Mark Scheme (Results)

January 2025

Pearson Edexcel International Advanced Level

In Chemistry (WCH15)
Paper 01 Transition Metals and Organic Nitrogen
Chemistry
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January 2025
Question Paper Log Number P78459A
Publications Code WCH15_01_2501_MS
All the material in this publication is copyright
© Pearson Education Ltd 2025
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.
Using the Mark Scheme

Examiners should look for qualities to reward rather than faults to penalise. This does NOT
mean giving credit for incorrect or inadequate answers, but it does mean allowing candidates
to be rewarded for answers showing correct application of principles and knowledge.
Examiners should therefore read carefully and consider every response: even if it is not what is
expected it may be worthy of credit.

The mark scheme gives examiners:

• an idea of the types of response expected
• how individual marks are to be awarded
• the total mark for each question
• examples of responses that should NOT receive credit.

/ means that the responses are alternatives and either answer should receive full credit.
( ) means that a phrase/word is not essential for the award of the mark, but helps the examiner
to get the sense of the expected answer.

Phrases/words in bold indicate that the meaning of the phrase or the actual word is essential
to the answer.
ecf/TE/cq (error carried forward) means that a wrong answer given in an earlier part of a
question is used correctly in answer to a later part of the same question.

Candidates must make their meaning clear to the examiner to gain the mark. Make sure that
the answer makes sense. Do not give credit for correct words/phrases which are put together
in a meaningless manner. Answers must be in the correct context.

Quality of Written Communication

Questions which involve the writing of continuous prose will expect candidates to:
• write legibly, with accurate use of spelling, grammar and punctuation in order to make the
meaning clear
• select and use a form and style of writing appropriate to purpose and to complex subject
matter
• organise information clearly and coherently, using specialist vocabulary when appropriate.

Full marks will be awarded if the candidate has demonstrated the above abilities. Questions
where QWC is likely to be particularly important are indicated (QWC) in the mark scheme, but
this does not preclude others.
Section A

Question
Answer Mark
Number
1 The only correct answer is A (C2H6) (1)

B is incorrect because CO has a lone pair of electrons which can be donated to the central metal ion

C is incorrect because the hydroxide ion has a lone pair of electrons which can be donated to the central metal ion

D is incorrect because butylamine has a lone pair of electrons which can be donated to the central metal ion

Question
Answer Mark
Number
2 The only correct answer is D (Cu2+) (1)

A is incorrect because the solution would be pale pink so will not absorb red

B is incorrect because the solution would be pink so will not absorb red

C is incorrect because the solution would be green so will not absorb green
 Question
Answer Mark
Number
3 (1)
The only correct answer is D (758 1646 3232 4950 7671)

A is incorrect because there is a big jump in the value between 1st and 2nd electron being removed so it is in Group
1

B is incorrect because there is a big jump in the value between 3rd and 4th electron being removed so it is in Group 3

C is incorrect because there is a big jump in the value between 2nd and 3rd electron being removed so it is in Group 2

Question
Answer Mark
Number
4 The only correct answer is A (blue) (1)

B is incorrect because the VO2+ ion is blue not green

C is incorrect because the VO2+ ion is blue not violet

D is incorrect because the VO2+ ion is blue not yellow

Question
Answer Mark
Number
5 (1)
The only correct answer is B (CO and NO are absorbed by the catalyst)

A is incorrect because the catalyst contains platinum

C is incorrect because after the reaction, desorption of CO2 and N2 takes place

D is incorrect because it is a heterogeneous catalytic reaction

 Question
Answer Mark
Number
6 (1)
The only correct answer is B (gold and titanium)

A is incorrect because scandium is not a transition metal

C is incorrect because neither scandium nor zinc are transition metals

D is incorrect because zinc is not a transition metal

Question
Answer Mark
Number
7 (1)
The only correct answer is C (tetrahedral and square planar)

