Alyn G McFarland Nora Henry CCEA A2 CHEMISTRY EXAM

CCEA A2CHEMISTRY EXAM PRACTICE

Alyn G McFarlandNora Henry

MARK SCHEMEThis mark scheme is for use in conjunction with the book Chemistry Exam Practice for CCEA A2, ISBN: 978 1 78073 254 1

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v2

2

Unit A2 1:

Further Physical and Inorganic Chemistry

Answers

11

4.1 LATTICE ENTHALPY

4.1 Lattice Enthalpy

1 C [1]

2 B [1]

Enthalpy of solution = lattice enthalpy + hydration enthalpy of calcium ion + 2 × hydration enthalpy of chloride ion = + 2237 + (–1650) + 2(–364) = –141 kJ mol–1. Most common error would be forgetting the 2 × for the enthalpy of hydration of the chloride ion. This would give answer C.

3 (a) the energy required to convert one mole of gaseous atoms into gaseous ions with a single positive charge. [2]

(b) enthalpy change when one mole of a compound is formed from its elements under standard conditions. [2]

(c) enthalpy change when one mole of gaseous atoms is formed from the element in its standard state. [2]

(d) enthalpy change when one mole of an ionic compound is converted to gaseous ions. [2]

4 (a) (i) Cs(g) Cs+(g) + e– [1](ii) Cs(s) + ½Cl

2(g) CsCl(s) [1]

(iii) Cl(g) + e– Cl–(g) [1](iv) CsCl(s) Cs+(g) + Cl–(g) [1]

(b) (i)

Cs+(g) + Cl(g) + e–

Cs+(g) + Cl–(g)Cs+(g) + ½Cl2(g) + e–

Cs(s) + ½Cl2(g)

Cs(g) + ½Cl2(g)

CsCl(s)

[1] for each correct level with state symbols [3](ii) lattice enthalpy = – (–433) + (+78) + (+380) + (+122) + (–364)

= + 649 kJ mol–1 [2](c) Cs+ ion larger ionic radius than other Group I metal ions [1]

weaker ionic bonding as ions not as close packed [1] [2]

5 (a) A, D, B, E, C, F all correct [2]; one error [1] [2]

(b) lattice enthalpy = – (–642) + (+150) + (+736) + (+1450) + (+242) + 2(–364)= + 2492 kJ mol–1 [3]

22


(c) –165 = + (+2492) + (–1891) + (2x)2x = –766x = –383 kJ mol–1 [3]

(d)

O

Mg

H

δ– δ– δ–

δ–

δ–

δ–δ– δ–δ+

δ+

δ+δ+

δ+

δ+δ+

δ+

δ+

δ+ δ+

δ+

δ+

δ+

δ+

δ+

2+

H O

H

H

O

H

H O OCl –

H

H

OH

H

O

H

H

O

H

H

H

H

[1] for each diagram with partial charges and bonds to ions bonds form between water molecules and ions [1]bond formation is exothermic [1] [4]

6 (a)

Ag(g) + ½F2(g)

Ag+(g) + ½F2(g) + e–

Ag+(g) + F(g) + e–

Ag+(g) + F–(g)

Ag(s) + ½F2(g)

AgF(s)

(b) first electron affinity = – (+79) – (+730) – (+286) + (–203) + (+943)= –355 kJ mol–1 [2]

(c) (i) 1 = lattice enthalpy of silver(I) fluoride [1]2 = hydration enthalpy of silver(I) ions [1]3 = hydration enthalpy of fluoride ions [1]4 = enthalpy of solution of silver(I) fluoride [1] [4]

(ii) + (+943) + (–464) + (–457) = + 22 kJ mol–1 [2]

33


7 (a)

2K+(g) + O2–(g)

2K+(g) + O–(g) + e–

2K+(g) + O(g) + 2e–

2K+(g) + ½O2(g) + 2e–

2K(g) + ½O2(g)

2K(s) + ½O2(g)

K2O(s)

–142+248

+180

–561

+844

+2531

[1] for each correct value [5](b) 2 × first ionisation energy

= – (+180) + (–561) + (+2531) – (+844) – (–142) – (+248) = +840first ionisation energy = 840 ÷ 2 = +420 kJ mol–1 [3]

(c) (i) sodium smaller ions in metallic lattice [1]stronger metallic bonding [1] [2]

(ii) atomic radius smaller for sodium [1]less shielding by inner electrons [1] [2]

(iii) lattice enthalpy = – (–416) + 2(+108) + 2(+500) + ½(+496) + (–142) + (+844) = +2582 kJ mol–1 [3]

K2O(s) -561

-142

+844

+2531

44


8 (a)

Mg2+(g) + O(g) + 2e–

Mg2+(g) + O2–(g)

Mg2+(g) + O–(g) + e–

Mg2+(g) + ½O2(g) + 2e–

Mg+(g) + ½O2(g) + e–

Mg(g) + ½O2(g)

Mg(s) + ½O2(g)

MgO(s)

[1] for each levelexothermic arrow for first electron affinity and endothermic for second [1]lattice enthalpy arrow [1] [4]

(b) enthalpy of formation = + (+150) + (+736) + (+1450) + (+248) + (–142) + (+844) – (+3888)= –602 kJ mol–1 [2]

55

4.2 ENTHALPY, ENTROPY AND FREE ENERGY

4.2 Enthalpy, Entropy and Free Energy

Entropy and predicting entropy changes

1 B [1]

All other reactions produce a gas which would increase entropy as gases are disordered. Reaction B only contains gases but the number of moles of gas decreases so entropy decreases.

2 C [1]

Reaction is exothermic and there is an increase in entropy.

3 C [1]

4 C [1]

ΔS = 189 – 70 = + 119 J K–1 mol–1

ΔG = 0 = ΔH – 373(+0.119) ΔH = 373 × 0.119 = +44.4 kJ mol–1

5 (a) ΔG < 0 [1](b) –56 = + 34 – Δ

fH(NO)

ΔfH(NO) = 34 + 56 = + 90 kJ mol–1 [2]

(c) when T = 794.3 K, ΔG = 0–56 – 794.3(ΔS) = 0ΔS = –56 ÷ 794.3 = – 0.0705 kJ K–1 mol–1

ΔS = – 70.5 J K–1 mol–1 [3](d) (i) a measure of disorder [1]

(ii) ΔS = –70.5 = 240 – (208 + 0.5x)–70.5 – 240 + 208 = –0.5x 0.5x = 102.5 so x = 205 J K–1 mol–1 [3]

6 (a) –602 kJ mol–1 [1](b) ΔH = TΔS

178 = T(0.160)T = 1112.5 K [2]

(c) –217 = 2(26.8) – 2(32.7) – xx = 205.2 J K–1 mol–1 [2]

(d) ΔG = ΔH – TΔS = –1204 – 298(–0.217) = –1139 kJ mol–1 [1]feasible as ΔG < 0 [1] [2]

(e) ΔS is positive and ΔH is negative [1]ΔG negative for all values of T [1] [2]

7 (a) ΔS = 42.7 – 32.7 = +10.0 J K–1 mol–1 [1](b) ΔH = TΔS

+8.95 = T(0.100)T = 895 K [2]

66

4.2 ENTHALPY, ENTROPY AND FREE ENERGY

8 (a) O2 is an element in its standard state [1]

(b) ΔS = 2(43.5) +4(240) + 205 – 2(193) = +866 J K–1 mol–1 [2](c) ΔH = 2(–155.2) + 4(+33.8) – 2(–302.9) = +430.6 kJ mol–1 [2](d) ΔG = ΔH – TΔS = +430.6 – 500(0.866) = –2.4 kJ mol–1 [1]

feasible as ΔG < 0 [1] [2](e) ΔG = 0 so ΔH = TΔS

T = ΔH ÷ ΔS = 430.6 ÷ 0.866 = 497 K [2]

9 (a) Ca2+ ion smaller/smaller ionic radius [1]ions more ordered/packed better together [1] [2]

(b) (i) ΔS = 70.4 + 189 – 99.7 = +159.7 J K–1 mol–1 [2](ii) ΔH – TΔS = 0 at 1060 K

ΔH = TΔS = 1060(0.1597) = +169.3 kJ mol–1 [3](iii) +169.3 = –533.5 + (–242.0) – x

x = – 944.8 kJ mol–1 [3](c) ΔS = 39.7 + 189 – 83.4 = +145.3 J K–1 mol–1

ΔH = –635.1 + (–242.0) – (–986.1) = +109 kJ mol–1

T = ΔH ÷ ΔS = 109 ÷ 0.1453 = 750 K [6](d) (i) size of cation increase down the group [1]

charge density of cation decreases down the group [1]larger cation less able to polarise and destabilise hydroxide ion [1] [3]

(ii) increases [1]

10 (a) ΔH = 6(–315.5) – 8(–46.0) = –1525 kJ mol–1 [2](b) ΔS = 191.4 +6(94.6) – (8(192.5) + 3(233.0))

= –1480 J K–1 mol–1 [2](c) ΔG = –1525 – 1500(–1.480) = +695 kJ mol–1 [1]

not feasible as ΔG > 0 [1] [2]

11 (a) ΔH = –111 – (–242) – (–75)= +206 kJ mol–1 [2](b) ΔG = ΔH – TΔS [1](c) –10 = +206 – 1000(ΔS)

ΔS = 216 ÷ 1000 [1] = + 0.216 kJ K–1 mol–1

= + 216 J K–1 mol–1 [3](d) + 216 = 198 + 3x – 189 – 186

3x = 393 x = 131 J K–1 mol–1 [3]

(e) T = ΔH ÷ ΔS = +206/0.216 = 954 K [2]

77

4.3 RATES OF REACTION

4.3 Rates of Reaction

Rate equations and associated graphs1 B [1]

The units of the rate constant are mol–1 dm3 s–1 as it is second order overall.

2 C [1]

The order of reaction with respect to W is 2 so the graph of rate against [W] should be non-linear.

3 C

Doubling the concentration of L and M increasing the rate by a factor of 8 means one of them has to be second order and the other is first order.

4 (a) from experiment 1 to experiment 2[Q] no changerate × 4(×4) = (×[P])2

×[P] = 2 [P] = 2.50 × 10–3 [2]

(b) from experiment 1 to experiment 3[P] × 1.6rate × 4.096(×4.096) = (1.6)2 × (×[Q])×[Q] = 1.6[Q] = 2.80 × 10–3 [2]

(c) from experiment 1 to experiment 4[P] × 2[Q] × 3.2× rate = (×2)2 × (×3.2) = ×12.8rate = 8.00 × 10–3 [2]

(d) 3 [1](e) k = 2.29 × 105 [1] mol–2 dm6 s–1 [1] [2]

Rate related to mechanism5 (a) the power to which the concentration of a reactant is raised in the rate equation [1]

(b) rate = k[A]2 [1](c) (i) the slowest step in the mechanism of a reaction [1]

(ii) step 1 [1]A is a reactant/B is not a reactant [1]order with respect to A is not zero/order with respect to B is zero [1] [3]

88


6 (a) order with respect to C4H

9Br = 1 [1]

order with respect to OH– = 1 [1] [2](b) rate = k[C

4H

9Br][OH–] [1]

(c) k = 1.92 [1] mol–1 dm3 s–1 [1] [2](d) rate constant increases as temperature increases [1](e) 1-bromobutane/1-bromo-2-methylpropane [1]

primary halogenoalkane [1]both halogenoalkane and OH– are not zero order so involved in rate determining step [1] [3]

Practical determination of rate and graphical analysis7 (a) measure gas volume of O

2 [1]

against time [1]plot volume against time [1]initial gradient = rate [1] [4]

(b) (i) 450 s [1]

(ii) tangent at 2.50 mol dm–3: gradient = 2.50

620 = 4.03 ×10–3 [1]

tangent at 1.00 mol dm3: gradient = 1.60

1000 = 1.60 × 10–3 [1]

(iii) [N2O

5] × 2.5 and rate × 2.5 [1]

order = 1 [1] [2](iv) rate = k [N

2O

5] [1]

(v) k = 1.60 × 10–3 [1] s–1 [1] (allow 1.60 – 1.61 × 10–3) [2]

8 (a) using known concentrations of iodine solution [1]plot calibration curve/absorbance against concentration of iodine [1]set up reaction with large excess of other reactants and record absorbance against time [1]convert absorbance to concentration (using calibration curve) [1]plot concentration of iodine against time [1]take tangents and measure the gradient [1]plot graph of rate against concentration and determine order from shape of graph [1] max [6]

(b) (i) zero order [1]constant gradient/straight line [1] [2]

(ii) rate = 1.00

400 = 2.50 × 10–3 mol dm–3 s–1 [1]

(iii) horizontal line [1]at 2.50 (× 10–3) on graph [1] [2]

(c) (i) rate = k[CH3COCH

3][H+] [1]

(ii) k = 0.104 [1] mol–1 dm3 s–1 [1] [2]

99


9 (a)

00.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

10 20 30 40 50 60 70 80 90 100

[A] /

mol

dm

–3

time / s

correctly drawn tangent at t = 0 s [1](b) gradient of tangent calculated from drawing

rate = 0.80

34 = 0.024 [1] mol dm–3 s–1 [1] [2]

(c) 0.40 = 20 s0.20 = 40 s [1]

(d) order = 1 [1](e) rate = k[A] [1]

(f) k = 0.024

0.80 = 0.03 [1] s–1 [1] [2]

(g) Step 1 as A is a reactant and present in rate equation [1]

1010

4.4 EQUILIBRIUM

4.4 Equilibrium

1 Answer is C [1]

moles at equilibrium: CH

3COOH = 0.35

CH3CH

2OOCCH

3 = H

2O = 0.65

Kc =

[CH3CH

2OOCCH

3][H

2O]

[CH3CH

2OH][CH

3COOH]

= (0.65)2

(0.35)2 = 3.4

2 Answer is D [1]

Kc = (2.40)2 = 5.76

3 (a) CH4 + 2H

2O s CO

2 + 4H2 [1]

(b) forward reaction is endothermic as K

c increases as temperature increases [1]

(c) (i) Kc =

[CO][H2]3

[CH4][H

2O]

[1]

(ii) CH4 = 0.09 [1]

H2O = 0.245 [1]

CO = 0.155 [1]H

2 = 0.465 [1] [4]

(iii) Kc =

[CO][H2]3

[CH4][H

2O]

= (0.155)(0.465)3

(0.09)(0.245) = 0.707 mol2 dm–6 [3]

4 (a) (i) Kc =

[HI]2

[H2][I

2]

[1]

(ii) concentrations cancel as (mol dm–3)2 / (mol dm–3)2 [1](iii) equilibrium moles:

H2 = I

2 = 0.4

HI = 3.00

Kc =

(3.00)2

(0.4)(0.4) = 56.3 [3]

(b) forward reaction exothermic so at lower temperatureK

c increases [1]

position of equilibrium moves to the right [1] [2](c) no effect on position of equilibrium or K

c [1]

same number of moles of gas on both sides [1]K

c not affected by changes in pressure [1] [3]

1111

4.4 EQUILIBRIUM

5 equilibrium molesmoles of CH

3COOH = 0.010 – 9.0 × 10–3 = 1.0 × 10–3

moles of C5H

10 = 0.020 – 9.0 × 10–3 = 0.011

moles of CH3COOC

5H

11 = 9.0 × 10–3

equilibrium concentrationsCH

3COOH = 1.67 × 10–3

C5H

10 = 0.0183

CH3COOC

5H

11 = 0.015

Kc =

[CH3COOC

5H

11]

[CH3COOH][C

5H

10]

Kc =

(0.015)

(1.67×10−3)(0.0183) = 490 mol–1 dm3 [4]

Answer is given to 2 significant figures as this is appropriate based on the data in the question.

