Download Soln Chem 2008Nov(9746)

08N-1
2008 Nov (9746) Paper 1
1.
5.
[08N P1 Q01 Moles]
B
[08N P1 Q05 Atomic Structure]
[04J P1 Q05] [93J P4 Q02]
N–
C
1× 6.02 ×1023
24000
At r.t.p., volume of 1 mol of gas = 24 dm3
1
⇒ mol of O2 in 1 cm3 =
mol
24 ×103
∴ no. of O2 molecules = nL
1
=
× 6.02 × 1023
24 ×103
Hence, option B. (ans)
N– has 8 electrons. On losing an electron, it
forms N which has 7 electrons (configuration
1s2 2s2 2px1 2py1 2pz1) and so, has a half-filled
set of p orbitals. (ans)
© Step-by-Step
6.
[08N P1 Q06 Bonding]
N atoms numbered
1 and 3
2 and 4
ionic
co-ordinate
© Step-by-Step
C
2.
[08N P1 Q02 Relative Masses] [99N P3 Q01]
A
Chlorophyll is a chlorin pigment, and has a
Mg2+ ion at the centre of the chlorin ring.
argon
Since density of 'atmospheric N2' is higher than
that of chemically pure N2, the gas that causes
this discrepancy would, therefore, be one of
higher mass than N2. [Mr : N2 = 28; Ar = 39.9;
He = 4; CH4 = 16; Ne = 20.0] (ans)
R
Element (X) in C has 3 valence electrons and
would lose all 3 valence electrons most easily
to achieve a stable octet configuration (forming
X3+ ion). (ans)
© Step-by-Step
4.
40
H
H
C H
C
C
C
C
C
C
In the decomposition, a neutron is converted to
a proton (⇒ mass number remains unchanged
since there is no change in total no. of neutrons
and protons) and an electron (⇒ atomic no.
increases by 1). Hence, the process is
40
described by 40
19 K → 20 Ca . (ans)
© Step-by-Step
R
C
R
O
© Step-by-Step
7.
[08N P1 Q07 Chemical Equilibria]
A
0.17
initial atm,
p
eqm atm,
p–a
K → 40Ca
R
C
2NO2(g)
[08N P1 Q04 Atomic Structure]
D
C
C
R
C
N
Mg2+
–
N
N
C
R
1s2 2s2 2p6 3s2 3p6 3d1 4s2
C
N–
C
H
C
C
H C
[08N P1 Q03 Atomic Structure]
C
C
C
N-1 and N-3 form
ionic bonds with
Mg2+, while N-2
and N-4 form
co-ordinate (or
dative covalent)
bonds with Mg2+.
(ans)
R
C
C
© Step-by-Step
3.
H
R
2NO(g) + O2(g)
0
a
0
½a
Since pressure at equilibrium is 20 % greater
than initial pressure,
⇒ a = 0.4 p
(p – a) + a + ½ a = 120
100 p
Hence, at equilibrium
2NO2(g)
eqm atm,
p – 0.4 p
= 0.6 p
2NO(g) + O2(g)
0.4 p
½ × 0.4 p
= 0.2 p
∴ mole fraction of O2 in equilibrium mixture,
0.2 p
x=
= 0.17 (ans)
0.6 p + 0.4 p + 0.2 p
© Step-by-Step
A-Level Solutions – Chemistry
08N-2
8.
12. [08N P1 Q12 Chemical Equilibria]
[08N P1 Q08 Solids]
copper
cations
D
C
iodine
molecules
Copper has a giant metallic structure – lattice
particles are metal cations. Iodine has a simple
molecular structure – lattice particles are
discrete I2 molecules. (ans)
© Step-by-Step
9.
When the equilibrium constant is independent
of temperature, it suggests that the reaction is
neither exothermic nor endothermic; i.e.
enthalpy change, ∆H = 0. (ans)
© Step-by-Step
13. [08N P1 Q13 Kinetics]
[08N P1 Q09 Energetics]
D
A
burning an element in oxygen
When an element is burnt in oxygen, the
enthalpy change (enthalpy change of
combustion, ∆Hc) is always negative
(i.e. exothermic). (ans)
© Step-by-Step
The enthalpy change is zero.
1
210
Since half-life, t½, of iodine-131 is 8 days,
80 days ⇒ 10 t½ have elapsed.
∴ fraction of isotope that remained = ( 12 )10
or 110 . (ans)
2
© Step-by-Step
10. [08N P1 Q10 Entropy]
14. [08N P1 Q14 Periodicity]
∆H
+
C
∆S
+
∆G
–
D
The phrase instant 'cold packs' suggests that
the reaction is endothermic; i.e. ∆H is positive.
When the pack is squeezed, NH4NO3(s)
dissolves in the water suggests that the reaction
is spontaneous; i.e. ∆G is negative. Dissolution
of NH4NO3(s) is accompanied by an increase
in entropy (less orderly); i.e. ∆S is positive.
