Download 2009 Nov (9746) Paper 1

09N-1
5.
2009 Nov (9746) Paper 1
1.
C
[09N P1 Q01 Alcohols]
moles of sodium
moles of gaseous hydrogen
5
2.5
A
O
CHO
–
+
CHOH
CHO Na
CHOH
CHO– Na+
CHOH
CHO– Na+
CHOH
CHO– Na+
CH2OH
CH2O– Na+
+
5
2
H2(g)
6.
(ans)
C
C
O
H
ethanoate
120
O–
o
O
N
nitrate
2
5
× 2.0 × 10–3 = 8.0 × 10–4
−4
10
× 1000) cm3
vol of KMnO4 = ( 8.00.×020
= 40 cm3 (ans)
7.
It is a non-conductor of electricity.
[09N P1 Q07 Gases]
D
proton number
electronic configuration
11
2.8
Since the ion is discharged at the cathode, it is
a positively charged ion (a cation). Thus, the
number of electrons as shown by the electronic
configuration must be less than the proton
number. Hence, option A. (ans)
© Step-by-Step
[09N P1 Q04 Ionisation Energies]
Co
To form Al 2+, ∆H = 1st I.E. + 2nd I.E.
= 577 + 1820
= 2397 kJ mol–1
1.6 ×10−3 × 8.31× (273 + 273)
Pa
3.0 ×10−3
Using ideal gas equation, pV = nRT
where p is in Pa, V is in m3, and T is in K.
1.6 ×10−3 × 8.31× (273 + 273)
∴ p = nRT =
Pa
V
3.0 ×10−3
(ans)
© Step-by-Step
8.
[09N P1 Q03 Atomic Structure]
[09N P1 Q08 Energetics] [03N P1 Q09]
C
Calcium ions have a lower enthalpy
change of hydration than magnesium ions.
MgC2O4 is soluble in water but not CaC2O4
suggests that CaC2O4 has a lower Ksp than
MgC2O4.
∆Hsol = ∆Hhyd – ∆Hlatt
2+
Since Ca is bigger (in size) than Mg2+, it has
a lower (less exothermic) ∆Hhyd and smaller
(less exothermic) lattice energy, ∆Hlatt.
However, ∆Hhyd decreases much more than
∆Hlatt (which remains almost unchanged due to
the large size of C2O42– ions). Hence, ∆Hsol is
less exothermic for CaC2O4, which accounts
for it being insoluble in water. (ans)
© Step-by-Step
To form Co2+, ∆H = 757 + 1640
= 2397 kJ mol–1 (ans)
© Step-by-Step
A-Level Solutions – Chemistry
(ans)
Dry hydrogen chloride has no reaction with dry
methylbenzene. Hence HCl exists as
molecules in the resultant solution and so, is a
non-conductor of electricity. (ans)
© Step-by-Step
A
120o
O–
© Step-by-Step
=
A
H
phenoxide
O
40 cm3
mol of C2O42– = 2 × mol of KHC2O4.H2C2O4
= 2 × 1.0 × 10–3 = 2.0 × 10–3 mol
mol of KMnO4 = 52 × mol of C2O42–
4.
H
109o
120o
[09N P1 Q06 Bonding]
B
2MnO4–(aq) + 5C2O42–(aq) + 16H+(aq)
→ 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
3.
O–
© Step-by-Step
[09N P1 Q02 Mole] [92J P4 Q02]
B
O–
C 120o
O–
carbonate
© Step-by-Step
2.
delocalised electrons
All four ions contain C=C double
bond(s) and so, have delocalised
π electrons.
Sugar, C6H12O6, has 5 –OH groups which
reacts with Na(s) to give hydrogen gas.
CHO
[09N P1 Q05 Bonding]
09N-2
9.
[09N P1 Q09 Entropy]
B
∆S
–
12. [09N P1 Q12 Chemical Equilibria]
B
∆G
+
In the purification process, pure water is
separated from that contaminated with
impurities (other molecules and ions) and so,
there is an increase in orderliness; i.e. ∆S < 0.
Since ∆G = ∆H – T ∆S and ∆H is zero, ∆G
and ∆S, therefore, must have opposite signs.
(ans)
© Step-by-Step
10. [09N P1 Q10 Entropy]
p
0
273
T/K
X2(g)
2X(g)
∆H +ve
As temperature increases, the position of
equilibrium shifts to the right favouring the
forward endothermic reaction so as to remove
some of the extra heat. As more X2(g)
decomposes to give X(g), pressure increases
(since 1 mol of X2 gives 2 mol of X ). Hence,
graph B. (ans)
© Step-by-Step
A
–66 J K–1
13. [09N P1 Q13 Enzymes]
–1
When water freezes, ∆H = –6.0 kJ mol .
At the freezing point, equilibrium exists and
so, ∆G = 0. Since ∆G = ∆H – T ∆S
–1
–1
.0
⇒ ∆S = ∆TH = (0−+6273
) × 1000 = –22.0 J K mol
mol of H2O =
m
Mr
=
54
18.0
= 3 mol
∴ entropy change = 3 × (–22.0)
= –66 J K–1 (ans)
© Step-by-Step
11. [09N P1 Q11 Electrochemical Cells]
C
Y only
C
At high ethanal concentrations all the
active sites in the enzyme molecules are
occupied by ethanal molecules.
The flattening off of the curve shows that rate
is constant, i.e. increase in [ethanal] has no
effect on the rate of reaction. At this point, the
enzyme is saturated with its substrate
(ethanal). Each enzyme active site has a
substrate bound to it, and all the enzyme
molecules are continuously catalysing the
conversion of substrate to product. (ans)
© Step-by-Step
2H+ + 2e–
H2
E o = 0.00 V
2+
–
Zn + 2e
Zn
E o = –0.76 V
Ecell o = ER o – EL o
= (0.00 – (–0.76)) V = +0.76 V
X: Increase in [Zn2+] shifts the position of
equilibrium to the right and so, EL o becomes
less negative. Hence, Ecell o < +0.76 V.
