A-Level Chemistry — Revision Lab

▶ 1. Formulae, Equations & Amounts of Substance AS1

Chemical Formulae

Empirical Formula

The simplest whole number ratio of atoms of each element in a compound. E.g. CH₂O for glucose.

Molecular Formula

The actual number of atoms of each element in one molecule of a compound. E.g. C₆H₁₂O₆ for glucose.

Structural Formula

Shows how atoms are bonded together in a molecule. E.g. CH₃CH₂OH for ethanol.

Balancing Equations

All equations must be balanced — the same number of each type of atom on both sides. Include state symbols: (s) (l) (g) (aq).

Balanced Equation

2Mg(s) + O₂(g) → 2MgO(s)

Ionic Equation — Neutralisation

H⁺(aq) + OH⁻(aq) → H₂O(l)

The Mole Concept

Avogadro's Constant

The number of particles in one mole of a substance: Nₐ = 6.022 × 10²³ mol⁻¹

Molar Mass (Mr)

The mass of one mole of a substance, in g mol⁻¹. Numerically equal to the relative formula mass.

Key Formulae

moles = mass / Mr

concentration (mol dm⁻³) = moles / volume (dm³)

moles of gas = volume (dm³) / 24.0 (at RTP)

PV = nRT (ideal gas equation)

% yield = (actual yield / theoretical yield) × 100

atom economy = (Mr of desired product / ΣMr of all products) × 100

Exam Tip

Always convert cm³ to dm³ by dividing by 1000. Convert cm³ to dm³: 250 cm³ = 0.250 dm³. This is the most common error in calculation questions.

Worked Calculations

What mass of NaOH is needed to make 250 cm³ of 0.100 mol dm⁻³ solution?

Step 1 — Convert volume Volume = 250 cm³ = 250 / 1000 = 0.250 dm³

Step 2 — Calculate moles n = c × V = 0.100 × 0.250 = 0.0250 mol

Step 3 — Find molar mass Mr(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹

Step 4 — Calculate mass mass = n × Mr = 0.0250 × 40.0 = 1.00 g

Find the empirical formula of a compound containing 40.0% C, 6.7% H, and 53.3% O by mass.

Step 1 — Divide by Ar C: 40.0 / 12.0 = 3.33
H: 6.7 / 1.0 = 6.7
O: 53.3 / 16.0 = 3.33

Step 2 — Divide by smallest C: 3.33 / 3.33 = 1
H: 6.7 / 3.33 = 2
O: 3.33 / 3.33 = 1

Step 3 — Write empirical formula Empirical formula = CH₂O

Titration: 25.0 cm³ of NaOH required 22.50 cm³ of 0.100 mol dm⁻³ HCl for neutralisation. Find the concentration of NaOH.

Step 1 — Write the balanced equation HCl + NaOH → NaCl + H₂O
Mole ratio = 1 : 1

Step 2 — Moles of HCl n(HCl) = c × V = 0.100 × (22.50/1000) = 0.00225 mol

Step 3 — Moles of NaOH (from ratio) n(NaOH) = 0.00225 mol (1:1 ratio)

Step 4 — Concentration of NaOH c = n / V = 0.00225 / 0.0250 = 0.0900 mol dm⁻³

Calculate the percentage yield if 5.40 g of Al produces 8.10 g of Al₂O₃ (4Al + 3O₂ → 2Al₂O₃).

Step 1 — Moles of Al n(Al) = 5.40 / 27.0 = 0.200 mol

Step 2 — Theoretical moles of Al₂O₃ From equation: 4 mol Al → 2 mol Al₂O₃
n(Al₂O₃) = 0.200 × (2/4) = 0.100 mol

Step 3 — Theoretical mass Mr(Al₂O₃) = (2×27.0) + (3×16.0) = 102.0
Theoretical mass = 0.100 × 102.0 = 10.2 g

Step 4 — Percentage yield % yield = (8.10 / 10.2) × 100 = 79.4%

Common Error

Forgetting to use the mole ratio from the balanced equation. Always write the equation first, then identify the ratio before calculating.

▶ 2. Atomic Structure AS1

Fundamental Particles

Particle	Relative Mass	Relative Charge	Location
Proton	1	+1	Nucleus
Neutron	1	0	Nucleus
Electron	1/1836	−1	Shells / orbitals

Isotopes

Atoms of the same element with the same number of protons but a different number of neutrons. They have identical chemical properties but different physical properties (e.g. density, rate of diffusion).

Mass Number (A)

The total number of protons and neutrons in the nucleus of an atom.

Atomic Number (Z)

The number of protons in the nucleus of an atom. Defines the element.

Electron Configuration

Electrons fill sub-shells in order of increasing energy: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p

s sub-shell: max 2 electrons, spherical shape
p sub-shell: max 6 electrons (3 orbitals), dumbbell shape
d sub-shell: max 10 electrons (5 orbitals)
f sub-shell: max 14 electrons (7 orbitals)

Example Electron Configurations

Na (Z=11): 1s² 2s² 2p⁶ 3s¹

Fe (Z=26): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²

Cu (Z=29): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ (anomalous — full d)

Cr (Z=24): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ (anomalous — half-full d)

Exam Tip

Cu and Cr are exceptions — a full or half-full d sub-shell provides extra stability. Learn these two anomalies.

Ionisation Energy

First Ionisation Energy

The energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.
X(g) → X⁺(g) + e⁻

Factors affecting ionisation energy:

Nuclear charge: more protons = stronger attraction = higher IE
Atomic radius: further from nucleus = weaker attraction = lower IE
Shielding: more inner shells = more shielding = lower IE
Electron pairing: paired electrons repel each other = slightly lower IE

Trends: IE increases across a period (more protons, same shielding) and decreases down a group (more shells, greater shielding).

Evidence from successive IEs: A large jump between successive ionisation energies indicates electrons being removed from a shell closer to the nucleus. This provides evidence for electron shells.

Exam Tip

The drop from Be to B (Group 2→3) is because B's outer electron is in 2p (higher energy, easier to remove than 2s). The drop from N to O is because O has a paired electron in 2p — electron-electron repulsion makes it easier to remove.

▶ 3. Bonding AS1

Types of Bonding

Ionic Bonding

The electrostatic attraction between oppositely charged ions, formed by the transfer of electrons from a metal to a non-metal. Forms a giant ionic lattice.

Covalent Bonding

A shared pair of electrons between two non-metal atoms, where both atoms contribute one electron to the shared pair.

Dative (Coordinate) Bonding

A covalent bond where both electrons in the shared pair come from the same atom (the donor). E.g. in NH₄⁺ (N donates a lone pair to H⁺) and CO (C donates to O).

Metallic Bonding

The electrostatic attraction between a lattice of positive metal ions and a sea of delocalised electrons.

Electronegativity & Bond Polarity

Electronegativity

The ability of an atom to attract the bonding pair of electrons in a covalent bond. Increases across a period and up a group. Fluorine is the most electronegative element (4.0 on Pauling scale).

If two atoms have different electronegativities, the bond is polar. The more electronegative atom carries a partial negative charge (δ⁻) and the other a partial positive charge (δ⁺).

Intermolecular Forces

London (dispersion) forces: present in ALL molecules. Caused by temporary dipoles from uneven electron distribution. Strength increases with number of electrons / molecular size.
Permanent dipole–dipole: between polar molecules. Stronger than London forces alone for similar-sized molecules.
Hydrogen bonding: a special strong dipole–dipole force. Occurs when H is bonded to N, O, or F (highly electronegative atoms with lone pairs). Explains high boiling points of H₂O, NH₃, HF.

Shapes of Molecules (VSEPR)

Electron pairs around a central atom repel each other and arrange themselves to be as far apart as possible. Lone pairs repel more than bonding pairs.

Bonding Pairs	Lone Pairs	Shape	Bond Angle	Example
2	0	Linear	180°	CO₂, BeCl₂
3	0	Trigonal planar	120°	BF₃, AlCl₃
4	0	Tetrahedral	109.5°	CH₄, NH₄⁺
3	1	Trigonal pyramidal	107°	NH₃
2	2	Bent / V-shaped	104.5°	H₂O
6	0	Octahedral	90°	SF₆

Common Error

Saying "water is bent because it has two lone pairs" — you must explain that lone pairs repel more than bonding pairs, compressing the bond angle from 109.5° to 104.5°.

▶ 4. Energetics AS1

Exothermic Reaction

A reaction that releases energy to the surroundings. ΔH is negative. Temperature of surroundings increases.

Endothermic Reaction

A reaction that absorbs energy from the surroundings. ΔH is positive. Temperature of surroundings decreases.

Standard Enthalpy of Formation (ΔH°f)

The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (298 K, 100 kPa).

Standard Enthalpy of Combustion (ΔH°c)

The enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions.

Standard Enthalpy of Neutralisation (ΔH°neut)

The enthalpy change when an acid and base react to form one mole of water under standard conditions.

Hess's Law

The total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same.

Hess's Law using formation enthalpies

ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants)

Hess's Law using combustion enthalpies

ΔH°rxn = ΣΔH°c(reactants) − ΣΔH°c(products)

Calorimetry

Heat energy equation

q = mcΔT

q = heat energy (J)

m = mass of solution (g) [usually assume density = 1 g cm⁻³]

c = specific heat capacity of water = 4.18 J g⁻¹ K⁻¹

ΔT = temperature change (K or °C)

50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. Temperature rises by 6.8 °C. Calculate ΔH°neut.

Step 1 — Calculate heat released Total volume = 100.0 cm³, so mass = 100.0 g
q = mcΔT = 100.0 × 4.18 × 6.8 = 2842 J = 2.842 kJ

Step 2 — Calculate moles of water formed n(HCl) = 1.00 × 0.0500 = 0.0500 mol
1:1 ratio, so 0.0500 mol H₂O formed

Step 3 — Calculate ΔH per mole ΔH = −q / n = −2.842 / 0.0500 = −56.8 kJ mol⁻¹
(Negative because exothermic)

Bond Enthalpy Calculations

Bond Enthalpy

The energy required to break one mole of a particular covalent bond in gaseous molecules, averaged over many different compounds.