6 (a) Kc =

1

8.00 = 0.354 mol dm–3 [2]

(b) (i) moles of H2 reacting = 3.00 – 0.30 = 2.70

equilibrium molesN

2 = x – 0.90

H2 = 0.30

NH3 = 1.80

equilibrium concentrationsN

2 = 5(x – 0.90)

H2 = 1.50

NH3 = 9.00 [3]

(ii) Kc =

[NH3]2

[N2][H

2]3 =

(9.00)2

(5(x − 0.90))(1.5)3 = 8.00

5(x – 0.90) = 81

8.00 × 3.375 = 3.00

x – 0.90 = 0.60x = 1.5 [4]

(iii) no effect on Kc as only affected by changes in temperature [1]

lower pressure so position of equilibrium moves to the left [1]4 moles of gas on left and 2 moles of gas on right [1] [3]

1212

4.4 EQUILIBRIUM

7 (a) Kc =

[I]2

[I2] [1]

(b) equilibrium moles of I2 = 1.25 ×10–3 – 6.00 × 10–4 = 6.50 × 10–4

equilibrium concentrationsI2 = 4.33 × 10–5

I = 8.00 × 10–5

Kc =

(8.00 × 10−5)2

(4.33 × 10−5) = 1.48 × 10–4 [4]

(c) (i) titration with thiosulfate/colorimetry using a calibration curve [1]

(ii) Kc =

[I]2

[I2] = 4.60 × 10–3 =

x2

0.0458x2 = 4.60 × 10–3 × 0.0458 = 2.11 × 10–4

x = 0.0145 mol dm–3 [3](d) endothermic as K

c increases with increasing temperature [1]

8 (a) 3.75 – 4 minutes [1]graphs level off [1] [2]

(b) initialN

2 = 1.00 (mol dm–3) [1] O

2 = 2.40 (mol dm–3) [1]

EquilibriumN

2 = 0.10 (mol dm–3) [1] O

2 = 0.60 (mol dm–3) [1] [4]

(c) 1.80 (mol dm–3) [1](d)

O2

N2

00.00

0.40

0.80

1.20

1.60

2.00

2.40

0.20

0.60

1.00

1.40

1.80

2.20

1 2 3 4 5 6 7

Con

cent

ratio

n / m

ol d

m–3

Time / min

starts at 0,0 [1] ends up at 1.80 mol dm–3 around 3.75 – 4 minutes [1] [2]

1313

4.4 EQUILIBRIUM

(e) Kc =

[NO2]2

[N2][O

2]2 [1] mol–1 dm3 [1] [2]

(f) Kc =

(1.80)2

(0.1)(0.6)2 = 90 [2]

(g) Kc would increase [1]

as forward reaction is endothermic [1] [2]

9 (a) (i) Kc =

[ICl]2

[I2][Cl

2] [1]

(ii) initial [I2] = [Cl

2] = 0.625 mol dm–3

equilibrium concentrations:[I

2] = [Cl

2] = 0.625 – x

[ICl] = 2x(2x)2

(0.625 − x)2 = 6.25 × 10–2

2x0.625 − x = 0.25

2x = 0.15625 – 0.25x2.25x = 0.15625x = 0.06940.625 – x = 0.625 – 0.0694 = 0.555 × 4 = 2.22 moles of I

2

or

initial moles I2 = Cl

2 = 2.50

equilibrium moles of I2 = Cl

2 = 2.50 – x

equilibrium moles of ICl = 2x(2x)2

(2.50 − x)2 = 6.25 × 10–2

2x2.50 − x

= 0.25

2x = 0.625 – 0.25x2.25x = 0.625x = 0.2782.50 – x = 2.50 – 0.278 = 2.22 moles of I

2 [4]

1414

4.4 EQUILIBRIUM

(b) (i)

I2

Cl2

ICl

00.00

0.40

0.80

1.20

0.20

0.60

1.00

1 2 3 4 5 6 7 8 9 10

Con

cent

ratio

n / m

ol d

m–3

Time / min

ICl starts at 0, 0 and ends at 1.04 (mol dm–3) [1]at around 7.5 – 8 minutes [1]Cl

2 starts at 0, 1.20 and ends at 0.68 (mol dm–3) [1]

at around 7.5 – 8 minutes [1] [4]

(ii) Kc =

(1.04)2

(0.28)(0.68) = 5.68 [1]

(iii) exothermic as Kc increases as temperature decreases / K

c decreases as temperature

increases [1]

1515

4.5 ACID-BASE EQUILIBRIA

4.5 Acid-Base Equilibria

Brønsted-Lowry acids and bases1 Answer is B [1]

2 CH3COOH CH

2ClCOOH CH

3COOH

2CH

2ClCOO–

base baseacid acid

+ +[1]

+

all boxes correct [1] [2]

3 (a) proton donor [1](b) HPO

4

2– + H2O H

2PO

4

– + OH– [1]dihydrogenphosphate(V) ion and hydroxide ion [1] [2]

(c) HPO4

2– + H2O PO

4

3– + H3O+ [1]

phosphate(V) ion and hydronium/hydroxonium/oxonium ion [1] [2]

4 (a) 2H2SO

4 + HNO

3 NO

2

+ + H3O+ + 2HSO

4

– [1](b) H

2SO

4 acid [1] and HSO

4

– is its conjugate base [1]HNO

3 base [1] and H

2NO

3

+ is its conjugate acid [1] [4](c) base as it is accepting a proton [1](d) sulfuric as it is donating a proton to nitric acid [1]

5 Answer is D [1]

pH calculations6 Answer is D [1]

A common error here is forgetting that barium hydroxide is Ba(OH)2 so has 2 moles of OH– per

mole of Ba(OH)2.

7 (a) Kw

= [H+][OH–] [1](b) in water [H+] = [OH–]

so [H+]2 = Kw

[H+] = 2.92 × 1014 = 1.71 × 10–7 mol dm–3

pH = – log10

[H+] = – log10

(1.71 × 10–14) = 6.77 [2](c) [H+] = 10(–6.15) = 7.08 × 10–7 mol dm–3

Kw

= [H+]2 = (7.08 ×10–7)2 = 5.01 ×10–13 [2]

8

Strong acid

Concentration of original strong acid

/ mol dm–3

Volume of strong acid used / cm3

Volume of water added

/ cm3

Concentration of diluted acid

/ mol dm–3

pH of diluted

acid

HCl 0.450 150 350 0.135 [1] 0.87 [1]

HNO3

0.597 100 1400 [1] 0.0398 [1] 1.40

H2SO

40.250 25 275 [1] 0.0208 [1] 1.38

[6]

1616


HCl:moles of HCl =

150 × 0.4501000

= 0.0675 mol

total new volume = 500 cm3

new [HCl] = 0.0675 × 2 = 0.135 mol dm–3

new [H+] = 0.135 mol dm3

new pH = – log10

[H+] = 0.87

HNO3:

new [H+] = 10–1.40 = 0.0398 mol dm–3

dilution factor = 0.597

0.0398 = 15

total volume required = 100 × 15 = 1500 cm3

volume of water required = 1500 – 100 = 1400 cm3

H2SO

4:

new [H+] = 10–1.38 = 0.04169 mol dm–3

new [H2SO

4] =

0.041692

= 0.0208 mol dm–3


0.0208 = 12

total volume required = 25 × 12 = 300 cm3

volume of water required = 300 – 25 = 275 cm3

9 Answer is D [1]

moles of H+ = 20.0 × 1.45

1000 × 2 = 0.058

moles of OH– = 45.0 × 2.40

1000 = 0.108

OH– in excessmoles of OH– remaining = 0.108 – 0.058 = 0.05total volume = 65.0 cm3

[OH–] = 0.0565.0

× 1000 = 0.0769 mol dm–3

[H+] = 1.00 × 10–14

0.0769 = 1.30 × 10–13 mol dm–3

pH = – log10

[H+] = – log10

(1.30 × 10–13) = 12.89

10 Answer is C [1]

old [HCl] = old [H+] = 10–1.75 = 0.0178 mol dm–3

new [HCl] = new [H+] = 10–2.45 = 3.55 × 10–3 mol dm–3


3.55 × 10–3 = 5

new total volume = 20 × 5 = 100 cm3

volume of water added = 100 – 20 = 80 cm3

1717


11 (a) BaO + H2O Ba(OH)

2 [1]

(b) moles of BaO = 3.15

153 = 0.02059

moles of Ba(OH)2 = 0.02059

[OH–] = 0.02059 × 4 × 2 = 0.1647 mol dm–3

[H+] = 1.00 × 10–14

0.1647 = 6.072 × 10–14 mol dm–3

pH = – log10

[H+] = – log10

(6.072 × 10–14) = 13.22 [5](c) magnesium oxide does not react with water/magnesium hydroxide low solubility in water [1]

12 (a) [Ca(OH)2] =

0.168

74 = 2.270 × 10–3 mol dm–3

[OH–] = 2.270 × 10–3 × 2 = 4.540 × 10–3 mol dm–3

[H+] = 1.00 × 10–14

4.540 × 10–3 = 2.202 × 10–12 mol dm–3

pH = – log10

[H+] = – log10

(2.202 × 10–12) = 11.66 [4]

(b) moles of Ca(OH)2 =

25.0 × 2.270 × 10−3

1000 = 5.675 × 10–5

moles of HCl = 5.0 × 0.0125

1000 = 6.25 × 10–5

moles of HCl required = 5.675 × 10–5 × 2 = 1.135 × 10–4 so Ca(OH)2 in excess

moles of Ca(OH)2 remaining = 5.675 × 10–5 –

6.25 × 10−5

2 = 2.550 × 10–5

total volume = 30.0 cm3

[OH–] = 2.550 × 10−5

30.0 × 1000 × 2 = 1.700 × 10–3 mol dm–3

[H+] = 1.00 × 10−14

1.700 × 10−3 = 5.882 ×10–12 mol dm–3

pH = – log10

[H+] = – log10

(5.882 × 10–12) = 11.23 [5]

(c) moles of Ca(OH)2 =

25.0 × 2.270 × 10−3

1000 = 5.675 × 10–5

moles of HCl = 15.0 × 0.0125

1000 = 1.875 × 10–4

moles of HCl required = 5.675 × 10–5 × 2 = 1.135 × 10–4 so HCl in excessmoles of HCl remaining = 1.875 × 10–4 – 1.135 × 10–4 = 7.400 × 10–5


[H+] = 7.400 × 10−5

40.0 × 1000 = 1.850 × 10–3 mol dm–3

pH = – log10

[H+] = – log10

(1.850 × 10–3) = 2.73 [5]

1818


13 (a) C6H

8O

7 [1]

(b) (i) C6H

8O

7 s C

6H

7O

7

– + H+

C6H

7O

7

– s C6H

6O

7

2– + H+

C6H

6O

7

2– s C6H

5O

7

3– + H+

all correct [2]1 error [1] [2]

(ii) moles of C6H

8O

7 =

0.355

192 = 1.849 × 10–3

[C6H

8O

7] = 1.849 × 10–3 × 8 = 0.01479 mol dm–3

Ka = 10–3.08 = 8.32 × 10–4 mol dm–3

[H+] = 8.32 × 10–4 × 0.01479 = 3.51 × 10–3 mol dm–3

pH = – log10

[H+] = – log10

(3.51 ×10–3) = 2.45 [4]

14 (a) (i) CH3CH

2COOH s CH

3CH

2COO– + H+ [1]

(ii) Ka =

[CH3CH

2COO−][H+]

[CH3CH

2COOH]

[1]

(iii) [H+] = 10–3.03 = 9.33 × 10–4 mol dm–3

Ka =

(9.33 × 10−4)2

6.45 × 10−2 = 1.35 × 10–5 mol [2]

(b) (i) CH3CH

2CH

2COOH + NaOH CH

3CH

2CH

2COONa + H

2O [1]

(ii) [H+] = 1.51 × 10–5 × 1.855 × 10–3 = 3.593 × 10–4 mol dm–3

pH = – log10

[H+] = – log10

(3.593 × 10–4) = 3.44 [2]

(iii) moles of CH3CH

2CH

2COOH =

25.0 × 8.55 × 10−3

1000 = 2.1375 × 10–4

moles of NaOH = 2.1375 × 10–4

volume of NaOH required = 2.1375 × 10−4 × 1000

1.20 × 10−2 = 17.8 cm3 [3]

(iv) dilution factor = 150

25 = 6

diluted acid concentration = 8.55 × 10−3

6 = 1.425 × 10–3

[H+] = [H+] = 1.51 × 10–5 × 1.425 × 10–3 = 1.467 × 10–4 mol dm–3

pH = – log10

[H+] = – log10

(1.467 × 10–4) = 3.83 [4]

15 (a) [H+] = 10–2.77 = 1.70 × 10–3 mol dm–3 [1](b) (i) HCOOH + NaOH HCOONa + H

2O [1]

(ii) moles of NaOH = 16.0 × 0.0258

1000 = 4.128 × 10–4

moles of H+ = 4.128 × 10–4

[H+] = 4.128 × 10–4 × 40 = 0.0165 mol dm–3 [3](iii) pH measures [H+] at equilibrium [1]

titration determines total [H+] [1] [2]

1919


Buffers16 Answer is C

In this buffer moles of acid = moles of salt formed so the [H+] = Ka.

pH = – log10

Ka.

17 (a) A solution which resist changes in pH on addition of small amounts of acid or alkali [1](b) lactate ion reacts with H+ and removes them maintaining pH [1]

CH3CH(OH)COO– + H+ CH

3CH(OH)COOH [1]

(explanation in terms of CH3CH(OH)COOH s CH

3CH(OH)COO– + H+ position of equilibrium

moving to the left to remove H+ ions will be accepted) [2](c) lactic acid reacts with OH– ions and removes them so maintaining pH [1]

CH3CH(OH)COOH + OH– CH

3CH(OH)COO– + H

2O [1]

(explanation in terms of CH3CH(OH)COOH s CH

3CH(OH)COO– + H+ position of equilibrium

moving to the right to replace the H+ ions which have reacted with OH– ions, i.e. H+ + OH– H2O

will be accepted) [2]

(d) moles of CH3CH(OH)COONa =

0.216

112 = 1.929 × 10–3

[CH3CH(OH)COONa] = 1.929 × 10–3 × 20 = 0.0386 mol dm–3

[H+] = K

a × [CH

3CH(OH)COOH]

[CH3CH(OH)COO−]

= 1.38 × 10−4 × 0.0125

0.0386 = 4.47 × 10–5 mol dm–3

pH = – log10

[H+] = – log10

(4.47 × 10–5) = 4.35 [4]

In this buffer calculation the solid sodium lactate added does not dilute the acid so the concentration of the acid is unchanged. The concentration, in mol dm–3, of the salt must be calculated and then used to calculate [H+].

18 (a) initial moles of acid (C6H

5COOH) =

35.0 × 0.250

1000 = 8.75 × 10–3

moles of salt (C6H

5COONa) = moles of NaOH =

15.0 × 0.250

1000 = 3.75 × 10–3

moles of acid remaining = 8.75 ×10–3 – 3.75 × 10–3 = 5.00 × 10–3

Ka = 10–4.19 = 6.46 ×10–5 mol dm–3

[H+] = K

a × [C

6H

5COOH]

[C6H

5COO−]

= 6.46 × 10−5 × 5.00 × 10−3

3.75 × 10−3 = 8.61 × 10–5 mol dm–3

pH = – log10

[H+] = – log10

(8.61 × 10–5) = 4.06 [5]

The pH of 4.06 is obtained using the rounded [H+] (8.61 × 10–3 mol dm–3) however using 8.613 × 10–5 mol dm–3 or the value in the calculator throughout gives 4.07 and both 4.06 and 4.07 would be accepted. The moles of the acid remaining and the moles of the salt may be used to calculate [H+]. Ideally you should convert these moles to concentration, in mol dm–3, by dividing by the total volume (50.0 cm3 here) and multiplying by 1000 but it gives the same value for [H+] but you must convert both of them but using the moles if faster. In 17(d) you had to calculate the concentration, in mol dm–3, of the sodium lactate as you were using the concentration of the acid in mol dm–3

(b) (i) moles of H+ = 5.00 × 0.0550

1000 = 2.75 × 10–4 [1]

(ii) moles of acid = 5.00 × 10–3 + 2.75 × 10–4 = 5.275 × 10–3 [1](iii) moles of benzoate ions = 3.75 ×10–3 – 2.75 × 10–4 = 3.475 ×10–3 [1]

2020


(iv) [H+] = K

a × [C

6H

5COOH]

[C6H

5COO−]

= 6.46 × 10−5 × 5.275 × 10−3

3.475 × 10−3 = 9.81 × 10–5 mol dm–3

pH = – log10

[H+] = – log10

(9.81 × 10–5) = 4.01 [2]

Whilst this type of calculation in (b) is beyond the specification, you should expect the unexpected at A2. However, you would be guided through it as shown here.

(c) C6H

5COOH + OH– C

6H

5COO– + H

2O [1]

benzoate ions react with OH– ions and remove them so maintaining pH [1](explanation in terms of C

6H

5COOH s C

6H

5COO– + H+ position of equilibrium moving to the right

to replace the H+ ions which have reacted with OH– ions, i.e. H+ + OH– H2O will be accepted) [2]

19 (a) nitrite ion reacts with H+ and removes them maintaining pH [1]NO

2

– + H+ HNO2 [1]

(explanation in terms of HNO2 s NO

2

– + H+ position of equilibrium moving to the left to remove H+ ions will be accepted) [2]

(b) initial moles of HNO2 =

25.0 × 0.150

1000 = 3.75 × 10–3

moles of NaOH = moles of = 15.0 × 0.100

1000 = 1.50 × 10–3

moles of acid remaining = 3.75 × 10–5 – 1.50 × 10–3 = 2.25 × 10–3

Ka = 10–3.40 = 3.98 ×10–4 mol dm–3

[H+] = K

a × [HNO

2]

[NO2]– =

3.98 × 10−4 × 2.25 × 10−3

1.50 × 10−3 = 5.97 × 10–4 mol dm–3

pH = – log10

[H+] = – log10

(5.97 × 10–4) = 3.22 [5](c) pH = pK

a so moles of acid remaining = moles of salt

moles of NaOH added = ½ × moles of acid addedmoles of acid added = 3.75 × 10–3

moles of NaOH required = 1.875 × 10–3


0.100 = 18.75 cm3 [3]

The key in the last part is understanding the pH = pKa so [H+] = [NO

2

–] and to achieve this the moles of acid remaining and moles of salt formed (= moles of NaOH added) must be equal.

20 (a) Ka =

[NH3][H+]

[NH4 ]+ [1]

(b) (i) NH3 + HCl NH

4Cl [1]

(ii) initial moles of NH3 =

50.0 × 0.1

1000 = 5 × 10–3

moles of HCl = moles of NH4Cl formed =

25.0 × 0.12

1000 = 3 × 10–3

moles of HCl remaining = 5 × 10–3 – 3 × 10–3 = 2 × 10–3

[H+] = K

a × [NH

4 ]

[NH3]

+

= 5.62 × 10−10 × 3 × 10−3

2 × 10−3 = 8.43 × 10–10 mol dm–3

pH = – log10

[H+] = – log10

(8.43 × 10–10) = 9.07 [5](c) NH

3 reacts with added H+/position of equilibrium moves to the left [1]

added H+ removed and pH maintained [1] [2]

2121


Titration curves and salts21 Answer is A [1]

A salt of a strong acid and a weak base (A) will have an acidic pH. A salt of a strong acid and a strong base (C) will have a neutral pH. A salt of a weak acid and a strong base (D) will have an alkaline pH. A salt of a weak acid and a weak base (B) could be acidic, alkaline or neutral depending on the parent acid and base but it would not be as acidic as A or as alkaline as D, if the concentrations are the same.

22 Answer is D [1]

Ethanoic acid is the weaker weak acid (higher pKa so lower K

a). A weaker weak acid will have a higher

concentration of the undissociated acid so when the anion is added to water it reacts to a greater extent giving a higher pH (A– + H

2O HA + OH–). The sodium ion does not affect the pH but the ammonium ion

will cause some acidity in the solution counteracting the effect of the anion.

