MODULE

5

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements

5.1 Rates, equilibrium and pH 5.1.1 Orders, rate equations and rate constants (page 13) 1

(a)

(b)

R: zero order S: 2nd order T: 1st order rate = k[S]2[T]

2

(a) (b) (c)

Rate quadruples Rate triples Rate increases by 27 times

3

k = 8.3 × 104 dm3 mol−1 s−1

5.1.2 Concentration–time graphs (page 15) 1

2

(a)

The graph plotted should appear as follows.

(b) (c) (d)

The graph is curved and would suggest the decomposition of H2O2 is first order. The concentration halves every 27 seconds, meaning t1/2 = 27 s. Because the half-life is constant, this confirms the decomposition of H2O2 is first order. 0.20 mol dm−3

(a)

1520 s

(b)

k

(c)

The concentration decreases rapidly, so this drug would need to be administered as a tablet with a coating that is slow to dissolve, so it reaches the intended location in the body before it begins to break down. Accept any other suitable answer; for instance, the drug could also be delivered intravenously.

1.82 × 10−3 s−1

5.1.3 Rate–concentration graphs (page 18) 1

(a) (b)

2

(a)

P: zero order Q: 2nd order rate = k[Q]2 R: 2nd order S: 1st order

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1

MODULE

5 (b)

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements Reactant R

Reactant S

3

(a)

The graph shows that the order with respect to X is first order. Because the order with respect to Y is zero, the rate equation would therefore be: rate = k[X]

(b)

The gradient of the graph is equal to k, so gradient =

(c)

k







.

0.02

s

5.1.4 Rate-determining step (page 20) 1

rate = k[X]2

2

(a) (b) (c)

The slowest step in the reaction mechanism of a multi-step reaction. The rate-determining step involves two molecules of NO. 2NO(g) → N2O2(g) N2O2(g) + O2(g) → NO2(g)

5.1.5 The effect of temperature on rate constants (page 22) 1

The first step is to convert the temperature to K and then calculate and lnk. Temperature/°C

Temperature/K

1/T (K−1)

Rate constant, k/dm3 mol−1 s−1

ln k

10 20 30 40 45

283 293 303 313 318

3.53 × 10−3 3.41 × 10−3 3.30 × 10−3 3.19 × 10−3 3.14 × 10−3

2.49 × 10−4 7.10 × 10−4 1.84 × 10−3 4.75 × 10−3 7.45 × 10−3

−8.29 −7.25 −6.30 −5.35 −4.90

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2

MODULE

5

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements

The second step is to plot the Arrhenius plot, which should appear as follows:

2

Using the points shown (any two points can be chosen) on the graph to calculate the gradient: . gradient = 8636.36 K

.

Therefore, 8636.36 K So: Ea = −R × −8636.36 K = −8.3145 J K−1 mol−1 × −8636.36 K = 71807.02 J mol−1 = 71.8 kJ mol−1 3

If ln A = 22.4 then A is equal to e22.4. Therefore A = 5.35 × 109

5.1.6 Equilibrium (page 25) 1

2

(a) (b)

(c)

2.00 − 0.43 = 1.57 mol [CH3COOH]: 0.43/V mol dm−3 [C2H5OH]: 1.43/V mol dm−3 [CH3COOC2H5]: 1.57/V mol dm−3 [H2O]: 1.57/V mol dm−3 Kc = 4.01

(a) (b)

Heterogeneous. There are solid and gaseous substances involved in the equilibrium. Kc = [CO2]

5.1.7 Equilibrium and Kp (page 27) 1

Kp =

2

(a)

.

Mole fraction PCl5 Mole fraction Cl2 Mole fraction PCl3

. .

.

.

.

4.0

10

0.81 . .

0.19

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3

MODULE

5

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements

(b)

Partial pressure = mole fraction × total pressure P PCl5 = 4.00 × 10−3 × 0.6 atm = 2.4 × 10−3 atm P Cl2 = 0.81 × 0.6 atm = 0.49 atm P PCl3 = 0.19 × 0.6 atm = 0.11 atm

(c)

Kp

(d)

Kp

.

.

22.46

.

5.1.8 Equilibrium constants and their significance (page 30) 1

Reaction A is exothermic; Kc decreases with increasing temperature. Reaction B is endothermic; Kc increases with increasing temperature.

2

(a)

Kc Kp

(b)

(c)

3





An increase in concentration of NO(g) increases the term on the bottom of the expression for both Kc and Kp. The equilibrium moves to restore Kc or Kp by increasing the top and decreasing the bottom – so the equilibrium moves from left to right. There are more concentration or partial pressure terms on the bottom of the expression. A decrease in pressure will decrease the term on the bottom of the expression by more than the top. The equilibrium moves to restore Kc or Kp by increasing the bottom and decreasing the top – so the equilibrium moves from right to left.

