Download 2011-A-Level-CH-H2-P3-soln

‘A’-Level H2 Chem Paper 3 (Nov 2011) Solutions
1 (a)
Oxidation number of iodine is +3
(b)
[1]
[1]
[1/2]
5 bond pairs and 1 lone pair = Square pyramidal
4 bond pairs and 2 lone pairs = Square planar
lp-lp>bp-lp>bp-bp
[1]
[1/2]
[1]
Examiners’ comments:
The dot-and-cross diagrams were generally drawn well, although several
candidates’ drawings were too compressed and confused for clear
interpretation. Some omitted the extra electron responsible for the negative
charge in ICl4-.
(c)(i)
Solve this using redox half-equation method.
H+ + HClO + 2e  Cl- + H2O
2I-  I2 + 2e
H+ + HClO + 2I- Cl- + H2O + I2
(c)(ii)
[1]
Amount of S2O32- = 0.006 x 0.0050 = 0.00003mol
Amount of I2 = 0.000015mol
[1]
Amount of HClO = Amount of N-chlorosuccinimide = 0.000015mol
[1]
Mr of N-chlorosuccinimide (C4H4NO2Cl) = 133.5
Mass of N-chlorosuccinimide = 133.5 x 0.000015 = 0.00200g (3s.f.)
[1]
(d)(i)
R-CONH2 + Br2  R-CONHBr + HBr
[1]
(d)(ii)
Oxidation number of Br = +1
[1]
(d)(iii)
Number of electrons around N atom in B = 6 electrons
[1]
(d)(iv)
The other product is CO32-
(e)
R-N=C=O + 2 OH-  R-NH2 + CO32-
[1]
Step V: Br2, AlBr3 dark
[7]
Step VI: KMnO4, dilute H2SO4, heat with reflux
Step VII: PCl5, room temperature / SOCl2, room temperature, pyridine / PCl3
heat.
Step VIII: NH3, room temperature
Structure C:
Structure D:
[1 mark per step / structure; except Structure D = 2 marks]
Examiners’ comments:
This was well-answered by the vast majority of candidates. Common incorrect
answers were using aqueous Br2 (with AlCl3 etc) for stage V and using HCl or
Cl2 or CH3COCl for stage VII, or using aqueous SOCl2.
2 (a)(i)
pV = nRT
n = (1.00 x 105 x 386 x 10-6) / (8.31 x 298) = 0.015587 mol
Mass of gases = 1.00 – 0.438 = 0.562 g
Average Mr = 0.562 / 0.015587 = 36.1 (1d.p.)
[1]
(a)(ii)
Carbon dioxide
[1]
(a)(iii)
Volume of CO2 = 193 cm3
Volume of other gas = 386 – 193 = 193 cm3
The two gases are in equal amounts.
Mr of CO2 = 44
Let Mr of unknown gas be x.
(44 + x)/2 = 36.1
Mr of unknown gas = 28.2
(a)(iv)
(b)(i)
It is carbon monoxide
[1]
E is CaC2O4.
[1]
Balanced equation: CaC2O4  CaO + CO + CO2
[1]
Lower pK1 value for malonic acid implies it is a stronger acid than ethanoic
acid.
The monoanion is stabilised by hydrogen bonding with the unionised –CO2H
group.
OR The monoanion is stabilised by an electron withdrawing –COOH group.
[1]
CH3CH2CO2- + H+
CH3CH2CO2H
(b)(ii)
Ethyl group is electron donating, destabilises the ethanoate ion by intensifying
negative charge. Thus malonic acid is stronger acid than ethanoic acid.
[1]
-
[1]
-
OOC-CH2-COOH
OOC-CH2-COO- + H+
It is harder to remove H+ from a monoanion hence it is less favoured, thus pK2
is higher.
(b)(iii)
pK1 = 2.85
[1]
K1 = [H+]2 / 0.10 = 10-2.85
[H+] = 0.01185
pH = 1.93 (3s.f.)
(b)(iv)
At initial point (0 cm3): pH is 1.93 (calculated in (b)(iii))
Volume of NaOH needed to neutralise malonic acid completely = 20 cm3
Since dibasic acid, two equivalence points (10 cm3 and 20 cm3)
pH at first maximum buffer capacity (5 cm3) = pK1 = 2.85
pH at second maximum buffer capacity (15 cm3) = pK2 = 5.70
pH at second equivalence point > 7 (basic salt formed)
pH at 30 cm3 = pH of NaOH = 13
pH
12.4
7
5.70
2.85
1.93
Vol of NaOH added / cm3
5
10
15
20
30
[1/2 mark for each underlined point and ½ mark for graph]
Examiners’ comments:
(b)(i) Although most candidates correctly deduced that malonic acid was a
stronger acid than ethanoic acid, many thought this was due to the
delocalisation of the negative charge in the monoanion over the two –CO2H
groups, or equated the greater acidity to the presence of two –CO2H groups.
Drawing the displayed structure of the monoanion, which few candidates did,
would not only have shown them that this delocalisation was impossible, but
might also have given them a hint towards the correct answer; stabilisation of
the monoanion by hydrogen bonding with the unionised –CO2H group.
