2
1
(a)
Samples of 2-bromo-2-methylpropane were dissolved in dilute aqueous ethanol (80%
ethanol and 20% water by volume) and reacted with sodium hydroxide solution.
Several experiments were carried out at constant temperature. The initial rate of
reaction was determined in each case.
Calculate a value for the rate constant. Hence write the rate equation for the reaction.
[3]
Expt
1
2
3
[(CH3)3CBr]
/ mol dm-3
0.020
0.020
0.040
[OH¯]
/ mol dm-3
0.010
0.020
0.030
Rate
/ mol dm-3 s-1
20.2
20.2
40.4
Compare Expt 1 and 2,
Keeping [(CH3)3CBr] constant, but double [OH¯], rate was constant.
Order of reaction wrt to [OH¯] is ZERO.
[With Working: 1m]
Compare Expt 1 and 3,
Triple [OH¯] will not affect rate. But double [(CH3)3CBr] and rate was double.
Order of reaction wrt to [(CH3)3CBr] is ONE. [With Working: 1m]
From Expt 1 data,
Rate = k [(CH3)3CBr]
20.2 = k (0.020)
k = 1010 s-1
[1m]
Rate = 1010 [(CH3)3CBr]
(b)
The hydrolysis of 2-bromopropane is now investigated. Samples were dissolved in
dilute aqueous ethanol (80% ethanol and 20% water by volume) and reacted with
sodium hydroxide solution. The rate equation is found to be as follows:
Rate = 0.24 x 10-5 [CH(CH3)2Br] + 4.7 x 10-5 [CH(CH3)2Br] [OH¯]
(i)
The rate equation obtained indicates that the hydrolysis of 2-bromopropane
exhibits a mixture of both first order and second order kinetics. Suggest why
this may be so.
[2]
2-bromopropane is a 2º alkyl halide. It poses some steric hindrance for the
nucleophile to approach bromoethane compared with a 1o RX, if the reaction
were to proceed by SN2. [1m]
On the other hand, if it were to proceed by SN1, 2o carbocation is not as stable
compared with a 3º alkyl halide. [1m]
As a result, 2-bromopropane undergoes both SN1 and SN2 reaction pathways,
simultaneously.
ACJC 2011
9647/03/Aug/11
Preliminary Examination
3
(ii)
By deriving an expression in terms of [OH¯], show that the % rate due to SN2
4.7[OH¯]
is 
[1]
X 100%
4.7[OH¯] + 0.24
%SN 2 =

4.7 x 10-5 [CH(CH3 )2Br] [OH¯]
x 100%
0.24 x 10-5 [CH(CH3 )2Br] + 4.7 x 10-5 [CH(CH3 )2Br] [OH¯]
= 
(iii)
SN 2
x 100%
SN1 + SN 2
4.7[OH¯]
X 100%
4.7[OH¯] + 0.24
[1m] It is alright if 100% is omitted.
Using the expression derived, calculate the % rate due to SN2 for various
[OH¯] of 0.01M, 0.1M and 1.0M.
[1]
At [OH¯] = 0.01M, % rate = 16.4%
At [OH¯] = 0.1M, % rate = 66.2%
At [OH¯] = 1M,
% rate = 95.1%
(iv)
[1m]
In light of your answer to (iii), state how the % rate due to SN2 depends on the
concentration of [OH¯] and explain why it varies in that manner.
[2]
SN2 is favoured by strong nucleophile in high concentration, whereas SN1 is
independent of the type or concentration of the nucleophile. [1m] Hence as
[OH¯] increases, SN2 is favoured and % rate due to SN2 increases. [1m]
(v)
By comparing the magnitude of two rate constants in the rate equation given
in (b), comment on the ease of hydrolysis by the two different pathways.
The Arrhenius Equation is given as follows:
[2]
k = A e–Ea / RT
SN2 component is characterized by a larger rate constant than the SN1
component. [1m]
At the same temperature, a larger rate constant means a smaller Ea value. For this
reaction, SN2 has a smaller Ea value than SN1. Hence reaction proceeds via
SN2 more than SN1. [1m]
ACJC 2010
9647/03/Aug/2010
[Turn over
4
(vi)
Draw a labelled Maxwell-Boltzmann Curve for this reaction to illustrate the
distribution of energy. Based on your answer to (v) indicate the activation
energies clearly on the diagram for the two different reaction pathways.
