## CHM 001: GENERAL CHEMISTRY
**Question 1**
(a)(i) State the Periodic Law. *(½ mark)*
(ii) State the group, period and block of the elements calcium (₂₀Ca) and bromine (₃₅Br). *[3 marks]*
(b) 4.0 g of NaCl is heated with manganese(IV) oxide in excess concentrated tetraoxosulphate(VI) acid. The chlorine gas liberated is then passed through KI solution to liberate iodine gas.
(i) Balance the equation for the reaction:
NaCl + H₂SO₄ + MnO₂ → MnSO₄ + NaHSO₄ + H₂O + Cl₂
(ii) Calculate the mass of MnO₂ used.
(iii) Provide an equation for the production of potassium chloride from the reaction of chlorine and potassium iodide.
(iv) Calculate the mass of iodine liberated.
[Na = 23, Cl = 35.5, H = 1, S = 32, Mn = 55, O = 16, I = 127] *[3½ marks]*
(c)(i) What is a precipitation reaction and why do precipitates form?
Write balanced molecular and net ionic equations for each of the following:
(ii) NH₄Cl(aq) + KOH(aq) →
(iii) Al(NO₃)₃(aq) + Ba(OH)₂(aq) → *[3 marks]*
---
**Question 2**
(a)(i) Define the term electronegativity.
(ii) Draw the Lewis structure for the following compounds: ICl₄⁻ and PCl₃.
(iii) Explain the difference between using Lewis structures and VSEPR to determine the shapes/structures of molecules. *[3 marks]*
(b)(i) Determine the empirical formula of a compound used as an explosive which contains C, H, N and O, if 50 g of the compound was found to contain 31.72 g of O, 7.93 g of C and 1.10 g of H. [Mᵣ of explosive = 227 g/mol]
(ii) Write an equation for the decomposition of the explosive obtained in (b)(i) if it decomposes into carbon dioxide, nitrogen, oxygen and water. *[4 marks]*
(c) Percentages of protein content in milk were recorded as: 54.01, 54.24, 54.05, 54.27, and 54.11. The accepted true value is 54.20. Calculate:
(i) Absolute error
(ii) Relative error
(iii) Mean deviation
(iv) Standard deviation *[3 marks]*
---
## CHM 002: PHYSICAL CHEMISTRY
**Question 3**
(a)(i) Define buffer solution.
(ii) Write the Ksp expressions for saturated solutions of AgCl and Fe(OH)₂. *[3 marks]*
(b)(i) Explain the term electrochemical cells.
(ii) Using equations only, explain the electrochemical process involved in the rusting of iron.
(iii) Derive a formula for the Nernst equation from the free energy relationship between a non-standard state and a standard state. *[3 marks]*
(c)(i) How does the neutron-to-proton ratio affect the stability of a radioactive element?
(ii) Positron emission from ¹¹C occurs with release of 3.45 × 10¹¹ J/mol energy. What is the mass change per mole for ¹¹C in the following nuclear reaction?
¹¹C → ¹¹B + ⁰e [c = 3 × 10⁸ ms⁻¹] *[2 marks]*
(d) What is the standard enthalpy of formation of ethene (C₂H₄), if the standard enthalpy of combustion is −1411 kJ mol⁻¹?
ΔHf°(CO₂) = −394 kJ mol⁻¹; ΔHf°(H₂O) = −286 kJ mol⁻¹ *[2 marks]*
---
**Question 4**
(a) State Faraday's second law of electrolysis. *[1 mark]*
(b) A direct current of 200 mA flows for 3 hours through three cells connected in series. They contain solutions of sodium chloride, copper(II) tetraoxosulphate(VI), and gold(III) trioxonitrate(V). Calculate the mass of metal deposited in each cell.
[Na = 23.0, Cu = 63.5, Au = 197] *[3 marks]*
(c)(i) Define solubility product.
(ii) State THREE factors that affect solubility product.
(iii) For the equilibrium: Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq), calculate the solubility product (Ksp) of Ag₂CrO₄ if the concentration of Ag⁺ is 1.5 × 10⁻⁴ M. *[3 marks]*
(d)(i) State TWO uses of radioactive isotopes.
(ii) The half-life of the isotope of uranium with mass number 234 is 3.5 × 10⁵ years. How long after the isolation of a sample of this isotope will only one-fifth of the original mass remain? *[3 marks]*
---
## CHM 003: INORGANIC CHEMISTRY
**Question 5**
(a) Outline FOUR differences between the most common allotropes of carbon. *[2 marks]*
(b) Give reasons for the following observations:
(i) Group IIA elements always show small positive electron affinity.
(ii) The beryllium ion is amphoteric but the magnesium ion is ionic.
(iii) Non-metallic properties of elements increase across a period.
(iv) Group IIA elements always form colourless compounds. *[2 marks]*
(c) Give the names of the following complexes:
(i) K₄[Fe(CN)₆]
(ii) [Co(SO₄)₂Cl₂]⁴⁻
(iii) [Ni(NH₂CH₂CH₂NH₂)₂Cl₂]²⁺ *[3 marks]*
(d) Show how the combustion of sulphur-containing fossil fuel causes acid rain in a polluted industrialised environment. (Name all products formed.) *[3 marks]*
---
**Question 6**
(a) The properties of lithium are quite different from those of other Group I elements but similar to those of magnesium. State SIX similarities between lithium and magnesium. *[3 marks]*
(b)(i) List the TWO major gases that cause acid rain and give their non-natural (artificial) sources.
(ii) What products are formed when each of the following reacts with red phosphorus?
I. Bromine
II. Chlorine *[4 marks]*
(c)(i) Give reasons why compounds of Sc³⁺ and Zn²⁺ are mostly white/colourless, whereas those of Cr³⁺ and Cu²⁺ are coloured.
(ii) List TWO distinct features of transition metals. *[3 marks]*
---
## CHM 004: ORGANIC CHEMISTRY
**Question 7**
(a) An alcohol has the molecular formula C₃H₈O. When warmed with an alkaline solution of iodine it forms a yellow precipitate.
(i) Give the reaction scheme and name the yellow precipitate.
(ii) Draw the displayed formula of the alcohol.
(iii) The first stage in the reaction of the alcohol with alkaline iodine solution is an oxidation reaction. Name the organic product of this first stage. *[3 marks]*
(b) A hydrocarbon X contains 92.3% carbon by mass. The relative molecular mass of X is 78.
(i) Calculate the empirical formula of X.
(ii) Calculate the molecular formula of X. *[4 marks]*
(c)(i) Draw the structures of the TWO geometrical isomers of 1,2-dibromoethene.
(ii) What feature of the double bond prevents one isomer from converting into the other?