A is incorrect because [CuCl4]2- is tetrahedral

B is incorrect because [CuCl4]2- is tetrahedral and Pt(NH3)2Cl2 is square planar

D is incorrect because Pt(NH3)2Cl2 is square planar

 Question
Answer Mark
Number
8 The only correct answer is B (Eօcell is proportional to ln K and ∆Stotal) (1)

A is incorrect because Eօcell is not proportional to ln ∆Stotal

C is incorrect because ln Eօcell is not proportional to K

D is incorrect because Eօcell is not proportional to ∆Ssurrounding

Question
Answer Mark
Number
9(a) The only correct answer is A (C6H5NH2 and NaNO2 ) (1)

B is incorrect because C6H5NO2 will not react in this way

C is incorrect because C6H5NO2 will not react in this way and the nitrate should be nitrite

D is incorrect because the nitrate should be nitrite

Question
Answer Mark
Number
9(b) The only correct answer is D (phenol dissolved in an alkaline solution at 5°C) (1)

A is incorrect because the reaction will not take place in acid and the temperature is too high

B is incorrect because the temperature is too high

C is incorrect because the reaction will not take place in acid

 Question
Answer Mark
Number
10 The only correct answer is C (dissolve in the minimum volume of hot solvent, filter to remove the insoluble (1)
impurities, then cool and filter to remove the soluble impurities)

A is incorrect because the hot solvent should be used first

B incorrect because there is no cold filtering

D is incorrect because the first filtration removes the insoluble impurities and the second filtration removes the
soluble impurities

Question
Answer Mark
Number
11 The only correct answer is B (27.0 g) (1)

A is incorrect because the mass of phenol has been multiplied by 0.8 and 0.75

C is incorrect because only the yield in the second step has been used

D is incorrect because only the yield in the first step has been used
 Question
Answer Mark
Number
12 The only correct answer is D (C6H12) (1)

A is incorrect because this is an empirical not molecular formula

B is incorrect because the molar mass of H2 has been used instead of H

C is incorrect because this is the closest formula using the ratio of mass of compound to mass of carbon dioxide

Question
Answer Mark
Number
13 The only correct answer is A (15.6 and 9.6 ) (1)

B is incorrect because 0.5 moles of O2 has been used not 0.65

C is incorrect because 0.9 moles of O2 has been used not 0.65 and the moles of C4H10 have not been multiplied by 4

D is incorrect because the volume of gaseous H2O has been used not CO2
 Question
Answer Mark
Number
14 The only correct answer is B (1)

A is incorrect because (poly)ethenol only has one OH in each repeat unit

C is incorrect because (poly)ethenol only has one OH in each repeat unit

D is incorrect because (poly)ethenol only has one OH in each repeat unit

Question
Answer Mark
Number
15 The only correct answer is A (H2N(CH2)6NH2 and HOOC(CH2)4COOH) (1)

B is incorrect because nitriles will not react in this way

C is incorrect because these monomers will not make this polymer

D is incorrect because the OH group will not react with the amine group
 Question
Answer Mark
Number
16(a) The only correct answer is C ( compound Y ) (1)

A is incorrect because it is not an amino acid

B is incorrect because it is not optically active

D is incorrect because it is not optically active

Question
Answer Mark
Number
16(b) The only correct answer is C ( compound Y) (1)

A is incorrect because it would give 3 proton NMR peaks and 3 13C peaks

B is incorrect because it would give 3 proton NMR peaks and 3 13C peaks

D is incorrect because it would give 3 proton NMR peaks and 3 13C peaks
 Question
Answer Mark
Number
17(a) The only correct answer is D (1)

A is incorrect because the amine group has not been protonated

B is incorrect because this is the zwitterion and would not be present at pH 2

C is incorrect because the amide oxygen would not be protonated

Question
Answer Mark
Number
17(b) The only correct answer is B (doublet) (1)

A is incorrect because on the adjacent C there is only one H so it would produce a doublet