23 (a) HCl + NaOH NaCl + H2O [1]

(b) (i) [H+] = 0.270 mol dm–3

pH = – log10

[H+] = – log10

(0.270) = 0.57 [2]

(ii) moles of HCl = 25.0 × 0.270

1000 = 6.75 × 10–3

moles of NaOH = 6.75 × 10–3


0.320 = 21.1 cm3 [3]

(iii)

0.00

2

4

6

8

10

12

14

1

3

5

7

9

11

13

5.0 10.0 15.0 20.0 25.0 30.0

pH

Volume of sodium hydroxide solution / cm3

starts around pH 0.57/0.6 [1]vertical region between pH 3 and 10 approximately [1]vertical region at approximately 21 cm3 [1] [3]

2222


24 (a) [H+] = 2 × 0.145 = 0.290 mol dm–3

pH = – log10

[H+] = – log10

(0.290) = 0.54 [2](b) [OH–] = 0.500 mol dm–3

[H+] = 1.00 × 10−14

0.500 = 2.00 × 10–14 mol dm–3

pH = – log10

[H+] = – log10

(2.00 × 10–14) = 13.70 [2]

(c) moles of H2SO

4 =

25.0 × 0.145

1000 = 3.625 × 10–3

moles of KOH = 3.625 × 10–3 × 2 = 7.25 × 10–3

volume of KOH = 7.25 × 10−3 × 1000

0.500 = 14.5 cm3 [3]

(d) moles of H2SO

4 = 3.625 × 10–3

moles of H+ = 7.25 × 10–3

moles of OH– = 14.0 × 0.500

1000 = 7.00 × 10–3

moles of H+ in excess = 7.25 × 10–3 – 7.00 × 10–3 = 2.50 × 10–4


new [H+] = 2.50 × 10−4

39.0 × 1000 = 6.41 × 10–3 mol dm–3

pH = – log10

[H+] = – log10

(6.41 × 10–3) = 2.19 [5](e) moles of H

2SO

4 = 3.625 × 10–3

moles of H+ = 7.25 × 10–3

moles of OH– = 15.0 × 0.500

1000 = 7.50 × 10–3

moles of OH– in excess = 7.50 × 10–3 – 7.25 × 10–3 = 2.50 × 10–4


new [OH–] = 2.50 × 10−4

40.0 × 1000 = 6.25 × 10–3 mol dm–3

[H+] = 1.00 × 10−14

6.25 × 10−3 = 1.60 × 10–12 mol dm–3

pH = – log10

[H+] = – log10

(1.60 × 10–12) = 11.80 [5]

2323


(f)

0.00

2

4

6

8

10

12

14

1

3

5

7

9

11

13

5.0 10.0 15.0 20.0 25.0 30.0

pH

Volume of potasium hydroxide solution added / cm3

starts around pH 0.54/0.5 [1]vertical region between pH 3 and 10 approximately [1]vertical region at approximately 14.5 cm3 [1] [3]

(g) phenolphthalein or methyl orange [1]indicator pH range colour change in pH range of vertical region [1] [2]

25 Answer is B [1]

Vertical region for this titration between 3 and 8/9 so only indicator with a colour change in this range would be bromophenol blue

2424


26 (a) 2.65 (accept 2.6 to 2.7) [1](b) 20 cm3 [1](c) at 10 cm3, pH = pK

a = 4.00 [1]

Ka = 10–4.00 = 1 × 10–4 mol dm–3 [1] [2]

(d) Ka =

[ClC6H

4COO−][H+]

[ClC6H

4COOH]

[1]

(e) [H+] = 10–2.65 = 2.24 × 10–3 mol dm–3

[acid] = [H+]2

Ka

= (2.24 × 10−3)2

1 × 10−4 = 0.0502 mol dm–3 [3]

(f) moles of ClC6H

5COOH =

25.0 × 0.0502

1000 = 1.255 × 10–3

moles of NaOH = 1.255 × 10–3

concentration of NaOH = 1.255 × 10−3 × 1000

20 = 0.0628 mol dm–3 [3]

(g) adding acid:4-chlorobenzoate ion reacts with H+ and removes them maintaining pH [1]ClC

6H

4COO– + H+ ClC

6H

4COOH [1]

(explanation in terms of ClC6H

4COOH s ClC

6H

4COO– + H+ position of equilibrium moving to the

left to remove H+ ions will be accepted)

adding alkali:4-chlorobenzoic acid reacts with OH– ions and removes them so maintaining pH [1]ClC

6H

4COOH + OH– ClC

6H

4COO– + H

2O [1]

(explanation in terms of ClC6H

4COOH s ClC

6H

4COO– + H+ position of equilibrium moving to

the right to replace the H+ ions which have reacted with OH– ions, i.e. H+ + OH– H2O will be

accepted) [4]

27 (a)

N NNCH

3

CH3

all N atoms indicated [1](b) (i) A CH

3COOH + NH

3 CH

3COONH

4 [1]

B CH3COOH + NaOH CH

3COONa + H

2O [1]

C HNO3 + NH

3 NH

4NO

3 [1]

D HNO3 + NaOH NaNO

3 + H

2O [1] [4]

(ii) C and D [1]indicator pH range colour change in pH range of vertical region / titrations involving a strong acid [1] [2]

(iii) CH3COO– + H

2O CH

3COOH + OH– [1]

OH– ions cause pH > 7 [1]sodium ion does not react with water [1] [3]

(iv) NH4

+ + H2O NH

3 + H

3O+ [1]

H3O+ ions cause pH < 7 [1]

nitrate ion does not react with water [1] [3](v) salt of a strong acid and a strong base [1]

both ions in the salt do not react with water [1] [2]

2525

4.6 ISOMERISM

4.6 Isomerism

Optical isomers

1 molecules which have the same molecular formula but a different structural formula. [2]

2

CH

H

H

C

H

O

C

H

H

CH

H

H

C

O

C H

H

H

propanal propanone [2]

3 (a)

CH

H

H

O C

O

C H

H

H

CH

H

H

C

H

H

O

O

C H

methyl ethanoate ethyl methanoate [2]

(b)

CH

H

H

C

OH

O

C

H

H

propanoic acid [1]

4 An atom which has four different atoms or groups attached [1]

5 (a) no [1](b) no [1](c) yes [1](d) no [1](e) yes [1](f) yes [1](g) yes [1]

6 Answer is D [1]

7 Answer is A [1]

8 (a)

HCOH

CH2

COOH

COOH

[1]

(b) molecules which exist as non-superimposable mirror images [1](c) hydroxyl carboxyl [1]

2626

4.6 ISOMERISM

9 (a)

HC

Cl

C2H

5

CH3

H3C

C

Cl

H5C

2

H

[2](b) pass plane polarised light through the sample [1]

each enantiomer will rotate the light in a different direction [1] [2](c) 1-chlorobutane/2-chloro-2-methylpropane /1-chloro-2-methylpropane [1]

does not contain a chiral centre/no atom with 4 different groups or atoms attached [1] [2]

10 (a) It contains a carbon which has 4 different atoms/ groups attached/ has a chiral centre. [1](b)

H3C

C

H

COOH

OH HOC

H

HOOC

CH3

[2] bond angle = 109.5° [1] [3]

(c) CH3COCOOH [1]

(d) pass plane polarised light through samples of each [1]it will be rotated by the single enantiomer [1]it will be unaffected by the racemate [1]the racemate contains equal amounts of each isomer and so the two opposite rotations cancel out [1] [4]

11 Answer is B [1]

12 (a) A sample which rotates the plane of plane polarised light. [1](b)

H3C

C

H

CH2CHO

OH HOC

H

OHCH2C

CH3

[2](c) a 50:50 mixture of the 2 optical isomers [1]

one isomer rotates plane polarised light one direction and the other isomer rotates it an equal amount the other direction [1] [2]

(d) the receptor sites are stereospecific [1]

13 (a) CH3CH

2CH(OH)CH

3 [1]

(b)

H3C

C

H

CH2CH

3

OH HOC

H

H3CH

2C

CH3

[2](c) light which vibrates in one plane [1](d) pass plane polarised light into a solution of each of the isomers [1]

one isomer rotates plane polarised light to the right and the other to the left [1] [2](e)

CH

H

H

C

H

OH

H

H

H

H

C C H CH

H

H

C

H

HO

H

H

C C H

[2]

2727

4.7 ALDEHYDES AND KETONES

4.7 Aldehydes and Ketones

Naming and Physical Properties1 (a) butanal [1]

(b) propanone [1](c) hexan-2-one[1](d) 2-methylbutanal [1](e) ethanal [1](f) 3-chloropentan-2-one [1](g) 1-bromo-4-methylpentan-2-one [1](h) 4-hydroxybutanal [1](i) pentane-2,,4-dione [1](j) 3,5-dichloro-4-methylhexanal [1] [10]

2 they can both form a hydrogen bond with water [1]between lone pair of the O of carbonyl and δ+ H of water [1] [2]

3 2-methylpropanal is branched [1]intermolecular forces between the molecules are weaker [1] [2]

4 cool the collection vessel in ice [1]

5 all have similar van der Waals’ forces between their molecules, as there is a similar Mr/similar number of electrons [1]the boiling point of the aldehyde is much higher than the boiling point of the alkane due to permanent dipole attractions between the carbonyl groups on neighbouring aldehyde molecules [1]the boiling point of the alcohol is much higher than the boiling point of the aldehyde due to it also having hydrogen bonds between the hydroxyl groups on neighbouring alcohol molecules [1] [3]

6

Hydrogen bondH C

H

O

C

H

H

O

H

H

δ+

δ+

δ+

δ–

δ–

[2]

Oxidation and reduction of aldehydes and ketones7 (a) E [1]

highest priority groups are on opposite sides of the double bond [1] [2](b) warm with Tollens’ reagent [1] silver mirror [1]

warm with Fehling’s solution [1] red ppt [1] [2](c) O

H + [O]

O

OH

[1]acidified potassium dichromate(VI) [1] [2]

2828


8 Answer is D [1]

9 Answer is B [1]

10 (a)

CH

H

H

C OH

H

H [1] ethanol [1]

(b)

CC

H

OH

H

H

H

C H

H

H [1] propan-2-ol [1]

(c)

H C

OH

O

C

H

H [1] ethanoic acid [1]

(d)

CH

H

H

C

H

HO

C H

[1] propanone [1]

11 (a) A carbonyl (aldehyde) and hydroxyl [1]B carbonyl (ketone) and hydroxyl [1] [2]

(b) (i)

CC

H

H

C

H

HA

HO

H

H

C

OH

O

C

H

H [1] B no reaction [1] [2]

(ii)

A

C

H

H

CC

H

H

C

OHHO

OO

C

H

H

B

C

H

H

CCH

H

H

C

OO

C H

H

H

[1] [1] [2](c) (i) Ag+ + e– Ag [1]

(ii) Cu2+ + e– Cu+ [1]

2929


12 (a) C butanone [1] D butan-2-ol [1] [2]

(b) lithium aluminium hydride/lithium tetrahydridoaluminate(III)/lithal [2](c) butanone [1]

13 (a)

Reagent Formula of metal/ ion before test

Formula of metal/ion after test

Tollens’ reagent Ag+ Ag

Fehling’s solution Cu2+ Cu+

[1] per row [2](b) freshly prepare Fehling’s solution by mixing 1 cm3 of Fehling’s solution A with 1 cm3 of Fehling’s

solution B [1]add a few drops of the liquid to 1 cm3 of Fehling’s solution reagent in a test tube [1] warm in a hot water bath [1]remains blue [1] [4]

Nucleophilic addition reactions

14 (a)

C

H

H H

CCH

H

H

C

HO

H C

H

H H

CCH

H

H CN

C

HOH

H+ HCN

[1] 2-hydroxy-2-methylbutanenitrile [1] [2]

(b) nucleophilic addition [1]

O CH3CH

2CCH

3

H+

CN

O–

CH3CH

2CCH

3

CN

OH

C

H3C

CN–

H3CH

2C

[4](c) the carbonyl group is planar CN– can attack from either side [1]

equal amounts of each enantiomer forms (racemate/racemic mixture) [1]rotates plane polarised light equally in opposite directions and effect cancels [1]

3030


15 (a)

Z-hex-2-enal E-hex-2-enal

C

HH

CHOCH3CH

2CH

2

C C

CHOH

HCH3CH

2CH

2

C

[1] [1] [2](b) CH

3CH

2 CH

2CH=CHCHO + Br

2 CH

3CH

2CH

2CHBrCHBrCHO [1]

(c)

CH

H

H

C

H

H

H

H

H

C

H H

CN

C OHC C

or CH3CH

2CH

2CH=CHCH(CN)OH [1]

(d) acidified potassium dichromate(VI)/Fehling’s solution/ Tollens’ reagent [1] CH

3CH

2CH=CHCOOH [1] [2]

16 (a) same molecular formula (C5H

10O) but different structural formula [1]

(b) pentan-2-one [1](c) nucleophilic addition [1]

2-hydroxy-2-methylpentanenitrile [1] [2](d) product is a racemic mixture/equal amounts of enantiomers [1]

one enantiomer rotates plane polarised light one direction and the other rotates it equally the opposite direct so no overall rotation [1]product from B has no chiral centre/optically inactive [1] [3]

(e) blue solution remains [1]

17 (a) C Oδ+ δ– [1]

(b) an ion or molecule, with a lone pair of electrons, that attacks regions of low electron density [2](c) the carbon δ+ is susceptible to attack by nucleophiles seeking such an electron

deficient centre [1](d) CH

3CHO + HCN CH

3CHOHCN [1]

2-hydroxypropanenitrile [1] [2](e) nucleophilic addition [1]

O

H+

CH3

O–

CH3

OH

C

H3C

CN–

H

H C CN H C CN

[3]

3131


Reaction with 2,4-dinitrophenylhydrazine18 (a)

C

H3C

H3C

H2OC

H3C

H3C

O +

+

H2N N

NO2

H

NO2

N N

NO2

H

NO2

[3](b) propanone-2,4-dinitrophenylhydrazone [1]

orange crystals/solid [1] [2](c) indicative points:

• place (5cm3) of 2,4-dinitrophenylhydrazine solution in a test tube/suitable container and add some drops of propanone • cool (the mixture in iced water) • filter off the crystals using suction filtration. • dry by sucking air over the crystals in the Büchner or in a low temperature oven/desiccator • determine the melting point• compare to data book to identify

5–6 indicative points = [5] – [6]3–4 indicative points = [3] – [4]1–2 indicative points = [1] – [2]

(d) dissolve the impure crystals in the minimum volume of hot solvent [1]filter when hot by gravity filtration, using a hot funnel, or fluted filter paper (to remove insoluble impurities) [1]allow filtrate to cool and crystallise filter off the crystals using suction filtration [1]solid must be soluble in hot solvent, but less soluble in cold solvent [1] [4]

(e) moles of product C9H

10O

4 = 10 ÷ 238 = 0.0420

mass of propanone = 0.0420 × 58 = 2.436 gvolume of propanone = 2.436 ÷ 0.790 = 3.084 cm3

for 80% yield 3.084 × 100 ÷ 80 = 3.855cm3 = 3.9 cm3 [4](f) place some solid in a capillary tube sealed at one end/melting point tube [1]

heat slowly (using melting point apparatus) [1]record the temperature at which the solid starts and finishes melting [1]repeat and average and compare the temperature(s) with known values in a data book [1] [4]

3232


19 (a)

C

CH3CH

2

CH3CH

2

H3C

H2OC

H3C

O +

+

H2N N

NO2

H

NO2

N N

NO2

H

NO2

[3](b)

C

H

H

H2OC

O +

+

H2N N

NO2

H

NO2

N N

NO2

H

NO2

[3]

3333

4.8 CARBOXYLIC ACIDS

4.8 Carboxylic Acids

Nomenclature and physical properties1 (a) 2-chlorobutanoic acid [1]

(b) methanoic acid [1](c) ethanoic acid [1](d) propanoic acid [1](e) butanedioic acid [1](f) 3-hydroxypentanic acid [1]

2 (a) C4H

7O

2Cl [1]

(b) CH2O

2 [1]

(c) C2H

4O

2 [1]

(d) C3H

6O

2 [1]

(e) C4H

6O

4 [1]

(f) C5H

10O

3 [1]

3 (a) (i)

CC

Cl

H

C

H

H

H

H

H

C

OH

O

C

H

H [1](ii)

CC

H

C

CH3

H

H

H

H

C

OH

O

C

H

[1](iii)

CCC

OH

H

H

H

H OH

O

[1](iv)

CCC

H

H

H

OH

H OH

O

[1](b) 3-hydroxypropanoic acid [1](c) 4-methylpent-2-enoic acid [1](d) it can form hydrogen bonds [1]

between the lone pair on O and the δ+ H in water [1] [2]

3434


4 Hydrogen bonds

H C

O

O

C

H

H

H

HH

O

OH

H

δ+ δ+

δ+

δ+

δ–

δ–

δ+

[2]

5 Same relative molecular mass, and so similar van der Waals’ forces between the molecules. [1]there are 2 hydrogen bonds between 2 molecules of ethanoic acid and only one between 2 molecules of propan-1-ol/hydrogen bonding occurs between two molecules of the acid which doubles the size of the molecule and increases the van der Waals’ forces [1] [2]

Preparation of acids

6 Answer is B [1]

7 (a) CH3CH

2CHO + [O] CH

3CH

2COOH [1]

(b) CH3CH

2CH

2OH+ 2[O] CH

3CH

2COOH + H

2O [1]

(c) CH3CH

2COOC

2H

5 + H

2O CH

3CH

2COOH + C

2H

5OH [1]

(d) CH3CH

2CN + 2H

2O + HCl CH

3CH

2COOH + NH

4Cl [1]

