1
TJC H2 Chemistry Preliminary Examination Paper 3 Answers
1
(a)
(i)
It acts as a Bronsted-Lowry base to abstract H+ from the carboxylic acid.[1]
(ii)
•
Electrophilic addition
δ+
δ-
Name of mechanism [1]
Mechanism [2]
Marker’s Comments:
A small number of students drew the primary carbocation intermediate without
realising that it is the less stable carbocation and that X would not be formed.
I
2
5
O
1
4
+
3
O
–
+
I
2
4
1
5
primary carbocation,
which is less stable
than the secondary
carbocation,
I
fast
3
O
NOT
I
O
O
product from
primary carbocation
product from
secondary carbocation
O
OH
(iii)
Energy
Ea2
Ea1
∆H0
+ IReaction Progress
Diagram & labelling:[1]
Ea1, Ea2 & ∆H0: [1]
9729 / TJC Prelims / 2022
O
2
Marker’s Comments:
•
Many students drew a single ‘hill’ energy profile despite drawing a two-step
mechanism in (a)(ii).
•
Most diagrams were poorly labelled. The following are common errors.
➢
Atoms were not balanced.
➢
4-pentenoic acid and HCO3− were labelled as reactants instead of the
4-pentenoate anion and I2.
➢
The arrow for Ea2 began from the same energy level as the reactants or as
Ea1 despite drawing the energy level of the intermediate higher than the
energy level of the reactant.
➢
The magnitudes for Ea1 and Ea2 are drawn equal or almost equal in a small
number of scripts.
(iv)

Marker’s Comments:
Common mistakes include
and
I
O
O
I
Do NOT assume that Step 2 always
produces a 5-membered ring.
O
O
This structure is also wrong, as
there are only 6 C atoms.
This structure is clearly wrong as
there are only 6 C atoms, while the
reactant from which it is formed has
7 C atoms.
(b)
Providing O2 forms Hb(O2)4(aq), which ✓ shifts the position of 1st equilibrium to the
right, ✓ removing Hb(aq) from the bloodstream.
By Le Chatelier’s Principle, the ✓ position of 2nd equilibrium will shift left to ✓ increase
amount of Hb(aq), thus reducing the amount of Hb(CO)4(aq) in the body.
2✓: [1]
Marker’s comments: Answer must focus on position of equilibrium of BOTH reactions and
how the amount of Hb(aq) changed, according to Le Chatelier’s Principle.
(c)
(i)
⇌ CO2 (g)
CO (g)
+ H2O (g)
Initial P / atm
0.0197
0.0394
0
0
∆ in P / atm
–0.0191
–0.0191
+0.0191
+0.0191
Eqm P / atm
0.0006
0.0203
0.0191
0.0191
9729 / TJC Prelims / 2022
+ H2 (g)
3
Working [1]
Kp =
(0.0191)(0.0191)
(0.0006)(0.0203)
= 30.0
Answer [1]
Marker’s comments: Read the question carefully for equilibrium partial pressure of
species (CO2 vs CO). Check calculations to avoid careless mistakes!!
(ii)
G (300) = – RT ln Kp (300)
= – (8.31)(300) ln (117000) = – 29090 J mol–1 [√]
G (900) = – RT ln Kp (900)
= – (8.31)(900) ln (1.978) = – 5101 J mol–1 [√]
G (300) = H – (300)S = – 29090 J mol–1 ----- (1) allow ecf [√]
G (900) = H – (900)S = – 5101 J mol–1 ----- (2) allow ecf [√]
(1) – (2):
S = – 40.0 J K–1 mol–1 (3 sf) allow ecf [1] include unit
Marker’s comments: Use the correct formula and substitute correct value
(temperature). Taking the difference of 2 G values is NOT S. There is no need to
find H using bond energy from Data Booklet as H will cancel out.
(iii)
The magnitude is small as there is no change in the amount of gaseous
molecules when reactants changed to products. Hence, there is little change in the
number of ways that energy can be distributed over the molecules.
Marker’s comments: S does not indicate spontaneity of reaction (G), answers
must focus on why S is small instead of the significance of small S.
(iv)
Boltzmann distribution diagram: [1]
Marker’s comments: Diagram must have proper axes labels and legends, correct
shading of area under curve. Curve must start at origin and peak of curve at higher
temp must be lower and shifted to the right.
At higher temperatures, the proportion of molecules with kinetic energy greater
than or equal to activation energy increases [√].
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4
Frequency of effective collision increases [√],
rate constant increases [√],
rate increases [√].
