S 7C
Name:_________________ ( )
Carmel Secondary School
Final Examination (04-05)
Chemistry (Paper 2)
Date: 23 Feb., 2005
Time allowed: 3 hrs.
Total no. of pages:10
Total mark: 100
Instructions:
1. There are TWO sections in this paper, Section A and Section B.
2. Section A carries 60 marks and Section B carries 40 marks.
3. Answer THREE questions from Section A and TWO questions from Section B.
4. Answers to questions in Section A and B are to be written in the single-lined
paper.
5. Some useful constants and a Periodic Table are respectively printed on page 10 of
this question paper.
Section A
Answer any THREE questions.
1 (a)
(i)
By means of a balanced chemical equation, including state symbols,
illustrate the term the average C-H bond energy in methane.
(ii) Hydrogen is used in large quantities in industry to convert nitrogen
into ammonia, for use in fertilizers. One method of manufacturing
hydrogen is to pass methane and stream over a heated nickel catalyst.
CH4(g) + H2O(g)
CO(g) + 3H2(g) H1=+206 kJ mol-1
(1) Use the value of H1 above, and bond energy values to calculate
the total bond energy in the carbon monoxide molecule.
(2) Suggest why the bond energy you have calculated in (i) is larger
than either of the carbon-oxygen bond energies given.
The carbon monoxide is further reacted with more steam over a
copper/zinc catalyst.
CO(g) + H2O(g)
CO2(g) + H2(g)
H2=+41 kJ mol-1
(3) Suggest and explain a method whereby the carbon dioxide could
be removed from the product gas stream.
(Bond energies: E(C-H) = + 412 kJ mol-1,
E(O-H) = + 463 kJ mol-1,
-1
E(H-H) = + 436 kJ mol ,
E(C-O) = + 437 kJ mol-1
E(C=O) = + 744 kJ mol-1 )
(10 marks)
(b)
(i)
(ii)
Explain how and why the presence of a catalyst affects the rate of a
chemical reaction.
State and explain how the rate constant k1 and k2 for the forward and
reverse reactions and the equilibrium constant Kc for the reaction
CHMS7Y04E1P2
P.1
H2(g) + I2(g)
2HI(g)
H is positive
would change,
(1) in the presence of a catalyst.
(2) with an increase in temperature.
(iii) Give a brief account of the use of catalysts in industry, referring to
the manufacture of ONE inorganic and ONE organic compound in
your answer.
(6 marks)
(c)
2 (a)
From the elements in the block Sc to Zn,
(i) give the formula of the ion of an element in the +2 oxidation state
which contains four unpaired electrons in the ground state and is
relatively easy to oxidize to the +3 oxidation state;
(ii) give the symbol for an element forming only one oxidation state, and
explain why only this single oxidation state is known;
(iii) give the formula of the ion containing a metal having the highest
oxidation state and state the oxidation state of the metal in this ion.
(4 marks)
(i)
(1) The following table lists the boiling points of some compounds
of hydrogen. By reference to the type and extent of relevant
intermolecular forces, explain as fully as you can the differences
in boiling points between A and B, between B and C, and
between A and D.
Compound
Methane
Ammonia
Water
Silane
A
B
C
D
Formula
CH4
NH3
H2O
SiH4
B.p. (oC)
-164
-33
100
-112
(2) Use the explanations you have given in (1) to predict boiling
points for hydrogen fluoride, HF, and for germanium hydride,
GeH4.
(ii)
Honey contains 20% of water by mass, all of which is bound of sugar
molecules by intermolecular forces.
Assume the sugar contained in honey consists entirely of glucose,
C6H12O6. A simplified structure is given below.
O
HO
OH
HO
OH
CHMS7Y04E1P2
P.2
OH
(1) Calculate the average number of water molecules bound to each
molecule of glucose in the honey, assuming the honey contains
glucose and water only.
(2) What type of intermolecular force is likely to be responsible for
the binding of water to glucose? Illustrate your answer with a
diagram based on the above figure of glucose, showing clearly
which atoms take part in the intermolecular interaction.
(12 marks)
(b)
The water of Lake Nakuru in the Kenyan rift valley contains dissolved
sodium carbonate and sodium hydrogencarbonate.
