Kinetics

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October 2021

Question Paper Log Number P67131A

Publications Code WCH15_01_2110_MS

All the material in this publication is copyright

© Pearson Education Ltd 2021

General Marking Guidance

All candidates must receive the same treatment. Examiners

must mark the first candidate in exactly the same way as

they mark the last.

Mark schemes should be applied positively. Candidates must

be rewarded for what they have shown they can do rather

than penalised for omissions.

Examiners should mark according to the mark scheme not

according to their perception of where the grade boundaries

may lie.

There is no ceiling on achievement. All marks on the mark

scheme should be used appropriately.

All the marks on the mark scheme are designed to be

awarded. Examiners should always award full marks if

deserved,

i.e.

if the

answer

matches

the

mark

scheme. Examiners should also be prepared to award zero

marks if the candidate’s response is not worthy of credit

according to the mark scheme.

Where some judgement is required, mark schemes will

provide the principles by which marks will be awarded and

exemplification may be limited.

When examiners are in doubt regarding the application of

the mark scheme to a candidate’s response, the team leader

must be consulted.

Crossed out work should be marked UNLESS the candidate

has replaced it with an alternative response.

Section A (Multiple Choice)

Question

number

Answer

Mark

1(a)

The only correct answer is D (Pt, Pt)

A

is incorrect because both electrodes should be made of platinum

B

is incorrect because both electrodes should be made of platinum

C

is incorrect because both electrodes should be made of platinum

1

Question

number

Answer

Mark

1(b)

The only correct answer is C (1.00 mol dm3 HCl(aq))

A

is incorrect because H3PO4 is not completely ionised

B

is incorrect because H2SO4 is not completely ionised

D

is incorrect because CH3COOH is not completely ionised

1

Question

number

Answer

Mark

1(c)

The only correct answer is B (17.91 g)

A

is incorrect because there should be only one mol of chromium ions per mol of dichromate ions

C

is incorrect because there should be only one mol of chromium ions per mol of dichromate ions

D

is incorrect because there should be only one mol of chromium ions per mol of dichromate ions

1

Question

number

Answer

Mark

1(d)

The only correct answer is A (H2SO4)

B

is incorrect because chloride ions could be oxidised

C

is incorrect because bromide ions would be oxidised

D

is incorrect because this would introduce additional chromium species into the mixture

1

Question

number

Answer

Mark

2

The only correct answer is C (Pt Fe2+, Fe3+∥ [MnO4 + 8H+], [Mn2+ + 4H2O] Pt)

A

is incorrect because both electrodes should be made of platinum

B

is incorrect because both electrodes should be made of platinum and the MnO4/ Mn2+ half-cell does not show
reduction

D

is incorrect because the MnO4/ Mn2+ half-cell does not show reduction

1

Question

number

Answer

Mark

3

The only correct answer is D (2Ag2+ Ag+ + Ag3+)

A

is incorrect because the disproportionation is not thermodynamically feasible

B

is incorrect because the disproportionation is not thermodynamically feasible

C

is incorrect because the disproportionation is not thermodynamically feasible

1

Question

number

Answer

Mark

4

The only correct answer is A (the cathode has a more positive potential than the anode)

B

is incorrect because oxidation always occurs at the anode

C

is incorrect because oxygen is reduced at the positive electrode

D

is incorrect because the overall reaction is the same under both acidic and alkaline conditions

1

Question

number

Answer

Mark

5

The only correct answer is B (carbon monoxide forms stronger dative covalent bonds with haemoglobin than does
oxygen)

A

is incorrect because carbon monoxide can be displaced from carboxyhaemoglobin

C

is incorrect because the formation of carboxyhaemoglobin does not lead to an increase in the entropy of the system

D

is incorrect because the difference in bond type does not fully explain the difference in dative covalent bond strength

1

Question

number

Answer

Mark

6

The only correct answer is D (pink solution  blue precipitate  yellow-brown solution)