C is incorrect because on the adjacent C there is only one H so it would produce a doublet

D is incorrect because on the adjacent C there is only one H so it would produce a doublet
 Section B

Question
Answer Additional Guidance Mark
Number
18(a)(i) An answer that makes reference to the following points: (4)

• A (saturated/concentrated solution of) potassium nitrate / (1) Allow sodium nitrate / NaNO3/ sodium chloride /
KNO3 NaCl/potassium chloride/KCl

• B platinum /Pt (1) Allow black Pt

• C (solution containing) iron(II) sulfate / FeSO4 and Accept iron nitrates and chlorides
iron(III) sulfate / Fe2(SO4)3 (1) Allow just Fe2+ and Fe3+
If name and formula are given both must be correct
but only penalise once.

• concentration 1 mol dm–3/ 1M with respect to Fe2+ and

Fe3+ One ion and its correct concentration this will score
or (1) 1 mark.
1 mol dm–3 of FeSO4 and 0.5 mol dm–3 Fe2(SO4)3 Ignore any reference to pressure
Ignore any state symbols
Question
Answer Additional Guidance Mark
Number
18(a)(ii) An answer that makes reference to the following point: (1)

• ions can flow through a salt bridge Allow the ions can move/pass
(but not through a wire) Allow ions cannot flow through the wire
Ignore to balance the ions
Ignore wire will interfere with the
reaction/products/cell

Do not award
electrons can flow through the salt bridge
ions can travel through the wire

Question
Answer Additional Guidance Mark
Number
18(b) An answer that makes reference to the following points: Example of equation (2)

• correct species (1) Zn + 2Fe3+ Zn2+ + 2Fe2+

• correct direction and balancing (1) Ignore state symbols even if incorrect
Allow ⇌ if Zn is on the LHS
Penalise uncanceled species, including Pt only once
Question
Answer Additional Guidance Mark
Number
18(c) An answer that makes reference to the following points: Standalone marks (2)

• the Eɵcell would increase/ become more positive (1)

• Zn2+(aq) + 2 e− ⇌ Zn(s) equilibrium would shift to the (1) Allow just (Zn cell) eqm shift to the left making it
LHS making the Zn2+: Zn cell more negative (and so more negative
Allow less positive/smaller
increasing the Eɵcell)
Or

as Zn2+ concentration has decreased the reaction

Zn + 2Fe3+  Zn2+ + 2Fe2+ will move to the right
Question
Answer Additional Guidance Mark
Number
19 An explanation that makes reference to the following points: (6)

• Step 1
LiAlH4/lithium aluminium hydride/ (1) Allow lithal
lithium tetrahydridoaluminate Do not award NaBH4
and
(dry) ether

• Step 2
KBr and conc H2SO4 or HBr or PBr3 / I2 and (red) P or PI3 (1) Accept phosphoric(V) acid for sulfuric acid
or HI / PCl5 or PCl3 Allow ≥ 50% for conc
Allow HCl
• compound X
Dependent on the reagent used in step 2.
bromoethane / C2H5Br Allow any type of formula
iodoethane / C2H5I
chloroethane / C2H5Cl (1)

• Step 3
magnesium and (dry) ether (1) Do not award if other reagents are added

• Grignard reagent (1) CH3CH2MgX/ CH3CH2-Mg-X

Do not award CH3CH2XMg
• Step 4 Dependent on compound X

dry ice/carbon dioxide/CO2 (and then hydrolyse using an

acid/water) (1) Allow CO2 and H+/acid

No TE
Ignore refluxing/any temperature throughout
Question
Answer Additional Guidance Mark
Number
20(a)(i) An answer that makes reference to the following points: Example of equation (2)

• equation with correct species (1) 2Cr3+ + 3H2O2 + H2O Cr2O72− + 8H+
Allow ⇌
• balanced (1) Allow multiples
Allow 1 mark for the correct equation with
additional uncanceled H+, H2O and electrons.