(e) CH3CH

2CN + H

2O + NaOH CH

3CH

2COONa + NH

3 [1]

CH3CH

2COONa + HCl CH

3CH

2COOH + NaCl [1] [2]

8 (a) CH3CH

2CH

2CH

2OH + 2[O] CH

3CH

2CH

2COOH + H

2O [1]

(b) indicative content:• add concentrated sulfuric acid to water in a pear shaped/round bottomed flask • swirl the solution and cool the flask to dissipate the heat and to prevent spitting • add potassium dichromate(VI) solution and swirl the mixture• add anti-bumping granules and attach a vertical condenser• add the alcohol slowly, down the condenser, to the acidified potassium dichromate(VI) solution, • cooling the reaction flask in a water bath.• heat the mixture under reflux• distil off the acid

6–7 indicative points = [5] – [6]4–5 indicative points = [3] – [4]2–3 indicative points = [1] – [2]

Reactions of acids

9 (a) CH3COOH s CH

3COO– + H+ [1]

(b) ethanoate [1](c) Yes the ions can move and carry charge [1]

10 (a) CH3CH

2COOH + NH

3 CH

3CH

2COONH

4 [1]

smell of ammonia disappears [1] [2](b) 2CH

3COOH + Na

2CO

3 2CH

3COONa + H

2O + CO

2 [1]

solid disappears, fizzing [1] [2](c) CH

3CH

2CH

2COOH + NaOH CH

3CH

2CH

2COONa + H

2O [1]

mixture warms up [1] [2]

3535


11 (a) ammonium propanoate [1](b) sodium ethanoate and water and carbon dioxide [1](c) sodium butanoate and water [1]

12 (a) KOH + CH3COOH CH

3COOK + H

2O [1]

(b) 2CH3COOH + Na

2CO

3 2CH

3COONa + H

2O + CO

2 [1]

(c) Mg + 2CH3CH

2COOH (CH

3CH

2COO)

2Mg + H

2 [1]

(d) HCOOH + NH3 HCOONH

4 [1]

13 (a) CH3COOH + PCl

5 CH

3COCl + POCl

3 + HCl [1]

(b) HOOCCH2CH

2COOH + 2PCl

5 ClOCCH

2CH

2COCl + 2POCl

3 + 2HCl [1]

(c) CH3CH

2COOH + 4[H] CH

3CH

2CH

2OH + H

2O [1]

14 ethanoyl chloride, hydrogen chloride, phosphorus trichloride oxide [1]bubbles/misty fumes [1] [2]

15 C2H

5COONH

4 [1] C

2H

5COONa [1]

C2H

5CH

2OH [1] [3]

16 (a)

+ + +

COOH

PCl5

OH

COCl

POCl3

HClOH

[1](b)

+ +

COOH

4[H]OH

CH2OH

H2O

OH

[1]

17 (a) A acidified potassium dichromate(VI) and heat [1]B acidified potassium dichromate(VI) and heat [1]C heat with dilute hydrochloric acid [1]D LiAlH

4 in dry ether [1]

E propan-1-ol concentrated sulfuric acid (catalyst); heat under reflux [1](b) A oxidation [1]

B oxidation [1]C hydrolysis [1]D reduction [1]E condensation/esterification [1]

3636

4.9 DERIVATES OF CARBOXYLIC ACIDS

4.9 Derivates of Carboxylic Acids

Nomenclature and physical properties1

Carboxylic acid

Alcohol Ester name Ester structural formulaEster molecular

formula

ethanoic acid ethanol ethyl ethanoate CH3COOCH

2CH

3C

4H

8O

2

ethanoic acid methanol methyl ethanoate CH3COOCH

3C

3H

6O

2

propanoic acid ethanol ethyl propanoate CH3CH

2COOCH

2CH

3C

5H

10O

2

butanoic acid propan-1-ol propyl butanoate CH3CH

2CH

2COOCH

2CH

2CH

3C

7H

14O

2

[1] each row [4]

2 (a) ethyl ethanoate [1](b) propyl methanoate [1](c) ethyl propanoate [1](d) methyl methanoate [1](e) ethyl propanoate [1](f) ethyl methanoate [1]

3 (a) hydrogen bonds form [1] between the lone pair on O of the ester and the δ+ H of water [1] [2]

(b) as the length of the hydrocarbon chain increases the boiling point decreases [1]hydrocarbon chain is non-polar [1] [2]

4 (a) 2CH3CH

2COOH + CaCO

3 (CH

3CH

2COO)

2Ca + H

2O + CO2 [1]

(b) (i) CH3COOCH

3 methyl ethanoate [1]

HCOOC2H

5 ethyl methanoate [1] [2]

(ii) between propanoic acid molecules there are strong hydrogen bonds [1]between ester molecules there are dipole-dipole attractions which are weaker [1] [2]

5 Answer is C [1]

Acyl chlorides6 (a) ethanoyl chloride [1]

(b) methanoyl chloride [1](c) butanoyl chloride [1]

7 (a) CH3COCl + H

2O CH

3COOH + HCl [1]

steamy/misty fumes, mixture warms up [1] [2](b) CH

3CH

2COCl + C

2H

5OH CH

3CH

2COOC

2H

5 + HCl [1]

steamy/misty fumes, mixture warms up [1] [2](c) dip a glass rod/ stopper in concentrated ammonia solution [1]

white fumes [1] [2](d) HOCH

2CH

2CH

2OH + 2CH

3COCl H

3CCOOH

2CH

2CH

2OOCCH

3 + 2HCl [1]

3737


Preparation of esters8 Answer is D [1]

9 (a) CH3COOH + CH

3OH s H

2O + CH

3COOCH3 [1]

(b) CH3CH

2CH

2CH

2COOH + CH

3CH

2OH s H

2O + CH

3CH

2CH

2CH

2COOCH

2CH

3 [1]

(c) CH3CH

2CH

2COOH + CH

3CH

2CH

2OH s CH

3CH

2CH

2COOCH

2CH

2CH

3 + H

2O [1]

10 (a) 3-methylbutanoic acid [1] methanol [1] [2](b)

CC

H

H

H

CH3

CH3OH

CH3

H

O

C + CC

H

H

H

CH3

O

CH3

H2O

CH3

O

C +

[1](c) (i) prevents the formation of large bubbles/promotes smooth boiling [1]

(ii) catalyst [1] dehydrating agent/removes water and promotes forward reaction [1] [2]

(iii) repeated boiling and condensing of a (reaction) mixture [1](iv) sand bath/electric heater [1]

no naked flames as the organic compounds are flammable [1] [2](d) (i) distillation [1]

mixture heated in a flask with still head containing a thermometer [1]water cooled condenser connected to the still head and suitable named collecting vessel [1] [3]

(ii) stopper and shake ester with portion of sodium carbonate solution in separating funnel [1]invert and release pressure [1]allow to settle and run off layers. Discard aqueous layer/keep organic layer [1] [3]

(iii) add a spatula of anhydrous magnesium sulfate/anhydrous sodium sulfate/anhydrous calcium chloride and swirl [1] add more of the drying agent until the liquid is clear/no longer cloudy [1]decant/filter off the liquid [1] [3]

(iv) distill [1] sharp boiling point/compare to data book [1] [2]

(v) side reactions/reaction is incomplete [1]

11 Answer is B [1]

moles of alcohol = 6.0 ÷ 74 = 0.0811 Moles ester butyl propanoate = 0.0811 theoretical yield = 0.0811 × 130 = 10.5 g % yield = 7.4 ÷ 10.5 × 100 = 70%

12 (a) CH3CH

2CH

2COOH + C

2H

5OH s CH

3CH

2CH

2COOC

2H

5 + H

2O [1]

concentrated sulfuric acid [1] [2](b) CH

3CH

2CH

2COCl + C

2H

5OH CH

3CH

2CH

2COOC

2H

5 + HCl [1]

(c) no catalyst needed [1] goes to completion/higher yield [1] less purification is needed [1] [3]

3838


13 (a) A = ethene [1]B = bromoethane [1]C = propanenitrile [1]D = propanoic acid [1]E = propanoyl chloride [1] [5]

(b) Reaction 1 = addition [1] Reaction 2 = substitution [1] [2]

(c) nucleophilic substitution [1](d) Reaction 2: NaCN/KCN [1]

Reaction 3: mineral acid/HCl [1] [2](e) phosphorus pentachloride/phosphorus(V) chloride [1]

14 (a) add sodium carbonate [1]with methyl ethanoate no reaction occurs [1] with propanoic acid bubbles of gas [1] [3]

(b) warm with acidified potassium dichromate(VI) solution [1]butan-2-ol orange to green [1]2 methylpropan-2-ol shows no reaction/stays orange [1] [3]

Fats and oils

15 (a) propane-1, 2, 3-triol [1]

OH

OH

OH

C

C

H

CH

H

H

H

[1] [2](b)

+ +

(CH2)

16CH

3

3CH3(CH

2)

16COOH

H2C C

O

(CH2)

16CH

3HC C

O

O

O

(CH2)

16CH

3H

2C

3H2O

CO

H2C

HC

OH

OH

H2C OH

O

[2]

3939


(c)

++

(CH2)

14CH

3

3CH3(CH

2)

14COONa

H2C C

O

(CH2)

14CH

3HC C

O

O

O

(CH2)

14CH

3H

2C

3NaOH

CO

H2C

HC

OH

OH

H2C OH

O

[2](d) heat with methanol [1]

acid catalyst [1]

++

(CH2)

14CH

3

3CH3(CH

2)

14COOCH

3

H2C C

O

(CH2)

14CH

3HC C

O

O

O

(CH2)

14CH

3H

2C

3CH3OH

CO

H2C

HC

OH

OH

H2C OH

O

[2](e) a fuel, similar to diesel, which is made from vegetable sources [1](f) a reaction where the alkyl group of an ester is exchanged with the alkyl group of an alcohol [2]

16 (a)

OH

OH

OH

C

C

H

CH

H

H

H

[1] glycerol [1] [2]

4040


(b)

(CH2)

7CH=CHCH

2CH=CH(CH

2)

4CH

3

(CH2)

7CH=CHCH

2CH=CH(CH

2)

4CH

3

(CH2)

7CH=CHCH

2CH=CH(CH

2)

4CH

3

H2C

H2C

HC

C

O

C

O

O

O

CO

O

[1](c) breaking up molecules by reaction with water [1]

aqueous sodium hydroxide [1] [2]

17 (a) (i)

++

(CH2)

16CH

3

3CH3(CH

2)

16COOK

C

C

C

H

H

H

H

C

OH

(CH2)

16CH

3C

O

O

O

(CH2)

16CH

3

3KOH

CO

C

C

C

H

H

H

H

H

OH

OH

OH

O

[2](ii) propane-1,2,3-triol [1]

(b) (i) C19

H38

O2 [1]

(ii) 2C19

H38

O2 +55O

2 38CO

2 + 38H

2O [2]

18 (a)

++

CH2OCOC

15H

31

CH2OCOC

15H

31

CHOCOC15

H31

CH2OH

CH2OH

CHOH 3C15

H31

COOH 3H2O

[1](b) It has no carbon-carbon double bonds/alkyl group is C

nH

2n+1 [1]

(c)

+ +

CH2OCOC

15H

31

CH2OCOC

15H

31

CHOCOC15

H31

CH2OH

CH2OH

CHOH 3C15

H31

COOCH3

3CH3OH

[1](d) C

15H

31COOCH

3 [1]

4141


19 (a)

+

+

CH2OOC(CH

2)

7CH=CHCH

2CH=CH(CH

2)

4CH

3

CH2OOC(CH

2)

7CH=CHCH

2CH=CH(CH

2)

4CH

3

CHOOC(CH2)

7CH=CHCH

2CH=CH(CH

2)

4CH

3

CH2OH

CH2OH

CHOH

3H2O

3CH3(CH

2)

4CH=CHCH

2CH=CH(CH

2)

7COOH

[2](b) contains carbon–carbon double bonds [1](c) CH

3OOC(CH

2)

7CH=CHCH

2CH=CH(CH

2)

4CH

3 [1]

(d)

+

CH2OOCC

17H

31

CHOOCC17

H33

CH2OOCC

17H

29

3CH3OH

+ C17

H31

COOCH3

CH2OH

CH2OH

CHOH + C17

H33

COOCH3

+ C17

H29

COOCH3

[2]

20 (a) can form hydrogen bonds between the δ+ H atom and lone pair of O of water [1]the long hydrocarbon chain in dodecanoic acid is non-polar [1] [2]

(b) solid disappears and fizzing [1](c) place some solid in a capillary tube sealed at one end [1]

heat slowly (using melting point apparatus) [1]record the temperature at which the solid starts and finishes melting [1]repeat and average the temperatures [1]if pure melting point is sharp/melting point should be same as that in data book [1] [4]

(d) C11

H23

COOH + 4[H] C11

H23

CH2OH + H

2O [1]

lithal/lithium tetrahydridoaluminate(III) [1] [2](e)

++

CH2OOCC

11H

23

CH2OOCC

11H

23

CHOOCC11

H23

CH2OH

CH2OH

CHOH 3C11

H23

COOH 3H2O

[1]

4242

4.10 AROMATIC CHEMISTRY

4.10 Aromatic Chemistry

Structure of benzene1 (a) molecular formula = C

6H

6 empirical formula = CH [1]

(b) planar [1]120° [1] [2]

(c) 6 [1](d) (i) methylbenzene [1]

(ii) nitrobenzene [1] (iii) 1,2-dichlorobenzene [1] (iv) ethylmethylbenzene [1] (v) phenylpropanone [1] (vi) 1-bromo-4-chlorobenzene [1] [6]

(e) each carbon also has one electron in a p orbital. The p orbitals overlap sideways to form a ring of charge above and below the plane [1] the 6 pi electrons spread across this ring and are delocalised [1] [2]

(f) CH3

[1](g) 12 [1] (h) pi electrons are spread over several atoms [1]

2 (a) bromine water [1](b) with benzene it remains orange; with cyclohexene it changes from orange to colourless [1]

3 Answer is C [1]

4 indicative content

bonding • each C has three (covalent) sigma bonds • one electron per carbon in a p orbital• p orbitals overalp overlap • delocalisation of electrons to form a pi ring

shape• planar • hexagon/6 carbon ring

type of reaction • addition causes delocalised ring to break • substitution maintains the stability of the ring

7–8 indicative points = [5] – [6]5–6 indicative points = [3] – [4]2–4 indicative points = [1] – [2] [6]

4343


5 Answer is A [1]

6 Answer is D [1]

7 Answer is A [1]

8 C=C in ethene is shorter [1]C-C bond in benzene is intermediate between that of a double and a single bond [1] [2]

Electrophilic substitution

9 Answer is C [1]

10 (a) electrophilic substitution [1](b) HNO

3 + 2H

2SO

4 NO

2

+ + 2HSO4

– + H3O+ [1]

CH3 CH3

NO2

+ ++

CH3

NO2

H

H+

NO2

[3]HSO

4

– + H+ H2SO

4 [1] [5]

11 (a) ester [1] (b) (i) COOCH

3COOCH

3

HNO3

+ + H2O

NO2

[1](ii) theoretical yield = (5.43 ÷ 60) × 100 = 9.05g

theoretical yield = 9.05 ÷ 181 = 0.05 moles0.05 moles of methyl benzoate minimum mass of methyl benzoate = 0.05 × 136 = 6.8 g [3]

(iii) concentrated nitric acid and concentrated sulfuric acid [1]HNO

3 + 2H

2SO

4 NO

2

+ + 2HSO4

– + H3O+ [1] [2]

(iv) nitronium [1](v) COOCH

3 COOCH3

NO2

+ ++

COOCH3

NO2

H H+

NO2

[3] electrophilic substitution [1] [4]

(d) (i) to remove impurities/purify the product [1] to ensure the hot solution would be saturated/crystals would form on cooling/minimise loss of product [1] [2]

(ii) Büchner funnel Büchner flask suction pump 3 correct = [2] marks, 2 correct = [1] mark [2]

4444


(iii) to remove soluble impurities [1](iv) suck air over crystals in the Büchner funnel/low temperature oven/desiccator [1](v) sharp melting point [1]

matches the data book value [1] [2](vi) water would cause the melting point to be lower [1]

12 (a) O

Cl

[1](b) (i) hydrogen atom on the benzene is replaced by an acyl group [1]

(ii) aluminium chloride [1]

CC

H

CH3

Cl

O

AlCl3

CH3

CC

H

CH3

O

CH3

+ AlCl4

+– +

[1]AlCl

4

– + H+ → AlCl3 + HCl [1] [3]

(iii) electrophilic substitution [1]

++ H+CC

H

CH3

O

CH3

+

CH C

H

CH3

O

CH3

+

CC

H

CH3

O CH3

[3]

15 Answer is C [1]

16 (a) phenylpropanone [1](b) electrophilic substitution [1]

AlCl3 CH

3CH

2COCl [1] [2]

(c) lithal/lithium tetrahydridoaluminate(III)/LiAlH4 [1]

secondary [1] [2](d) C as it has a chiral centre attached to the benzene ring [1]

17 Answer is B [1]

4545

Unit A2 2:

Analytical, Transition Metals, Electrochemistry and Organic

Nitrogen Chemistry

Answers

4646

5.1 MASS SPECTROMETRY

5.1 Mass Spectrometry

1 (a) ethene and carbon monoxide [1](b) M

r C

2H

4 = (2 × 12.0000) + (4 × 1.0078) = 28.0312 [1]

Mr CO = (12.0000 + 15.9949) = 27.9949 [1]

C2H

4 value matches the molecular ion peak [1] [3]

2 Answer is D [1]

(12.0000 × 10) + (14 × 1.00788 ) + 15.9949

3 (a) (relative) abundance [1](b) mass/charge ratio [1](c) 46 [1](d) a peak produced by an ion formed by the removal of one electron from a molecule [2](e) peak of greatest abundance in a mass spectrum [1]

31 [1] [2]