2√: [1]
Marker’s comment: answer must comment on how proportion of molecules’ energy
≥ Ea changes and how rate constant changes.
2
(a)
(i)
G: CO2
A: CuO
B: Cu2O
C: [Cu(H2O)6]2+
1 tick for each answer.
4 ticks − [2]
2 to 3 ticks − [1]
(ii)
Oxidation of copper in F, CuI = +1
Electronic configuration of Cu+: [Ar] 3d10
The d orbitals of Cu+ ion are fully filled.
No electrons can undergo d-d transition from the lower energy group of d orbitals
to the higher energy group. OR No d-d electronic transition can occur. [1]
Thus, solid F is expected to be white OR grey-white. [1]
(b)
✓ When NH3(aq) is added dropwise to C, a blue precipitate D, Cu(OH)2(H2O)4 OR
Cu(OH)2, is formed. NH3 acts as a base. (Note: Not ligand exchange)
✓ NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq) (Note: Not a full arrow)
✓ When excess NH3(aq) is added, E, which contains the dark blue complex,
[Cu(NH3)4(H2O)2]2+, is formed NH3 acts as a ligand here.
✓ By Le Chatelier’s Principle, concentration of [Cu(H2O)6]2+ decreases, and position of
equilibrium (1) shifts left, causing D to dissolve.
4 ticks − [2]
2 to 3 ticks − [1]
[Cu(H2O)6]2+(aq) + 2OH−(aq) ⇌ Cu(OH)2(H2O)4(s) + 2H2O(l) −−−− (1)
OR
Cu2+(aq) + 2OH−(aq) ⇌ Cu(OH)2(s) −−−− (1) [1]
[Cu(H2O)6]2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l) [1]
(Note: equation does not start with Cu(OH)2(H2O)4)
Marker’s comments: Be familiar with the formula of the species and check that the
equations are balanced.
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5
(c)
2+
N
N
2+
N
OH2
H2O
N
Cu
Cu
N
OH2
N
N
OH2
N
[1]
Y
[1]
Z
(trans-isomer of X)
Correct labels for Y and Z − [1]
Marker’s comments: Remember to include the charge and the diagram should
show O bonded to Cu.
(d)
𝐾𝑠𝑝 𝑜𝑓 𝐹𝑒(𝑂𝐻)2 = [𝐹𝑒 2+ ][𝑂𝐻 − ]2
8.0 × 10−16 = (0.100)[𝑂𝐻 − ]2
[𝑂𝐻 − ] = 8.944 × 10−8 𝑚𝑜𝑙 𝑑𝑚−3
Calculation of [OH−] at max separation − [1]
 pH at maximum separation = 14 – [– log10(8.944  10–8)] = 6.95 [1]
Explanation:
Cu(OH)2 would first be precipitated as a lower concentration of OH- is needed for
precipitation. Maximum separation is achieved when the maximum amount of Cu(OH)2 has
precipitated i.e. just before Fe(OH)2 is precipitated i.e. a saturated solution of Fe(OH)2.
(e)
(i)
step
1
reagents & conditions
compound
LiAlH4 in dry ether, r.t.p.
OR NaBH4 in ethanol, r.t.p.
OR H2, Ni, high temperature and pressure
[1]
OR H2, Pd or Pt, r.t.p.
[1]
2
P
excess conc H2SO4, 170 C
OR Al2O3, 400 C
[1]
[1]
Q
3
−
cold, conc H2SO4, followed by H2O(l), heat
OR H2O(g), conc H3PO4, high temperature and
pressure
[1]
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6
Marker’s Comments:
•
A number of students used ‘alcoholic KOH, heat’ to eliminate H2O from alcohol.
Note that ‘alcoholic KOH’, heat, is used to eliminate a HX from a halogenoalkane
(RX).
•
Many students got the number of C atoms wrong when drawing the skeletal
formula of P. Do be careful when drawing skeletal formulae, and make sure the
number of C atoms is correct.
terminal C atom
+H2O
OH
instead of
OH
9 C atoms
(ii)
WRONG!
8 C atoms;
missing 1 C atom
CORRECT
9 C atoms
Appropriate chemical test: [1]
Observation for both compounds: [1]
Possible Tests:
To
each
compound
separately, add
1
Observations
II (ketone)
2,4Orange ppt
dinitrophenylhydrazine.
R (secondary alcohol)
No ppt
Accept 2,4-DNPH OR
Brady’s reagent.
