Worked Solutions to 2012 A-level H2 Chemistry Paper 1
1 There are 4 OH grp directly attached to the ring. Hence only 4 OH- grp will be replaced with Cl atom.
2
Molecular formula of chlorinated glucose = C
6
H
8
Cl
4
O
2
Empirical formula of chlorinated glucose = C
3
H
4
Cl
2
O
1
Answer: C
Initial / mol
Fe +
1
2Fe
5
3+  3Fe 2+
0
Change / mol -1 -2 +3
Equilibrium / mol 0 3 3
Answer: C
3 Cr: [Ar] 3d 5 4s 1
Ge: [Ar] 4s 2 4p 2
S :
Sc:
[Ne] 3s 2 3p 5
[Ar] 3d 1 4s 2
Answer: D
4 Element belongs to Group II
Largest energy difference between the 2 nd and 3 rd IE. Removal of the 3 rd electron is from an inner quantum shell which requires more energy
Answer: B
5 AFalse. (Electronegativity: F > O, hence H bonding forms b/w the HF molecule is stronger than that between H
2
O molecules)
BFalse. (Strength of VWD depends on size of electron cloud; H bonding is the predominant factor to account for the differences in the boiling point.)
CTrue. (There is an average of 2 H-bonds per H
2
O molecule due to the presence of 2 lone pairs. Hence, although O is not as electronegative as F, the strength of H-bonding existing in the structure is stronger.
DFalse. (Electronegativity: F > O, HF has a greater permanent dipole than H
2
O)
Answer: C
6 ΔS <0 (decrease in entropy as change in state from gaseous  aqueous)
ΔH <0 (forward reaction is exothermic as hydrogen bonds are formed)
7
Answer: A
Expt 1: (2Cl – (aq)

Cl
2
(g) + 2e
Expt 2: 2H
2
O(l)

O
2
(g) + 4H +
– ) x 2
(aq) + 4e –
Amt of Cl
2
= 100/24000 = 1/240 mol
Amt of electrons = 1/120 mol
Q = I x t
Since t is constant in both of the experiments
Q α I
Amount of electricity passed in expt 1 = 1/120 x F
1/120 F α I
1

Worked Solutions to 2012 A-level H2 Chemistry Paper 1
Amt of electricity needed in expt 2 = 200/24000 x 4F= 1/30 F
Hence current needed is 4 I
Answer: D
8
Initial / mol
2H
2
+
2
CO
1
 CH
3
OH
0
Change / mol -x -x/2 +x/2
Equilibrium / mol 2-x 1-x/2 x/2
Answer: C
9 Weak Acid - Strong Base titration
The pH at equivalence point is > 7 due to the hydrolysis of CH
3
COCOONa salt.
Answer: A
10 Stomach juice, pH =1
[H + ] = 0.1 mol dm -3
Since HA is a weak acid, it dissociates partially.
HA ↔ H + + A - high [H + ] will suppress the dissociation of aspirin thus [HA] > [A ]
Answer: D
11 HCl + NaOH  NaCl + H
2
O
Amt of excess HCl = 0.15 mol
HCl  H+ + Cl pH = -lg [HCl] = 2.82
Answer: D
12 Uranium: 1  ½  ¼
Lead: 0  ½  ¾


[ R ]
[ R ]
0 t



1
2 n
[ R ] t
= 1
[ R ]
0
= 4 n = 2
Time taken = 2 x 4.5 x 10 9 years = 9.0 x 10 9 years
Answer: C
13 For all first order reaction, the half - life is constant at a given temperature.
Since temperature is constant, no change in half –life. Same % of hydrogen peroxide would decompose.
Answer: B
14 Catalysed reaction involved 2 steps  2 Ea
Addition of catalyst can decrease the required E
A
but does not alter the ΔH
2

