Key to Chemistry Paper I, Mock Examination (2008-09)
1.
(a) (i)
ALE&T08.705a
The energy levels inside an atom of hydrogen are not evenly distributed.
(ii) Ionization energy of hydrogen = hvL
=( 6.626 × 10-34) × (3.275 × 1015) × (6.02 × 1023)
= 1307000 (J mol-1)
= 1307 kJ mol-1
(b) (i)
H2O can form two hydrogen bonds per molecule while NH3 and HF can form only
one hydrogen bond per molecule. Thus, the boiling point of water is higher than
those of NH3 and HF.
Besides, as F is more electronegative than N, the intermolecular hydrogen bond
formed between HF molecules is stronger than that between NH3 molecules. Thus,
the boiling point of HF is higher than that of NH3.
(ii) For molecules with similar structures, their boiling points depend on their
molecular sizes (masses). As the molecular size (mass) of ethanol is greater than
that of methanol, the boiling point of ethanol is higher.
(c) (i)
Mol. % of N2O4 = 46
Mol. % of NO2 = 54
PN 2O4 = 700 x 46% = 322
PNO2 = 700 x 54% = 378
Kp =
378 2
= 443.7 (kPa)
322
(ii) N2O4(g)
2NO2(g)
ΔH = positive
Initially the colour of the mixture becomes lighter instantaneously because there is
an expansion in volume. Then, the colour of the mixture turns from light brown to
dark brown.
The above reaction is endothermic, according to Le Chatelier’s Principle, the
equilibrium will shift to the right when the temperature is increased. More
nitrogen dioxide, which is responsible for the brown colour of the mixture, will be
formed in order to reduce the increase in temperature.
(d) In general, ions in solution have higher entropies than solids.
For the process NH4Cl(s)  NH4+(aq) + Cl -(aq)
As the dissolving process will increase the disorder of the system, that means ΔS > 0.
Since ΔG = ΔH – TΔS, though the ΔH for this process is endothermic, but its value is
small, on the other hand, ΔS is positive, a positive ΔH can be compensated by the
positiveTΔS and the overall ΔG for dissolving NH4Cl(s) is negative, so NH4Cl(s)
dissolves spontaneously under standard conditions.
P.1
2.
(a) (i)
Let rate = k[A]x[B]y[C]z
6×10-6 = k (3x) (0.1y) (0.01z)
(1)
12×0-6 = k (3x) (0.2y) (0.01z)
(2)
16×10-6= k (4x)(0.2y) (0.01z)
(3)
8×10-6 = k (4x) (0.1y) (0.02z)
(4)
(2) ÷ (1):
2 = 2y
y=1
4
4
(3) ÷ (2):
= ( )x
3
3
x=1
(3) ÷ (4):
1
2 = 2y ( )z
2
1
2 = 2 ( )z
2
1
1 = ( )z
2
z=0
Rate law: Rate = k[A][B]
(ii) From expt.1: 6×10-6 = k (31)(0.11)(0.010)
k = 2 × 10-5 (dm3 mol-1 s-1)
(iii) z = 0 implies that the reaction is zero order with respect to C which means that the
rate of reaction is independent of reactant C.
(iv) 1.
An increase in temperature cause an in increase in kinetic energy, therefore,
frequency of collision increase.
2.
An increase in temperature causes an increase in the internal energy of
molecule. Therefore, the number of molecules which possess energy greater
than or equal to the activation energy increases.
As a result, more effective collisions between the reactants and this will bring
about considerable increase in the rate of reaction.
(b)
P.2
2.
(c) (i)
(Covalent radius is defined as the half of the internuclear distance between two
like atoms bound only by a single covalent bond.)
r: covalent radius
(Van der Waals' radius is the half of the internuclear distance between atomic
nuclei of adjacent molecules in solid state.)
R: Van der Waals’ radius
(b) Since fluorine is the most electronegative atom, it does not exhibit any positive
oxidation number, it always shows an oxidation number of -1 in its compounds.
Moreover, fluorine has no low-lying, vacant d orbitals in its outermost shell. it
cannot expand its octet. Consequently, it does not show any higher oxidation
numbers.