Bond enthalpy calculation

ΔH = Σ(bonds broken) − Σ(bonds formed)

Breaking bonds = endothermic (+)

Making bonds = exothermic (−)

Using bond enthalpies, calculate ΔH for: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). Bond enthalpies: C–H = 413, O=O = 498, C=O = 805, O–H = 464 kJ mol⁻¹.

Step 1 — Bonds broken 4 × C–H = 4 × 413 = 1652 kJ
2 × O=O = 2 × 498 = 996 kJ
Total broken = 2648 kJ

Step 2 — Bonds formed 2 × C=O = 2 × 805 = 1610 kJ
4 × O–H = 4 × 464 = 1856 kJ
Total formed = 3466 kJ

Step 3 — Calculate ΔH ΔH = 2648 − 3466 = −818 kJ mol⁻¹
(Exothermic — more energy released making bonds than needed to break them)

Exam Tip

Bond enthalpy calculations only give approximate values because bond enthalpies are averages and only apply to gases. Hess's Law gives more accurate results.

▶ 5. Kinetics AS1

Rate of Reaction

The change in concentration of a reactant or product per unit time. Units: mol dm⁻³ s⁻¹.

Measuring Rate

Gas collection: measure volume of gas produced over time (gas syringe)
Mass loss: measure decrease in mass as gas escapes (balance)
Colour change: colorimetry or disappearing cross (Na₂S₂O₃/HCl)
Titration: withdraw samples at intervals, quench, and titrate (clock methods)

Collision Theory

For a reaction to occur, particles must collide with sufficient energy (≥ activation energy) and correct orientation.

Activation Energy (Ea)

The minimum energy that colliding particles must have for a reaction to occur.

Factors affecting rate:

Concentration: more particles per unit volume → more frequent collisions → faster rate
Temperature: particles move faster → more frequent collisions AND more particles have energy ≥ Ea → faster rate
Surface area: smaller pieces → more surface exposed → more collisions → faster rate
Catalyst: provides an alternative reaction pathway with lower Ea → more particles have sufficient energy

Maxwell-Boltzmann Distribution

A graph showing the distribution of molecular kinetic energies at a given temperature.

No molecules have zero energy (curve starts at origin)
Most probable energy is at the peak
The curve is asymptotic — it never touches the x-axis at high energies
Area under the curve = total number of molecules (constant)
Higher temperature: peak shifts right and lowers; more molecules exceed Ea
Catalyst: Ea moves left (lower); more molecules exceed new lower Ea. Curve shape does NOT change.

Common Error

Saying a catalyst "gives particles more energy" — it does NOT. A catalyst provides an alternative pathway with a lower activation energy. The distribution curve is unchanged.

▶ 6. Group 7 — Halogens AS1

Physical Properties & Trends

Halogen	Colour	State (RTP)	Electronegativity
F₂	Pale yellow	Gas	4.0
Cl₂	Yellow-green	Gas	3.0
Br₂	Red-brown	Liquid	2.8
I₂	Grey/purple vapour	Solid	2.5

Trends down the group: boiling point increases (more electrons → stronger London forces), electronegativity decreases, oxidising ability decreases, reactivity decreases.

Displacement Reactions

A more reactive halogen displaces a less reactive halide from solution.

Displacement

Cl₂(aq) + 2KBr(aq) → 2KCl(aq) + Br₂(aq) [solution turns orange]

Cl₂(aq) + 2KI(aq) → 2KCl(aq) + I₂(aq) [solution turns brown]

Br₂(aq) + 2KI(aq) → 2KBr(aq) + I₂(aq) [solution turns brown]

Disproportionation with Alkalis

Disproportionation

A reaction where the same element is simultaneously oxidised and reduced.

Chlorine with cold dilute NaOH

Cl₂(aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l)

Cl: 0 → −1 (reduced) and 0 → +1 (oxidised)

Test for Halide Ions

Add dilute HNO₃ (to remove carbonates/hydroxides) then AgNO₃(aq):

Halide	Precipitate	Colour	Solubility in NH₃
Cl⁻	AgCl	White	Soluble in dilute NH₃
Br⁻	AgBr	Cream	Soluble in conc. NH₃
I⁻	AgI	Yellow	Insoluble in NH₃

▶ 7. Group 2 — Alkaline Earth Metals AS1

Reactions

Reaction with water (increases in vigour down the group)

Mg(s) + 2H₂O(l) → Mg(OH)₂(aq) + H₂(g) [very slow with cold water]

Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g) [steady fizzing]

Ba(s) + 2H₂O(l) → Ba(OH)₂(aq) + H₂(g) [vigorous]

Reaction with oxygen

2Mg(s) + O₂(g) → 2MgO(s) [bright white flame]

2Ca(s) + O₂(g) → 2CaO(s)

Reaction with chlorine

Ca(s) + Cl₂(g) → CaCl₂(s)

Trend in reactivity: increases down the group because atomic radius increases, so outer electrons are further from the nucleus, have more shielding, and are more easily lost.

Solubility Trends

Hydroxides: solubility increases down the group. Mg(OH)₂ is sparingly soluble; Ba(OH)₂ is soluble.
Sulfates: solubility decreases down the group. MgSO₄ is soluble; BaSO₄ is insoluble (used in barium meal X-rays).

Exam Tip

Remember: "Hydroxides Help — they get more soluble down the group" and "Sulfates Sink — they get less soluble."

▶ 8. Equilibrium AS2

Dynamic Equilibrium

A state in a closed system where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant.

Le Chatelier's Principle

If a system at equilibrium is subjected to a change, the position of equilibrium shifts to oppose the change.

Change	Effect on Equilibrium
Increase concentration of reactant	Shifts right (towards products)
Decrease concentration of product	Shifts right (towards products)
Increase pressure (gaseous)	Shifts to side with fewer moles of gas
Increase temperature	Shifts in endothermic direction
Add catalyst	No change in position — reaches equilibrium faster

Kc — Equilibrium Constant

For: aA + bB ⇌ cC + dD

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Kc only changes with temperature. Concentration changes and catalysts do NOT affect Kc.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), at equilibrium [N₂] = 0.50, [H₂] = 0.30, [NH₃] = 0.20 mol dm⁻³. Calculate Kc.

Step 1 — Write Kc expression Kc = [NH₃]² / ([N₂][H₂]³)

Step 2 — Substitute values Kc = (0.20)² / (0.50 × (0.30)³)
Kc = 0.040 / (0.50 × 0.027)
Kc = 0.040 / 0.0135

Step 3 — Calculate Kc = 2.96 dm⁶ mol⁻²
(Units: mol² dm⁻⁶ / (mol dm⁻³ × mol³ dm⁻⁹) = dm⁶ mol⁻²)

Exam Tip

Always work out the units of Kc by cancelling — they depend on the stoichiometry. If total moles of products = total moles of reactants, Kc has no units.

▶ 9. Acid-Base Chemistry AS2

Brønsted-Lowry Acid

A proton (H⁺) donor.

Brønsted-Lowry Base

A proton (H⁺) acceptor.

Strong Acid

An acid that fully ionises (dissociates) in aqueous solution. E.g. HCl, HNO₃, H₂SO₄.

Weak Acid

An acid that partially ionises in aqueous solution. E.g. CH₃COOH, H₂CO₃.

pH Calculations

Key Equations

pH = −log₁₀[H⁺]

[H⁺] = 10⁻ᵖᴴ

Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ (at 25°C)

Ka = [H⁺][A⁻] / [HA]

pKa = −log₁₀(Ka)

Calculate the pH of 0.0500 mol dm⁻³ HCl (strong acid).

Step 1 — Identify [H⁺] HCl is a strong monobasic acid, so fully ionises:
[H⁺] = 0.0500 mol dm⁻³

Step 2 — Calculate pH pH = −log₁₀(0.0500) = 1.30

Calculate the pH of 0.100 mol dm⁻³ NaOH (strong base) at 25°C.

Step 1 — Find [OH⁻] NaOH fully ionises: [OH⁻] = 0.100 mol dm⁻³

Step 2 — Find [H⁺] using Kw [H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 0.100 = 1.00 × 10⁻¹³ mol dm⁻³

Step 3 — Calculate pH pH = −log₁₀(1.00 × 10⁻¹³) = 13.0

Calculate the pH of 0.100 mol dm⁻³ CH₃COOH (Ka = 1.74 × 10⁻⁵ mol dm⁻³).

Step 1 — Write Ka expression Ka = [H⁺][CH₃COO⁻] / [CH₃COOH]

Step 2 — Assume [H⁺] = [CH₃COO⁻] and [HA] ≈ initial conc. Ka = [H⁺]² / 0.100
[H⁺]² = Ka × c = 1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶

Step 3 — Solve for [H⁺] and pH [H⁺] = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³
pH = −log₁₀(1.32 × 10⁻³) = 2.88

Buffer Solutions

Buffer Solution

A solution that resists changes in pH when small amounts of acid or base are added. Made from a weak acid and its conjugate base (e.g. CH₃COOH + CH₃COONa).

How a buffer works:

If acid (H⁺) is added: the conjugate base (A⁻) reacts with H⁺ to form HA, removing the added acid.
If base (OH⁻) is added: the weak acid (HA) reacts with OH⁻ to form A⁻ and H₂O, removing the added base.

Henderson-Hasselbalch Equation

pH = pKa + log₁₀([A⁻] / [HA])

Titration Curves

Strong acid + strong base: vertical section at pH 7, use any indicator
Strong acid + weak base: equivalence point below pH 7, use methyl orange
Weak acid + strong base: equivalence point above pH 7, use phenolphthalein
Weak acid + weak base: no sharp pH change — cannot use an indicator accurately

Exam Tip

The indicator must change colour within the vertical (steep) section of the titration curve. Phenolphthalein changes at pH 8-10, methyl orange at pH 3-5.

▶ 10. Redox AS2

Oxidation

Loss of electrons / increase in oxidation number. OIL RIG: Oxidation Is Loss.

Reduction

Gain of electrons / decrease in oxidation number. RIG: Reduction Is Gain.

