OCR A Level Chemistry A (H432)
Chemistry B (H433)
PAG 6: Synthesis of an organic solid
Exam Questions and Mark Scheme
OCR A Level Chemistry A (H432)
... [Show More] Chemistry B (H433)
PAG 6: Synthesis of an organic solid
Exam Questions and Mark Scheme
1. A solid organic compound can be purified by recrystallisation.
Which statement(s) about recrystallisation is/are true?
1 The organic compound is more soluble in hot solvent.
2 The hot solution is cooled before the purified organic compound is collected.
3 The melting point of the purified organic compound is lower than the impure compound.
A 1, 2 and 3
B Only 1 and 2
C Only 2 and 3
D Only 1
Your answer
[1]
2. When are insoluble impurities removed during recrystallisation?
A when the hot solution is filtered
B as the solution cools
C when the crystals are filtered off
D when the crystals are washed
Your answer [1]
2(a). Benzoic acid, C6H5COOH, is added to some foods as a preservative.
A student prepares benzoic acid as outlined below.
The student mixes 4.00 cm3 of phenylmethanol, C6H5CH2OH, (density = 1.04 g cm−3) with
Step 1 sodium carbonate and aqueous potassium manganate(VII), as an oxidising agent. The
mixture is heated under reflux.
Step 2
The resulting mixture is cooled and then acidified with concentrated HCl . Impure crystals
of benzoic acid appear.
Step 3 The student recrystallises the impure crystals to obtain 1.59 g of pure benzoic acid.
In Step 1, sodium carbonate, Na2CO3, makes the reaction mixture alkaline.
Write an ionic equation to show how carbonate ions form an alkaline solution in water.
[1]
(b). In Step 2, explain why the mixture must be acidified so that crystals of benzoic acid appear.
[1]
(c). Write the overall equation for the preparation of benzoic acid from phenylmethanol.
Use [O] for the oxidising agent.
[1]
(d). Calculate the percentage yield of benzoic acid.
Give your answer to 3 significant figures.
percentage yield = ............................................................. % [3]
(e). In Step 3, describe how the student can recrystallise the impure crystals to obtain pure benzoic
acid.
[2]
3(a). * Aspirin can be made by refluxing 2-hydroxybenzoic acid with ethanoic anhydride, (CH3CO)2O, in
the presence of a concentrated acid catalyst.
• Aspirin is much more soluble in hot water than in cold water.
• The percentage yield of aspirin from this method is 90.0%.
Outline a plan to prepare 8.10 g of pure aspirin.
In your answer, include the mass of 2-hydroxybenzoic acid needed and the purification steps from
the hot reaction mixture.
[6]
(b). The Rf values and melting point ranges of 2-hydroxybenzoic acid and pure aspirin are shown in the
table.
Compound Rf Melting point range / °C
2-Hydroxybenzoic acid 0.30 158–161
Pure aspirin 0.75 138–140
i. A student analyses the purity of their impure aspirin by thin-layer chromatography (TLC).
From the results the student concludes that the impure aspirin is contaminated with a small
amount of unreacted 2-hydroxybenzoic acid.
Draw spots on the chromatogram below to show how the student arrived at this conclusion.
[2]
ii. Predict the melting point range of the impure aspirin.
[1]
4(a). Aspirin is a medicine that reduces fever and relieves pain.
Some students prepare a sample of aspirin from salicylic acid.
salicylic acid
Before the students start the preparation, they test the salicylic acid with iron(III) chloride.
What colour would they see?
[1]
(b). The students then make aspirin by warming 6.0 g of salicylic acid with 10 cm3 of ethanoic anhydride in
the presence of concentrated sulfuric acid.
HOOCC6H4OH
+
salicylic acid
(CH3CO)2O
ethanoic
anhydride
→
HOOCC6H4OCOCH3
+
aspirin
i.
Balance the equation by writing the structural formula of the other product on the dotted line. [1]
ii. The density of ethanoic anhydride is 1.1 g cm−3.
Calculate the amount (in moles) of ethanoic anhydride used.
amount of ethanoic anhydride = mol [
iii. Which is in excess, the salicylic acid or the ethanoic anhydride?
[2]
(c). The students pour their hot solution into water and aspirin crystallises out as the water cools.
The students then look for a suitable solvent to recrystallise the aspirin.
i. State the properties of a suitable solvent for recrystallisation.
[1]
ii. Name a method for testing the purity of the aspirin formed.
