Free flashcards · AQA A Level · 7405

Chemistry flashcards

319 cards covering every topic of the 7405 syllabus (For teaching from September 2015). Study one topic or mix them all, turn each card, and mark yourself honestly.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 319 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

AS Topic 1: Atomic structure

Relative mass and relative charge of a proton, a neutron and an electron: Proton: mass 1, charge +1. Neutron: mass 1, charge 0. Electron: mass 1/1840 (often given as 1/1836), charge −1.
Atomic number and mass number: Atomic number (proton number): the number of protons in the nucleus. Mass number: the number of protons plus neutrons.
What did Rutherford’s gold foil experiment show, and which model did it replace?: Most alpha particles passed straight through, a few were deflected and a tiny fraction bounced back. So the atom has a tiny, dense, positive nucleus and is mostly empty space. It replaced the plum pudding model.
What did Bohr and Chadwick each add to the model of the atom?: Bohr: electrons occupy fixed energy levels (shells). Chadwick (1932): the nucleus also contains neutrons, neutral particles with a mass similar to the proton.
How many protons, neutrons and electrons are in ²⁴Mg²⁺ (atomic number 12)?: 12 protons, 12 neutrons (24 − 12) and 10 electrons (two lost to give the 2+ charge).
Isotopes, and why isotopes of an element react in the same way: Atoms of the same element with different numbers of neutrons. Chemistry depends on the electron arrangement, which is the same, so their chemical properties are identical.
The five stages of time-of-flight (TOF) mass spectrometry, in order: Ionisation, acceleration, ion drift, detection, data analysis.
Electron impact vs electrospray ionisation: Electron impact: a high-energy electron knocks an electron out, forming X⁺ (elements and low-Mr compounds). Electrospray: the sample gains a proton, forming XH⁺ (large molecules that would otherwise fragment).
In TOF mass spectrometry, why do lighter ions reach the detector first?: All ions are accelerated to the same kinetic energy. Since KE = ½mv², lighter ions have a higher speed, so they cross the drift region in less time.
An electrospray spectrum shows a molecular ion peak at m/z 181. What is the Mr?: 180. Electrospray forms XH⁺, so the m/z is one more than the Mr: subtract 1.
Magnesium is 79.0% ²⁴Mg, 10.0% ²⁵Mg and 11.0% ²⁶Mg. Calculate its Ar.: (24 × 79.0 + 25 × 10.0 + 26 × 11.0) ÷ 100 = 2432 ÷ 100 = 24.3
Electron configurations of Fe and Fe²⁺: Fe: [Ar]3d⁶4s². Fe²⁺: [Ar]3d⁶. The 4s electrons are lost first, even though 4s fills before 3d.
The two Period 4 exceptions to the filling order: Cr = [Ar]3d⁵4s¹ and Cu = [Ar]3d¹⁰4s¹, because a half-filled or filled d sub-shell gives extra stability.
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⁻
Why does first ionisation energy decrease down a group?: More shells, so a larger atomic radius and more shielding. The outer electron is held less tightly, which outweighs the increase in nuclear charge.
Why is the first ionisation energy of Al lower than that of Mg?: Aluminium’s outer electron is in a 3p sub-shell, which is higher in energy and slightly shielded by the 3s electrons, so it is easier to remove.
Why is the first ionisation energy of S lower than that of P?: Sulfur has two electrons paired in one 3p orbital, and the repulsion between them makes one easier to remove.
How do successive ionisation energies show the group of an element?: A large jump occurs when an electron is removed from a shell closer to the nucleus. The number of electrons removed before the first big jump is the number of outer electrons, which gives the group.