Oxidation Number Rules

Elements in their standard state: 0
Simple ions: equal to charge (Na⁺ = +1, Cl⁻ = −1)
Oxygen: usually −2 (except in peroxides −1, and OF₂ +2)
Hydrogen: usually +1 (except in metal hydrides −1)
Fluorine: always −1
Sum of oxidation numbers = overall charge of species

Half Equations

Writing half equations (in acidic solution)

1. Write the species being oxidised or reduced

2. Balance atoms other than H and O

3. Balance O using H₂O

4. Balance H using H⁺

5. Balance charge using electrons (e⁻)

Example Half Equations

Oxidation: Fe²⁺ → Fe³⁺ + e⁻

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Exam Tip

To combine half equations, multiply so the electrons cancel. The total electrons lost must equal the total electrons gained.

▶ 11. Introduction to Organic Chemistry AS2

Homologous Series & Functional Groups

Homologous Series	Functional Group	General Formula	Example
Alkanes	C–C, C–H only	CₙH₂ₙ₊₂	CH₄ (methane)
Alkenes	C=C	CₙH₂ₙ	C₂H₄ (ethene)
Alcohols	–OH	CₙH₂ₙ₊₁OH	C₂H₅OH (ethanol)
Halogenoalkanes	C–X (X = halogen)	CₙH₂ₙ₊₁X	CH₃Cl (chloromethane)
Aldehydes	–CHO	CₙH₂ₙ₊₁CHO	CH₃CHO (ethanal)
Ketones	>C=O	CₙH₂ₙ₊₁COCₘH₂ₘ₊₁	CH₃COCH₃ (propanone)
Carboxylic acids	–COOH	CₙH₂ₙ₊₁COOH	CH₃COOH (ethanoic acid)
Esters	–COO–	CₙH₂ₙ₊₁COOCₘH₂ₘ₊₁	CH₃COOCH₃

Naming (IUPAC)

Find the longest carbon chain → gives root name (meth-, eth-, prop-, but-, pent-, hex-)
Number from the end nearest the functional group
Name branches as alkyl groups (methyl, ethyl)
Use prefixes di-, tri- for multiple identical groups

Types of Isomerism

Structural Isomers

Compounds with the same molecular formula but different structural arrangements. Types: chain isomerism, position isomerism, functional group isomerism.

Chain isomerism: different carbon chain arrangements (e.g. butane vs methylpropane)
Position isomerism: same functional group at different positions (e.g. butan-1-ol vs butan-2-ol)
Functional group isomerism: different functional group, same formula (e.g. ethanol C₂H₅OH vs methoxymethane CH₃OCH₃)

▶ 12. Alkanes AS2

Alkanes are saturated hydrocarbons — only C–C and C–H single bonds (all sigma bonds). Tetrahedral geometry around each C, bond angle 109.5°.

Combustion

Complete combustion

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

Incomplete combustion (limited O₂)

CH₄(g) + 1½O₂(g) → CO(g) + 2H₂O(l) [toxic carbon monoxide]

CH₄(g) + O₂(g) → C(s) + 2H₂O(l) [soot/carbon]

Free Radical Substitution

Mechanism: Free Radical Substitution of Methane with Chlorine

Initiation: Cl₂ → 2Cl• (UV light causes homolytic fission of Cl–Cl bond)

Propagation 1: Cl• + CH₄ → CH₃• + HCl (Cl radical abstracts H from methane)

Propagation 2: CH₃• + Cl₂ → CH₃Cl + Cl• (methyl radical reacts with Cl₂, regenerating Cl•)

Termination: Cl• + Cl• → Cl₂ | CH₃• + CH₃• → C₂H₆ | CH₃• + Cl• → CH₃Cl

Exam Tip

The major product is chloromethane (CH₃Cl) but further substitution can occur to give CH₂Cl₂, CHCl₃, and CCl₄. This is a limitation — the reaction gives a mixture of products.

▶ 13. Alkenes AS2

Alkenes are unsaturated — they contain a C=C double bond (one sigma + one pi bond). Trigonal planar around C=C, bond angle 120°.

Electrophilic Addition Reactions

Addition of Br₂ (test for unsaturation)

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br (1,2-dibromoethane)

Observation: bromine water decolourises (orange → colourless)

Addition of HBr

CH₂=CH₂ + HBr → CH₃CH₂Br (bromoethane)

Addition of H₂O (acid catalyst, e.g. H₃PO₄)

CH₂=CH₂ + H₂O → CH₃CH₂OH (ethanol)

Addition of H₂ (Ni catalyst, hydrogenation)

CH₂=CH₂ + H₂ → CH₃CH₃ (ethane)

Markovnikov's Rule

When HX adds to an unsymmetrical alkene, the hydrogen adds to the carbon already carrying the most hydrogen atoms. This produces the major product.

Example: propene + HBr

CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (major — 2-bromopropane)

→ CH₃CH₂CH₂Br (minor — 1-bromopropane)

E/Z (Geometric) Isomerism

Occurs when there is restricted rotation around C=C and each carbon of the double bond has two different groups attached. Uses Cahn-Ingold-Prelog priority rules (higher atomic number = higher priority).

Z (zusammen): higher priority groups on the Same side
E (entgegen): higher priority groups on opposite sides

Addition Polymerisation

Poly(ethene)

n CH₂=CH₂ → −(CH₂−CH₂)ₙ−

The double bond opens and monomers join together. No other product — no atoms are lost.

▶ 14. Halogenoalkanes AS2

Nucleophilic Substitution

Nucleophile

An electron-pair donor that is attracted to an electron-deficient (δ⁺) carbon atom. Examples: OH⁻, CN⁻, NH₃, H₂O.

SN2 — Primary Halogenoalkanes

One-step mechanism. Nucleophile attacks the δ⁺ carbon from the opposite side to the leaving group. Simultaneous bond-making and bond-breaking. Rate = k[RX][Nu⁻]

SN1 — Tertiary Halogenoalkanes

Step 1: C–X bond breaks heterolytically → carbocation + X⁻ (slow, rate-determining)

Step 2: Nucleophile attacks the carbocation (fast). Rate = k[RX]

Hydrolysis with NaOH(aq) — makes an alcohol

CH₃CH₂Br + NaOH → CH₃CH₂OH + NaBr

Rates of Hydrolysis

Bond strength: C–F > C–Cl > C–Br > C–I. Therefore C–I is broken most easily and iodoalkanes react fastest.

Elimination

With hot ethanolic NaOH (NaOH dissolved in ethanol), halogenoalkanes undergo elimination to form an alkene + HX.

Elimination

CH₃CH₂Br + NaOH (ethanol, heat) → CH₂=CH₂ + NaBr + H₂O

Exam Tip

The same reagent (NaOH) gives different products depending on conditions: aqueous NaOH → substitution (alcohol); ethanolic NaOH → elimination (alkene).

CFCs and the Ozone Layer

CFCs (chlorofluorocarbons) are stable, non-toxic, non-flammable — historically used as refrigerants and propellants. UV radiation in the stratosphere breaks C–Cl bonds, releasing Cl• radicals that catalytically destroy ozone:

Ozone depletion

Cl• + O₃ → ClO• + O₂

ClO• + O₃ → Cl• + 2O₂

Overall: 2O₃ → 3O₂ (Cl• is regenerated — acts as catalyst)

▶ 15. Alcohols AS2

Classification

Primary (1°): –OH attached to a carbon bonded to one other carbon. E.g. CH₃CH₂OH
Secondary (2°): –OH attached to a carbon bonded to two other carbons. E.g. CH₃CH(OH)CH₃
Tertiary (3°): –OH attached to a carbon bonded to three other carbons. E.g. (CH₃)₃COH

Oxidation with Acidified K₂Cr₂O₇

Acidified potassium dichromate(VI) is the oxidising agent. Colour change: orange → green (Cr₂O₇²⁻ → Cr³⁺).

Alcohol Type	Distillation	Product	Reflux	Product
Primary	Distil immediately	Aldehyde (–CHO)	Heat under reflux	Carboxylic acid (–COOH)
Secondary	—		Heat under reflux	Ketone (>C=O)
Tertiary	No reaction — resistant to oxidation

Oxidation examples ([O] = oxidising agent)

CH₃CH₂OH + [O] → CH₃CHO + H₂O (primary → aldehyde)

CH₃CHO + [O] → CH₃COOH (aldehyde → carboxylic acid)

CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O (secondary → ketone)

Dehydration

Heating with concentrated H₂SO₄ or H₃PO₄ removes H₂O to form an alkene (elimination).

Dehydration

CH₃CH₂OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, 170°C)

Esterification

Ester formation (conc. H₂SO₄ catalyst)

CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O

(ethanoic acid + ethanol → ethyl ethanoate + water)

Biofuels

Ethanol can be produced by fermentation of sugars (renewable) or hydration of ethene (from crude oil). Bioethanol is carbon-neutral in principle — CO₂ released on combustion was absorbed during crop growth.

▶ 16. Lattice Enthalpy & Born-Haber Cycles A2-1

Lattice Enthalpy of Formation

The enthalpy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions. Always exothermic (negative). E.g. Na⁺(g) + Cl⁻(g) → NaCl(s)

Lattice Enthalpy of Dissociation

The enthalpy change when one mole of an ionic compound is completely separated into gaseous ions. Always endothermic (positive).

Born-Haber Cycle Steps

Enthalpy of atomisation (ΔHₐₜ): forming 1 mol of gaseous atoms from element in standard state
First ionisation energy (IE₁): M(g) → M⁺(g) + e⁻
Second ionisation energy (IE₂): M⁺(g) → M²⁺(g) + e⁻
First electron affinity (EA₁): X(g) + e⁻ → X⁻(g) (usually exothermic)
Second electron affinity (EA₂): X⁻(g) + e⁻ → X²⁻(g) (always endothermic)
Lattice enthalpy: gaseous ions → ionic solid

Born-Haber Cycle (Hess's Law)

ΔH°f = ΔHₐₜ(M) + ΔHₐₜ(X) + IE₁(M) + EA₁(X) + ΔH_lattice

Calculate the lattice enthalpy of NaCl given: ΔH°f(NaCl) = −411 kJ mol⁻¹, ΔHₐₜ(Na) = +107, ΔHₐₜ(½Cl₂) = +122, IE₁(Na) = +496, EA₁(Cl) = −349 kJ mol⁻¹.