Y: Increase in [H+] shifts the position of
equilibrium to the right and so, ER o becomes
more positive (i.e. ER o > 0 V).
Hence, Ecell o > +0.76 V. (ans)
© Step-by-Step
14. [09N P1 Q14 Periodicity]
B
Water vapour is produced on
decomposition.
2Al(OH)3 → Al2O3 + 3H2O
The water vapour produced when Al(OH)3
decomposes acts as an extinguisher and so,
helps to delay the spread of flames in the event
of a fire. (ans)
© Step-by-Step
15. [09N P1 Q15 Transition Elements]
D
A Cu atom has two more occupied
electron shells than a Mg2+ ion.
Cu
1s2 2s2 2p6 3s2 3p6 3d10 4s1
2+
Mg 1s2 2s2 2p6
Hence, a Cu atom has two more occupied
electron shells (3rd and 4th quantum shells) than
a Mg2+ ion. (ans)
© Step-by-Step
A-Level Solutions – Chemistry
09N-3
16. [09N P1 Q16 Redox]
after reaction with
potassium
dichromate(VI)
after reaction with
potassium iodide
0
–2
C
•
19. [09N P1 Q19 Bonding]
Orange Cr2O72– turns green suggests that it is
reduced to Cr3+. Hence, H2O2 is oxidised to O2
and so, oxidation number of oxygen after the
reaction is zero.
E o /V
+1.33
–0.68
Cr2O72– + 14H + + 6e– → 2Cr3+ + 7H2O
H2O2 → O2 + 2H + + 2e–
B
118°
..
..
Each nitrogen atom in di-imine
has one lone pair and two bond
pairs. Hence, the bond angle at
each nitrogen atom is 118°. (ans)
N
N
o
H 118 H
di-imine
© Step-by-Step
20. [09N P1 Q20 Isomerism]
carvone
1
D
Q
3
Cr2O72– + 3H2O2 + 8H+ → 2Cr3+ + 3O2 + 7H2O
E
•
o
cell
KI solution changes from colourless to brown
suggests that I– is oxidised to I2. Hence, H2O2
is reduced to H2O and so, oxidation number of
oxygen after the reaction is –2.
E o /V
–0.54
+1.77
2I– → I2 + 2e–
H2O2 + 2H + + 2e– → 2H2O
–
+
2I + H2O2 + 2H → I2 + 2H2O
Since E
o
cell
E
o
cell
= +1.23
> 0, ∴ reaction is feasible. (ans)
© Step-by-Step
Co, Ni, Cu, Zn
The graph shows a steady increase in 1st I.E.,
which suggests that the elements are in the dblock or transition elements. From the Data
Booklet, the 1st I.E. (in kJ mol–1) of Co, Ni, Cu
and Zn are 575, 736, 745 and 908 respectively
which agrees with the trend shown. (ans)
© Step-by-Step
18. [09N P1 Q18 Redox Potentials]
D
C
Cu2+ + e– → Cu+
2I– → I2 + 2e–
E
E
E
o
H
C
*
C
*CH
HO
*
H
CH2
carvone
CH
H2C
CH2
H
H3C
C
CH2
*
C
CH2
H
CH3
Q
Hence, carvone has 1 chiral centre while Q has
3 chiral centres (marked by *). (ans)
© Step-by-Step
21. [09N P1 Q21 Alkenes]
Bromine acts as an electrophile.
The reaction involves electrophilic addition of
Br2 across C=C bond, where the δ+ end of the
polarised δ+Br–Brδ– acts as an electrophile.
(ans)
© Step-by-Step
22. [09N P1 Q22 Halogen Derivatives]
D
nucleophilic substitution
Cl
When Cu2+(aq) and I–(aq) are mixed, the brown
precipitate formed is Cu2I2(s) in brown I2(aq).
On adding S2O32–(aq), the brown colour
disappears which suggests that I2 is reduced to
I–. The white precipitate that remains is Cu2I2.
o
o
cell
= +0.15 V
= –0.54 V
= –0.39 V
(ans)
© Step-by-Step
A-Level Solutions – Chemistry
C
C
H2C
H3C
CH3
The reaction involves nucleophilic substitution
of halogenoalkane, ClCH2CO2H, with
–0.39 V
2Cu2+ + 2I– → 2Cu+ + I2
O
A
17. [09N P1 Q17 Transition Elements]
B
CH3
= (+1.33 – 0.68) V
= +0.65V (> 0 ∴ reaction is feasible).
acting as the nucleophile. (ans)
–
O
Cl
© Step-by-Step
09N-4
23. [09N P1 Q23 Halogen Derivatives] [90J P1 Q25]
CH3
D
The final product has the structure shown and
so, has 4 double bonds.
HO
Cl
CH3
O
O
O
C–C
OH
CH3
Little or no precipitate was seen when S was
boiled under reflux with ethanolic silver nitrate
suggests that S does not undergo hydrolysis.
Hence, S is an aryl halide (option D). (ans)
(ans)
O
© Step-by-Step
© Step-by-Step
26. [09N P1 Q26 Aldehydes]
24. [09N P1 Q24 Phenol]
D
It reacts with Br2(aq) to incorporate up to 4
atoms of bromine in each molecule.
A
CH2=CHCO2H
CH2=CHCH=O
CH3
K2Cr2O7/H+
heat
CH2=CHCO2H
Q
(CH2)5CH3
O
With NaBH4, only the aldehyde group is
reduced to a 1° alcohol, which is then oxidised
by K2Cr2O7/H+ to a carboxylic acid. (ans)
CH3 CH3
ajulemic acid
With Br2(aq), both alkene and phenol reacts to
give the
Br CO2H
following
Br
OH
product.
Br
CH3
CH3
(CH2)5CH3
O
Br CH3 CH3
(ans)
© Step-by-Step
25. [09N P1 Q25 Ketones]
B
4
ketone
3° alcohol
ketone
HO
O
CH3
ketone
O
CH2=CHCH2OH
P
phenol
OH
CH3
then H2O
carboxylic acid
CO2H
alkene
NaBH4 in CH3OH,
CH3
O
CCH2OH
1° alcohol
cortisone
alkene
When cortisone is reacted with H2/Pt, the
alkene is hydrogenated while the ketone groups
are reduced to the corresponding 2° alcohols.