AS Topic 2: Amount of substance

Relative atomic mass (Ar): The weighted mean mass of an atom of an element, accounting for isotopic abundance, on a scale where an atom of carbon-12 has a mass of exactly 12.
Relative molecular mass (Mr): The weighted mean mass of a molecule on the carbon-12 scale; the sum of the Ar values of its atoms. Used for covalently bonded molecules.
Relative formula mass is used for…: Ionic compounds (e.g. NaCl) — use this term, not Mr.
Why do Ar and Mr have no units?: They are ratios: masses relative to one-twelfth of the mass of a carbon-12 atom, not actual masses.
The three mole equations: moles = mass ÷ Mr moles = concentration (mol dm⁻³) × volume (dm³) number of particles = moles × Avogadro constant
The Avogadro constant: The number of particles in one mole of a substance, about 6.02 × 10²³ mol⁻¹. You do not need to recall its value; it is given when needed.
How many molecules are in 8.8 g of CO₂ (Mr 44)? Take L = 6.02 × 10²³ mol⁻¹.: 8.8 ÷ 44 = 0.20 mol; 0.20 × 6.02 × 10²³ = 1.2 × 10²³ molecules (2 s.f.).
How many ions are in 250 cm³ of 0.20 mol dm⁻³ NaCl? Take L = 6.02 × 10²³ mol⁻¹.: 0.20 × 0.250 = 0.050 mol NaCl → 3.0 × 10²² formula units. Each gives 2 ions, so 6.0 × 10²² ions.
To how many significant figures should a calculated answer be given?: To the number of significant figures in the least precise (least accurate) piece of data used.
The ideal gas equation and its SI units: pV = nRT p in Pa, V in m³, n in mol, T in K, R in J K⁻¹ mol⁻¹ (8.31, given in the question).
Convert to SI: 600 cm³, 2.5 dm³, 101 kPa, 25 °C: 600 cm³ = 6.00 × 10⁻⁴ m³ 2.5 dm³ = 2.5 × 10⁻³ m³ 101 kPa = 1.01 × 10⁵ Pa 25 °C = 298 K
0.0250 mol of argon is in 600 cm³ at 300 K. Find the pressure in kPa (R = 8.31 J K⁻¹ mol⁻¹).: p = nRT ÷ V = (0.0250 × 8.31 × 300) ÷ (6.00 × 10⁻⁴) = 1.04 × 10⁵ Pa = 104 kPa
How do you find the Mr of a volatile liquid with a heated gas syringe?: Inject a weighed mass of liquid into a syringe above its boiling point and read the vapour volume. Then M = mRT ÷ pV, with T the syringe temperature and p the room pressure.
Classic unit trap when converting cm³ for pV = nRT: cm³ to m³ is × 10⁻⁶. Dividing by 1000 only gives dm³.
Empirical formula and molecular formula: Empirical: the simplest whole-number ratio of atoms of each element in a compound. Molecular: the actual number of atoms of each element in the formula.
Method: empirical formula from percentage composition: Treat percentages as masses in 100 g. Divide each by Ar, divide all by the smallest, then clear fractions (× 2 for .5, × 3 for .33 or .67, × 4 for .25 or .75).
A carbide is 75.0% Al and 25.0% C by mass (Al 27.0, C 12.0). Find its empirical formula.: Al 75.0 ÷ 27.0 = 2.778; C 25.0 ÷ 12.0 = 2.083. Ratio 1.333 : 1, × 3 gives 4 : 3. Al₄C₃.
How do you get the molecular formula from the empirical formula?: Multiplier = Mr ÷ empirical formula mass. Multiply every subscript by it. Composition alone cannot do this: CF₂ and C₂F₄ have the same percentages.
Finding an oxide’s formula: how do you know the reaction is complete?: Heat to constant mass. Then mass of oxygen = mass of oxide − mass of metal.

AS Topic 3: Bonding

Ionic bonding: Electrostatic attraction between oppositely charged ions in a lattice.
Co-ordinate (dative covalent) bond: A shared pair of electrons with both electrons supplied by one atom. Drawn as an arrow from the donor atom to the acceptor. Examples: NH₄⁺, H₃O⁺, NH₃BF₃.
Metallic bonding, and what makes it stronger: Attraction between positive metal ions in a lattice and the delocalised electrons around them. Stronger with a bigger ion charge, smaller ions, or more delocalised electrons per atom.
Electronegativity: The ability of an atom to draw the bonding pair of electrons towards itself in a covalent bond. It increases across a period and decreases down a group; fluorine is the most electronegative.
Order of repulsion between electron pairs, and its effect: Lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Each lone pair squeezes the bond angles.
Shapes and bond angles: CH₄ or NH₄⁺, NH₃, H₂O: NH₄⁺/CH₄: tetrahedral, 109.5°. NH₃: trigonal pyramidal, about 107°. H₂O: bent, about 104.5°. Each lone pair reduces the angle.
Shapes of SF₆, XeF₄, PCl₅ and ClF₃: SF₆: octahedral, 90°. XeF₄: square planar (2 lone pairs opposite each other). PCl₅: trigonal bipyramidal, 120° and 90°. ClF₃: T-shaped (2 lone pairs).
Why is CH₂Cl₂ polar but BF₃ non-polar?: Both have polar bonds. BF₃ is symmetrical (trigonal planar) so the dipoles cancel. CH₂Cl₂ is tetrahedral with two different outer atoms, so its dipoles do not cancel.
The three intermolecular forces, weakest to strongest: Induced dipole–dipole (van der Waals, London) between all molecules; permanent dipole–dipole between polar molecules; hydrogen bonding (H on N, O or F to a lone pair on N, O or F).
Diamond vs graphite: bonding and conductivity: Diamond: each C makes 4 covalent bonds (tetrahedral); no free electrons, so no conduction. Graphite: each C makes 3 bonds in hexagonal layers; one delocalised electron per C, so it conducts. Weak forces between layers let them slide.
What is overcome when a simple molecular solid such as iodine melts?: Only the weak intermolecular forces (van der Waals) between molecules. The covalent bonds inside the molecules do not break.
Why is ice less dense than liquid water?: In ice each H₂O molecule is hydrogen bonded to four others in an open lattice, holding the molecules further apart than in the liquid.