2
solid PCl5
No white fumes
White fumes (of HCl)
3
Na(s)
No effervescence
Effervescence.
H2 gas evolved “pops”
with a lighted splint
4
K2Cr2O7(aq), dil H2SO4 Orange solution does Orange
solution
and heat in a water bath. not turn green.
turns green.
Accept
K2Cr2O7.
5
acidified
KMnO4(aq), dil H2SO4 Purple solution is not Purple solution
and heat in a water bath. decolourised.
decolourised.
Accept
KMnO4.
is
acidified
Marker’s Comments:
•
Note that R contains the structure −CH(OH)CH3, and is also oxidised by hot,
alkaline I2(aq) to form yellow ppt of CHI3. Hence, this test cannot be used to
distinguish between R and II.
•
Compound II is not an aldehyde. Thus, Tollens’ or Fehling’s test cannot be used
to distinguish it from R.
9729 / TJC Prelims / 2022
7
3
(a)
(i)
Cream ppt is AgBr, Mr = 187.8
Amt of ppt formed = 0.392/187.8 = 0.002087 mol
Amt of Si3O2Br8 used = 0.192/755.5 = 2.541 x 10-4 mol (8.2 Br atoms)
Amt of Si2OBr6 used = 0.192/551.6 = 3.481 x 10-4 mol (5.99 Br atoms)
Hence, the oxybromide is Si2OBr6 [1]
working [1]
Marker’s comments: Calculate the number of moles of Br or Br-, not Br2.
(ii)
Cream ppt AgBr is sparingly soluble in water and exists in equilibrium with its
aqueous ions
AgBr(s) ⇌ Ag+(aq) + Br-(aq) … (1)
In the presence of aqueous NH3, free Ag+ forms a complex ion, [Ag(NH3)2]+.
However since AgBr has moderate Ksp value, ionic product [Ag+][Br-]
decreases but remains larger than Ksp(AgBr), so the ppt is remains insoluble.
[1]
When cyanide ions are added, ✓ Ag+ forms a more stable complex due to
larger Kstab value. Hence ✓ conc of Ag+ decrease to a greater extent. [1]
By Le Chatelier’s principle, ✓ position of equilibrium (1) shifts to right to
increase conc of Ag+ ions. ✓ Ionic product decreases below Ksp, hence cream
ppt dissolves. [1]
Marker’s comments: Common errors include:
(b)
(i)
•
Mixing up Ksp and Kstab
•
Inability to identify the ppt as AgBr. Some students thought the ppt were the
complexes.
•
Missing out the comparison between ionic product and Ksp
 A zwitterion is a dipolar compound that contains both cationic and anionic
groups and is electrically neutral. [1]
1 physical property [1]:
High melting point: in the solid state, dipeptide exist as zwitterions in which
both the anion and cation are held in the same unit (an internal salt). A large
amount of energy is required to overcome the strong electrostatic forces of
attraction or ionic bonds between the zwitterions.
Soluble in water: formation of ion-dipole interactions between water and
zwitterions release sufficient energy to overcome the ionic bonds between the
zwitterions and hydrogen bonds between water molecules.
Marker’s comments: It is vague to refer to a zwitterion as being “neutral”, compared
to “electrically neutral”.
The command word in the question is “explain”, so it is insufficient to only state the
physical property. Be careful in the use of term “zwitterion” vs “molecule”.
(ii)
At 1st equivalence point, the α-COOH of Phe is deprotonated, giving a
zwitterion
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8
Hence isoelectric point, pI = ½ (2.18+8.14) = 5.16 [1] (accept between 2.18 and
8.14)
Structure of zwitterion is
[1]
Marker’s comments:
Common mistakes include:
• missing H atoms on the α carbons
• having an extra O in the peptide bond (-COONH-)
• quoting one of the pKa values as the isoelectric pH
(iii)
E1 hydrolyse the peptide bond at the carboxylic acid end of phe, so 2 possible
structures are:
val-gln-phe-cys-phe-gln-lys-asp
cys-phe-val-gln-phe-gln-lys-asp
E2 hydrolyse the peptide bond at the amino end of gln, so 2 possible
structure are:
cys-phe-val-gln-lys-asp-gln-phe
cys-phe-val-gln-phe-gln-lys-asp
logical reasoning / working [1]
Hence the correct sequence of amino acids in Y is
 cys-phe-val-gln-phe-gln-lys-asp [1]
(c)
CH3COCl, room temp [1]
Marker’s comments: Know the conditions well, reactions with acyl chloride do not require
heating or catalyst.