(iii) Provide the name and structure of any monosubstituted benzene compound. *[3 marks]*
---
**Question 8**
(a) Hydrocarbons of high molecular mass are subjected to a process known as cracking.
(i) Define the term cracking.
(ii) What conditions are used for cracking?
(iii) Differentiate between cracking and reforming. *[3 marks]*
(b) The following scheme shows some reactions of toluene:
- Toluene + conc. H₂SO₄ → V
- Toluene + Reagent I → Methylchlorobenzene (W)
- Toluene + Reagent II → Benzyl chloride (X)
- Toluene + Reagent III → Benzoic acid (Y)
- Benzoic acid (Y) + Reagent IV → Benzoyl chloride (Z)
(i) Suggest reagents and conditions for reactions I to IV.
(ii) Draw the structure of V.
(iii) Give the IUPAC name of V, W, Y and Z. *[5 marks]*
(c) B is a saturated unbranched hydrocarbon. When treated with chlorine under free radical conditions it produces a mixture of chlorinated compounds including C, C₆H₁₂Cl₂. Dehydrochlorination of C produces D, C₆H₁₀Cl. When D is oxidised by hot concentrated KMnO₄, two compounds, CO₂ and a carboxylic acid, are formed in equimolar amounts.
(i) Suggest a reagent and condition for the conversion of C to D.
(ii) Suggest a structure for compound D.
(iii) Deduce the structure of compound B. *[2 marks]*
---
---
# ANSWERS
## CHM 001: GENERAL CHEMISTRY
### Question 1(a)(i) — Periodic Law
The physical and chemical properties of elements are periodic functions of their atomic numbers; when elements are arranged in order of increasing atomic number, elements with similar properties recur at regular intervals.
---
### Question 1(a)(ii) — Group, Period and Block
| Element | Atomic Number | Group | Period | Block |
|---|---|---|---|---|
| Calcium (Ca) | 20 | Group 2 | Period 4 | s-block |
| Bromine (Br) | 35 | Group 17 | Period 4 | p-block |
**Reasoning:** Ca electron configuration: [Ar] 4s² — last electron enters the 4s sub-shell (s-block), second column → Group 2, fourth shell → Period 4. Br configuration: [Ar] 3d¹⁰ 4p⁵ — last electron enters 4p (p-block), 7th group of p-block → Group 17, fourth shell → Period 4.
---
### Question 1(b)(i) — Balanced Equation
$$2\text{NaCl} + 2\text{H}_2\text{SO}_4 + \text{MnO}_2 \rightarrow \text{MnSO}_4 + \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} + \text{Cl}_2$$
*(In this reaction, MnO₂ oxidises Cl⁻ to Cl₂ while being reduced from Mn⁴⁺ to Mn²⁺; concentrated H₂SO₄ provides the acid medium.)*
---
### Question 1(b)(ii) — Mass of MnO₂ Used
Molar mass of NaCl = 23 + 35.5 = **58.5 g/mol**
$$n(\text{NaCl}) = \frac{4.0}{58.5} = 0.0684 \text{ mol}$$
From the balanced equation: **2 mol NaCl : 1 mol MnO₂**
$$n(\text{MnO}_2) = \frac{0.0684}{2} = 0.0342 \text{ mol}$$
Molar mass of MnO₂ = 55 + 2(16) = **87 g/mol**
$$\text{Mass of MnO}_2 = 0.0342 \times 87 = \boxed{2.98 \text{ g}}$$
---
### Question 1(b)(iii) — Reaction of Cl₂ with KI
$$\text{Cl}_2 + 2\text{KI} \rightarrow 2\text{KCl} + \text{I}_2$$
Chlorine (a stronger oxidising agent) displaces iodine from potassium iodide solution.
---
### Question 1(b)(iv) — Mass of Iodine Liberated
From the balanced equation: **1 mol Cl₂ : 1 mol I₂**
From the original equation: **2 mol NaCl : 1 mol Cl₂**, so:
$$n(\text{Cl}_2) = \frac{n(\text{NaCl})}{2} = \frac{0.0684}{2} = 0.0342 \text{ mol}$$
$$n(\text{I}_2) = 0.0342 \text{ mol}$$
Molar mass of I₂ = 2 × 127 = **254 g/mol**
$$\text{Mass of I}_2 = 0.0342 \times 254 = \boxed{8.69 \text{ g}}$$
---
### Question 1(c)(i) — Precipitation Reaction
A **precipitation reaction** is a reaction between two aqueous ionic solutions that produces an insoluble solid product called a **precipitate**, which separates from the solution.
Precipitates form because the product of the concentrations of the ions in solution (the ion product, Q) exceeds the solubility product constant (Ksp) of that ionic compound, making the salt thermodynamically unstable in solution and causing it to crystallise out as a solid.
---
### Question 1(c)(ii) — NH₄Cl(aq) + KOH(aq)
**Balanced molecular equation:**
$$\text{NH}_4\text{Cl(aq)} + \text{KOH(aq)} \rightarrow \text{NH}_3\text{(g)} + \text{H}_2\text{O(l)} + \text{KCl(aq)}$$
**Net ionic equation:**
$$\text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{NH}_3\text{(g)} + \text{H}_2\text{O(l)}$$
*(KCl, K⁺, and Cl⁻ are spectator ions; no precipitate forms here — this is an acid-base reaction producing gaseous ammonia.)*
---
### Question 1(c)(iii) — Al(NO₃)₃(aq) + Ba(OH)₂(aq)
**Balanced molecular equation:**
$$2\text{Al(NO}_3)_3\text{(aq)} + 3\text{Ba(OH)}_2\text{(aq)} \rightarrow 2\text{Al(OH)}_3\text{(s)} + 3\text{Ba(NO}_3)_2\text{(aq)}$$
**Net ionic equation:**
$$\text{Al}^{3+}\text{(aq)} + 3\text{OH}^-\text{(aq)} \rightarrow \text{Al(OH)}_3\text{(s)↓}$$
*(The precipitate is white aluminium hydroxide. Ba²⁺ and NO₃⁻ are spectator ions.)*
---
### Question 2(a)(i) — Electronegativity
Electronegativity is a measure of the tendency of an atom to attract the shared pair of electrons in a covalent bond towards itself. It increases across a period (increasing nuclear charge, decreasing atomic radius) and decreases down a group (increasing atomic radius and electron shielding). The Pauling scale is the most commonly used measure.