A change in pressure changes the top of the Kc or Kp expression by the same as the bottom of the expression. The relative concentrations still give the correct Kc or Kp value, so the system is still at equilibrium and does not move.

5.1.9 Brønsted–Lowry acids and bases (page 32) HNO3(aq) → H+(aq) + NO3−(aq) H2CrO4(aq) → H+(aq) + HCrO4−(aq) HCrO4−(aq) → H+(aq) + CrO42−(aq)

1

(a) (b)

2

HNO3 is monobasic as it can release one proton, H2CrO4 is dibasic as it can release two protons.

3

Ammonia is not acting as a Brønsted–Lowry base as it is not accepting a proton in this reaction. This can be seen as it does not have any extra H attached to it in the final product.

4

(a) (b) (c)

acid 1: HIO3 acid 1: CH3COOH acid 1: H2O

base 1: IO3− base 1: CH3COO− base 1: OH−

acid 2: H3O+ acid 2: H3O+ acid 2: NH4+

base 2: H2O base 2: H2O base 2: NH3

5.1.10 Acid–base reactions and Ka (page 35) 1

(a) (b) (c) (d) (e)

full: H2SO4(aq) + MgCO3(s) → MgSO4(aq) + CO2(g) + H2O(l) ionic: 2H+(aq) + MgCO3(s) → Mg2+(aq) + CO2(g) + H2O(l) full: H2SO4(aq) + K2CO3(aq) → K2SO4(aq) + CO2(g) + H2O(l) ionic: 2H+(aq) + CO32−(aq) → CO2(g) + H2O(l) full: 2HCl(aq) + CaO(s) → CaCl2(aq) + H2O(l) ionic: 2H+(aq) + CaO(s) → Ca2+(aq) + H2O(l) full: HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l) ionic: H+(aq) + OH−(aq) → H2O(l) full: 6HCl(aq) + 2Al(s) → 2AlCl3(aq) + 3H2(g) ionic: 6H+(aq) + 2Al(s) → 2Al3+(aq) + 3H2(g)

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4

MODULE 2

5

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements

(a)

Ka =

(b)

Ka =

3

(a) (b) (c)

pKa = 0.64 pKa = 2.60 pKa = 10.32

4

(a) (b) (c)

Ka = 1.26 × 10−3 mol dm−3 Ka = 6.31 × 10−8 mol dm−3 Ka = 2.51 × 10−11 mol dm−3



5.1.11 Calculating pH of strong and weak acids (page 38) 1

(a) (b) (c) (d)

2.48 3.33 11.62 −0.36

2

(a) (b) (c) (d)

2.95 × 10−7 mol dm−3 1.35 × 10−3 mol dm−3 2.63 × 10−10 mol dm−3 1.32 mol dm−3

3

(a) (b)

2.48 4.39

4

(a) (b)

Ka = 7.02 × 10−7 mol dm−3; pKa = 6.15 Ka = 5.34 × 10−10 mol dm−3; pKa = 9.27

5.1.12 The ionisation of water and Kw (page 41) 1

(a) (b) (c)

[OH−(aq)] = 10−8 mol dm−3 [OH−(aq)] = 10−12 mol dm−3 [OH−(aq)] = 10−3 mol dm−3

2

(a) (b) (c)

[H+(aq)] = 10−1 mol dm−3 [H+(aq)] = 10−10 mol dm−3 [H+(aq)] = 10−4 mol dm−3

3

(a) (b)

pH = 11.70 pH = 12.55

4

(a) (b)

2.69 × 10−2 mol dm−3 6.61 × 10−1 mol dm−3

5.1.13 Buffers (page 44) 1

The buffer equilibrium system is: CH3CH2COOH(aq) ⇌ H+(aq) + CH3CH2COO−(aq) Added acid, H+(aq), reacts with the conjugate base, CH3CH2COO−(aq): H+(aq) + CH3CH2COO−(aq) → CH3CH2COOH(aq) Added alkali, OH−(aq), reacts with the H+(aq): H+(aq) + OH−(aq) → H2O(l) H+(aq) ions are replaced by the equilibrium moving to the right.

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5

MODULE 2

5

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements

[H+(aq)] = Ka ×

.

. .

3.7 × 10−4 mol dm−3

pH = −log [H+(aq)] = −log (3.7 × 10−4) = 3.43 3

An increase in H+(aq) ions in the blood is removed by HCO3−(aq), forming carbonic acid: H+(aq) + HCO3−(aq) → H2CO3(aq) The carbonic acid is converted into aqueous carbon dioxide by an enzyme. In the lungs the dissolved carbon dioxide is converted into carbon dioxide gas, which is then exhaled.