Worked Solutions to 2012 A-level H2 Chemistry Paper 1
Answer: A
15 [Cu(H
2
O)
6
] 2+ (aq) + conc HCl)  [CuCl
4
] 2 (aq)
Different ligands have different effects on the relative energies of d-orbitals of a particular ion
 Changes in the energy gap b/w d-orbitals
In both of the complexes, oxidation state of Cu is +2.
Answer: C
16 Mg(OH)
2
is only sparingly soluble.
It is more correct to write Mg(OH)
2
(s) as the major product.
Option B is true; If excess water is added, limewater Ca(OH)
2
(aq) is formed, slaked lime
Ca(OH)
2
(s) will be formed is limited water is added instead.
Answer: D
17 Transition element has high densities and melting point, relatively low 1st and 2nd ionisation energies
Answer: D
18 With hot aqueous NaOH
3C l
2
0
Answer: C disproportionate l (aq) + NaC l O
3
(aq) + 3H
2
O( l )
-1 +5
19
20 sp hybridised – Cx and Cy has 2 sigma bonds sp 2 hybridised – C i and C j
has 3 sigma bonds sp 3 hybridised – Cz has 3 sigma bonds
ATrue. (

bond is formed by side on overlap of p orbitals.)
BTrue. (σ bond is formed b/w H atom and Cx via 1s-2sp overlap.)
CFalse.
DTrue. (σ bond is formed b/w Cz and Ci via 2sp 2 –sp 3 overlap )
Answer: C
There are 3 different H atoms Cl radical can substitute.
Answer: C
3

Worked Solutions to 2012 A-level H2 Chemistry Paper 1
21
In the above conversion, only the ketone is reduced to secondary alcohol.
The alkene and carboxylic acid remains unconverted.
H
2
/ Ni and H
2
/ Pt will convert alkene as well.
H
2
O / H + is not a reducing agent.
Only NaBH
4
will reduce ketones.
Answer: D
22 Upon reaction of cholesterol with cold dilute acidified KMnO
4
the following compound will be formed:
W: The above ring is now saturated.
X: cyclohexene ring is not broken open.
The following structure shows the structure of cholesterol with the chiral centres indicated:
4

Worked Solutions to 2012 A-level H2 Chemistry Paper 1
Original no. of chiral centres = 8
New no. of chiral centres = 10.
Hence statement Z is correct.
Answer: B
23 In and SN2 reaction, it’s a one step reaction where the electron rich anionic nucleophile attacks the electron deficient carbon. Hence the charges on N, C and L are:
Answer: B
24 Replacement for CFCs will be compounds that do not contain chlorine or bromine atoms in the molecular formula i.e. option A, B and C. option D is not a halogenoalkane and will not have
CFC properties.
Answer: D
25 The following shows the functional groups that will react with NaOH (aq) under reflux.
4 phenol (neutralisation) + 1 acid (neutralisation) + 1 ester (basic hydrolysis) = 6 moles of
NaOH (aq)
26
Answer: C
A.
Addition of Br
2
(aq) produces (CH
3
)
2
C(OH) – C * HBr – CH
2
– CH
2
– CO – CH
3
with one chiral centre.
B.
Prolonged heating with acidified KMnO
4
produces: CH
3
COCH
3
and
5

Worked Solutions to 2012 A-level H2 Chemistry Paper 1
27 A secondary structure (alpha helix) is formed from the folding of a polypeptide chain by hydrogen bonding between O of C=O and H of N-H of peptide linkages.
Disulfide bridges and van der waals forces occur in the tertiary structure only.
Answer: C
28
HOOC-CH
2
CH
2
COCH
3
C.
Reduction with NaBH
4
produces: (CH
3
)
2
C = CH – CH
2
– CH
2
– CH(OH) – CH
3
with molecular formula: C
8
H
16
O
D.
Warming with alkaline aqueous iodine produces:
(CH
3
)
2
C = CH – CH
2
– CH
2
– COO and CHI
3
.
Answer: A
Purpose of adding NaOH: To liberate the amine from the ammonium salt as the concentrated
HCl causes phenylammonium salt to be formed.
Answer: B
29 The amide is neutral as the lone pair of electrons on the nitrogen atom is withdrawn by the electronegative oxygen atom (both N and O are attached to the same carbon atom), resulting in the lone pair of electrons delocalising over the O-C-N bond (forming a resonance structure).
This makes it unavailable for protonation. Hence, amides are neutral and amines are more basic than amides.
Answer: B
30 The shown mechanism is that of electrophilic substitution. The complete mechanism is as shown below:
6