---
### Question 2(a)(ii) — Lewis Structures
**ICl₄⁻ (Iodine tetrachloride anion):**
- Total valence electrons: I(7) + 4×Cl(7) + 1(charge) = 36 electrons
- Central atom I forms 4 single bonds to Cl atoms → 8 electrons used in bonding
- Remaining: 28 electrons → 3 lone pairs on each Cl (12 pairs) + **2 lone pairs on I**
- Geometry: 6 electron domains (4 bonding + 2 lone pairs) → **octahedral electron geometry; square planar molecular shape**
```
:Cl:
|
:Cl: — I — :Cl: (with 2 lone pairs above and below I)
|
:Cl:
```
**PCl₃ (Phosphorus trichloride):**
- Total valence electrons: P(5) + 3×Cl(7) = 26 electrons
- Central atom P forms 3 single bonds to Cl atoms → 6 electrons used in bonding
- Remaining: 20 electrons → 3 lone pairs on each Cl (9 pairs) + **1 lone pair on P**
- Geometry: 4 electron domains (3 bonding + 1 lone pair) → **tetrahedral electron geometry; trigonal pyramidal molecular shape**
```
:Cl — P — Cl:
|
:Cl:
(lone pair on P pointing upward)
```
---
### Question 2(a)(iii) — Lewis Structures vs VSEPR
| Aspect | Lewis Structures | VSEPR Theory |
|---|---|---|
| Primary purpose | Shows the arrangement of all bonding and non-bonding (lone pair) electrons around atoms in a molecule | Predicts the three-dimensional shape/geometry of a molecule |
| Information provided | Identifies which atoms are bonded, bond order (single/double/triple), and location of lone pairs | Uses the number and type of electron domains (bonding + lone pairs) around the central atom to predict bond angles and molecular shape |
| Shape prediction | Does not directly predict 3D geometry — a Lewis structure is a 2D representation | Directly predicts 3D molecular geometry based on electron pair repulsion (lone pairs repel more strongly than bonding pairs) |
| Relationship | Lewis structure is the starting point — the number and type of electron pairs it reveals are then used as the input for VSEPR analysis | VSEPR is applied after drawing the Lewis structure to determine the actual shape |
---
### Question 2(b)(i) — Empirical Formula of Explosive
**Given (in 50 g sample):** O = 31.72 g; C = 7.93 g; H = 1.10 g
**Mass of N** = 50 − (31.72 + 7.93 + 1.10) = **9.25 g**
**Molar ratios:**
| Element | Mass (g) | Molar Mass | Moles | Ratio (÷ 0.661) |
|---|---|---|---|---|
| C | 7.93 | 12 | 0.661 | 1.00 |
| H | 1.10 | 1 | 1.100 | 1.66 ≈ 5/3 |
| N | 9.25 | 14 | 0.661 | 1.00 |
| O | 31.72 | 16 | 1.983 | 3.00 |
Multiply all ratios by 3 to clear the fraction: C = 3, H = 5, N = 3, O = 9
$$\boxed{\text{Empirical formula: } \text{C}_3\text{H}_5\text{N}_3\text{O}_9}$$
**Verification:** Molar mass of C₃H₅N₃O₉ = 3(12) + 5(1) + 3(14) + 9(16) = 36 + 5 + 42 + 144 = **227 g/mol** ✓ (matches given Mᵣ, so empirical formula = molecular formula — this is **nitroglycerin**.)
---
### Question 2(b)(ii) — Decomposition Equation
$$4\text{C}_3\text{H}_5\text{N}_3\text{O}_9 \rightarrow 12\text{CO}_2 + 10\text{H}_2\text{O} + 6\text{N}_2 + \text{O}_2$$
*(Coefficients multiplied by 4 to give whole numbers; the reaction is highly exothermic and explosive.)*
---
### Question 2(c) — Statistical Analysis
**Data:** 54.01, 54.24, 54.05, 54.27, 54.11 | **True value** = 54.20
**Step 1 — Calculate the mean:**
$$\bar{x} = \frac{54.01 + 54.24 + 54.05 + 54.27 + 54.11}{5} = \frac{270.68}{5} = 54.136$$
**(i) Absolute Error:**
$$\text{Absolute error} = |\bar{x} - \text{True value}| = |54.136 - 54.20| = \boxed{0.064}$$
*(Note: the absolute error uses the mean, not a single measurement.)*
**(ii) Relative Error:**
$$\text{Relative error} = \frac{\text{Absolute error}}{\text{True value}} \times 100 = \frac{0.064}{54.20} \times 100 = \boxed{0.118\%}$$
**(iii) Mean Deviation:**
Individual deviations from the mean (|xᵢ − x̄|):
| Value | Deviation from mean |
|---|---|
| 54.01 | |54.01 − 54.136| = 0.126 |
| 54.24 | |54.24 − 54.136| = 0.104 |
| 54.05 | |54.05 − 54.136| = 0.086 |
| 54.27 | |54.27 − 54.136| = 0.134 |
| 54.11 | |54.11 − 54.136| = 0.026 |
$$\text{Mean deviation} = \frac{0.126 + 0.104 + 0.086 + 0.134 + 0.026}{5} = \frac{0.476}{5} = \boxed{0.0952}$$
**(iv) Standard Deviation:**
| Value | (xᵢ − x̄)² |
|---|---|
| 54.01 | (−0.126)² = 0.015876 |
| 54.24 | (+0.104)² = 0.010816 |
| 54.05 | (−0.086)² = 0.007396 |
| 54.27 | (+0.134)² = 0.017956 |
| 54.11 | (−0.026)² = 0.000676 |
$$\sum(x_i - \bar{x})^2 = 0.05272$$
Using **sample standard deviation** (n − 1 = 4):
$$s = \sqrt{\frac{0.05272}{4}} = \sqrt{0.01318} = \boxed{0.1148}$$
---
## CHM 002: PHYSICAL CHEMISTRY
### Question 3(a)(i) — Buffer Solution
A buffer solution is a solution that resists significant changes in pH upon the addition of small amounts of strong acid or strong base, or upon dilution. It typically consists of a weak acid and its conjugate base (acidic buffer, e.g., CH₃COOH/CH₃COO⁻Na⁺) or a weak base and its conjugate acid (basic buffer, e.g., NH₃/NH₄Cl).
---
### Question 3(a)(ii) — Ksp Expressions
For a sparingly soluble salt dissolving at equilibrium, Ksp = product of ion concentrations raised to the power of their stoichiometric coefficients (pure solid omitted):
$$\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \qquad K_{sp} = [\text{Ag}^+][\text{Cl}^-]$$
$$\text{Fe(OH)}_2\text{(s)} \rightleftharpoons \text{Fe}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \qquad K_{sp} = [\text{Fe}^{2+}][\text{OH}^-]^2$$
---
### Question 3(b)(i) — Electrochemical Cells
Electrochemical cells are devices in which chemical reactions are used to produce electrical energy (galvanic/voltaic cells) or, conversely, electrical energy is used to drive non-spontaneous chemical reactions (electrolytic cells). They consist of two electrodes (anode and cathode) connected by a conductor, with an electrolyte solution providing ionic conduction between them.