[3]



(vii)
Ea,SN2 < Ea,SN1 : [1m]
Legend and Axis Labels: [1m]
Shape of Curve: [1m]
On the same axes, draw the energy profile illustrating both the SN1 kinetics
and the SN2 kinetics components observed for this reaction. Indicate the
activation energies clearly on the diagram, for the two reaction pathways.
[2]
CH(CH3)2Br + OH¯
EaSN2
CH(CH3)2OH + Br¯


ACJC 2011
Ea,SN2 < Ea,SN1 : [1m]
Shape of Curve, Axis Labels : [1m]
9647/03/Aug/11
Preliminary Examination
5
Should be exothermic.
(viii) Explain how the rate constant will change if CH(CH3)2Cl is used instead of
CH(CH3)2Br.
[2]
As C–Cl bond is stronger than C–Br, it is harder for the C–Cl bond to break.
Since reaction involves breaking the C-Hal bond, the activation energies for
forming the carbocation and transition state will be higher for both SN1 and SN2
pathways. Hence the rate constant will decrease. [2m]
No explanation: [0]
(ix)
Draw the structures of two possible organic by-products that may be formed
in this reaction, other than CH(CH3)2(OH). Briefly account for their formations.
[2]
H3C
O
CH2 CH3
C
CH3
H2C
H
C
CH3
H
Structures: [½m] each

Acid base reaction between ethanol and NaOH leads to the formation of a
small amount of C2H5O¯ which competes with OH¯ as a nucleophile.
[½m]

OH¯ also brings about elimination as it also functions as a strong base.
[½m]
[Total: 20 marks]
2
Oxalic acid is an organic compound with the formula H2C2O4. This colourless solid is a
dicarboxylic acid. In terms of acid strength, it is about 3,000 times stronger than acetic acid. Its
conjugate base, known as oxalate (C2O42−), is a reducing agent as well as a chelating agent for
metal cations.
Oxalic acid dissociates in water according to the following equations
HOOC-COOH + H2O
HOOC-COO- + H3O+
HOOC-COO- + H2O
-
(a)
(i)
OOC-COO- + H3O+
I
Ka1= 5.6 x 10-2 mol dm-3
II
Ka2 = 5.4 x 10-5 mol dm-3
Explain why the value of Ka1 is larger than Ka2.
[1]
HOOC-COOH is a stronger acid than HOOC-COO- , it is more difficult to
remove a H+ from HOOC-COO-, a negative ion. [1]
(ii)
ACJC 2010
Write expressions for acid dissociation constants for equation I and II above. [2]
9647/03/Aug/2010
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6
Ka1=
(b)
(i)
[H 3 O  ][HOOC  COO  ]
[HOOC  COOH]
Ka2 =
[H 3 O  ][  OOC  COO  ]
[HOOC  COO  ]
The neutralization between oxalic acid and sodium hydroxide corresponds to the
two equations given
HOOC-COOH + NaOH → HOOC-COONa + H2O
step 1
HOOC-COONa+ + NaOH→ NaOOC-COONa + H2O step 2
A 25 cm3 sample of oxalic acid of concentration 0.100 mol dm-3 was titrated with
sodium hydroxide of concentration 0.100 mol dm-3.
The titration curve for the above titration was given below.
pH
Y
X
25
50
Vol of NaOH/cm3
At the first equivalent point, X, the species formed is HOOC-COO-(aq), which is
both an acid and a base where the relevant equilibriums are:
HOOC-COO-(aq) + H2O(l)
H3O+(aq) +
(COO)22-(aq)
HOOC-COO (aq) + H2O(l)
OH (aq) +
(COOH)2 (aq)
+
It can be shown that in such an instance that the [H ] at the first equivalent point,
X, can be given by expression,
[H3O+] =
K a1 K a 2
Using the expression, determine the pH value at point X.