Iodine exhibits a wide range of oxidation numbers, because it has low-lying,
vacant d orbitals in its outermost shell, it can expand its octet. By promoting
electrons into these vacant d orbitals, each iodine atom can form more bonds,
(covalent bonds) giving rise to the higher oxidation numbers.
3.
(a) (i)
B
(ii) A
(iii) A
(iv) C
P.3
3.
(b) (i)
Benzene consists of 6 C-H bonds, 3C-C and 3 C=C.
6H(g) + 6C(g)  C6H6(g) ΔH1 =-( 6×413 + 3×348 + 3×612) = -5358
3H2(g)  6H(g)
ΔH2 = 6×218 =1308
6C(s)  6C(g)
ΔH3 = 6×715 = 4290
C6H6(g)  C6H6(l)
ΔH4 = -31
-----------------------------------------------------------------6C(s) + 3H2(g)  C6H6(l) ΔHf[C6H6(l)]
ΔHf[C6H6(l)] = (-5358) + 1308 + 4290 + (-31)
= 209 (kJ mol-1)
(ii) This prediction is greater than the experimental one (49) by 160 kJ mol-1.
Reason 1: The bond energy given is an average value only.
Reason 2: It shows that benzene is more stable than just a ring with alternating
double and single carbon-carbon bonds as assumed.
i.e. The molecule is stabilized by the delocalization of π electrons.
(c) The solubility of the Group II hydroxides increases down the group. Barium hydroxide
is the most soluble, therefore, it has the strongest basic strength.
Solubility depends on hydration energy and lattice energy. But the solubility of Group
II hydroxide (OH- is a small anion) is affected more by lattice enthalpy than the
hydration enthalpy.
Down the group, the cationic size increase, hence, the resulting lattice energy decreases.
Therefore solubility increase down the group and the basic strength increases down the
group.
4.
(a) (i) ∵ The haloalkane gives the molecular ion peak at m/e = 84,
∴ The molecule does not contain bromine (atomic mass = 79.9) and iodine
(atomic mass = 126.9).
∵ There is no peak at m/e = 19.
∴The molecule does not contain fluorine(atomic mass = 19)
∵ Molecular ion peak – Base peak = 84-49 = 35 (atomic mass of Cl)
Conclusion: The molecule contains chlorine.
(ii) (I)
m/e = 49 is due to CH235Cl+
m/e = 51 is due to CH237Cl+.
(II) m/e = 84 is due to [CH235Cl35Cl]+
m/e = 86 is due to [CH235Cl37Cl]+
m/e = 88 is due to [CH237Cl37Cl]+
(iii) W is CH2Cl2
P.4
4. (b)
(i)
HO
O
O
C
C
NH2
H2N
and
OH
(ii) There are weak van der Waals’ force (instantaneous dipole induced-dipole
attraction) between polyethene chains, but there are stronger intermolecular
hydrogen bonds between Kevlar polymer chains. This hydrogen bond network
causes the chains to interlock with one another. Therefore, Kevlar has a higher
tensile strength then polyethene.
(iii) As the C-C linkage in polyethene is non-polar, it does not react with alkalis.
However, the polar amide linkage in Kevlar will be hydrolysed in alkalis and
produce -NH2 and –COONa, resulting in the degradation of the chain.
(iv) Acidic condition (or enzyme).
(v) Thermoplastics.
Because there is no cross linkage inside the polymer chain. The main
intermolecular force between Kevlar polymer chains is hydrogen bond, hydrogen
bonds can be broken on strong heating.
5.
OH
(a) (i)
C
CH3
CN
2-hydroxyl-2phenylpropanenitrile
(ii)
O
O
C
C
CH3
CN
-
H
CN
CH3
CN
OH
C
CH3 (major product)
+
CN
CN
(iii) No. L obtained from the reaction is optically inactive.
It is because the phenylethanone J has a trigonal planar structure and the CN- have
equal chances to attack from above and below. Hence, a racemic mixture is
formed. The optical activity of one enantiomer is cancelled by another
enantiomer.