---
### Question 3(b)(ii) — Electrochemical Process in Rusting of Iron
**At the anode (oxidation — iron surface):**
$$\text{Fe(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + 2e^-$$
**At the cathode (reduction — in the presence of water and dissolved oxygen):**
$$\text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4e^- \rightarrow 4\text{OH}^-\text{(aq)}$$
**Overall cell reaction:**
$$2\text{Fe(s)} + \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + 4\text{OH}^-\text{(aq)}$$
**Further oxidation to rust:**
$$4\text{Fe}^{2+} + \text{O}_2 + 8\text{OH}^- \rightarrow 2\text{Fe}_2\text{O}_3 + 4\text{H}_2\text{O}$$
$$\text{Fe}_2\text{O}_3 + x\text{H}_2\text{O} \rightarrow \text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O} \quad \text{(hydrated iron(III) oxide = rust)}$$
---
### Question 3(b)(iii) — Derivation of the Nernst Equation
Starting from the relationship between Gibbs free energy and the reaction quotient Q:
$$\Delta G = \Delta G^\circ + RT\ln Q \quad \text{...(1)}$$
The relationship between free energy and cell potential is:
$$\Delta G = -nFE \quad \text{and} \quad \Delta G^\circ = -nFE^\circ \quad \text{...(2)}$$
Substituting (2) into (1):
$$-nFE = -nFE^\circ + RT\ln Q$$
Dividing throughout by −nF:
$$\boxed{E = E^\circ - \frac{RT}{nF}\ln Q}$$
This is the **Nernst equation**. At 25°C (298 K), substituting R = 8.314 J mol⁻¹K⁻¹ and converting ln to log₁₀:
$$E = E^\circ - \frac{0.0592}{n}\log Q$$
---
### Question 3(c)(i) — Neutron-to-Proton Ratio and Nuclear Stability
The **neutron-to-proton (n/p) ratio** is a key determinant of nuclear stability:
- For **light nuclei** (Z < 20), the stable n/p ratio is approximately **1:1**. Neutrons help hold the nucleus together by contributing to the strong nuclear force without adding to electrostatic repulsion between protons.
- As Z increases, more neutrons are needed to stabilise the nucleus against the growing repulsion between protons; the stable n/p ratio rises toward **1.5:1** for heavy elements.
- If n/p is **too high** (neutron-rich), the nucleus undergoes **β⁻ decay** (a neutron converts to a proton, emitting an electron and antineutrino), reducing the ratio.
- If n/p is **too low** (proton-rich), the nucleus undergoes **β⁺ (positron) emission** or **electron capture**, increasing the ratio.
- Very heavy nuclei (Z > 82) are unstable regardless of n/p ratio and undergo **α decay** to reduce both proton and neutron numbers.
---
### Question 3(c)(ii) — Mass Change for ¹¹C Positron Emission
Using Einstein's mass-energy equivalence: **E = mc²**
$$m = \frac{E}{c^2} = \frac{3.45 \times 10^{11} \text{ J mol}^{-1}}{(3 \times 10^8 \text{ m s}^{-1})^2} = \frac{3.45 \times 10^{11}}{9 \times 10^{16}}$$
$$m = 3.833 \times 10^{-6} \text{ kg mol}^{-1} = 3.833 \times 10^{-3} \text{ g mol}^{-1}$$
$$\boxed{m \approx 3.83 \times 10^{-3} \text{ g mol}^{-1} \approx 3.83 \text{ mg mol}^{-1}}$$
This is the mass converted to energy (mass defect) per mole of ¹¹C undergoing positron emission.
---
### Question 3(d) — Standard Enthalpy of Formation of Ethene
**Combustion reaction of ethene:**
$$\text{C}_2\text{H}_4\text{(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \qquad \Delta H_{comb}^\circ = -1411 \text{ kJ mol}^{-1}$$
**Applying Hess's Law:**
$$\Delta H_{comb}^\circ = \left[2\Delta H_f^\circ(\text{CO}_2) + 2\Delta H_f^\circ(\text{H}_2\text{O})\right] - \left[\Delta H_f^\circ(\text{C}_2\text{H}_4) + 3\Delta H_f^\circ(\text{O}_2)\right]$$
Note: ΔHf°(O₂) = 0 (element in standard state)
$$-1411 = [2(-394) + 2(-286)] - [\Delta H_f^\circ(\text{C}_2\text{H}_4) + 0]$$
$$-1411 = [-788 - 572] - \Delta H_f^\circ(\text{C}_2\text{H}_4)$$
$$-1411 = -1360 - \Delta H_f^\circ(\text{C}_2\text{H}_4)$$
$$\Delta H_f^\circ(\text{C}_2\text{H}_4) = -1360 + 1411 = \boxed{+51 \text{ kJ mol}^{-1}}$$
The positive value indicates that the formation of ethene from its elements is endothermic.
---
### Question 4(a) — Faraday's Second Law of Electrolysis
When the same quantity of electricity passes through different electrolytic cells connected in series, the masses of different substances deposited or liberated at the electrodes are directly proportional to their **chemical equivalent weights** (molar mass divided by the number of electrons transferred, i.e., M/n).
---
### Question 4(b) — Mass of Metal Deposited
**Total charge passed:**
$$Q = I \times t = 0.200 \text{ A} \times (3 \times 3600 \text{ s}) = 0.200 \times 10800 = 2160 \text{ C}$$
**Moles of electrons transferred:**
$$n(e^-) = \frac{Q}{F} = \frac{2160}{96500} = 0.02238 \text{ mol}$$
**Cell 1 — Sodium chloride (Na⁺ + e⁻ → Na; n = 1):**
$$n(\text{Na}) = 0.02238 \text{ mol}$$
$$m(\text{Na}) = 0.02238 \times 23.0 = \boxed{0.515 \text{ g}}$$
**Cell 2 — Copper(II) sulphate (Cu²⁺ + 2e⁻ → Cu; n = 2):**
$$n(\text{Cu}) = \frac{0.02238}{2} = 0.01119 \text{ mol}$$
$$m(\text{Cu}) = 0.01119 \times 63.5 = \boxed{0.711 \text{ g}}$$
**Cell 3 — Gold(III) nitrate (Au³⁺ + 3e⁻ → Au; n = 3):**
$$n(\text{Au}) = \frac{0.02238}{3} = 0.00746 \text{ mol}$$
$$m(\text{Au}) = 0.00746 \times 197 = \boxed{1.470 \text{ g}}$$
---
### Question 4(c)(i) — Solubility Product
The solubility product (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound in water at a given temperature. It equals the product of the concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the dissolution equation. It applies only to saturated solutions of sparingly soluble salts.
