## CHM 001: GENERAL CHEMISTRY
### Question 1
**(a)** The phase diagram shown is for a one-component system, Z.
(i) What is T꜀?
(ii) Under what condition will Z sublime?
(iii) What phase does Z exist as at 298 K and 10⁵ Pa? [3 marks]
---
**(b)** The atomic mass of a naturally occurring element Y is 55.8. The masses of the isotopes of the element are ⁵⁴Y and ⁵⁷Y.
(i) Calculate the percentage abundance of each isotope.
(ii) Deduce the isotopic mass ratio. [3 marks]
---
**(c)** Balance the redox reaction below in alkaline medium and identify the oxidizing and reducing agents.
I⁻ + MnO₄⁻ → IO₃⁻ + MnO₂ [4 marks]
**[Total = 10 marks]**
---
### Question 2
**(a)** In an experiment, 15.0 g of methanol and 10.0 g of carbon(II) oxide were placed in a reaction vessel. [C = 12, H = 1, O = 16]
$$\text{CH}_3\text{OH}(l) + \text{CO}(g) \rightarrow \text{CH}_3\text{COOH}(l)$$
(i) Determine which reactant is the limiting reactant.
(ii) From (i), calculate the theoretical yield of acetic acid.
(iii) If the actual yield is 19.1 g, what is the percentage yield? [4 marks]
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**(b)**
(i) Distinguish between a polar and a non-polar molecule.
(ii) State which of the following molecules is/are polar or non-polar:
(a) CCl₄ (b) HF (c) CO₂ (d) CHCl₃ [3 marks]
---
**(c)**
(i) Define the term degeneracy as it applies to atomic orbitals.
(ii) Give the values of the azimuthal and magnetic quantum numbers of the electrons in an atom when n = 4. [3 marks]
**[Total = 10 marks]**
---
---
## CHM 002: PHYSICAL CHEMISTRY
### Question 3
**(a)**
(i) What are fuel cells?
(ii) With the aid of equations, state the reactions that occur at the anode and cathode of an H₂/O₂ fuel cell. [3 marks]
---
**(b)** Calculate the standard enthalpy change for the reaction:
$$\text{CH}_4(g) + \text{Cl}_2(g) \rightarrow \text{CH}_3\text{Cl}(g) + \text{HCl}(g)$$
given the following bond enthalpies at 298 K:
| Bond | ΔH / kJ mol⁻¹ |
|---|---|
| C–H | +411 |
| Cl–Cl | +243 |
| C–Cl | +327 |
| H–Cl | +431 |
[3 marks]
---
**(c)** The decay series of ²³⁸₉₂U is shown below:
$$^{238}_{92}\text{U} \xrightarrow{\text{2α, 2β}} \text{Z} \xrightarrow{A} ^{222}_{86}\text{Rn} \xrightarrow{Q} ^{211}_{82}\text{Pb}$$
(i) State the mass number and atomic number of Z.
(ii) Identify the types of radiation represented by A and Q. [3 marks]
**[Total = 10 marks]**
---
### Question 4
**(a)** 2 g of He, 3 g of N₂, and 4 g of Ar were introduced into a 15 dm³ vessel at 100 °C. [He = 4.0, N = 14.0, Ar = 39.9, 1 atm = 101,325 N m⁻²]
(i) What are the mole fractions of He, N₂, and Ar in the system?
(ii) Calculate the total pressure of the system and hence the partial pressures of each gas in the vessel. [4 marks]
---
**(b)**
(i) Use the dissociation constants below to arrange the following acids in order of increasing acidity:
| Acid | Kₐ (mol dm⁻³) |
|---|---|
| CH₃CH₂COOH | 1.259 × 10⁻⁵ |
| CH₃CHClCOOH | 1.585 × 10⁻³ |
| CH₃CCl₂COOH | 3.982 × 10⁻² |
| CH₂ClCH₂COOH | 7.943 × 10⁻⁵ |
(ii) Suggest a reason for your arrangement. [2 marks]
---
**(c)**
(i) Differentiate between ideal and non-ideal solutions.
(ii) How does the dissolution of a non-volatile solute affect the following colligative properties?
(a) Vapour pressure
(b) Freezing point [4 marks]
**[Total = 10 marks]**
---
---
## CHM 003: INORGANIC CHEMISTRY
### Question 5
**(a)**
(i) Use balanced chemical equations to illustrate what happens when the following compounds are added to water:
(a) NaCl (b) SO₃ (c) Al₂O₃ (d) Na₂O
(ii) Predict the pH of the solutions formed. [4 marks]
---
**(b)**
(i) Explain, giving reasons, the trend in the solubility of Group 2 sulphates.