[1]
(d). After recrystallisation, the students obtained 3.1 g of aspirin.
What value for the percentage yield does this give?
yield = % [2]
5. Two students aim to prepare a sample of phenyl benzoate using the reaction below.
*The students purified the phenyl benzoate by recrystallisation from a suitable solvent.
Describe and explain the steps in the purification of an organic solid by recrystallisation, including
the properties needed of a suitable solvent.
[6]
END OF QUESTION PAPER
Mark scheme
Question Answer/Indicative content Marks Guidance
1
B
1
Examiner’s Comments
Many candidates correctly selected B.
Option A proved a good distractor,
presumably as candidates linked melting
point to recrystallisation without fully
interpreting statement 3.
Total 1
2
A
1
Examiner’s Comments
Over half answered this correctly. The
spread of incorrect responses suggests
that there is confusion between the terms
soluble and insoluble.
Total 1
2 a
2– – –
CO3 + H2O → OH + HCO3
OR
CO3 + H2O → 2OH + CO2 ✓
2– –
1 ALLOW
2– –
CO3 + 2H2O → 2OH + H2CO3
IGNORE state symbols
ALLOW inclusion of Na+ as spectator
ion, e.g.
2Na+ + CO3
2– + H2O → 2OH– + 2Na+ +
CO2
IGNORE
Na2CO3 + H2O → 2NaOH + CO2
Ionic equation required
IGNORE equation with H+ or H3O+
2– + –
e.g. CO3 + H → OH + CO2
Question asks for reaction with H2O
Examiner’s Comments
This equation presented problems for
many candidates, despite the question
asking for an equation between
carbonate ions and water. An acceptable
equation had to be ionic and needed to
produce OH– (for the alkaline solution)
and either HCO3
– or CO2.
Many candidates wrote an equation with
H+ instead of H2O, with lower ability
candidates showing the carbonate ion
with the wrong charge as CO3
–.
Many candidates wrote full equations
despite the question asking for an ionic
equation. Candidates do need to read
the instructions in the question.
b
Acid/H+/HCl reacts with OR protonates
• benzoate / C6H5COO–
• carboxylate / salt
(to form benzoic acid) ✓
1
ALLOW suitable equation, e.g.
C6H5COO– + H+ → C6H5COOH
IGNORE responses purely in terms of
neutralisation of alkali, e.g. Acid/H+/HCl
neutralises / reacts with/removes alkali /
OH– / CO3
2– / Na2CO3
Examiner’s Comments
Candidates found this part extremely
difficult. The question was aimed to
stretch and challenge.
Many candidates followed on directly
from part (a), stating in simple terms that
the alkaline solution needed to be
neutralised to remove hydroxide ions.
However, candidates were expected to
recognise that the alkaline conditions
would lead to benzoate ions rather than
benzoic acid being present in the
mixture. The mixture is acidified to
protonate the benzoate. The hint in the
question was about making the benzoic
acid appearing when acid is added.
c
C6H5CH2OH + 2[O] → C6H5COOH +
H2O ✓
1
ALLOW molecular, structural, displayed
formulae, etc
e.g. molecular:
C7H8O + 2[O] → C7H6O2 + H2O
Examiner’s Comments
This part discriminated well with many
candidates being able to write a correct
equation using their knowledge of the
oxidation of alcohols. Mistakes usually
resulted in the balancing with either [O]
instead of 2[O] or 2H2O instead of H2O.
Written equations always need to be
checked for the atoms balancing.
d
FIRST CHECK THE ANSWER ON
ANSWER LINE
If answer = 33.8 OR 33.9 (%) award 3
marks
Theoretical moles
n(C6H5COOH) OR n(C6H5CH2OH)
OR 0.0385….. (mol)
✓
Actual moles
n(C6H5COOH OR 0.013(0)
) ….
(mol) ✓
= 33.8% OR
33.9
(3 sig fig) ✓
Answer depends on some
intermediate roundings to 3SF
3 ALLOW ECF for each step
Calculator = 0.03851851852
Calculator = 0.01303278689
Alternative method using mass
1. Theoretical moles = 0.0385 mol
2. Mass = 0.0385 × 122.0 = 4.70 g
3
.
Common errors
35.2% → 2 marks
•
(no use of density)
36.5 OR 36.6% → 2 marks
•
(÷ density instead of × density)
Examiner’s Comments
Candidates are well practised with
percentage yield calculations with about
half obtaining the correct percentage
yield of 33.8 or 33.9% to secure all 3
marks. Many were able to secure partial
credit for incorrect answers, provided
that the working was laid out clearly.