[H3O+] =
[1]
5.6 x10 2 x 5.4 x10 5 = 1.73 x 10-3
pH =2.75
(ii)
ACJC 2011
Calculate the pH value at the second equivalent point, Y, given that the [OH-] can
be assumed to be entirely due to the hydrolysis:
OOC-COO- + H2O
HOOC-COO- + OH[3]
9647/03/Aug/11
Preliminary Examination
7
Kb2 =
[OH  ][HOOC  COO  ]
,
[  OOC  COO  ]
K w [OH  ][HOOC  COO  ]
[OH  ] 2
=
=
K a2
[  OOC  COO  ]
[  OOC  COO  ]
[-OOC-COO-] = 25/1000 x 0.1 (25+50)/1000 = 0.0333 mol dm-3
1.85 x 10-10 =
(c)
(d)
[1]
[1]
[OH  ] 2
,
[OH-] = 2.48 x 10-6 mol dm-3 , pOH =5.61
0.0333
pH =8.39
[1]
The concentration of oxalic acid in a solution can be determined by an acid-base
titration. State another way by which it’s concentration can be determined
volumetrically.
[1]
Another method will to titrate an acidified solution of the oxalic acid with
potassium manganate (VII) through redox titration.
Oxalate, the salt from oxalic acid, is able to combine chemically with certain metals
commonly found in the human body, it is also able to bond chemically, behaving as
bidentate chelating agents, to transition elements, such as iron to form complex ion. A
chelating agent is a ligand that is attached to a central metal ion by bonds from two or
more donor atoms.
[Fe(H2O)6]3+ + 3 C2O42-
[Fe(C2O4)3]3- + 6H2O Kstab = 5.00 x 104
(i)
What do you understand by the term “bidentate” ligand.
(ii)
bidentate ligand is ligand which can form two dative bonds with the central
atom or ion [1]
Suggest a reason why the stability constant of the above is greater than 1.
[1]
[1]
Bidentate or polydentate ligands, bind more strongly with multiple dative
bonds and hence form more stable complexes than Fe(H2O)63+. [1]
(iii) Draw the structure of the complex ion, [Fe(C2O4)3] 3State the type of bonds formed between the ligands and the metal ion.
Hence suggest the shape for this complex
ACJC 2010
9647/03/Aug/2010
[3]
[Turn over
8
O
C
O
O
C
C
O
O
O
Fe
O
C O
O
O
C
C
O
O
[1]
Bond between ligand and metal: dative bonds [1]
Shape: octahedral [1]
(e)
Oxalic acid was one of the products formed when an aromatic organic compound,
A, with molecular formula C10H10O2 undergoes oxidation with acidified
manganate(VII) to form another organic product, B, with the molecular formula
C8H8O2. No other organic compound was formed in the oxidation. Compound B
reacts readily with 2 mole of Br2(aq) to form compound E, C8H6O2Br2.
Compounds A and B are both soluble in NaOH and both A and B reacts with 2,4 DNPH. Compound A reacts with acidified dichromate to give an acid, C, C10H10O3.
Compound C reacts with SOCl2 to form a sweet-smelling compound D, C10H8O2.
(i)
Deduce the structures of A,B,C,D and E. Explain your deductions
Compound A:
[7]
Compound C :
HO
H
H
HO
C C C
O
:
ACJC 2011
HO
H
C C C
CH3
O
9647/03/Aug/11
CH3
Preliminary Examination
9
Compound B:
Compound D:
HO
O
O
C
O C
H C
CH3
C
CH3
Compound E
HO
Br
O C
CH3
Br
[1] for structure A and C,
[1] for structure B
[1] for structure D and [1] for structure E
Deductions:
- A and B are soluble in NaOH  A and B consists of carboxylic acids
or phenol/B reacts with 1 mole of Br2(aq)  B contains phenol. [1]
- A and B reacts with 2,4-DNPH  A and B contains the carbonyl
functional group.
[1]
- A reacts with acidified dichromate to form an acid,  A contains the
aldehyde.