(ii) Arrange the Group 2 sulphates in order of increasing solubility. [3 marks]
---
**(c)**
(i) List four greenhouse gases.
(ii) State the source and environmental impact of any two gases listed in (c)(i). [3 marks]
**[Total = 10 marks]**
---
### Question 6
**(a)**
(i) State two properties of water that differ from those of the other Group 16 hydrides.
(ii) Why are the first ionization energies of d-block metals greater than those of s-block metals? [4 marks]
---
**(b)**
(i) List three types of hydrides and give an example of each.
(ii) Give balanced chemical equations for the reaction of two types of hydrides with water. [5 marks]
---
**(c)** Name the ore from which aluminium is extracted. [1 mark]
**[Total = 10 marks]**
---
---
## CHM 004: ORGANIC CHEMISTRY
### Question 7
**(a)** Give two differences between nucleophilic substitution unimolecular (SN1) and nucleophilic substitution bimolecular (SN2) reactions. [2 marks]
---
**(b)** There are several isomers with the molecular formula C₅H₁₂O.
(i) Provide the structures and names of three unbranched isomers of the alkanol with formula C₅H₁₂O.
(ii) Which isomer(s) is/are chiral?
(iii) Write a chemical equation for the reaction of each compound in (b)(i) with excess acidified K₂Cr₂O₇ and provide names for the products formed. [5 marks]
---
**(c)** The structures of two natural amino acids, glycine and alanine, are given.
(i) Draw the structure of the compound formed when glycine is combined with alanine.
(ii) What class of compound is formed in (i)?
(iii) What type of bond is obtained?
(iv) Draw the zwitterionic forms of glycine and alanine. [3 marks]
**[Total = 10 marks]**
---
### Question 8
**(a)** An organic compound F has the empirical formula C₃H₆O and a vapour density of 43.
(i) What is the molecular formula of F?
(ii) If F does not react with Fehling's or Tollens' reagent but gives a yellow precipitate when reacted with aqueous iodine, draw the structure and give the IUPAC name of F.
(iii) Write a reaction equation for the reduction of F with NaBH₄. [3 marks]
---
**(b)** Methylbenzene can react with chlorine in two ways. The products depend on the reaction conditions.
(i) Draw the structures of products A (formed under UV light) and B (formed with AlCl₃ catalyst).
(ii) Predict the products for the reaction of A and B with NaOH.
(iii) Write a balanced equation and name the product(s) formed when:
(a) CH₃CH₂COCH₂CH₂CH₃ reacts with LiAlH₄
(b) Benzene reacts with CH₃Cl
(c) CH₃CH₂OH reacts with CH₃CH₂CH₂COOH [3 marks]
---
**(c)** Arrange the following carboxylic acids in order of increasing acidity and justify your answer:
3-chloropentanoic acid; pentanoic acid; 2,2-dichloropentanoic acid; 3,3-dichloropentanoic acid; 2-chloropentanoic acid; 4-chloropentanoic acid. [2 marks]
**[Total = 10 marks]**
---
---
# ANSWERS
## CHM 001: GENERAL CHEMISTRY
### Answer 1
**(a)**
**(i)** T꜀ is the **critical temperature** — the temperature above which the substance cannot be liquefied regardless of the pressure applied. Above T꜀, the distinction between liquid and gas phases disappears and Z exists only as a supercritical fluid.
**(ii)** Z will sublime when the temperature and pressure conditions fall **below the triple point** — specifically along the solid–vapour boundary curve. At any pressure below the triple-point pressure, heating Z will cause it to pass directly from solid to vapour without passing through the liquid phase.
**(iii)** From the phase diagram, the intersection of 298 K and 10⁵ Pa falls in the **liquid** region (above the vapour–liquid boundary and below the solid–liquid boundary at that temperature). Z therefore exists as a **liquid** at 298 K and 10⁵ Pa.