4 Answer is D [1]

5 Answer is C [1]

6 (a) 29 [1](b) M+1 peak [1]

due to presence of one 13C atom [1] [2](c) (i) a positively charged ion produced when the molecular ion breaks apart [1]

(ii) 31 = CH3O+ [1]

57 = CH3CH

2CO+ [1] [2]

7 any two possible fragments, e.g. CH

3+ at m/z 15

C2H

5+ at m/z 29

C3H

7+ at m/z 43

C4H

9+ at m/z 57

OH+ = 17

CH2OH+ 31 or CH

3CH

2CH

2CH

2O+ at m/z = 73 [2]

8 from IR absorption, A contains O–H/alcohol [1]C H O 70.59/12 13.72 15.69/16 5.8825 13.72 0.9806 [1]empirical formula = C

6H

14O [1]

Mr = 102 empirical formula is the molecular formulae C

6H

14O = 102 [1]

OH

OHOH

[2] for 3 structures [1] for 2 structures [6]

4747


9 (a) butanedioic acid [1](b) (i) 56 [1]

(ii) a peak produced by a molecular ion with an increased mass due to the presence of one carbon-13 atom [1]

(iii) 119 [1](iv) 45 COOH+ [1]

73 HOOCCH2CH

2+ [1] [2]

10 (a) 46 [1](b) a peak produced by a molecular ion with an increased mass due to the presence of one

carbon-13 atom [1](c) CH

2OH+ [1]

(d) 15 CH3+ [1]

17 OH+ [1]

29 CH3CH

2+ [1] [3]

(e) C2H

5OH+ [1]

11 (a) 31 CH2OH+ [1]

57 C4H

9+ [1] [2]

(b) 59 (CH3)

2 COH+ [1]

74 C4H

9OH+ [1] [2]

(c) spectrum A [1]C

4H

9+ peak present/CH

2OH+ peak present [1] [2]

12 (a)

butan-1-ol ethoxyethane

CH

H

H

C

HH

HH

C

H

H

C OH CH

H

H

C

H

H

O

H

H

C

H

H

C H

(b) CH2OH+ [1]

(c) 29 C2H

5+ [1]

45 C2H

5+ [1]

57 C2H

5OCH

2+ [1] [3]

(d) spectrum 2 [1]ethoxyethane cannot have a peak at 31 [1] [2]

4848


13

Compound m/z Fragment ion

butanone 29 C2H

5+

butanone 43 CH3CO+

ethyl methanoate 29 C2H

5+ or HCO+

ethyl methanoate 45 C2H

5O+ or HCOO+

ethyl methanoate 74 HCOOC2H

5+

CH3CH

2CONH

257 CH

3CH

2CO+

CH3CH

2CONH

244 CONH

2+

[1] per row [7]

14 (a) due to the isotopes of Cl-35 and Cl-37 [1](b) 35 35Cl+ [1]

63 CH3CH37Cl+ [1] [2]

4949

5.2 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

5.2 Nuclear Magnetic Resonance Spectroscopy

1 A [1]

2 (a) 1,4-dichloro-2,2-dimethylbutane [1](b) 4 [1](c) triplet [1]

it has two hydrogens on adjacent carbon (n+ 1) [1] [2](d) a high resolution spectrum is one which shows the spin-spin splitting pattern and a low

resolution nmr spectrum does not [1]

3 (a) 2 [1](b) triplet [1]

as it has two hydrogens on adjacent carbon [1] [2](c) area under the peaks which is proportional to the relative number of hydrogens atoms

in each environment [1](d) 3:2 [1]

4 (a) CDCl3/CD

2Cl

2/C

6D

6/CCl

4/tetramethylsilane [1]

(b) CH3CH

2CH

2CH

2OH 5 [1]

(CH3)

3COH 2 [1] [2]

(c)

CH3

Si

CH3

CH3

CH3

[1]tetramethylsilane [1] [2]

(d) any two from:sharp strong signal as all 12 hydrogens are in the same environment [1]unreactive/volatile liquid easily removed and recovered [1]produces a peak to the extreme right away from most other peaks [1] [2]

5 two peaks as 2 chemically equivalent environments [1] no splitting/singlets as no adjacent protons [1]integration is 1:1 same ratio of each type of hydrogen [1] [3]

6

chemical shift / ppm5 3 14 2

1

3

0

2 singlets [1]integration 1:3 [1] [2]

7 Answer is C [1]

8 Answer is A [1]

9 Answer is C [1]

5050


10 (a)

CH

H

H

C

O

O

C

H

H C H

H

H[1]

CH

H

H O

O

C

C H

H

H

C

H

H[1] [2]

(b)

CH3C

CH3

CH3

C

[1]

O

O CH3

CH3C

CH3

CH3

C

[1]

O

C CH3

[2]

11 (a) 3 peaks [1] quartet triplet singlet [1]integration 3:2:1 [1] [3]

(b) 2 peaks [1] quartet triplet [1] integration 3:2 [1] [3]

12 indicative pointsinfrared • absorption at 1000-1300 shows C–O present • absorption at 1650-1800 shows C=O present • absorption at 2850-3000 shows C–H presentnmr• there are 3 peaks which indicates 3 different environments of chemically equivalent hydrogen atoms• the integration ratio = 0.8:1.2:1.2 – simplest whole number ratio is 2:3:3 • the quartet at chemical shift 4 must be due to a CH

2 group with an adjacent CH

3

• the singlet at chemical shift 2 must be due to a CH3 with no H atoms bonded to adjacent

carbon atoms• the quartet at chemical shift 1.2 must be due to a CH

3 with an adjacent CH

2

• quartet and triplet suggest CH2CH

3 group

• structure:

CH

H

H

C O

O

C H

H

H

C

H

H


13 Answer is B [1]

5151


14 (a) due to TMS standard [1](b) proton(s) adjacent to CH

2/2H atoms bonded to adjacent carbon [1]

split into n+1 (2+1=) 3 peaks [1] [2] (c) proton(s) adjacent to a methyl/CH

3 group/3H atoms bonded to adjacent carbon [1]

split into n+1 (3+1 =) 4 peaks [1] [2](d) protons adjacent to nitrogen which is deshielding/withdraws electrons [1](e) an extra peak [1]

integration would show 3:2:1:3 [1] some/all the peaks would change their chemical shift [1] max [2]

(f) a signal which appears as 3 lines in the approximate intensity ratio 1:2:1 [2]

5252

5.3 VOLUMETRIC ANALYSIS

5.3 Volumetric Analysis

Iodine-thiosulfate titrations1 Answer is C [1]

2 (a) blue-black to colourless [1](b) H

2O

2 + 2H+ + 2I– I

2 + 2H

2O [2]

(c) using titrations 2 and 4 [1] concordant results/within 0.1 cm3 [1]given to 3 significant figures [1] [3]

(d) moles of thiosulfate = 17.9 × 0.245

100 = 4.3855 × 10–3

moles of I2 =

4.3855 × 10−3

2 = 2.19275 × 10–3

moles of H2O

2 = 2.19275 × 10–3

concentration of H2O

2 = 2.19275 × 10–3 × 40 = 0.0877 (mol dm–3) [4]

(e) ratio H2O

2:H

2SO

4 = 1:1

moles of H2SO

4 required = 2.19275 ×10–3

volume of 1.50 mol dm–3 H2SO

4 required =

2.19275 × 10−3 × 10001.50

= 1.46 cm3 [3]

(f) react with H2O

2 and required to form triiodide ion or /allow iodine to remain in solution [1]

3 (a) indicative points• weigh out accurately 2.89 g of solid KIO

3

• dissolve in suitable volume of deionised water• transfer to a 250 cm3 volumetric flask• wash all apparatus with deionised water with washings going into volumetric flask• add deionised water until the bottom of the meniscus is on the line/mark• stopper and invert flask to mix


(b) (i) oxidation: 2I– I2 + 2e– [1]

reduction: 2IO3– + 12H+ + 10e– I

2 + 6H

2O [1] [2]

(ii) oxidation: 2S2O

32– S

4O

62– + 2e– [1]

reduction: I2 + 2e– 2I– [1] [2]

(iii) it forms an irreversible complex with the iodine when large amount of iodine present [1]

(c) moles of IO3– =

25.0 × 0.05401000

= 1.35 × 10–3

moles of S2O

32– required = 1.35 × 10–3 × 6 = 8.10 × 10–3

volume of Na2S

2O

3 required =

8.10 × 10−3 × 10000.445

= 18.2 cm3 [3]

5353


4 (a) (i) colourless to brown [1]iodine produced in reaction between iodate(V) ions and iodide ions [1]IO

3– + 6H+ + 5I– 3I

2 + 3H

2O [1] [3]

(ii) chloride in HCl could be oxidised [1]sulfur in H

2SO

4 in highest oxidation state [1] [2]

(b) (i) starch [1]blue-black to colourless [1] [2]

(ii) 2S2O

32– + I

2 S

4O

62– + 2I– [1]

(iii) moles of IO3– =

25.0 × 0.01501000

= 3.75 × 10–4

moles of S2O

32– = 3.75 × 10–6 × 6 = 2.25 × 10–3

moles of S2O

32– in 1.00 dm3 =

2.25 × 10−3

22.5 × 1000 = 0.1

RFM of Na2S

2O

3.xH

2O =

24.80.1

= 248

xH2O = 248 – 158 = 90

x = 9018

= 5 [5]

5 Answer is C [1]

moles of S2O

32– =

12.5 × 0.1001000

= 1.25 × 10–3

moles of H2O

2 =

1.25 × 10−3

2 = 6.25 × 10–4

molarity of diluted H2O

2 solution = 6.25 × 10–4 × 40 = 0.0250 M

molarity of original solution = 0.025 × 25 = 0.625 M

6 (a)

Rough titration

Titration 1 Titration 2 Titration 3

Initial burette reading /cm3

Final burette reading /cm3 24.20 42.60

Titre /cm3 12.70 11.85

(b) estimate the end point/faster subsequent accurate titrations [1](c) lower concentration of sodium thiosulfate [1]

higher titre [1] [2](d) 11.55 cm3 [1]

titres from titrations 1 and 3 as they are concordant [1] [2](e) 3

26 [1]

5454


(f) moles of S2O

32– =

11.55 × 0.6401000

= 7.392 × 10–3

moles of IO3– in 25.0 cm3 =

7.392 × 10−3

6 = 1.232 × 10–3

moles of IO3– in 250.0 cm3 = 1.232 × 10–3 × 10 = 0.01232

RFM of MIO3 =

2.440.01232

= 198

RAM of M = 198 – 175 = 23

Identity of M = Na/sodium [5]

7 (a) Cu + 4H+ + 2NO3– Cu2+ + 2NO

2 + 2H

2O [1]

(b) 2NO3– + 4H+ + 2I– I

2 + 2NO

2 + 2H

2O [1]

(c) 2Cu2+ + 4I– 2CuI + I2 [1]

(d) creamy white colour is due to precipitate of CuI present [1]

(e) moles of S2O

32– =

14.2 × 0.1251000

= 1.775 × 10–3

moles of I2 =

1.775 × 10−3

2 = 8.875 × 10–4

moles of Cu2+ = moles of Cu = 8.875 × 10–4 × 2 = 1.775 × 10–3

mass of Cu = 1.775 × 10–3 × 64 = 0.1136 g

% Cu = 0.11360.245

× 100 = 46.4 % [5]

Manganate(VII) titrations8 Answer is B [1]

9 (a) graduated pipette with pi pump [1]rinse pipette with iron(II) sulfate solution [1]pipette 10.0 cm3 of the solution into a 250 cm3 volumetric flask [1]add deionised water until the bottom of the meniscus is on the line or mark [1]stopper flask and invert to mix [1] [5]

(b)

Initial burette reading / cm3

Final burette reading / cm3 Titre / cm3

Rough titration 0.0 22.5 22.5

Accurate titration 1 22.5 44.1 21.6

Accurate titration 2 20.7 42.4 21.7

[1](ii) 21.7 cm3 (allow 21.65 cm3) [1]

5555


(iii) moles of = 21.7 × 0.0175

1000 = 3.7975 × 10–4

moles of Fe2+ in 25.0 cm3 = 3.7975 × 10–4 × 5 = 1.899 × 10–3

moles of Fe2+ in 250.0 cm3 = 0.01899

moles of Fe2+ in original 10.0 cm3 = 0.01899

molarity of FeSO4 solution = 0.01899 × 100 = 1.90 M [5]

(iv) concentration in g dm–3 = 1.90 × 152 = 289 g dm–3

solubility in g/100 g water = 28.9 g/100 g water [2]

10 (a) 5Fe2+ + MnO4– + 8H+ 5Fe3+ + Mn2+ + 4H

2O [1]

(b) colourless to pink [1]

(c) moles of MnO4– =

17.85 × 0.01501000

= 2.6775 × 10–4

moles of Fe2+ = 2.6775 × 10–4 × 5 = 1.339 × 10–3

moles of Fe2+ in 250 cm3 = 0.01339

mass of FeSO4 = 0.01339 × 152 = 2.035 g

% Fe2SO

4 =

2.0352.75

× 100 = 74 % [5]

11 (a) 5Fe2+ + MnO4– + 8H+ 5Fe3+ + Mn2+ + 4H

2O [1]

(b) 5C2O

42– + 2MnO

4– + 16H+ 10CO

2 + 2Mn2+ + 8H

2O [1]

(c) 5FeC2O

4 + 3MnO

4– + 24H+ 5Fe3+ + 10CO

2 + 3Mn2+ + 12H

2O [2]

(d) (i) purple colour of [1]last drop of MnO

4– added not decolourised [1] [2]

(ii) moles of MnO4– =

17.7 × 0.001201000

= 2.124 × 10–5

moles of FeC2O

4 in 25 cm3 =

2.124 × 10−5

3 × 5 = 3.54 × 10–5

[FeC2O

4] = 0.001416 mol dm–3

Concentration of FeC2O

4 = 0.001416 × 144

= 0.204 g dm–3 = 204 mg dm–3 [4]

12 (a) moles of MnO4– =

20.8 × 0.01301000

= 2.704 × 10–4

moles of Fe2+ in 25 cm3 = 2.704 × 10–4 × 5 = 1.352 × 10–3

moles of Fe2+ in 1 dm3 = 1.352 × 10–3 × 40 = 0.05408 mol dm–3

moles of (NH4)

2 Fe(SO

4)

2.xH

2O = 0.05408

RFM = 19.6

0.05408 = 362.4

RFM of (NH4)

2Fe(SO

4)

2 = 284

RFM due to xH2O = 362.4 – 284 = 78.4

x = 78.418

= 4.4 [5]

(b) average degree of hydration across the sample [1]

5656


(c) moles of (NH4)

2Fe(SO

4)

2.6H

2O =

19.6392

= 0.05

moles of Fe2+ in 25.0 cm3 = 0.0540

= 1.25 × 10–3

moles of MnO4– required =

1.25 × 10−3

5 = 2.5 × 10–4

volume of MnO4– required =

2.5 × 10−4 × 10000.0130

= 19.2 cm3 [4]

Back titrations13 (a) colourless to pink [1]

(b) Mg + 2HCl MgCl2 + H

2 [1]

2Al + 6HCl 2AlCl3 + 3H

2 [1]

Ca + 2HCl CaCl2 + H

2 [1]

Zn + 2HCl ZnCl2 + H

2 [1] [4]

(c)

Metal Mean titre / cm3 Percentage of metal in sample / %

Magnesium 19.9 78.4 [1]

Aluminium 12.0 80.1 [1]

Calcium 29.1 [1] 75.2

Zinc 32.8 [1] 86.7

Method for Mg

moles of NaOH = moles of HCl in 25 cm3 = 19.9 × 0.075

1000 = 1.4925 × 10–3

moles of HCl in 1 dm3 = 1.4925 × 10–3 × 40 = 0.0597

moles of HCl added initially = 50.0 × 2.50

1000 = 0.125

moles of HCl which reacted with the metal = 0.125 – 0.0597 = 0.0653

moles of metal = 0.0653

2 = 0.03265

mass of metal = 0.03265 × 24 = 0.7836 g

% Mg = 0.7836

1.00 × 100 = 78.4 %

Method for Al

moles of NaOH = moles of HCl in 25 cm3 = 12.0 × 0.075

1000 = 9.00 × 10–4

moles of HCl in 1 dm3 = 1.4925 × 10–4 × 40 = 0.0360


1000 = 0.125

moles of HCl which reacted with the metal = 0.125 – 0.0360 = 0.0890


3 = 0.02967

mass of metal = 0.02967 × 27 = 0.801 g

% Mg = 0.8011.00

× 100 = 80.1 %

5757


Method for Camass of metal = 0.752 g


40 = 0.0188

moles of HCl which reacted with metal = 0.188 × 2 = 0.0376


1000 = 0.125

moles of HCl left over = 0.125 – 0.0376 = 0.0874

moles of HCl in 25 cm3 = moles of NaOH required = 0.0874

40 = 2.185 × 10–3

volume of NaOH = 2.185 × 10−3 × 1000

0.0750 = 29.1 cm3

Method for Znmass of metal = 0.867 g


65 = 0.0133

moles of HCl which reacted with metal = 0.133 × 2 = 0.0266


1000 = 0.125

moles of HCl left over = 0.125 – 0.0266 = 0.0984


40 = 2.46 × 10–3

volume of NaOH = 2.46 × 10−3 × 1000

0.0750 = 32.8 cm3

(d) calcium also reacts with water [1]

14 (a) CaCO3 + 2HCl CaCl

2 + CO

2 + H

2O [1]

NaOH + HCl NaCl + H2O [1] [2]

(b) colourless to pink [1](c) 13.9 cm3 [1] (allow 13.85)

titrations 1 and 3 are concordant [1] [2]