5.
(b) (i)
A
(ii) B
(iii) B
(iv) C
P.5
H
OH H
O
C
C
C
C
H
H
H
6. (a)
H
OH
Or
HO
CH3 H
O
C
C
C
H
H
OH
O
HB
CH3 H
O
C
C
C
H
H
O
CH3 H
O
C
C
C
H
H
2 units of polymer
(b) Plastics which will repeatedly soften and harden as they are heated and cooled are
called thermoplastics.
(c)
Advantages
disadvantages
1. Totally biodegradable;
1. Unsuitable for long-term uses;
2. Made from renewable resources;
2. Cannot be recycled;
3. Safe incineration;
3. Unsuitable for food packaging;
4. Does not pollute the environment;
4. More expensive (at present).
5. No waste disposal problem.
Any THREE
Any THREE
(d) 2C4H8O3 + 9O2 8CO2 + 8H2O
(e) Greatly increases the surface area, therefore the rate of oxidation is increased.
(f). Advantages might include: cheapness and availability of starting materials, no great
capital outlay for a factory, can be done in any agricultural area, low transport costs, no
oil needed, conservation of foreign exchange, provides work, more
environmentally acceptable. Disadvantages: takes up land and uses labour that
could be used for growing food, might cost more than conventional plastics.
7.
(a) 1.
2.
3.
4.
5.
Dissolve the mixture (contains A and B) in ethoxyethane (ether).
The mixture is shaken with NaHCO3(aq) in a separating funnel.
Two layers are found in the separating funnel.
To the ethereal layer:
a. 4-methylphenol(A) is in the ethereal layer.
b. Dry the ethereal layer with anhydrous magnesium sulphate.
c. Filter off MgSO4.
d. Distill off ether to collect pure 4-methylphenol(A).
To the aqueous layer:
a. Add dilute HCl(aq).
b. Extract the aqueous layer with ether(ethoxyethane).
c. Collect the etheral layer and 4-methylbenzoic acid(B) is in the ethereal
extract.
d. Dry the ethereal layer with anhydrous magnesium sulphate.
e. Filter off MgSO4.
f. Distill off ether to collect pure 4-methylbenzoic acid(B).
P.6
7.
(b) Ketones such as propanone react with 2,4-dinitrophenylhydrazine to form condensation
products, these condensation products have sharp characteristic melting points They
can be purified by recrystallization. After washing and drying, their melting points can
then be determined. These values can be compared with that from data book for the
purpose of identification of the original ketone.
NO2
H
O
CH3
C
CH3
N
N
H
H
NO2
NO2
CH3
C
CH3
N
N
NO2
H
(c) (i)
(ii) Molar mass of NaOH = 40
No. of moles of NaOH =
4
= 0.1
40
Molar mass of phenylmethanol = 108
2.43
No. of moles of phenylmethanol =
= 0.0225 (limiting reagent)
108
Molar mass of C6H5CHO = 106
5. 3
No. of moles of C6H5CHO =
= 0.05
106
The percentage yield of phenylmethanol =
8.
(a) (i)
0.0225
= 90%
0.05
2
Mole ratio of Cu2+ (aq) to 1,2-diaminoethane = 1 : 2
(ii)
P.7
8.
(b) 1.
A known volume (e.g. 25.0 cm3) and known concentration (e.g. 0.10 mol dm-3) of
ethanoic acid is poured into a polystyrene cup.
2.
The initial temperatures of ethanoic acid and potassium hydroxide are recorded.
3.
The same volume (e.g. 25.0 cm3) and concentration (e.g. 0.10 mol dm-3) of
potassium hydroxide is added to the acid.
4.
The reaction mixture is stirred and the highest temperature is recorded.
5.
The heat evolved by the reaction can be calculated by using the following
equation:
Heat evolved (ΔH) = m1s1ΔT (solution) + m2s2ΔT (polystyrene cup)
6.
By working out the number of moles of water formed in the reaction, the enthalpy
change of neutralization can be found.