A

is incorrect because CoCl2(aq) is a pink solution

B

is incorrect because the blue precipitate dissolves in excess aqueous ammonia to form a yellow-brown solution

C

is incorrect because CoCl2(aq) is a pink solution

1

Question

number

Answer

Mark

7

The only correct answer is B ([Ni(EDTA)]2)

A

is incorrect because chloride ions act as monodentate ligands

C

is incorrect because ethanedioate ions act as bidentate ligands

D

is incorrect because 1,2-diaminoethane molecules act as bidentate ligands

1

Question

number

Answer

Mark

8

The only correct answer is D (none of the products are harmful to the environment)

A

is incorrect because the reactions occurring in catalytic converters involve heterogeneous catalysis

B

is incorrect because carbon monoxide is adsorbed onto the surface of the catalyst

C

is incorrect because nitrogen is desorbed from the surface of the catalyst

1

Question

number

Answer

Mark

9

The only correct answer is C (Mn2+)

A

is incorrect because MnO4 ions are neither a product nor a catalyst in this reaction

B

is incorrect because H+ ions are neither a product nor a catalyst in this reaction

D

is incorrect because CO2 is not a catalyst in this reaction

1

Question

number

Answer

Mark

10

The only correct answer is A (bothFe2+(aq) and Fe3+(aq) catalyse the reaction)

B

is incorrect because both Fe2+(aq) and Fe3+(aq) catalyse the reaction

C

is incorrect because both Fe2+(aq) and Fe3+(aq) catalyse the reaction

D

is incorrect because both Fe2+(aq) and Fe3+(aq) catalyse the reaction

1

Question

number

Answer

Mark

11

The only correct answer is D (

)

A

is incorrect because this product is formed by the substitution of one chlorine atom in CHCl3

B

is incorrect because this product is formed by the substitution of two chlorine atoms in CHCl3

C

is incorrect because this product is formed by the substitution of all three chlorine atoms in CHCl3

1

Question

number

Answer

Mark

12

The only correct answer is C (330.7)

A

is incorrect because this is the molar mass of bromobenzene

B

is incorrect because this is the molar mass of the monosubstituted product

D

is incorrect because this is the molar mass of the fully substituted product

1

Question

number

Answer

Mark

13

The only correct answer is C (CH3CH2CH2NH2 > NH3 > C6H5NH2)

A

is incorrect because C6H5NH2 is the weakest base in the sequence

B

is incorrect because CH3CH2CH2NH2 is a stronger base than NH3

D

is incorrect because this shows the order of increasing basicity

1

Question

number

Answer

Mark

14

The only correct answer is B (

)

A

is incorrect because this amine could be prepared by the reduction of butanenitrile

C

is incorrect because this amine could be prepared by the reduction of 2-methylpropanenitrile

D

is incorrect because this amine could be prepared by the reduction of 2,2-dimethylpropanenitrile

1

Question

number

Answer

Mark

15
The only correct answer is B (4)

A

is incorrect because the repeat unit of the polymer is formed from four different amino acids

C

is incorrect because the repeat unit of the polymer is formed from four different amino acids

D

is incorrect because the repeat unit of the polymer is formed from four different amino acids

1

Question

number

Answer

Mark

16
The only correct answer is D (carbon dioxide giving carboxylic acids )

A

is incorrect because Grignard reagents react with water giving alkanes

B

is incorrect because Grignard reagents react with methanal giving primary alcohols

C

is incorrect because Grignard reagents react with ketones giving tertiary alcohols only

1

Question

number

Answer

Mark

17

The only correct answer is C (will be lower than the true value)

A

is incorrect because using a sample that is impure would cause the value to be lower

B

is incorrect because using a sample that is impure would cause the value to be lower

D

is incorrect because using a sample that is impure would cause the value to be lower

1

Total for Section A = 20 marks

Section B

Question

Number
Answer

Additional guidance

Mark

18(a)

any indication that A contains FeCl2/iron(II) chloride (1)

working to show that A is a tetrahydrate (1)