Step 1 — Apply Hess's Law ΔH°f = ΔHₐₜ(Na) + ΔHₐₜ(Cl) + IE₁(Na) + EA₁(Cl) + ΔH_lattice

Step 2 — Substitute −411 = +107 + 122 + 496 + (−349) + ΔH_lattice
−411 = +376 + ΔH_lattice

Step 3 — Solve ΔH_lattice = −411 − 376 = −787 kJ mol⁻¹
(Lattice enthalpy of formation — exothermic)

Enthalpy of Solution & Hydration

Enthalpy of Solution (ΔH_sol)

The enthalpy change when one mole of an ionic solid dissolves in sufficient water to form an infinitely dilute solution.

Enthalpy of Hydration (ΔH_hyd)

The enthalpy change when one mole of gaseous ions is surrounded by water molecules. Always exothermic.

Relationship

ΔH_sol = −ΔH_lattice(dissociation) + ΔH_hyd(cation) + ΔH_hyd(anion)

Or: ΔH_sol = ΔH_lattice(formation) + ΔH_hyd(cation) + ΔH_hyd(anion)

Exam Tip

If experimental lattice enthalpy differs significantly from theoretical (calculated from purely ionic model), it suggests significant covalent character in the bonding. This is common with polarising cations (small, highly charged) and polarisable anions (large).

▶ 17. Entropy & Free Energy A2-1

Entropy (S)

A measure of the disorder or randomness of a system. Units: J K⁻¹ mol⁻¹. Gases have highest entropy, solids lowest. S(gas) > S(liquid) > S(solid).

Entropy Change

ΔS = ΣS°(products) − ΣS°(reactants)

Gibbs Free Energy

Gibbs Free Energy Equation

ΔG = ΔH − TΔS

ΔG < 0 → reaction is feasible (thermodynamically spontaneous)

ΔG > 0 → reaction is not feasible

ΔG = 0 → system is at equilibrium

Exam Tip

"Feasible" does NOT mean the reaction will happen quickly — it may have a very high activation energy. ΔG tells us about thermodynamic feasibility, not kinetic feasibility.

For a reaction, ΔH = −100 kJ mol⁻¹ and ΔS = −200 J K⁻¹ mol⁻¹. At what temperature does the reaction become non-feasible?

Step 1 — At the boundary, ΔG = 0 0 = ΔH − TΔS
T = ΔH / ΔS

Step 2 — Convert units (ΔS to kJ) ΔS = −200 J K⁻¹ mol⁻¹ = −0.200 kJ K⁻¹ mol⁻¹

Step 3 — Calculate T T = −100 / −0.200 = 500 K
Above 500 K, the TΔS term dominates, ΔG becomes positive, and the reaction is no longer feasible.

▶ 18. Kinetics (A2) A2-1

Rate Equations

Rate Equation

Rate = k[A]ᵐ[B]ⁿ

k = rate constant

m = order with respect to A

n = order with respect to B

Overall order = m + n

Determining Order from Data

Compare experiments where only one concentration changes:

If doubling [A] has no effect on rate → order 0 with respect to A
If doubling [A] doubles the rate → order 1 (first order)
If doubling [A] quadruples the rate → order 2 (second order)

From the data: Exp 1: [A]=0.10, [B]=0.10, rate=2.0×10⁻³. Exp 2: [A]=0.20, [B]=0.10, rate=8.0×10⁻³. Exp 3: [A]=0.10, [B]=0.20, rate=4.0×10⁻³. Find the rate equation and k.

Step 1 — Find order w.r.t. A (compare Exp 1 & 2) [A] doubles (0.10→0.20), [B] constant
Rate × 4 (2.0→8.0 × 10⁻³)
2ᵐ = 4, so m = 2 (second order in A)

Step 2 — Find order w.r.t. B (compare Exp 1 & 3) [B] doubles (0.10→0.20), [A] constant
Rate × 2 (2.0→4.0 × 10⁻³)
2ⁿ = 2, so n = 1 (first order in B)

Step 3 — Write rate equation Rate = k[A]²[B]

Step 4 — Calculate k k = rate / ([A]²[B]) = 2.0×10⁻³ / (0.10² × 0.10)
k = 2.0×10⁻³ / 1.0×10⁻³ = 2.0 mol⁻² dm⁶ s⁻¹

Half-Life

For a first-order reaction, the half-life is constant regardless of initial concentration:

First-order half-life

t½ = ln 2 / k = 0.693 / k

Arrhenius Equation

k = Ae^(−Ea/RT)

ln k = −Ea/RT + ln A

Plot ln k vs 1/T → straight line, gradient = −Ea/R

R = 8.314 J K⁻¹ mol⁻¹

Exam Tip

The rate equation can only be determined experimentally — you cannot deduce it from the balanced equation. The orders in the rate equation give clues about the mechanism (the rate-determining step).

▶ 19. Equilibrium (A2) — Kp A2-1

Kp for Gaseous Equilibria

Key Relationships

Mole fraction of A: χ_A = moles of A / total moles

Partial pressure: p_A = χ_A × P_total

For: aA(g) + bB(g) ⇌ cC(g) + dD(g)

Kp = (p_C)ᶜ(p_D)ᵈ / (p_A)ᵃ(p_B)ᵇ

N₂O₄(g) ⇌ 2NO₂(g). At equilibrium: 0.20 mol N₂O₄ and 0.80 mol NO₂. Total pressure = 200 kPa. Calculate Kp.

Step 1 — Calculate mole fractions Total moles = 0.20 + 0.80 = 1.00
χ(N₂O₄) = 0.20/1.00 = 0.20
χ(NO₂) = 0.80/1.00 = 0.80

Step 2 — Calculate partial pressures p(N₂O₄) = 0.20 × 200 = 40 kPa
p(NO₂) = 0.80 × 200 = 160 kPa

Step 3 — Calculate Kp Kp = p(NO₂)² / p(N₂O₄) = 160² / 40 = 25600 / 40 = 640 kPa

▶ 20. Electrochemistry A2-1

Standard Electrode Potential (E°)

The EMF of a half-cell compared to a standard hydrogen electrode (SHE) under standard conditions (298 K, 100 kPa, 1.00 mol dm⁻³). SHE is assigned E° = 0.00 V.

Predicting Feasibility

Cell EMF

E°_cell = E°(reduction, cathode) − E°(reduction, anode)

If E°_cell > 0 → reaction is feasible

The more positive half-cell is reduced (cathode)

The more negative half-cell is oxidised (anode)

Calculate E°_cell for: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V.

Step 1 — Identify cathode and anode Cu²⁺/Cu is more positive → cathode (reduction)
Zn²⁺/Zn is more negative → anode (oxidation)

Step 2 — Calculate E°_cell E°_cell = E°(cathode) − E°(anode) = +0.34 − (−0.76) = +1.10 V
Positive → reaction is feasible

Fuel Cells

Hydrogen fuel cells produce electricity from H₂ and O₂, with water as the only product. More efficient than combustion engines. Zero emissions at point of use.

Hydrogen fuel cell

Anode: H₂(g) → 2H⁺(aq) + 2e⁻

Cathode: O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)

Overall: 2H₂(g) + O₂(g) → 2H₂O(l)

▶ 21. Carbonyl Compounds A2-1

Aldehydes vs Ketones

Aldehydes (–CHO): carbonyl at end of chain. Can be oxidised further to carboxylic acids.
Ketones (>C=O): carbonyl in middle of chain. Cannot be oxidised further.

Nucleophilic Addition (with HCN)

Mechanism: Nucleophilic Addition of HCN

Step 1: CN⁻ (nucleophile) attacks the δ⁺ carbon of C=O

Step 2: Electrons from C=O shift onto O⁻

Step 3: O⁻ is protonated by HCN to form –OH, regenerating CN⁻

Product: A hydroxynitrile (cyanohydrin)

Distinguishing Aldehydes from Ketones

Test	Reagent	Aldehyde Result	Ketone Result
Tollens' reagent	[Ag(NH₃)₂]⁺	Silver mirror formed	No reaction
Fehling's solution	Cu²⁺ in alkaline tartrate	Red precipitate (Cu₂O)	No reaction
2,4-DNPH	Brady's reagent	Orange precipitate	Orange precipitate

2,4-DNPH tests for the presence of a carbonyl group (both aldehydes and ketones). Tollens' and Fehling's distinguish between them.

Reduction

Reduction with NaBH₄ (sodium borohydride)

Aldehyde + NaBH₄ → primary alcohol

Ketone + NaBH₄ → secondary alcohol

CH₃CHO + 2[H] → CH₃CH₂OH

CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃

▶ 22. Aromatic Chemistry A2-1

Benzene Structure

Benzene (C₆H₆) has a planar hexagonal ring with delocalised pi electrons above and below the ring. All C–C bonds are equal length (between single and double). This is more stable than the Kekule structure (three separate C=C) — evidence: enthalpy of hydrogenation is less exothermic than expected.

Electrophilic Substitution

Benzene undergoes substitution (not addition) to preserve the stable delocalised ring.

Nitration of Benzene

Reagents: conc. HNO₃ + conc. H₂SO₄, below 55°C

Electrophile generation: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O

Step 1: NO₂⁺ attacks the delocalised ring, forming a Wheland intermediate

Step 2: H⁺ is lost, restoring aromaticity. Product: nitrobenzene (C₆H₅NO₂)

Friedel-Crafts Acylation

Reagents: acyl chloride (RCOCl) + AlCl₃ catalyst

Electrophile: RCO⁺ (acylium ion) formed: RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻

Product: a phenyl ketone (aryl ketone)

Friedel-Crafts Alkylation

Reagents: halogenoalkane (RCl) + AlCl₃ catalyst

Electrophile: R⁺ (carbocation) formed: RCl + AlCl₃ → R⁺ + AlCl₄⁻

Product: an alkylbenzene

Halogenation of Benzene

Reagents: Cl₂ (or Br₂) + AlCl₃ (or FeBr₃) catalyst (halogen carrier)

Electrophile: Cl⁺ formed: Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻

Product: chlorobenzene (C₆H₅Cl) + HCl

Phenol

Phenol (C₆H₅OH) is more acidic than alcohols because the phenoxide ion (C₆H₅O⁻) is stabilised by delocalisation of the negative charge into the benzene ring. Phenol reacts with bromine water without a catalyst (ring is activated by the –OH group), forming 2,4,6-tribromophenol (white precipitate).