(b)(ii) One or two candidates realised that the stabilising hydrogen bonding
would be destroyed by the ionisation of the second –CO2H group, and hence
would cause the pKa to be higher than expected. Most candidates who gained
credit here did so by realising that the removal of H+ from a species that
already carries a negative charge would be electrostatically unfavourable.
(b)(iv) A significant number of candidates only drew the titration curve for a
monobasic acid. Key points looked for were that the curve started at pH =
1.93, that there were two neutralisation points, at V = 10 cm3 and V = 20 cm3,
that there were two ‘buffer’ regions, centred at pH values corresponding to
the two given pKa values and that the curve flattened off between pH 12 and
13.
(c)(i)
Structure for F: O=C=C=C=O
[1]
Shape: Linear with respect to all atoms
[1]
(c)(ii)
[1]
With NH3:
[1]
With HCl:
(d)
To form HO2C-CO2H:
BrCH2CH2Br
[2]
HOCH2CH2OH
NaOH (aq), heat with reflux
K2Cr2O7, dilute H2SO4,
heat with reflux
HO2C-CO2H
[1 mark for first step including intermediate, 1 mark for second step]
To form HO2C-CH2CH2-CO2H:
[2]
Ethanolic KCN, heat with reflux
BrCH2CH2Br
NCCH2CH2CN
dilute HCl, heat with
reflux
HO2C-CH2CH2-CO2H
[1 mark for first step including intermediate, 1 mark for second step]
Examiners’ comments:
The NaOH had to be aqueous, not ethanolic, in order to substitute OH for Br-;
and for the change from Br to CN, NaCN without HCN is required. More care
could have been taken by some in the drawing of the intermediates.
CNCH2CH2CN represents an iso-nitrile, rather than the correct dinitrile
NCCH2CH2CN. It should also be noted that oxidising ethanediol with hot
acidified KMnO4 would result in the production of CO2 rather than ethanedioic
acid. Acidified K2Cr2O7 would be the reagent of choice.
3 (a)(i)
[Ca2+] = 0.0008 / 40 = 0.0002 mol dm-3
[Mg2+] = 0.0049 / 24.3 = 0.0002 mol dm-3
[Cl-] = 0.0071 / 35.5 = 0.0002 mol dm-3
[HCO3-] = 0.0366 / 61 = 0.0006 mol dm-3
[1 mark if all calculations correct]
Salts: CaCl2, MgCl2, Ca(HCO3)2, Mg(HCO3)2
[1 mark for both correct chlorides, 1 mark for both correct
hydrogencarbonates]
[4]
Relative amounts: CaCl2 : MgCl2 : Ca(HCO3)2 : Mg(HCO3)2 = 1 : 1 : 3 : 3 [1 mark]
(a)(ii)
CaCO3
[1]
The anion must be divalent because these tend to be insoluble, as opposed to
the monovalent anions which tend to be soluble. As the sample of water was
boiled, CO2(g) escaped and the equilibrium below shifted left:
CO2(g) + H2O(l) + CO32-(aq)
2HCO3-(aq)
CaCO3 is more insoluble than MgCO3 as the salts become more insoluble down
(a)(iii)
the group. Ksp of CaCO3 is smaller than that of MgCO3.
[1]
The increase in [CO32-] will increase the ionic products of both CaCO3 and
MgCO3. Since Ksp of CaCO3 is lower, ionic product of CaCO3 will exceed its Ksp
first, hence precipitating first.
[1]
The rock was contained calcium ions, magnesium ions, carbonate ions and
chloride ions. Carbon dioxide in the atmosphere dissolves in rainwater and
shifts the equilibrium in (a)(ii) to the right to form hydrogencarbonate ions.
Chloride ions simply dissolve in the water.
[1]
[1]
Examiners’ comments:
(a)(ii) Majority of candidates suggested either a chloride or a
hydrogencarbonate as the first solid to be precipitated, despite the former
being described in bullet point 3 as being soluble, and the latter in bullet point
4 as not existing in the solid state. Arguments couched in terms of water
boiling off and hence decreasing its concentration are spurious.
(a)(iii) Once again, the information given in the bullet points was sometimes
ignored. A significant proportion of candidates suggested the (solid) rock was
composed of hydrogencarbonates. An explanation of the formation of
hydrogencarbonates in the mineral water should have explained the shifting
of equilibrium to the right due to the presence of atmospheric CO2 dissolved
in the rainwater.
(b)(i)
6Ca(PO4)3F + SiO2 + H2O  SiF4 + 2HF + 3CaO + 9Ca3(PO4)2
[1]
a = 1, b = 1, c = 1, d = 2, e = 3, f = 9
(b)(ii)
(b)(iii)
Ksp = [Ca2+]3 [PO43-]2
[1]
Units for Ksp = mol5 dm-15
[1]
Let solubility of Ca3(PO4)2 be x.