Example of calculation:

Ignore (A contains) Fe2+

Ignore [FeCl4]2

mass of water = 198.8  (55.8 + 2  35.5)

= 72.0 (g)

moles of water = 72.0 ÷ 18.0 = 4

A is FeCl2.4H2O/iron(II) chloride tetrahydrate

Allow FeCl2(H2O)4

2

Question

Number
Answer

Additional guidance

Mark

18(b)

[Fe(H2O)6]2+

Allow [Fe(OH)(H2O)5]+ / [Fe(Cl)(H2O)5]+

Ignore omission of square brackets

Ignore name even if incorrect

1

Question

Number
Answer

Additional guidance

Mark

18(c)

A diagram showing the octahedral shape

Example of diagram:

Accept arrows for dative covalent bonds

Allow CN for C≡N

Do not award KCN/HCN for C≡N

Do not award M for Fe

Ignore connectivity of CN ligands

Ignore lone pairs

Ignore omission of square brackets

Ignore all charges

Do not award diagrams with no 3D shape

1

Question

Number
Answer

Additional guidance

Mark

18(d)

moles of K, Fe (1)

moles of C and N (1)

calculation of K:Fe:C:N mole ratio

and

empirical formula (1)

Example of calculation:

mol K = 35.6 ÷ 39.1 = 0.91049

Allow 0.91282 from Ar value of 39

mol Fe = 17.0 ÷ 55.8 = 0.30466

Allow 0.30357 from Ar value of 56

Ignore SF

mol C = 21.9 ÷ 12.0 = 1.8250

mol N = 25.5 ÷ 14.0 = 1.8214

Ignore SF except 1 SF

K : Fe : C : N

0.91049 : 0.30466 : 1.825 : 1.8214

3 : 1 : 6 : 6

empirical formula is K3FeC6N6

Allow K3Fe(CN)6

Allow elements in any order

TE on moles of K, Fe, C and N provided empirical formula is

closest whole number ratio

Correct answer with no working scores (3)

3

Question

Number
Answer

Additional guidance

Mark

18(e)

An equation including:

[Fe(CN)6]4 reactant and [Fe(CN)6]3 product (1)

rest of equation and balancing (1)

Example of equation:

2[Fe(CN)6]4 + Cl2 2[Fe(CN)6]3 + 2Cl

Allow K4[Fe(CN)6] reactant and K3[Fe(CN)6] product

Ignore omission of square brackets

M2 dependent on M1

Allow multiples

Ignore state symbols even if incorrect

Do not award uncancelled K+ spectator ions

If no other mark awarded, award (1) for any multiple of the

following equation:

2Fe2+ + Cl2 2Fe3+ + 2Cl

2

Question

Number
Answer

Additional guidance

Mark

18(f)

A completed table showing:

correct identification for reaction 2 (1)

correct identification for reaction 3 (1)

Example of completed table:

Allow any form of positive identification, including crosses

Ignore any form of negative identification

Do not award more than 1 box ticked in each row

2

(Total for Question 18 = 11 marks)

Question

Number
Answer

Additional guidance

Mark

19(a)

suitable test: bromine water/Br2(aq) (1)

result of test: decolourises (from orange) with

Dewar structure (and no change with benzene) (1)

Allow bromine / Br2((l)) / Br2 in organic solvent

Do not award Br/Br

Accept potassium manganate((VII))/KMnO4and acidified/

named acid/H+

Do not award hydrogenation

Do not award combustion

Allow does not decolourise with benzene

Allow brown/orange/yellow for colour of bromine water

Allow red/brown/orange for colour of bromine

Allow pink/purple for colour of potassium manganate((VII))

Ignore colour fades for decolourises

Ignore reference to addition/substitution

Do not award any reference to decolourisation/reaction with

benzene

2

Question

Number
Answer

Additional guidance

Mark

19(b)

similarity: (both compounds have) one (NMR) peak(1)

difference: expected chemical shift values (1)