AS Topic 4: Energetics

Enthalpy change, ΔH, and standard conditions: ΔH is the heat energy change measured at constant pressure. Standard conditions: 100 kPa and a stated temperature, e.g. ΔH⦵₂₉₈ for 298 K.
Standard enthalpy of combustion, ΔcH⦵: The enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions, all reactants and products in their standard states.
Standard enthalpy of formation, ΔfH⦵: The enthalpy change when 1 mole of a compound is formed from its elements under standard conditions, all reactants and products in their standard states.
Hess’s law: The enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.
Mean bond enthalpy: The enthalpy change to break 1 mole of a stated covalent bond in gaseous molecules, averaged over a range of compounds.
Calorimetry: the two equations, with units: q = mcΔT (m in g of water or solution, c in J g⁻¹ K⁻¹, ΔT in K; q in J). ΔH = −q ÷ n, with q in kJ and n the moles of the limiting reactant.
Hess’s law shortcuts with formation and with combustion data: Formation: ΔH = ΣΔfH(products) − ΣΔfH(reactants). Combustion: ΔH = ΣΔcH(reactants) − ΣΔcH(products).
Required practical 2: why extrapolate the cooling line back to the time of mixing?: To estimate the maximum temperature change, correcting for heat lost to the surroundings while the temperature was rising.
Why does a bond-enthalpy ΔH differ from a Hess’s law value?: Mean bond enthalpies are averages over many compounds, not the values in these molecules; and the method assumes all species are gases, ignoring enthalpy changes of state.
What is ΔfH⦵ of O₂(g), and of Br₂(g)?: O₂(g): zero, as an element in its standard state. Br₂(g): not zero, because bromine’s standard state is liquid.

AS Topic 5: Kinetics

Activation energy, Ea: The minimum energy that colliding particles must have for a reaction to occur.
Rate of reaction: The change in amount or concentration of a reactant or product per unit time.
Catalyst, and how it works: A substance that increases the rate of a chemical reaction without being changed in chemical composition or amount. It provides an alternative reaction route of lower activation energy.
Maxwell–Boltzmann distribution: three drawing rules: Starts at the origin; the curve never meets the energy axis at high energy; the area under it equals the total number of molecules.
How does the curve change at a higher temperature?: The peak is lower and further right, with the same area. Many more molecules have E ≥ Ea.
Main reason a small temperature rise greatly increases rate: A much larger proportion of molecules have energy ≥ Ea, so many more collisions are successful. Slightly more frequent collisions is only a minor reason.
Raising concentration or gas pressure: what changes and what does not?: Collision frequency rises (more particles per unit volume). The proportion of particles with E ≥ Ea does not change, and the Maxwell–Boltzmann curve does not move.
Catalyst on a Maxwell–Boltzmann diagram: One curve, unchanged. A lower Ea(cat) marked to the left of Ea, so a greater proportion of molecules have enough energy to react.
Required practical 3 (thiosulfate + acid): what measures the rate?: 1/t, where t is the time for the sulfur precipitate to hide a cross. The same amount of sulfur forms each run, so rate ∝ 1/t.