▶ 23. Amines, Amino Acids & Polymers A2-1

Amines

Primary amine: RNH₂ (one alkyl/aryl group on N)
Secondary amine: R₂NH (two groups on N)
Tertiary amine: R₃N (three groups on N)

Preparation: Halogenoalkane + excess NH₃ in ethanol, heat in sealed tube.

Formation of primary amine

CH₃Br + 2NH₃ → CH₃NH₂ + NH₄Br

Amines are bases — the lone pair on N can accept a proton. Aliphatic amines are stronger bases than ammonia (alkyl groups are electron-donating). Aromatic amines (e.g. phenylamine) are weaker bases — the lone pair is delocalised into the ring.

Amino Acids

Amino acids have both –NH₂ (basic) and –COOH (acidic) groups. In solution, they exist as zwitterions at the isoelectric point: ⁺NH₃CH(R)COO⁻.

Addition Polymerisation

Alkene monomers join by opening their C=C double bonds. No other molecule is lost — all atoms from the monomer end up in the polymer.

General Addition Polymerisation

nCH₂=CHX → –(CH₂–CHX)ₙ–

Monomer	Polymer	Uses
Ethene (CH₂=CH₂)	Poly(ethene) / polyethylene	Bags, bottles, packaging
Propene (CH₂=CHCH₃)	Poly(propene) / polypropylene	Rope, carpets, containers
Chloroethene (CH₂=CHCl)	Poly(chloroethene) / PVC	Pipes, window frames
Tetrafluoroethene (CF₂=CF₂)	PTFE (Teflon)	Non-stick coatings
Phenylethene (CH₂=CHC₆H₅)	Polystyrene	Insulation, packaging

Addition polymers are non-biodegradable — the C–C backbone is chemically inert and resistant to hydrolysis, oxidation, and microbial attack.

Condensation Polymerisation

Monomers join by losing a small molecule (usually H₂O). Two different functional groups react. The monomer must be bifunctional (have a reactive group at each end).

Polyesters

Formed from a diol + dicarboxylic acid (or from a hydroxy acid). An ester link (–COO–) is formed with loss of H₂O at each join.

Terylene (PET) Formation

nHOOC–C₆H₄–COOH + nHO–CH₂CH₂–OH →

–(OC–C₆H₄–COO–CH₂CH₂–O)ₙ– + 2nH₂O

(benzene-1,4-dicarboxylic acid + ethane-1,2-diol → PET)

Polyamides (Nylons)

Formed from a diamine + dicarboxylic acid (or from an amino acid). An amide link (–CONH–) is formed with loss of H₂O.

Nylon-6,6 Formation

nHOOC–(CH₂)₄–COOH + nH₂N–(CH₂)₆–NH₂ →

–(OC–(CH₂)₄–CONH–(CH₂)₆–NH)ₙ– + 2nH₂O

(hexanedioic acid + hexane-1,6-diamine → Nylon-6,6)

Polypeptides/Proteins: amino acids linked by peptide bonds (–CONH–) — natural polyamides

Biodegradable Polymers & Disposal

Condensation polymers are often biodegradable because the ester/amide bonds can be hydrolysed by water, acid/base, or enzymes.

Biodegradable polymers: e.g. polylactic acid (PLA) from corn starch — contains ester links, broken down by microorganisms
Photodegradable polymers: break down when exposed to UV light

Disposal issues with polymers:

Landfill: addition polymers are non-biodegradable, persist for hundreds of years. Takes up space.
Incineration: burning produces CO₂ (greenhouse gas). PVC produces toxic HCl fumes. Requires energy.
Recycling: must be sorted by polymer type. Thermoplastics can be melted and remoulded. Thermosets cannot.
Feedstock recycling: cracking waste polymers back into monomers or useful hydrocarbons.

Exam Tip

To identify the monomer from a polymer: for addition polymers, find the repeat unit and add back the C=C double bond. For condensation polymers, break the ester/amide link and add back –OH and –H (i.e. add H₂O across the link).

▶ 24. Organic Analysis & Synthesis A2-1

Infrared Spectroscopy (IR)

Bond	Wavenumber / cm⁻¹	Notes
O–H (alcohol)	3200–3550	Broad
O–H (carboxylic acid)	2500–3300	Very broad
N–H	3300–3500	Medium, may show 2 peaks (primary amine)
C–H	2850–3100	Strong
C=O	1680–1750	Strong, sharp
C=C	1620–1680	Medium
C–O	1000–1300	Strong

Mass Spectrometry

M⁺ peak (molecular ion): gives the relative molecular mass
Fragmentation: molecule breaks into pieces. Common losses: 15 (CH₃), 17 (OH), 29 (CHO or C₂H₅), 45 (OC₂H₅ or CHO₂)
Base peak: the most abundant fragment (tallest peak, set to 100%)
(M+1) peak: due to ¹³C isotope — can help determine molecular formula

Interpreting Mass Spectra

Start from the M⁺ peak (molecular ion) and work backwards. Subtract common fragment masses from M⁺ to identify what was lost. E.g. if M⁺ = 74 and a peak at 59, then 74 − 59 = 15 → loss of CH₃.

m/z Lost	Fragment Lost	Suggests
15	CH₃	Methyl group present
17	OH	Hydroxyl group / alcohol
18	H₂O	Alcohol or carboxylic acid (dehydration)
28	CO or C₂H₄	Carbonyl or ethene loss
29	CHO or C₂H₅	Aldehyde or ethyl group
31	CH₃O	Methoxy group
44	CO₂	Carboxyl group
45	OC₂H₅	Ethoxy group (ester)

NMR Spectroscopy

Nuclear Magnetic Resonance (NMR) identifies the environment of hydrogen atoms (¹H NMR) or carbon atoms (¹³C NMR) in a molecule.

¹H NMR — Key Concepts

Chemical shift (δ): position of peak on x-axis, measured in ppm. Different environments give different shifts.
Integration (peak area): ratio of areas under peaks = ratio of H atoms in each environment
Splitting pattern (multiplicity): a peak splits into (n + 1) peaks, where n = number of H atoms on adjacent carbon(s)

Proton Environment	Chemical Shift / ppm
R–CH₃ (alkyl)	0.7–1.6
R–CH₂–R	1.2–1.8
R–CH₂–Cl/Br	3.0–4.0
R–O–CH₃ (methoxy)	3.3–3.7
R–CHO (aldehyde)	9.0–10.0
R–COOH (acid)	10.0–12.5
Ar–H (aromatic)	6.5–8.0
R–OH (alcohol)	1.0–6.0 (variable)
R–NH₂	1.0–5.0 (variable)

Splitting Patterns (n + 1 Rule)

Singlet (s) = 0 adjacent H; Doublet (d) = 1 adjacent H; Triplet (t) = 2 adjacent H; Quartet (q) = 3 adjacent H. O–H and N–H protons do NOT cause splitting and appear as singlets (exchangeable with D₂O).

Exam Tip

D₂O shake: adding D₂O to the NMR sample causes O–H and N–H peaks to disappear (H replaced by D). This confirms the presence of –OH or –NH groups.

¹³C NMR

Each peak represents a different carbon environment
Number of peaks = number of unique carbon environments (accounts for molecular symmetry)
No splitting pattern in ¹³C NMR (proton-decoupled)

Combined Analytical Problems

In exams you may be asked to identify an unknown compound using multiple techniques together:

Step 1 — Mass spec: M⁺ peak gives molecular mass. Fragmentation suggests functional groups.
Step 2 — IR spectrum: identify functional groups from absorption peaks (O–H, C=O, N–H etc.).
Step 3 — ¹H NMR: number of H environments, integration ratios, and splitting patterns narrow down the structure.
Step 4 — ¹³C NMR: number of carbon environments confirms symmetry and structure.
Step 5 — Combine all data to propose a structural formula consistent with all spectra.

An organic compound has M⁺ = 60, a broad O–H absorption at 2500–3300 cm⁻¹, a strong C=O at 1710 cm⁻¹, a singlet (3H) at δ 2.1 and a singlet (1H) at δ 11.4 in the ¹H NMR. Identify the compound.

Mr = 60. Broad O–H + C=O → carboxylic acid.
Singlet 3H at δ 2.1 → CH₃ group with no adjacent H.
Singlet 1H at δ 11.4 → COOH proton.
Answer: CH₃COOH (ethanoic acid, Mr = 60)

Multi-Step Organic Synthesis Routes

Key conversions to know:

Alkene → alcohol (H₂O/H⁺ or H₃PO₄)
Alkene → halogenoalkane (HX or X₂)
Halogenoalkane → alcohol (aq NaOH, reflux)
Halogenoalkane → amine (excess NH₃ in ethanol)
Halogenoalkane → nitrile (KCN in ethanol — extends carbon chain)
Alcohol → alkene (conc. H₂SO₄, heat)
Alcohol → aldehyde/ketone/acid (K₂Cr₂O₇/H₂SO₄)
Aldehyde → carboxylic acid (K₂Cr₂O₇/H₂SO₄, reflux)
Carboxylic acid + alcohol → ester (conc. H₂SO₄ catalyst)
Nitrile → carboxylic acid (acid hydrolysis)

Exam Tip

When asked for a synthesis route, give: reagent, conditions, and name the type of reaction for each step. If you need to increase the carbon chain length, use KCN to form a nitrile, then hydrolyse.

▶ 28. Period 3 Elements & Their Oxides A2-1

Reactions of Period 3 Elements with Oxygen

All Period 3 elements (Na to Cl) react with oxygen, though reactivity and vigour vary. Argon does not react (noble gas).

Period 3 Element + O₂ Reactions

4Na(s) + O₂(g) → 2Na₂O(s) (vigorous, yellow flame)

2Mg(s) + O₂(g) → 2MgO(s) (very vigorous, bright white flame)

4Al(s) + 3O₂(g) → 2Al₂O₃(s) (burns brightly once oxide layer breached)

Si(s) + O₂(g) → SiO₂(s) (reacts slowly, needs high temperature)

P₄(s) + 5O₂(g) → P₄O₁₀(s) (spontaneous in air, white smoke)

S(s) + O₂(g) → SO₂(g) (blue flame)

Reactions of Period 3 Elements with Water

Period 3 Element + H₂O Reactions

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g) (vigorous, floats, fizzes)

Mg(s) + 2H₂O(l) → Mg(OH)₂(aq) + H₂(g) (very slow with cold water)

Mg(s) + H₂O(g) → MgO(s) + H₂(g) (reacts with steam)

Al does not react with cold water due to its protective oxide layer. Si, P, S and Cl do not react with water in the same way (though Cl dissolves to form a mixture of acids).