The following
equilibrium exists:
HCO3-(aq)
H+(aq) + CO32-(aq)
(i) Explain how this solution acts as a buffer on the addition of either
acid or alkali.
[CO32 (aq )]
(ii) The pH of Lake Nakuru is 10.3 and the ratio
is 0.958.
[ HCO3 ]
Calculate the equilibrium constant for the above reaction.
(iii) When 10.0 cm3 of lake water were titrated with 0.20 mol dm-3 HCl,
22.0 cm3 of acid were required to neutralize all carbonate and
hydrogencarbonate ions according to the following equations:
H+(aq) + HCO3-(aq)  H2O(l) + CO2(g)
2H+(aq) + CO32-(aq)  H2O(l) + CO2(g)
Calculate the total number of moles of acid used, and thus, by using the ratio
quoted in part (ii), calculate [HCO3-(aq)] and [CO32-(aq)] in the lake.
(8 marks)
3 (a)
(b)
(i) Helium has a triple point of 1.0 K, 0.05 x 105 Pa. Moreover, solid
helium has the same density as liquid helium.
Sketch and label the phase diagram of helium.
Explain the slope of the solid-liquid equilibrium line on your phase
diagram.
(ii)What is the meaning of “Triple point”?
(iii)State and explain the difference between the heating of solid helium on
the moon and on the earth.
(7 marks)
(i)
From the s-block elements (Group I and II), give the symbol for the
element which reacts most vigorously with water and give an
equation for the reaction.
Suggest why this element reacts most vigorously with water.
CHMS7Y04E1P2
P.3
(c)
4.
(a)
(ii)
How do the solubilities of the hydroxides in Group II vary down the
group?
(3 marks)
(i)
Give an account of the redox chemistry of the anions of nitrogen and
sulphur. Your answer should include reference to the chemistry of
NO2-, NO3-, S2-, SO32-, S2O32- and S2O82-. You should consider the
relative stabilities of the various oxidation states exhibited by
nitrogen and sulphur, and include a range of reactions to illustrate
your account. Marks will be gained for properly balanced equations.
(ii)
When a heavy metal nitrate is heated to produce the metal oxide,
nitrogen dioxide and oxygen, the nitrogen dioxide and oxygen are
always produced in the ratio of 4:1. By considering the oxidation
state changes of the nitrogen and the oxygen, show why this should
be so, and hence write an equation for the thermal decomposition of
aluminium nitrate.
(10 marks)
This question concerns the elements boron, aluminium and gallium, which
are found in Group 3 of the Periodic Table. Their electronic structures are:
Boron
Aluminium
Gallium
(i)
(ii)
(b)
1s22s22p1
1s22s22p63s23p1
1s22s22p63s23p63d104s24p1
(1) State the acid-base character of boron oxide and of aluminium
oxide.
(2) Why is the acid-base character of aluminium oxide of industrial
importance?
(1) Sketch the molecular shapes of BCl3 and Al2Cl6 at the
temperature of vaporization.
(2) Suggest why boron does not form B2Cl6 at room temperature.
(6 marks)
State Le Chatelier’s principle.
Many commercially important processes involve equilibrium
reactions. Give TWO examples of such processes, naming each
process and giving the appropriate equation(s).
(iii) In the Mond process for the purification of nickel, carbon monoxide
is passed over impure nickel at 50oC to form, at this temperature,
gaseous nickel tetracarbonyl, Ni(CO)4. This vapour is then passed
over pure nickel pellets at 230oC. When the nickel tetracarbonyl
decomposes, pure nickel will be deposited.
(i)
(ii)
CHMS7Y04E1P2
P.4
(1) Write the equation for the formation of nickel tetracarbonyl.
(2) Give the expression for Kc for the reaction in (1).
(3) If the concentration of CO(g) in an equilibrium mixture is
doubled, calculate the change in the nickel tetracarbonyl
concentration.
(4) Hence suggest a reason why a new Canadian plant operates at
elevated pressure of 20 atmospheres in the first stage, compared
with the atmospheric pressure of older plants.