Some responses showed a simple
percentage of the two masses with no
consideration of moles or molar masses.
Such a response received no credit.
e Dissolve in the minimum quantity of 2 ALLOW any solvent
hot water/solvent ✓
Cool
AND
Filter
AND
(leave to) dry ✓
All three needed
DO NOT ALLOW use of drying agent
(e.g. MgSO4)
IGNORE
• Initial filtering
• hot filtration to remove insoluble
impurities
Examiner’s Comments
Many candidates produced thorough
responses, showing that they had
encountered recrystallisation as a
technique in their practical work.
Most candidates were aware that the
impure product is dissolved in a
minimum volume of hot solvent, although
‘minimum’ was sometimes omitted.
The subsequent stages were sometimes
incomplete or in the wrong order. Many
were aware that the hot solution can be
passed through fluted filter paper to
remove solid impurities. (This is beyond
the specification requirements for A Level
but good practice).
Most candidates were aware of the need
to filter (usually under reduced pressure)
but the necessary cooling stage to form
the crystals was sometimes omitted.
Finally, many responses omitted the
need to dry the crystals. Candidates did
sometimes dry the crystals by adding an
anhydrous salt (e.g. CaCl2 or MgSO4), a
clear confusion with drying an organic
liquid. Others described the purification
of an organic liquid for their response,
including use of a separating funnel,
drying and distillation.
Total 8
3
a
Level 3 (5–6 marks)
Correctly calculates mass of 2-
hydroxybenzoic acid.
AND
Outlines full details of the two steps to
obtain a pure sample of aspirin from
the hot reaction mixture
• Calculation shows all relevant
steps.
• Purification steps are detailed
and clear, in the correct order,
using appropriate scientific
terms, e.g. filter under reduced
pressure / using a Buchner
flask; dissolve in the minimum
volume of solvent.
Level 2 (3–4 marks)
Attempts a calculation which is mostly
correct
AND
Some details of steps to obtain impure
aspirin from the hot reaction mixture
and recrystallisation
• Calculation can be followed but
lacks clarity.
• Purification steps lack detail,
e.g. filter without reduced
pressure; dissolve without
minimum volume of solvent.
Level 1 (1–2 marks)
Attempts to calculate the mass of B
using mole approach but makes little
progress with only 1 step correct.
AND
Few or imprecise details about steps to
obtain impure aspirin from hot reaction
mixture and recrystallisation
• Calculation is difficult to follow
and lacks clarity
• Purification steps are unclear
with few scientific terms and
little detail, e.g. just ‘filter and
crystallise’.
0 marks: No response or no response
worthy of credit.
6 Indicative scientific points, with
bulleted elements, may include:
1. Mass of 2-hydroxybenzoic acid
•
• n(2-hydroxybenzoic acid) needed
• Mass = 0.0500 × 138 = 6.9(0) g
2. Purification
Impure aspirin from hot reaction
mixture
• Cool reaction mixture
• Filter product under reduced
pressure
Recrystallisation of impure aspirin:
• Dissolve impure solid in
minimum volume of hot water /
solvent
• Cool solution and filter solid
• Wash with cold water / solvent
and dry
NOTE Filtration of hot solution to remove
solid particles is not required.
b
i
Pure aspirin and 2-hydroxybenzoic
acid correct
Impure aspirin with 2 spots in line with
aspirin and 2-hydroxybenzoic acid
spots
AND 2-hydroxybenzoic acid spot
fainter than aspirin spot
2 Check measurements on diagram using
online measuring tool.
Distance from baseline to top of spot for
aspirin
= 70–80% of baseline → solvent front
Distance from baseline to top of spot for
aspirin
= 25–35% of baseline → solvent front
ii
Melting point range between 130–
140°C
AND
Range ≥ 2°C
1
Range that starts <138 and finishes
≤140
Total 9
2
a
violet/purple/mauve/lilac ✓
1
ALLOW anything which suggests a
mixture of red and blue (This is a
definition of purple)
IGNORE reference to starting colour
Examiner’s Comments
The majority of candidates recognised
this test for phenols but a few stated the
colour would be green rather than purple.
b i CH3COOH ✓ 1 ALLOW any unambiguous structural
formula
Examiner’s Comments
This was not well answered. Many
candidates gave a molecular rather than
unambiguous structural formula. H2O
was a common incorrect response.
ii
(10 × 1.1/102) = 0.11 ✓
1
Evidence of calculation NOT required
ALLOW 0.1078…… OR 0.108
Examiner’s Comments
Evidence of calculation was not required
for this mark. Higher ability candidates
generally answered correctly (0.11) but
many did not to multiply by 10 cm3.