Carboxylic acid and amine undergo neutralisation to give salt, rather than condensation to
give amide.
(d)
(i)
[1]
Marker’s comments: Some students leave the protecting group or R in the answer.
9729 / TJC Prelims / 2022
9
(ii)
Amine X is less basic than 2-aminopentane. [1]
The presence of the electron withdrawing oxygen atom reduces the electron
density on the nitrogen atom in amine X. The lone pair electrons are less
available to accept / dative bond with H+.[1]
Marker’s comments: 2-aminopentane is a 1° amine and has only one branched
alkyl group (not 2 alkyl groups).
(e)
1 mark for each step + charges + lone pair e and curly arrows
Marker’s comments: Read the description in the question carefully. Common mistakes
include:
(f)
•
attacking the N atom (rather than the C atom stated in the question)
•
not breaking the pi bond in first step
•
placing the + charge on the wrong atom
•
missing out the H+ in the products
(i)
CH3OH + H2O → CO2 + 6H+ + 6e
Marker’s comments:
(ii)
•
Be familiar with the ion-electron half equation method when balancing halfequations: Apple Of His/Her Eye. Do not use [O] to balance O atoms.
•
As CO2 is a product, reaction medium is assumed to be acidic.
 +0.40 V (any value less positive than +0.45 V. Accept only 1 value, reject if range
is given instead)
At the anode, both CH3OH and anion intermediate B can be oxidised. Since B
 is preferentially oxidised at the anode instead of CH3OH, its Eo value has to
be less positive, implying that B is more easily oxidised than CH3OH at the
anode.
Marker’s comments: Read the question carefully.
•
The question is focusing on oxidation reaction at the anode. Methanol is present
as the solvent (which is mentioned at the start of the question on p.14).
•
Context is electrolysis, not galvanic cell. It is thus inappropriate to explain the
need for Eo > 0.
cell
(g)
Total charge, Q = It = 3600  0.1 = 360 C
Amt of electrons used for electrolysis = 360/96500  0.95 = 3.54410-3 mol [1]
Amt of A produced = 3.54410-3 / 2 = 1.77210-3 mol
Mr of compound A = 145
9729 / TJC Prelims / 2022
10
Mass of A produced = 1.77210-3  145 = 0.257 g [1]
Marker’s comment: Time is in seconds, do not forget/ignore the efficiency of the reaction,
use correct mole ratio of electrons and A (1A ≡ 2e) from flow chart in qn.
4
(a)
(i)
High resistance voltmeter
V
Zn(s)
Salt bridge
1 mol dm-3
Zn2+(aq)
T = 298 K
Pt(s)
1 mol dm-3 Sn2+(aq)
and 1 mol dm-3 Sn4+(aq)
✓ voltmeter labelled (or circled V)
✓ Zn electrode and Pt electrode
✓ 1 mol dm-3 Zn2+
✓ 1 mol dm-3 each for Sn2+ and Sn4+
✓ 298 K
Note: Salt bridge must be labelled
Marker’s comment: A galvanic cell diagram comprises two half-cells under
standard conditions. Please be familiar with the 3 different types of half-cells.
Note: (not in the context of this question) If there are H+ or OH– in the half-equation,
it must be indicated in the cell diagram as well.
(ii)
EƟZn2+/Zn = -0.76V = EƟoxid
EƟSn4+/ Sn2+ = +0.15V = EƟred
EƟcell = EƟred – EƟoxid
= +0.15 – ( -0.76)
= + 0.91 V [1]
E values either quoted or shown in working. Sign must be shown.
(iii)
Formation of insoluble zinc carbonate causes the [Zn2+] to decrease. The
position of equilibrium of Zn2++ 2e- ⇌ Zn shifts to the left, favouring oxidation.
[1]
EƟZn2+/Zn or Eoxid becomes more negative, hence Ecell becomes more positive.
[1]
Note: Discussion in terms of Zn(OH)2 is also accepted as zinc carbonate exists as a
mixture of ZnCO3 and Zn(OH)2.
9729 / TJC Prelims / 2022
11
Marker’s comment:
(iv)
•
Be familiar with the terms “electrode potential” and “cell potential”. They are not
the same.
•
Carbonate ions react with zinc ions, not zinc. Important to recognise the
formation of insoluble ZnCO3, before explaining the decrease in [Zn2+].
ZnCO3(s) → ZnO(s) + CO2(g)
ΔH is positive for endothermic reaction.