Question
Answer Additional Guidance Mark
Number
20(a)(ii) An answer that makes reference to the following points: (2)

suitable metal
• Mg / V / Zn / Fe / Ni / Cu (1) No other metals will score and no TE on other
metals
correct Eө cell
• Mg = (+) 3.7 (V), V = (+) 2.51(V), Zn = (+) 2.09(V),
Fe = (+)1.77(V), Ni = (+) 1.58 (V), Cu = (+) 0.99 (V) (1)
Question
Answer Additional Guidance Mark
Number
20(a)(iii) An answer that makes reference to the following points: (2)

• [Cr (NH3)6 ] 3+ (1) Ignore omission of square brackets

Ignore (aq)

• ligand exchange (1) Allow ligand substitution / replacement

Question
Answer Additional Guidance Mark
Number
20(a)(iv) An answer that makes reference to the following points: (2)

• not redox because the oxidation number of chromium has Allow just ‘no as the oxidation number of
not changed (1) chromium has not changed’

• oxidation number is 6/+6/6+/VI in both Cr2O72– and (1)

CrO42–

Question
Answer Additional Guidance Mark
Number
20(b)(i) An answer that makes reference to the following point: (1)

• KCr(SO4)2 Allow ions in any order

Ignore correct charges on some/all of the ions
Question
Answer Additional Guidance Mark
Number
20(b)(ii) Example of calculation (4)

• M1 calculation of molar mass of KCr(SO4)2 (1) 39.1+ 52 + (32.1 × 2) + (16 × 8) = 283.3 (g mol−3)
Allow TE from formula in (b)(i)

• M2 moles of KCr(SO4)2 (1) 56.74 ÷ 283.3 = 0.200 (mol)

Allow TE from M1

• M3 moles of water (1) (100 −56.74)(= 43.26) ÷ 18 = 2.403 (mol)

Allow fractions

• M4 calculation of no of moles of water of crystallisation (1) 2.403 ÷ 0.200 = 12

An alternative route using mass Correct answer with some working scores 4

• M2 = 283.3 ÷ 56.74 × 100 = 499.3 (g)

• M3 = 499.3 – 283.3 ÷ = 216 (g)

• M4 = 216 ÷ 18 = 12
Question
Answer Additional Guidance Mark
Number
20(c)(i) Example of calculation (2)

• calculation of g dm-3 (1) 1.345 × 10–7 (mol dm-3) × 52.0 = 6.994 × 10–6
(g dm–3)

• calculation of g cm-3 6.994 × 10-6 (g dm–3) ÷ 103 = 6.994 ×10−9

and
calculation of ppb (1) = 6.994 ×10−9 ×109 = 6.994 (ppb)
(which is greater than 4 ppb)

Or
• calculation of mol cm-3 (1) 1.345 × 10–7 (mol dm-3) ÷ 103 = 1.345 × 10–10

• calculation of g cm-3 1.345 × 10–10 × 52.0 = 6.994 ×10−9

and
calculation of ppb (1) = 6.994 ×10−9 ×109 = 6.994 (ppb)
(which is greater than 4 ppb)
Other units can be used provided they are consistent and a
comparison made. Ignore SF
699.4 (ppm) exceeds 400 (ppm)
6.994 ×10−9 (g cm-3 ) exceeds 4.00 10−9 (g cm-3) Correct answer with no working scores 2

Alternative comparison using mol dm-3

• 4 ×10−9 ÷ 52 = 7.692 10−11 mol dm-3 (1)

• 7.692 10−11 × 1000 = 7.692 10−8 which is smaller than

1.345 × 10−7 mol dm-3 (1)
Question
Answer Additional Guidance Mark
Number
20(c)(ii) An answer that makes reference to the following points: (2)

• (EDTA4– ) is a hexadentate ligand/can form 6 dative Allow multidentate/ polydentate