Ignore any reference to IR and/or 13C NMR

Allow (both compounds have) one proton environment

Ignore just same number of peaks

Ignore references to relative peak area/integration/splitting

Do not award any other number of peaks

chemical shift for benzene within range of 6.4 to 8.4 ppm

(actual value is 7.3 ppm)

and

chemical shift for Ladenburg structure within range of

0 to 2.3 ppm (actual value is 2.3 ppm)

Allow any range or value within the above ranges

Ignore just benzene would have a higher chemical shift than

Ladenburg structure or reverse argument

Do not award additional incorrect chemical shifts

2

Question

Number
Answer

Additional guidance

Mark

19(c)

An explanation that makes reference to the following

points:

showed that all CC bonds are the same length in

benzene (1)

in Kekulé structure the C=C bonds would be

shorter than the CC bonds (or reverse argument) (1)

Ignore any reference to:

CH bonds

bond strength/bond angle

delocalised electrons

Dewar/Ladenburg structures

Allow showed benzene is a regular hexagon

Allow showed benzene contains only one type of

carbon-carbon bond

Allow benzene bond lengths are in between C=C and CC

Ignore just benzene has no C=C bonds

Do not award benzene bond lengths are longer than CC/

shorter than C=C

Allow Kekulé structure would have shown two different

lengths/types of carbon-carbon bond

Allow Kekulé structure would have alternating carbon-carbon

bond lengths

Ignore just Kekulé has C=C bonds

Do not award CC bonds would be shorter than the C=C bonds

If no other mark awarded, just bond lengths equal in benzene

but different in Kekulé scores (1)

2

Question

Number
Answer

Additional guidance

Mark

19(d)(i)

A diagram showing:

correct relative stabilities (1)

two or three numerical differences in enthalpy

with appropriate arrows (1)

Example of diagram:

Allow names for structures

If three values and arrows are given they must all be correct to

score M2

Allow slight imprecision in start and end of arrows in M2

Ignore any x-axis label

Do not award double headed arrows in M2

Do not award incorrect sign in M2

2

Question

Number
Answer

Additional guidance

Mark

19(d)(ii)

An answer that makes reference to the following:

pi bonds are weaker/more reactive/require less

energy to break (than sigma bonds)

or

fewer bonds must break to convert the

Dewar structure to benzene

Ignore just Dewar structure has pi/double bonds/is unsaturated

Ignore just Dewar structure has weaker bonds

Do not award C=C/double bonds weaker/require less energy to

break (than CC/single bonds)

Accept reverse argument

Allow any specified numbers to indicate fewer bonds must break

Ignore fewer new bonds must form

Ignore Dewar structure is more similar to benzene

Ignore carbon atoms already in a ring/hexagon

Ignore any reference to intermolecular forces

Do not award (higher) ring strain in Ladenburg

Do not award smaller difference in enthalpy

Do not award Ladenburg structure more stable

1

Question

Number
Answer

Additional guidance

Mark

19(e)

An explanation that makes reference to the following

points:

E-hexa-1,4-diene

twice the hydrogenation enthalpy (of hex-3-ene)

as two (isolated) C=C bonds (1)

E-hexa-1,3-diene

less exothermic/more stable (by 22 kJ mol1 than

E-hexa-1,4-diene

and

as some delocalisation of pi-bond(s) (1)

Allow double bond for C=C throughout

Accept 118  2 (=236) as two C=C bonds

Allow twice the hydrogenation enthalpy as no delocalisation of

pi-bond(s)

Accept less negative

Allow more positive

Allow some delocalisation of double bond(s)

Allow double bonds/p-orbitals are conjugated

Allow double bonds/p-orbitals are close enough to overlap

Ignore just C=C are close

Ignore just delocalisation of electrons

Ignore electron density more spread out

Ignore resonance stabilised

2

Question

Number
Answer

Additional guidance

Mark

19(f)(i)

skeletal formula of any one isomer (1)

skeletal formulae of second and third isomers (1)