---
### Question 4(c)(ii) — Three Factors Affecting Solubility Product
1. **Temperature:** Ksp is temperature-dependent; for most salts, solubility (and hence Ksp) increases with temperature; for a few (e.g., CaSO₄), it decreases.
2. **Common ion effect:** The presence of a common ion (one already in the equilibrium) suppresses dissolution, reducing the apparent solubility of the salt (though Ksp itself remains constant at constant temperature).
3. **Presence of complexing agents:** Ligands that form soluble complexes with the cation (e.g., NH₃ with Cu²⁺ or Ag⁺) effectively remove the cation from the equilibrium, driving dissolution forward and increasing apparent solubility beyond what Ksp alone would predict.
---
### Question 4(c)(iii) — Ksp of Ag₂CrO₄
$$\text{Ag}_2\text{CrO}_4\text{(s)} \rightleftharpoons 2\text{Ag}^+\text{(aq)} + \text{CrO}_4^{2-}\text{(aq)}$$
$$K_{sp} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}]$$
Given [Ag⁺] = 1.5 × 10⁻⁴ M. From stoichiometry, each formula unit produces 2 Ag⁺ and 1 CrO₄²⁻, so:
$$[\text{CrO}_4^{2-}] = \frac{[\text{Ag}^+]}{2} = \frac{1.5 \times 10^{-4}}{2} = 7.5 \times 10^{-5} \text{ M}$$
$$K_{sp} = (1.5 \times 10^{-4})^2 \times (7.5 \times 10^{-5})$$
$$K_{sp} = 2.25 \times 10^{-8} \times 7.5 \times 10^{-5} = \boxed{1.69 \times 10^{-12}}$$
---
### Question 4(d)(i) — Two Uses of Radioactive Isotopes
1. **Medical diagnosis and treatment:** Radioactive isotopes such as ⁹⁹ᵐTc are used as tracers in nuclear medicine imaging (e.g., PET scans, gamma scintigraphy) to diagnose cancer, heart disease, and organ function; ¹³¹I is used to treat thyroid cancer and hyperthyroidism.
2. **Carbon dating and archaeological dating:** ¹⁴C (radiocarbon) is used to determine the age of organic materials up to approximately 50,000 years old; other isotopes (e.g., ²³⁸U/²⁰⁶Pb) are used for geological and cosmological dating.
*(Other acceptable answers: sterilisation of medical equipment by gamma irradiation; food irradiation to kill pathogens; industrial radiography to detect structural flaws in metals; nuclear power generation.)*
---
### Question 4(d)(ii) — Radioactive Decay of ²³⁴U
**Given:** t½ = 3.5 × 10⁵ years; N/N₀ = 1/5
Using the radioactive decay equation:
$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/t_{1/2}}$$
$$\frac{1}{5} = \left(\frac{1}{2}\right)^{t / 3.5 \times 10^5}$$
Taking logarithms of both sides:
$$\ln\left(\frac{1}{5}\right) = \frac{t}{3.5 \times 10^5} \times \ln\left(\frac{1}{2}\right)$$
$$\frac{t}{3.5 \times 10^5} = \frac{\ln(1/5)}{\ln(1/2)} = \frac{-\ln 5}{-\ln 2} = \frac{\log 5}{\log 2} = \frac{0.6990}{0.3010} = 2.322$$
$$t = 2.322 \times 3.5 \times 10^5 = \boxed{8.13 \times 10^5 \text{ years}}$$
---
## CHM 003: INORGANIC CHEMISTRY
### Question 5(a) — Four Differences Between Carbon Allotropes
| Property | Diamond | Graphite | Graphene | Buckminsterfullerene (C₆₀) |
|---|---|---|---|---|
| Structure | 3D tetrahedral network; each C bonded to 4 others by sp³ bonds | Layered 2D hexagonal sheets; each C bonded to 3 others by sp² bonds, layers held by weak van der Waals forces | Single atomic layer of graphite; 2D hexagonal lattice | Spherical cage of 60 C atoms arranged in pentagons and hexagons |
| Hardness | Hardest known natural substance | Very soft and slippery (layers slide easily) | Extremely strong in plane (strongest known material by tensile strength) | Relatively soft molecular solid |
| Electrical conductivity | Electrical insulator (all valence electrons in σ bonds) | Good electrical conductor (delocalised π electrons between layers) | Excellent conductor (outstanding electron mobility) | Semiconductor; poor conductor in bulk |
| Appearance / transparency | Transparent, colourless crystalline solid | Opaque, grey-black solid | Transparent single layer; appears nearly invisible | Dark brown to black solid powder |
---
### Question 5(b) — Reasons for Observations
**(i) Group IIA elements always show small positive electron affinity:**
Group IIA elements (alkaline earth metals) have completely filled s-subshells in their valence shell (ns²). The added electron would enter the next available higher-energy p-suborbital, which is less stable and farther from the nucleus. Additionally, the fully filled s-subshell creates extra electron–electron repulsion. Therefore, the energy released on gaining an electron is very small (or the process may even be endothermic), resulting in a small positive electron affinity.
**(ii) Be²⁺ is amphoteric but Mg²⁺ forms ionic compounds:**
Be²⁺ has an exceptionally high charge density (charge-to-radius ratio) due to beryllium's very small ionic radius (31 pm). This high polarising power causes Be²⁺ to strongly distort the electron clouds of surrounding anions and to interact covalently with both OH⁻ and H⁺, enabling it to act both as an acid and as a base (amphoteric character). Mg²⁺ is larger (72 pm) with a lower charge density; it has insufficient polarising power to form predominantly covalent bonds and therefore exists as a straightforward ionic species in its compounds.
**(iii) Non-metallic properties increase across a period:**
Across a period, the nuclear charge (number of protons) increases while the number of electron shells remains constant. This results in increasing **electronegativity** (greater tendency to attract electrons) and **ionisation energy**, and decreasing **atomic radius**. Elements with high electronegativity and ionisation energy are more likely to gain electrons than to lose them — the defining characteristic of non-metals. Therefore non-metallic character increases from left to right across a period.