When the product is oxidised by warm
acidified KMnO4, both the 1° and 2° alcohols
are oxidised to carboxylic acid and ketones
respectively.
© Step-by-Step
27. [09N P1 Q27 Carboxylic Acids]
C
2
1
4
3
Acidity: CH2ClCO2H > CH2BrCO2H >
CH3CH2CO2H > C6H5OH
Carboxylic acids (1, 2, 4) are stronger acids
than phenol because the carboxylate ion,
RCO2–, is stabilised to a greater
O
extent by delocalisation of the R C
negative charge over the C atom
O
and both O atoms.
-
Chloroethanoic acid is the strongest acid (its
acidity being enhanced by the presence of an
electron-withdrawing Cl atom). Br is less
electronegative than Cl and so, bromoethanoic
acid is less acidic than chloroethanoic acid.
Propanoic acid is the least acidic among the
three carboxylic acid due to the presence of an
electron-donating ethyl group. (ans)
© Step-by-Step
A-Level Solutions – Chemistry
09N-5
28. [09N P1 Q28 Amines] [84N P3 Q29]
C
31. [09N P1 Q31 Formulae]
shaking the mixture with dilute aqueous
acid
1
2
3
Amine is basic and so, dissolves in aqueous
acid whereas benzene is insoluble (forms an
immiscible layer), which can then be separated
from the aqueous layer by using a separating
funnel. (ans)
mole ratio C : H = 54.5 : 9.1 = 4.54 : 9.1
12.0 1.0
=1:2
© Step-by-Step
1
2
3
29. [09N P1 Q29 Amino Acids]
NH3+
C
CH3CH2CH2CO2H
OHCCH2CH2CH2OH
CH3CH=CHCH2SH
molecular formula
elements
C : H ratio
C4H8O2
C4H8O2
C4H8S
C, H, O
C, H, O
C, H, S
1:2
1:2
1:2
Hence, option A. (ans)
HO2CCH2CH2CH
CO2H
© Step-by-Step
When insulin is heated in HCl, it undergoes
acid hydrolysis (peptide linkage breaks to give
the corresponding carboxylic acid and amines).
32. [09N P1 Q32 Solids]
1
OH
CH2OH
CH2
CH
CH3 CH3
CH2CO NH2
The hydrolysis products are:
CH2OH
•
C–C–C bond angle is smaller in diamond
(109.5°, tetrahedral) than in graphite (120°,
trigonal planar).
•
C–C bond is shorter in graphite (0.142 nm;
sp2–sp2 overlap) than in diamond (0.154 nm;
sp3–sp3 overlap and so, has more p-character).
•
Diamond has a giant covalent structure and all
C–C covalent bonds are of the same strength.
Graphite has a layered structure. The C–C
covalent bonds within each layer are of the
same strength (but weak van der Waals' forces
exist between the layers). (ans)
CH2
–CO NHCHCO NHCHCO NHCHCO NHCHCO NH–
+
H3NCHCO2H
+
OH
+
H3NCHCO2N
CH2
CH
H3NCHCO2H
CH3 CH3
CH2
+
H3NCHCO2H
The C–C–C bond angle between nearest
neighbours is smaller in diamond than in
graphite.
CH2CO2H
© Step-by-Step
(ans)
33. [09N P1 Q33 Electrochemical Cells]
© Step-by-Step
1
30. [09N P1 Q30 Proteins]
B
377
The hydrolysis of peptide X gives 5 amino
acids suggests that X has 4 peptide linkages (or
4 mol of H2O are lost in forming peptide X).
∴ Mr of X = (2 × 75) + 89 + (2 × 105) – (4 × 18)
= 377 (ans)
The electrode potential of electrode 1
becomes more negative as the
concentration of ethanol increases.
•
As [ethanol] increases at electrode 1, position
of equilibrium shifts to the right to remove
some of the extra ethanol. Oxidation occurs
and so, the electrode potential becomes more
negative.
•
At electrode 2, oxidation number of H remains
as +1. Hence, hydrogen is not reduced.
•
Oxygen is reduced only at electrode 2 (O.N.
decreases from 0 in O2 to –2 in H2O) but not at
electrode 1 (O.N. remains as –2). (ans)
© Step-by-Step
© Step-by-Step
A-Level Solutions – Chemistry
09N-6
34. [09N P1 Q34 Kinetics]
1
2
•
•
37. [09N P1 Q37 Carbonyl Compounds]
The half-life is 2.0 × 10–4 s.
The reaction is first order with respect to
[•OH].
CH3
The same time (2.0 × 10–4 s) is taken for the
relative [•OH] to decrease from 10.0 to 5.0 and
then, from 5.0 to 2.5. This suggests that the
half-life is 2.0 × 10–4 s. It also suggests that the
reaction is first order with respect to [•OH]
(since half-life is constant).
There is insufficient data given to deduce the
overall order of reaction. (ans)
•
•
•
•
The sp2 hybridised carbon (C=O) becomes sp3
hybridised in the product formed.
CH3
•
CH3 H
C
H
H
sp2
C
sp3
H
+
+ NO2
NO2
sp3
planar
(sp2 hybridised
C atoms)
(ans)
Silver(I) ions act as an oxidising agent.
When AgNO3(aq) reacts with FeCl2(aq) , a
grey precipitate of Ag(s) forms which does not
dissolve in NH3(aq). Hence, Ag+ is reduced to
Ag(s) which suggests that Ag+ ions act as an
oxidising agent.
When AgNO3(aq) reacts with MgCl2(aq), a
white precipitate of AgCl(s) forms which
dissolves in NH3(aq) due to the formation of
complex ion, [Ag(NH3)2]+; i.e. NH3 complexes
with Ag+ ion (and not with Ag metal).