Hence, option C. (ans)
© Step-by-Step
11. [08N P1 Q11 Ionic Equilibria] [93N P4 Q10]
D
Ka
0
V
HCl (produced when the chloride of R is
hydrolysed) gives a white precipitate of AgCl
with AgNO3(aq).
Ag+(aq) + Cl –(aq) → AgCl(s)
mol of Cl – from chloride of R = mol of Ag+
= 0.30 × 100
1000
= 0.030 mol
⇒ 0.010 mol of chloride of R ≡ 0.030 mol of Cl –
or 1 mol of chloride of R ≡ 3 mol of Cl –
However, the low b.p. (76 °C) suggests that the
chloride of R is not an ionic compound, i.e. R
is not a Group III element. Hence, R is a
Group V element which forms two chlorides,
RCl3 and RCl5. RCl3 gives 3 mol of HCl when
mixed with water. (ans)
© Step-by-Step
V
Ka, the acid dissociation constant, is only
affected by changes in temperature. It is not
affected by changes in volume, V. (ans)
© Step-by-Step
A-Level Solutions – Chemistry
08N-3
15. [08N P1 Q15 Group II]
A
18. [08N P1 Q18 Transition Elements]
Some of the barium hydroxide had reacted
with carbon dioxide in the air to from solid
barium carbonate.
In the titration, the reaction is
Ba(OH)2.8H2O + 2HCl → BaCl2 + 10H2O
The titres were lower than expected suggests
that the solution contains less Ba(OH)2 than
expected. This is because when the prepared
solution was left in an open beaker, some
Ba(OH)2 has reacted with CO2(g) in the air to
form solid BaCO3 (in an acid-base reaction).
Ba(OH)2 + CO2 → BaCO3 + H2O (ans)
© Step-by-Step
16. [08N P1 Q16 Group VII]
C
•
•
boiling point
increases
C
Fe2O3
The formation of chromium(VI) compounds
from chromium(III) involves oxidation.
Al2O3: Al 3+ + 3e– → Al E o = –1.66 V
CuO: Cu2+ + 2e– → Cu E o = +0.34 V
Fe2O3: Fe3+ + e– → Fe2+ E o = +0.77 V
ZnO: Zn2+ + 2e– → Zn E o = –0.76 V
From the list, the best oxidising agent is Fe2O3
(most positive E o value). (ans)
© Step-by-Step
19. [08N P1 Q19 Hydrocarbons]
A
It is an sp2–sp2 overlap.
Each carbon atom in buta-1,3-diene is sp2
hybridised. Hence, the covalent bond
between C2 and C3 are formed by the overlap
of sp2 hybrid orbitals. (ans)
electron affinity
less negative
© Step-by-Step
Boiling point of Group VII elements increases
from Cl2 to I2 due to stronger intermolecular
van der Waals' forces as the number of
electrons increases from Cl2 to I2.
From Cl to I, electron affinity becomes less
negative due to the increase in atomic size and
hence, weaker attraction for the additional
electron. (ans)
© Step-by-Step
17. [08N P1 Q17 Transition Elements]
B
20. [08N P1 Q20 Alkenes]
B
HBr
Br2, hv
Option B will not give a good yield of 1,2dibromocyclohexane because the second step
involves a free-radical substitution reaction
which is not stereospecific and so, a mixture of
substitution products is obtained. (ans)
© Step-by-Step
1s2 2s2 2p6 3s2 3p6 3d6
21. [08N P1 Q21 Alkanes]
6
2
26Fe has electronic configuration: [Ar] 3d 4s
(where [Ar] is 1s2 2s2 2p6 3s2 3p6). In forming
ions, electrons are removed from the 4s orbitals
first before the 3d orbitals.
∴ Electronic structure of Fe2+ is [Ar] 3d6 4s0.
(ans)
© Step-by-Step
C
CH3CH2CH3 + Cl •
The relevant bond energies (in kJ mol–1) are:
C–Cl
C–H
Bond energy
340
410
C–Cl bond is weaker than C–H bond. Hence,
homolytic fission of C–Cl bond occurs and
products formed are CH3CH2CH3 and Cl •.
(ans)
© Step-by-Step
A-Level Solutions – Chemistry
08N-4
22. [08N P1 Q22 Halogen Derivatives]
A
26. [08N P1 Q26 Carbonyl Compounds]
BrCH2CH2CH2CH2Br
B
Compound X is a halogenoalkane,
BrCH2CH2CH2CH2Br.