Trend in Properties Across Period 3

Property	Na → Mg → Al	Si	P → S → Cl → Ar
Structure	Giant metallic	Giant covalent (macromolecular)	Simple molecular
Melting point	Increases (Na → Al) as metallic bonding strengthens	Very high (strong covalent bonds)	Low (weak intermolecular forces)
Electrical conductivity	Good conductors (metallic)	Semiconductor	Non-conductors
Bonding in oxide	Ionic	Covalent (macromolecular)	Covalent (simple molecular)

Period 3 Oxides — Acid-Base Character

There is a clear trend across Period 3: basic oxides → amphoteric oxide → acidic oxides.

Oxide	Bonding	Character	Reaction with Water
Na₂O	Ionic	Basic	Na₂O + H₂O → 2NaOH (pH ~14)
MgO	Ionic	Basic (slightly soluble)	MgO + H₂O → Mg(OH)₂ (pH ~9)
Al₂O₃	Ionic/covalent	Amphoteric	Insoluble in water
SiO₂	Giant covalent	Weakly acidic	Insoluble in water
P₄O₁₀	Simple covalent	Acidic	P₄O₁₀ + 6H₂O → 4H₃PO₄ (pH ~1)
SO₂ / SO₃	Simple covalent	Acidic	SO₂ + H₂O → H₂SO₃; SO₃ + H₂O → H₂SO₄ (pH ~1)

Amphoteric Behaviour of Al₂O₃

Al₂O₃ reacts with both acids and bases:

Al₂O₃ as a Base (with acid)

Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l)

Al₂O₃ as an Acid (with base)

Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq)

Exam Tip

The trend in oxide character from basic → amphoteric → acidic mirrors the change from metallic to non-metallic character across the period. Remember: metal oxides are basic, non-metal oxides are acidic.

▶ 29. Organic Mechanisms — In Depth A2-1

Curly Arrow Conventions

A curly arrow shows the movement of an electron pair. The arrow starts from where the electrons are (lone pair or bond) and points to where they are going. A half-headed (fishhook) arrow shows movement of a single electron (radical mechanisms).

Nucleophilic Substitution — S_N1 vs S_N2

S_N2 Mechanism (bimolecular)

One step: The nucleophile attacks the carbon from the opposite side to the leaving group (backside attack). Bond making and bond breaking happen simultaneously.

Curly arrows: Arrow from lone pair on nucleophile (e.g. OH⁻) to C; arrow from C–X bond to X (leaving group)

Transition state: [HO---C---Br]⁻ (5 groups around C, pentacoordinate)

Stereochemistry: Inversion of configuration (Walden inversion)

Favoured by: primary halogenoalkanes, strong nucleophile (e.g. OH⁻, CN⁻), good polar aprotic solvent
Rate: Rate = k[halogenoalkane][nucleophile] — rate depends on both species

S_N1 Mechanism (unimolecular)

Step 1 (slow, rate-determining): The C–X bond breaks heterolytically to form a carbocation. Curly arrow from C–X bond to X.

Step 2 (fast): The nucleophile attacks the planar carbocation from either side. Curly arrow from lone pair on nucleophile to C⁺.

Stereochemistry: Racemic mixture produced (attack from both sides equally likely)

Favoured by: tertiary halogenoalkanes (stable carbocation), weak nucleophile, polar protic solvent
Rate: Rate = k[halogenoalkane] — rate depends only on the halogenoalkane

Common Error

Secondary halogenoalkanes react by a mixture of S_N1 and S_N2. Don't assume one mechanism dominates unless told otherwise. The mechanism depends on conditions.

Elimination — E1 vs E2

Elimination reactions remove HX from a halogenoalkane to form an alkene. The base removes an H from a carbon adjacent to the C bearing the leaving group.

E2 Mechanism (bimolecular elimination)

One step: Strong base (e.g. ethanolic KOH) abstracts an H from the carbon adjacent to C–X. The C–H bond pair forms the C=C double bond while the C–X bond breaks simultaneously.

Curly arrows: Arrow from base to H; arrow from C–H bond to form C=C; arrow from C–X bond to X

Favoured by: strong base, high temperature, primary/secondary halogenoalkanes

E1 Mechanism (unimolecular elimination)

Step 1 (slow): C–X bond breaks heterolytically → carbocation + X⁻

Step 2 (fast): Base removes H from adjacent C; electron pair forms C=C

Favoured by: weak base, tertiary halogenoalkanes, polar protic solvent

Exam Tip

Substitution vs Elimination? Aqueous NaOH/KOH (dilute) favours substitution. Ethanolic NaOH/KOH (concentrated, hot) favours elimination. The solvent and conditions matter!

Electrophilic Addition to Alkenes

Addition of HBr to an Alkene

Step 1: The electron-rich C=C attacks the δ⁺ H of H–Br. Curly arrow from C=C bond to H. The H–Br bond breaks heterolytically (arrow from H–Br bond to Br). A carbocation intermediate forms.

Step 2: Br⁻ attacks the carbocation. Curly arrow from lone pair on Br⁻ to C⁺.

Markovnikov's Rule

When HX adds to an unsymmetrical alkene, the H adds to the carbon already bonded to more hydrogens (the major product goes via the more stable carbocation). E.g. CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (major, via secondary carbocation) not CH₃CH₂CH₂Br.

Electrophilic Substitution of Benzene

Benzene undergoes substitution (not addition) to preserve its stable delocalised ring. An electrophile replaces one H atom.

Nitration of Benzene

Generate electrophile: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O (nitronium ion)

Step 1: Curly arrow from delocalised ring to NO₂⁺. An intermediate carbocation forms (positive charge on ring, H and NO₂ both attached).

Step 2: A base (HSO₄⁻) removes H⁺ from the intermediate. Curly arrow from C–H bond back into ring, restoring aromaticity.

Conditions: Conc. HNO₃ + conc. H₂SO₄ (catalyst), 50°C. Product: nitrobenzene C₆H₅NO₂

Halogenation of Benzene (Bromination)

Generate electrophile: Br₂ + AlBr₃ → Br⁺ + AlBr₄⁻ (or use FeBr₃ as halogen carrier)

Step 1: Curly arrow from ring to Br⁺. Intermediate cation forms.

Step 2: AlBr₄⁻ removes H⁺. Ring aromaticity restored.

Conditions: Br₂ + AlBr₃ catalyst, reflux. Product: bromobenzene C₆H₅Br + HBr

Friedel-Crafts Alkylation

Generate electrophile: CH₃Cl + AlCl₃ → CH₃⁺ + AlCl₄⁻

Step 1: Ring attacks CH₃⁺ electrophile → intermediate cation

Step 2: AlCl₄⁻ removes H⁺ → methylbenzene + HCl + AlCl₃ (catalyst regenerated)

Friedel-Crafts Acylation

Generate electrophile: CH₃COCl + AlCl₃ → CH₃CO⁺ (acylium ion) + AlCl₄⁻

Step 1: Ring attacks acylium ion → intermediate

Step 2: H⁺ lost → phenylethanone (methyl phenyl ketone) + HCl

Advantage over alkylation: only monosubstitution occurs (C=O deactivates ring)

Nucleophilic Addition to Carbonyls

The C=O bond is polar (C is δ⁺, O is δ⁻). Nucleophiles attack the electrophilic carbon.

Nucleophilic Addition of HCN to a Carbonyl

Step 1: CN⁻ (nucleophile) attacks δ⁺ carbon of C=O. Curly arrow from lone pair on C of CN⁻ to the carbonyl C. The C=O π bond breaks — arrow from C=O bond to O. Oxygen gains a negative charge.

Step 2: The O⁻ intermediate is protonated by HCN (or H⁺). Curly arrow from H–CN bond or lone pair on the O. Product is a hydroxynitrile (cyanohydrin).

Conditions: HCN / NaCN in dilute H₂SO₄ (generates CN⁻ in situ). SAFETY: HCN is extremely toxic — use NaCN + acid, fume cupboard.

Overall equation

CH₃COCH₃ + HCN → CH₃C(OH)(CN)CH₃

(propanone → 2-hydroxy-2-methylpropanenitrile)

Exam Tip

When drawing mechanisms in exams: (1) always show curly arrows starting from a lone pair or bond, (2) show the arrow head pointing to where electrons move to, (3) show all charges and lone pairs on intermediates, (4) show all bonds being made and broken. Marks are awarded for each correct curly arrow.

Draw the mechanism for the reaction of 2-bromopropane with aqueous NaOH. State whether this is more likely to proceed by S_N1 or S_N2, and explain your answer.

2-bromopropane is a secondary halogenoalkane.
With aqueous NaOH (a strong nucleophile), both S_N1 and S_N2 are possible.
S_N2 pathway: OH⁻ attacks from the back of C–Br, bond forms to OH while C–Br breaks simultaneously.
Product: CH₃CH(OH)CH₃ (propan-2-ol) + Br⁻.
(In practice, a mixture of both mechanisms operates for secondary substrates.)

▶ 25. Transition Metals A2-2

Transition Metal

A d-block element that forms at least one stable ion with a partially filled d sub-shell. (Sc and Zn are d-block but NOT transition metals.)

Properties of Transition Metals

Variable oxidation states: because 3d and 4s sub-shells are close in energy
Form coloured ions: due to d-d electron transitions when light is absorbed
Catalytic activity: can change oxidation state or provide a surface for adsorption. E.g. Fe in Haber process, V₂O₅ in Contact process, Mn in MnO₂ decomposition of H₂O₂
Form complex ions: ions surrounded by ligands (coordinate bonding)

Complex Ions

Ligand

A molecule or ion that donates a lone pair of electrons to a central metal ion to form a dative covalent (coordinate) bond. E.g. H₂O, NH₃, Cl⁻, CN⁻.

Coordination Number

The total number of dative bonds formed between ligands and the central metal ion.

