1]
(c) 3-dodecylbenzenesulfonic acid can be prepared from benzenesulfonic acid.
benzenesulfonic acid 3-dodecylbenzenesulfonic acid
SO3H SO3H
C12H25
Suggest the reagents and conditions and name the mechanism for this reaction.
reagents and conditions ……
mechanism ……
[2]
(d) When concentrated sulfuric acid is added to water, dissociation takes place in two stages.
stage 1 H2SO4 H+ + HSO4–
stage 2 HSO4– H+ + SO42– Ka2 = 1.0 × 10–2 mol dm–3
Ka2 is the acid dissociation constant for stage 2.
(i) Write the expression for the acid dissociation constant Ka2.
Ka2 =
[1]
(ii) H2SO4 is considered a strong acid whereas HSO4– is considered a weak acid.
Suggest how the magnitude of the acid dissociation constant for stage 1 compares to Ka2.
. …… [1]
(e) Benzoic acid, C6H5CO2H, is a weak acid. A solution of 0.0250 mol dm–3 benzoic acid has a pH of 2.90.
Calculate the Ka of benzoic acid.
Ka = …… mol dm–3
[2]
[Total: 14]
Mark scheme
7(a)(i) M1: reduction / hydrogenation 2
M2: H2 + Ni / Pt catalyst
7(a)(ii) M1: benzene (120°) and cyclohexane (109.5°) 2
M2: as π-bonds are transformed into σ-bonds
7(b)(i) 3
M1: first curly arrow to the sulfur atom
M2: intermediate shown
M3: 2nd curly arrow and H+ formed / lost
7(b)(ii) HSO4– + H+ → H2SO4 1
7(c) M1: C12H25Br and halogen carrier e.g. Al Br3 (+ heat) 2
M2: electrophilic substitution
7(d)(i) [H+ ][SO42− ] 1
K a2 =
[HSO4 − ]
7(d)(ii) Ka of H2SO4 is larger than Ka2 1
7(e) M1: [H+] = 10–2.90 = 1.26 × 10–3 2
M2: Ka = [1.26 × 10–3]2 / 0.025 = 6.3 × 10–5 (mol dm–3)
(a) The mass spectrum of compound X, C5H10O2, is recorded.
The peak heights of the M and M+1 peaks are 22.65 and 1.25 respectively.
(i) Use these data to show that there are five carbon atoms present in one molecule of X.
Show your working.
[1]
(ii) The mass spectrum has a peak at m/e = 57.
Complete the equation to show the fragmentation of X to produce this peak.
[C5H10O2]+ …… + ……
[2]
(b) State the use of TMS and CDCl 3 in NMR spectroscopy.
● TMS ……
● CDCl 3 ……
[1]
(c) The proton NMR spectrum of compound X, C5H10O2, is shown.
3
2
3
2
5 4 3 2 1 0 δ / ppm
(i) By considering both the relative peak areas and their δ values, use the Data Booklet to
● deduce the part of the molecule that produces the peak at δ 2.2,
. ……
● deduce the part of the molecule that produces the peaks at δ 1.2 and δ 3.5,
. ……
● deduce the part of the molecule that produces the peak at δ 4.0.
. ……
[3]
(ii) When reacted with aqueous alkaline iodine, X produces a yellow precipitate.
Use this information and your answers to (c)(i) to suggest a structure for X.
[1]
(d) Compound W is an ester with the molecular formula C5H10O2.
The proton NMR spectrum of W contains only two peaks.
The relative areas of these two peaks are in the ratio 9 : 1.
Suggest a structure for this ester, W.
[1]
(e) Compound V is a carboxylic acid which contains a chiral centre. It also has the molecular formula C5H10O2.
(i) Explain what is meant by the term chiral centre.
. ……
. …… [1]
(ii) Suggest a structure for V.