(d) moles NaOH = moles of HCl in 25 cm3 = 13.9 × 0.350

1000 = 4.865 × 10–3

moles HCl in 250 cm3 = 4.865 × 10–3 × 10 = 0.04865


1000 = 0.075

moles of HCl which reacted with CaCO3 = 0.075 – 0.04865 = 0.02635

moles of CaCO3 =

0.026352

= 0.013175

mass of CaCO3 = 0.013175 × 100 = 1.3175 g

mass of CaCO3 in 1 tablet =

1.31755

= 0.2635 g = 264 mg [5]

5858


15 moles NaOH = moles of HCl in 25 cm3 = 15.6 × 0.100

1000 = 1.56 × 10–3

moles HCl in 250 cm3 = 1.56 × 10–3 × 10 = 0.0156


1000 = 0.0500


moles of CaCO3 =

0.03442

= 0.0172

mass of CaCO3 = 0.0172 × 100 = 1.72 g


1.722.00

= 86 % [5]

16 (a) Ba(OH)2 + 2HCl BaCl

2 + 2H

2O [1]

(b) moles NaOH = moles of HCl in 25 cm3 = 27.95 × 0.500

1000 = 0.013975


1000 = 0.0375

moles of HCl which reacted with Ba(OH)2= 0.0375 – 0.013975 = 0.023525

moles of Ba(OH)2 =

0.0235252

= 0.0117625

mass of Ba(OH)2 = 0.0117625 × 171 = 2.011 g

mass of Ba(OH)2 in 1 tablet =

2.0112.25

= 89.4 % [5]

17 (a) 2Cr + 3O2 Cr

2O

3 [1]

(b) Cr2O

3 + 6HCl 2CrCl

3 + 3H

2O [1]

(c) colourless to green [1](d) green colour diluted so it does not interfere with the colour change of the indicator [1](e) mass of Cr which reacts = 4.046 g

moles of Cr which reacts = 4.046

52 = 0.0778

moles of Cr2O

3 formed =

0.077812

= 0.0389

moles of HCl which react = 0.0389 × 6 = 0.2334


1000 = 0.350

moles of HCl remaining = 0.350 – 0.2334 = 0.1166


80 = 1.4575 × 10–3


0.145 = 10.1 cm3 [5]

5959


(f) moles of NaOH = moles of HCl in 25 cm3 = 17.4 × 0.145

1000 = 2.523 × 10–3

moles of HCl in 2 dm3 = 2.523 × 10–3 × 80 = 0.20184


1000 = 0.350

moles of HCl which reacted with Cr2O

3 = 0.350 – 0.20184 = 0.14816

moles of Cr2O

3 =

0.148166

= 0.02469

moles of Cr = 0.02469 × 2 = 0.04938

mass of Cr which reacted = 0.04938 × 52 = 2.568 g

% of Cr which reacted = 2.5684.25

× 100 = 60.4 % [5]

6060

5.4 CHROMATOGRAPHY

5.4 Chromatography

1 Answer is D [1]

2 (a)

SpotDistance moved by

spot in solvent 1 / mmDistance moved by

spot in solvent 2 / mmRf value solvent 1 Rf value solvent 2

W 30 1630

122 = 0.246

1687

= 0.184

X 72 2072

122 = 0.590

2087

= 0.230

Y 90 4390

122 = 0.738

4387

= 0.494

Z 40 8040

122 = 0.328

8087

= 0.920

[2] per column [8]

(b) indicative points• draw two base lines in pencil close to edges of paper• spot sample onto the intersection of the lines using a capillary tube• allow to dry and repeat to make spot concentrated• place paper in tank containing a small amount of solvent and cover with lid• allow solvent to run up over the spots until almost at top of the paper• remove and mark position of solvent front; allow to dry• rotate the paper through 90 ° and repeat run in second solvent


(c) solvent 1 more polar than solvent 2 [1]Y greater R

f than Z in solvent 1 / Z greater R

f than Y in solvent 2 [1] [2]

3

Chromatography Mobile phase Stationary phase

Gas-liquid chromatographyinert carrier gas/

nitrogen/helium etc [1]film of liquid on solid

support

TLC solvent [1] silica/alumina [1]

Paper chromatography solvent [1]water bonded to

cellulose on the paper [4]

6161

5.4 CHROMATOGRAPHY

4 (a) pencil does not run in the solvent/does not interfere with the results [1](b)

Dis

tanc

e / m

m A

Solvent front

D

B

C

0

10

20

30

40

50

60

70

solvent front marked at 65 mmspot A at 40 mmspot B at 12.5 mmspot C at 57.5 mmspot D at 20 mm [3]

(c) dye C [1]runs furthest in non-polar solvent [1] [2]

5 (a) avoid contamination of TLC plate from hands [1](b) not to dissolve the sample/remain below the sample [1](c) provide better separation of components [1](d) toxic solvent [1](e) compounds are colourless so position viewed under UV [1]

(f) (i) Rf =

682

= 0.0732 [2]

(ii) moves the least [1]more attracted to polar stationary phase [1] [2]

(iii) provide better separation of the components/prevent spots overlapping [1]

6 (a) stationary phase: liquid/solvent on solid support [1]mobile phase: inert carrier gas [1]sample enters as a gas [1]components partition between liquid and gas phase [1] [4]

(b) some components may have the same retention time under these conditions [1](c) each component analysed by mass spectrometry [1]

identified as pure substance/mixture by comparison with database of known spectra [1] [2](d) total integration = 8.3

% N = 0.78.3

× 100 = 8.4 % [2]

6262

5.4 CHROMATOGRAPHY

7 (a) mobile phase = solvent [1]stationary phase = silica [1] [2]

(b) amino acid are colourless and ninhydrin stains them [1](c) UV/iodine [1](d) measure the distance moved by the spot to the centre of the spot [1]

measure the distance moved by the solvent [1]both from base line/origin [1]

Rf =

distance moved by spotdistance moved by solvent

[1] [4]

(e)

Solvent front

Origin

lys

cysval

tyrphe

lys at approximately 10 from base linecys at approximately 23 from base lineval at approximately 27 from base linetyr at approximately 34 from base linephe at approximately 39 from base line [3]

(f) lysine most polar and phenylalanine least polar [1] solvent relatively non-polar and silica stationary phase polar [1]lysine binds to stationary phase/less soluble in solvent but phenylalanine more soluble in solvent [1] [3]

6363

5.5 TRANSITION METALS

5.5 Transition Metals

General properties of transition metals and complexes1 Answer is B [1]

2 Answer is A [1]

3 Answer is C [1]

4 Answer is B [1]

5 (a)

Ni

OH2

OH2

2+OH2

OH2

H2O

H2O

[1]octahedral [1] [2]

(b) [Ni(H2NCH

2CH

2NH

2)

3]2+ + edta4– [Ni(edta)]2– + 3H

2NCH

2CH

2NH

2 [1]

(c) 1,2-diaminoethane [1](d) 6 [1](e) +3 [1](f) (i) same number and type of co-ordinate/covalent bonds broken and formed [1]

Ni—N co-ordinate/covalent bonds [1] [2](ii) increase in entropy [1]

4 species in solution on left and 7 on right [1] [2](g) (i) two chloro ligands beside each other [1]

(ii)

Co

NH3

Cl

+NH3

NH3

Cl

H3N

[1]

6 (a) co-ordination number changes from 6 to 4 [1]chloro ligands too large to have 6 around the central ion [1] [2]

(b) (i) green solution formed [1](ii) NiSO

4 + 6H

2O [Ni(H

2O)

6]2+ + SO

42– [1]

(iii) Ag+(aq) + Cl–(aq) AgCl(s) [2]

(iv) moles of NiSO4.6H

2O = moles of [Ni(H

2O)

6]2+ in 100 cm3 =

3.30263

= 0.01255

moles of [Ni(H2O)

6]2+ in 10 cm3 = 1.255 × 10–3

moles of HCl added = moles of Cl– added = 10 × 1.5

1000 = 0.015

moles of Cl– which reacted = 1.255 × 10–3 × 4 = 0.00502

moles of Cl– remaining = 0.015 – 0.00502 = 9.98 × 10–3

moles of AgNO3 required = 9.98 × 10–3

volume of AgNO3 =

9.98 × 10−3 × 10000.25

= 40 cm3 [5]

(v) white (precipitate) [1]moles of AgCl formed = 9.98 × 10–3

mass of AgCl formed = 9.98 × 10–3 × 143.5 = 1.432 g = 1400 mg [2] [3]

6464


7 (a) a ligand which uses two lone pairs of electrons to form two co-ordinate bonds with a central metal atom or ion in a complex [2]

(b) aminoethanoate (ion) [1](c) [Cu(H

2NCH

2COO)

3]– [1]

(d) –

OFe

NH2

O

NH2 NH

2

CH2

H2

O

CH2

C

C

C

CO

O

O

[1]

8 (a) [Fe(H2O)

6]3+ + 2C

2O

42– [Fe(H

2O)

2(C

2O

4)

2]– + 4H

2O [1]

(b) ethanedioate (ion) [1](c) large ion so three do not fit around the central ion [1](d) same number and type of co-ordinate/covalent bonds broken and formed [1]

Fe—O co-ordinate/covalent bonds [1]increase in entropy [1]3 species in solution on left and 5 on right [1] [4]

(e) 6 [1]

9 (a) hexaaquacopper(II) (ion) [1](b) N less electronegative than O [1]

lone pair more readily donated from N/more stable complex formed [1] [2](c) high conc of Cl– [1] pushes equilibrium to right hand side [1] [2](d) blue solution [1] changes to a deep/dark blue solution [1] [2](e)

Complex Shape Bond angle / ° Co-ordination number

[Cu(H2O)

6]2+ octahedral 90 6

[CuCl4]2– tetrahedral 109.5 4

[1] per row [2]

10 (a) element which forms at least one stable ion with a partially filled d-subshell [1](b) a central metal atom or ion with ligands bonded by co-ordinate bonds [1](c) (i) oxidation state of Fe changes from +2 to +3 = oxidation [1]

oxidation state of Ir changes from +4 to +3 = reduction [1]oxidation and reduction occur in the same reaction = redox [1] [3]

(ii) K2IrCl

6 [1]

(iii) hexachloroiridate(III) (ion) [1]

6565


(d) (i) [AuCl2]– = linear [1]

[AuCl4]– = tetrahedral [1] [2]

(ii) disproportionation = oxidation and reduction of the same element in the same reaction [1]oxidation state of Au changes from +1 in [AuCl

2]– to 0 in Au = reduction [1]

oxidation state of Au changes from +1 in [AuCl2]– to +3 in [AuCl

4]– = oxidation [1] [3]

11 (a) 2KCl + PdCl2 K

2[PdCl

4] [1]

(b)

PdCl

2– 2–

90°

ClCl

ClNi

ClCl

109.5°

[1] [1]

Cl

Cl

[2](c) (i) [NiCl

4]2– + 3H

2NCH

2CH

2NH

2 [Ni(H

2NCH

2CH

2NH

2)

3]2+ + 4Cl– [2]

(ii) [NiCl4]2– + edta4– [Ni(edta)]2– + 4Cl– [2]

(iii)

Ligand Monodentate Bidentate HexadentateCN–

H2O

Cl–

1,2-diaminoethane

edta4–

all correct [2]; one error [1]; more than one error [0] [2]

Colour of transition metal ions and their identification12 Answer is B [1]

13 Answer is D [1]

14 (a) CrCl3.6H

2O [1]

(b) dissolve in water and add dilute nitric acid/dissolve in dilute nitric acid [1]add silver nitrate solution [1]white precipitate [1]add aqueous ammonia and white precipitate is soluble forming a colourless solution [1] [4]

(c) indicative points• make solutions of both solids• add sodium hydroxide solution until in excess to a sample of the solution• Cr3+ – green-blue precipitate which is soluble in excess• Ni2+ – green precipitate which is insoluble in excess• add aqueous ammonia until in excess to a sample of the solution• Cr3+ – green-blue precipitate• Ni2+ – green precipitate which is soluble in excess forming a blue solution• Cr3+ + 3OH– Cr(OH)

3

• Ni2+ + 2OH– Ni(OH)2

• Cr(OH)3 + 3OH– [Cr(OH)

6]3–

• Ni(OH)2 + 6NH

3 [Ni(NH

3)

6]2+ + 2OH–

6666



(d) moles of CrCl3.6H

2O = moles of [Cr(H

2O)

6]3+ =

0.450266.5

= 1.689 × 10–3

moles of Cl– = moles of AgCl = 1.689 × 10–3 × 2 = 3.378 × 10–3

mass AgCl = 3.378 × 10–3 × 143.5 = 0.485 g = 485 mg [3]

15 (a) A = nickel(II) sulfate/nickel sulfate [1]B = iron(III) chloride [1] [2]

(b) (i) Ni(OH)2 [1]

(ii) [Ni(NH3)

6]2+ [1]

(iii) BaSO4 [1]

(iv) [Fe(H2O)

6]3+ [1]

(v) Fe(OH)3 [1]

(vi) AgCl [1](vii) [Ag(NH

3)

2]+ [1]

(c) Fe3+ + 3OH– Fe(OH)3 [1]

(d) green precipitate [1]which remains on addition of excess sodium hydroxide solution [1] [2]

16 Answer is C [1]

17 (a)

Transition metal ion

Observation on addition of a few drops of ammonia solution

Observation on addition of excess ammonia solution

Mn2+

Fe2+ ppt is insoluble [1]

Fe3+ brown ppt [1] ppt is insoluble [1]

Cr3+ green-blue ppt [1] ppt is insoluble [1]

Ni2+ green ppt [1]ppt is soluble forming

a blue solution [1]

Co2+ blue ppt [1]ppt is soluble forming a yellow solution [1]

Cu2+

each error [–1] [4](b) Cu2+ + 2OH– Cu(OH)

2 [1]

(c) Cu(OH)2 + 4NH

3 + 2H

2O [Cu(NH

3)

4(H

2O)

2]2+ + 2OH– [2]

(d) 2Mn(OH)2 + ½O

2 Mn

2O

3.2H

2O [2]

6767


(e) (i) green solution to yellow [1]yellow solution to orange [1] [2]

(ii) Cr(OH)3 + H

2O CrO

42– + 5H+ + 3e–

(or Cr3+ + 4H2O CrO

42– + 8H+ + 3e–) [2]

(iii) 2Cr(OH)3 + 3H

2O

2 2CrO

42– + 4H+ + 4H

2O

(or 2Cr3+ + 2H2O + 3H

2O

2 2CrO

42– + 10H+) [2]

(iv) 2CrO42– + 2H+ Cr

2O

72– + H

2O [1]

Vanadium chemistry18 Answer is A [1]

Use a table to answer this style of question

Oxidation Ag Ag+ Fe Fe2+ Sn Sn2+ SO2 SO

42–

Oxidation potential / V –0.80 V +0.44 –0.14 –0.17

Vanadium reduction

+5 +4 + 1.00 V

+4 +3 + 0.32 V ×

+3 +2 – 0.26 V × × ×

Once all the information is in the table add together the oxidation potential value (negative of the reduction electrode potential given) and the vanadium reduction electrode potential and if you get a positive value then it is feasible () and if it is negative then it is not feasible (×). Note that iron would reduce vanadium from +5 to +2 whereas both tin and sulfur dioxide would reduce vanadium from +5 to +3.

19 (a)

Vanadium compound Oxidation state of vanadium Colour of compound

NH4VO

3+5 yellow

VSO4

+2 violet

VCl3

+3 green

VO(NO3)

2+4 blue

[1] per row [4](b) (i) VO

3– + 2H+ VO

2+ + H

2O [1]

(ii) vanadium oxidation state remains at +5 [1](c) (i) 2VO

2Cl + 8HCl + 3Zn 3ZnCl

2 + 2VCl

2 + 4H

2O [2]

(ii) initial colour = yellowfinal colour = violet [1]

6868


20 (a) SO2 + ½O

2 SO3 [1]

(b) yellow [1](c) vanadium(V) oxide is solid and reactants/SO

2 + O

2 are gases [1]

heterogeneous catalyst in different state to reactants [1] [2](d) (i) V

2O

5 + 2HNO

3 2VO

2NO

3 + H

2O [2]

(ii) V2O

5 + 6NaOH 2Na

3VO

4 + 3H

2O [2]

(iii) VO43– + 6H+ + e– VO2+ + 3H

2O [1]

(iv) intermediate between the yellow and blue [1]

(v) 2Na3VO

4 + 3Fe + 8H

2SO

4 3Na

2SO

4 + 3FeSO

4 + 2VSO

4 + 8H

2O [1]

(vi) VO2Cl = yellow

VOSO4 = blue

VSO4 = violet

V2(SO

4)

3 = green

all correct [2]; 1 error [1]; more than 1 error [0] [2]

21 (a) (i) Pb2+ [1]E Pb2+/Pb > E V3+/V2+ /EMF = +0.13 V [1]E Pb2+/Pb < E VO2+/V3+/EMF = – 0.45 V [1] [3]

Reduction I2 2I– Br

2 2Br– Ag+ Ag Pb2+ Pb

Reduction potential / V +0.54 +1.09 +0.80 -0.13

Vanadium oxidation

+2 +3 + 0.26 V

+3 +4 – 0.32 V ×

+4 +5 – 1.00 V × × ×

(ii) violet to green [1](iii) 2V2+ + Pb2+ 2V3+ + Pb [1]

(b) (i) 3Cl2 + 2V2+ + 4H

2O 2VO

2+ + 6Cl– + 8H+ [2]

(ii) yellow [1](iii) +2 +3 EMF = +1.36 – (–0.26) = +1.62 V [1]

+3 +4 EMF = +1.36 – (+0.32) = +1.04 V [1] +4 +5 EMF = +1.36 – (+1.00) = +0.36 V [1] [3]

6969

5.6 ELECTRODE POTENTIALS

5.6 Electrode Potentials

Electrochemical cells1 Answer is D [1]

2 Answer is C [1]

3 Answer is C [1]

4 (a)