H
ΔHneu =
No. of moles of water formed
(b) Mass of borax needed to prepare 250 cm3 of about 0.1M borax solution is:
250
0.1 ×
× 381.4 = 9.535 (g)
1000
Procedure:
1.
Accurately weigh out, using a clean weighing bottle, about 9.5 g of borax by an
electronic balance.
Record the reading (m1)
2.
Transfer the solid borax to a clean 250 cm3 beaker. Reweigh the weighing bottle.
Record the reading again (m2).
3.
Dissolve the solid borax in the beaker using some distilled water.
4.
Transfer the solution to a clean 250.0 cm3 volumetric flask using a clean filter
funnel.
5.
Rinse the beaker and filter funnel with distilled water three times and transfer all the
liquid into the volumetric flask.
6.
When the water level is near the mark, add distilled water drops by drops until the
meniscus is just about the mark.
7.
Shake the solution.
m1  m2
[8. The concentration of the borax solution is 381.4 ]
1000
25
P.8
No.9 (Essay)
Van der Waal’s forces is a general term. It consists of three types of intermolecular attractions:
1.
Dipole-dipole attractions.
2.
Dipole-induced dipole attractions.
3.
Instantaneous dipole-induced dipole attractions.
1.
Dipole-dipole attractions.
Molecules are classified into two types:
(a) Non-polar molecules.
(b) Polar molecules.
The arise of polarity in a molecule is mainly due to
(i)
the difference of electronegativity between the bonding atom and the dipole moment of
the whole molecule which is non-zero. For example,
(ii) the presence of unpaired electron. Usually a molecule with lone pair is polar. For
example,
Polar molecules such as HCI have permanent dipole moments. They tend to orientate
themselves in such a way that the attractive forces between molecules are maximized while
repulsive forces are minimized.
The attraction between the δ+ and δ- of the permanent dipoles of neighbouring molecules is
a type of van der Waals' forces known as the dipole-dipole attractions.
P.9
The fact that polar molecules are held to each other more strongly than non -ploar
molecules of similar molecular mass, as shown in the following example:
O
C
H3C
CH3 CH2 CH2 CH3
CH3
Polar: b.p.higher (50 oC)
Non-polar: b.p.lower (0 oC)
Dipole-dipole forces are typically only about 1% as strong as covalent or ionic bonds,
and they rapidly become weaker as the distance between the dipoles increases. At low
pressure in the gas phase, where the molecules are apart, these forces are relatively
unimportant. However, at room temperature and atmospheric pressure theses forces
are important. For example, chloroform is a liquid at room temperature due to these
forces.
Hydrogen bond, which is basically a dipole-dipole interaction, is much stronger and
can be intermolecular or intramolecular in nature, it is normally excluded from the
term van der Waal’s force.
2.
Dipole-induced dipole attractions
When a non-polar molecule approaches a polar molecule with a permanent dipole, a
dipole will be induced in the non-polar molecule. The dipole induced will be in opposite
orientation to that of the polar molecule, This induced dipole can then interact with the
permanent dipole and the two molecules are attracted together.
The strength of the dipole-induced dipole attractions depends on two factors:
(i)
The magnitude of dipole moment:
The larger the dipole moment, the stronger the dipole-induced dipole attractions.
(ii) The polarizability of the non-polar molecule:
The larger the non-polar molecule, the easier the molecule to be polarized, the
stronger the dipole-induced dipole attractions
The dipole-induced dipole attractions are generally weaker than the dipole-dipole
attractions.
P.10
3.
Instantaneous dipole-induced dipole attractions.
Non-polar molecules do not have any permanent dipole moment. It seems that there
does not exist any attractive forces between the molecules. However, non-polar
molecules such as H2, C12 and even noble gases can be solidified at low temperature
and under high pressure. Some sort of attractive forces must be present which hold the
gas atoms together to form the solid crystal.
Take argon as an example. The electron cloud distribution is generally symmetrical
around the argon nucleus. However, due to the instant mobility of the electron cloud,
its position fluctuates all the time. At any particular instant, it is likely to be
concentrated on one side of the atom and there is more electron clo ud on that side
of the atom than on the other. Thus, the atom possesses an electric dipole at that
particular instant. It is known as the instantaneous dipole because it only exists
during a very short time interval.