---
**(b)**
**(i)** Let the fractional abundance of ⁵⁴Y = x, so the fractional abundance of ⁵⁷Y = (1 − x).
$$54x + 57(1 - x) = 55.8$$
$$54x + 57 - 57x = 55.8$$
$$-3x = -1.2$$
$$x = 0.4$$
$$\boxed{^{54}\text{Y} = 40\%, \quad ^{57}\text{Y} = 60\%}$$
**(ii)** The isotopic mass ratio is:
$$^{54}\text{Y} : ^{57}\text{Y} = 54 : 57 = \boxed{18 : 19}$$
---
**(c)** Balancing I⁻ + MnO₄⁻ → IO₃⁻ + MnO₂ in alkaline medium using the half-reaction method:
**Oxidation half-reaction (I⁻ → IO₃⁻):**
$$\text{I}^- \rightarrow \text{IO}_3^-$$
Balance O by adding H₂O; balance H by adding H⁺; then convert to alkaline by adding OH⁻:
$$\text{I}^- + 3\text{H}_2\text{O} \rightarrow \text{IO}_3^- + 6\text{H}^+ + 6e^-$$
In alkaline medium (add 6 OH⁻ to each side):
$$\text{I}^- + 6\text{OH}^- \rightarrow \text{IO}_3^- + 3\text{H}_2\text{O} + 6e^-$$
**Reduction half-reaction (MnO₄⁻ → MnO₂):**
$$\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$$
**Equalize electrons** (multiply oxidation × 1 and reduction × 2):
$$\text{I}^- + 6\text{OH}^- \rightarrow \text{IO}_3^- + 3\text{H}_2\text{O} + 6e^-$$
$$2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6e^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^-$$
**Add and simplify:**
$$\boxed{\text{I}^- + 2\text{MnO}_4^- + \text{H}_2\text{O} \rightarrow \text{IO}_3^- + 2\text{MnO}_2 + 2\text{OH}^-}$$
- **Oxidizing agent:** MnO₄⁻ (Mn is reduced from +7 to +4)
- **Reducing agent:** I⁻ (I is oxidized from −1 to +5)
---
### Answer 2
**(a)**
**Molar masses:** CH₃OH = 12 + 4 + 16 = 32 g/mol; CO = 28 g/mol; CH₃COOH = 60 g/mol
**(i) Limiting reactant:**
$$n(\text{CH}_3\text{OH}) = \frac{15.0}{32} = 0.469 \text{ mol}$$
$$n(\text{CO}) = \frac{10.0}{28} = 0.357 \text{ mol}$$
The stoichiometry is 1:1, so the reactant present in fewer moles is the limiting reactant.
$$\boxed{\text{CO is the limiting reactant}}$$
**(ii) Theoretical yield:**
1 mol CO produces 1 mol CH₃COOH, so:
$$\text{Theoretical yield} = 0.357 \times 60 = \boxed{21.4 \text{ g}}$$
**(iii) Percentage yield:**
$$\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{19.1}{21.4} \times 100 = \boxed{89.3\%}$$
---
**(b)**
**(i)**
- A **polar molecule** has an uneven distribution of electron density due to a net dipole moment resulting from polar bonds that do not cancel. It has a positive and a negative end (partial charges δ+ and δ−).
- A **non-polar molecule** has either non-polar bonds, or polar bonds arranged symmetrically so that the individual bond dipoles cancel out, resulting in zero net dipole moment.
**(ii)**
| Molecule | Polar / Non-polar | Reason |
|---|---|---|
| (a) CCl₄ | **Non-polar** | Four C–Cl bonds arranged tetrahedrally; dipoles cancel symmetrically |
| (b) HF | **Polar** | Large electronegativity difference between H and F; significant dipole |
| (c) CO₂ | **Non-polar** | Linear geometry; two C=O dipoles cancel exactly |
| (d) CHCl₃ | **Polar** | Tetrahedral but asymmetric (one H, three Cl); dipoles do not cancel |
---
**(c)**
**(i)** **Degeneracy** in atomic orbitals refers to the condition where two or more orbitals have the **same energy**. Orbitals are said to be degenerate when they possess equal energy, such as the three 2p orbitals (2p_x, 2p_y, 2p_z) in an isolated atom.
**(ii)** For n = 4, the azimuthal quantum number l can take values 0, 1, 2, 3.
For each value of l, the magnetic quantum number m ranges from −l to +l:
| l | Subshell | m values | Number of orbitals |
|---|---|---|---|
| 0 | 4s | 0 | 1 |
| 1 | 4p | −1, 0, +1 | 3 |
| 2 | 4d | −2, −1, 0, +1, +2 | 5 |
| 3 | 4f | −3, −2, −1, 0, +1, +2, +3 | 7 |
Total orbitals for n = 4: **16 orbitals**; l = 0, 1, 2, 3; m = −3 to +3 (depending on l).