Platinum

Salt bridge

Solution containing Fe2+

and Fe3+, both 1 mol dm–3

Platinum

H2 gas at

100 kPa

1 mol dm–3 H+

complete circuit with salt bridge at 298 K [1] and external circuit with voltmeter [1]left hand cell with H

2 gas at 100 kPa [1]

into a solution of H+ at 1 mol dm–3 (e.g. 1 mol dm–3 hydrochloric acid) [1]platinum electrodes in both solutions [1]right hand cell with Fe2+ and Fe3+ both 1 mol dm–3 [1] [6]

(b) (i) 2Fe + 3Cl2 2FeCl

3 [2]

(ii) E (Cl2/Cl–) > E (Fe2+/Fe)/EMF = + 1.80 V [1]

E (Cl2/Cl–) > E (Fe3+/Fe2+)/EMF = + 0.59 V [1] [2]

(iii) with bromine: iron(III) bromide [1]with iodine: iron(II) iodide [1]E (Br

2/Br–) > E (Fe2+/Fe)/EMF = +1.53 V [1]

E (Br2/Br–) > E (Fe3+/Fe2+)/EMF = +0.32 V [1]

E (I2/I–) > E (Fe2+/Fe)/EMF = +0.98 V [1]

E (I2/I–) < E (Fe3+/Fe2+)/EMF = –0.23 V [1] [6]

(c) (i) oxidation state of O changes from –2 in H2O to 0 in O

2 [1]

oxidation state of Cl changes from 0 in Cl2 to –1 in HCl [1]

redox is oxidation and reduction occurring in the same reaction [1] [3](ii) Pt|Cl

2|Cl–||H

2O|O

2|H+|Pt

EMF = + 1.36 – (+1.23) = +0.13 V [3](iii) chlorine removed so potentially harmful bacteria no longer killed [1]

7070


5 (a) salt bridge [1]completes the circuit without introducing a metal/electrical connection between the two half cells or electrodes/maintains the balance of anions and cations [1] [2]

(b) chloride ions would form a complex with Cu2+ ions [1](c) on left: Cu Cu2+ + 2e– [1]

lower concentration of Cu2+ [1]on right Cu2+ + 2e– Cu [1]higher concentration of Cu2+ [1] [4]

(d) Cu|Cu2+||Cu2+|Cu [2](e) the concentrations of Cu2+ in the two half cells would be equal [1](f) (i) 0 V/zero [1]

(ii) 0.03 = 0.34 – xx = +0.31 V [2]

6 (a) salt bridge [1]completes the circuit without introducing a metal/electrical connection between the two half cells or electrodes/maintains the balance of anions and cations [1] [2]

(b) B = copper [1]C = platinum [1] [2]

(c) D = Cu2+ ions (e.g. copper(II) sulfate/copper(II) nitrate) [1]1 mol dm–3 [1] E = Fe2+ and Fe3+ ions (e.g. iron(II) nitrate and iron(III) nitrate) [1] both 1 mol dm–3 [1] [4]

(d) Cu|Cu2+||Fe3+, Fe2+|Pt [2](e) EMF = + 0.77 – (+0.34) = +0.43 V [2](f) copper electrode [1]

oxidation/loss of electrons occurring [1] [2]

7 (a) S4O

82– + 2I– I

2 + 2SO

42– [1]

(b) EMF = 2.01 – (+0.54) = +1.47 V [2](c) two negative ions reacting [1]

repel each other [1] [2](d) (i) iron(II) used in step 1 and reformed in step 2 [1]

(ii) Step 1: EMF = 2.01 – (+0.77) = +1.24 V [1]Step 2: EMF = 0.77 – (+0.54) = +0.23 V [1] [2]

(iii) steps 1 and 2 can occur in either order [1](iv) Eu2+ and S

2O

82– react/EMF is +2.36 V [1]

Eu3+ and I– do not react/EMF is –0.89 V [1] [2]

8 (a) gas electrode/platinum electrode [1]H

2 at 100 kPa [1]

1 mol dm–3 H+ ions in solution [1]298 K [1] [4]

(b) 2H+(aq) + 2e– H2(g) [2]

(c) (i) yellow solution changes to green [1]Fe3+ reduced to Fe2+ [1]E

(Fe3+/Fe2+) > E (H+/H

2) or EMF = + 0.77 [1] [3]

(ii) Pt|H2|H+||Zn2+|Zn [2]

7171


(iii) E (Au+/Au) > E (O2/H

2O) or EMF = +0.45 V [1]

Au+ oxidises water/water reduces Au+ [1]4Au+ + 2H

2O 4Au + 4H+ + O

2 [1]

solid gold precipitate [1]bubbles of a colourless gas (oxygen) [1] [5]

(d) (i) electron acceptor [1](ii) largest E [1]

most readily reduced [1] [2]

Commercial cells9 (a) EMF = +0.57 – (– 3.03) = +3.60 V [2]

(b) Li + CoO2 Li+[CoO

2]– [1]

(c) oxidation state of cobalt in CoO2 = +4 [1]

oxidation state of cobalt in Li+[CoO2]– = +3 [1]

reduction as oxidation state decreasing [1] [3](d) Li+[CoO

2]– Li + CoO

2 [1]

(e) Li|Li+||Li+, CoO2|LiCoO

2|Pt [2]

(allow , or | between components of right hand cell) (f) lithium would react with water [1]

10 (a) oxidation = H2(g) + 2OH– 2H

2O + 2e– [1]

(allow 2H2(g) + 4OH– 4H

2O + 4e–)

reduction = O2(g) + 2H

2O + 4e– 4OH– [1] [2]

(b) 1.23 = 0.40 – xx = – 0.83 V [2]

(c) 2H2(g) + O

2(g) 2H

2O(l) [2]

(d) inert [1]good electrical conductivity [1] [2]

(e) no CO2 produced [1]

only product is water which is non-polluting [1]H

2 available from electrolysis of water [1] max [2]

11 (a) EMF = +1.23 – (+0.03) = + 1.20 V voltage = 1.2 V [2](b) CH

3OH + 1½O

2 CO

2 + 2H

2O [2]

(c) electrode 2 as reduction occurs/gain of electrons [1](d) require a constant supply of fuel [1]

12 (a) lithium cell/lead acid cell/nickel-metal hydride [1](b) oxidation state of Cd changes from +2 in Cd(OH)

2 to 0 in Cd [1]

oxidation state of Ni changes from +2 in Ni(OH)2 to +3 in NiO(OH) [1] [2]

(c) (i) oxidation: Cd + 2OH– Cd(OH)2 + 2e– [1]

reduction: NiO(OH) + H2O + e– Ni(OH)

2 + OH– [1] [2]

(ii) 1.4 = x – (–0.88)x = +0.52 V [2]

(iii) cadmium electrode [1]oxidation occurs/loss of electrons [1] [2]

(iv) provide OH– ions [1]

7272

5.7 AMINES

5.7 Amines

Nomenclature and physical properties1 (a) ethylamine/aminoethane/ethanamine [1]

(b) methylpropylamine/N-methylpropan-1-amine [1](c) phenylamine [1](d) N-methylethylamine/N-methylethanamine [1](e) 2-aminopropane/propan-2-amine [1](f) 2,4-diaminopentane/pentane-2,4-diamine [1](g) 1,3-diaminopropane/propane-1,3-diamine [1](h) diethylamine/N-ethylethanamine [1](i) butan-2-amine/2-aminobutane [1](j) diethylpropylamine/N,N-diethylpropan-1-amine [1] [10]

2 (a) primary amine = only one carbon atom directly bonded to the nitrogen atom and therefore has the (–NH

2) group [1]

secondary amine = two carbon atoms directly bonded to the nitrogen atom, i.e. –NH [1] [2](b)

Amine1°

(primary)2°

(secondary)3°

(tertiary)

methylamine

ethylamine

dimethylamine

phenylamine

triethylamine

[1] per correct [5]

3 (a) NH3C

CH3

H [1]N dimethylamine/N-methylmethamine [1] [2]

(b) for the secondary amine the nitrogen in the middle of the chain [1](makes the dipole less) and the hydrogen bonds between molecules are less strong and so less energy is needed to break them [1] [2]

(c) hydrogen bonds form [1]between the lone pair on N of amine and δ+ hydrogen of water/between lone pair on O of water and δ+ hydrogen of amine [1] [2]

4 (a) propylamine/1-aminopropane/propan-1-amine [1](b) N

tertiary

H3C

CH3

CH3

CH

primary

H3C

NH2

CH3

N

secondary

H3C

H

CH2CH

3

[3](c) A can form strong hydrogen bonds between molecules [1]

tertiary amines cannot form hydrogen bonds between molecules [1] [2]

7373

5.7 AMINES

Preparation of amines5 (a) CH

3CH

2OH + NH

3 CH

3CH

2NH

2 + H

2O [1]

(b) (i) ammonia [1]

heated in a sealed glass tube at 100 °C [1] [2]

(ii) CH3CH

2Br + NH

3 CH

3CH

2NH

2 + HBr [1]

6 (a) CH3CH

2CH

2CN + 4[H] CH

3CH

2CH

2CH

2NH

2 [1]

butylamine/1-aminobutane/butan-1-amine [1] [2]

(b) CH3CH

2CH

2I + NH

3 CH

3CH

2CH

2NH

2 + HI [1]

propylamine/1-aminopropane/propan-1-amine [1] [2]

(c) C6H

5NO

2 + 6[H] C

6H

5NH

2 + 2H

2O [1]

phenylamine [1] [2]

7 (a) lithium tetrahydridoaluminate(III)/lithal [1]in ether [1] [2]

(b) tin and concentrated hydrochloric acid [1]heat under reflux [1] [2]

8 (a) C is CH3CH

2CN/propanenitrile [1]

(b) route 1: KCN/potassium cyanide/NaCN/sodium cyanide [1]LiAlH

4/lithium tetrahydridoaluminate(III)/lithal [1] [2]

route 2: NH

3/ammonia [1] [3]

9 (a) concentrated sulfuric acid and concentrated nitric acid [1](b) HNO

3 + 2H

2SO

4 2HSO

4– + NO

2+ + H

3O+ [1]

nitronium ion [1] [2](c) (i) reduction [1]

tin and concentrated hydrochloric acid [1] [2](ii) phenylammonium chloride/C

6H

5NH

3Cl produced which is soluble [1]

sodium hydroxide added to liberate the amine [1] [2]

10 (a) methylbenzene [1](b)

NO2

CH3

accept any position of NO2 [1]

(c) reducing agent: tin and concentrated hydrochloric acid [1]

6[H]+NO

2

CH3

2H2O+

NH2

CH3

[2](d) ammonia/NH

3 [1]

7474

5.7 AMINES

Amines as bases11 (a) CH

3CH

2NH

2 + HCl CH

2CH

2NH

3Cl [1]

ethylammonium chloride [1] [2]

(b) 2CH3NH

2 + H

2SO

4 (CH

3NH

3)

2SO

4 [1]

methylammonium sulfate [1] [2]

(c) C6H

5NH

2 + HNO

3 C

6H

5NH

3NO

3 [1]

phenylammonium nitrate [1] [2]

(d) CH3CH

2NH

3Cl + NaOH CH

3CH

2NH

2 + NaCl + H

2O [1]

ethylamine/aminoethane/ethanamine [1] [2]

(e) 2C6H

5NH

2 + H

2SO

4 (C

6H

5NH

3)

2SO

4 [1]

phenylammonium sulfate [1] [2]

12 (a) ethylamine [1]the electron donating alkyl group releases electrons to the N atom so the lone pair on N atom is more able to accept a proton [1] [2]

(b) ammonia [1]in phenylamine the lone pair of the N atom becomes delocalised into the pi system so the electron density on the N atom and the lone pair is less available to accept a proton [1] [2]

(c) butylamine [1]larger electron donating alkyl group releases electrons to the N atom so the lone pair is more able to accept a proton [1] [2]

13 propylamine strongest then ammonia then phenylamine weakest [1]phenylamine lone pair on N (less available) delocalised into ring [1]lone pair in propylamine is more available [1]propyl group increases electron density (on N)/is an electron donating group [1] [4]

14 Answer is D [1]

15 lone pair on the nitrogen on side chain is more available/more able to be donated [1]lone pair on nitrogen in ring is delocalised into ring [1] [2]

16 (a) tin and concentrated HCl [1] sodium hydroxide [1] [2]

(b) 1,3-diaminobenzene ethane-1,2-diamine

[1] [1]

NH2

NH2

CH2N

H

H

C NH2

H

H

[2](c) both have a lone pair of electrons on the nitrogen [1]

that can bond to a hydrogen ion/proton [1] [2](d) aromatic ring is electron withdrawing/lone pair delocalised into ring [1]

the lone pair of electrons on nitrogen less available [1] [2]

7575

5.7 AMINES

Reactions of amines17 (a) C

6H

5NH

2 + CH

3COCl CH

3CONHC

6H

5 + HCl [1]

N-phenylethanamide [1] [2]

(b) CH3CH

2NH

2 + CH

3CH

2COCl CH

3CH

2CONHCH

2CH

3 + HCl [1]

N-ethylpropanamide [1] [2]

(c) C6H

5NH

2 + HNO

2 + HCl C

6H

5N

2Cl + 2H

2O [1]

benzene diazonium chloride [1] [2]

(d) CH3NH

2 + HNO

2 CH

3OH + N

2 + H

2O [1]

methanol [1] [2]

(e) C6H

5N

2Cl + C

6H

5OH C

6H

5N=NC

6H

5OH + HCl [1]

4-hydroxyazobenzene/azo dye [1] [2]

18 (a)

NH2

HClHNO2

+ N N Cl+ –

2H2O++

[2]

(b) dissolve phenylamine in (dilute) hydrochloric acid [1]below 10 °C [1]add sodium nitrite [1] [3]

(c) a reaction in which two benzene rings are linked together through an azo(–N=N–) group [1](d)

N

N N

OH NaOHN Cl+ – + +

OH H2O+ NaCl+

[2]

(e) < 10 °C [1]yellow ppt [1] [2]

19 (a) NO

2N N

+

[1]

(b) C12

H9N

3O

4 [1]

259 [1] [2]

(c) moles of G = 28

259 = 0.10811

75% 100% 0.10811 ÷ 75 × 100 = 0.1441 (1:1) number of moles of 4-nitrophenylamine = 0.1441 mass of 4-nitrophenylamine = 0.1441 × 138 = 20 g [3]

7676

5.7 AMINES

(d)

NO2N N OH

HO

[1](e) extensive delocalisation (of electrons) [1]

energy levels close together (so absorb in visible region) [1]less energy required for electron transitions [1]electrons move to higher energy level removing a colour [1] [4]

(f) sodium nitrite and hydrochloric acid [1]<10 °C [1] [2]

20 (a) hydrogen bonds form [1]between lone pair on N atom and δ+ hydrogen atom in water or lone pair of O atom of water with δ+ hydrogen atom of NH

2 [1] [2]

(b) H2N(CH

2)

4NH

2 + 2CH

3COCl H

3CCONH(CH

2)

4NHCOCH

3 + 2HCl [2]

(c) find the melting point of the solid [1]compare with data book/actual melting point [1] [2]

(d) H2N(CH

2)

4NH

2 + 2HNO

2 HO(CH

2)

4OH + 2H

2O + 2N

2 [2]

21 Answer is A [1]

moles of 4-hydroxyphenylamine = 10.9109 = 0.1

moles of paracetamol with 80% yield = 0.08

mass of paracetamol = 0.08 × 151 = 12.1 g

22 (a) A = nitrobenzene [1]B = phenylammonium chloride [1]C = phenylamine [1]D = methylbenzene [1] [4]

(b)

Step Type of reaction Reagents

1nitration/electrophilic

substitution concentrated nitric acid and concentration sulfuric acid

2 reductiontin and concentrated hydrochloric

acid

4 radical substitution chlorine

5 nucleophilic substitution ammonia

[1] per row [4]

(c) sodium hydroxide [1](d) sodium nitrite and hydrochloric acid [1]

< 10 °C [1] [2]

7777

5.7 AMINES

(e)

N N N(CH3)

2

N2Cl

HCl+

N(CH3)

2+

equation [1]circle azo group [1] [2]

7878

5.8 AMIDES

5.8 Amides

1 (a) ethanamide [1](b) pentanamide [1](c) 3-methylbutanamide [1](d) 2-bromobutanamide [1](e) N-methylpropanamide [1](f) N-phenylethanamide [1](g) propanamide [1](h) N,N-dimethylmethanamide [1]

2 hydrogen bonds form [1]between the lone pair of one molecule and the δ+ H atom of the other [1] [2]

3 both have similar van der Waals’ forces between molecules [1]propanamide has stronger hydrogen bonding between molecules which takes more energy to break [1] [2]

4 (a) breaking up molecules by reaction with water [1](b) CH

3CONH

2 + HCl + H

2O CH

3COOH + NH

4Cl [1]

ethanoic acid [1] [2]

(c) CH3CONH

2 + NaOH CH

3COONa + NH

3 [1]

sodium ethanoate [1] [2]

(d) 2CH3CH

2CONH

2 + H

2SO

4 + 2H

2O 2CH

3CH

2COOH + (NH

4)

2SO

4 [1]

propanoic acid [1] [2]

5 (a) a reaction which involves the elimination of water from the amide [1](b) phosphorus pentoxide/phosphorus(V) oxide [1](c) CH

3CH

2CONH

2 CH

3CH

2CN + H

2O [1]

propanenitrile [1] [2]

6 Answer is A [1]

7 (a) 2CH3CH

2COOH + (NH

4)

2CO

3 2CH

3CH

2COONH

4 + H

2O + CO

2 [2]

solid disappears [1] fizzing [1] [4](b) ammonium propanoate water + propanamide [1]

CH3CH

2COONH

4 H

2O + CH

3CH

2CONH

2 [1] [2]

8 (a)

CC

H

OH

OH

H

C

[1]