5.1.14 Neutralisation – titration curves (page 47) 1

The end point relates to the action of the indicator during the titration. The end point occurs when there are equal concentrations of the weak acid and its conjugate base. The end point is accompanied by a visible colour change, half way between the colour of the weak acid and the colour of the conjugate base. The equivalence point relates to the actual reaction between the acid and the base. It is the point in a titration at which the volume of one solution has reacted exactly with the volume of the second solution. The equivalence point can have a pH value that is different to the end point of an indicator.

2

Bromocresol green: strong acid–strong base and strong acid–weak base. Thymolphthalein: strong acid–strong base and weak acid–strong base.

Practice questions (page 50) 1

B

2

C

3

A

4

D

5

Refer to official mark scheme for OCR past paper F325, June 2010; this is Q3. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

6

Refer to official mark scheme for OCR past paper F325, June 2012; this is Q2. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

7

Refer to official mark scheme for OCR past paper F325, June 2012; this is Q3. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

5.2 Energy 5.2.1 Lattice enthalpy (page 57) 1

(a) (b) (c)

Li(g) → Li+(g) + e− Ca(s) + Cl2(g) → CaCl2(s) Ca2+(g) + 2Cl−(g) → CaCl2(s)

2

(a)

(i) Mg(s) + Br2(l) → MgBr2(s) (ii) Mg2+(g) + 2Br−(g) → MgBr2(s) enthalpy change of formation; lattice enthalpy

(b) 3

Energy is required to break the bond between the two chlorine atoms in a chlorine molecule.

4

Lattice enthalpies cannot be measured directly because it is impossible to form one mole of an ionic lattice from gaseous ions experimentally.

5.2.2 Born–Haber cycle calculations (page 59) 1

∆LEH = − 435 − 81 − 121 − 403 + 346 = −694 kJ mol−1

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6

MODULE

5

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements

5.2.3 Further Born–Haber cycle calculations (page 61) 1

(a)

(b)

−642 = +150 + 242 + 736 + 1450 + (−2 × ∆EAH ) + (−2493) ∆EAH = −363.5 kJ mol−1

5.2.4 Enthalpy change of solution and hydration (page 65) 1

As the ionic radius increases, the attraction between the ions decreases. This leads to the lattice enthalpy becoming less negative and the ionic bonds becoming weaker. As ionic charge increases, attraction between ions increases. Therefore the lattice enthalpy becomes more negative and the ionic bonds become stronger.

2

Small ions have a greater attraction for water molecules than large ions so more energy is released when they become hydrated. Na+ is smaller than Rb+ (it is higher up in group 1, has fewer occupied shells and hence has a smaller radius), so the hydration enthalpy of Na+ ions will be more exothermic than that of Rb+ ions.

3

(a)

Lattice enthalpy; enthalpy change of hydration of sodium ions; enthalpy change of solution of sodium fluoride.

(b) (c)

F−(g) F−(aq) ∆hydH = −457 kJ mol−1

5.2.5 Entropy (page 68) 1

(a) (b) (c)

positive negative negative

2

(a) (b)

−323 J K−1 mol−1 −217 J K−1 mol−1

5.2.6 Free energy (page 70) 1

−806 kJ mol−1

2

347 °C

5.2.7 Redox (page 72) 1

(a) (b)

2Al + 3Cu2+ → 2Al3+ + 3Cu 10Br− + 2MnO4− + 16H+ → 5Br2 + 2Mn2+ + 8H2O

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7

MODULE 2

5 (a) (b)

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements 2HBr + H2SO4 → Br2 + SO2 + 2H2O 3H2S + 2HNO3 → 3S + 2NO + 4H2O

5.2.8 Redox titrations (page 76) 1

Amount of Fe2+ = 3.75 × 10−3 mol Amount of MnO4− = 7.50 × 10−4 mol Concentration of MnO4− = 2.31 × 10−2 mol dm−3

2

Amount of MnO4− = 4.46 × 10−4 mol Amount of Fe2+ = 2.23 × 10−3 mol Amount of Fe2+ in 250 cm3 = 2.23 × 10−2 mol Mass of Fe = 1.244 g % Fe = 49.8%

5.2.9 Standard electrode potentials (page 79) 1

(a) (b)

2

It consists of 1 mol dm−3 hydrochloric acid as the source of H+(aq), hydrogen gas, H2(g), at 100 kPa pressure, and an inert platinum electrode. (i) Along a wire (ii) Through a salt bridge

The copper half cell would have been set up so that copper metal was in contact with its ions: a strip of copper would have been placed in a solution such as CuSO4(aq), at a concentration of 1 mol dm−3 and a temperature of 298 K. The copper metal would have been the electrode and would have been connected by wire to a voltmeter and then to the hydrogen half cell, via a platinum electrode. The standard hydrogen half cell would consist of 1 mol dm−3 hydrochloric acid and hydrogen gas at 100 kPa. The two half cells would be connected via a salt bridge. This would consist of a piece of filter paper soaked in an aqueous solution of an ionic substance such as KNO3.