[1]
3 The lac repressor is a DNA-binding protein which inhibits formation of proteins involved in
the metabolism of lactose in bacteria. It is active in the absence of lactose, ensuring that
the bacterium only invests energy metabolism of lactose when lactose is present. When
lactose becomes available, it is converted into allolactose, which inhibits the lac
repressor's DNA binding ability.
Structurally, the lac repressor is a homotetramer made up of four identical polypeptides,
each consisting of 347 amino acid residues.
(a)
(i)
(ii)
ACJC 2010
State the highest level of protein structure in the lac repressor.
Quarternary structure. [1]
One of the most common features in the secondary structure of proteins is the
α-helix. Draw a diagram of the α-helix, showing the bonding which maintains
the structure.
[2]
9647/03/Aug/2010
[Turn over
10
Either
(b)
(i)
or
[1] for showing hydrogen bonding between amide groups in adjacent
coils of the alpha helix
An 18 residue section from the lac repressor was digested using two types of
enzymes. The enzymatic hydrolysis gave rise to these fragments.
Enzyme 1
Enzyme 2
Gly-Trp-Leu-Ala-Glu
Lys-Tyr-Trp-Leu
Val-Asp-Asp
Glu-Met
Val-Arg
Asp-Phe
Tyr-Trp-Leu-Val-Arg
Glu-Glu-Met-Lys
Val-Asp
Asp-Asp-Phe
Gly-Trp-Leu-Ala
Suggest the primary sequence of this 18 residue polypeptide.
Gly-Trp-Leu-Ala-Glu-Glu-Met-Lys-Tyr-Trp-Leu-Val-Arg-Val-Asp-Asp-Asp-Phe
[1] Answer + [1] working
(ii)
Another segment of tripeptide, Thr-Glu-Lys, was subjected briefly to acidic
hydrolysis which produced individual amino acids as well as various peptides
due to partial hydrolysis. The resulting mixture buffered at pH 8 was separated
in an electric field using electrophoresis.
pKa = 2.19
pKa = 2.18
pKa = 2.09
O
O
H2N
Amino acid
CH
C
CH
OH
OH
H 2N
CH3
CH
C
OH
CH 2
CH 2
C
O
OH
pKa = 4.25
ACJC 2011
9647/03/Aug/11
Preliminary Examination
11
O
Mr = 147
Mr = 131
H2N
CH
C
OH
CH 2
CH 2
CH 2
CH 2
NH 2
Mr = 146
Abbreviation
Thr
Glu
Lys
pI
5.60
3.22
9.74
Match each of the three amino acids to the spots A, B and C.
Cathode
A
Start
D
B
C
Anode
A – Lys; B – Thr; C – Glu.
([2] If all three correct; OR [1] If two are correct)
(iii) Given that longer peptide strands show greater resistance when migrating
through the electrophoresis plate, suggest a structure of the substance at spot
D when the electrophoresis was buffered at pH 8.
O
O
H2 N
CH
C
CH 2
CH
CH 3
OH
H
N
CH
C
O
H
N
CH
CH 2
CH 2
CH 2
CH 2
C
O-
O
C
O-
CH 2
CH 2
NH 2
[1]
(iv) Threonine (Thr) can be used as a buffer at pH 7. Show with the aid of
equations how it serves as a buffer at this pH.
When a small amount of strong acid is added,
ACJC 2010
9647/03/Aug/2010
[7]
[Turn over
12
O
+H N
3
O
O-
CH C
+
HC
H2N
OH-
OH
O-
CH C
HC
CH3
OH
CH3
[1]
When a small amount of strong base is added,
O
H2 N
CH C
HC
O
O-
+
H+
+H N
3
OH
HC
CH 3
(c)
CH C
O-
OH
CH3
[1]
lac repressor can be purified using affinity chromatography. In this technique, the
natural substrate of the lac repressor, a specific DNA sequence, is covalently
attached to a chromatographic support.
When a mixture of proteins is poured into the chromatographic column, lac
repressor proteins bind specifically to its natural substrate, while other proteins are
washed through. This step is pH sensitive and a suitable buffer is used to maintain
optimal binding pH.