2AgNO3 + BaCl2 → 2AgCl(s) + Ba(NO3)2
AgCl(s) + 2NH3 → [Ag(NH3)2]+(aq) + Cl –
(ans)
CH3
⊕
Benzene undergoes electrophilic substitution
reaction. In the intermediate, the reactive
carbon atom is sp3 hybridised (tetrahedral).
+
36. [09N P1 Q36 Group VII]
•
3
CN sp
H
CH3
C C
+ H+
CH3
H
•
O–
The sp2 hybridised carbon (C=C) becomes sp3
hybridised in the carbocation formed.
© Step-by-Step
•
C
CH3
sp2
There is no metallic bonding in hydroxyapatite.
(ans)
1
CH3
CH3
C O + CN–
Sr2+ ions can easily replace Ca2+ ions because
its size (0.113 nm) is nearly the same as that of
Ca2+ ions (0.099 nm).
Group II hydroxides become more soluble
down the group. Hence, Sr(OH)2 is more
soluble than Ca(OH)2 and so, precipitates less
readily.
+ NO2+
3
35. [09N P1 Q35 Group II] [06J P1 Q35]
Strontium ions are nearly the same size as
calcium ions and so may easily replace
them in the hydroxyapatite.
H
CH3
+ H+
C C
CH3
H
2
© Step-by-Step
1
CH3
C O + CN–
1
© Step-by-Step
38. [09N P1 Q38 Carbonyl Compounds]
1
CH3CD2ND2
D
OD
2
•
CH3C≡N
D2 / Pt
O
•
CH3CD2ND2
D
D2 / Pt
OD
(ans)
© Step-by-Step
© Step-by-Step
A-Level Solutions – Chemistry
09N-7
39. [09N P1 Q39 Esters]
1
40. [09N P1 Q40 Amino Acids]
2
COCl
N
CH3
+ CH3CH2OH
The identity of the reactants used to produce
pepthidine could be determined from the
products of hydrolysis of the ester.
3
H3N+CHRCO2H + OH–
•
H3N+CHRCO2– + OH–
from carboxylic acid
or acyl chloride
from alcohol
K2
H3N+CHRCO2– + H2O
H2NCHRCO2– + H2O
•
The point in the titration curve when the slope
is at its maximum in the centre corresponds to
the isoelectric point. Hence, the species
present is H3N+CHRCO2– which has no net
charge.
•
H3N+CHRCO2– is the most common species
present at the isoelectric point, which need not
be at pH 7. Isoelectric point ranges from pH
5.5 to 6.2 depending on the nature of the amino
acid. (ans)
Hence, option A.
COCl
N
CH3
K1
At pH = pK1, which is equivalent to the point
of half-neutralisation in an acid-base titration,
[H3N+CHRCO2H] = [H3N+CHRCO2–].
CO 2CH2CH3
N
CH3
Equal concentrations of H3N+CHRCO2H
and H3N+CHRCO2– are present at
pH = pK1.
There is no net charge on the amino acid at
the point when the slope of the curve is at
a maximum at its centre.
+ CH3CH2OH
© Step-by-Step
CO2CH2CH3
N
CH3
[09N P1 MCQ Key]
+ HCl
pepthidine
(ans)
© Step-by-Step
Q.
Key
Q.
Key
Q.
Key
Q.
Key
1
2
3
4
5
A
B
A
A
C
11
12
13
14
15
C
B
C
B
D
21
22
23
24
25
A
D
D
D
B
31
32
33
34
35
A
D
D
B
D
6
7
8
9
10
B
D
C
B
A
16
17
18
19
20
C
B
D
B
D
26
27
28
29
30
A
C
C
C
B
36
37
38
39
40
D
A
B
D
C
© Step-by-Step
A-Level Solutions – Chemistry
09N-8
2009 Nov (9746) Paper 2
2.
1.
(a) Metals 3 and 4 are Zn and Cu respectively.
(ans)
[09N P2 Q01 Energetics / Solids]
(a) The lattice energy of MgO is the heat energy
evolved when one mole of crystalline MgO(s)
is formed from its separate gaseous ions,
Mg2+(g) and O2–(g), under standard conditions.
Mg2+(g) + O2–(g) → MgO(s) ∆H = L.E. (ans)
(b)
(b) Cu2+ + 2e– → Cu
2H+ + 2e– → H2
Zn2+ + 2e– → Zn
Al 3+ + 3e– → Al
Mg2+ + 2e– → Mg
•
q+ q−
(i) L.E. ∝
(r+ + r− )
ionic radii/nm: Cl –, 0.181; Br –, 0.195; I –, 0.216
Lattice energies decrease from NaCl to NaI due
to increase in ionic size of the X– anions (Cl – <
Br – < I –) while the charge remains the same in
each case (Na+ Cl –, Na+ Br –, Na+ I –), which
results in weaker electrostatic forces of
attraction between Na+ and X– ions from NaCl
to NaI. (ans)
q+ q−
(r+ + r− )
ionic radii/nm: Mg2+, 0.065; Na+, 0.095
(ii) L.E. ∝
Lattice energy of MgO is considerably larger
than those of sodium halides due to the higher
charge (charge on Mg2+ O2– is double that on
sodium halides, Na+ X –) and smaller ionic radii
(Mg2+ < Na+; O2– < X –), which results in much
stronger electrostatic forces of attraction
between Mg2+ and O2– ions than that between
Na+ and X – ions. (ans)
(c) AgI is not fully ionic but has considerable
percentage of covalent character because I –
ion (0.216 nm) is much larger than F – ion
(0.136 nm). Hence, I – anion is readily
polarised by the Ag+ ion leading to electron
density between the Ag+ and I – ions; i.e.
