All solutions here are SUGGESTED. Mr. Lee will hold no liability for any errors. Comments are entirely personal opinions.[Please do not ask me how many marks for A. The bell curve is something that is out of your control so there is no point in estimating your grade based on the number of marks you have lost. No one can tell you any accurate information on this. I hope you can learn from your mistakes here and do not make the same mistakes in Paper 1.]
1 (a) (i) Amount of menthol = 1.32 x 10-2 x 2 = 0.0264 mol
Mass of menthol = 0.0264 mol x (10 x 12 + 20 + 16)
= 4.1184 g
Percentage mass = 4.1184 / 10.0 x 100%
= 41.2% (3 sf)
(ii) 3 chiral centres, 23 = 8 optical isomers
(iii) The catalyst is in a solid state and it functions as a heterogeneous catalyst as it is in a different phase than menthone (liquid state) and hydrogen (gaseous state). Menthone and hydrogen will undergo adsorption at the active sites of the surface of the catalyst, forming weak bonds between the reactant molecules and the catalyst and this causes the bonds in menthone and hydrogen to be weakened. This provides an alternative pathway which has a lower activation energy. The reactant molecules are now in closer proximity with one another and hence, the frequency of effective collisions between menthone and hydrogen gas increases and this increases the speed of reaction.
(iv) Iron might be suitable as a catalyst. Iron is a transition metal with incompletely filled 3d orbitals.
(c) All three isomers undergo electrophilic addition with bromine water. Hence, they contain carbon-carbon double bond.
All three isomers contain carbonyl group as they undergo condensation reaction with 2,4-dinitrophenylhydrazine.
All three isomers do not contain aliphatic aldehyde group.
All three isomers undergo reduction with hydrogen gas to form menthol with molecular formula C10H20O. Hence, all three isomers contain two functional groups that can be reduced by hydrogen gas as 4 hydrogen atoms are added.
Menthol contains a secondary alcohol, hence, the 3 isomers contain a ketone group. The other functional group that is reduced must be carbon-carbon double bond.
When reacted with hot concentrated KMnO4, strong oxidation or oxidative cleavage occurs.
A gives D, which is a ketone, and E which contains two ketone groups (one ketone group is originally present in A).
B gives F, which contains two ketone groups (one ketone group is originally present in B). B contains a terminal carbon-carbon double bond.
C contains a carbon-carbon double bond in a ring. G contains a carboxylic acid group and two ketone groups (one ketone group is originally present in C).
D, F and G contains CH3C=O group as they undergo positive iodoform test (oxidation), forming yellow precipitate of CHI3.
(a) The volatilities of the halogens decrease from chlorine to iodine (i.e. melting and boiling point increase from chlorine to iodine).
From chlorine to iodine, the number of electrons increases. Hence the size of the electron cloud increases which increases the strength of van der Waals’ forces of attraction between the halogen molecules. Hence, the volatilities decrease from chlorine to iodine.
(b) 3Cl2(g) + 6OH–(aq) → 5Cl–(aq) + ClO3–(aq) + 3H2O(l)
The oxidation number of chlorine increases from 0 in Cl2 to +5 in ClO3–
The oxidation number of chlorine decreases from 0 in Cl2 to -1 in Cl–
(c) (i) The bond energy for H–F is 562 kJ mol-1 while the bond energy for H–Cl to H–I decreases from 431 to 366 to 299 kJ mol-1. H–F bond has the highest bond energy so the bond strength is the strongest which require the most energy to break. Hence, HF does not dissociate completely in water to produce hydrogen ions.
(ii) pH of HCl = – lg (0.50) = 0.301
Ka = [H+]2 / [HF] [H+]2 = 5.6 x 10-4 x 0.50 = 2.8 x 10-4
pH = – lg [(2.8 x 10-4)1/2] = 1.78
(d) When aqueous silver nitrate is added to chloride ions, a white precipitate of silver chloride (AgCl) is formed.
Ag+(aq) + Cl–(aq) → AgCl(s)
When aqueous ammonia is added to silver chloride, the white precipitate dissolves to form a colourless solution of [Ag(NH3)2]+
AgCl(s) ⇌ Ag+(aq) + Cl–(aq)
Ag+(aq) + 2NH3(aq) → [Ag(NH3)2]+
When aqueous silver nitrate is added to iodide ions, a yellow precipitate of silver iodide (AgI) is formed.