(covalent) bonds/multiple dative (covalent) bonds with (1) Allow coordinate bonds
the Cr3+ Ignore just you need fewer EDTA4– ions than
diaminoethane molecules

• (EDTA4– ) complex is more stable than (1) Allow leads to a (large) increase in total
(bidentate complexes)/ is a chelating agent/ traps the entropy/entropy of the system
Cr3+/can wrap around the Cr3+ (so the Cr3+ can be Allow there is an increase in disorder
removed from the blood) Ignore just EDTA4– is more stable
Question
Answer Additional Guidance Mark
Number
21(a)(i) An answer that makes reference to the following point: (1)
The electrons in the doubled orbit must be pointing
• in opposite directions. They can be in any of the 3d
orbitals.
Allow half headed arrows or a combination of both

Question
Answer Additional Guidance Mark
Number
21(a)(ii) An answer that makes reference to the following points: (3)

• (the oxygen in the air) oxidises the Fe2+ (1) Allow it is oxidised
Ignore reacts with oxygen

• (forming the brown) Fe3+ (1) Allow iron(III) sulfate forms

Allow any mention of Fe3+

• Fe3+ is more stable (than Fe2+) due to half-filled d- Allow reverse argument Fe2+ has a pair of electrons
subshell/ due to half full (d ) orbitals (1) in one orbital that repel each other and so an
electron is easily lost
Or
pair of electrons in one orbital that repel each other
and so it is less stable

Ignore a half-filled d orbital

Ignore half-filled d shell
Question
Answer Additional Guidance Mark
Number
21(b) Example of calculation (6)

• M1 calculation of moles of MnO4– in the titre (1) 17.70 × 0.00740 ÷ 1000 = 1.3098 × 10–4 /
0.00013098 (mol)

• M2 calculation of moles of Fe2+ in 25 cm3 of solution (1) 1.3098 × 10–4 / 0.00013098 × 5 = 6.549 × 10–4
/0.0006549 (mol)

• M3 calculation of moles of Fe2+ in 250 cm3 of solution (1) 6.549 × 10–4 × 10 = 6.549 × 10–3 /0.006549 (mol)

• M4 calculation of mass of FeSO4 in 250 cm3 of solution (1) 6.549 × 10–3 × 151.9 = 0.99479(g)

• M5 calculation of % FeSO4 in the moss killer (1) 0.99479 ÷ 6.42 × 100 = 15.495 (%)

M6 answer to 2 or 3 SF (1) 15.5 (%)/15(%)

This is not a standalone mark it can only be
Marks are for the processes are shown and they may be in a awarded if there has been an attempt to calculate a
different order %
M2 = × 5
M3 = × 10 Ignore intermediate rounding and incorrect
M4 = × 151.9 truncating
M5 = ÷ 6.42 × 100 TE throughout
The correct answer with or without working scores
6
Question
Answer Additional Guidance Mark
Number
21(c) An answer that makes reference to the following points: (2)

• the iron(II) ions are surrounded by water ligands/exist as (1) Allow [Fe(H2O)6]2+
an aqua complex

• which are polarised by the iron(II) ions so lose protons (1) Allow just protons are lost/ deprotonation takes
(to water molecules) place
Allow any balanced equation showing
deprotonation
[Fe(H2O)6]2+ [Fe(H2O)5OH]+ + H+
M2 is dependent on M1 or near miss as it must be
clear that the protons are coming from the complex
Question
Answer Additional Guidance Mark
Number
*22 This question assesses the student’s ability to show a coherent and Guidance on how the mark scheme should be applied. (6)
logically structured answer with linkages and fully sustained reasoning.
The mark for indicative content should be added to
Marks are awarded for indicative content and for how the answer is the mark for lines of reasoning. For example, a
structured and shows lines of reasoning. response with five indicative marking points that is
partially structured with some linkages and lines of
The following table shows how the marks should be awarded for reasoning scores 4 marks (3 marks for indicative
indicative content. content and 1 mark for partial structure and some
Number of indicative marking Number of marks awarded for linkages and lines of reasoning).
points seen in answer indicative marking points
6 4 If there were no linkages between the points, then the
5-4 3 same indicative marking points would yield an overall
3-2 2 score of 3 marks (3 marks for indicative content and
1 1 no marks for linkages).
0 0