(i) Suggest two types of bonding that could take place between the following
amino acid side chains of the lac repressor and DNA nucleotides.
NH 2
N
N
N
HO
H
H
O
H
H
O
O
N
O
H 2N
CH
C
O
O
H
N
CH
C
H
N
CH
C
OH
H
P
CH 2
O
O-
CH 2
CH 2
OH
CH 2
NH
O
CH 2
N
O
CH 2
O
H
H
O
H
H
O
NH 2
H
P
O-
OH
DNA sequence
ACJC 2011
Section of peptide sequence from lac repressor
9647/03/Aug/11
Preliminary Examination
13
[1] for each of the following bonding types; Max [2])
Fnc group on DNA
Type of Bonds
Fnc group on aa residues
nucleotide
Ionic bonding
Phosphate group
Amino side chain (of lysine)
Van der Waals
Benzene side chain (of
Heterocyclic arene
forces
phenylalanine)
O atom on phosphate,
ribose grp or heterocyclic
Hydroxyl H atom (of serine)
Hydrogen
arene
bonding
H atom on ribose grp or
Hydroxyl O atom (of serine)
heterocyclic arene
After washing away impurities, the target molecule is eluted (i.e. released from the
chromatographic column) using one of a variety of methods, such as changing the
pH within the chromatographic column. This weakens the protein binding and
causes pure lac repressor protein to be washed through.
(ii)
For a protein that binds fully to the covalently bonded substrate, determine the
pH for 70% of the protein to be eluted given that
Protein-Substrate complex (aq) + H+ (aq)
Protein-H+ (aq) + Substrate (s)
5
-1
Kb = 8.81 x 10 mol dm3
Kb =
[Protein  H ]
[P  S complex] [H  ]

For [Protein  H ] = 0.7,
[P  S complex]
Kb = 0.7 / [H+]
[H+] = 7.94 x 10-7 mol dm-3 [1]
pH = -lg (7.94 x 10-7) = 6.10 [1]
(iii) When pH in the chromatographic column is adjusted to elute the purified
protein, care must be taken to maintain very mild pH conditions.
Suggest how the protein structure is affected when pH for elution is too high
(e.g. pH 12) and explain your answer using lysine (Lys) and glutamic acid
(Glu).
Denaturation of the protein occurs, affecting ionic bonding in tertiary and
quaternary structure [1].
Side chain of lysine becomes uncharged at pH 12, hence salt bridges
responsible for ter. and quat. structure are disrupted [1].
(iv) Uncoiling of the protein structure is favoured when temperature is higher than
physiological conditions. With the aid of the equation,
ΔG = ΔH – TΔS
suggest why this process is favoured at higher temperature such as 70 °C.
[8]
Although ΔH is positive [1] when Van der Waals forces and hydrogen bonds
are overcomed at 70 °C, this is offset by the increase in entropy (ΔS) [1] of
the protein when it uncoils.
ACJC 2010
9647/03/Aug/2010
[Turn over
14
O
(d)
S
R
OH
Sulfinic acids are reactive species with structure similar to carboxylic acid.
Unlike carboxylic acids, the analogous sulfinic acids are chiral.
Deduce the shape of the molecule using and draw the pair of optical isomers.
In sulphinic acid, S has 3 bond pairs and 1 lone pair. [1]
S has a trigonal pyramidal shape. [1]
S
S
O
R
[3]
O
OH
R
OH
[1]
[Total: 20m]
4
A chemist design an ion-specific probe for measuring [Ag +] in a NaCl solution saturated with
AgCl using the following set- up.
voltmeter
Pt wire
Ag wire
Paste of Hg2Cl2 in
Hg
Salt Bridge
Saturated KCl
solution
NaCl solution
saturated with AgCl
(a)
(i)
Given the following standard half reactions.
Hg2Cl2(s) +2e
Ag+(aq) + e
2Hg(s) + 2Cl-(aq)
Ag(s)
Eøred = +0.24 V
Eøred = +0.80 V
Obtain an overall balanced equation, including state symbols for the cell above
and state the direction of the flow of electron in the cell.