ΔS is positive due to the increase in number of moles of gases, hence greater
disorder.
ΔG = ΔH – TΔS is negative at high temperature. [1]
Marker’s comment:
(v)
•
Thermal decomposition goes to 100% completion. It is erroneous to include a
reversible arrow in the balanced equation.
•
When discussing the impact of temperature on the spontaneity of the reaction,
need to recognise whether ΔH and ΔS are > or < 0, and explain why.
✓ Ionic radii: Zn2+ - 0.074nm, Ca2+ - 0.099nm
✓ Order of thermal stability ZnCO3 < CaCO3
2 points [1]
Zn2+ has the highest charge density. Zn2+ polarises the carbonate ion and
weakens the C-O bond to the greater extent. Less energy is needed to break
the C-O bond. [1]
(b)
(i)
No. of moles of MnO4– reacted with Sn2+ =
𝟏𝟕.𝟒𝟎
×
𝟏𝟎𝟎𝟎
𝟎. 𝟎𝟏𝟓 = 2.61 x 10-4 mol
𝟓
No. of moles of Sn2+ = 𝟐 × 𝟐. 𝟔𝟏 × 𝟏𝟎−𝟒 = 6.53 x 10-4 mol
No. of moles of MnO4- reacted with Sn2+ reduced from Sn4+
=
𝟐𝟔.𝟏𝟎
× 𝟎. 𝟎𝟏𝟓
𝟏𝟎𝟎𝟎
- 2.61 x 10-4
= 1.31 x 10-4 mol
𝟓
No. of moles of Sn4+ = = 𝟐 × 𝟏. 𝟑𝟏 × 𝟏𝟎−𝟒 = 3.28 x 10-4 mol
working [1]
Alternative
volume of MnO4- reacted with Sn2+ originally present : Sn2+ reduced from Sn4+
= 17.40 : (26.10 – 17.40)
= 17.40 : 8.70 [1]
Sn2+/ Sn4+ =
𝟔.𝟓𝟑 𝐱 𝟏𝟎−𝟒 𝟐
=
𝟑.𝟐𝟖 𝐱 𝟏𝟎−𝟒 𝟏
[1] or use volume ratio
Formula of G is Sn3O4 [1]
9729 / TJC Prelims / 2022
12
Marker’s comment:
(ii)
•
When Zn is added to another portion of H, Sn4+ that is present will be reduced
to Sn2+. During titration, MnO4– reacts with both sources of Sn2+ i.e. (i) Sn2+
originally present and (ii) Sn2+ from the reduction of Sn4+.
•
If only Sn2+ is present, formula of oxide will be SnO. If only Sn4+ is present,
formula of the oxide is SnO2. However, question mentioned that tin in both
oxidation states are present
SnO2 + 4HCl → SnCl4 + 2H2O [1]
SnO2 + 2NaOH → Na2SnO3 + H2O [1]
(c)
(i)
[1]
(ii)
✓ J has a high C:H ratio, J contains benzene.
✓ J undergoes electrophilic substitution with aqueous bromine to form K. J
is a phenylamine (not phenol since J does not contain O).
✓ 1 of the 2,4,6 positions of the phenylamine contains a substituent and is
not available for substitution by Br.
✓ J undergoes elimination with alcoholic NaOH to form L and M which are
alkenes.
✓ L and M undergoes oxidation/ oxidative cleavage with KMnO4. L contains
a terminal alkene.
✓ Following the reaction scheme in Fig. 4.1, N is propanone.
✓ J undergoes nucleophilic substitution to form P, -NH2 and the bromoalkyl
substituents should be adjacent to each other on the benzene.
Every 2 √ [1]
max [3]
Marker’s comments: Many students had forgotten the writing skills for deductive
question. Some common mistakes and misconceptions observed:
i)
did not comment on the ratio of C:H
ii)
reaction with Br2 gives white ppt is electrophilic substitution reaction and not
electrophilic addition. Conclusions should be phenylamine and not benzene
or alkene.
iii) J reacts with alcoholic NaOH is elimination and not alkaline hydrolysis
reaction.
iv) Heating J gives P, many thought it was a decomposition reaction. The
students did not relate to the halogenoalkane and phenylamine functional
groups present in J would lead to nucleophilic substitution reaction.
v)
Very few students could deduce the organic structures of J to P.
9729 / TJC Prelims / 2022
13
J√
K√
L√
M√
P [1]
structures [3]
5
(a)
(i)
√ For graph of G against T, y–intercept gives H and the gradient gives –S.