The following table shows how the marks should be awarded for In general it would be expected that
structure and lines of reasoning 5 or 6 indicative points would get 2 reasoning marks
3 or 4 indicative points would get 1 reasoning mark
Number of marks awarded for 0, 1 or 2 indicative points would get zero reasoning
structure of answer and marks
sustained lines of reasoning
Answer shows a coherent logical 2 If there is any incorrect chemistry, deduct mark(s)
structure with linkages and fully from the reasoning. If no reasoning mark(s) awarded
sustained lines of reasoning do not deduct mark(s).
demonstrated throughout
Comment: Look for the indicative marking points
Answer is partially structured with 1
first, then consider the mark for the structure of the
some linkages and lines of
answer and sustained line of reasoning
reasoning
Answer has no linkages between 0
points and is unstructured

Indicative content
Similarity
• IP1
both reactions are electrophilic substitution

Differences
• IP2 Allow C6H6 + Br+ C6H5Br + H+
C6H6 + Br2 C6H5Br + HBr Can be shown by a correct mechanism

• IP3 Allow C6H5OH + 3Br2 C6H2Br3OH + 3HBr

Ignore state symbols

Two correct aromatic products identified regardless

of the equations will score ONE IP for IP2 and IP3.
If name and formula are given both must be correct.

• IP4
benzene requires a named catalyst e.g AlBr3/ FeBr3 (and heat) Allow AlCl3/ Fe + Br2
This can be shown via an equation
Do not award bromine water
• IP5
phenol reacts with bromine water/ Allow phenol reacts with bromine without a
phenol reacts (with bromine) at room temperature catalyst
Ignore milder conditions
• IP6 Ignore just quicker/ easier to react
phenol is more reactive because the lone pair of electrons on
the oxygen atom (the lone pair on the OH will not score) are Allow any indication that the lone pair on the O of
delocalised into the ring (making phenol more susceptible to the phenol becomes delocalised within the ring
electrophilic attack)
Ignore the electron pair on the OH becomes
delocalised within the ring
 Section C

Question
Answer Additional Guidance Mark
Number
23(a) An answer that makes reference to the following point: (1)
C13H20N2O2
• correct formula Allow any order and non-subscripts

Question
Answer Additional Guidance Mark
Number
23(b)(i) An answer that makes reference to the following points: Standalone marks (2)

• CH3Cl/ chloromethane (1) Allow CH3Br/ bromomethane /CH3l /iodomethane

• AlCl3 (1) Allow AlBr3 / FeBr3 / FeCl3

Question
Answer Additional Guidance Mark
Number
23(b)(ii) A description that makes reference to the following points: (5)
HNO3 + H2SO4 → NO2+ + HSO4− + H2O
• equation to show the formation of the electrophile (1) Or
HNO3 + 2H2SO4 → NO2+ + 2HSO4– + H3O+
Or
HNO3 + H2SO4 → H2NO3+ + HSO4–
and H2NO3+ → NO2+ + H2O

• curly arrow from anywhere on the central ring to positive (1) Allow curly arrow from anywhere in the hexagon
nitrogen Do not award if the arrow is heading to the O

• structure of intermediate (1) Horseshoe facing the bottom tetrahedral carbon and
covering at least three carbon atoms. Some part of
the positive charge in the horseshoe

• curly arrow from C-H bond to reform the ring (1)

• equation showing regeneration of catalyst (1) HSO4− + H+ → H2SO4

Allow M5 as part of mechanism, with curly arrow
Example of mechanism from oxygen of HSO4− to H on benzene ring