**(iv) Group IIA elements always form colourless compounds:**
Colour in ionic compounds arises from the absorption of specific wavelengths of visible light, which occurs when there are electrons capable of undergoing d–d transitions or charge-transfer transitions between closely spaced energy levels. Group IIA metal ions (Mg²⁺, Ca²⁺, Ba²⁺, etc.) have completely empty d-orbitals (electron configuration of the preceding noble gas); they therefore have **no unpaired electrons and no available d–d transitions**. Their compounds absorb no visible light and appear colourless or white.
---
### Question 5(c) — Names of Coordination Complexes
**(i) K₄[Fe(CN)₆]:**
- K⁺ is the counter cation; [Fe(CN)₆]⁴⁻ is the complex anion
- Fe oxidation state: charge = −(4) from K₄, so complex is 4−; 6 CN⁻ contribute −6; Fe must be +2
- CN⁻ is **cyano**; 6 ligands → **hexacyano**; iron(II) in anion → **ferrate(II)**
**Name: Potassium hexacyanoferrate(II)**
**(ii) [Co(SO₄)₂Cl₂]⁴⁻:**
- Complex anion; overall charge = −4
- SO₄²⁻ (sulfato): 2 × (−2) = −4; Cl⁻ (chlorido): 2 × (−1) = −2; total ligand charge = −6
- Co oxidation state: x + (−6) = −4 → x = +2
**Name: Dichloridobis(sulfato)cobaltate(II)**
**(iii) [Ni(NH₂CH₂CH₂NH₂)₂Cl₂]²⁺:**
- NH₂CH₂CH₂NH₂ is **ethylenediamine (en)** — a bidentate ligand; 2 ligands → **bis(ethylenediamine)**
- Cl⁻: 2 × (−1) = −2; en is neutral; overall charge = +2
- Ni oxidation state: x + 0 + (−2) = +2 → x = +4
**Name: Dichloridobis(ethylenediamine)nickel(IV)²⁺**
*(Note: If the charge is considered as written, Ni = +4; some sources interpret this complex differently depending on context.)*
---
### Question 5(d) — How Sulphur-Containing Fossil Fuels Cause Acid Rain
**Step 1 — Combustion of sulphur in fossil fuel:**
$$\text{S(s)} + \text{O}_2\text{(g)} \rightarrow \text{SO}_2\text{(g)}$$
*(Sulphur dioxide is the primary product of sulphur combustion.)*
**Step 2 — Further oxidation of SO₂ in the atmosphere:**
$$2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{SO}_3\text{(g)}$$
*(This reaction is slow in the atmosphere but catalysed by particulate matter, ozone, or nitrogen oxides.)*
**Step 3 — Reaction with atmospheric water to form acid rain:**
$$\text{SO}_2\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_3\text{(aq)} \quad \text{(sulphurous acid)}$$
$$\text{SO}_3\text{(g)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_4\text{(aq)} \quad \text{(sulphuric acid — primary acid in acid rain)}$$
**Products formed:** SO₂(g), SO₃(g), H₂SO₃(aq) — sulphurous acid, and H₂SO₄(aq) — sulphuric acid (the most significant acidic component of acid rain, with pH typically between 4 and 5). This acid rain damages ecosystems, corrodes metals and stonework, and acidifies soils and freshwater bodies.
---
### Question 6(a) — Six Similarities Between Lithium and Magnesium (Diagonal Relationship)
1. Both Li and Mg react directly with nitrogen gas to form **nitrides** (Li₃N and Mg₃N₂ respectively), unlike other Group I elements which do not readily form nitrides.
2. Both form **oxide** as the main product of combustion in air (Li₂O and MgO), rather than the peroxides or superoxides formed by heavier Group I metals.
3. Both have relatively **high melting and boiling points** compared to their respective group neighbours, due to stronger metallic bonding arising from their small ionic/atomic radii and high charge density.
4. Both form **covalent or partially covalent compounds** with a degree of polarisation — LiCl and MgCl₂ show significant covalent character compared to NaCl, due to their high charge-to-size ratios.
5. Their **hydroxides** (LiOH and Mg(OH)₂) are sparingly soluble in water and decompose on heating to form the oxide, unlike the highly soluble and thermally stable hydroxides of heavier alkali metals.
6. Both form **organometallic compounds** (e.g., butyl lithium RLi and Grignard reagents RMgX) of similar reactivity and synthetic utility, reflecting comparable polarising abilities and charge densities.
---
### Question 6(b)(i) — Major Gases Causing Acid Rain and Their Artificial Sources
**Two major gases:**
| Gas | Artificial (Anthropogenic) Sources |
|---|---|
| Sulphur dioxide (SO₂) | Combustion of sulphur-containing fossil fuels (coal, petroleum) in power stations and industrial plants; smelting of sulphide ores (e.g., PbS, ZnS, FeS₂) in the metallurgical industry |
| Nitrogen dioxide (NO₂) / nitrogen oxides (NOₓ) | High-temperature combustion in internal combustion engines (vehicles) and jet engines, where atmospheric N₂ and O₂ react; emissions from thermal power stations, industrial furnaces, and agricultural use of nitrogen fertilisers |
---
### Question 6(b)(ii) — Products of Reactions with Red Phosphorus
When halogens react with red phosphorus (excess phosphorus conditions), phosphorus trihalides are the primary products:
**I. Bromine + red phosphorus:**
$$2\text{P(s)} + 3\text{Br}_2\text{(l)} \rightarrow 2\text{PBr}_3\text{(l)}$$
**Product: Phosphorus tribromide (PBr₃)**
**II. Chlorine + red phosphorus:**
$$2\text{P(s)} + 3\text{Cl}_2\text{(g)} \rightarrow 2\text{PCl}_3\text{(l)}$$
**Product: Phosphorus trichloride (PCl₃)**
*(With excess chlorine, PCl₅ is also formed: PCl₃ + Cl₂ → PCl₅. These reactions are used in organic chemistry to convert alcohols to alkyl halides via the intermediate phosphorus trihalide.)*
---
### Question 6(c)(i) — Colour in Transition Metal Compounds
**Sc³⁺ and Zn²⁺ are colourless/white:**
- Sc³⁺ has the electron configuration [Ar] — it has lost all three valence electrons, leaving a completely **empty d-subshell** (d⁰ configuration).
- Zn²⁺ has the electron configuration [Ar] 3d¹⁰ — it has a completely **filled d-subshell** (d¹⁰ configuration).
- In both cases, there are **no partially filled d-orbitals** and therefore no possibility of d–d electronic transitions within the visible light range. No visible light is absorbed and the compounds appear colourless/white.
**Cr³⁺ and Cu²⁺ are coloured:**
- Cr³⁺ has the configuration [Ar] 3d³ — three unpaired electrons in a **partially filled d-subshell**.
- Cu²⁺ has the configuration [Ar] 3d⁹ — a **partially filled d-subshell** with one unpaired electron.