(9 marks)
(c)
In the hydrolysis of CH3Br, which is a second order reaction, the rate
of reaction at 60oC was found to be 8.13x10-5 mol dm-3 s-1 when the
concentration of hydroxide ion and CH3Br were 0.2 and 0.05 mol
dm-3 respectively.
(i) Give the rate equation for the reaction and calculate the rate constant
at 60oC.
(ii) At 80oC the rate constant was measured as 50.0 x 10-3 mol-1 s-1.
Calculate the activation energy, E for the hydrolysis using the
E
Arrhenius’ equation which may be written as ln k = ln A - A
RT
(R = 8.31 J K-1 mol-1)
(5 marks)
CHMS7Y04E1P2
P.5
Section B
Answer any TWO questions.
5.
(a) Many organic reactions can be classified as either substitution, elimination,
oxidation or reduction reactions. Their mechanisms can also be classified
as either nucleophilic, electrophilic or free radical.
Suggest reagents and conditions for each of the following transformations,
and classify them as far as you can according to the above groupings.
(i)
(ii)
(iii)
(iv)
CH3CH=CH2  CH3CHBrCH3
CH3CH2CH3  CH3CHBrCH3
CH3CHBrCH3  CH3CH(OH)CH3
CH3COCH3  CH3CH(OH)CH3
(8 marks)
(b)
(c)
For a smoother-running car engine, it is important that the fuel/air mixture
can be fully compressed before it is ignited. In order to achieve this,
tetraethyl-lead, Pb(C2H5)4, has been added to petrol for many years.
(i) (1) Suggest what sort of chemical bond exists between the lead and
each ethyl group.
(2) What is the shape of the tetraethyl-lead molecule?
(ii) At the temperature of a car engine, the tetraethyl-lead molecule
breaks up into two main components. Suggest their identities (in
words or formulae).
(iii) Write a balanced equation for the complete combustion of tetraethyllead.
(iv) (1) Which one of the combustion products in (iii) is likely to cause
damage to the moving parts of the car engine?
(2) In order to prevent this damage, 1,2-dibromoethane is added to
leaded petrols.
Suggest the formula of the substance containing bromine which
is now emitted in the exhaust fumes.
(3) Why is this exhaust emission undesirable?
(v) In 1972, before unleaded petrols were available, 1.0 x 1012 liters
(dm3) of petrol were used by vehicles. On average, each liter of
petrol contained 0.60 g of tetraethyl-lead.
Calculate, in kg, the mass of lead that was emitted from car exhaust
pipes in that year.
(8 marks)
Indicate how you could distinguish between each of the following pairs of
substances using only simple chemical tests, giving the results of the tests in
each case. Account briefly for the observed differences for each pair.
CHMS7Y04E1P2
P.6
(i)
CH3COCH2CH2CH3 and CH3CH2COCH2CH3
(ii)
H3C
6. (a)
CH2NH2
(4 marks)
and
NH2
Discuss the bonding forces in each of the following substances and
show how the magnitude of these forces between particles is reflected
in physical properties.
(i) ethanol
(ii) cis- and trans-butenedioic acids
(iii) ethane
(6 marks)
(b)
A secondary alcohol P, (C4H10O) can be dehydrated to give two
isomeric alkenes, Q and R, one of which (Q) can exhibit geometrical
isomerism. Q reacts with hydrogen bromide to give S, which can
undergo the following sequence of reactions.
H
H
C
C
H
Br
H
H
C
C
H
H
CN
H
C
C
H
COOH (U)
H
H
C
C
H
CO2C2H5 (V)
step 1
Q
CH3
CH3
(S)
step 2
CH3
CH3
(T)
step 3
CH3
CH3
step 4
CH3
(i)
(ii)
CH3
Name a reagent which could bring about dehydration of P.
(1) Draw the structure of P.
(2) Draw the structures of the geometrical isomers of Q.
CHMS7Y04E1P2
P.7
(3) Explain why Q can exist as geometric isomers.
(iii) Give the reagents and conditions required for steps 2, 3 and 4.
(iv) (1) Show the mechanism of the reaction of Q with hydrogen bromide.
(2) Step 2 proceeds by an SN2 mechanism. Show this mechanism in
full.