Example of correct skeletal formulae:

(1,2-isomer) (1,3-isomer) (1,4-isomer)

Allow Kekulé benzene ring

Allow structural/displayed CH3 and CH3CO groups

If no other mark awarded, 1,2-, 1,3- and 1,4-isomers with

incorrect side chain/cyclohexane ring scores (1)

2

Question

Number
Answer

Additional guidance

Mark

19(f)(ii)

(identification of X as) 1,4-isomer (1)

(7 peaks consistent with) 7 carbon environments (1)

Allow any form of identification, including (f)(i) annotation

Allow just ‘1,4’ or ‘para’

M2 dependent on a structure containing 7 carbon

environments

Accept 1,2-isomer and/or 1,3-isomer have 9 carbon

environments/would have 9 peaks

Allow (4 arene peaks consistent with) 4 arene carbon

environments

Allow 1,2-isomer and/or 1,3-isomer have 6 arene carbon

environments/would have 6 arene peaks

2

Question

Number
Answer

Additional guidance

Mark

19(f)(iii)

A mechanism including:

curly arrow from on or within circle

to C+ of CH3CO+ (1)

structure of intermediate ion (1)

curly arrow from CH bond to within ring

and

correct product (1)

balanced equation for regeneration of catalyst (1)

Example of mechanism:

Allow Kekulé benzene ring

‘Horseshoe’ facing tetrahedral carbon and covering at least three

carbons with some part of positive sign within ‘horseshoe’

Allow methyl at 1,2- or 1,3-positions

Ignore missing methyl substituent

Do not award dotted/dashed CH/CC bonds unless 3D structure

Allow 1,2- or 1,3-product from corresponding intermediate

Do not award missing methyl substituent

AlCl4 + H+ AlCl3 + HCl

4

(Total for Question 19 = 19 marks)

Question

Number
Answer

Additional guidance

Mark

20(a)

An answer that makes reference to one of the

following points:

variable oxidation state/oxidation number

or

(easily) oxidised and reduced (back to

original oxidation state)

or

(easily) donate and accept electrons

(from other molecules/species)

Ignore references to ionisation energy

Ignore partially full d orbital(s)/d subshell

Ignore more than one stable ion

Ignore references to heterogeneous catalysis/adsorption

Ignore references to alternative reaction pathways/activation energy

Allow can change oxidation state/oxidation number

Allow have different oxidation state(s)/oxidation number(s)

Ignore variable valency

Allow just lose and gain electrons (easily)

1

Question

Number
Answer

Additional guidance

Mark

20(b)(i)

An answer that makes reference to the following points:

monodentate: forms a single/one dative (covalent) bond (1)

ligand: (a species with a) lone pair (of electrons) that can form

a dative (covalent) bond to a (central transition) metal (ion) (1)

Accept coordinate for dative throughout

Accept donates a single/one lone pair

Allow occupies a single/one coordination site

2

Question

Number
Answer

Additional guidance

Mark

20(b)(ii)

A completed diagram showing:

three adjacent THF/Cl ligands

Expected diagram:

Ignore lone pairs

Ignore 1 charge on Cl ligands

1

Question

Number
Answer

Additional guidance

Mark

20(c)(i) yellow to (permanent pale) green

Ignore qualifiers, eg pale

Ignore precipitate

Do not award green to yellow

Do not award any combination of yellow and green, eg yellowy-green

Do not award any other colour

1

Question

Number
Answer

Additional guidance

Mark

20(c)(ii)

A calculation including:

moles of Ti3+ in titre (1)

moles of Mg(NO3)2.6H2O in 100 cm3 (1)

moles of NO3 in 10.00 cm3 (1)

Ti3+ : NO3 mol ratio (1)

final oxidation state of nitrogen (1)