electrons are incompletely transferred in
forming Ag+ and I – ions. (ans)
(d)
(i) Co-ordination number is the maximum number
of ions that can be placed around another ion
of opposite charge in a crystal lattice; i.e. the
number of its nearest neighbours. (ans)
[09N P2 Q02 Transition Elements]
E
E
E
E
E
o
o
o
o
o
= +0.34 V
= 0.00 V
= –0.76 V
= –1.66 V
= –2.38 V
The solid residue is copper because
E o (Cu2+/Cu) is positive, so Cu has no reaction
with acids (i.e. cannot reduce H+ ). (ans)
(c)
(i) Sulfuric acid acts as an oxidising agent. (ans)
(ii) SO42– + 4H+ + 2e– → SO2 + 2H2O
Zn → Zn2+ + 2e–
∴ overall equation is
SO42– + Zn + 4H+ → Zn2+ + SO2 + 2H2O (ans)
(d)
(i) Al(OH)3 (ans)
(ii) Al 3+(aq) + 3OH–(aq) → Al(OH)3(s) (ans)
Al(OH)3(s) + OH–(aq) → Al(OH)4–(aq)
(ans)
aluminate
(e) Zn [Ar] 3d10 4s2
Like Group II elements, Zn has a fully filled
outer s-orbital. Hence, Zn shows only +2
oxidation state in its compounds; i.e. Zn forms
Zn2+ only. (ans)
(f)
(i) Zn(NO3)2 → ZnO + 2NO2 +
1
2
O2 (ans)
(ii) Zn2+ (0.074 nm) is larger than Mg2+ (0.065 nm)
and so, has smaller polarising power which
results in the NO3– anion being less polarised.
Hence, Zn(NO3)2 is relatively more stable to
heat and so, decomposes at a higher
temperature than Mg(NO3)2. (ans)
(g) NH3 + H2O
NH4+ + OH–
Zn2+(aq) + 2OH–(aq) → Zn(OH)2(s)
Zn(OH)2 + 4NH3 → [Zn(NH3)4]2+ + 2OH–
(ans)
© Step-by-Step
(ii) Ionic radii: Cs+ > K+ > Na+
The much larger ionic size of Cs+ ion allows
more Cl – ions to be packed around it. Hence
co-ordination number in CsCl lattice is larger
than those in NaCl and KCl. (ans)
© Step-by-Step
A-Level Solutions – Chemistry
09N-9
3.
4.
[09N P2 Q03 Carboxylic Acids]
(a) Benzoic acid and ethanoic acid are stronger
than carbonic acid. (ans)
[A stronger acid has a larger Ka value.]
(b) C6H5OH(aq)
C6H5O–(aq) + H+(aq)
[C H O− ] [H+ ]
Ka = 6 5
[C6H5OH]
Since the degree of dissociation, α , is very
small, [C6H5OH]eqm ≈ [C6H5OH]0
[H+ ]2
since [C6H5O–] = [H+]
∴ Ka =
[C6H5OH]
[H+] =
Ka × [C6H5OH]
(a)
(i) A, C and D (ans)
[no silver mirror with Tollens' reagent
⇒ not aldehyde]
(ii) A and D (ans)
[no ppt with 2,4-dinitrophenylhydrazine
⇒ not carbonyl compound ]
(iii) A and D (ans)
[molecular formula shows HBr added
⇒ C=C bond present ]
(iv) A and D (ans)
[two Na in molecular formula of product
⇒ two –OH groups present ]
= 1.3 ×10−10 × 0.10
= 3.605 × 10–6 mol dm–3
∴ pH = –log10 [H+]
= –log10 (3.605 × 10–6) = 5.44 (ans)
(c) C6H5OH is more acidic (larger Ka) than
CH3OH because the electron-withdrawing
benzene ring weakens the O–H bond, thereby
causing phenol to dissociate readily to give H+
ions. In addition, the anion C6H5O– is
stabilised by delocalisation of the charge over
the ring. CH3OH is less acidic (smaller Ka)
because the negative charge in CH3O– is
intensified by the electron-donating methyl
group making it less stable. (ans)
(d) observation: reddish-brown Br2(aq)
decolourised, steamy fumes of HBr evolved
and white precipitate formed.
structural formula
of product:
(v) D (ans)
[two Na in molecular formula of product
⇒ two acidic groups present ]
(b)
(i) Reaction (a)(v). (ans)
(ii)
(c)
CO2–
CO2–
[ Product formed is a salt of
benzene-1,4-dicarboxylic acid
because the oxidation is carried
out in alkaline medium.]
(ans)
Br
(d)
Br
(white ppt)
(ans)
(e)
(i) C6H5CO2H + CH3OH → C6H5CO2CH3 + H2O
Benzoic acid is heated with methanol under
reflux in the presence of a little conc. H2SO4 as
catalyst to give methyl benzoate. (ans)
(ii) CH3CO2H + SOCl2 → CH3COCl + SO2 + HCl
CH3COCl + C6H5OH → CH3CO2C6H5 + HCl
Ethanoic acid is first converted to ethanoyl
chloride by reacting with SOCl2. Ethanoyl
chloride is then reacted with phenol to give
phenyl ethanoate. (ans)
© Step-by-Step
A-Level Solutions – Chemistry
O– Na+
CH=CHCO2– Na+ (ans)
OH
Br
[09N P2 Q04 Carboxylic Acids]
A produces
D produces
CO2– Na+
OH
CH=CHOH
CH=CHCO2– Na+
[Phenol has no reaction with NaHCO3.]
(ans)
© Step-by-Step
09N-10
5.
[09N P2 Q05 Esters]
(a)
(i) mole ratio C : H : O =
•
70.6
12.0
:
:
5.9
1.0
23.5
16.0
= 5.88 : 5.90 : 1.47
= 4 : 4 : 1
∴ empirical formula of E is C4H4O. (ans)
CO2 dissolves in water to give carbonic acid,
H2CO3, which is a stronger acid than phenol
but a weaker acid than both ethanoic acid and
benzoic acid. Hence only sodium phenoxide
reacts with H2CO3 to give phenol. Sodium
ethanoate and sodium benzoate have no
reaction with H2CO3. (ans)
© Step-by-Step
(ii) Using the ideal gas equation,
pV = nRT = m RT
(since n = m )
Mr
Mr
0.344 × 8.31× (21 + 273)
Mr of E = mRT =
101.7 ×103 × 60.4 ×10−6
pV
= 136.8
Let the molecular formula of E be C4nH4nOn.