(v) State, giving an explanation, whether or not S would be chiral.
(14 marks)
7 (a)
The compound A (pentyl 4-methoxycinnzmate) is used in some sunscreen creams to protect the skin from burning in ultra-violet rays from
the sun. The formula of A is:
O
CH3
O
CH
CH
C
OCH2(CH2)3CH3
(i)
(1) Name three functional groups present in A.
(2) Describe a test to show the presence of unsaturation in this
molecule. Give reagents, conditions and an equation for this
reaction as well as the observable result.
(ii) Give the structures of the organic products formed when A undergoes
the following reactions.
(1) Heating under reflux with alkaline KMnO4, followed by
acidification.
(2) Heating with hydrogen in the presence of a nickel catalyst.
(3) Heating under reflux with aqueous sodium hydroxide.
(iii) Two isomeric forms of the compound A exist. Give the structures of
these two isomers, state the type of isomerism and explain briefly
how it occurs.
(iv) Explain briefly what environmental changes are occurring which
may require an increase in the use of such sun-screen creams in the
future.
(11 marks)
(b)
On hydrolysis, a tripeptide produced the following amino acids in equimolecular amounts.
Amino acid B
Amino acid C
Amino acid D
NH2
CH
COOH
NH2
CH
CH2
(CH2)4
NH2
NH2
CH
COOH
(CH2)2
COOH
OH
CHMS7Y04E1P2
COOH
P.8
(i)
In how many different ways can these three amino acids be coupled
by peptide bonds to form a tripeptide?
(ii) Draw the structural formula of one such tripeptide.
Indicate what happens when the tripeptide is dissolved, without
hydrolysis, in
(1) dilute aqueous sodium hydroxide;
(2) dilute hydrochloric acid.
(iii) Describe simple tests to distinguish amino acids B, C and D.
(6 marks)
(c)
Suggest a synthetic route, in not more than three steps, for the
transformation of propylbenzene to N,N-diethylbenzamide.
O
C
CH2CH3
N
CH2CH3
CH2CH3
(3 marks)
End of paper
CHMS7Y04E1P2
P.9
Useful Constants
Gas constant, R = 8.31 J K-1 mol-1
Faraday constant, F = 9.65 x 104 C mol-1
Avogadro constant, L = 6.02 x 1023 mol-1
Plank constant, h = 6.63 x 10-34 Js
Speed of light in vacuum, c = 3.00 x 108 ms-1
Ionic product of water at 298 K, Kw = 1.00 x 10-14 mol2 dm-6
Specific heat capacity of water = 4.18 J g-1 K-1
Bond
C=C
C=O
C C
C N
O-H
C-H
O-H
N-H
Characteristic Infra-red Absorption Wavenumber Ranges
Compound type
Wavenumber range /cm-1
Alkenes
1610 to 1680
Aldehydes, ketones, carboxylic acids, esters
1680 to 1750
Alkynes
2070 to 2250
Nitriles
2200 to 2280
Acids (hydrogen-bonded)
2500 to 3300
Alkanes, alkenes, arenas
2840 to 3095
Alcohols, phenols (hydrogen-bonded)
3230 to 3670
Amines
3350 to 3500
CHMS7Y04E1P2
P.10
S 7C
Name:_________________ ( )
1(a) (i)
(ii)
CH4(g)  C(g) + 4H(g)
1
BE(C—H) = H
4
CH4(g) + H2O(g)
(1)
Carmel Secondary School
Final Examination (04-05)
Chemistry (Paper 1)
Answer
H
Date: 23 Feb., 2005
Time allowed: 3 hrs.
Total no. of pages:10
Total marks: 100
(1)
(1)
CO(g) + 3H(g)
H1 = +206 kJ/mol
(1)
BE (reactants) = 4  BE (C—H) + 2  BE (O—H)
= (4  412) + (2  463) = 2 574 kJ
(1)
(1)
BE (products) = 1  BE (CO) + 3  BE (H—H)
= BE (CO) + (3  436)
= BE (CO) + 1 308
H (reaction) = BE (reactants) – BE (products)
+206 = 2 574 – [BE (CO) + 1 308]
BE (CO) = +1 060 kJ/mol
(1)
(1)
(2)
Bond in CO is CO, a triple bond. It is much stronger than C—O and
C=O.