Since the gradient for reaction 2 is positive, S is negative.
√ This is because there is a decrease in disorder due to decrease in the number of
moles of gas molecules from 1 to 0.
√ Since H and S are both negative, –TS becomes more positive with
increasing temperature and hence the graph for reaction 2 becomes more
positive
3 √ = 2m; 2 √ = 1m
Marker’s comment: For those who attempted this question were able to deduce the
sign of ∆S.
(ii)
To prevent the hot titanium extracted from oxidising to TiO2
Note: traces of oxygen in the titanium tend to make the metal brittle.
Marker’s comment: Students should avoid giving general answers such as argon is
inert, no reaction with take place.
(iii)
½Ti + Cl2
½TiCl4
G at 1110 K = –350
Mg + Cl2
MgCl2
G at 1110 K = –465
G = –465 – (–350)
= –115 kJ mol–1
Marker’s comment: Some students were careless in reading the value for ∆G at
1100 K.
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14
(iv)
1870 K (at T higher than 1870 K, G > 0)
Accept 1860-1880 K
Marker’s comment: Well-answered except for some careless mistake.
Hfo(CoCl2)
(b)
Co(s)
+
Hatomo
Cl2(g)
CoCl2(s)
bond energy
of Cl-Cl
Co(g)
2Cl(g)
Hlatto
1st I.E.
-e
2(1st E.A.)
+2e
Co+(g)
2nd I.E.
-e
Co2+(g)
+
2Cl−(g)
OR
By Hess’ Law,
∆Hf o(CoCl2) = +427 + 244 + 757 + 1640 + 2(-364) + (-2624)
= -284 kJ mol-1
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1m – labelled Born-Haber cycle (axis, zero point, all enthalpy terms/value)
1m – balanced equations with state symbols
1m – calculated value with sign and units (do not award if no units/sign)
Marker’s comment: Most students were able to calculate and construct the cycle
well.
(c)
[1]
+ NH3 + KCN
CoCl2,
dilute HCl, heat,
followed by careful
neutralisation with
dilute NaOH
CH3CN, rt
[1]
Marker’s comment: The structure of aldehyde was erroneously given as
instead of
(d)
Observation
Type of reaction
Deduction
Q does not react with 2,4–
DNPH.
√ does not undergo
condensation
√ Q does not contain aldehyde or
ketone functional group.
Q reacts with alkaline
I2(aq) to give yellow ppt.
√ oxidation
√ Q contains CH3CH(OH)–
(reject if answer suggested
methyl ketone structure)
Q reacts with concentrated √ (Intramolecular)
H2SO4 to give R, which is a condensation
sweet–smelling liquid
Q reacts with concentrated
H2SO4 to give S and T,
with the formula C5H8O2
√ elimination
√ R is an ester.
√ Q contains both the –OH and –
COOH group
√ S and T contains C=C with √
two different substituents about
each sp2 C
(S and T exhibits cis-trans
isomerism)
S and T reacts with KMnO4 √ oxidation
to give U, C3H4O4 and
ethanoic acid
1 mole of U reacts with
excess Na2CO3 to give 1
mole of CO2.
√ The C=C in S/T is cleaved,
there is no terminal C=C as no
CO2 is produced.
√ acid–base reaction
OH
U contains 2
C O
√ U is a dicarboxylic acid.
9729 / TJC Prelims / 2022
16
Marker’s comment:
•
•
•
•
•
Carbonyl group ≠ carbonyl compound
Q does not undergo acidic hydrolysis with conc. H2SO4. Please note that acidic
hydrolysis requires aqueous H2SO4 (2021 A level paper has a similar question)
Please be specific when describing the presence of C=C bond. Stating that “a double
bond” is present is ambiguous as there are various double bonds e.g. C=O, C=N
A species that exhibits cis-trans has two different substituents about each sp2 C,
which is an important deduction
Please note that CO32– reacts with two moles of H+ to form H2CO3, which exists in
equilibrium with CO2 + H2O
OH
CH3CCH2CH2COOH
O
H3C
O
H
Q
H
CH3
C
C
S
R
H
CH2CO2H
CH3
H
C
C
H
CH2CO2H
T
HOOCCH2COOH
U
Structures: [1] each for Q, R, and U; [1] for both S and T (120○ about sp2 C)
Deductions:
[5]: 13 – 14 √
[4]: 10 – 12 √
[3]: 7 – 9 √
[2]: 4 – 6 √
[1]: 2 – 3 √
9729 / TJC Prelims / 2022