---
---
## CHM 002: PHYSICAL CHEMISTRY
### Answer 3
**(a)**
**(i)** Fuel cells are electrochemical devices that convert the chemical energy of a fuel (such as hydrogen) directly into electrical energy through an electrochemical reaction, without combustion. They operate continuously as long as fuel and oxidant are supplied.
**(ii) H₂/O₂ fuel cell electrode reactions:**
**Anode (oxidation):**
$$\text{H}_2(g) + 2\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) + 2e^-$$
**Cathode (reduction):**
$$\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq)$$
**Overall cell reaction:**
$$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$$
---
**(b)**
**Bonds broken (endothermic, +):**
- 1 × C–H: +411 kJ/mol
- 1 × Cl–Cl: +243 kJ/mol
- Total energy absorbed = **+654 kJ/mol**
**Bonds formed (exothermic, −):**
- 1 × C–Cl: +327 kJ/mol
- 1 × H–Cl: +431 kJ/mol
- Total energy released = **−758 kJ/mol**
$$\Delta H_{\text{rxn}} = \text{Energy absorbed} - \text{Energy released} = 654 - 758 = \boxed{-104 \text{ kJ mol}^{-1}}$$
The reaction is exothermic.
---
**(c)**
**(i) Finding Z (after 2α and 2β decays from ²³⁸₉₂U):**
Each α decay: mass number −4, atomic number −2
Each β decay: mass number unchanged, atomic number +1
After 2α decays: Mass = 238 − 8 = 230; Atomic number = 92 − 4 = 88
After 2β decays: Mass = 230; Atomic number = 88 + 2 = **90**
$$\boxed{Z: \text{ Mass number} = 230, \text{ Atomic number} = 90 \quad (^{230}_{90}\text{Th})}$$
**(ii) Identifying A and Q:**
From Z (²³⁰₉₀Th) → ²²²₈₆Rn via decay A:
- Mass change: 230 − 222 = 8 → loss of 2 α particles
- Atomic number change: 90 − 86 = 4 → consistent with 2 α decays
$$\boxed{A = \text{two alpha (α) decays}}$$
From ²²²₈₆Rn → ²¹¹₈₂Pb via decay Q:
- Mass change: 222 − 211 = 11 → loss of ~2–3 α particles (with possible β emissions)
- Atomic number change: 86 − 82 = 4 → net loss of 4 protons
A mass loss of 11 with an atomic number decrease of 4 is consistent with **3 α decays and 2 β decays** (3α: −12 mass, −6 atomic; 2β: 0 mass, +2 atomic; net: −12+1= adjustment — the series from Rn-222 to Pb-211 involves a sequence of **alpha and beta decays**).
$$\boxed{Q = \text{a series of alpha (α) and beta (β) decays}}$$
---
### Answer 4
**(a)**
**Moles of each gas:**
$$n(\text{He}) = \frac{2}{4.0} = 0.500 \text{ mol}$$
$$n(\text{N}_2) = \frac{3}{28.0} = 0.107 \text{ mol}$$
$$n(\text{Ar}) = \frac{4}{39.9} = 0.100 \text{ mol}$$
$$n_{\text{total}} = 0.500 + 0.107 + 0.100 = 0.707 \text{ mol}$$
**(i) Mole fractions:**
$$x_{\text{He}} = \frac{0.500}{0.707} = \boxed{0.707}$$
$$x_{\text{N}_2} = \frac{0.107}{0.707} = \boxed{0.151}$$
$$x_{\text{Ar}} = \frac{0.100}{0.707} = \boxed{0.141}$$
**(ii) Total pressure (using PV = nRT):**
T = 100°C = 373 K; V = 15 dm³ = 15 L; R = 0.08206 L·atm/mol·K
$$P_{\text{total}} = \frac{n_{\text{total}}RT}{V} = \frac{0.707 \times 0.08206 \times 373}{15} = \frac{21.66}{15} = \boxed{1.444 \text{ atm}}$$
In N m⁻²: 1.444 × 101,325 = **146,313 N m⁻² ≈ 1.46 × 10⁵ Pa**
**Partial pressures:**
$$P_{\text{He}} = x_{\text{He}} \times P_{\text{total}} = 0.707 \times 1.444 = \boxed{1.021 \text{ atm}}$$
$$P_{\text{N}_2} = 0.151 \times 1.444 = \boxed{0.218 \text{ atm}}$$
$$P_{\text{Ar}} = 0.141 \times 1.444 = \boxed{0.204 \text{ atm}}$$
---
**(b)**
**(i) Order of increasing acidity (smallest Kₐ = weakest acid):**
$$\text{CH}_3\text{CH}_2\text{COOH} < \text{CH}_2\text{ClCH}_2\text{COOH} < \text{CH}_3\text{CHClCOOH} < \text{CH}_3\text{CCl}_2\text{COOH}$$
(Kₐ: 1.259×10⁻⁵ < 7.943×10⁻⁵ < 1.585×10⁻³ < 3.982×10⁻²)
**(ii) Reason:**
The chlorine atom is highly electronegative and exerts an **inductive (electron-withdrawing) effect** that stabilises the carboxylate anion (conjugate base) formed after proton loss, thereby increasing acidity. The closer the Cl atom is to the –COOH group, the stronger the inductive effect and hence the greater the acidity. Multiple Cl atoms on the same carbon (gem-disubstitution) amplify this effect further. Therefore, acidity increases as the number of Cl atoms increases and as their proximity to the carboxyl group increases.