- In the presence of ligands (in complex ions), the d-orbitals split into two sets of different energies (crystal field splitting). Electrons can be promoted from the lower to the higher energy d-orbitals by absorbing photons of specific visible wavelengths (**d–d transitions**). The complementary colour of the absorbed light is transmitted, giving the compound its characteristic colour.
---
### Question 6(c)(ii) — Two Distinct Features of Transition Metals
1. **Variable oxidation states:** Transition metals can form stable ions and compounds in multiple oxidation states (e.g., iron exists as Fe²⁺ and Fe³⁺; manganese ranges from +2 to +7) because the 3d and 4s electrons have similar energies and can be removed sequentially. This gives rise to diverse redox chemistry.
2. **Formation of coloured compounds and complex ions:** Transition metals have partially filled d-orbitals that can bond with Lewis base ligands (neutral molecules or anions) to form coordination complexes. d–d electron transitions within these complexes absorb visible light, producing characteristic colours (e.g., [Cu(H₂O)₆]²⁺ is blue; [Fe(SCN)]²⁺ is blood-red).
---
## CHM 004: ORGANIC CHEMISTRY
### Question 7(a)(i) — Iodoform Reaction: Reaction Scheme and Yellow Precipitate
The alcohol C₃H₈O that gives a positive iodoform test must contain the **CH₃CH(OH)−** group, identifying it as **propan-2-ol** (isopropanol).
**Stage 1 — Oxidation by alkaline iodine:**
$$\text{CH}_3\text{CH(OH)CH}_3 \xrightarrow{\text{I}_2/\text{OH}^-} \text{CH}_3\text{COCH}_3 \text{ (propanone)} + \text{H}_2\text{O}$$
**Stage 2 — Iodination of the methyl ketone:**
$$\text{CH}_3\text{COCH}_3 + 3\text{I}_2 + 3\text{OH}^- \rightarrow \text{CI}_3\text{COCH}_3 + 3\text{I}^- + 3\text{H}_2\text{O}$$
**Stage 3 — Cleavage of the triiodomethyl ketone:**
$$\text{CI}_3\text{COCH}_3 + \text{OH}^- \rightarrow \text{CHI}_3 + \text{CH}_3\text{COO}^-$$
**Overall equation:**
$$\text{CH}_3\text{CH(OH)CH}_3 + 4\text{I}_2 + 6\text{OH}^- \rightarrow \text{CHI}_3 + \text{CH}_3\text{COO}^- + 5\text{I}^- + 5\text{H}_2\text{O}$$
**Yellow precipitate: Iodoform (CHI₃, triiodomethane)** — a pale yellow crystalline solid with a characteristic antiseptic odour.
---
### Question 7(a)(ii) — Displayed Formula of Propan-2-ol
```
H H H
| | |
H — C — C — C — H
| | |
H O–H H
```
Or more clearly:
$$\text{CH}_3 - \overset{|}{\overset{\displaystyle\text{OH}}{\text{CH}}} - \text{CH}_3$$
The hydroxyl group (−OH) is on the **central** (second) carbon atom.
---
### Question 7(a)(iii) — Organic Product of First Stage Oxidation
The first stage is the oxidation of propan-2-ol (a secondary alcohol) to a ketone:
**Product: Propanone (acetone, CH₃COCH₃)**
Secondary alcohols are oxidised to **ketones** (not carboxylic acids) because the central carbon has no hydrogen directly bonded to the carbonyl carbon after oxidation.
---
### Question 7(b)(i) — Empirical Formula of Hydrocarbon X
Assume 100 g of X: C = 92.3 g; H = 100 − 92.3 = 7.7 g
$$n(\text{C}) = \frac{92.3}{12} = 7.692 \text{ mol}; \quad n(\text{H}) = \frac{7.7}{1} = 7.7 \text{ mol}$$
$$\text{Ratio C:H} = \frac{7.692}{7.692} : \frac{7.7}{7.692} = 1 : 1.001 \approx 1:1$$
$$\boxed{\text{Empirical formula: CH}}$$
---
### Question 7(b)(ii) — Molecular Formula of X
Empirical formula mass of CH = 12 + 1 = **13 g/mol**
$$n = \frac{M_r}{\text{Empirical formula mass}} = \frac{78}{13} = 6$$
$$\boxed{\text{Molecular formula: C}_6\text{H}_6}$$
This is **benzene** — consistent with the molecular mass of 78 g/mol and the high carbon content characteristic of aromatic compounds.
---
### Question 7(c)(i) — Geometrical Isomers of 1,2-Dibromoethene
*(Note: The question states 1,2-dibromoethene — i.e., BrCH=CHBr, which does show geometrical isomerism due to the C=C double bond. The original document incorrectly stated "1,2-dibromoethane" which has only single bonds and shows no geometrical isomerism.)*
**cis-1,2-dibromoethene (Z-isomer):** Both Br atoms on the same side of the double bond.
```
Br Br
\ /
C == C
/ \
H H
```
**trans-1,2-dibromoethene (E-isomer):** Br atoms on opposite sides of the double bond.
```
Br H
\ /
C == C
/ \
H Br
```
---
### Question 7(c)(ii) — Why the Double Bond Prevents Interconversion
The C=C double bond consists of a **σ (sigma) bond** and a **π (pi) bond**. The π bond is formed by the lateral (side-on) overlap of p-orbitals above and below the plane of the molecule. For interconversion between cis and trans isomers, the molecule would need to **rotate about the C=C bond**. Such rotation would require breaking the π bond (which has a bond energy of approximately 264 kJ mol⁻¹), destroying the lateral p-orbital overlap. At ordinary temperatures, there is insufficient thermal energy to break the π bond, so rotation is effectively prevented and the two geometrical isomers are stable and distinct compounds.
---
### Question 7(c)(iii) — Monosubstituted Benzene Compound
**Name: Chlorobenzene (C₆H₅Cl) — IUPAC name: chlorobenzene**
**Structure:**
```
Cl
|
___C___
/ \
CH CH
|| ||
CH CH
\___C___/
|
H
```
Or in simplified notation: a benzene ring with one hydrogen replaced by a Cl substituent at any position on the ring (all positions are equivalent for monosubstitution).
*(Other acceptable answers: nitrobenzene, C₆H₅NO₂; methylbenzene/toluene, C₆H₅CH₃; phenol, C₆H₅OH.)*
---
### Question 8(a)(i) — Definition of Cracking
Cracking is the industrial process by which large, high-molecular-mass hydrocarbon molecules (typically long-chain alkanes from the higher fractions of petroleum distillation) are broken down into smaller, more useful hydrocarbon molecules — including shorter-chain alkanes, **alkenes** (which are particularly valuable as feedstocks for the petrochemical industry and for polymer production), and hydrogen — by the application of heat, pressure, and/or catalysts.