5.2.10 Standard cell potentials (page 83) 1

(a)

(b)

2

(a)

(b)

(i) 1.02 V (ii) 0.30 V (iii) 2.10 V Cu(s) + Cl2(g) → Cu2+(aq) + 2Cl−(aq) 3Fe2+(aq) + 2Cr(s) → 3Fe(s) + 2Cr3+(aq) 2Cr(s) + 3Cl2(g) → 2Cr3+(aq) + 6Cl−(aq) Ni(s) with Fe3+(aq); Ni(s) with Br2(aq); Ni(s) with O2(g) and H+(aq) Fe2+(aq) with Br2(aq); Fe2+(aq) with O2(g) and H+(aq) Br−(aq) with O2(g) and H+(aq) Ni(s) + 2Fe3+(aq) → Ni2+(aq) + 2Fe2+(aq) Ni(s) + Br2(aq) → Ni2+(aq) + 2Br−(aq) 2Ni(s) + O2(g) + 4H+(aq) → 2Ni2+(aq) + 2H2O(l) 2Fe2+(aq) + Br2(aq) → 2Fe3+(aq) + 2Br−(aq) 4Fe2+(aq) + O2(g) + 4H+(aq) → 4Fe3+(aq) + 2H2O(l) 4Br−(aq) + O2(g) + 4H+(aq) → 2Br2(aq) + 2H2O(l)

Practice questions (page 86) 1

D

2

A

3

B

4

C

5

Refer to official mark scheme for OCR past paper F325, January 2011; this is Q4. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

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8

MODULE

5

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements

6

Refer to official mark scheme for OCR past paper F325, June 2010; this is Q1. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

7

(a)

(b)

∆G is the Gibbs’ free energy change (measured in kJ mol−1) and predicts the feasibility of the reaction [1] ∆H is the enthalpy change for the reaction (in kJ mol−1) [1] ∆S is the entropy change for the reaction (measured in J mol−1 K−1) [1] T is temperature in K Both units and definition needed for each mark. (i) Endothermic reactions have a +∆H value [1]. For an endothermic reaction to be feasible ∆G must be negative [1]. This means that if ∆S is positive and T is sufficiently large then T∆S will have a larger magnitude than ∆H and thus at high T an endothermic reaction will become feasible. e.g. CaCO3 → CaO + CO2 ∆H = +178.4 kJ mol−1 But ∆S for reaction is +160.3 J mol−1 K−1 (0.1603 kJ mol−1 K−1) ∆

(c)

.

Reaction becomes feasible when ∆G = 0, or when T . Input values = 1113 K ∆ . (or 840 °C) [2 marks for worked example] (ii) If ∆H is positive [1] and ∆S is negative [1] then no value of T will ever allow ∆G to become negative. e.g. 2CO → 2C + O2 The enthalpy change for two moles of CO is + 221 kJ mol−1 and the entropy change for this reaction is −292.7 J mol−1 K−1. Thus no matter what the value of T, ∆G can never be negative [2 marks for worked example]. The ∆H value for N2 + O2 → 2NO is −180.4 kJ mol−1 for 2 moles of NO formed. Thus the reactants are thermodynamically unstable with respect to the products [1]. However, because the activation energy for this reaction is very high [1] the rate of this reaction is effectively zero and thus the reaction mixture is kinetically stable [1].

5.3 Transition metals 5.3.1 Transition metals (page 92) 1

A d-block element is one in which electrons are filling d-orbitals and the highest energy sub-shell is a d-subshell. A transition element is a d-block element which forms at least one ion with an incomplete d-sub-shell.

2

(a) (b) (c) (d)

1s22s22p63s23p64s13d5 1s22s22p63s23p63d5 1s22s22p63s23p6 1s22s22p63s23p64s23d2

5.3.2 Transition metal compounds (page 96) 1

They form coloured compounds; have variable oxidation states; can be used as catalysts.

2

Colourless because the Zn2+ ion does not have a partially filled d-orbital.

3

+5 and +4

4

A catalyst increases the rate of a chemical reaction without undergoing any permanent change itself. It does this by providing an alternative route of lower activation energy.

5

e.g. The Haber process uses iron as a catalyst; N2(g) + 3H2(g) ⇌ 2NH3(g)

5.3.3 Transition metals and complex ions (page 99) 1

(a) (b)

[CoCl4]2− [Fe(H2O)5Cl]2+

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9

MODULE

5

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements

2

A water molecule has a lone pair of electrons which can be donated to the central metal ion. The copper accepts these electrons and a coordinate bond is formed. A ligand is an electron-pair donor to a metal ion.

3

The coordination number is the total number of coordinate bonds formed between the central metal ion and any ligands. In the complex ion [Co(H2O)6]2+ there are six coordinate bonds, one from each of the water ligands surrounding Co2+ – the coordination number is 6.