Coordination No.	Shape	Example
6	Octahedral	[Cu(H₂O)₆]²⁺, [Fe(CN)₆]³⁻
4	Tetrahedral	[CuCl₄]²⁻, [CoCl₄]²⁻
4	Square planar	[Pt(NH₃)₂Cl₂] (cisplatin)
2	Linear	[Ag(NH₃)₂]⁺

Electronic Configuration of d-Block Elements

d-block elements fill the 3d sub-shell. Note the exceptions: Cr is [Ar] 3d⁵ 4s¹ (not 3d⁴ 4s²) and Cu is [Ar] 3d¹⁰ 4s¹ (not 3d⁹ 4s²) — half-filled and fully filled d sub-shells are more stable.

Element	Symbol	Electron Configuration	Common Oxidation States
Titanium	Ti	[Ar] 3d² 4s²	+2, +3, +4
Vanadium	V	[Ar] 3d³ 4s²	+2, +3, +4, +5
Chromium	Cr	[Ar] 3d⁵ 4s¹	+2, +3, +6
Manganese	Mn	[Ar] 3d⁵ 4s²	+2, +4, +7
Iron	Fe	[Ar] 3d⁶ 4s²	+2, +3
Cobalt	Co	[Ar] 3d⁷ 4s²	+2, +3
Nickel	Ni	[Ar] 3d⁸ 4s²	+2
Copper	Cu	[Ar] 3d¹⁰ 4s¹	+1, +2

When forming ions, the 4s electrons are lost first (even though 4s fills first). E.g. Fe²⁺ = [Ar] 3d⁶, Fe³⁺ = [Ar] 3d⁵.

Why Coloured Ions?

In a complex ion, the d orbitals split into two energy levels. When visible light is absorbed, an electron is promoted from the lower to the higher d orbitals — this is a d-d transition. The colour observed is the complementary colour of the light absorbed.

The colour depends on: the metal ion, its oxidation state, the ligand, and the coordination number
Ions with empty d orbitals (d⁰, e.g. Sc³⁺) or full d orbitals (d¹⁰, e.g. Zn²⁺, Cu⁺) are colourless — no d-d transition is possible

Common Ion Colours

Ion	Formula	Colour
Copper(II) aqua	[Cu(H₂O)₆]²⁺	Blue
Copper(II) ammonia	[Cu(NH₃)₄(H₂O)₂]²⁺	Deep blue
Copper(II) chloro	[CuCl₄]²⁻	Yellow/green
Iron(II)	[Fe(H₂O)₆]²⁺	Pale green
Iron(III)	[Fe(H₂O)₆]³⁺	Yellow/brown
Chromium(III)	[Cr(H₂O)₆]³⁺	Green
Chromium(III) ammonia	[Cr(NH₃)₆]³⁺	Purple
Cobalt(II)	[Co(H₂O)₆]²⁺	Pink
Cobalt(II) chloro	[CoCl₄]²⁻	Blue
Manganese(II)	[Mn(H₂O)₆]²⁺	Very pale pink
Dichromate	Cr₂O₇²⁻	Orange
Permanganate	MnO₄⁻	Purple
Vanadium(V)	VO₂⁺	Yellow
Vanadium(IV)	VO²⁺	Blue
Vanadium(III)	V³⁺	Green
Vanadium(II)	V²⁺	Violet

Ligand Types

Ligand	Type	Lone Pairs Donated
H₂O, NH₃, Cl⁻, OH⁻, CN⁻	Monodentate	1 per ligand
H₂NCH₂CH₂NH₂ (ethane-1,2-diamine, "en")	Bidentate	2 per ligand
EDTA⁴⁻	Hexadentate	6 per ligand

Bidentate and polydentate ligands form more stable complexes — the chelate effect. Replacement of monodentate by polydentate ligands increases entropy (more free molecules released).

Ligand Substitution Reactions

Ligands can be exchanged in a complex. If the new ligand is similar in size to the old one, the coordination number stays the same. If ligand size changes significantly, the coordination number and shape may change.

H₂O → NH₃ substitution (no shape change — similar size ligands)

[Cu(H₂O)₆]²⁺ + 4NH₃ → [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O

Blue → Deep blue (still octahedral)

H₂O → Cl⁻ substitution (shape change — Cl⁻ is larger)

[Cu(H₂O)₆]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 6H₂O

Blue octahedral → Yellow/green tetrahedral

H₂O → Cl⁻ substitution (cobalt)

[Co(H₂O)₆]²⁺ + 4Cl⁻ → [CoCl₄]²⁻ + 6H₂O

Pink octahedral → Blue tetrahedral

Catalytic Activity — Examples

Transition metals act as catalysts because they can:

Change oxidation state, providing an alternative reaction pathway with lower activation energy (homogeneous catalysis)
Provide a surface for adsorption of reactant molecules, weakening bonds (heterogeneous catalysis)

Catalyst	Process	Type
Fe	Haber process: N₂ + 3H₂ ⇌ 2NH₃	Heterogeneous
V₂O₅	Contact process: 2SO₂ + O₂ ⇌ 2SO₃	Heterogeneous
MnO₂	Decomposition of H₂O₂ → H₂O + ½O₂	Heterogeneous
Ni	Hydrogenation of alkenes	Heterogeneous
Fe²⁺/Fe³⁺	Reaction of S₂O₈²⁻ with I⁻ ions	Homogeneous
Mn²⁺ (autocatalysis)	MnO₄⁻ / C₂O₄²⁻ titration	Homogeneous

Important Compounds

Copper Compounds

CuSO₄·5H₂O: blue crystals; anhydrous CuSO₄ is white — used as test for water
Cu₂O: red solid (Cu in +1 state); used in Fehling's/Benedict's test for reducing sugars

Iron Compounds

FeSO₄ (iron(II) sulfate): pale green solution. Oxidised to Fe³⁺ by acidified KMnO₄ or K₂Cr₂O₇
FeCl₃ (iron(III) chloride): yellow/brown. Used as catalyst for halogenation of benzene

Manganese Compounds

KMnO₄ (potassium permanganate): purple; powerful oxidising agent. Reduced to Mn²⁺ (colourless/pale pink) in acid
MnO₂ (manganese(IV) oxide): black solid; catalyst for H₂O₂ decomposition

Permanganate Titration (in acid)

MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)

Uses of Transition Metals

Iron: steel production, construction, Haber process catalyst
Copper: electrical wiring (excellent conductor), plumbing, alloys (brass, bronze)
Titanium: aircraft, medical implants (lightweight, strong, corrosion-resistant)
Platinum: catalytic converters in cars, jewellery
Nickel: hydrogenation catalyst, stainless steel alloy

Exam Tip

Remember: Sc and Zn are d-block but NOT transition metals. Sc³⁺ has no d electrons (d⁰) and Zn²⁺ has a full d sub-shell (d¹⁰). Neither can form coloured ions or show variable oxidation states in stable compounds.

▶ 26. Reactions of Ions in Aqueous Solution A2-2

Reactions with NaOH(aq)

Ion	NaOH (few drops)	Excess NaOH
Cu²⁺	Blue precipitate Cu(OH)₂	No change (insoluble)
Fe²⁺	Green precipitate Fe(OH)₂	No change
Fe³⁺	Brown precipitate Fe(OH)₃	No change
Cr³⁺	Green precipitate Cr(OH)₃	Dissolves → green solution [Cr(OH)₄]⁻
Al³⁺	White precipitate Al(OH)₃	Dissolves → colourless [Al(OH)₄]⁻

Reactions with NH₃(aq)

Ion	NH₃ (few drops)	Excess NH₃
Cu²⁺	Blue precipitate Cu(OH)₂	Dissolves → deep blue [Cu(NH₃)₄(H₂O)₂]²⁺
Fe²⁺	Green precipitate Fe(OH)₂	No change
Fe³⁺	Brown precipitate Fe(OH)₃	No change
Co²⁺	Blue precipitate	Dissolves → yellow/brown [Co(NH₃)₆]²⁺

▶ 27. Further Organic — Optical Isomerism & Chromatography A2-2

Optical Isomerism

Chiral Centre

A carbon atom bonded to four different groups. Chiral molecules exist as two non-superimposable mirror images (enantiomers).

Enantiomers have identical physical properties but rotate plane-polarised light in opposite directions. A racemic mixture (50:50 mix of enantiomers) shows no overall rotation.

Chromatography

TLC (Thin Layer Chromatography): stationary phase is silica/alumina on a plate. Rf = distance moved by spot / distance moved by solvent front.
Paper chromatography: stationary phase is water held in paper fibres.
Column chromatography: used for separation on a larger scale.
Gas chromatography (GC): used to separate and identify volatile compounds. Area under peak = amount of substance.

▶ 30. Environmental Chemistry A2-2

Ozone Depletion

Ozone (O₃) in the stratosphere absorbs harmful UV radiation, protecting life on Earth. CFCs (chlorofluorocarbons, e.g. CCl₃F) were used as refrigerants, aerosol propellants, and foam-blowing agents.

Radical Mechanism for Ozone Depletion

Initiation: UV light breaks the C–Cl bond homolytically:
CCl₃F → CCl₂F· + Cl· (photodissociation, produces chlorine radical)

Propagation step 1: Cl· + O₃ → ClO· + O₂ (chlorine radical destroys ozone)

Propagation step 2: ClO· + O· → Cl· + O₂ (chlorine radical regenerated — acts as catalyst)

Overall: O₃ + O· → 2O₂ (ozone is destroyed, Cl is regenerated)

A single chlorine radical can destroy thousands of ozone molecules before being removed. This is why even small amounts of CFCs cause significant damage.

Montreal Protocol (1987): international agreement to phase out CFCs
Alternatives: HFCs (hydrofluorocarbons) — no C–Cl bonds, so don't produce Cl radicals. However, HFCs are potent greenhouse gases.
NO_x radicals from aircraft also destroy ozone by a similar catalytic cycle: NO + O₃ → NO₂ + O₂; NO₂ + O → NO + O₂

Greenhouse Effect & Climate Change Chemistry

Greenhouse Effect

Greenhouse gases absorb infrared radiation emitted by the Earth's surface and re-emit it in all directions, warming the atmosphere. This is a natural process, but enhanced by human activity.