CH3COCH3
HCN
CH3 OH
C
CH3 CN
Reaction of CH3COCH3 (a ketone) with HCN
involves the use of a homogeneous catalyst,
NaCN(aq) – the catalyst is in the same physical
state as the reactants. (ans)
BrCH2CH2CH2CH2Br
KCN
in ethanol
NCCH2CH2CH2CH2CN
© Step-by-Step
reduction
27. [08N P1 Q27 Acid Derivatives]
(ans)
H2N(CH2)6NH2
OH
© Step-by-Step
OCH3
23. [08N P1 Q23 Halogen Derivatives]
C
[92J P4 Q24]
A
CH2–CH=CH2
0
PCBs contain only aryl Cl (i.e. Cl attached
directly to benzene ring) which are resistant to
hydrolysis. Hence, no Cl atoms will be
removed by hydrolysis. (ans)
CH3COCl reacts with the phenol functional
group to give the corresponding ester.
OH
OCH3
© Step-by-Step
24. [08N P1 Q24 Alcohols]
C
CH3COCl
CH2–CH=CH2 (ans)
CH2–CH=CH2
© Step-by-Step
CH3(CH2)10O(CH2CH2O)10H
A possible formula of the non-ionic detergent
made by the reaction of excess epoxyethane
with a C11 alcohol is
CH3(CH2)10O(CH2CH2O)10H
C11 alcohol
excess
epoxyethane
(ans)
© Step-by-Step
25. [08N P1 Q25 Alcohols] [03J P1 Q29]
28. [08N P1 Q28 Esters]
C
(CH3)2CHCO2CH2CH3
Acid hydrolysis of ester X gives a carboxylic
acid and an alcohol. The yellow precipitate
obtained with alkaline aqueous iodine (positive
iodoform test) suggests that the alcohol
contains a CH3CH(OH)– group. Hence, X is
(CH3)2CHCO2CH2CH3.
Hydrolysis of ester:
reagent Y
dilute NaOH
X (CH3)2CHC
OCH2CH3
When ethanol is heated with an excess of
concentrated H2SO4 (Y) at 170 oC, it undergoes
dehydration to give ethene.
H H
H H
C
C
H
c.H2SO4
heat
O
solution Z
concentrated H2SO4
C
H
OCOCH3
OCH3
H
C
C
acid hydrolysis
O
(CH3)2CHC
+ CH3CH2OH
OH
H + H2O
Iodoform test:
H OH
The impure ethene is then bubbled into dilute
NaOH (Z) to remove any excess acid. (ans)
© Step-by-Step
CH3CH2OH + 4I2 + NaOH
CHI3 + HCO2– Na+ + 5HI
yellow ppt
(ans)
© Step-by-Step
A-Level Solutions – Chemistry
08N-5
29. [08N P1 Q29 Amino Acids]
C
33. [08N P1 Q33 Electrochemistry]
total number of
positive charges
total number of
negative charges
2
0
H+ + e–
At pH 2 (acidic), the basic –NH2 group would
react with the H+ ions present to give a salt.
The product formed is
CH2
CH
NH3+
NH3+
3+
Fe + e
D
1 2
3
HO2CCH2CH(CH3)NH2
1
cell
= ER – EL
= +0.77 – 0.00
= +0.77 V
34. [08N P1 Q34 Kinetics] [96N P4 Q34]
1
2
The rate equation for the reaction can be
written: rate = k [H2O2] [I–].
The reaction is zero order with respect to
acid.
Comparing expt 1 and 2: [I–] and [H+]
constant, when [H2O2] triples, the initial rate is
also tripled. ∴ first order w.r.t. H2O2.
Comparing expt 2 and 3: [H2O2] and [H+]
constant, when [I–] doubles, the initial rate is
also doubled. ∴ first order w.r.t. I–.
CO2, COS and COSe are all linear in shape.
Down Group VI, electronegativity of the
elements decreases from O to Se.
i.e. electronegativity: O > S > Se
32. [08N P1 Q32 Energetics]
= +0.77 V
o
© Step-by-Step
The C=S bond is more polar than the
C=Se bond.
The C=O bond is more polar than the
C=S bond.
© Step-by-Step
= 0.00 V
Under standard conditions, concentrations of
Fe3+ and Fe2+ in the right hand solution should
be 1.0 mol dm–3. (ans)
(ans)
∴ bond polarity:
C=O > C=S > C=Se
This accounts for the overall polarity of COS
being less than that of COSe. (ans)
E
o
•
31. [08N P1 Q31 Bonding]
2
o
o
At the left hand electrode, oxidation takes
place and so, it is the negative electrode.
1 H → H+ + e–
2
2
© Step-by-Step
1
Fe
2+
E
•
HO2CCH2CH(CH3)NH2
Proteins in the body are built from α - amino
acids (or 2-aminocarboxylic acid). The amino
acid in D is a 3-aminocarboxylic acids.
–
H2
E
© Step-by-Step
30. [08N P1 Q30 Amino Acids]
o
1
2
•
CO2H
(ans)
E o cell = 0.77 V.
The left hand electrode is the negative
electrode.
1
2
Comparing expt 3 and 4: [H2O2] and [I–]
constant, when [H+] doubles, the initial rate
remains unchanged. ∴ zero order w.r.t. H+.