5.3.4 Stereoisomerism in complex ions (page 103) 1

cis–trans

2

Stereoisomers are molecules or complexes with the same structural formula but a different spatial arrangement of these atoms.

3

5.3.5 Ligand substitution in complexes (page 105) 1

This is a reaction in which a ligand in a complex ion is replaced by another ligand.

2

(a) (b)

Cu(OH)2 precipitates Cu2+(aq) + 2OH−(aq) → Cu(OH)2(s) [Cu(NH3)4(H2O)2]2+

5.3.6 Ligand substitution and precipitation reactions (page 108) 1

Four (one for each of the 4 haem groups).

2

Fe(OH)3; +3

3

Ni2+(aq) + 2OH−(aq) → Ni(OH)2(s)

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10

MODULE

5

OCR A level Chemistry – Answers to Student Book 2 questions

Physical chemistry and transition elements

5.3.7 Redox reactions (page 113) 1

(a) (b)

A reaction in which one species is oxidised at the same time as another species is reduced. +7

2

Amount of Fe2+ = 3.75 × 10−3 mol Amount of MnO4− = 7.50 × 10−4 mol Concentration of MnO4− = 2.31 × 10−2 mol dm−3

3

Amount of MnO4− = 4.46 × 10−4 mol Amount of Fe2+ = 2.23 × 10−3 mol Amount of Fe2+ in 250 cm3 = 2.23 × 10−2 mol Mass of Fe = 1.244 g; %Fe = 49.8%

4

900 cm3

5.3.8 Testing for ions (page 115) 1

Na2CO3(aq) + 2HNO3(aq) → H2O(aq) + CO2(g) + 2NaNO3(aq)

2

(a)

(b)

3

Add an aqueous solution of silver nitrate:  If a white precipitate of silver chloride is produced it must be potassium chloride.  If a yellow precipitate of silver iodide is produced it must be potassium iodide.  If the colour is hard to distinguish, first add dilute ammonia, then concentrated ammonia.  Potassium chloride is soluble in dilute ammonia.  Potassium iodide is insoluble in both dilute ammonia and concentrated ammonia Add sodium hydroxide solution to both compounds and warm gently. One of the compounds will produce a distinctive smelling gas that turns damp red litmus paper blue: this will tell you that the sample contained ammonium ions and must therefore be ammonium hydroxide. The other sample is potassium hydroxide.

Add excess sodium hydroxide to both samples. The sample containing chromium(III) ions will re-dissolve while the sample containing iron(II) ions will remain insoluble.

Practice questions (page 118) 1

B

2

C

3

B

4

Refer to official mark scheme for OCR past paper F325, February 2012; this is Q3. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

5

Refer to official mark scheme for OCR past paper F325, February 2012; this is Q5. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

6

Refer to official mark scheme for OCR past paper F325, June 2010; this is Q6. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

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11

MODULE

6

OCR A level Chemistry – Answers to Student Book 2 questions

Organic chemistry and analysis

6.1 Aromatic compounds, carbonyls and acids 6.1.1 Benzene and its structure (page 123) 1

He knew the empirical and molecular formula of benzene. He then observed the products of the reactions of benzene and noticed a pattern in the types of structural isomers made. Using this information, he generated a structure that matched his observations.

2

X-ray diffraction techniques showed that the Kekulé structure did not work as the bond lengths were all the same.

6.1.2 Naming aromatic compounds (page 125) 1

(a) (b)

nitrobenzene bromobenzene

2

chlorobenzene:

3

There is one isomer of ethylbenzene.

4

There are three isomers of chloroethylbenzene.

6.1.3 Electrophilic substitution (page 127) 1

Heated at 50 °C, concentrated sulfuric acid

2

Benzene, chlorine and iron, or iron(III) chloride or aluminium chloride

3

6.1.4 Halogenation and Friedel–Crafts (page 129) 1

No observable change as benzene is not able to induce a dipole in the non-polar bromine molecule and therefore electrophilic substitution does not occur.

2

It was possible to add alkyl chains to aromatic compounds.

3

Acylation product (ketone) is less reactive than the reactant. Alkylation product is more reactive than the reactant and therefore multiple substitution reactions are likely.

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1

MODULE

6

OCR A level Chemistry – Answers to Student Book 2 questions

Organic chemistry and analysis

6.1.5 Phenols (page 131) 1

The hydroxyl group is not directly attached to the aromatic ring.

2

Phenol is more reactive than benzene as the electrons from the oxygen atom are donated to the aromatic ring allowing it to induce a dipole in a bromine molecule.