[2]
2Ag+(aq) + 2Cl-(aq) + 2 Hg(s) → Hg2Cl2(s) + 2Ag(s)
From Pt to Ag
ACJC 2011
9647/03/Aug/11
[1]
[1]
Preliminary Examination
15
(ii)
An engineer wish to use the probe to analyse an ore sample. After pretreating the
sample, the chemist measured the cell voltage, Ecell as 0.53V.
The Nernst equation can be used to measure the concentration of silver ions using
the probe,
Ecell = Eøcell –
0.0592
log10 K
n
Where n is the no of moles of electrons transferred in the overall reaction and K is
the equilibrium constant for the overall equation in a(i).
Assuming the concentration of Cl- is so high that it is essentially constant, use
the Nernst equation to calculate the concentration of silver ions in the ore sample [3]
K = 1/ [Ag+]2 since [Cl-] is high and effectively constant)
n=2
0.0592
0.53 = 0.56 –
log10 1/ [Ag+]2 [Cl-]2
n
[Ag+] = 0.311 mol dm-3
(iii)
[1]
[1]
[1]
Given that,
AgCl(s) + e
Ag(s) + Cl-(aq) Eøred = +0.22 V
Using the half equation above and any other relevant data from the Data Booklet,
derive the Eøcell for the overall reaction: Ag+(aq) + Cl-(aq)
AgCl(s) and use
the equation given in a(ii), calculate a value of the Ksp for silver chloride at
equilibrium, if Ecell of any process at equilibrium is 0V.
[3]
Ag+(aq) + Cl-(aq)
AgCl(s)
Eøcell = +0.80 – 0.22 = 0.58V
[1]
Given that for a process at equilibrium, the Ecell = 0V
K = 1 / Ksp(AgCl) and n = 1
0.0592
0 = 0.58 log10 1/Ksp
n
1/Ksp = 6.27 x 109
Ksp = 1.59 x10-10 mol2dm-6
(b)
(i)
ACJC 2010
[1]
[1]
To prepare the saturated solution of AgCl and NaCl. AgCl is added to a 200 cm 3
solution of NaCl of concentration of 1 x 10-3 mol dm-3. Using the Ksp value obtained in
a(iii), predict the maximum mass of AgCl to be added to obtain a saturated solution.
If you had not been able to obtain a value for the Ksp in a(iii), assume a value of 1.59
x 10-10 mol2 dm-6
[2]
Ksp = [Ag+][Cl–] = 1.59 x 10–10
Let s be the solubility of AgCl in the given NaCl solution.
(s )(s+ 1 x 10-3) = 1.59 x 10–10
9647/03/Aug/2010
[Turn over
16
Assume s << 1.0 x 10–3,
s = 1.59 x 10–7 mol dm–3 [1]
Maximum mass of AgCl that could dissolve
= 1.59 x 10–7 x 0.200 x 143.5
= 4.56 x 10–6 g
[1]
(ii)
The concentration of chloride ions from silver chloride in (b)(i) is less than the
square root of the Ksp value of silver chloride. Explain why this is so.
[1]
AgCl(s) ⇌ Ag+(aq) + Cl–(aq)
Cl-(aq) from NaCl
caused the above equilibrium to shift left by Le Chatelier’s Principle.
Thus, the solubility of AgCl is suppressed due to common ion effect. [1]
(iii)
Suggest why when aqueous sodium thiosulfate was added to the resulting
solution in (b)(i), more silver chloride was able to dissolve.
[3]
AgCl(s) ⇌ Ag+(aq) + Cl–(aq)
S2O32– forms a complex with Ag+.
[1]
This decreases [Ag+].
[1]
Hence, the above equilibrium shifts right which causes more AgCl to
dissolve.
[1]
(c)
In 1999, researchers in Israel reported a new type of alkaline battery, called the “super
iron” battery. This battery used the same anode reaction as an ordinary alkaline battery.
The overall equation for the cell was found to be:
2 FeO42-(aq) + 8H2O(l) + 3 Zn(s) → 3Zn(OH)2(s) + 2 Fe(OH)3(s) + 4 OH-(aq)
Construct
(i)
Construct the half-equations, with state symbols, for each electrode reaction in
alkaline conditions.