Instantaneous dipole
The instantaneous dipole will induce a dipole in the neighbouring atom by attracting
opposite charges. If the negative end of the dipole is pointing towards a
neighbouring atom, the induced dipole will then have its positive end pointing
towards the negative pole of that dipole. That makes the instantaneous dipole attract
the induced dipole.
It is the type of weak attraction that exists between atoms of noble gases and
between non-polar molecules.
They are much weaker than dipole-dipole
attractions.
P.11
4.
The strength of instantaneous dipole-induced dipole force depends on
(a) the size of molecules: greater molecules result in stronger instantaneous dipole-induced
dipole force.
Reason: As the molecular size increases, the number of electrons increases and the nuclei
have weaker control over their outer electrons. The electron cloud tends to have a greater
momentary fluctuation, resulting in a larger instantaneous dipole.
For small molecules such as H2, N2 and CO2, the force are relatively small, and so we
expect all of these substances to be gases at 25 oC, However, as the size of the moecule
increases, the instantaneous dipole-induced dipole forces become larger and many of
these large molecules can exist as solids at 25 oC.
e.g.1: Boiling point: I2>Br2>Cl2>F2
e.g.2: Boiling point: CBr4>CCl4>CH4
e.g.3: Hydrocarbons up to 4 carbons are gases, hydrocarbons with 5 to 17 carbons are
liquids, and the heavier hydrocarbons are solids.
(b) Shape of molecules
(i)
the symmetry of molecules: more symmetrical molecules exhibit stronger
instantaneous dipole-induced dipole force.
Reasons: They have larger contact surface area and shorter distance.
For example, trans alkenes have a higher melting point than its cis isomers.
(ii) surface area of the molecules: molecules with greater surface exhibit stronger
instantaneous dipole-induced dipole force.
For example:
H
Boiling point
5.
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
CH3
H
Higher (36 oC)
CH3
C
CH3
CH3
Lower (10 oC)
Instantaneous dipole-induced dipole force in molecular crystals
Molecular crystals like P4, S8 and dry ice are solids. Dry ice contain discrete CO2 held together
by weak instantaneous dipole-induced dipole force. These substances are generally
characterized by strong covalent bonds within molecules, but relatively weak forces between
molecules.
Dry ice crystal
CO2 molecules can interact through instantaneous dipole-induced dipole force
P.12
No.10 (Essay)
O
Organic compounds with carbonyl group
are called carbonyl compounds. Aldehydes
C
and ketones are carbonyl compounds as they contain the carbonyl group.
O
The general formula for an aldehyde is
Where R = H, alkyl or aryl group.
H,
C
R
O
e.g.
H ethanal
CH3 C
Where R and R’= alkyl or aryl group.
O
The general formula for a ketone is
R
C
R' ,
R and R’ may or may not be the same.
O
e.g.
CH3CH2
C
CH3 butanone
The carbonyl carbon is sp2 hybridized, with its 3 attached atoms lying in the same plane, forming
a trigonal planar structure, the bond angles are approximately 120o.
Within the carbonyl group, as oxygen is more electronegative than carbon, the bonding electrons
of the C=O bond are drawn towards the more electronegative oxygen atom. Therefore, the
carbonyl oxygen bears a partial negative charge and the carbonyl carbon bears a partial positive
charge.
A.
Preparation of carbonyl compounds
Oxidation of alcohols.
Aldehydes and ketones can be prepared by oxidation of primary alcohols and secondary
alcohols respectively using oxidizing agents such as acidified potassium dichromate.
H
H
H
H
H
C
C
C
H
H
H
H
H
H
C
C
C
H
H
CH3
OH
OH
H
H
O
C
C
C
H
H
K2Cr2O7 / H3O+
H
K2Cr2O7 / H3O+
H
P.13
H
H
O
C
C
C
H
H
H
CH3
B.
Physical properties of carbonyl compounds
1.