∴ rate equation is rate = k [H2O2] [I–].
•
rate
[H 2O 2 ][I- ]
2 ×10-6
From expt 1, k =
0.010 × 0.010
= 0.02 dm3 mol–1 s–1
rate constant, k =
Hence, k ≠ 2.0 × 10–1 dm3 mol–1 s–1 (ans)
NH3(g) + HCl(g) → NH4Cl(s)
© Step-by-Step
At 298 K (or 25 °C), carbon exists in the solid
state, C(s), which therefore rules out option 2.
Option 3 is ruled out because at 298 K, water
exists as H2O(l) and not, H2O(g). (ans)
© Step-by-Step
A-Level Solutions – Chemistry
08N-6
35. [08N P1 Q35 Periodicity]
1
δ–
The cation has a greater nuclear charge
than the anion.
Group II cation has charge 2+, while Group
VII anion has charge 1–. Hence, configuration
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 suggests that the
Group II element has (36+2) = 38 electrons (or
proton no. 38), while the Group VII element
has (36 – 1) = 35 electrons (or proton no. 35).
∴ The smaller size of the cation is due to the
greater nuclear charge which results in the
same number of electrons being attracted more
strongly by the nucleus. (ans)
3
–
X2(aq) + 2e
Y2(aq) + 2e–
Z2(aq) + 2e–
•
–
2X (aq)
2Y –(aq)
2Z –(aq)
C
OH + Hal–
The energy difference
between X and Z
represents the activation energy. (ans)
© Step-by-Step
38. [08N P1 Q38 Alcohols]
1
2
•
C
+1.36 V
+1.07 V
+0.54 V
•
The reaction is an example of nucleophilic
substitution.
Between X and Y the C–Hal bond will be
lengthening.
The energy difference between X and Z
represents the activation energy.
Halogenoalkanes undergo nucleophilic
substitution. The energy profile shows a onestage process involving the formation of an
intermediate Z. Between X and Y the C–Hal
bond will be lengthening as C–OH is gradually
formed.
OH
With excess hot conc. KMnO4, the 1° alcohol
group, 2° alcohol group and C=C are oxidised
to give the product:
O
C
CH3
O
CH3
O
CO2H
OH
H
H
H
HO2C O
37. [08N P1 Q37 Halogen Derivatives]
[02J P1 Q39] [93J P4 Q37]
3
CH3
CH2OH
O
© Step-by-Step
2
Hydrocortisone has seven chiral centres
(indicated by * ):
O
2Z –(aq) + Y2(aq) → Z2(aq) + 2Y –(aq)
Ecell o = ER o – EL o = +1.07 – (+0.54)
= +0.53 V (> 0, ∴ reaction is feasible)
Hence the reaction occurs. (ans)
1
The hydrocortisone molecule has 7 chiral
centres.
When treated with an excess of hot
concentrated acidified KMnO4,
hydrocortisone will produce a compound
with 2 carboxylic acid groups.
*
*
CH3 H *
*
*
* H *H
E o value becomes less positive suggests a
decrease in oxidising power : X2 > Y2 > Z2.
OH
C
fast
HO
∴ reducing power increases: X – < Y – < Z –.
•
Θ
Hal
..
36. [08N P1 Q36 Group VII]
There is an increase in reducing power in
the sequence X –, Y – and Z –.
The reaction 2Z –(aq) + Y2(aq) → Z2(aq) +
2Y –(aq) occurs.
slow
C
OH–
© Step-by-Step
2
δ+
Hal
(which has 2 carboxylic acid groups).
•
With hot acidified K2Cr2O7, only the alcohol
groups are oxidised. The product formed is:
O
C
CH3
O
CH3
H
CO2H
OH
H
H
O
(which has 3 carbonyl groups). (ans)
© Step-by-Step
A-Level Solutions – Chemistry
08N-7
39. [08N P1 Q39 Aldehydes]
H
OCH3
2
O
3
•
40. [08N P1 Q40 Proteins]
O
–CH2CONH2
2
–CH2CH2CH2NHC
3
–CH2
NH2
CH3CH=NNH
NH
Option 2 involves the reaction of aldehyde
group with CH3O– nucleophile (from CH3OH).
H
O
–
OCH3
O
OH O
OH HO
N
H
H
OCH3
In each of the above, the amino group forms
hydrogen bonding and hence, the cross-link to
stabilise the tertiary structure of a protein.
(ans)
- H2O
O
•
1
O
H
OCH3
Option 3 involves the reaction of aldehyde
group with phenylhydrazine.
H
CH3
O + H2 N N
C
H
- H2O
H
CH3
C
H
N
N
(ans)
© Step-by-Step
© Step-by-Step
[08N P1 MCQ Key]
Q.
Key
Q.
Key
Q.
Key
Q.