Ag+(aq) + l–(aq) → Agl(s)
When aqueous ammonia is added to silver iodide, the yellow precipitate is insoluble in aqueous ammonia.
(e) (i) The value of pV remains constant.
(ii) pV = nRT
12.0 x 105 x V = 0.40 x 8.31 x 300
V = 8.31 x 10-4 m3
(iii) 9.26 x 105
8.88 x 105
pV = 9.11 x 105 Pa dm3
V = 0.759 dm3
(iv) There is presence of permanent dipole-permanent dipole interactions between hydrogen chloride molecules due to the net dipole moment present as the chlorine atom is electronegative. These intermolecular forces of attraction are significant and will deviate from the ideal gas properties. Hence, the molecules are closer together and occupy a smaller volume as compared to an ideal gas.
(f) (i) ΔG = ΔH – TΔS
0 = +16.8 – 188ΔS
ΔS = +0.089362 kJ K-1 mol-1 ≈ +0.0894 kJ K-1 mol-1 (3 sf)
The entropy change is positive because there is an increase in disorderness when the liquid state of hydrogen chloride is changed to the gaseous state as the molecules have an increased number of ways of arranging themselves in the gaseous state.
(ii) ΔG = +16.8 – (298)(0.089362) = –9.83 kJ mol-1 (3 sf)
A negative sign for ΔG means that the reaction is feasible and can take place spontaneously.
(a) (i) Magnesium burns with a brilliant white flame when it reacts with oxygen to form magnesium oxide.
2Mg(s) + O2(g) → 2MgO(s)
Calcium burns with a brick-red flame when it reacts with oxygen to form calcium oxide.
2Ca(s) + O2(g) → 2CaO(s)
(ii) Magnesium oxide is sparingly soluble in water, forming magnesium hydroxide while calcium oxide is partially soluble in water, forming aqueous calcium hydroxide which is alkaline in water.
MgO(s) + H2O(l) → Mg(OH)2(aq)
CaO(s) + H2O(l) → Ca(OH)2(aq)
(b) (i) 2.0 x 10-1 mol dm-3
(ii) Ca(OH)2 ⇌ Ca2+(aq) + 2OH–(aq)
Concentration of OH– = 2.5 x 10-2 x 2 = 5.0 x 10-2 mol dm-3
pOH = – lg (5.0 x 10-2) = 1.30
pH = 14 – 1.30 = 12.7
(iii) Ksp = [Mg2+][OH–]2
Ksp = (s) (2s)2 = (1.6 x 10-4)(3.2 x 10-4)2 = 1.64 x 10-11 mol3 dm-9
(iv) White precipitate of magnesium hydroxide would be observed. Barium hydroxide is more soluble in water than magnesium hydroxide. Hence, the concentration of hydroxide ions increases. Common ion effect occurs and hence, the solubility of magnesium hydroxide decreases so less magnesium hydroxide is able to dissolve in water.
(c) (i) Aspartate and glutamate
(ii) The alpha helix is held in place due to hydrogen bonding formed between the N–H group of each amino acid and the fourth C=O group following it along the chain.
(iii) H2SO4(aq), heat under reflux / NaOH(aq), heat under reflux
(ii) Step 4: Reagent: excess concentrated ethanolic NH3
Condition: heat in sealed tube
Step 5: Reagent: H2SO4(aq)
Condition: heat under reflux
(a) Step 1: hydrolysis
Step 2: dehydration
Step 3: reduction
(b) C6H8O(l) + 15/2 O2(g) → 6CO2(g) + 4H2O(l)
Energy taken in for bond breaking
= 8 (C – H) + 3 (C – C) + 2 (C = C) + 2 (C – O) + 15/2 (O = O)
= 8(410) + 3(350) + 2(610) + 2(360) + 15/2 (496)
= 9990 kJ mol-1
Energy released for bond formation
= 12 (C = O) + 8 (O – H)
= 12(805) + 8(460)
= 13340 kJ mol-1
Enthalpy change of combustion = 9990 – 13340 = –3350 kJ mol-1
(c) Q = mcΔT = (200)(4.18)(32) = 26752 J
100% heat = 26752 / 80 x 100 = 33440 J
Amount of DMF = 1.00 / 96 = 0.01042 mol
Experimental enthalpy change of combustion
= 33.440 / 0.01042 = –3209 kJ mol-1 ≈ –3210 kJ mol-1
The enthalpy change of combustion obtained in (b) uses bond energies from the data booklet which are average values of bond energies obtained from a range of molecules containing that bond. In addition, some of the heat energy released from the burning of DMF is lost through the heating of the container itself or through the surroundings. Hence, the experimental value is less exothermic. Hence, there is a slight difference in both values.
