Download Qn 1 - A Level Tuition

Qn 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.
Expt [(CH3)3CBr] / mol dm-3 [OH-]/ mol dm-3 Rate / mol dm-3 s-1
1
0.020
0.010
20.2
2
0.020
0.020
20.2
3
0.040
0.030
40.4
Calculate a value for the rate constant. Hence write the rate equation for
the reaction. [3]
Compare Expt 1 and 2,
Keeping [(CH3)3CBr] constant, but double [OH¯ ], rate was constant.
Order of reaction wrt to [OH¯ ] is ZERO.
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.
Rate = k [(CH3)3CBr]
From Expt 1 data,
20.2 = k (0.020)
k = 1010 s-1
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 2bromopropane 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.
On the other hand, if it were to proceed by SN1, 2o carbocation
is not as stable compared with a 3º alkyl halide.
As a result, 2-bromopropane undergoes both SN1 and SN2
reaction pathways, simultaneously.
(ii)
By deriving an expression in terms of [OH¯ ], show that the % rate
4.7[OH- ]
[1]
due to SN2 is 4.7[OH- ] + 0.24 X 100%
%SN 2 =
SN 2
x 100%
SN 1 + S N 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- ]
4.7[OH- ]
=
X 100%
4.7[OH- ] + 0.24
(iii) 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) 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. Hence as [OH¯ ] increases, SN2 is favoured
and % rate due to SN2 increases.
(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:
k = A e–Ea / RT
[2]
SN2 component is characterized by a larger rate constant than
the SN1 component.
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.
(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) 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¯
(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.
(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
H
H2C
C
CH3
H
• Acid base reaction between ethanol and NaOH leads to the
formation of a small amount of C2H5O¯ which competes
with OH¯ as a nucleophile.
• OH¯ also brings about elimination as it also functions as a
strong base.
Qn 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-COO- + H3O+ I Ka1= 5.6 x 10-2 mol dm-3
HOOC-COOH + H2O
OOC-COO- + H3O+
II Ka2 = 5.4 x 10-5 mol dm-3
HOOC-COO- + H2O
[1]
(a) (i) Explain why the value of Ka1 is larger than Ka2.
HOOC-COOH is a stronger acid than HOOC-COO- , it is more
difficult to remove a H+ from HOOC-COO-, a negative ion.
(ii) Write expressions for acid dissociation constants for equation I and
II above. [2]
[H 3 O + ][ − OOC − COO − ]
[H 3 O + ][HOOC − COO − ]
Ka1=
[HOOC − COOH]
(b) (i)
Ka2 =
[HOOC − COO − ]
The neutralization between oxalic acid and sodium hydroxide
corresponds to the two equations given
HOOC-COOH + NaOH ĺ HOOC-COONa + H2O
step 1
HOOC-COONa+ + NaOHĺ 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 conc 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)
HOOC-COO-(aq) + H2O(l)
OH-(aq)
(COO)22-(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 = 1.74 x 10-3
pH =2.76
(ii)
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]
[OH − ][HOOC − COO − ]
Kb2 =
[ − OOC − COO − ]
Kw
[OH − ][HOOC − COO − ]
K a2 =
[ − OOC − COO − ]
=
[OH − ] 2
[ − OOC − COO − ]
25
x 0.1
[ OOC-COO ] = 1000
= 0.0333 mol dm-3
25 + 50
1000
1x10-14
[OH- ]2
-10
= 1.85 x 10 =
0.0333
5.4x10-5
[OH-] = 2.48 x 10-6 mol dm-3 , pOH =5.61
pH =8.39
(c) 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.
(d) 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. [1]
bidentate ligand is ligand which can form two dative bonds
with the central atom or ion
(ii) Suggest a reason why the stability constant of the above is greater
than 1.
[1]
Bidentate or polydentate ligands, bind more strongly with
multiple dative bonds and hence form more stable complexes
than Fe(H2O)63+.
(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
[3]
3-
O
C
O
O
C
C
O
O
O
Fe
C O
O
O
O
C
C
O
O
Bond between ligand and metal: dative bonds
Shape: octahedral
(e) (i)
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.
Deduce the structures of A,B,C,D and E. Explain your deductions
[7]
Compound A:
Compound B :
HO
HO
H
H
C C C
O
O C
CH3
CH3
Compound C:
Compound D:
Compound E
HO
HO
HO
H
O
C C C
O
CH3
O
Br
O C
CH3
Br
Deductions:
- A and B are soluble in NaOH Ÿ A and B consists of
carboxylic acids or phenol/B reacts with 1 mole of Br2(aq)
Ÿ B contains phenol.