(1)
(3)
Pass mixture of CO2 and H2 over aq. KOH.
Acidic gas CO2 dissolves in alkaline KOH:
CO2 + 2OH–  CO32– + H2O
(1)
(1)
(b)(i) The presence of catalyst will increase the rate of reaction by providing a low Ea
path for the reaction to occur.
(2)
(ii)
k1
k–1
K
(1) catalyst
increase
increase
unchanged
(1)
(2) Increase in temperature
increase
increase
increase
(1)
less than k1
(iii) Manufacture of NH3 from N2 and H2 (Haber’s process)
Catalyst — finely divided iron
Hydrogenation of ethene to give ethane
Catalyst — finely divided nickel
(c)
Cr2+

[Ar] 3d4
(i)
(ii)
Cr3+
[Ar] 3d3
Sc

Sc3+
[Ar] 3d1 4s2
[Ar]
Sr3+ has a stable octet configuration.
CHMS7Y04E1P2
P.11
(1)
(1)
(1)
(1)
(1)
(iii)
2.
(a)
MnO4– (+7)
(1)
(i)(1) A: CH4, non-polar molecule
(1)
Discrete molecules held together by weak instantaneous dipole — induced
dipoles  very low bp.
(1)
D: SiH4 similar to A
Weak intermolecular force (instantaneous dipole — induced dipoles) 
very low bp.
(1)
Since SiH4 has more electrons than CH4, the strength of intermolecular
forces is stronger.  bp of it is higher than CH4.
(1)
B: NH3 — polar molecule with highly electronegative atom N. Strong Hbonding between NH3 molecules  bp is higher than A and D.
(1)
C: H2O — polar molecule has highly electronegative atom O. H-bonding
between H2O molecules is stronger than NH3 because O is more
electronegative than N.
(1)
Hence bp of H2O is higher than NH3.
(2) bp of H2O
(forms more H
bonds than HF)
>
bp of HF
> bp of NH3
(F more
electronegative than N)
bp of GeH4 > bp of SiH4
GeH4 has more electrons.
 stronger Van der Waal’s forces.
(ii)
(1)
% by mass
Mr
Rel. no. of moles
Ratio
(2)
(2)
C6H12O6 H2O
80
20
180
18
80
20
180
18
1 : 2.5
(2)
(1)
H-bonds
(1)
CHMS7Y04E1P2
P.12
2.
(b)
(i)
(ii)
CO32– remove additional H+ from added acid:
CO32– + H+
HCO3–
–
HCO3 removes additional OH– from added alkali:
HCO3– + OH–  CO32– + H2O
[CO3 2 – ]
K = [H+]
[HCO3 – ]
(1)
(1)
pH = 10.3  [H+] = 5.012  10–11 mol/dm3
(1)
K
= (5.012  10–11)  0.958
= 4.80  10–11 mol/dm3
(1)
22
(iii) moles HCl =
 0.2 = 0.004 4
(1)
1 000
Let
[HCO3–] = a mol/dm3
 [CO32–] = 0.958a
 [H+] = a
[H+] = 2  0.958a
10
10
moles [H+] =
a
moles [H+] =
 2  0.958a
1 000
1 000
(1)
10
Total [H+] =
a(1 + 2  0.958)= 0.004 4
(1)
1 000
a = [HCO3–] = 0.151 mol/dm3
 [CO32–] = 0.145 mol/dm3
(1)
3
(a)(i)
(2)
The solid-liquid equilibrium line is vertical since the volume of solid and liquid
(1)
He is the same for a fixed mass of He, i.e. pressure has no effect on mp.
(1)
(ii) Three phases coexist and at equilibrium.
(1)
(iii) The pressure on the moon is lower than the triple point. Therefore, solid will
sublime.
(1)
However, the pressure on the earth is higher than the triple point. Solid will
change to liquid first and then to vapour.
(1)
CHMS7Y04E1P2
P.13
2Cs + 2H2O  2CsOH + H2
Cs is the most metallic of all the Groups I and II elements.