Mr of C4nH4nOn = 136.8
4n(12.0) + 4n(1.0) + n(16.0) = 136.8
⇒ n =2
∴ molecular formula of E is C8H8O2. (ans)
(b)
(i)
H
O
O
C
H C C
H
H
O C H
O
H
structure E1
structure E2
(ans)
(ii)
E1 produces
H
O
O
H C C
H
E2 produces
C
O– Na+
O– Na+
+
O– Na+
+ CH3OH
(ans)
(c)
mixture from E1
produces
H
O
H C C
H
+
mixture from E2
produces
O– Na+
no reaction
OH
(ans)
A-Level Solutions – Chemistry
09N-11
2009 Nov (9746) Paper 3
1.
[09N P3 Q01 Proteins / Chem Equilibria]
(ii) The aggregation of Hb-S molecules is due to
the presence of van der Waals’ forces (arising
from induced dipoles) between the non-polar
side-chain of valine residue. (ans)
(a) The haemoglobin molecule is a tetramer
consisting of four polypeptide chains (two αchains and two β-chains, called subunits) each
with its own haem group. This is a quaternary
structure of protein.
(iii) Molecules of normal haemoglobin do not
attract each other because of the repulsion
between haemoglobin chains due to the
negatively charged glutamate. (ans)
•
(c)
•
•
•
(b)
(i)
In each polypeptide chain, the amino acids are
bonded together by peptide linkages (covalent
bonds). The sequence of amino acids in each
polypeptide chain gives the primary structure
of protein.
The secondary structure refers to the detailed
configurations of the polypeptide chains – how
the chains may be coiled or folded to give the
α - helix or β -pleated sheets. The structures
are stabilised by hydrogen bonds between the
N–H group of one amino acid unit and the
C=O group of another along the main chain.
The tertiary structure refers to the overall 3-D
shape of the protein involving folding or
coiling of the chains. It shows how protein
molecules are arranged in relation to each
other. The four types of R-group interactions
that hold the tertiary structure in its necessary
shape are hydrogen bonds between polar Rgroups, ionic bonds between charged Rgroups, hydrophobic interactions between nonpolar R-groups, and disulfide linkages
(covalent bonds).
The quaternary structure of proteins refers to
the spatial arrangement of its subunits. It
shows how the individually folded subunits are
packed together. The quaternary structure is
stabilised mainly by hydrophobic interactions
between non-polar areas on the surface of the
individual polypeptide chains (subunits). (ans)
H
N
O
CH
C
CH2
A-Level Solutions – Chemistry
(ii) Kc =
[Hb(O2 )4 ]
[Hb] (7.6 ×10-6 )4
=
1
(7.6 ×10-6 )4
since [Hb] = [Hb(O2)4]
= 3.00 × 1020 dm12 mol–4 (ans)
(iii) 99% of Hb converted to Hb(O2)4
⇒
[Hb(O2 ) 4 ]
= 99
[Hb]
1
[Hb(O2 )4 ]
[Hb] [O2 ]4
3.00 × 1020 = 99 4
[O 2 ]
4
[O2] = 3.30 × 10–19
since
Kc =
∴ [O2] =
4
3.30 ×10−19
= 2.40 × 10–5 mol dm–3 (ans)
(d) Mb(aq) + O2(aq)
Kc =
1 × 106 =
∴
MbO2(aq)
[MbO 2 ]
[Mb] [O2 ]
[MbO2 ]
[Mb] (7.6 ×10-6 )
[MbO 2 ]
= 7.6
[Mb]
or [MbO2] = 7.6[Mb]
∴ % of MbO2 =
[MbO2 ]
× 100
[Mb] + [MbO2 ]
=
7.6[Mb]
× 100
[Mb] + 7.6[Mb]
© Step-by-Step
O
O–
[Hb(O2 )4 ]
dm12 mol–4 (ans)
[Hb] [O2 ]4
= 88.4 % (ans)
CH2
C
(i) Kc =
(ans)
09N-12
2. [09N P3 Q02 Alcohols / Energetics]
(a)
(i) Let the oxidation no. of C in CH3OH be x.
so, x + 3(+1) + (–2) + 1 = 0
or x = –2
(ii)
O
H
C
A
H
H
HO
C
B
OH
OH
∆H
2CO2 + 3H2O
-1367
By Hess' Law,
∆H = – (–182) + (–1367)
= –1185 kJ mol–1 (ans)
C
C
O2 + H2O
CH3CH2OH + 3O2
O
O
5
2
-182
∴ oxidation no. of C in CH3OH = –2 (ans)
(ii)
CH3CHO +
(ans)
(iii) To convert methanol to A:
reagent: acidified K2Cr2O7(aq)
condition: distil (ans)
(iii) CH3CO2H + 2O2 → 2CO2 + 2H2O
CH3CHO + 52 O2 → 2CO2 + 2H2O
∆Hc for CH3CO2H is expected to be less
exothermic than that for CH3CHO because
more energy is needed to break the strong C–O
and O–H bonds in CH3CO2H while the same
amount of energy is released when bonds are
formed (since same amount of CO2 and H2O
are formed in each case). (ans)
To convert methanol to B:
reagent: acidified K2Cr2O7(aq)
condition: heat under reflux (ans)
To convert methanol to C:
reagent: acidified KMnO4(aq)
condition: heat under reflux (ans)
(d)
F
D
(i) CH3(CH2)3OH CH3CHCH2CH3
E
(b)
(i) heat evolved = mc∆T = (200 × 4.18 × 30.0) J
= 25080 J
Mr of C2H5OH = 2(12.0) + 6(1.0) + 16.0 = 46.0
mol of ethanol = m = 1.50 = 0.0326 mol
Mr
46.0
∴ ∆Hc = heat evolved = 25080 J mol–1
mol of ethanol 0.0326
= –769 kJ mol–1 (ans)
(ii) ∆Hc value calculated in (i) is much less than
the true value of ∆Hc because of heat lost to the
surroundings and to the copper can; i.e. not all
the heat from the burning ethanol are used to
heat the water. (ans)
CH3CHCH2OH
OH
G
CH3
CH3–C–CH3
OH
CH3
(ans)
(ii) D and E are primary alcohols, F is a secondary
alcohol, and G is a tertiary alcohol. (ans)
*
(iii) F is chiral (chiral
CH3CHCH
2CH3
(ans)
carbon indicated with *).