Greenhouse gases absorb IR radiation because their bonds vibrate at frequencies matching IR wavelengths. A molecule must have a changing dipole moment during vibration to absorb IR — this is why CO₂ and H₂O are greenhouse gases but N₂ and O₂ are not.

Greenhouse Gas	Formula	Sources	Relative Effect
Carbon dioxide	CO₂	Burning fossil fuels, deforestation	1 (reference)
Methane	CH₄	Agriculture (livestock, rice), landfill, natural gas	~25× CO₂
Water vapour	H₂O	Evaporation (natural)	Variable
Nitrous oxide	N₂O	Fertilisers, combustion	~300× CO₂
CFCs/HFCs	Various	Refrigerants, industry	~1000–10000× CO₂

Key Combustion Equations

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

C₈H₁₈(l) + 12½O₂(g) → 8CO₂(g) + 9H₂O(g) (octane — petrol)

C(s) + O₂(g) → CO₂(g) (coal combustion)

Green Chemistry Principles

Green chemistry aims to reduce the environmental impact of chemical processes. The 12 principles include:

Prevention: better to prevent waste than treat it after
Atom economy: maximise incorporation of all atoms into the desired product
Less hazardous synthesis: use and generate substances with little or no toxicity
Use of renewable feedstocks: raw materials from plant-based sources where possible
Catalysis: use catalysts rather than stoichiometric reagents (reduces waste, lowers energy)
Reduce derivatives: minimise unnecessary protection/deprotection steps
Energy efficiency: conduct reactions at ambient temperature and pressure where possible

Atom Economy

The percentage of atoms in the reactants that end up in the desired product. A measure of efficiency — higher is better.

Atom Economy Formula

% atom economy = (Mr of desired product / ΣMr of all products) × 100

Addition reactions have 100% atom economy (only one product formed)
Substitution and elimination reactions always have <100% atom economy (by-products formed)

Calculate the atom economy for making ethanol by fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

Step 1 — Identify Mr values Mr(C₆H₁₂O₆) = 180. Desired product = 2 × C₂H₅OH = 2 × 46 = 92.
All products: 2 × C₂H₅OH + 2 × CO₂ = 92 + 88 = 180.

Step 2 — Calculate atom economy % atom economy = (92 / 180) × 100 = 51.1%
Nearly half the mass of the reactant becomes CO₂ (waste).

Sustainability in Chemical Processes

Use of catalysts: reduces activation energy → less energy needed, lower temperatures
Recycling unreacted starting materials: in reversible reactions, unreacted feed is separated and recirculated
Using renewable energy to power chemical plants (solar, wind)
Carbon capture and storage (CCS): capturing CO₂ from industrial processes and storing underground
Biofuels: ethanol from fermentation of biomass — carbon-neutral (CO₂ absorbed during growth equals CO₂ released on burning)
Hydrogen fuel cells: 2H₂ + O₂ → 2H₂O — only product is water; zero emissions at point of use

Exam Tip

Atom economy and percentage yield are different things! Atom economy is a theoretical measure based on the equation. Percentage yield measures how much desired product was actually obtained in practice. A reaction can have 100% atom economy but low percentage yield.

▶ Titration Technique Practical

Step-by-Step Titration

Rinse the burette with the solution it will contain (e.g. acid), then fill and record the initial reading to the nearest 0.05 cm³.
Use a pipette (rinsed with the solution) to transfer exactly 25.0 cm³ of the other solution into a conical flask.
Add 2–3 drops of a suitable indicator (not too much — it affects the result).
Perform a rough titration: add acid quickly until the colour changes. Record the final burette reading.
Repeat accurately: add acid dropwise near the endpoint. Swirl the flask continuously.
Record concordant results (within 0.10 cm³ of each other) and calculate the mean titre.

Exam Tip

Use a conical flask (not a beaker) — easier to swirl. Rinse the conical flask with distilled water only (not the solution) — this does not affect the number of moles. A white tile under the flask helps detect colour change.

▶ Calorimetry Technique Practical

Simple Calorimetry

Measure a known volume of solution (e.g. 50 cm³) into a polystyrene cup (insulating).
Record the initial temperature every minute for 3 minutes (establishes baseline).
Add the reactant (e.g. solid, or second solution) at a noted time.
Stir continuously and record temperature every minute for several minutes.
Plot temperature vs time and extrapolate back to the time of mixing to find the true ΔT.
Calculate q = mcΔT, then calculate ΔH per mole.

Sources of error: heat loss to surroundings, incomplete reactions, heat capacity of the container not accounted for, imprecise temperature readings.

▶ Distillation & Reflux Practical

Distillation

Used to separate and collect a liquid product. The mixture is heated, the vapour rises through a condenser, and the distillate is collected in a separate flask. Used to collect an aldehyde before it is further oxidised.

Reflux

Used to heat a reaction mixture for a prolonged period without losing volatile reactants or products. A vertical condenser is fitted above the flask — vapour condenses and drips back in. Used for oxidation of primary alcohols to carboxylic acids, and for hydrolysis of halogenoalkanes.

▶ Tests for Gases Practical

Gas	Test	Positive Result
H₂	Burning splint	Squeaky pop
O₂	Glowing splint	Relights
CO₂	Bubble through limewater Ca(OH)₂(aq)	Turns milky (white precipitate of CaCO₃)
Cl₂	Damp litmus paper	Bleaches (turns white)
NH₃	Damp red litmus paper	Turns blue (alkaline gas)

▶ Tests for Ions Practical

Flame Tests (Cations)

Ion	Flame Colour
Li⁺	Crimson red
Na⁺	Yellow/orange
K⁺	Lilac
Ca²⁺	Orange-red
Ba²⁺	Green
Cu²⁺	Blue-green

Test for Halide Ions

Add dilute HNO₃, then AgNO₃(aq): Cl⁻ → white AgCl; Br⁻ → cream AgBr; I⁻ → yellow AgI.

Test for Sulfate Ions

Add dilute HCl, then BaCl₂(aq). White precipitate of BaSO₄ confirms SO₄²⁻.

Sulfate test

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) [white precipitate]

Test for Carbonate Ions

Add dilute HCl. Effervescence (bubbles). Gas turns limewater milky → CO₂ confirmed → CO₃²⁻ present.

▶ Key Equations & Constants Data

Amounts

n = m / Mr (moles = mass / molar mass)

c = n / V (concentration = moles / volume in dm³)

PV = nRT (R = 8.314 J K⁻¹ mol⁻¹)

n = V / 24.0 (molar volume at RTP, dm³)

Energetics

q = mcΔT (c = 4.18 J g⁻¹ K⁻¹ for water)

ΔH = Σ(bonds broken) − Σ(bonds formed)

ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants)

ΔG = ΔH − TΔS

Acid-Base

pH = −log₁₀[H⁺]

[H⁺] = 10⁻ᵖᴴ

Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ (25°C)

Ka = [H⁺][A⁻] / [HA]

pH = pKa + log₁₀([A⁻]/[HA])

Kinetics

Rate = k[A]ᵐ[B]ⁿ

t½ = 0.693 / k (first order)

k = Ae^(−Ea/RT)

ln k = −Ea/(RT) + ln A

Equilibrium

Kc = [products]ⁿ / [reactants]ᵐ

Kp = (partial pressures of products) / (partial pressures of reactants)

p_A = χ_A × P_total

Electrochemistry

E°_cell = E°(cathode) − E°(anode)

% yield = (actual / theoretical) × 100

Atom economy = (Mr desired / ΣMr all products) × 100

▶ Key Element Data Data

Element	Symbol	Z	Ar
Hydrogen	H	1	1.0
Carbon	C	6	12.0
Nitrogen	N	7	14.0
Oxygen	O	8	16.0
Fluorine	F	9	19.0
Sodium	Na	11	23.0
Magnesium	Mg	12	24.3
Aluminium	Al	13	27.0
Silicon	Si	14	28.1
Phosphorus	P	15	31.0
Sulfur	S	16	32.1
Chlorine	Cl	17	35.5
Potassium	K	19	39.1
Calcium	Ca	20	40.1
Iron	Fe	26	55.8
Copper	Cu	29	63.5
Zinc	Zn	30	65.4
Bromine	Br	35	79.9
Silver	Ag	47	107.9
Iodine	I	53	126.9
Barium	Ba	56	137.3

▶ Standard Electrode Potentials Data

Half-Equation (Reduction)	E° / V
Li⁺ + e⁻ → Li	−3.04
K⁺ + e⁻ → K	−2.92
Ca²⁺ + 2e⁻ → Ca	−2.87
Na⁺ + e⁻ → Na	−2.71
Mg²⁺ + 2e⁻ → Mg	−2.37
Al³⁺ + 3e⁻ → Al	−1.66
Zn²⁺ + 2e⁻ → Zn	−0.76
Fe²⁺ + 2e⁻ → Fe	−0.44
2H⁺ + 2e⁻ → H₂	0.00
Cu²⁺ + 2e⁻ → Cu	+0.34
I₂ + 2e⁻ → 2I⁻	+0.54
Ag⁺ + e⁻ → Ag	+0.80
Br₂ + 2e⁻ → 2Br⁻	+1.07
Cl₂ + 2e⁻ → 2Cl⁻	+1.36
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O	+1.51
F₂ + 2e⁻ → 2F⁻	+2.87

▶ IR Absorption Table Data

Bond	Wavenumber / cm⁻¹	Absorption Type
O–H (alcohol, free)	3580–3670	Sharp
O–H (alcohol, H-bonded)	3200–3550	Broad
O–H (carboxylic acid)	2500–3300	Very broad
N–H (amine)	3300–3500	Medium
C–H (alkane)	2850–2960	Strong
C–H (alkene)	3010–3100	Medium
C≡N	2200–2260	Medium
C=O (aldehyde/ketone)	1680–1750	Strong, sharp
C=O (carboxylic acid)	1700–1725	Strong
C=O (ester)	1735–1750	Strong
C=C (alkene)	1620–1680	Medium
C–O	1000–1300	Strong

▶ Reactivity Series Data

Most reactive → least reactive:

Reactivity Series of Metals

K > Na > Ca > Mg > Al > Zn > Fe > Sn > Pb > H > Cu > Ag > Au > Pt

A-Level Chemistry — CCEA GCE