Melting and boiling point:
The carbonyl group is a polar group, hence, aldehydes and ketones have higher boiling
points and melting points than hydrocarbons of similar relative molecular masses.
Besides, molecules of aldehydes and ketones are held together by dipole-dipole
interactions, while those of alcohols are held together by intermolecular hydrogen
bonds. As a result, they have lower boiling points and melting points than the
corresponding alcohols.
2.
Solubility:
The carbonyl oxygen atom allows molecules of aldehydes and ketones to form strong
hydrogen bonds with water molecules. As a result, aldehydes and ketones of low
molecular masses show appreciable solubilities in water.
For example, propanone and ethanal are soluble in water in all proportions.
C.
Infra-red spectra of carbonyl compounds
The infrared spectra of carbonyl compounds show a strong band between 1680 cm-1 to 1750
cm-1, due to C=O vibration.
D.
Chemical properties of carbonyl compounds
1.
Nucleophilic addition
As the carbonyl carbon bears a partial positive charge, the carbonyl group is susceptible
to nucleophilic attack.
Addition of hydrogen cyanide
O
OH
+
C
CH3CH2
HCN
CH3CH2
H
C
H
CN
Mechanism:
O
O
C
C
CH3CH2
H
CN
CH3CH2
-
H
CN
H
CN
OH
CH3CH2 C
H (major product)
+
CN
CN
In general, aldehydes are more reactive in nucleophilic addition than ketones. This is
due to both the inductive effect and steric effect
P.14
Stereochemical aspect of addition across the C=O bond:
Because the carbonyl carbon is sp2 hybridized, nucleophilic attack can come from
above or below the plane. Since addition can occur at both sides of the plane at equal
rates, both enantiomers are formed in exactly the same amount, resulting in a racemic
mixture of products.
2.
Addition-Elimination reactions
Addition-Elimination reactions involve the first addition of two molecules to form an
unstable intermediate. It is then followed by elimination of water.
(i)
Reaction with hydroxylamine (NH2OH)
O
C
CH3
+ H2N
OH
C
N OH
CH3
(ii) Reaction with 2,4-dinitrophenylhydrazine
NO2
H
O
CH3CH2CH2
C
CH3
N
N
H
H
NO2
NO2
CH3CH2CH2
C
CH3
N
N
NO2
H
The condensation product has a sharp characteristic melting point. It can be
purified by recrystallization from ethanol. After washing and drying, its melting
point can then be determined, the value can be compared with that from data book
for the purpose of identification of the original aldehyde or ketone.
3.
Oxidations
Aldehydes can be oxidized to carboxylic acids by strong oxidizing agents such as
potassium manganate(VII) and potassium dichromate(VI).
H
H
H
O
C
C
C
H
H
+
KMnO4 / H3O
H
H
heat
H
H
O
C
C
C
H
H
OH
Ketones do not undergo oxidation readily. It requires more drastic conditions to bring
about the cleavage of the C-C bond.
P.15
4.
Reductions
Both aldehydes and ketones undergo reduction reaction, forming primary and
secondary alcohols respectively.
The two reducing agents are LiAlH4 and NaBH4.
(i)
Reaction with lithium tetrahydridoaluminate.
H
H C
H
H
C
O
C
H
C
H
+
H3O
2.
H
H
1. LiAlH4/ dry ether
H
H
OH H
H C
C
C
C
H
H
H
H
H
Secondary alcohol
(ii) Reaction with sodium tetrahydridoborate.
H
H
E.
H
H
C
C
C
H
H
H
O
C
H
NaBH4
H
H
H
H
H
C
C
C
C
OH
H H H H
Primary alcohol
Uses of Carbonyl Compounds
1.
As raw materials for making plastics.
Carbonyl compounds such as methanal and propanone are important raw material in
the manufacture of plastics.
e.g. Methanal is used in the manufacture of urea-methanal.
Propanone is used in the manufacture of perspex.
2.
As Solvents.
Propanone is an important solvent used in industry and in the laboratory as it can
dissolve a variety of organic compounds.
P.16