Key
1
2
3
4
5
B
A
C
D
C
11
12
13
14
15
D
C
D
D
A
21
22
23
24
25
C
A
A
C
C
31
32
33
34
35
B
D
B
B
D
6
7
8
9
10
C
A
D
A
C
16
17
18
19
20
C
B
C
A
B
26
27
28
29
30
B
C
C
C
D
36
37
38
39
40
C
A
B
C
A
© Step-by-Step
A-Level Solutions – Chemistry
08N-8
2008 Nov (9746) Paper 2
2.
1.
(a)
(i) Ksp = [Ag+] [Br–] (ans)
[08N P2 Q01 Hydroxy Compounds]
(a)
(i)
(ii) [Ag+] = [Br–] = [AgBr]
H H
∴ Ksp = (7.1 × 10–7) × (7.1 × 10–7)
H O C C O H
= 5.0 × 10–13 mol2 dm–6 (1 d.p.) (ans)
(ans)
H H
(ii)
(b)
(i) ∆G
H H δ–
δ+
H O C C O H
H H
hydrogen
bonding
[08N P2 Q02 Solubility Product]
o
δ– H H
(ans)
H H
(ii) ∆G
(b)
(i) Gas A is ethene, C2H4. (ans)
∆S
o
o
(ii) The purple acidified KMnO4 solution is
decolourised. (ans)
(c) The three functional groups are:
• phenol
• secondary alcohol, and
• primary amine. (ans)
(d)
(i)
CH(OH)CH2NH2
Br
Br
Br
(ii)
= ∆H o – T ∆S o
o
o
= ∆H − ∆G
T
−66.0 − (−55.3)
= − 10.7
=
298
298
= – 0.03591 kJ mol–1 K–1
= –35.9 J mol–1 K–1 (1 d.p.) (ans)
(iii) ∆S o is negative suggests that there is a
decrease in entropy due to the formation of an
ordered lattice as the precipitate forms. (ans)
(c) ∆G
[Reaction of phenol ]
OH
o
ppt =
2.303RT log Ksp
= 2.303 × 8.31 × 298 × log 1.006
= +14.8 J mol–1 (1 d.p.)
Since ∆G o ppt is positive, precipitation will not
occur, i.e. AgF is soluble in water. (ans)
(ans)
OH
2.303RT log Ksp
= 2.303 × 8.31 × 298 × log (2.0 × 10–10)
= –55310 J mol–1
= –55.3 kJ mol–1 (1 d.p.) (ans)
H O C C O H
δ+
ppt =
CH(OH)CH2NH2
© Step-by-Step
[Acid-base reaction
of phenol ]
O– Na+
O Na+
–
(iii)
O
C
CH2NH3+
[Oxidation of
secondary alcohol ]
OH
OH
(iv)
(ans)
(ans)
CH(OH)CH2NH3+ Cl –
OH
OH
[Acid-base reaction
of amine]
(ans)
© Step-by-Step
A-Level Solutions – Chemistry
08N-9
3.
4.
[08N P2 Q03 Carbonyl Compounds]
(a) Products are
CH3
CH3
(a) Cu
1s2 2s2 2p6 3s2 3p6 3d10 4s1
Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9 (ans)
CO2H
CHCO2H and
CO2H
(b)
(i) B is Ag(NH3)2Cl. (ans)
(ans)
(ii) CuCl2(aq): [Cu(H2O)6]2+ (ans)
(b)
(i) Ecell o = ER o – EL o
= +1.52 – (+0.70)
= +0.82V (ans)
(ii) 5 HO
5O
[08N P2 Q04 Transition Elements]
C:
(c)
(i) D
OH + 2MnO4– + 6H+
O + 2Mn2+ + 8H2O
E
(ans)
(ii) D
E
[Cu(NH3)4(H2O)2]2+. (ans)
CuCl2 + 2HCl → H2CuCl4 (ans)
CuCl2 + Cu + 2HCl → 2HCuCl2 (ans)
[CuCl4]2– (ans)
[CuCl2]– (ans)
(iii) Ligand exchange occurs. (ans)
(iii) To reduce quinone, the reducing agent must
have a E o value less positive than +0.70 V.