3

C6H5OH + KOH → C6H5O−K+ + H2O

6.1.6 Electrophilic substitution in aromatic compounds (page 133) 1

Room temperature in the presence of bromine water

2

3-chloronitrobenzene

3

2,4,6-triiodophenol

6.1.7 Reactions of carbonyl compounds (page 135) 1

propan-1-ol

2

potassium dichromate and sulfuric acid

3

It acts as a catalyst to make the C=O more polar and allow the cyanide ion to act as an effective nucleophile.

6.1.8 Characteristic tests for carbonyl compounds (page 137) 1

React each sample with Brady’s reagent. Then filter, recrystallise and dry the precipitate. Measure the melting point and use a data table to determine the sample chemical.

2

Tollens' reagent is a weak oxidising reagent. The aldehyde functional group is oxidised to a carboxylic acid, while the silver ions are reduced to silver metal creating the silver mirror.

3

Mix Brady’s reagent with each sample; the sample with no reaction is propanoic acid. (Also give credit for testing pH as propanoic acid is acidic whereas propanal and propanone are not.) Then filter, recrystallise and dry the precipitate. Measure the melting point and use a data table to determine the sample chemical. Alternatively take fresh samples of the two unidentified chemicals and mix with Tollens' reagent. Propanone will have no reaction, propanal will have a silver mirror present.

6.1.9 Carboxylic acids (page 139) 1

(a) (b) (c)

butanoic acid 2-methylpropanoic acid 2,3-dimethyloctanoic acid

2

NaHCO3 + CH3COOH → CO2 + H2O + CH3COONa

6.1.10 Esters (page 141) 1

This is a reversible reaction. Removing the product prevents equilibrium from being established and maximises the yield.

2

This allows the preparation of esters from phenols and has a higher yield than using a carboxylic acid.

3

Historically, the alkaline hydrolysis of a triglyceride ester to make soap. Now describes general alkaline hydrolysis of esters.

6.1.11 Acyl chlorides (page 143) 1

ethanoic acid and sulfur dichloride oxide

2

A reaction involving acyl chloride will be non-reversible which improves the yield and can be used to make esters from phenols.

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MODULE

6

OCR A level Chemistry – Answers to Student Book 2 questions

Organic chemistry and analysis

Practice questions (page 146) 1

C

2

A

3

D

4

B

5

Refer to official mark scheme for OCR past paper F324, January 2010; this is Q1. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

6

(a)

(b) 7

(a) (b)

(c)

8

(a)

(b)

(i)

(added to benzene) an equimolar mixture of concentrated sulfuric acid [1] and nitric acid [1] keeping the reaction mixture at 50 °C or below by cooling on ice if necessary [1] (ii) FeBr3 (catalyst) is prepared in situ by reacting bromine [1] with iron [1]. The mixture is heated under reflux [1]. (iii) A Friedel–Crafts catalyst [1] is prepared (e.g. AlCl3) [1] C2H5Cl [1] or other suitable to generate ethyl carbocation). To prevent further nitration the reaction mixture should not be allowed to get above 50 °C (cool on ice) [1] C6H6(l) + 3H2(g) → C6H12(l) correct reactants and products [1] balanced [1] The enthalpy of hydrogenation of cyclohexene to cyclohexane is −120 kJ mol−1 [1] On the basis of a Kekulé structure (of 3 alternating single and 3 double bonds one might expect the enthalpy of hydrogenation to be 3 × (−120) = −360 kJ mol−1 [1]. The greater thermodynamic stability of benzene is explained by a resonance delocalisation [1] of the 6 π-electrons [1] so that each C–C bond has an intermediate bond length and electron density between the C–C and the C=C bond [1] making it less susceptible to electrophilic attack [1] [Any 5] To distinguish between benzene and cyclohexene bromine water [1] used (away from a UV light source) would go from orange to colourless with cyclohexene [1] but not with benzene. The fact that benzene does not react with bromine water means that there is a lower localised electron density in isolated C=C bonds; rather the π-electrons are spread evenly around the ring [1] A = phenylethanone [1] Friedel–Crafts acylation [1] Ethanoyl chloride + AlCl3 catalyst [1] (Alternative: ethanoyl bromide + FeBr3) Q = chlorobenzene [1] Cl2 gas [1] bubbled through benzene in presence of AlCl3 [1] J = methyl benzene [1] CH3Cl [1] AlCl3 [1] (Alternative: CH3Br and FeBr3) React phenylethanone with 2,4-dinitrophenyl hydrazine [1] Collect orange precipitate by filtration and recrystallise [1] Specific melting point of DNP adduct will enable dinitrophenylhydrazone derivative to be identified [1]

6.2 Nitrogen compounds, polymers and synthesis 6.2.1 Basicity and the preparation of amines (page 153) 1

CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br

2

ethylamine

3

tetraethylammonium bromide

4

Ammonia is too volatile and would be lost from the system before it had a chance to react.

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MODULE

6

OCR A level Chemistry – Answers to Student Book 2 questions

Organic chemistry and analysis

6.2.2 Reactions of amino acids (page 155) 1

A compound where –COOH and –NH2 functional groups are present on the same carbon atom.