---
**(c)**
**(i)**
- An **ideal solution** obeys Raoult's law exactly over the entire range of composition. Solute–solvent interactions are identical in strength to solute–solute and solvent–solvent interactions, so there is no enthalpy or volume change on mixing.
- A **non-ideal solution** deviates from Raoult's law because solute–solvent interactions differ in strength from those in the pure components, leading to either positive deviation (weaker interactions → higher vapour pressure than predicted) or negative deviation (stronger interactions → lower vapour pressure than predicted).
**(ii)**
**(a) Vapour pressure lowering:**
Dissolving a non-volatile solute in a solvent lowers the vapour pressure of the solvent. This occurs because solute particles occupy surface sites, reducing the number of solvent molecules able to escape into the vapour phase. According to Raoult's law: P = x_solvent × P°. The greater the solute concentration (mole fraction), the greater the vapour pressure lowering.
**(b) Freezing point depression:**
The presence of a non-volatile solute disrupts the formation of the ordered crystal lattice of the solvent. As a result, a lower temperature is needed to achieve the degree of ordering required for solidification. The freezing point is depressed by: ΔTf = Kf × m, where Kf is the cryoscopic constant and m is the molality of the solution.
---
---
## CHM 003: INORGANIC CHEMISTRY
### Answer 5
**(a)**
**(i) Equations and (ii) pH predictions:**
**(a) NaCl — dissolves (no hydrolysis):**
$$\text{NaCl}(s) \xrightarrow{\text{H}_2\text{O}} \text{Na}^+(aq) + \text{Cl}^-(aq)$$
Na⁺ and Cl⁻ are spectator ions from a strong acid (HCl) and strong base (NaOH). **pH = 7** (neutral)
**(b) SO₃ — acidic oxide reacts vigorously:**
$$\text{SO}_3(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_4(aq)$$
Strong diprotic acid formed. **pH < 7** (strongly acidic)
**(c) Al₂O₃ — amphoteric oxide, sparingly soluble:**
$$\text{Al}_2\text{O}_3(s) + 3\text{H}_2\text{O}(l) \rightarrow 2\text{Al(OH)}_3(s) \text{ (very slight reaction)}$$
Al₂O₃ is largely insoluble in water; the resulting suspension is near neutral to very slightly acidic due to slight hydrolysis of Al³⁺. **pH ≈ 7** (effectively insoluble/neutral)
**(d) Na₂O — basic oxide reacts vigorously:**
$$\text{Na}_2\text{O}(s) + \text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq)$$
Strong base formed. **pH > 7** (strongly alkaline)
---
**(b)**
**(i)** The solubility of Group 2 sulphates **decreases down the group** (from Be to Ba). This trend is explained by the relative magnitudes of lattice enthalpy and hydration enthalpy:
- As the cation size increases down the group (Be²⁺ < Mg²⁺ < Ca²⁺ < Sr²⁺ < Ba²⁺), the **hydration enthalpy decreases** more rapidly than the **lattice enthalpy** decreases.
- For smaller cations (Be²⁺, Mg²⁺), hydration enthalpy is large enough to overcome lattice energy, favouring dissolution.
- For larger cations (Ba²⁺), the lattice energy dominates over the hydration energy, so the compound becomes increasingly insoluble.
**(ii) Order of increasing solubility:**
$$\text{BaSO}_4 < \text{SrSO}_4 < \text{CaSO}_4 < \text{MgSO}_4 < \text{BeSO}_4$$
---
**(c)**
**(i) Four greenhouse gases:**
1. Carbon dioxide (CO₂)
2. Methane (CH₄)
3. Nitrous oxide (N₂O)
4. Water vapour (H₂O)
*(Chlorofluorocarbons/CFCs and ozone are also accepted)*
**(ii) Source and environmental impact of two gases:**
**Carbon dioxide (CO₂):**
- *Source:* Combustion of fossil fuels (coal, oil, natural gas), deforestation, cement production, and respiration.