SO42– + 4H+ + 2e– → SO2 + 2H2O
E
o
= +0.17 V
∴ SO2(g) is a suitable reducing agent. (ans)
(c) Reagent used is NaBH4. (ans)
(d)
(i) reagent: 2,4-dinitrophenylhydrazine
observation: orange precipitate formed. (ans)
(ii) reagent: neutral FeCl3 solution
observation: violet/purple coloration. (ans)
(iii) reagent: aqueous Br2 solution
observation: brown Br2(aq) decolourised
(with quinol, white ppt and steamy fumes of
HBr are also observed) (ans)
© Step-by-Step
(iv) tetrahedral in shape. (ans)
(v) +1 oxidation state. (ans)
(d)
(i) reduction occurs. (ans)
(ii) F is CuCl. (ans)
(e)
(i) mole ratio Cu : F : K = 21.5 : 38.7 : 39.8
63.5 19.0 39.1
= 0.338 : 2.036 : 1.018
= 1:6:3
∴ empirical formula of G is CuF6K3. (ans)
(ii) +3 oxidation state. (ans)
(f)
(i) Cu+ 1s2 2s2 2p6 3s2 3p6 3d10
Cu(I) in E and F has no empty or partially
filled d-orbitals. Hence, d-d* electron
transitions cannot occur which accounts for
both E and F being colourless. (ans)
(ii) Cu3+ 1s2 2s2 2p6 3s2 3p6 3d8
Cu(III) in G has empty d-orbitals. Hence, d-d*
electron transitions can take place and the light
energy not absorbed is seen as the colour of the
complex. (ans)
© Step-by-Step
A-Level Solutions – Chemistry
08N-10
2008 Nov (9746) Paper 3
1.
[08N P3 Q01 Group VII / Halogenoalkanes]
(a) Bond energy (of dissociation) is the energy
required to break one mole of a covalent bond
between two atoms in the gaseous state. (ans)
(b)
(i) This reaction can be carried out by heating the
hydrogen halides. (ans)
[or by inserting a hot wire/glass rod into a test
tube of the gas.]
(ii) Down Group VII, the hydrogen halides, HX, is
increasingly easy to decompose. (ans)
(d)
(i) boiling point: C2H5Cl < C2H5Br < C2H5I
Boiling point increases from C2H5Cl to C2H5I
due to stronger intermolecular van der Waals'
forces as the number of electrons increases
from C2H5Cl to C2H5I. (ans)
(ii) bond polarity: C–Cl > C–Br > C–I
Bond polarity decreases from C–Cl to C–I due
to decrease in electronegativity from Cl to I.
(ans)
(iii) reactivity towards nucleophilic reagent:
C2H5Cl < C2H5Br < C2H5I
Reactivity towards nucleophilic reagent
increases from C2H5Cl to C2H5I due to the
decrease in strength of C–X bond. (ans)
H2 + Cl2
(iii) 2HCl
∆H = 2 × EH–Cl – EH–H – ECl–Cl
(e) P is a tertiary bromide (chiral carbon indicated
by *)
CH
= (2 × 431) – 436 – 244
= +182 kJ mol–1 (ans)
2HI
3
CH3CH2 *C CHCH3
∆H = 2 × EH–I – EH–H – EI–I
The four alkenes are:
= (2 × 299) – 436 – 151
= +11 kJ mol–1 (ans)
H
CH3
H
δ+ δ–
slow
C
H
C⊕
H
C
H
H
X
+ X–
H
H
C⊕
H
C
H
H
X
+ X–
fast
CH3
II
H
CH2CH3
CH3
CH3
C=C
C=C
CH(CH3)2
CH3CH2
X
C
H
H
C
X
H
CH3
IV
I and II are geometrical isomers. (ans)
NB. Q is a tertiary alcohol since it does not react
with hot, acidified Na2Cr2O7(aq); i.e. not oxidised.
Hence, P is a tertiary alkyl halide (chiral carbon
indicated with *). HBr can be eliminated from P in
three possible ways:
H CH3 CH3
CH3C *C
C CH3
carbocation
H
CH3
III
H
X––X
C=C
CH(CH3)2
H
Mechanism: electrophilic addition with the
positive end of the polarised X δ+––X δ– as
electrophile.
C
CH(CH3)2
H
I
(c) Electrophilic addition. (ans)
H
CH3
C=C
(iv) Down the group, less energy is needed for the
reaction. As the size of the halogen increases,
the H–X bond becomes longer and weaker and
so, breaks more easily. (ans)
•
(ans)
Br CH3
H2 + I 2
H Br
H
(i) (ii)
(iii)
(i)
(iii)
H
CH3
C=C
CH3
CH(CH3)2
exists as geometrical isomers
(ii)
(ans)
H
CH3
CH3
C=C
CH2CH3
CH3CH2
CH3
C=C
H
CH(CH3)2
© Step-by-Step
A-Level Solutions – Chemistry
08N-11
2.
