PMT
A-level
CHEMISTRY
(7405/2)
Paper 2: Organic and Physical Chemistry
Mark scheme
Specimen paper
PMT
MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Mark schemes are prepared by the Lead Assessment Writer and considered, together with the
relevant questions, by a panel of subject teachers. This mark scheme includes any amendments
made at the standardisation events which all associates participate in and is the scheme which was
used by them in this examination. The standardisation process ensures that the mark scheme covers
the students’ responses to questions and that every associate understands and applies it in the same
correct way. As preparation for standardisation each associate analyses a number of students’
scripts. Alternative answers not already covered by the mark scheme are discussed and legislated
for. If, after the standardisation process, associates encounter unusual answers which have not been
raised they are required to refer these to the Lead Assessment Writer.
It must be stressed that a mark scheme is a working document, in many cases further developed and
expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark
schemes on the basis of one year’s document should be avoided; whilst the guiding principles of
assessment remain constant, details will change, depending on the content of a particular
examination paper.
Further copies of this mark scheme are available from aqa.org.uk
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
01.1 Consider experiments 1 and 2: [B constant]
[A] increases × 3: rate increases by 32 therefore 2nd order with 1 AO3 1a
respect to A
Consider experiments 2 and 3:
[A] increases × 2: rate should increase × 22 but only increases × 2
Therefore, halving [B] halves rate and so 1st order with respect to B 1 AO3 1a
Rate equation: rate = k[A]2[B] 1 AO3 1b
01.2 rate = k [C]2[D] therefore k = rate / [C]2[D] 1 AO2h
k= 7.2 × 10 -4 = 57.0 1 AO2h Allow consequential marking on incorrect
transcription
(1.9 × 10 −2 )2 × (3.5 × 10 −2 )
–2
mol dm s
+6 –1 1 AO2h Any order
–2 2 –2 –3 –3 –1
01.3 rate = 57.0 × (3.6 × 10 ) × 5.4 × 10 = 3.99 × 10 (mol dm s ) 1 AO2h
OR
–2 2 –2
Their k × (3.6 × 10 ) × 5.4 × 10
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
01.4 Reaction occurs when molecules have E≥E a 1 AO1a
Raising T by 10 °C causes many more molecules to have this E 1 AO1a
Whereas doubling [E] only doubles the number with this E 1 AO1a
01.5 E a = RT(lnA – lnk)/1000 1 AO1b Mark is for rearrangement of equation and factor of
1000 used correctly to convert J into kJ
–1
E a = 8.31 × 300 (23.97 – (–5.03))/1000 = 72.3 (kJ mol ) 1 AO1b
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
02.1 Gradient drawn on graph 1 AO3 1a Line must touch the curve at 0.012 but must not
cross the curve.
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
02.2 Extended response
–3
Stage 1: Rate of reaction when concentration = 0.0120 mol dm
From the tangent
Change in [butadiene] = –0.0160 – 0 and change in time = 7800 – 0 1 AO3 1a
–6
Gradient = –(0.0160 – 0)/(7800 – 0) = –2.05 × 10
Rate = 2.05 × 10–6 (mol dm–3 s–1) 1 AO3 1a
Stage 2: Comparison of rates and concentrations
Marking points in stage 2 can be in either order
–6 –6
Initial rate/rate at 0.0120 = (4.57 × 10 )/(2.05 × 10 ) = 2.23 1 AO3 1a
Inital concentration/concentration at point where tangent drawn =
0.018/0.012 = 1.5 1 AO3 1a
Stage 3: Deduction of order
2
If order is 2, rate should increase by factor of (1.5) = 2.25 this is 1 AO3 1b
approximately equal to 2.23 therefore order is 2nd with respect to
butadiene
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
03.1 2,2,4-trimethylpentane 1 AO1a
03.2 5 1 AO2b
03.3 C 20 H 42 C 8 H 18 + 2C 3 H 6 + 3C 2 H 4 1 AO2b
03.4 Mainly alkenes formed 1 AO1b
03.5 4 (monochloro isomers) 1 AO2b
CH3 H CH3
1 AO2a
H3C C C C CH3
H Cl CH3
03.6 1 AO2a
Cl
Cl
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
03.7 C 8 H 17 35Cl = 96.0 + 17.0 + 35.0 = 148.0 1 AO1b Both required
and C 8 H 17 37Cl = 96.0 + 17.0 + 37.0 = 150.0
M r of this C 8 H 17 Cl (1.5 × 148.0) + (1.0 × 150.0) = 148.8 1 AO1b
2.5 2.5
03.8 24.6 2.56 72.8 = 2.05 : 2.56 : 2.05
12 1 35.5
Simplest ratio = 2.05 : 2.56 : 2.05
2.05 2.05 2.05
= 1 : 1.25 : 1 1 AO2b
Whole number ratio (× 4) = 4 : 5 : 4 1 AO2b