(b) ammonium propenoate [1]CH

2CHCOONH

4 [1] [2]

(c) hydrogen bonds form [1]between the lone pair of electrons on O atom of amide and δ+ hydrogen atom of H

2O or

between lone pair of electrons on O atom of H2O and δ+ hydrogen atom of NH

2 [1] [2]

7979

5.8 AMIDES

(d) (i) P4O

10/P

2O

5 [1]

(ii) dehydration/elimination [1](e) (i) CH

2CHCONH

2 + Br

2 CH

2BrCHBrCONH

2 [1]

orange solution decolourised [1] [2](ii) CH

2CHCONH

2 + NaOH CH

2CHCOONa + NH

3 [1]

bubbles [1] [2]

9 2CH3CH

2CH

2COOH + (NH

4)

2CO

3 2CH

3CH

2CH

2COONH

4 + H

2O + CO

2 [1]

CH3CH

2CH

2COONH

4 H

2O + CH

3CH

2CH

2CONH

2 [1] [2]

10 (a) A = sodium hydroxide/NaOH [1]B = hydrochloric acid/HCl/dilute acid [1]C = phosphorus pentoxide/phosphorus(V) oxide/P

4O

10 [1] [3]

(b) CH3CH

2CH

2NH

2 + CH

3COCl CH

3CONHCH

2CH

2CH

3 + HCl [1]

8080

5.9 AMINO ACIDS

5.9 Amino Acids

1 (a) (i) a sample which rotates the plane of plane polarised light [2](ii)

C*

NH2

COOH

(CH2)

4NH

2H

[2](iii) 2,6-diaminohexanoic acid [1]

(b) strong ionic attractions between oppositely charged ions [1]require substantial energy to break [1] [2]

(c) H2N(CH

2)

4CH(NH

2)COO– [1]

(d)

H3N

H

(CH2)

4

H3N

C

O

C+

+

OH

[1]

(e)

H2N

H

(CH2)

4

H2N

C

H

(CH2)

4

H2N

C

H

N

O

C

O

C OH

[1](f) any two from:

H2N

H

CH3

C

H

(CH2)

4

H2N

C

H

N

O

C

O

C OH

H2N

H

CH3

C

H

(CH2)

4

NH2

C

H

N

O

C

O

C OH

H2N

H

CH3

C

H

(CH2)

4

H2N

C

H

N

O

C

O

C OH

[2]

8181

5.9 AMINO ACIDS

(g) moles of N = 10.514

= 0.75

moles of H = 5.3

moles of C = 36.112

= 3.01

moles of O = 48.116

= 3.01

C4H

7O

4N [2]

2 (a)

H2N

H

CH3

C COOH

[1]

(b)

CH3N

H

CH

COO

H3C CH

3

+ –

[1](c)

C

NH2

H

COOH

(CH3)

2HC

C

H2N

H

CH(CH3)

2

HOOC

[2](d) ions which have a permanent positive and negative charge but which are neutral overall [2]

(e)

H3N

+ –

H

CH CH3

[1]

CH3

H3C

C

H

C

H

N

O

C COOH3N

+ –

H

CHCH3

[1]

CH3

H3C

C

H

C

H

N

O

C COO

[2]

(f)

CH3

H

C

H

N

O

C COOHH2N

H

CHCH3

CH3

H3C

C

H

C

H

N

O

C

[2]

3 Answer is B [1]

8282

5.9 AMINO ACIDS

4 (a) hydrolysis [1](b) 2-aminopropanoic acid [1](c)

CH3N

H

CHOH

COO

CH3

+ –

[1]

(d)

CH3N

H

COOH

CH3

+

[1]

(e)

CH2N

H

C

O

O

CH3

CH3

(or protonated form) [1]

(f) 2+

H3C CH

3

Cu

N

H H H H

O–

OO–

NCHCH

CC

O [1]dark blue solution [1] [2]

(g) HOOCCH(CH3)NH

2 + HNO

2 HOOCCH(CH

3)OH + N

2 + H

2O [1]

bubbles/fizzing/effervescence [1] [2](h) 2HOOCCH(CH

3)NH

2 + Na

2CO

3 2NaOOCCH(CH

3)NH

2 + CO

2 + H

2O [1]

bubbles/fizzing/effervescence [1] [2](i) zwitterion forms ionic interactions with polar water molecule [1]

5 (a) sequence of amino acids joined by peptide links in the chain [1](b) (i) A = alpha helix [1]

B = beta pleated sheet [1] [2](ii) hydrogen bonds [1]

form between a lone pair of electrons on an O atom and a δ+ H [1] [2](c) the bending/folding of the secondary structure to give a precise 3D shape held together by

hydrogen bonding/disulfide bridges/ionic interactions/van der Waals’ forces [2]

8383

5.9 AMINO ACIDS

(d) (i) the substrate induces a change of shape of the active site of the enzyme [1]the reaction occurs in active site with a path of lower activation energy [1] [2]

(ii) enzymes lose their activity at (low) or high temperature [1]enzymes lose their activity at low or high pH [1]enzyme structure is denatured [1]with heat hydrogen bonds of tertiary structure break/pH change breaks the ionic attractions of tertiary structure [1] [4]

(iii) enzymes break down proteins/carbohydrates/ stains on clothes [1]reaction can occur at lower temperature/ less agitation required [1] [2]

(iv) the site on the surface of the enzyme into which the substrate fits [1]

6

CHH3N

CH2

OH

C

O

O

(Na )– +

CHH2N

CH2

CH2

C

C

O

O

[1] [2]

O

(Na )– +

O(Na )–+

[3]

7 (a) 4 [1](b)

R

H

C COOHH2N

where R = H, CH3, CH

2OH or CH

2C

6H

5 [1]

(c)

N

H H H

H

CH2

OH

C

O

C

H2C

N C

H H

HO

C N C

H CH3

HO

C

O

CN C

any one of circles shown [1]

(d) hydrogen bonding [1](e) hydrolysis [1]

dilute acid/alkali e.g. hydrochloric acid or sodium hydroxide solution [1] [2]

8 (a) HOOCCH2CH

2CH(NH

2)COOH + Na

2CO

3 NaOOCCH

2CH

2CH(NH

2)COONa + H

2O + CO

2 [2]

(b) the NH2 and COOH group attached to the same carbon [1]

(c) 2-aminopentanedioic acid [1]

9 Answer is C [1]

8484

5.10 POLYMERS

5.10 Polymers

1 (a) (i)

(CH2)

4(CH

2)

6

O

C

O H H H H

C N N (CH2)

4(CH

2)

6

O

C

O

C N N

[2](ii) polyamide [1](iii) polymers formed by the elimination of small molecules such as water or hydrogen chloride

when monomers bond together [2](iv)

(CH2)

4C

Cl

OO

Cl

C

[1]

(b) (i)

CHO

H

H

H

H

C OH

[1]

(ii) ethane-1,2-diol [1] condensation [1] [2]

(c) (i) 2 [1](ii)

C

OH

OO

HO

C H2N NH

2

[2](iii) Kevlar as there are hydrogen bonds between molecules which require more energy

to break [1] Terylene only has weaker permanent dipole–dipole attractions/van der Waals’ forces [1] [2]

(d) (i) 3-methylpent-2-ene [1](ii) addition [1](iii) Terylene is biodegradable [1]

the ester bond can be hydrolysed [1] [2]

2 (a) butane-1,4-diamine /1,4-diamiobutane [1](b) (CH

2)

4(CH

2)

4N

H H O

N C

O

C

[1](c) hydrogen bonds form [1]

between the lone pair on the O and the δ+ H on an adjacent chain [1] [2](d) polyamide [1]

amide link can be hydrolysed by microorganisms [1] [2]

8585

5.10 POLYMERS

3 (a) H3CC

6H

4CH

3 + 6[O] HOOCC

6H

4COOH + 2H

2O [1]

(b) HOOCC6H

4COOH + 2CH

3OH H

3COOCC

6H

4COOCH

3 + 2H

2O [1]

(c) (i)

C

O

C O OCH2

CH2

O

[1]

(ii) ester bond can be hydrolysed [1]add acid/alkali [1] [2]

4 (a) polyamide [1] condensation [1] [2]

(b) propene [1](c) CH

2CH

2CH

2CH

2O

O

C CH2CH

2CH

2CH

2O

Oor

C [1](d)

C

CH3

CH3

CH3

H

C

[1]

(e) polyamides have hydrogen bonds between the chains which are stronger [1]than the (weaker) van der Waals’ forces between polyalkenes [1] [2]

5 (a) (i) H2N(CH

2)

6NH

2 [1]

HOOC(CH2)

8COOH [1] [2]

(ii) HOOCC6H

4COOH [1]

HOCH2CH

2OH [1] [2]

(iii)

C CHO OH

N NH

H H

H

Cor CCl Cl

O O O O

[2]

(iv)

C

O

C

H2N NH

2

HO OH

O

[2]

8686

5.10 POLYMERS

(b) (i) polyamide [1] (ii) polyester [1] (iii) polyamide [1] (iv) polyamide [1]

6 (a)

C

O

C N

H

N

H O

C

O

C N

H

N

H

C

O

N

HO

[1](b) benzene-1,4-dicarboxylic acid [1]

benzene-1,4-diamine [1] [2](c) 2 [1]

7 (a)

N

+

HCl+

H2N NH

2C

H2N

Cl

OO

Cl

C

Cl

O

CC

H O [2]

(b) amide group [1]is hydrolysed by the action microorganisms [1] [2]

8787

5.11 CHEMISTRY IN MEDICINE

5.11 Chemistry in Medicine

1 Answer is B [1]

2 Answer is C [1]

3 Answer is D [1]

4 (a) barium compounds are toxic [1]

(b) moles of NaOH = moles of HCl in 25.0 cm3 = 19.4 × 0.0960

1000 = 1.8624 × 10–3

moles of HCl in 250 cm3 = 1.8624 × 10–3 × 10 = 0.018624


1000 = 0.05375

moles of HCl which reacted with Mg(OH)2

= 0.05375 – 0.018624 = 0.035126

moles of Mg(OH)2 =

0.0351262

= 0.017563

mass of Mg(OH)2 = 0.017563 × 58 = 1.019 g

% Mg(OH)2 =

1.0191.25

× 100 = 81.5 % [5]

(c) mass of CaCO3 = 0.925 × 1.25 = 1.15625 g

moles of CaCO3 =

1.15625100

= 0.0115625

moles of HCl which reacted with CaCO3 = 0.0115625 × 2 = 0.023125


1000 = 0.05375

moles of HCl left over 0.05375 – 0.023125 = 0.030625

moles of HCl in 25.0 cm3 = moles of NaOH = 0.030625

10 = 3.0625 × 10–3


0.0960 = 31.9 cm3 [5]

5 (a) decanedioic acid [1](b) (i) HOOC(CH

2)

8COOH + Na

2CO

3 NaOOC(CH

2)

8COONa + CO

2 + H

2O [1]

(ii) ionic/COO– [1]can interact with water molecules more readily than unionised acid [1] [2]

(c) (i) C7H

6O

3 [1]

(ii) moles required = 3 × 10–4

mass required = 3 × 10–4 × 138 = 0.0414 g

volume of 17 % solution = 0.0414

17 × 100 = 0.24 cm3 [3]

8888


6 (a) cisplatin [1]anticancer drug [1] [2]

(b) Pt

Cl

NH3

Cl

H3N [1]

(c) shape: square planar [1]bond angle: 90° [1]co-ordination number = 4 [1] [3]

(d) (i) the process by which a double stranded DNA molecule is copied to produce two identical DNA molecules [2]

(ii) [Pt(NH3)

2Cl

2] + H

2O [Pt(NH

3)

2(H

2O)Cl]+ + Cl– [2]

(iii) C5H

5N

5O [1]

(e) K2PtCl

4/K

2[PtCl

4] [1]

7 (a)

CN

CH2COONaNaOOCH

2C

NaOOCH2C CH

2COONa

H

H

C

H

H

N

CH2N

H

H

C

H

H

NH2

++ 4H2O

+ 4NH3

+4NaCN 4HCHO

[2](b) (i) sequesters Ca2+ [1]

prevent blood from clotting [1] [2](ii) C

10H

13O

8N

2K

3 [1]

(iii) mass of tripotassium edta = 1.5 × 4 = 6 mg

moles of tripotassium edta = 0.006406

= 1.48 × 10–5 mol [2]

(iv) A ligand which uses many lone pairs of electrons to form more than two co-ordinate bonds with a central metal atom or ion in a complex [2]

(c) (i)

C

O

C

OH

CHO

O

OH

C

O

OH

H2C

H2C [1]

(ii) C6H

8O

7 + 3NaOH C

6H

5O

7Na

3 + 3H

2O [2]

8989


8 (a) 4 [1](b) haem bonds to protein/globin by co-ordinate bond [1]

haem and 4 protein molecules form haemoglobin complex [1]O

2 binds reversibly to the iron/iron(II) ion in haem [1] [3]

(c) CO binds irreversibly to iron in haemoglobin [1]forming carboxyhaemoglobin [1]lower O

2 carrying capacity [1] [3]

9 (a) C

O

O

O

C CH3

OH

[1](b)

C

+

C

C ++

O

O

O

C 2NaOHCH3

OH O

OH H2OH

3C

ONaO

ONa

[2](c) (i) more soluble in ethanol than water [1]

(ii) some impurities in the tablets insoluble in ethanol [1](iii) increase rate of hydrolysis [1]

(iv) moles of aspirin = 0.9180

= 5 × 10–3

moles of NaOH added = 25.0 × 0.5

1000 = 0.0125

moles of NaOH which reacted = 5 × 10–3 × 2 = 0.01moles of NaOH left over = 0.0125 – 0.01 = 2.5 × 103

moles of NaOH left over in 25.0 cm3 = mole of HCl = 2.5 × 10−3

10 = 2.5 ×10–4

volume of HCl required = 2.5 × 10−4 × 1000

0.0125 = 20.0 cm3 [5]

(v) phenolphthalein [1] or methyl orange [1]pink to colourless [1] or yellow to red [1] [2]

(vi) hydrolysis not complete so more NaOH present than expected [1]

10 (a) reacts with water/becomes aquated [1]ligands bind to nitrogen atoms in guanine [1]inter or intra strand links [1]prevent DNA replication [1]cell death [1] max [4]

(b) fewer side effects [1](c) [Pt(NH

3)

2Cl

2] [1]

(d) same group in Periodic Table/directly above Pt in Periodic Table/similar electronic configuration [1]

9090


11 (a) (i) use up all 2-hydroxybenzoic acid [1]excess ethanoic anhydride can be hydrolysed [1] [2]

(ii) moles of 2-hydroxybenzoic acid = 0.500138

= 3.623 × 10–3

moles of ethanoic anhydride required = 3.623 × 10–3

mass of ethanoic anhydride required = 3.623 × 10–3 × 102 = 0.370 g

volume of ethanoic anhydride = 0.3701.08

= 0.342 cm3 [3]

(iii) moles of aspirin = 3.623 × 10–3

mass of aspirin = 3.623 × 10–3 ×180 = 0.652 g = 652 mg [3]

(iv) % yield = 485652

× 100 = 74.4 % [1]

(iv) loss by mechanical transfer/not all recrystallised/side reactions/reaction not complete [1]

(b) % atom economy = 180240

× 100 = 75 % [1]

(c) more soluble in water [1]

12 (a) remove an insoluble substances which may react [1](b) titre values within 0.1 cm3 (allow 0.2 cm3) [1](c) CaCO

3 + 2HCl CaCl

2 + CO

2 + H

2O [1]

NaOH + HCl NaCl + H2O [1] [2]

(d) moles of NaOH = moles of HCl in 25.0 cm3 = 13.1 × 0.250

1000 = 3.275 × 10–3

moles of HCl in 250 cm3 = 3.275 × 10–3 × 10 = 0.03275


1000 = 0.075


moles of CaCO3 =

0.042252

= 0.021125

mass of CaCO3 = 0.021125 ×100 = 2.1125 g


2.11255

= 0.4225 g = 423 mg [5]

13 (a) gas-liquid chromatography attached to a mass spectrometer [1](b) mobile phase = inert carrier gas [1]

stationary phase = liquid on solid support [1]mixture injected with carrier gas [1]components separate based on solubility in liquid [1]components identified by mass spectrometry by comparison with known spectra [1] [5]

9191


14 (a) (i)

C C

+ +

O OH O

O

O

C

C

O

O

C O

CH3

OH

OH

H3C

H3C

C

HO

O

H3C

[2](ii) ethanoyl chloride too reactive [1]

ethanoic acid reaction not complete [1] [2](iii) 2-hydroxybenzoic acid [1](iv) H

3PO

4 [1]

catalyst [1] [2](v) hydrolyse any unreacted ethanoic anhydride [1](vi) remove any soluble impurities [1](vii) remove any acid [1](viii) mass of ethanoic anhydride = 1.08 × 5 = 5.4 g

moles of ethanoic anhydride = 5.4102

= 0.0529

moles of salicylic acid = 2.0138

= 0.0145

salicylic acid limitingmoles of aspirin = 0.0145mass of aspirin = 0.0145 ×180 = 2.61 g

% yield = 2.242.61

× 100 = 85.8 % (85.9 % with no rounding) [5]

(b) (i) draw pencil line and × about 1 cm from the bottom of the plate [1]using a capillary tube spot the mixture onto the line [1]allow to dry and repeat [1] [3]

(ii) place in a developing tank with ethyl ethanoate to a depth of < 1 cm andcover with a lid [1]allow to run and remove plate when solvent near the top [1]mark solvent front [1] spray with potassium manganate(VII) solution/view under UV [1]

calculate Rf values using R

f =

distance moved by spotdistance moved by solvent

[1]

(iii) compare to known values for pure samples [1]no spot for salicylic acid [1] [2]

(iv) oxidation [1]

Documents

Alyn G McFarland Nora Henry CCEA A2 CHEMISTRY EXAM