Answers correctly rounded to 2 or more
significant figures were accepted.
iii
(Amount salicylic acid = 6/138) = 0.043
(1 or more sf) (mol) ✓
(amount ethanoic anhydride = 0.11) so
ethanoic anhydride ( in excess.) ✓
2
Second mark dependent on evidence of
calculation (ECF).
ALLOW ECF from 21bii
Examiner’s Comments
This question could only be answered by
calculating the amount of salicylic acid.
The concept of one reactant being in
excess seems well known but some
candidates justified a choice of ethanoic
anhydride from a practical perspective
e.g. ‘There must be an excess of
anhydride to react all the salicylic acid’.
This response was not accepted in the
context of the question.
c
i
(solute/aspirin/solid) soluble in hot
AND less soluble/insoluble in cold ✓
1
Examiner’s Comments
Higher attaining candidates recognised
the need for differential solubility in hot
and cold solvent but fewer than half of all
candidates gained this mark.
ii
melting point ✓
1
ALLOW GLC or HPLC or TLC
DO NOT ALLOW just “chromatography”.
NOT “titration”
Examiner’s Comments
Half of the responses were correct.
Higher attaining candidates tended to
choose melting point but TLC was also
common. Measurement of boiling point
was quite often incorrectly given.
d
CHECK ANSWER ON ANSWER LINE
If answer lies within range 39% to
40.1% award 2 marks (2sf or more)
✓✓
OR
100% yield = 0.043(5) × 180 OR 7.74g
OR 7.83 (if 0.435 used)✓
% yield = 3.1 × 100/100% yield
calculated above ✓
2 Alternative:
Moles aspirin formed = 3.1/180 OR
0.017(2) ✓
% yield = moles of aspirin calculated
above × 100/0.043(5) ✓
ALLOW calculations dependent on ECF
from 21biii {i.e. either incorrect value for
moles of salicylic acid OR answers
based on moles of ethanoic anhydride
(calculated in 21bii) if salicylic acid said
to be in excess – this gives yield =16%}
Examiner’s Comments
Higher attaining candidates correctly
calculated a yield in the allowed range
(39 to 40.1%) but it was evident that
many candidates did not understand the
concept of yield. Lower attaining
candidates frequently divided 3.1 by the
mass of salicylic acid (or by the total
mass of reagents).
Total 9
3 Level 3 (5–6 marks)
Description and detailed explanation of
most stages in the recrystallization.
AND
Specifies some of the desirable
properties of a suitable solvent.
There is a well-developed line of
reasoning which is clear and logically
structured. The information presented
is relevant and substantiated.
Level 2 (3–4 marks)
Description of most stages in the
recrystallisation with a limited
explanation of the steps.
AND
Specifies some of the desirable
properties of a suitable solvent.
There is a line of reasoning presented
with some structure. The information
presented is relevant and supported by
some evidence.
6 Indicative scientific points may
include:
Description of steps in
recrystallisation
• dissolves organic solid in a
minimum volume of (suitable) hot
solvent
• filters hot solution
• cools solution
• re-filters
• allows crystals to dry
Explanation of steps
• vacuum filtration used so solution
remains hot
• insoluble impurities removed
• crystals formed when solution
cools vacuum filtered again;
soluble impurities go through in
solution
Level 1 (1–2 marks)
Limited description of stages in the
recrystallisation with detailed
explanation of some of the stages.
AND
Specifies some of the desirable
properties of a suitable solvent.
The information is basic and
communicated in an unstructured way.
The information is supported by limited
evidence and the relationship to the
evidence may not be clear.
0 marks
No response or no response worthy of
credit
• washed with cold solvent
(minimises further loss of product
but ensures any soluble
impurities left washed through)
• Crystals left to either air dry or in
oven at a temperature lower than
mp of solid
Desirable properties of a suitable
solvent
• solid should be very soluble in
solvent chosen at high temps, but
not very soluble at low temps
• soluble impurities remain in
solution even at low temps
Total 6 [Show Less]