MF = C 8 H 10 Cl 8 1 AO2b
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
04.1 3-methylbutan-2-ol 1 AO1a
04.2 1 AO2g Allow (CH 3 ) 2 CHCOCH 3
CH3
H3C C C CH3
H O
04.3 Elimination 1 AO1a
04.4 CH3 1 AO2g Allow (CH 3 ) 2 C=CHCH 3
H3C C C CH3
CH3 1 AO2g Allow (CH 3 ) 2 CHCH=CH 2
H3C C C CH2
H H
H
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
04.5 Position 1 AO1a
04.6 CBA 1 AO3 1b
04.7 CH3 1 AO2g Allow (CH 3 ) 2 C(OH)CH 2 CH 3
H3C C CH2CH3
OH
04.8 CH3 1 AO2e Allow (CH 3 ) 3 CCH 2 OH
H3C C CH2OH
CH3
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
05.1 Secondary 1 AO1a
05.2 Nitrogen and oxygen are very electronegative 1 AO1a
Therefore, C=O and N–H are polar 1 AO1a
Which results in the formation of a hydrogen bond between O and H 1 AO1a
In which a lone pair of electrons on an oxygen atom is strongly 1 AO1a
attracted to the δ+H
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
06.1 CH3 NH2 1 AO2a
H CHOH (CH2)4
H2N C C N C C N C COOH
H O H H O H H
glycine threonine lysine
06.2 1 AO2a
H
H3N C COO
06.3 CH3 H 1 AO2a Allow
H3C N C C OH (Br )
(CH3)3N CH2 COOH (Br )
CH3 H O
06.4 2-amino-3-hydroxybutanoic acid 1 AO2a
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
06.5 1 AO2a
NH3
(CH2)4
H3N C COOH
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
07.1 H CH3 AO1a
1
C C
CH3 Cll
Addition 1 AO1a
07.2 H H 1 AO2e
HO C C OH
CH3 CH3
O CH3 H O O CH3 H O
1 AO2e
HO C C C C OH OR Cl C C C C Cl
CH3 H CH3 H
07.3 Q is biodegradable 1 AO2g
Polar C=O group or δ+ C in Q (but not in P) 1 AO2c
Therefore, can be attacked by nucleophiles (leading to breakdown) 1 AO2c
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
08.1 2-deoxyribose 1 AO1a
08.2 Base A 1 AO3 1b If Base B stated, allow 1 mark only for response
including hydrogen bonding
Top N–H forms hydrogen bonds to lone pair on O of guanine 1 AO2a
The lone pair of electrons on N bonds to H–N of guanine 1 AO2a
A lone pair of electrons on O bonds to lower H–N of guanine 1 AO2a Allow all 4 marks for a correct diagram showing the
hydrogen bonding
Students could also answer this question using
labels on the diagram
08.3 Allow either of the nitrogen atoms with a lone pair NOT involved in 1 AO2a
bonding to cytosine
08.4 Use in very small amounts / target the application to the tumour 1 AO2e
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
09.1 (nucleophilic) addition-elimination 1 AO1a Not electrophilic addition-elimination
Allow C 6 H 5 or benzene ring
M2
M3 Allow attack by :NH 2 C 6 H 5
O
O O M2 not allowed independent of M1, but allow M1 for
CH3 C CH3 C Cl CH3 C correct attack on C+
Cl NH M3 for correct structure with charges but lone pair on
C6H5 N H
C6H5 4 AO2a O is part of M4
C6H5NH2 H
M4 (for three arrows and lone pair) can be shown in
M1 RNH2 more than one structure
M4 for 3 arrows and lp
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
09.2 The minimum quantity of hot water was used:
To ensure the hot solution would be saturated / crystals would form on 1 AO1b
cooling
The flask was left to cool before crystals were filtered off:
Yield lower if warm / solubility higher if warm 1 AO1b
The crystals were compressed in the funnel:
Air passes through the sample not just round it 1 AO1b Allow better drying but not water squeezed out
A little cold water was poured through the crystals:
To wash away soluble impurities 1 AO1b
09.3 Water 1 AO3 1b Do not allow unreacted reagents
Press the sample of crystals between filter papers 1 AO3 2b Allow give the sample time to dry in air
09.4 M r product = 135.0 1 AO2h
Expected mass = 5.05 × 135.0 = 7.33 g 1 AO2h
93.0
Percentage yield = 4.82 × 100 = 65.75 = 65.8(%)
7.33 1 AO1b Answer must be given to this precision
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
09.5 1 AO2c
NHCOCH3 NHCOCH3
+ NO2+ + H+
NO2
OR
+ +
C 6 H 5 NHCOCH 3 + NO 2 → C 6 H 4 (NHCOCH 3 )NO 2 + H
09.6 Electrophilic substitution 1 AO1a
09.7 Hydrolysis 1 AO3 1a
09.8 Sn/HCl 1 AO1b Ignore acid concentration; allow Fe/HCl
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
10 IR Extended response
–1
M1 Absorption at 3360 cm shows OH alcohol present 1 AO3 1a Deduction of correct structure without explanation
scores maximum of 4 marks as this does not show a
NMR clear, coherent line of reasoning.