[2]
Anode: Zn(s)+2OH(aq) →Zn(OH)2(s) + 2e
Cathode: FeO42(aq) + 4H2O (l) + 3e → Fe(OH)3 (s) + 5OH(aq)
(ii)
A “super-iron” battery should last longer than an ordinary alkaline battery of
the same size and weight.
Calculate the quantity of charge released by the reduction of 10.0 g of K2FeO4 to
Fe(OH)3.
[3]
10.0
Amount of FeO42 =
= 0.0505 mol
2(39.1)  55.8  4(16)
Amount of e = 3(0.0505) mol
Q = 3(0.0505)  96500 = 1.46  104 C (missing units 1)
ACJC 2011
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Preliminary Examination
17
(iii)
Suggest a reason why the “super-iron” battery can last longer, given that for a
normal alkaline battery, the reaction at the cathode is:
2MnO2(s) + 2NH4+(aq) + 2e → 2NH3(aq) + 2MnO(OH) (s)
[1]
Storage capacity of alkaline batteries are cathode limited and hence
the cathode in the “super iron” battery accepts 3 moles of electrons - more
electrons than the normal alkaline battery.
1 FeO42 Ξ 3e vs 1 MnO2 Ξ 1e
5
(a)
Boron is used in the manufacture of boron steel and boron carbide is used for shielding in
nuclear reactors and as control rods for nuclear reactors. Boron reacts with hydrogen to
form a series of hydrides. Upon thermal decomposition, the simplest of the hydrides when
analysed, is found to comprise 78.3% B and 21.7% H.
Find the empirical formula of this hydride.
[1]
Element
%
Ar
%/Ar
Ratio
B
73.8
10.8
6.83
1
H
21.7
1.0
21.7
3
The empirical formula of this hydride is BH3.
[1]
(b)
This hydride actually exists in the form of a gaseous dimer (diborane). This dimer is
‘unusual’ in the way some of the hydrogen atoms are involved in bonding. The structure of
diborane is similar to Al2Cl6 as shown below.
(i)
(ii)
[8]
Unlike Al2Cl6, what do you think is ‘unusual’ about this dimer?
H has only one electron but can form 2 covalent bonds. (3 center-2 electrons)
[1]
Using the above data and any relevant data where applicable from the Data Booklet,
construct an energy cycle to calculate the bond energy for the bridging Hb – B bond.
ΔHf B2H6(g) = +36.4 kJ mol-1
ΔHf BH3(g) = +89.2 kJ mol-1
Bond energy of B – Ha = +340 kJ mol-1
ΔH atomisation B(s) = +563 kJ mol-1
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2B(s)
+
3H2(g)
2B(g)
+
6H(g)

B2H6(g)
36.4 = 2(536) + 3(436) - 4x – 4(340)
Bond energy for (Hb – B) = 259.4 kJ mol-1
[1] for energy cycle
[1] for working; [1] for answer
[3]
(iii) In the light of your answer to (b)(ii) do you then expect the B – Ha bonds to be of the
same length as that of B – Hb bond? Explain any difference that is expected.
No. One electron on Hb is shared with 2 B whereas in Ha is shared with 1 B.
[1]
(iv) Despite the unusual characteristics, BH3 tends to exist as a dimer instead of a
monomer. Given the following data to calculate the ΔGo for the dimerisation, show
that B2H6 is energetically more favourable than BH3.
So BH3(g) = 188.2 J mol-1 K-1
So B2H6(g) = 232.1 J mol-1 K-1
2 BH3  B2H6
ΔHor = -142 kJ mol-1
ΔSor = -144.3 J K-1
ΔGor = -142 – 298(-144.3/1000) = -99.0 kJ mol-1
[2]
(v)
(c)
(d)
Would you expect AlH3 to be able to exist in a form similar to diborane? Explain.
No. Even though aluminium is in the same group as boron, size of
aluminium atom is bigger than boron. This makes the formation of 3 center2 electrons not viable.