---
### Question 8(a)(ii) — Conditions for Cracking
| Type | Temperature | Catalyst | Pressure |
|---|---|---|---|
| **Catalytic cracking** | 450–550°C | Zeolite (aluminosilicate) or aluminium oxide/silica | Moderate (slightly above atmospheric) |
| **Thermal (steam) cracking** | 700–900°C | No catalyst (steam diluent used) | Low to moderate |
The primary commercial process is **fluid catalytic cracking (FCC)**, using zeolite catalysts, which produces high yields of branched alkanes (for high-octane petrol) and alkenes (for chemical feedstocks).
---
### Question 8(a)(iii) — Cracking vs Reforming
| Feature | Cracking | Reforming |
|---|---|---|
| Purpose | To **break** large hydrocarbon molecules into smaller, more volatile and more useful ones | To **restructure** the carbon skeleton of hydrocarbons without significantly reducing carbon number |
| Feed | Long-chain, high-molecular-mass alkanes (heavy fractions) | Straight-chain alkanes (naphtha fraction, C₆–C₁₀) |
| Products | Shorter-chain alkanes, alkenes (ethene, propene), hydrogen | Branched-chain alkanes, cycloalkanes, aromatic hydrocarbons (benzene, toluene), hydrogen |
| Chemical change | C–C bond cleavage; molecular mass decreases | Isomerisation, cyclisation, aromatisation; molecular mass approximately unchanged |
| Purpose in industry | Increase yield of petrol and produce chemical feedstocks | Increase the octane rating of petrol; produce aromatic feedstocks |
---
### Question 8(b)(i) — Reagents and Conditions for Reactions I to IV
| Reaction | Reagent and Conditions |
|---|---|
| **I** (Toluene → Methylchlorobenzene W, ring chlorination) | Cl₂ (g) with anhydrous FeCl₃ or AlCl₃ as Lewis acid catalyst; room temperature; electrophilic aromatic substitution |
| **II** (Toluene → Benzyl chloride X, side-chain chlorination) | Cl₂ (g) under UV light (ultraviolet radiation) or heat; free radical substitution |
| **III** (Toluene → Benzoic acid Y, oxidation of methyl group) | Acidified or alkaline KMnO₄ (potassium permanganate); heat under reflux |
| **IV** (Benzoic acid Y → Benzoyl chloride Z) | SOCl₂ (thionyl chloride); warm/reflux; or PCl₅ (phosphorus pentachloride) |
---
### Question 8(b)(ii) — Structure of V (Sulphonation Product)
V is the product of toluene reacting with concentrated H₂SO₄ — electrophilic aromatic **sulphonation**:
$$\text{C}_6\text{H}_5\text{CH}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CH}_3\text{C}_6\text{H}_4\text{SO}_3\text{H} + \text{H}_2\text{O}$$
The SO₃H group is introduced onto the benzene ring (predominantly at the para position):
```
CH₃
|
___C___
/ \
CH CH
|| ||
CH CH
\___C___/
|
SO₃H
```
**Structure: 4-methylbenzenesulphonic acid (para-toluenesulphonic acid, p-TsOH)**
---
### Question 8(b)(iii) — IUPAC Names of V, W, Y and Z
| Compound | Description | IUPAC Name |
|---|---|---|
| V | Sulphonation product of toluene | 4-methylbenzenesulphonic acid *(or 2-methylbenzenesulphonic acid for ortho product)* |
| W | Ring chlorination product of toluene | 2-chloro-1-methylbenzene (2-chlorotoluene) *or* 4-chloro-1-methylbenzene (4-chlorotoluene) |
| Y | Oxidation product of toluene | Benzoic acid *(IUPAC: benzenecarboxylic acid)* |
| Z | Acid chloride from benzoic acid | Benzoyl chloride *(IUPAC: benzenecarbonyl chloride)* |
---
### Question 8(c)(i) — Conversion of C to D (Dehydrochlorination)
**Reagent and condition:** Concentrated **KOH dissolved in ethanol** (alcoholic KOH); heat under reflux.
The reaction proceeds by an **E2 elimination mechanism** in which the strong base (OH⁻) abstracts a β-hydrogen while the C–Cl bond breaks simultaneously, forming the C=C double bond in D (an alkene or cycloalkene).
$$\text{C}_6\text{H}_{12}\text{Cl}_2 \xrightarrow{\text{alc. KOH, heat}} \text{C}_6\text{H}_{10}\text{Cl} + \text{HCl}$$
*(If a second equivalent of KOH is used, a further elimination may occur to give C₆H₁₀.)*
---
### Question 8(c)(ii) — Structure of Compound D
D has the molecular formula **C₆H₁₀** and upon oxidation with hot concentrated KMnO₄ yields **CO₂ and a carboxylic acid in equimolar amounts**.
The formation of CO₂ from KMnO₄ oxidation indicates a **terminal or vinyl-type carbon** (=CH₂ group), while the carboxylic acid comes from an internal carbon. The equimolar ratio with a 6-carbon starting alkene suggests:
D is most likely **1-methylcyclopent-1-ene** or an acyclic diene, but given the saturated starting compound B is unbranched and C₆H₁₂Cl₂, the most consistent interpretation is:
D is **hex-1-ene** (CH₂=CHCH₂CH₂CH₂CH₃) or, more precisely given the equimolar CO₂:RCOOH ratio from oxidative cleavage of a terminal alkene:
$$\text{CH}_2=\text{CH}(\text{CH}_2)_3\text{CH}_3 \xrightarrow{\text{hot conc. KMnO}_4} \text{CO}_2 + \text{CH}_3(\text{CH}_2)_3\text{COOH (pentanoic acid)}$$
**Structure of D: CH₂=CH−CH₂−CH₂−CH₂−CH₃ (hex-1-ene)**
---
### Question 8(c)(iii) — Structure of Compound B
B is a **saturated unbranched hydrocarbon** that, upon free radical chlorination, yields C₆H₁₂Cl₂ (a dichloride). The 6-carbon framework and unbranched structure point to:
**B = hexane (CH₃CH₂CH₂CH₂CH₂CH₃)**
$$\text{Structure: } \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$$
Free radical chlorination of hexane produces various positional isomers of hexyl chloride, and further chlorination gives dichlorohexane (C₆H₁₂Cl₂ = compound C). Dehydrochlorination of C then produces hexene (D, C₆H₁₀ after losing HCl, or C₆H₁₀ if a diene is formed from a dichloro compound by double elimination).