2

Ensure the pH is lower than the isoelectric point.

3

Increase the pH to around pH 13. This can be achieved by adding an alkali such as sodium hydroxide. A proton is donated to the hydroxide ion leaving the amino acid as a negative ion.

4

The product of a neutralisation reaction where ammonia or an amine group is acting as a base.

5

Amino acids react with both acids and bases.

6.2.3 Amides (page 157) 1

RC(O)–

2

Both have a nitrogen atom directly attached to an acyl group. Primary amides have two hydrogen atoms attached to the nitrogen, secondary amides have only one hydrogen attached to the nitrogen.

3

Butanamide is a primary amide. The displayed formula is shown below.

4

A condensation polymer made of a dicarboxylic acid monomer and a diamine monomer. These functional groups form amide links between each monomer.

5

The alkyl group (ethyl) is attached to a nitrogen atom.

6.2.4 Chirality (page 159) 1

(a)

(b)

2

Butan-1-ol does not have optical activity because it does not contain a chiral carbon.

3

8

6.2.5 Condensation polymers (page 163) 1

(a) (b)

Carboxylic acid + alcohol Amine + carboxylic acid

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MODULE

6

2

(a) (b)

3

(a)

OCR A level Chemistry – Answers to Student Book 2 questions

Organic chemistry and analysis

Condensation polymerisation

Addition polymerisation

Condensation polymerisation

One monomer One product (the polymer)

Two monomers Two products (the polymer and a small molecule – often water) Polymerises around functional groups other than C=C (e.g. COOH)

Polymerises around C=C

(b)

They are both produced from monomers. They both contain identifiable repeat units.

6.2.6 Hydrolysis of polymers (page 165) 1

(a) (b)

diol and dicarboxylic acid diammonium salt and dicarboxylic acid

2

(a) (b)

diol and dicarboxylic acid salt dicarboxylic acid salt and diamine

3

(a) (b)

The rate of reaction for the hydrolysis of amides is very slow. Polyesters are hydrolysed with alkali very quickly. Nylon is very resistant to hydrolysis in alkali.

6.2.7 Extending carbon chain length (page 167) 1

(a) (b)

cyanide nitrile

2

Ethanol is used rather than water to ensure the nucleophile is cyanide and not hydroxide. KCN is used rather than HCN as it is easier to handle and less dangerous.

3

(a)

The cyanide ion can attack the carbonyl group from both sides. This produces optical isomers.

(b)

Methanal will not produce a chiral compound with 4 different groups attached to the central carbon. Two of the groups will be hydrogen atoms.

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MODULE

6

OCR A level Chemistry – Answers to Student Book 2 questions

Organic chemistry and analysis

6.2.8 Reactions of nitriles (page 169) 1

(a) (b)

CH3CN + 2H2 → CH3CH2NH2 CH3CN + 4[H] → CH3CH2NH2

2

(a) (b)

Propanenitrile, dilute hydrochloric acid Heated reflux

6.2.9 Substitution reactions in aromatic compounds (page 171) 1

(a) (b)

Electron pair acceptor and halogen carrier Iron(III) chloride and aluminium(III) chloride

(a)

They are both electrophilic substitution reactions. They both require a strong Lewis acid catalyst. Alkylation adds an R group. Acylation adds RCO– group.

2

3

(b)

6.2.10 Practical skills for organic synthesis (page 173) 1

If you had a large quantity of solid, you would use a Büchner funnel. If you had a small amount of solid, you would use a Hirsch funnel.

2

(a) (b)

The organic product will have the highest solubility at high temperatures. As solubility is lower at lower temperatures this reduces the amount of desired product that would dissolve and be lost into the filtrate.

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6

MODULE 3

6

OCR A level Chemistry – Answers to Student Book 2 questions

Organic chemistry and analysis

All pure substances have distinct melting points. A pure substance will have a melting point that corresponds with the data from a trusted chemical data book. The more impure the substance is, the greater the melting point range will be.

6.2.11 Synthetic routes in organic synthesis (page 177) 1

A series of reactions designed to create a specific target molecule.

2

Sodium borohydride would react with water to form boric acid.

3

4

Practice questions (page 180) 1

B

2

A

3

B

4

D

5

Refer to official mark scheme for OCR past paper F324, January 2010; this is Q5. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

6

Refer to official mark scheme for OCR past paper F324, June 2011; this is Q2. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

7

(a)

2-amino-3-phenylpropanoic acid

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MODULE

6 (b)

OCR A level Chemistry – Answers to Student Book 2 questions

Organic chemistry and analysis (i)

A zwitterion is an internal salt with no charge, which is formed by the donation of a proton from a carboxylic acid functional group to the amine functional group in an amino acid. For example, phenylalanine has both positive and negative charges (ammonium cation and carboxylic acid anion), as shown in the diagram below.