Ksp = (3x)3 (2x)2
[1]
x = 2.473 x 10-6
[Ca2+] = 3x = 7.42 x 10-6 mol dm-3
[1]
(c)(i)
PCl3 can be made by heating phosphorus in the presence of a small amount of
chlorine, while PCl5 is made by heating phosphorus in excess chlorine.
[1]
(c)(ii)
G: POCl3
[1]
Shape: Tetrahedral with respect to P atom
[1]
(c)(iii)
4 (a)
H: SO2
[1]
PCl3 + SO3  POCl3 + SO2
[1]
PCl5 + H2O  POCl3 + 2HCl
[1]
Secondary structure of protein is formed by folding of polypeptide chain by
hydrogen bonding [1]between O of C=O and H of N-H of the peptide linkages
[1].
[6]
Tertiary structure of protein is the unique 3D arrangement due to the
interactions between the R-groups (side chains) of alpha amino acids [1] of
the polypeptide backbone. These interactions can be van der waals’ forces,
hydrogen bonding, ionic bonding, disulfide linkages and ion-dipole
interactions [1].
Quarternary structure of proteins is formed when two or more polypeptide
chains [1] interact with one another through side-chain interactions similar to
tertiary structure [1].
Examiners’ comments:
Incorrect responses were given by not specifying that the hydrogen bonding
which occurs in the secondary structure takes places between the C=O and NH groups of peptide groups in the polypeptide backbone (rather than
between the side-chains of the R groups). To gain credit for quarternary
structure, candidates needed to specify that more than one polypeptide chain
was involved.
(b)
(c)(i)
As the chains coalesce, bonds are being formed in secondary, tertiary and
quarternary structures. Bond formation is exothermic so ΔH is negative.
[1]
As the chains come together to form a molecule, the degree of disorderliness
decreases and ΔS is negative.
[1]
Since the coalescence is spontaneous, ΔG is negative as magnitude of ΔH is
greater than magnitude of TΔS
[1]

In the isolated gas phase atoms, all 5 d orbitals of the metal ion are
degenerate

In forming complexes, the d orbitals split into 2 groups (non-degenerate)
with an energy gap between them.

When a d-electron from the lower energy group is promoted to the higher
energy group, energy from the visible region is absorbed (green). The light
energy not absorbed will be seen (reflected) as the colour of the complex
[1]
[1]
(c)(ii)
(red).
High spin state:
[1]
Low spin state:
[1]
(c)(iii)
Electrons prefer to occupy orbitals singly to minimise inter-electron repulsion.
[1]
(c)(iv)
Oxyhaemoglobin, having a low spin state, has a larger energy gap between its
d-orbitals. The electrons do not occupy the higher energy levels because the
energy gap is greater than the repulsion energy, hence rather be paired up.
[1]
Dilute HCl / NaOH(aq)
[1]
Heat for several hours
[1]
(d)(i)
(d)(ii)
5(a)
[1]
met-asp-gly-ser-ala-gly-glu-ser-lys-tyr
[2]
H+ + OH-
[4]
H2O
CH3CH2OH
CH3CH2O- + H+
+ H+
[1 mark for all 3 dissociation equations]
phenol is the most acidic, followed by water and ethanol is the least acidic. [1]
Phenoxide ion is the most stable as the negative charge delocalises into the
benzene ring [1] and does not accept H+ readily.
Ethoxide ion is the least stable as the electron-donating ethyl group intensifies
the negative charge on O atom [1] and accepts H+ readily.
(b)
Dilute HNO3, room temperature
[1]
(c)
4-nitrophenol is more acidic than phenol as the electron-withdrawing nitro
group disperses the negative charge too.
[1]
(d)(i)
Step I: Sn with conc HCl, heat with reflux, followed by careful addition of
NaOH(aq)
[1]
Step II: NaOH(aq) or Na(s)
[1]
[1]
Step IV: CH3COCl, room temperature
(d)(ii)
Step I: Reduction
[1]
Step II: Neutralisation (if NaOH(aq) used) or Redox (if Na used)
[1]
Step III: Nucleophilic substitution
[1]
Step IV: Nucleophilic substitution or acylation or condensation
[1]
(d)(iii)
Phenoxide ion (K) is a better nucleophile than J.
[1]
(e)
Cl2 (aq)
[1]
Examiners’ comments:
Many candidates incorrectly specified Cl2 + AlCl3 rather than the milder
Cl2(aq) which is used for activated phenolic rings.
(f)(i)
3Cl2 + 6NaOH  5NaCl + NaClO3 + 3H2O
(f)(ii)
Amount of Fe2+ = 0.00565 mol
Fe2+  Fe3+ + e
[1]
[1]
ClOx- + 2x H+ + 2x e  Cl- + x H2O
Mole ratio of ClOx- : Fe2+ = 1 : 2x
[1]
Amount of ClOx- = (0.15 / (74.6 + 16x))
(1 / 2x) = [0.15 / (74.6 + 16x)] / 0.00565
[1]
x=2
[1]
4Fe2+ + KClO2 + 4H+  4Fe3+ + KCl + 2H2O