(b) (i)
(c)
(ii)
Group II metallic hydroxides become more soluble  down the group. (1)
(i)
NO2– — reducing and oxidising properties
As a reducing agent, it is oxidised from +3 to +5 (NO3–).
e.g. 2MnO4– + 5NO2– + 6H+  2Mn2+ + 5NO3– + 3H2O
As an oxidising agent, it is reduced from +3 to +2 (NO).
e.g. 2I– + 2NO2– + 4H+  I2 + 2NO + 2H2O
•
•
•
(2)
NO3– — very weak oxidising property, oxidised by strong reducing agents.
e.g. nascent H2 (Devorda alloy + NaOH).
NO3– is reduced from +5 to –3 (NH3).
NO3– + 4Zn + 7OH– + 6H2O  NH3 + 4Zn(OH)42–
(2)
S2– — (H2S) reducing; oxidised from –2 to 0 (sulphur)
e.g. 2H2S + SO32– + 2H+  3H2O + 3S
(1)
SO32– – reducing property; oxidised from +4 to +6 (SO42–)
e.g. OCl– + SO32–  Cl– + SO32–
(1)
S2O32– — reducing; oxidised to S4O62– or SO42–
e.g. 2S2O32– + I2  2I– + S4O62–
e.g. 4Cl2 + S2O32– + 5H2O  8Cl– + 2SO42– + 10H+
(1)
S2O82– — strong oxidising agent; reducing to SO42–(+6)
e.g. Mn2+ +S2O82– + 2I–  I2 + 2SO42–
(ii) e.g. Mg(NO3)2  MgO + 4NO2 + O2
Decrease in oxidation no. of one N: +5 to +4 = –1
Increase in oxidation no. of two O: 2(–2) to 0 = +4
 Ratio NO2 to O2 = 4 : 1
4.
(a)
(1)
(1)
(i)
(ii)
(1)
(2)
(1)
B2O3 — acidic
Al2O3 — amphoteric
(2)
Al2O3 dissolves in both acids and alkalis. Thus, aluminium can be
separated from other impurties.
(1)
(1)
CHMS7Y04E1P2
(1/2)
(1/2)
(2)
P.14
(2)
(2)
(b)
Le Chatelier’s principle states that if a system in equilibrium is subjected to
a change, the equilibrium position of the system will shift in a way to
minimize the effect of the change.
(2)
Haber’s Process:
N2(g) + 3H2(g)
2NH3(g)
(1)
Manufacture of H2SO4:
SO2(g) + O2(g)
SO3(g)
SO3(g) + H2SO4(l)
H2S2O7(l)
H2S2O7(g) + H2O(l)  H2SO4(l)
(2)
(i)
(ii)
(iii)(1)
(2)
(3)
(c)
BCl3 does not dimerise at room temperature because of the
small size of the B atom which may not be able to fit 4 big Cl
atoms around it in the dimer.
(2)
(i)
Ni(s) + 4CO(g)
Ni(CO)4(g)
(1)
[Ni(CO)4 ]
Kc =
(1)
[CO]4
[CO] is doubled.
[Ni(CO) 4] increases by 24 = 16 times.
(1)
At 20 atm, [CO] increases. This shifts equilibrium in the direction of
Ni(CO)4 and increase the yield of Ni(CO)4.
(1)
The rate equation:
rate = k[CH3Br][OH-]
8.13x10-5 = k (0.2)(0.05)
 k = 8.13 x 10-3 mol-1dm3s-1
ln 8.13 x 10-3 = ln A – Ea/(8.31 x 333.16) ----- (*)
ln 50.0 x 10-3 = ln A – Ea/(8.31 x 353.16) ---- (**)
(**)-(*)
Ea = 88.8 kJ mol-1
(ii)
(1)
(2)
(1)
(1)
Section B
5.
(a)
(i)
(ii)
(iii)
CH3CH=CH2  CH3CHBrCH3
HBr(l) at room temperature
Electrophilic addition reaction
(1)
(1)
CH3CH2CH3  CH3CHBrCH3
Br2(g), UV light
Free radical substitution
(1)
(1)
CH3CHBrCH3  CH3CH(OH)CH3
Aqueous NaOH, reflux
(1)
CHMS7Y04E1P2
P.15
Nucleophilic substitution
(1)
CH3COCH3  CH3CH(OH)CH3
LiAlH4 (anhydrous) in ether, room temperature.