OH
(iv) F contains CH3CH(OH)– group and so, reacts
with alkaline aqueous iodine to give CHI3 and
CH3CH2CO2–. (ans)
(v) F (a secondary alcohol) is oxidised to give a
ketone, CH3CH2COCH3 which is a non-acidic
organic product. (ans)
© Step-by-Step
(c)
(i) CH3CH2OH + ½O2 → CH3CHO + H2O
Bonds broken (∆H )
1 C–C
5 C–H
1 C–O
1 O–H
½ O=O
(+350)
5(+410)
(+360)
(+460)
½ (+496)
+3468
Bonds formed (∆H )
1 C–C
4 C–H
1 C=O
2 O–H
(–350)
4(–410)
1(–740)
2 (–460)
–3650
∆Hc = +3468 – 3650
= –182 kJ mol–1 (ans)
A-Level Solutions – Chemistry
09N-13
3.
(d)
[09N P3 Q03 Periodicity / Kinetics]
(a) Magnesium burns in O2 with a brilliant white
flame to give the oxide, MgO (white residue).
2Mg + O2 → 2MgO
•
P4 burns in O2 with a pale bluish-green flame
to give the oxides, P4O6 or P4O10 (white solid).
P4 + 3O2 → P4O6 ; P4 + 5O2 → P4O10
•
Sulfur burns with a blue flame to give SO2,
which is oxidised to SO3 in excess oxygen.
S + O2 → SO2 ; 2SO2 + O2 → 2SO3 (ans)
(b) P4O6/P4O10 (covalent oxides) dissolve readily in
water to give strongly acidic solutions, which
turn universal indicator red.
P4O6 (s) + 6H2O(l) → 4H3PO3(aq)
P4O10 (s) + 6H2O(l) → 4H3PO4(aq) (ans)
•
SO2/SO3 (covalent oxides) dissolve readily in
water to give acidic solutions, which turn
universal indicator red.
SO2(g) + H2O(l) → H2SO3(aq)
SO3(g) + H2O(l) → H2SO4(aq) (ans)
1 . The faster the reaction, the
time
shorter is the time taken for the blue colour to
appear. (ans)
(i) rate ∝
(ii)
× 2.25 × 10–3
(ii) 2MnO4 + 5H2O2 + 6H → 2Mn2+ + 5O2 + 8H2O
mol of H2O2 in 25.0 cm3 =
5
2
× mol of MnO4–
=
5
2
× 0.0200 ×
•
Compare expt 2 and 3: [H2O2] and [H+]
15
= 1.5 times,
constant, when [I–] increases by 10
0.0149
0.0100
–
= 1.5 times.
10.0
1000
Let rate equation be rate = k [H2O2] [I–] [H+]n
From expt 1, 0.0303 = k (15)(10)(5)n
(1)
n
From expt 2, 0.0100 = k (5)(10)(10)
(2)
0.0303 = k (15)(10)(5)n
0.0100 k (5)(10)(10)n
⇒
n=0
∴ Reaction is zero order w.r.t. H+. (ans)
(1)
,
(2)
(iii) rate = k [H2O2] [I–] (ans)
= 5.00 × 10–4 mol
–4
100
∴ mol of H2O2 in 100 cm = 25
.0 × 5.00 × 10
units of k =
units of rate
(units of concentration) 2
s-1
=
(mol dm-3 )2
= 0.00200 mol (ans)
(iii) Na2O2 + 2H2O → 2NaOH + H2O2
mol of Na2O2 = mol of H2O2
= 0.00200 mol (ans)
= dm6 mol–2 s–1 (ans)
[ Also accepted: usual second-order units of
dm3 mol–1 s–1.]
mol of NaOH from Na2O2 = 2 × mol of H2O2
= 2 × 0.00200 mol
= 0.00400 mol
∴ mol of NaOH from Na2O = 0.00900 – 0.00400
= 0.00500 mol
Na2O + H2O → 2NaOH
∴ mol of Na2O = 12 × mol of NaOH
× 0.00500 = 0.00250 mol
(ans)
A-Level Solutions – Chemistry
(ans)
Compare expt 2 and 4: [I–] and [H+] constant,
when [H2O2] doubles, the initial rate is also
doubled.
∴ Reaction is first order w.r.t. H2O2. (ans)
3
1
2
0.0303
0.0100
0.0149
0.0200
•
•
+
=
33
100
67
50
/ s–1
∴ Reaction is first order w.r.t. I . (ans)
= 0.00900 mol (ans)
–
1
2
3
4
1
time
Since the total volume of reaction mixture is
constant (100 cm3), the volume of each reagent
used is proportional to its concentration.
= 2.25 × 10–3 mol
100
25.0
time taken /s
the initial rate increases by
(c)
(i) NaOH + HCl → NaCl + H2O
mol of NaOH in 25.0 cm3 = mol of HCl
22.5
= 0.100 × 1000
∴ mol of NaOH in 100 cm3 =
expt no.
© Step-by-Step
4.
[09N P3 Q04 Bonding / Amines]
(a) In an ideal gas, the gas particles have
negligible size/volume, negligible
intermolecular forces of attraction, and the
collisions of gas particles are perfectly elastic.