- *Impact:* Primary driver of global warming. CO₂ absorbs outgoing infrared radiation from the Earth's surface and re-emits it, trapping heat in the atmosphere (greenhouse effect), leading to rising global temperatures, sea level rise, and climate change.
**Methane (CH₄):**
- *Source:* Enteric fermentation in ruminant livestock, decomposition of organic matter in landfills, rice paddies, natural gas leaks, and wetlands.
- *Impact:* About 25–30 times more potent than CO₂ as a greenhouse gas over a 100-year period. Contributes significantly to tropospheric ozone formation and atmospheric warming.
---
### Answer 6
**(a)**
**(i) Two anomalous properties of water compared to other Group 16 hydrides (H₂S, H₂Se, H₂Te):**
1. **Abnormally high boiling point:** Water boils at 100°C, far above the expected trend for Group 16 hydrides. This is due to extensive **hydrogen bonding** between water molecules (O–H···O), which requires much more energy to overcome than the van der Waals forces in H₂S, H₂Se, and H₂Te.
2. **Maximum density at 4°C (anomalous expansion on freezing):** Unlike most substances, water is less dense as a solid (ice) than as a liquid. Hydrogen bonding in ice creates an open hexagonal lattice structure, making ice float on water — an anomaly not shared by other Group 16 hydrides.
**(ii)** The first ionization energies of d-block metals are generally greater than those of s-block metals because:
- D-block metals have **smaller atomic radii** than s-block metals in the same period, due to the poor shielding provided by d-electrons and the increased effective nuclear charge experienced by the outer electrons.
- The outermost electrons of d-block metals are therefore held more tightly, requiring more energy to remove. Additionally, the d-electrons provide poor shielding of the nucleus compared to s and p electrons, increasing the effective nuclear charge felt by the valence electrons.
---
**(b)**
**(i) Three types of hydrides with examples:**
| Type | Description | Example |
|---|---|---|
| **Ionic (saline) hydrides** | Formed by active metals (Groups 1 and 2) by transfer of electrons; contain H⁻ ion | NaH, CaH₂ |
| **Covalent (molecular) hydrides** | Formed by non-metals; H shares electrons with the element | H₂O, NH₃, HCl |
| **Metallic (interstitial) hydrides** | Formed by d- and f-block metals; H atoms occupy interstitial sites in the metal lattice | TiH₂, PdH₀.₆ |
**(ii) Reactions with water:**
**Ionic hydride (NaH) with water:**
$$\text{NaH}(s) + \text{H}_2\text{O}(l) \rightarrow \text{NaOH}(aq) + \text{H}_2(g)$$
**Covalent hydride (reacting example — nitrogen trifluoride analogue; using NH₃ which is a covalent hydride that reacts with water):**
$$\text{NH}_3(g) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq)$$
Or for a non-metal hydride that hydrolyses fully (e.g. HCl):
$$\text{HCl}(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_3\text{O}^+(aq) + \text{Cl}^-(aq)$$
---
**(c)**
Aluminium is extracted from **bauxite** (Al₂O₃·2H₂O — hydrated aluminium oxide). The pure oxide (Al₂O₃, corundum) obtained after purification is then electrolytically reduced in the **Hall–Héroult process**.
---
---
## CHM 004: ORGANIC CHEMISTRY
### Answer 7
**(a) Differences between SN1 and SN2:**
| Feature | SN1 | SN2 |
|---|---|---|
| **Mechanism** | Two-step: carbocation intermediate forms first, then nucleophile attacks | One-step: concerted; nucleophile attacks as the leaving group departs simultaneously |
| **Rate-determining step / Kinetics** | Unimolecular — rate depends only on the concentration of the substrate: rate = k[substrate] | Bimolecular — rate depends on both substrate and nucleophile: rate = k[substrate][nucleophile] |
| **Stereochemistry** | Racemisation (mixture of products) due to planar carbocation intermediate | Inversion of configuration (Walden inversion) at the carbon centre |
| **Substrate preference** | Favoured by tertiary substrates (stable carbocation) | Favoured by primary substrates (less steric hindrance) |
*(Any two of the above differences are sufficient for the 2 marks.)*
---
**(b) Unbranched isomers of C₅H₁₂O (alkanols):**
The three unbranched (straight-chain) isomers differ in the position of the –OH group:
**(i)**
**1. Pentan-1-ol:**
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$$
(Primary alcohol — OH on C1)
**2. Pentan-2-ol:**
$$\text{CH}_3\text{CH}(\text{OH})\text{CH}_2\text{CH}_2\text{CH}_3$$
(Secondary alcohol — OH on C2)
**3. Pentan-3-ol:**
$$\text{CH}_3\text{CH}_2\text{CH}(\text{OH})\text{CH}_2\text{CH}_3$$
(Secondary alcohol — OH on C3)
**(ii) Chiral isomers:**
**Pentan-2-ol** is chiral — C2 bears four different groups (–OH, –H, –CH₃, –CH₂CH₂CH₃). It has a non-superimposable mirror image.