(ii)

In a chiral centre, the chiral α-carbon has 4 different substituents, as shown in the diagram below.

(iii)

A condensation reaction between phenylalanine molecules will result in the formation of amide bond(s). If allowed to react further, poly(phenylalanine) will be formed.

6.3 Analysis 6.3.1 Chromatography (page 185) 1

0.5

2

The retention time, read from the x-axis of the chromatogram, can be used to identify the substance. The area under the peak shows the relative concentration compared to the other components in the mixture.

3

Gas chromatography allows you to separate different volatile parts from a mixture. These different parts need to be analysed via another technique.

6.3.2 Tests for organic functional groups (page 186) 1

It contains bromine atoms and a phenol group.

2

It is a weak acid as it partially ionises in solution releasing a proton. However, it is such a weak acid that there is not a high enough concentration of protons in solution to react with the carbonate ion.

3

(a) (b)

Both compounds would be oxidised producing the same result. You will get a colour change from orange to green. Distil the alcohol with acidified potassium dichromate. This should oxidise a primary alcohol to an aldehyde and a secondary alcohol to a ketone. React the product of the distillation with Fehling’s solution or Tollens’ reagent. No reaction will be seen if the substance is a secondary alcohol. Red precipitate with Fehling’s or silver mirror with Tollens’ reagent will be seen if the substance is a primary alcohol (and so produced an aldehyde on distillation).

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MODULE

6

OCR A level Chemistry – Answers to Student Book 2 questions

Organic chemistry and analysis

6.3.3 Introduction to nuclear magnetic resonance (page 188) 1

(a) (b)

D 2O C6D6

2

Parts per million (ppm)

3

It allows you to create a reference value against which samples can be compared.

4

Water would give a line in a proton NMR that would interfere with the spectrum of the sample. Water does not contain carbon and so would not produce a line in a carbon-13 NMR spectrum.

6.3.4 Carbon-13 NMR spectroscopy (page 192) 1

tetramethylsilane

2

carbon-13

3

(a) (b)

(c) (d)

CH3CH2CH2CHO; CH3CH2COCH3; (CH3)2CHCHO CH3CH2CH2CHO has four carbon environments CH3CH2COCH3 has four carbon environments (CH3)2CHCHO has three carbon environments In the carbon-13 NMR spectrum, there are three peaks. Therefore, there are three carbon environments. Therefore (CH3)2CHCHO produced the spectrum. The other two isomers cannot be distinguished as they both have four carbon environments and similar chemical shift values.

6.3.5 Proton NMR spectroscopy (page 196) 1

Hydrogen atoms bonded to the same carbon. They experience the same magnetic field in the NMR spectrometer.

2

A CH3CH2− section in the molecule

3

(a) (b)

A: CH–Br; B: (CH3)2–C C: –OCH2; D: CH3CO; E: CH3–C

4

(a) (b) (c) (d)

CH3–CH2 CH3–CH CH2–CH CH2–CH2

6.3.6 NMR spectra of –OH and –NH protons (page 198) 1

To remove the peaks caused by –OH and –NH groups in the molecule

2

(a) (b)

Chemical reactivity Physical properties (e.g. melting and boiling points, density)

3

(a)

Without D2O: three peaks, in the ratio 1(HO) : 2(CH2) : 1(COOH) With D2O: one peak only (CH2) Without D2O: three peaks, in the ratio 2(H2N) : 2(CH2) : 1(COOH) With D2O: one peak only (CH2) Without D2O: four peaks, in the ratio 3(CH3) : 1(CH) : 1(HO) : 1(COOH) With D2O: two peaks, in the ratio 3(CH3) : 1(CH)

(b) (c) 4

COOH at δ = 11.0 ppm; NH2 at δ = 5.2 ppm; CH at δ = 3.8 ppm and CH3 at δ = 1.2 ppm Structure = CH3CHNH2COOH

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MODULE

6

OCR A level Chemistry – Answers to Student Book 2 questions

Organic chemistry and analysis

6.3.7 Combined techniques (page 201) 1

IR spectra can identify multiple functional groups in a molecule without changing the molecule in the process.

2

A single test is rarely conclusive enough to deduce the full structure of a molecule, e.g. IR spectra can only give data on functional groups.

3

From left to right: 2(OCH2) : 1(OH) : 2(CH2) : 3(CH3)

Practice questions (page 204) 1

A

2

C

3

D

4

B

5

Refer to official mark scheme for OCR past paper F324, January 2010; this is Q3. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

6

Refer to official mark scheme for OCR past paper F324, June 2011; this is Q4. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

7

Refer to official mark scheme for OCR past paper F324, January 2013; this is Q1. All past papers and mark schemes are available on the OCR website, www.ocr.org.uk.

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