Worked Solutions to 2012 A-level H2 Chemistry Paper 1
31
Answer : A
1.
Graphite is made up of sp 2 hybrid orbitals. Hence the unhybridised p orbital will form delocalized electrons which travels perpendicular to the plane.
2.
Each sp 2 carbon is connected by sigma bonds to three other carbons. Hence, each carbon has three closest neighbours.
3.
The IUPAC definition limits valence to the maximum number of univalent atoms that
may combine with the atom, that is the maximum number of valence bonds that is possible for the given element. For carbon, the valency of each carbon atom is 4.
Three out of four valence electrons of each carbon atom are used in bond formation with three other carbon atoms while the fourth electron is free to move in the structure of graphite.
Answer : B (Option 1, 2)
32 Since Q (charge) = I (current) x t (time)
The mass of copper formed is only dependent on the charge, current and time.
Answer : D (option 1 only)
33
34

Comparing experiment 1 and 2, when [1-bromobutane] x 2, rate x 2. Hence order wrt
[1-bromobutane] = 1.
Comparing experiment 4 and 5, when [H
2
S] x 1.5 , rate x 1.5. Hence order wrt [H
2
S] =
1.
Hence, 1 mole of 1 bromobutane and 1 mole of H
2
S are involved in the rds.

Since r = k [1-bromobutane][H
2
S]

Hence using experiment 1, k = 1.5 x 10 -5 / (0.1) 2 = 1.5 x 10 -3 .
Units of k are: mol -1 dm +3 s -1
Answer : A (option 1, 2 and 3)
X Y Cl
2
(X)
:Cl
2
(Y)
Mg + Cl
2
 MgCl
2
H
2
+ Cl
2
 2HCl
 with cold aqueous NaOH at 15°C
Na + ½ Cl
2
 NaCl
KBr + ½ Cl
2
 KCl + ½ Br
2
2 : 1
2 : 1
1: 1
7

Worked Solutions to 2012 A-level H2 Chemistry Paper 1
C l
2 disproportionate l (aq) + NaC l O(aq) + H
2
O( l )
0 -1 +1
35
36
37
38
39
 with hot aqueous NaOH at 70°C
3C l
2 disproportionate l (aq) + NaC l O
3
(aq) + 3H
2
O( l )
0
-1 +5
Answer : B (option 1, 2)
1.
At the anode, 2Cl  Cl
2
+ 2e. Oxidation number of chlorine changes from -1 to 0.
2.
H
2
+ Cl
2
 2HCl. Oxidation number of chlorine changes from 0 to -1.
3.
NaCl + H
2
SO
4
 NaHSO
4
+ HCl. There is no change in oxidation number of chlorine.
Answer : B (Option 1, 2)
1.
Calcium only exist in one oxidation state i.e. +2.
2.
Even though Ca 2+ can be reduced to Ca. However, due to the very negative electrode potential (-2.87 V) of Ca 2+ /Ca, this makes it highly unlikely for Ca 2+ to be an oxidising agent.
3.
Cationic radii of both the calcium and transition metal ion have a smaller ionic radii than the atomic radii. This is because cations have a higher effective nuclear charge
(due to lesser shielding effect) than their respective metal atoms.
Answer : B (option 1, 2)
1.
Under alcoholic NaOH conditions, CH
3
CH
2
CH
2
Br could undergo elimination to form
CH
3
CH=CH
2
.
2.
CH
3
CH
2
CH
2
OH cannot react with NaOH (aq) via and acid base reaction to form
CH
3
CH
2
CH
2
O as the alcohol is not sufficiently acidic.
3.
Compound cannot be formed unless concentrated sulfuric acid at 0 o C followed by the presence of steam is present.
Answer : D (Option 1)
1.
Cl
2
(g) / light can react with any aliphatic compound with C-H present via free radical substitution.
2.
SOCl
2
can react with alcohol to form chloroalkanes.
3.
H
2
/ Pt can reduce ketone to secondary alcohol.
Answer : A (option 1, 2 and 3)
1.
In CH
2
=CHBr, the lone pair of electrons on Br is delocalised into the pi bond hence causing the C-Br bond to be strengthened and hence resistant to hydrolysis.
2.
As nucleophiles are electron rich, the electron rich pi bond might repel the
8

Worked Solutions to 2012 A-level H2 Chemistry Paper 1 nucleophile. (akin to benzene repelling the nucleophile).
3.
Rotation of C-Br bond is irrelevant in ease of nucleophilic substitution.
Answer : B (1 and 2)
40 This is a pattern question, the synthesis of methylbenzene follows the pattern:
1.
2.
3.
Answer : B (options 1, 2 only)
END
1,3 dimethylbenzene
9