Pentan-1-ol and pentan-3-ol are **not chiral** (pentan-3-ol has a plane of symmetry; C3 bears two identical –CH₂CH₃ groups).
**(iii) Oxidation with excess acidified K₂Cr₂O₇:**
**Pentan-1-ol** (primary) → oxidised to **pentanoic acid:**
$$\text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{OH} \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+} \text{CH}_3(\text{CH}_2)_3\text{COOH}$$
Product: **Pentanoic acid**
**Pentan-2-ol** (secondary) → oxidised to **pentan-2-one:**
$$\text{CH}_3\text{CH}(\text{OH})\text{CH}_2\text{CH}_2\text{CH}_3 \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+} \text{CH}_3\text{CO}\text{CH}_2\text{CH}_2\text{CH}_3$$
Product: **Pentan-2-one**
**Pentan-3-ol** (secondary) → oxidised to **pentan-3-one:**
$$\text{CH}_3\text{CH}_2\text{CH}(\text{OH})\text{CH}_2\text{CH}_3 \xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+} \text{CH}_3\text{CH}_2\text{CO}\text{CH}_2\text{CH}_3$$
Product: **Pentan-3-one**
---
**(c) Amino acids — glycine (H₂NCH₂COOH) and alanine (H₂NCH(CH₃)COOH):**
**(i) Dipeptide (glycine + alanine):**
The –COOH of glycine condenses with the –NH₂ of alanine (or vice versa):
$$\text{H}_2\text{N-CH}_2\text{-CO-NH-CH(CH}_3\text{)-COOH}$$
(Gly-Ala dipeptide; the bond shown is –CO–NH– i.e. the peptide bond)
**(ii) Class of compound:** **Dipeptide** (a type of amide / peptide)
**(iii) Type of bond:** **Peptide bond** (an amide bond, –CO–NH–), formed by condensation (elimination of water)
**(iv) Zwitterionic forms:**
In a zwitterion, the amine group is protonated (–NH₃⁺) and the carboxyl group is deprotonated (–COO⁻):
**Glycine zwitterion:**
$$^+\text{H}_3\text{N-CH}_2\text{-COO}^-$$
**Alanine zwitterion:**
$$^+\text{H}_3\text{N-CH(CH}_3\text{)-COO}^-$$
---
### Answer 8
**(a)**
**(i) Molecular formula of F:**
$$M_r = 2 \times \text{vapour density} = 2 \times 43 = 86 \text{ g/mol}$$
Empirical formula C₃H₆O has mass = (3×12) + (6×1) + 16 = 36 + 6 + 16 = **58 g/mol**
$$n = \frac{86}{58} \approx 1.48$$
Since this does not give a whole number, re-examine: trying C₄H₈O₂ (M = 88) or C₅H₁₀O (M = 86).
**C₅H₁₀O** has M = (5×12)+(10×1)+16 = 60+10+16 = **86 g/mol** ✓
Note: C₅H₁₀O fits the empirical formula C₃H₆O when simplified? → C₅H₁₀O simplifies to C₁H₂O (ratio 5:10:1), not C₃H₆O. The correct interpretation is that the molecular formula that satisfies M = 86 and contains the C:H:O ratio 3:6:1 is:
Empirical unit mass = 58; 86/58 is not integral, so consider rounding: the closest molecular formula with C:H:O in ratio 3:6:1 and M near 86 is **C₃H₆O × 1 = 58** (no), so the molecular formula is best determined as:
$$\boxed{\text{Molecular formula: } \text{C}_5\text{H}_{10}\text{O}}$$
*(The empirical formula C₃H₆O with M = 86 is approximate; the compound with M = 86 that is a methyl ketone and gives positive iodoform is pentan-2-one, C₅H₁₀O)*
**(ii)** F does **not** react with Fehling's or Tollens' reagent → F is **not** an aldehyde (rules out RCHO). F gives a **yellow precipitate with I₂/NaOH** → positive **iodoform test** → F contains a CH₃CO– group → F is a **methyl ketone**.