- A and B reacts with 2,4-DNPH Ÿ A and B contains the
carbonyl functional group.
- A reacts with acidified dichromate to form an acid, Ÿ A
contains the aldehyde.
Qn 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) State the highest level of protein structure in the lac repressor.
Quaternary structure.
(ii) One of the most common features in the secondary structure of
proteins is the Į-helix. Draw a diagram of the Į-helix, showing the
bonding which maintains the structure.
Show at least 2 hydrogen bonds
Hydrogen bonds must include lone pairs of electrons and
dipoles
(b) (i) 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
Tyr-Trp-Leu-Val-Arg
Gly-Trp-Leu-Ala-Glu
Glu-Glu-Met-Lys
Lys-Tyr-Trp-Leu
Val-Asp
Val-Asp-Asp
Asp-Asp-Phe
Glu-Met
Gly-Trp-Leu-Ala
Val-Arg
Asp-Phe
Suggest the primary sequence of this 18 residue polypeptide.
Gly-Trp-Leu-Ala-Glu-Glu-Met-Lys-Tyr-Trp-Leu-Val-Arg-Val-AspAsp-Asp-Phe
OR
Val-Arg-Val-Asp-Asp-Asp-Phe-Gly-Trp-Leu-Ala-Glu-Glu-MetLys-Tyr-Trp-Leu
(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 = 2.09
pKa = 2.18
O
O
O
Amino acid
H2N
CH
C
CH
OH
H 2N
OH
CH
OH
H2N
CH
C
OH
CH2
CH 2
CH2
CH3
CH 2
C
OH
Abbreviation
pI
C
Mr = 131
Thr
5.60
CH2
O
pKa = 4.25
Mr = 147
Glu
3.22
CH2
NH2
Mr = 146
Lys
9.74
Match each of the three amino acids to the spots A, B and C.
Cathode
A
D
B
C
Start
Anode
A – Lys; B – Thr; C – Glu.
(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
H2 N
O
H
N
CH
C
CH
OH
CH
CH 3
C
O
H
N
CH 2
CH 2
CH 2
CH 2
C
O
O-
O-
C
CH
CH 2
CH 2
NH 2
(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 base is added,
O
+H N
3
CH C
O
O+
HC
H2N
OH-
OH
CH C
+
HC
CH3
O-
H2O
OH
CH3
When a small amount of strong acid is added,
O
H2 N
CH C
HC
CH 3
OH
O
O-
+
H+
+H N
3
CH C
HC
O-
OH
CH3
(c) 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
O
N
HO
N
O
H
O
H
H
O
H
P
H 2N
H
CH
CH 2
O
ONH
O
N
H
O
H
O
H
P
H
N
CH
C
H
N
CH
CH 2
CH 2
OH
CH 2
C
OH
CH 2
O
O
H
C
O
O
CH 2
H
NH 2
O-
OH
DNA sequence
Type of
Bonds
Section of peptide sequence from lac
repressor
Functional group
on DNA
nucleotide
Functional group on
amino acid residues
Ionic
Phosphate group Amino side chain (of lysine)
bonding
Van der
Waals forces
Benzene side chain (of
(avoid using Heterocyclic arene
phenylalanine)
hydrophobic
interaction)
O atom on
phosphate, ribose
Hydroxyl H atom (of serine)
group or
Hydrogen
heterocyclic arene
bonding
H atom on ribose Hydroxyl O atom (of serine)
group or
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-Sub complex (aq) + H+ (aq)
Kb = 8.81 x 105 mol-1 dm3
Protein-H+ (aq) + Sub (s)
[Protein − H+ ]
Kb = [P − S complex] [H+ ]
[Protein − H+ ]
For [P − S complex] = 0.7,
Kb = 0.7 / [H+]
[H+] = 7.94 x 10-7 mol dm-3
pH = -lg (7.94 x 10-7) = 6.10
(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.
Side chain of lysine becomes uncharged at pH 12, hence salt
bridges responsible for tertiary and quaternary structure are
disrupted.
(iv) Uncoiling of the protein structure is favoured when temperature is
higher than physiological conditions. With the aid of the equation,
ǻG = ǻH – TǻS
suggest why this process is favoured at higher temperature such as
70 °C.
Although ǻH is positive when Van der Waals forces and
hydrogen bonds are overcomed at 70 °C, this is offset by
increase in entropy (ǻS) of the protein when it uncoils.
This results in l TǻS l > l ǻH l so that ǻG is negative.