(d) (i) R–CH2OH + H+ → R–CH2OH2+ (fast)
R–CH2OH2+ → R–CH2+ + H2O (slow)
R–CH2+ + Cl– → R–CH2Cl (fast)
The second step (R–CH2OH2+ → R–CH2+ + H2O) is the rate determining step.
(ii) Reagents: K2Cr2O7(aq), H2SO4(aq)
Condition: heat under reflux
[KMnO4(aq), H2SO4(aq) is not accepted due to oxidative cleavage]
(e) (i) Ester
(iii) Reagents: concentrated sulfuric acid
Condition: heat under reflux
(f) Name of mechanism: Free radical substitution
Cl–Cl → 2 Cl·
Cl· + R–CH3 → R–CH2· + HCl
R–CH2· + Cl2 → R–CH2Cl + Cl·
Cl· + Cl· → Cl2
R–CH2· + R–CH2· → R–CH2CH2–R
R–CH2· + Cl· → R–CH2Cl
(a) (i) Proton number is the number of protons in the nucleus of the atom.
Nucleon number is the number of protons and neutrons in the nucleus of the atom.
(ii) Let the relative abundance of 7Li be k
Relative abundance of 6Li = 1 – k
k x 7.016 + (1 – k)(6.015) = 6.942
7.016k + 6.015 – 6.015k = 6.942
1.001k + 6.015 = 6.942
1.001k = 0.927
k = 0.9261
Therefore, relative percentage abundance of 7Li = 92.61%
Relative percentage abundance of 6Li = 7.39%
(iii) X = 3He Y = 7Li
(b) (i) Ionic bonding. The lithium atom is able to transfer its valence electron to the carbon atom in graphite which has one unpaired electron, forming positive Li+ cation and anionic graphite. The bonding between the positive Li+ cation and anionic graphite is ionic. [answer is edited]
(ii) Before discharge: +4
After the cell is totally discharged: +3
(iii) The shape is tetrahedral with respect to boron. There are 4 bond pairs of electrons and 0 lone pair of electrons for boron. Hence, the shape is tetrahedral.
(iv) Cold, alkaline KMnO4(aq)
(c) (i) 2Li2O2 + 2CO2 → 2Li2CO3 + O2
(ii) Lithium has similar chemical properties as magnesium because the atomic radius and electronegativity are similar. Lithium is smallest in size for Group I metal and hence, lithium ion has the largest charge density for Group I metal ions. Therefore, it is able to polarise and distort the electron cloud of the carbonate anion to a larger extent and the distorted electron cloud of the carbonate anion is more readily decomposed by heat energy. Hence, the thermal stability of lithium carbonate is low and can be easily decomposed.
(d) (i) CH3CH2CH2CHO + CH3CH2Br
CH3CH2CHO + CH3CH2CH2Br
(ii) (CH3CH2)2CO + CH3Br
CH3CH2COCH3 + CH3CH2Br
(e) Only P and Q will turn orange acidified potassium dichromate(VI) to green. R, being a tertiary alcohol will not be oxidised and will not turn orange K2Cr2O7 to green.
After oxidation, 2,4-dinitrophenylhydrazine can be added. For Q, after oxidation, the product is a ketone and will form an orange precipitate with 2,4-dinitrophenylhydrazine. For P, after oxidation, the product is a carboxylic acid and will not form an orange precipitate.
Please do let me know of any mistakes or typing errors that I made while rushing this. Much appreciated and thanks!