For C₅H₁₀O as a methyl ketone:
**Structure:**
$$\text{CH}_3 - \text{C}(=\text{O}) - \text{CH}_2\text{CH}_2\text{CH}_3$$
**IUPAC name: Pentan-2-one**
**(iii) Reduction of pentan-2-one with NaBH₄:**
NaBH₄ reduces ketones to secondary alcohols:
$$\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 + \text{NaBH}_4 \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{CH}_3$$
Product: **Pentan-2-ol**
---
**(b)**
**(i) Structures of A and B:**
**A** — formed under UV light (free radical substitution on the **side chain**):
$$\text{C}_6\text{H}_5\text{CH}_2\text{Cl}$$
**Benzyl chloride** (chloromethyl)benzene — the methyl group is chlorinated on the benzylic carbon.
**B** — formed with AlCl₃ catalyst (electrophilic aromatic substitution on the **ring**):
$$\text{ClC}_6\text{H}_4\text{CH}_3$$
Mixture of **2-chlorotoluene** (ortho) and **4-chlorotoluene** (para) — Cl is introduced onto the benzene ring, predominantly at ortho/para positions due to the directing effect of –CH₃.
**(ii) Reaction of A and B with NaOH:**
**A (benzyl chloride) + NaOH(aq):** Nucleophilic substitution
$$\text{C}_6\text{H}_5\text{CH}_2\text{Cl} + \text{NaOH}(aq) \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{NaCl}$$
Product: **Benzyl alcohol**
**B (chlorotoluene) + NaOH(aq):** Aryl halides are generally resistant to nucleophilic substitution under mild conditions. No reaction occurs with dilute NaOH at room temperature. Under harsh conditions (high T, high P):
$$\text{ClC}_6\text{H}_4\text{CH}_3 + \text{NaOH} \xrightarrow{\text{high T, P}} \text{HOC}_6\text{H}_4\text{CH}_3 + \text{NaCl}$$
Product: **Cresol (methylphenol)**
**(iii)**
**(a)** CH₃CH₂COCH₂CH₂CH₃ + LiAlH₄ → **CH₃CH₂CH(OH)CH₂CH₂CH₃** (then H₂O workup)
$$\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_2\text{CH}_3 \xrightarrow{\text{1. LiAlH}_4 \text{ 2. H}_2\text{O}} \text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_2\text{CH}_3$$
Product: **Hexan-3-ol** (LiAlH₄ reduces ketones to secondary alcohols)
**(b)** Benzene + CH₃Cl (with AlCl₃ — Friedel–Crafts alkylation):
$$\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl}$$
Product: **Methylbenzene (toluene)**
**(c)** CH₃CH₂OH + CH₃CH₂CH₂COOH (esterification with acid catalyst):
$$\text{CH}_3\text{CH}_2\text{OH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} \underset{\Delta}{\overset{\text{H}^+}{\rightleftharpoons}} \text{CH}_3\text{CH}_2\text{CH}_2\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$
Product: **Ethyl butanoate** (a fruity-smelling ester)
---
**(c) Increasing acidity of substituted pentanoic acids:**
**Principle:** Chlorine is electron-withdrawing by induction (−I effect), stabilising the carboxylate anion and increasing acidity. The effect is strongest when Cl is closest to the –COOH group (α > β > γ > δ position) and is amplified by multiple Cl atoms.
**Order of increasing acidity:**
$$\text{pentanoic acid} < \text{4-chloropentanoic acid} < \text{3-chloropentanoic acid} < \text{3,3-dichloropentanoic acid} < \text{2-chloropentanoic acid} < \text{2,2-dichloropentanoic acid}$$
**Justification:**
- **Pentanoic acid** (no Cl): no inductive effect; weakest acid.
- **4-chloropentanoic acid**: Cl is far from –COOH (δ-position); inductive effect minimal.
- **3-chloropentanoic acid**: Cl is one carbon closer (γ-position); stronger effect.
- **3,3-dichloropentanoic acid**: two Cl atoms at γ-position; cumulative effect greater than one Cl at γ.
- **2-chloropentanoic acid**: Cl is at α-position (adjacent to –COOH); strong inductive stabilisation.
- **2,2-dichloropentanoic acid**: two Cl atoms at α-position; maximum inductive stabilisation → strongest acid.