(ans)
09N-14
(b)
(i) NH3 has a much higher boiling point than
CH4. This is because the intermolecular
hydrogen bonding in NH3 is.. much
stronger than the weak
N
δ–
δ+
: N
intermolecular van
H
H
H ′′′′′′′′′
der Waals' forces in
hydrogen bonding
CH4, and so requires
much more energy to overcome. (ans)
H
H
or ∆S
o
= T ∆S
= ∆H
o
[OH − ]2
0.100
[OH–]2 = (0.100)(6.4 × 10–4)
H
(iii) ∆G o = ∆H o – T ∆S o
At –33°C, ∆G o = 0.0 kJ mol–1
o
+
[C2H5 NH3 ][OH − ]
[C2 H5 NH 2 ]
Since the degree of dissociation, α , is very
small, [C2H5NH2]eqm ≈ [C2H5NH2]0
Kb =
(ii) When methane is liquefied, the entropy of the
sample decreases (∆S < 0) because CH4
molecules in the liquid state display more order
than in the gaseous state. (ans)
∴ ∆H
Kb =
⇒ [OH–] = 0.00800 mol dm–3
∴ pOH = –log10 [OH–]
= –log10 (0.00800) = 2.10
∴ pH = 14 – pOH
= 14 – 2.10 = 11.9 (ans)
(iii) C2H5NH2 + (CH3)2CHCOCl
Product:
O
o
/T
(CH3)2CHC
= (+23.3 × 103) ÷ (–33 + 273 )
= +97.1 J K–1 mol–1 (ans)
(c)
(i) Order of base strength:
ethylamine > ammonia > phenylamine
+ H2O
NH3 + H2O
CH3CH2NH2 + H2O
•
•
•
+ HCl
N
C2H5
(ans)
H
•
C6H5NH2 + Br2(aq)
Product:
NH2
Br
NH3+
NH2
C2H5NH3+ + OH–
(ii) C2H5NH2 + H2O
Br
+ 3HBr
+ OH –
NH4+ + OH –
CH3CH2NH3+ + OH –
Ethylamine is a stronger base than NH3
because the electron-donating ethyl group
(CH3CH2−) releases electrons towards the N
atom, making the lone pair on the N atom more
available to accept a proton than that in NH3.
Phenylamine is a weaker base than NH3
because the lone pair of electrons on the N
atom is delocalised over the benzene ring, and
this makes it less available to accept a proton
than that in NH3.
Besides, the charge on CH3CH2NH3+ is more
dispersed than for the smaller NH4+ ion,
making CH3CH2NH3+ ion more stable.
Whereas, the charge on C6H5NH3+ is
intensified, making C6H5NH3+ ion less stable.
(ans)
(ans)
Br
(d)
(i) Nucleophilic substitution. (ans)
(ii) J is 1,5-dibromooctane. (ans)
CH2
CH2
CH2
CH2
CH
Br Br
CH2CH2CH3
(ans)
(iii) The four isomeric alkenes are:
H
H
CH2=CHCH2
H
C=C
C=C
CH2=CHCH2
CH2CH2CH3
cis isomer
CH2CH2CH3
H
trans isomer
H CH2=CHCH2CH2
H
C=C
CH2=CHCH2CH2
CH2CH3
cis isomer
H
C=C
CH2CH3
H
trans isomer
(ans)
© Step-by-Step
A-Level Solutions – Chemistry
09N-15
5.
[09N P3 Q05 Group VII / Arenes]
•
(a) Cl2 is a yellowish-green gas, Br2 is a dark red
liquid and I2 is a black solid. (ans)
CH3
Volatility decreases down the group (from Cl2
to I2) due to increasing number of electrons in
the molecules (as the halogen molecule
becomes larger) and hence, increasing
intermolecular van der Waals' forces. (ans)
–
(b) When Cl ion reacts with concentrated H2SO4,
only steamy fumes of HCl is produced.
H2SO4 + Cl – → HCl(g) + HSO4– (ans)
•
•
When Br – ion reacts with concentrated H2SO4,
steamy fumes of HBr together with some
orange-brown fumes of Br2 are produced.
H2SO4 + Br – → HBr(g) + HSO4–
H2SO4 + 2HBr → Br2 + SO2 + 2H2O
Some of the HBr produced is oxidised by
concentrated H2SO4 to Br2. (ans)
When I – ion reacts with concentrated H2SO4,
purple vapour of I2 are produced together with
some steamy fumes of HI.
H2SO4 + I – → HI(g) + HSO4–
H2SO4 + 8HI → 4I2 + H2S + 4H2O
Most of the HI produced is oxidised by
concentrated H2SO4 to I2. (ans)
(c) Phenol is nitrated more easily than
methylbenzene; the nitration reaction does not
require the strong electrophile NO2+ produced
by the reaction of HNO3 with H2SO4. This is
because the –OH group in phenol has an
electron-donating effect due to the interaction
of the unshared pair of electrons on the O atom
with the delocalised π orbitals of the benzene
ring, thereby making the ring more electronrich and so, is more susceptible to electrophilic
attack; i.e. –OH group activates the ring
toward electrophilic substitution. (ans)
(d)
K
L
M
N
P
CH3
CH2Cl
CH2CN
CH2CO2H
CH2OH
With Cl2 and AlCl3, methylbenzene undergoes
electrophilic substitution to give K, which is
either 2-chloro or 4-chloro methylbenzene.
CH3
+ 2Cl2
CH3
Cl
AlCl3
+
+ 2HCl
Cl
[Take K as (4-chloro)methylbenzene.]
•
With more Cl2 in the presence of light, K
undergoes free-radical substitution to give L.
CH3
Cl
Cl2
light
CH2Cl
Cl
K
•
L
When heated with NaCN in ethanol, L
undergoes nucleophilic substitution to give M.
NaCN
CH2Cl
Cl
L
•
M
When heated with dilute H2SO4, M undergoes
acid hydrolysis to give a carboxylic acid N.
Cl
CH2CN
H2O/H+
reflux
N
When heated with NaOH(aq), L undergoes
nucleophilic substitution to give P.
CH2Cl
Cl
NaOH
heat
CH2OH
Cl
L
•
CH2CO2H
Cl
M
•
CH2CN
Cl
ethanol
P
When a mixture of P and N is heated with a
small amount of c. H2SO4, esterification occurs
to give an ester Q.
CH2CO2H + Cl
Cl
N
c. H2SO4
heat
CH2OH
P
O
Cl
Cl
Cl
Cl
Cl
Cl
O
Cl
CH2C
Q
A-Level Solutions – Chemistry
OCH2
CH2C
Q
OCH2
Cl
(ans)
© Step-by-Step
Cl