(e)
(i) P2O5 + 3H2O → 2H3PO4 (ans)
[08N P3 Q02 Electrolysis / Alkanes]
(a) Products at the cathode: H2(g) and NaOH(aq)
(ans)
Overall reaction is:
2RCO2– Na+ + 2H2O → H2 + R–R + 2CO2
+ 2NaOH (ans)
(b) charge, Q = I × t = (2.0 × 40 × 60) C
= 4800 C
= 4800 F = 0.0497 F
96500
CO2 + 2NaOH → Na2CO3 + H2O (ans)
(ii) mass of H2O = 1.55 g
1.55
mol of H2O = m =
= 1.55
M r 2(1.0)+ 16.0
18.0
= 0.0861 mol (ans)
•
2RCO2– Na+ → R–R + 2CO2 + 2e–
= 0.0689 mol (ans)
From equation, 2 F ≡ 1 R–R
or 0.0497 F ≡ ( 12 × 0.0497) mol of R–R
(iii) mol of H in D = 2 × mol of H2O
= 2 × 0.0861 = 0.172 mol
mol of C in D = mol of CO2
= 0.0689 mol
∴ mol of C2H6 = ( 12 × 0.0497) mol
Mr of C2H6 = (2 × 12.0) + (6 × 1.0) = 30.0
∴ mass of C2H6 = ( 12 × 0.0497) × 30.0
∴ H : C ratio in D = 0.172 : 0.0689
= 2.5 : 1
= 5:2
= 0.746 g (ans)
(c) Since A is a mono-carboxylic acid,
mol of A = mol of NaOH
= cV
•
molecular formula of D is C4H10. (ans)
(f) Possible structures:
11.4
= 0.100 × 1000
= 1.14 × 10–3 mol
∴ mass of 1.14 × 10–3 mol of A = 0.100 g
mass of 1 mol of A =
mass of CO2 = 3.03 g
3.03
mol of CO2 = m =
= 3.03
M r 12.0+ 2(16.0)
44.0
CH3CH2CH2CH2CH2CH3
CH3CH2CO2H
alkane E
acid B
0.100 = 87.7 g
1.14 ×10−3
(ans)
(g) Structures of F, G and H:
Mr of A = 87.7 (ans)
O
•
CH3
Possible structure of A: CH3CH2CH2C
F is CH3CH CHCH3 ; G is CH3CH C CH3
OH
CH3
(ans)
(d) Using the ideal gas equation,
(since n = m )
pV = nRT = m RT
Mr
Mr
0
.
20
×
8
.
31
×
380
mRT
Mr of C =
=
1.01×105 × 87 ×10−6
pV
CH3 Br
CH3
H is CH3CH CHCH2Br
*
CH3
(ans)
•
F is a symmetrical alkane since it gives only
two isomeric monobromo compounds when
reacted with Br2 under u.v. light; i.e. H atoms
in only two different chemical environments.
•
H has a chiral carbon atom indicated by *.
= 71.9 (ans)
•
CH3
Molecular formula of C is C5H12. (ans)
(ans)
© Step-by-Step
A-Level Solutions – Chemistry
08N-12
3.
[08N P3 Q03 Acid Derivatives]
(a)
(i) Reaction I: HCN + trace amount of NaOH;
10 – 20 °C (ans)
(ii) Reaction II: H2O/H+ [or dilute H2SO4(aq)];
heat (ans)
(e)
(i) Reaction III: aq. NaOH, heat;
followed by acidification (say,
with aq. H2SO4). (ans)
(ii) Structures for K and L:
O
CH3CHC
(b) HA
H+ + A –
(where HA ≡ lactic acid)
From the above equation, [H+] = [A–]
and since the degree of dissociation is very
small, [HA]eqm = [HA]initial
Ka =
[H + ][A − ]
[HA]
[H + ]2
0.20
[ H+]2 = (0.20)(1.38 × 10–4)
Cl
[H+] = 5.25 × 10–3 mol dm–3
∴ pH = –log10 [H+]
•
(ii) pH = pKa + log10 [salt]
[acid]
= –log10 Ka + log10 [salt]
[acid]
= –log10 (1.38 × 10–4) + log10 00..30
20
NH2
N C2H5
H
L
(ans)
In J, the Cl atom attached to C=O (i.e. acyl
chloride) is much more reactive due to the
electron-withdrawing effect of C=O group.
(ans)
(f) The two chemical tests are:
1. Add 2,4-dinitrophenylhydrazine to each
compound separately and warm.
= –log10 (5.25 × 10–3) = 2.28 (ans)
(c)
(i) A buffer solution is a solution whose pH
remains almost unchanged on dilution or when
small amounts of acid or base are added to it.
(ans)
N C2H5
H
K
Ka =
⇒
O
CH3CHC
Compound M gives an orange precipitate. No
such precipitate is observed with N. (ans)
2.
Add Tollens' reagent to each compound
separately and warm.
Compound M gives a silver mirror. No such
silver mirror is observed with N. (ans)
[or warm with Fehling's solution – only M gives a
red precipitate; or add Na2CO3 – only N gives
brisk effervescence of CO2(g); or add Na(s) – only
N gives effervescence of H2(g).]
© Step-by-Step
= 4.04 (ans)
(iii) CH3CH(OH)CO2– + H+ → CH3CH(OH)CO2H
(added)
(ans)
(d)
(i) Ester functional group is present in PLA. (ans)
(ii) Hydrolysis might occur during biodegradation
of PLA. (ans)
A-Level Solutions – Chemistry
08N-13
4.