M2 There are 4 peaks which indicates 4 different environments of 1 AO3 1a Maximum of 6 marks if no structure given OR
hydrogen if coherent logic not displayed in the explanations of
M3 The integration ratio = 1.6 : 0.4 : 1.2 : 2.4 1 AO3 1a how two of OH, CH 3 and CH 2 CH 3 are identified.
The simplest whole number ratio is 4 : 1 : 3 : 6
M4 The singlet (integ 1) must be caused by H in OH alcohol 1 AO3 1a
M5 The singlet (integ 3) must be due to a CH 3 group with no 1 AO3 1b
adjacent H
M6 Quartet + triplet suggest CH 2 CH 3 group 1 AO3 1b
M7 Integration 4 and integration 6 indicates two equivalent 1 AO3 1b
CH 2 CH 3 groups
CH2CH3
1 AO3 1b
H3C C OH
M8
CH2CH3
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
11.1 CH 3 CH 2 COCH 3 + 2[H] CH 3 CH 2 CH(OH)CH 3 1 AO1b
11.2 This question is marked using levels of response. Refer to the Mark 6 1 Indicative Chemistry content
Scheme Instructions for Examiners for guidance on how to mark this AO1a
Stage 1: Formation of product
question.
Level 3 All stages are covered and the explanation of each stage is • Nucleophilic attack
5 • Planar carbonyl group
generally correct and virtually complete.
5–6 marks –
• H attacks from either side (stated or drawn)
AO2a
Answer is communicated coherently and shows a logical
progression from stage 1 to stage 2 then stage 3. Stage 2: Nature of product
Level 2 All stages are covered but the explanation of each stage may
be incomplete or may contain inaccuracies OR two stages • Product of step 1 shown
3–4 marks • This exists in two chiral forms (stated or
are covered and the explanations are generally correct and
virtually complete. drawn)
• Equal amounts of each enantiomer/racemic
Answer is mainly coherent and shows progression from stage mixture formed
1 to stage 3.
Level 1 Two stages are covered but the explanation of each stage Stage 3: Optical activity
may be incomplete or may contain inaccuracies, OR only one
1–2 marks • Optical isomers/enantiomers rotate the plane
stage is covered but the explanation is generally correct and of polarised light equally in opposite directions
virtually complete.
• With a racemic/equal mixture the effects
cancel
Answer includes isolated statements but these are not
presented in a logical order or show confused reasoning.
Level 0
Insufficient correct chemistry to gain a mark.
0 marks
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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 - SPECIMEN
Question Marking guidance Mark AO Comments
12.1 HBr OR HCl OR H 2 SO 4 1 AO1b Allow HI or HY
12.2 Electrophilic addition 1 AO1a
H M3 H
H M1 H 4 AO2a Allow consequential marking on acid in 12.1 and
C C H3C C CH2CH3 H3C C CH2CH3 allow use of HY
H CH2CH3 Br
H
M4 Br
M2 Br
12.3 The major product exists as a pair of enantiomers 1 AO2a
The third isomer is 1-bromobutane (minor product) 1 AO2a
Because it is obtained via primary carbocation 1 AO2a
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