Reduction of ketone to alcohol.
(1)
(1)
(i)
(1) Covalent
(2) Tetrahedral
(1)
(1)
(ii)
Pb and butane
(1)
(iii)
Pb(C2H5)4 + 14O2  PbO2 + 8CO2 + 10H2O
(1)
(iv)
(1) PbO2(s)
(1)
(2) PbBr2
(1)
(3) PbBr2 is volatile and discharged into atmosphere.
—
Poison catalyst in catalytic converter
—
Lead is toxic and cause anaemia and brain damage
(1)
(iv)
(b)
(iv)
(c)
207

= 1  1012  
 0.6 = 3.85  1011 g
323

8
= 3.85  10 kg
Mass Pb
(1)
(i)
and
Warm with aq. NaOH and I2. Cooling of Pentan-2-one (not pentan-3-one) gives
yellow crystals of CHI3.
(2)
(ii)
Add aq. Bromine.
4-methylphenylamine decolorises Br2 and forms a white ppt but not the other
compound.
(2)
6.(a) (i)
Ethanol
CHMS7Y04E1P2
P.16
•
•
Strong hydrogen bonding are formed due to highly polar O—H group 
boiling point is higher than simple molecular compounds with similar
number of electrons.
(1)
Soluble in water (forms hydrogen bonds with water), soluble in organic
solvents (forms dipole-induced dipole).
(1)
(ii)
Cis and trans-butenedioic acids
• Cis-isomer has overall dipole moment, trans-isomer has zero dipole moment
 cis-isomer is polar and trans-isomer is non-polar
 cis-isomer has higher boiling point than trans.
(1)
• Cis-isomer binds together with hydrogen bonds as well as Van der Waal’s
forces and thus has a high boiling point.
(1)
• Cis-isomer is more soluble in water than trans-isomer and both are expected
to be soluble in organic solvents.
(iii)
Ethane
• Non polar molecules bind together with weak temporary dipole-induced
dipole forces  low boiling point.
(1)
• Insoluble in water (forms only weak dipole-induced dipoles with water) but
soluble in organic liquids.
(1)
(b)
(i)
Concentrated H2SO4
(1)
(ii)(1)
(P)
(1)
(2)
trans
cis
(2)
(3) Two different groups/atoms bonded to each sp2 carbon restricted the
rotation about C=C.
(1)
(iii) Alcoholic KCN – reflux
— step 2
aq. HCl, reflux
— step 3
Ethanol, conc. H2SO4 catalyst, reflux — step 4
CHMS7Y04E1P2
P.17
(1)
(1)
(1)
(iv)(1) HBr  H+ + Br+
(2)
(2)
(v)
7
(a)
(i)
(2)
S would not be chiral as a racemic mixture containing equimolar amounts
of d and l forms of S would be formed. The addition of HBr to a planar
carbon can occur from opposite sides.
(2)
(1) Ether, alkene, ester
(1)
(2) Add aq. Br2 at room temperature in the dark.
Decolorisation of Br2 occurs.
(1)
(1)
(1)
CHMS7Y04E1P2
P.18
(ii)(1)
(1)
(2)
(1)
(3)
(1)
(iii)
trans
cis
Geometric isomerism
(2)
(1)
(iv)
Depletion in ozone layer by reaction with free radicals (formed from CFCs
released into the upper atmosphere) occurs.
(1)
(b)
(i)
6
(1)
(ii)
(1)
(1)
(1)
CHMS7Y04E1P2
P.19
(2)
(1)
(iii) Add universal indicator.
Aq. B has the highest pH, turns indicator blue.
Aq. D has the lowest pH, turns indicator red.
Aq. C has a pH between that of B and D.
(1)
(1)
(c)
O
CH2CH3
C
KMnO4
O
OH
C
PCl5
H3O +
CHMS7Y04E1P2
O
Cl
HN(CH2CH3)2
NaOH
P.20
C
N
CH2CH3
CH2CH3
(3)