Example of calculation:

moles of Ti3+ = 0.085  20.70 = 0.0017595 / 1.7595  103

1000

moles of Mg(NO3)2.6H2O = 0.75 = 0.0029263 / 2.9263  103

256.3

moles of NO3 = 0.0029263  2 = 0.00058525 / 5.8525  104

10

TE on moles Mg(NO3)2.6H2O

Ignore SF except 1 SF in M1, M2 and M3

Ignore truncation of moles in M1, M2 and M3, eg 0.0005852

Ti3+ : NO3 mol ratio = 0.0017595 : 0.00058525

= 3 : 1

TE on moles Ti3+ and moles NO3

final oxidation state of nitrogen = (+)2

TE on mol ratio provided final oxidation state of nitrogen is

between 3 and +4, eg

Ti3+ : NO3 mol ratio = 0.0017595 : 0.00029263

= 6 : 1

final oxidation state of nitrogen = 1

Do not award incorrect oxidation state of N in NO3

Correct answer with no working scores (1)

5

Question

Number
Answer

Additional guidance

Mark

20(c)(iii)

An equation including:

selection of correct nitrogen half-equation (1)

balanced ionic equation from chosen half-equations (1)

Example of equation:

3Ti3+ + H2O + NO3 3TiO2+ + 2H+ + NO

TE on (c)(ii) provided +3 or +4 oxidation state

Allow multiples

Allow reversible arrow

Ignore state symbols even if incorrect

2

Question

Number
Answer

Additional guidance

Mark

20(c)(iv)

An answer that makes reference to the following point:

(calculation of) Eocell value

Example of calculation:

(Eocell = 0.96  0.10 =) (+)0.86 (V)

TE on ionic equation from (c)(iii):

(+)0.7(0) (V) for Ti3+ + NO3 TiO2+ + NO2

(+)0.84 (V) for 2Ti3+ + H2O + NO3 2TiO2+ + H+ + HNO2

1

Question

Number
Answer

Additional guidance

Mark

20(c)(v)

An answer that makes reference to the following point:

(heat is to) speed up/increase rate of reaction

Allow to ensure fast oxidation of Ti3+

Allow to provide activation energy/Ea

Allow (reaction has a) high activation energy/Ea

Ignore just to provide (more) energy

Ignore to increase collision frequency

Ignore to ensure complete reaction

Ignore any reference to thermodynamic feasibility

1

Question

Number
Answer

Additional guidance

Mark

20(c)(vi)

This question assesses a student’s ability to show a coherent and

logically structured answer with linkages and fully-sustained

reasoning.

Marks are awarded for indicative content and for how the answer

is structured and shows lines of reasoning.

The following table shows how the marks should be awarded for

indicative content.

Number of indicative marking

points seen in answer

Number of marks awarded

for indicative marking points

6

4

5-4

3

3-2

2

1

1

0

0

The following table shows how the marks should be awarded for

structure and lines of reasoning.

Number of marks awarded

for structure and sustained

lines of reasoning

Answer shows a coherent and

logical structure with linkages and

fully sustained lines of reasoning

demonstrated throughout.

2

Answer is partially structured

with some linkages and lines of

reasoning.

1

Answer has no linkages between

points and is unstructured.

0

The mark for indicative content should be added to the

mark for lines of reasoning. For example, an answer

with five indicative marking points that is partially

structured with some linkages and lines of reasoning

scores 4 marks (3 marks for indicative content and 1

mark for partial structure and some linkages and lines of

reasoning).

If there are no linkages between points, the same five

indicative marking points would yield an overall score

of 3 marks (3 marks for indicative content and no marks

for linkages).

If there is any incorrect chemistry, deduct mark(s) from

the reasoning. If no reasoning mark(s) awarded, do not

deduct mark(s).

Comment: Look for the indicative marking points first,

then consider the mark for the structure of the answer

and sustained line of reasoning.