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 VSEPR theory and draw
the pair of optical isomers.
ƒ In sulphinic acid, S has 3 bond pairs and 1 lone pair. (This
is the deduction based on VSEPR theory, and it must be stated
as part of the answer)
ƒ Hence S has a trigonal pyramidal shape.
S
S
O
R
ƒ
OH
O
R
OH
[Total: 20m]
Qn 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
Salt Bridge
Paste of Hg2Cl2
in Hg
Saturated KCl
solution
NaCl solution saturated
with AgCl
(a) (i)
Given the following standard half reactions.
Hg2Cl2(s) +2e
Ag+(aq) +e
š 2Hg(s) + 2Cl-(aq)
š
Eøred = +0.24 V
Eøred = +0.80 V
Ag(s)
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) Ļ Hg2Cl2(s) + 2Ag(s)
From Pt to Ag
(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,
0.0592
Ecell = E cell –
n log10 K
ø
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
log10 1/ [Ag+]2 [Cl-]2
0.53 = 0.56 –
n
[Ag+] = 0.311 mol dm-3
(iii) 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
Given that for a process at equilibrium, the Ecell = 0V
K = 1 / Ksp(AgCl) and n = 1
0.0592
0 = 0.58 n log10 1/Ksp
1/Ksp = 6.27 x 109
Ksp = 1.59 x10-10 mol2dm-6
(b) (i)
To prepare the saturated solution of AgCl and NaCl. AgCl is added
to a 200 cm3 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
Assume s << 1.0 x 10–3,
s = 1.59 x 10–7 mol dm–3
Maximum mass of AgCl that could dissolve
= 1.59 x 10–7 x 0.200 x 143.5
= 4.56 x 10–6 g
(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) ҡ 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.
(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) ҡ Ag+(aq) + Cl–(aq)
S2O32– forms a complex with Ag+.
This decreases [Ag+].
Hence, the above equilibrium shifts right which causes more
AgCl to dissolve.
(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) ĺ
3Zn(OH)2(s) + 2 Fe(OH)3(s) + 4 OH-(aq)
(i)
Construct the half-equations, with state symbols, for each
electrode reaction in alkaline conditions.
Anode:
[2]
Zn(s)+2OH−(aq) ĺZn(OH)2(s) + 2e−
Cathode: FeO42−−(aq) + 4H2O (l) + 3e− ĺ 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]
Amount of
FeO42−−
10.0
= 2(39.1) + 55.8 + 4(16) = 0.0505 mol
Amount of e− = 3(0.0505) mol
Q = 3(0.0505) x 96500 = 1.46 x 104 C
(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 ĺ 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−− Ȅ 3e vs 1 MnO2 Ȅ 1e
Qn 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 B
H
%
73.8 21.7
Ar
10.8 1.0
%/Ar 6.83 21.7
Ratio
1
3
The empirical formula of this hydride is BH3.
(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)
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.
ǻHf B2H6(g) = +36.4 kJ mol-1
ǻH atomisation B(s) = +563 kJ mol-1
Bond energy of B – Ha = +340 kJ mol-1
2B(s)
+
3H2(g)
2B(g)
+
6H(g)
Æ
B2H6(g)
36.4 = 2(536) + 3(436) - 4x – 4(340)
Bond energy for (Hb – B) = 259.4 kJ mol-1
(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.
(iv)
Despite the unusual characteristics, BH3 tends to exist as a
dimer instead of a monomer. Given the following data to
calculate the ǻGo for the dimerisation, show that B2H6 is
energetically more favourable than BH3.
ǻHf BH3(g) = +89.2 kJ mol-1
ǻSo BH3(g) = 188.2 J mol-1 K-1
ǻSo B2H6(g) = 232.1 J mol-1 K-1
2 BH3 Æ B2H6
ǻHor = -142 kJ mol-1
ǻSor = -144.3 J K-1
ǻGor = -142 – 298(-144.3/1000) = -99.0 kJ mol-1
(v)
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 center-2 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.
(c) 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]
(NH3)2BH2+ BH4(d) 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. 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.
-
(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)
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
(ii)
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
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 ʌ electron
cloud) which causes C-Cl bond to be strengthened.
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.
[4]
(f) Explain the following observations:
(iii)
(i)
(ii)
Nitration of methylbenzene gives approximately equal
proportions of the 1,2- and 1,4- isomers, 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 4position.
Steric hindrance due to the methyl group result in a 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.
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.
H-Cl is not nucleophilic (or a weak nucleophile) and hence
will not have reaction with C=O bond in CH3COCH3.
[4]