[1]
[Total: 11]
Mark scheme
8(a)(i) no. of carbons = 100 × 1.25 / (22.65 × 1.1) (= 5.02) 1
8(a)(ii) M1: C2H5O 2
M2: C3H5O+ (positive sign required for m / e = 57 fragment)
8(b) TMS: Reference 1
CDCl3: Solvent
8(c)(i) M1: CH3CO 3
M2: CH3CH2O
M3: (CO)CH2O
8(c)(ii) CH3COCH2OCH2CH3 1
8(d) HCO2C(CH3)3 1
8(e)(i) this is a (carbon) atom which has four different atoms or groups attached to it 1
8(e)(ii) CH3CH2CH(CH3)COOH 1
(a) Organochlorine compounds can undergo hydrolysis.
R–Cl + H2O R–OH + HCl
State and explain the relative rates of hydrolysis of the following compounds.
CH3CH2Cl CH3COCl C6H5Cl
. ……
. ……
. ……
. ……
. …… [3]
(b)
Epibatidine is a naturally occurring organochlorine compound.
epibatidine
Cl
H
N N
(i) Epibatidine is a weak base.
State what is meant by the term weak base.
. ……
. …… [1]
A molecule of epibatidine contains two nitrogen atoms, both of which can act as a base.
(ii) Epibatidine reacts with HCl (aq).
Complete the structure to suggest the product formed in this reaction.
Cl
H
N N
[1]
(c) Polyamides, such as nylon-6, can be prepared from a monomer that contains both an amine and an acyl chloride functional group.
nylon-6 monomer
O
H 2N (CH2)5 C
Cl
(i) When the nylon-6 monomer is hydrolysed, bonds are broken and formed.
By considering the two steps in the mechanism of the reaction, complete the table by placing one tick () in each row to indicate the types of bonds broken and formed during the mechanism.
σ bonds only π bonds only both σ and π bonds bonds broken bonds formed
[1]
(ii) Draw two repeat units of nylon-6. The amide bond should be shown fully displayed.
[2]
(d) An addition polymer made from two different alkene monomers is called a co-polymer. A section of a polyalkene co-polymer is shown.
CH3 H C2H5 CH3 CH3 H
C C C C C C
Cl H H H Cl H
Draw the structure of the two alkene monomers which produce this co-polymer.
[2]
(e) Explain why polyamides normally biodegrade more readily than polyalkenes.
. ……
. …… [1]
(f) The alkene phenylethene can be prepared from benzene in three steps.
O
step 1
step 2
H phenylethene
step 3
(i) Deduce the identity of compound H and draw its structure in the box. [1]
(ii) Suggest reagents and conditions for each of the steps 1–3.
step 1 . ……
step 2 . ……
step 3 . ……
[3]
[Total: 15]
Mark scheme
9(a) M1: CH3COCl > CH3CH2Cl > C6H5Cl 3
M2 & M3 any two from:
• in C6H5Cl (no hydrolysis) C-Cl bond is part of delocalised system
OR p-orbital on Cl overlaps with π system OR electrons from Cl overlap with π system
• CH3COCl carbon in C-Cl bond is more electron deficient since it is also attached to an oxygen atom (ora)
or C-Cl bond strength is weakest in CH3COCl (ora)
• CH3CH2Cl carbon in C-Cl bond strengthened by positive inductive effect of alkyl group
9(b)(i) partially ionised and proton acceptor 1
9(b)(ii) 1
or
9(c)(i) 1
σ-bonds only π-bonds only both σ- and π-bonds
bonds broken 9
bonds formed 9
Both ticks correct
9(c)(ii) 2
M1: amide link
M2: rest of the structure
9(d) 2
or CH3CCl=CH2 and C2H5CH=CHCH3 each correct structure scores one mark
9(e) C-C bonds are non-polar / polyalkenes cannot be hydrolysed and polyamides can be broken down by hydrolysis 1
9(f)(i) 1
9(f)(ii) M1: step 1: CH3COCl + AlCl3 [1] 3
M2: step 2: NaBH4 / LiAlH4 [1]
M3: step 3: conc. H2SO4, heat [1]
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