6

Indicative points:

IP1: identification of [Ti(H2O)6]3+ complex ion

IP2: partially filled d-subshell/d-orbital(s) (in Ti3+)

IP3: splitting in energy of d-subshell/d-orbitals by

water/ligands

IP4: absorption of light/photon/(electromagnetic) radiation

and electronic transition

IP5: origin of observed colour of complex ion

IP6: clearer colour change at end-point with indicator

Allow [Ti(H2O)5Cl]2+ / [Ti(H2O)4Cl2]+

Accept incomplete for partially filled

Accept (Ti3+ is) 1s22s22p63s23p63d1(4s0)

Allow (Ti3+ is) (3)d1

water/ligands and

split the energy of the d-subshell/d-orbitals

Allow ligands cause d-d splitting

Do not award d-orbital (singular)

(visible) light/photon/(electromagnetic) radiation is

absorbed and

promoting electrons from lower to higher energy

Allow light etc causes d-d transitions

colour due to reflected/transmitted light

Allow colour due to wavelengths/frequencies of light

that are not absorbed

Allow complementary colour observed

Do not award any reference to emission of light

Accept reverse argument

Allow colours are more intense/distinct/sharp/strong

Allow concentration (of [Ti(H2O)6]3+/TiCl3) too low to

accurately determine end-point in absence of indicator

Ignore just easier to determine end-point

Ignore just more accurate/precise

Ignore mention of specific colours, even if incorrect

Do not award reference to acid-base colour change

(Total for Question 20 = 20 marks)
TOTAL FOR SECTION B = 50 MARKS

Section C

Question

Number
Answer

Additional guidance

Mark

21(a)

A completed mechanism showing:

curly half-arrows to show homolytic fission of OH bond

Example of completed mechanism:

Accept curly half-arrows originating from

opposite sides of the OH bond

Left-hand curly half-arrow must terminate

between T● and H

Right hand curly half-arrow must terminate on

or near to O atom of HO

1

Question

Number
Answer

Additional guidance

Mark

21(b)

A completed mechanism showing:

curly arrow from lone pair on Se to correct C of C=C (1)

curly arrow from C=C bond to CC bond

and

curly arrow from C=O bond to O (1)

Example of completed mechanism:

Penalise curly half-arrows once only

Do not award curly arrow from negative charge on Se

Ignore (+)C=C() dipole

Do not award (-)C=C() dipole

Do not award full charge on either carbon of C=C bond

Do not award incorrect (-)C=O(+) dipole

2

Question

Number
Answer

Additional guidance

Mark

21(c)

A completed table showing:

two or three correct answers (1)

four correct answers (2)

Example of completed table:

2

Question

Number
Answer

Additional guidance

Mark

21(d)(i)

K2Cr2O7and H2SO4 (1)

heat/reflux (1)

If name and formula given, both must be correct to score M1

Accept names (eg sodium dichromate((VI)) and sulfuric acid)

Allow Cr2O72and H+ / acidified dichromate

Ignore concentration of acid

Do not award KMnO4 for K2Cr2O7

Do not award HCl for H2SO4

M2 dependent on some mention of dichromate (or manganate) oxidising agent

Ignore distillation

2

Question

Number
Answer

Additional guidance

Mark

21(d)(ii)

correct structure for 2-aminobenzoic acid

Example of correct structure:

Accept hydrochloride salt (NH3Cl)

Allow protonated amine group (NH3+)

Allow any correct combination of skeletal, structural or

displayed formulae

Allow Kekulé benzene

Ignore connectivity

Ignore name, even if incorrect

1

Question

Number
Answer

Additional guidance

Mark

21(d)(iii) NaNO2/sodium nitrite/sodium nitrate(III)

and

HCl/hydrochloric acid

Allow HNO2/nitrous acid

Allow H+and NO2

Ignore conditions, including concentration of HCl

Ignore H2O

Do not award NaNO3/sodium nitrate

1

Question

Number
Answer

Additional guidance

Mark

21(d)(iv)

correct structure for N,N-dimethylphenylamine

Example of correct structure:

Allow any correct combination of skeletal, structural or

displayed formulae

Allow Kekulé benzene

Ignore quaternary salt (N(CH3)3+)

Ignore name, even if incorrect

1

Question

Number
Answer

Additional guidance

Mark

21(d)(v)

An explanation including:

effect of temperature higher than 5C (1)

effect of temperature lower than 5C (1)

(diazonium/it) decomposes / reacts with water / forms a phenol /

undergoes nucleophilic substitution (above 5C)

Ignore byproducts form / side reactions occur / yield too low (above 5C)

(rate of reaction) too slow (below 5C)

Allow just slows down (below 5C)

Ignore insufficient energy for reaction to occur (below 5C)

Ignore any reference to activation energy/collision frequency

Ignore freezes (at 0C and below)

2

Question

Number
Answer

Additional guidance

Mark

21(e)(i)

A completed equation showing:

correct balancing of propanone and

sodium ethanoate (1)

correct balancing of sodium hydroxide

and water (1)

Example of equation:

M2 dependent on M1

2

Question

Number
Answer

Additional guidance

Mark

21(e)(ii)

M1: molar masses

M(2-nitrobenzaldehyde)

and

M(indigotin) (1)

Then, for M2 and M3, either:

Method 1 (M2 and M3)

moles indigotin in 10.0 g

and

indigotin:2-nitrobenzaldehyde mol ratio (1)

moles 2-nitrobenzaldehyde required

and

mass 2-nitrobenzaldehyde required (1)

Correct answer with no working scores (3)

Example of calculation:

M(2-nitrobenzaldehyde) = 7  12.0 + 5  1.0 + 1  14.0 + 3  16.0

= 151(.0 g mol1)

M(indigotin) = 16  12.0 + 10  1.0 + 2  14.0 + 2  16.0

= 262(.0 g mol1)

Allow truncation of mass/moles throughout, eg 0.03816 for 0.038168

Ignore SF except 1 SF in M2 and M3

moles indigotin in 10.0 g = 10.0 = 0.038168

262

TE on M(indigotin)

moles 2-nitrobenzaldehyde = 2  0.038168 (= 0.076336)

TE on moles indigotin

moles 2-nitrobenzaldehyde required = 100  0.076336

85

= 0.089807

mass 2-nitrobenzaldehyde required = 0.089807 × 151

= 13.561

= 14 (g)

TE on moles 2-nitrobenzaldehyde

TE on M(2-nitrobenzaldehyde)

3

OR

Method 2 (M2 and M3)

mass indigotin if 100% yield

and

moles indigotin if 100% yield (1)

indigotin:2-nitrobenzaldehyde mol ratio

and

mass 2-nitrobenzaldehyde (1)

mass indigotin if 100% yield = 100  10.0

85

= 11.765 (g)

moles indigotin if 100% yield = 11.765 = 0.044903

262

TE on M(indigotin)

moles 2-nitrobenzaldehyde = 2  0.044903 (= 0.089807)

TE on moles indigotin

mass 2-nitrobenzaldehyde = 0.089807 × 151

= 13.561

= 14 (g)

TE on moles 2-nitrobenzaldehyde

TE on M(2-nitrobenzaldehyde)

Question

Number
Answer

Additional guidance

Mark

21(f)(i)

A drawing showing:

hydrolysed ester linkage

and

correct carbon frame and amine group (1)

deprotonated carboxylic acid and phenol groups (1)

Example of drawing:

Allow ONa

Do not award ONa

Allow E isomer

Allow Kekulé benzene

Allow any correct combination of skeletal,

structural or displayed formulae

2

Question

Number
Answer

Additional guidance

Mark

21(f)(ii)

A drawing showing:

correct condensation product

Example of drawing:

Allow Kekulé benzene

Allow any correct combination of skeletal,

structural or displayed formulae

Allow diacylated product, ie

1

Total for Section C = 20 marks
Total for Paper = 90 marks

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