OR
Yes. Aluminium is in the same group as boron. Since they are in the same
group, they are likely to react in the same way.
[1]
When diborane is reacted with ammonia, a nucleophilic substitution reaction occurs
resulting in a compound, B2H12N2 which is found to have electrical conductivity in the
molten state. Given that one mol of diborane reacted with 2 moles of ammonia,
Suggest a possible structural formula for this compound.
[1]
Diborane is a powerful electrophilic reducing agent and is used in the reduction of certain
functional groups. It attacks sites with a high electron density such as N atom in cyanides
and nitriles and O in carbonyl compounds. However it is a highly reactive gas which
catches fire spontaneously in air. Thus its direct use is restricted. Hence many organic
syntheses (involving reduction) make use of the safer solid hydrides as reducing reagents.
ACJC 2011
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Preliminary Examination
19
A typical example is sodium borohydride, NaBH4, which is commonly used to reduce
carbonyl compounds. However, it cannot be used in reduction of alkenes to alkanes.
Suggest a reason for this.
[1]
NaBH4 provide H- which behaves as a nucleophile. Therefore, NaBH4 is not suitable
for the reduction of alkenes to alkanes.
[1]
(e)
In practice, R-X (where X = Cl or Br) can react with ammonia to form an alkylammonium
salt where the amine formed in the reaction mixture will react with HX. An alkali has to be
added to the mixture to liberate the free amine. The reaction scheme for the formation of
free amine is as follows:
R - X + NH3
alcohol, sealed tube
heat
R - NH3+ X-
NaOH (aq)
R - NH2 + H2O + NaX
Consider the three organic halogen compounds, A, B and C shown below
CH3CH2Cl
CH2=CHCl
CH2=CHCH2Cl
A
B
C
(i)
(ii)
State the type of hybridization of the carbon atom of the C-X bond for all three
halogen compounds.
A – sp3
B – sp2
C – sp3
[1]
Separate samples of A, B and C were warmed initially with aqueous ethanolic sliver
nitrate and left to stand. A white precipitate was expected to appear in one or more
of the samples.
The difference in reactivity between the different organic halogen compounds and the
nucleophile can be owed to the different strength of the C-X bond. With reference to
your answer to (i), which sample will yield a precipitate and why?
Samples A and C will result in ppt
[1]
More energy is needed to break the C-Cl bond in B where carbon is sp2
hybridised. This is due to the double bond character (overlapping of p orbital of
Cl with π electron cloud) which causes C-Cl bond to be strengthened.
[1]
(iii) A can be converted to a primary amine by reacting it with ammonia is in a sealed
tube. However for a good yield of the amine, A must be used in limiting amount.
However when ethanoyl chloride is reacted with ammonia to form the primary amide,
a good yield is obtained even if the ethanoyl chloride is in excess. Explain.
A must be used in limiting amount and ammonia must be used in excess for
the formation of primary amine to prevent multiple alkylation and that will
ensure good yield of amine.
[1]
Excess ethanoyl chloride is not necessary for the formation of amide with
ammonia as amide is not nucleophilic due to electron withdrawal by C=O
group.
[1]
ACJC 2010
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20
[4]
(f)
Explain the following observations:
(i)
Nitration of methylbenzene gives approximately equal proportions of the 1,2- and 1,4isomers, but nitration under the same conditions of the compound C6H5C(CH3)3 gives
90% of the 1,4-isomer.
Although attack of the 2-position is twice as likely as the 4-position.
Steric hindrance due to the methyl group result in a 1:1.
[1]
The very bulky C(CH3)3 group in C6H5C(CH3)3 poses a greater steric hindrance
to substitution at 2-position much further.
[1]
(ii)
HCl (g) will add to the C=C bond in CH2=CH2 but not to the C=O bond in CH3COCH3.
Adding of HCl (g) to C=C bond is by electrophilic addition where H in HCl which
has a partial positive charge is the electrophile.
[1]
H-Cl is not nucleophilic (or a weak nucleophile) and hence will not have
reaction with C=O bond in CH3COCH3.
[1]
[4]
ACJC 2011
9647/03/Aug/11
Preliminary Examination