If the NO2 is attached in a different position

penalise M3 only. Likewise, if benzene is used M3
is penalised.
Question
Answer Additional Guidance Mark
Number
23(b)(iii) An answer that makes reference to the following points: (2)

• esterification Allow addition- elimination

(1) Allow condensation
Do not award condensation polymerisation
•

(1) Do not award

Accept HO for H-O

Allow displayed /structural formulae
Do not award molecular formula
Penalise -H-O connectivity

Question
Answer Additional Guidance Mark
Number
23(b)(iv) An answer that makes reference to the following point: (1)

• Sn/tin and (concentrated) HCl/ hydrochloric acid Ignore tin is a catalyst

Do not award dilute HCl
Question
Answer Additional Guidance Mark
Number
23(c)(i) A description that makes reference to the following points: (3)

• alkyl groups attached (to the N) are electron releasing/ (1) Allow positively inductive for electron releasing
donating

• benzene ring attached (to the N) is electron withdrawing/ (1)

lone pair gets incorporated into (the delocalised electrons
of ) the benzene ring

• the basicity of the alkyl N is greater because the lone pair Allow basicity greater due to higher electron
is more available to accept/attract a proton/form a dative density (on the alkyl N)
covalent bond Allow basicity greater as it forms stronger bonds
Or with proton

the basicity of the N attached to the benzene ring is Allow basicity lower due to lower electron density
weaker as the lone pair is less available to accept/attract a (1) (on the aryl N)
proton Allow basicity lower as it forms weaker bonds with
proton
Question
Answer Additional Guidance Mark
Number
23(c)(ii) A description that makes reference to the following points: (1)

Allow just NH+ with no covalent bond

Allow normal covalent bond
Allow + charge anywhere adjacent to the NH or the
NH3 or outside brackets if drawn
Ignore lack or position of Cl−
Allow HCl added to the other NH2 if in (c)(i) the
NH2 is thought to be more basic
Allow HCl added to any N if they have not said
which is more basic in (c)(i)

Question
Answer Additional Guidance Mark
Number
23(c)(iii) An answer that makes reference to the following points: (2)

• procaine hydrogen chloride is ionic (1)

• the ions are hydrated by the water (and the compound is (1) Allow ion-dipole interaction
more soluble)
Ignore reference to any type of intermolecular
bonds
Question
Answer Additional Guidance Mark
Number
23(d) Example of calculation (3)

• mass of lidocaine in 1.5 cm3 (1) (1.5 ÷ 2.2) × 0.044 (g) = 0.03 / 3 x10–2 (g)

• mol of lidocaine in 1.5 cm3 (1) 0.03 (g) ÷ 234 = 1.2821 × 10–4 / 0.00012821 (mol)

• molecules of lidocaine in 1.5 cm3 (1) 1.2821 × 10–4 × 6.02 × 1023 = 7.7179 × 1019
Alternative 1
• mol of lidocaine in 2.2 cm3 cartridge. (1) 0.044 ÷ 234 = 1.88 × 10–4/ 0.00188 (mol)

• mol of lidocaine in 1.5 cm3 (1) 1.88 × 10–4 × 1.5 ÷ 2.2 == 1.282 × 10–4 (mol)

• molecules of lidocaine in 1.5 cm3 (1) 1.2821 × 10–4 × 6.02 × 1023 = 7.7179 × 1019

Alternative 2
(1) 0.044 ÷ 234 = 1.88 × 10–4/ 0.000188 (mol)
• mol of lidocaine in 2.2 cm3 cartridge.
(1) 1.88 × 10–4 × 6.02 × 1023 = 1.13176 × 1020
• molecules of lidocaine in 2.2 cm3 cartridge
(1) 1.13176 × 1020 × 1.5 ÷ 2.2 = 7.7179 × 1019
• molecules of lidocaine in 1.5 cm3

Ignore intermediate rounding

Ignore SF except 1SF in final answer
Correct answer with or without working scores 3
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