[08N P3 Q04 Bonding / Energetics]
q+ ⋅q,
(r+ + r- )
the lattice energies of the oxides of Group II
elements decrease down the group as the size
of the cation, M2+, increases down the group.
(ans)
(ii) Since lattice energy ∝
(a) molecular formula of vapour: Al2Cl6 (ans)
Cl
Cl
Cl
Al
Al
Cl
Cl
Cl
Al2Cl6 dimer
(ans)
(b) With a few drops of water, steamy white fumes
of HCl(g) evolved and a white solid, Al2O3(s),
remains.
2AlCl3(s) + 3H2O(l) → Al2O3(s) + 6HCl(g)
[or Al2Cl6 + 6H2O → 2Al(OH)3 + 6HCl ]
•
With a large amount of water, AlCl3 undergoes
hydrolysis to give a weakly acidic solution.
AlCl3 + 6H2O → [Al(H2O)6]3+ + 3Cl –
[Al(H2O)6]3+ → [Al(H2O)5(OH)]2+ + H+ (ans)
(c) When Na2O(s) is added to a solution of litmus
in water, a blue solution is obtained as Na2O(s)
dissolves readily in water to give an alkaline
solution (pH ≈ 13). (ans)
Na2O(s) + H2O(l) → 2NaOH(aq) (ans)
(d)
(i)
3 Mg
2+
3–
2
N
(ans)
(ii) Mg3N2 + 6H2O → 2NH3 + 3Mg(OH)2 (ans)
•
q+ ⋅q,
(r+ + r- )
the lattice energy of Mg3N2 is larger than that
of MgO due to the bigger charge of N3– ions
(compared to O2– ions). (ans)
(iii) Since lattice energy ∝
The white insoluble solid is Mg(OH)2.
Mr of Mg3N2 = (3 × 24.3) + (2 × 14.0) = 100.9
Mr of Mg(OH)2 = 24.3 + 2 (16.0 + 1.0) = 58.3
mol of Mg(OH)2 = 3 × mol of Mg3N2
= 3 × 2.0 = 0.0595 mol
100.9
mass of Mg(OH)2 = 0.0595 × 58.3
= 3.47 g (ans)
© Step-by-Step
5.
[08N P3 Q05 Group II / Acid Derivatives]
(a) Two reactions in which iron or its compounds
behaves as a catalyst:
1.
N2(g) + 3H2(g) → 2NH3(g)
with finely divided iron, Fe(s), as
heterogeneous catalyst.
2.
2I–(aq) + S2O82–(aq) → I2(aq) + 2SO42–(aq)
with Fe3+(aq) as homogeneous catalyst. (ans)
(b) Mg(NO3)2.6H2O → MgO + 2NO2 +
•
ionic charge – the bigger the ionic charge, the
bigger (more exothermic) is the lattice energy;
•
ionic radius – the smaller the ionic radius, the
bigger (more exothermic) is the lattice energy;
•
arrangement of ions in the crystal (crystal
structure) – but this effect is small.
A-Level Solutions – Chemistry
O2
+ 6H2O (ans)
(c)
(i) HNO3 + 2H2SO4 → NO2+ + 2HSO4– + H3O+
(ans)
(ii) Type of reaction: nitration via electrophilic
substitution mechanism with NO2+ as the
electrophile.
Step 1: Production of electrophile, NO2+.
HNO3 + 2H2SO4 → NO2+ + 2HSO4– + H3O+
Step 2: The NO2+ electrophile is attracted to
the delocalised electron system in benzene to
form an unstable intermediate.
H
+
NO2
[or Mg3N2 + 3H2O → 2NH3 + 3MgO
in which case, mass of MgO = 2.40 g]
(e)
(i) Factors affecting magnitude of lattice energy:
1
2
slow
+
NO2
carbocation intermediate
Step 3: Expulsion of a proton from the
intermediate and the stable delocalised system
of benzene is retained.
H
+
NO2 + HSO4–
fast
NO2
+ H2SO4
(ans)
08N-14
(d) Synthesis of T from R:
(iii) Structural formulae of compounds formed:
•
OCH2CH2CH3
cold HCl (aq) :
R
O
c. HNO3
O2N
C
CH2
HO2C
CH2
CH
CH
N
H
NH3+ Cl –
OCH2CH2CH3
NO2
•
S
OCH3
C
O
(ans)
CH3COCl :
(i) Sn + HCl (small amount)
(ii) aq. NaOH
O
C
CH2
O2N
HO2C
OCH2CH2CH3
CH
N
NH2
(ans)
T
(e)
(i) The four functional groups in aspartame are:
carboxylic acid, amine, amide and ester. (ans)
CH2
CH
N
H
H
OCH3
C
O
C
O
CH3
(ans)
© Step-by-Step
(ii) The three compounds are:
CH3OH,
O
Na+ –O2C
C
CH2
CH
O– Na+
and
NH2
CH2
O– Na+
CH
H2N
C
O
(ans)
A-Level Solutions – Chemistry