AS Topic 6: Chemical equilibria, Le Chatelier's principle and Kc

Dynamic equilibrium: In a closed system, the forward and reverse reactions proceed at equal rates and the concentrations of reactants and products remain constant (not equal).
Le Chatelier’s principle: If a factor affecting a system at equilibrium is changed, the position of equilibrium shifts to oppose the change.
Effect of increasing pressure on an equilibrium: The position shifts to the side with fewer moles of gas, reducing the pressure. No shift if both sides have the same number of moles of gas.
Effect of increasing temperature on an equilibrium: The position shifts in the endothermic direction, absorbing heat.
Effect of a catalyst on an equilibrium: No shift in position and no change in yield. It speeds up the forward and reverse reactions equally, so equilibrium is reached faster.
Why are moderate temperatures used for an exothermic industrial equilibrium such as the Haber process?: Low temperature gives a higher yield but too slow a rate. A moderate temperature is a compromise: an acceptable yield at an acceptable rate.
Units of Kc for 2A(g) ⇌ B(g): Kc = [B] ÷ [A]², so units are (mol dm⁻³) ÷ (mol dm⁻³)² = mol⁻¹ dm³.
What changes the value of Kc?: Only temperature. Concentration, pressure and catalysts change the position or rate, not Kc.
Forward reaction exothermic, temperature raised: effect on Kc: Equilibrium shifts left, so Kc decreases (and so does the yield).
2A ⇌ B: start with 1.00 mol A in 2.00 dm³; 0.200 mol B at equilibrium. Find Kc.: A used 0.400 mol, A left 0.600 mol. [A] = 0.300, [B] = 0.100 mol dm⁻³. Kc = 0.100 ÷ 0.300² = 1.11 mol⁻¹ dm³.

AS Topic 7: Oxidation, reduction and redox equations

Oxidation and reduction (electron definitions): Oxidation: the process of electron loss. Reduction: the process of electron gain.
Oxidising agent and reducing agent: Oxidising agent: an electron acceptor (it is reduced). Reducing agent: an electron donor (it is oxidised).
Oxidation state of H in NaH, and of O in H₂O₂: H in metal hydrides: −1. O in peroxides: −1.
Oxidation state of Cl in ClO₃⁻: x + 3(−2) = −1, so x = +5.
Oxidation states of Mn in MnO₄⁻ and Cr in Cr₂O₇²⁻: Mn +7 in MnO₄⁻. Cr +6 in Cr₂O₇²⁻.
Steps to balance a half-equation in acid: Balance the element that changes oxidation state; balance O with H₂O; balance H with H⁺; balance charge with e⁻.
Half-equation: NO₃⁻ to NO₂ in acid: NO₃⁻ + 2H⁺ + e⁻ → NO₂ + H₂O (N goes from +5 to +4: reduction).
How do you combine two half-equations into an overall equation?: Scale them so electrons lost = electrons gained, add, then cancel electrons and any H⁺ and H₂O that appear on both sides. Check atoms and charge.

A Level Topic 8: Thermodynamics

Enthalpy of atomisation, ΔatH⦵: The enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state, e.g. ½Cl₂(g) → Cl(g). Always positive.
First and second electron affinity of oxygen: equations and signs: First: O(g) + e⁻ → O⁻(g), exothermic. Second: O⁻(g) + e⁻ → O²⁻(g), endothermic (adding an electron to a negative ion).
Lattice enthalpy of formation vs lattice enthalpy of dissociation: Formation: 1 mole of solid ionic compound formed from its gaseous ions (negative). Dissociation: 1 mole of solid separated into its gaseous ions (positive). Same size, opposite sign.
Enthalpy of solution from lattice and hydration enthalpies: ΔsolH = lattice enthalpy of dissociation + Σ ΔhydH of all the ions (count each ion).
Born–Haber value more exothermic than the perfect ionic model value: what does it show?: Covalent character. The cation polarises the anion’s electron cloud, giving extra attraction. Greatest with a small, highly charged cation and a large anion.
When is ΔS positive?: When a solid becomes liquid or gas, a liquid becomes gas, or the number of moles of gas increases.
Gibbs free-energy equation and the feasibility condition: ΔG = ΔH − TΔS (T in K, ΔS in kJ K⁻¹ mol⁻¹ to match ΔH). A reaction is feasible when ΔG ≤ 0.
Temperature at which a reaction becomes feasible: Set ΔG = 0, so T = ΔH ÷ ΔS. With ΔH and ΔS both positive it is feasible above this T; both negative, below it.
Graph of ΔG against T: gradient and intercept: Gradient = −ΔS. Intercept at T = 0 = ΔH. It crosses ΔG = 0 where the reaction becomes feasible.
Classic unit trap in ΔG = ΔH − TΔS: Entropies are in J K⁻¹ mol⁻¹ but ΔH is in kJ mol⁻¹: divide ΔS by 1000 first.

A Level Topic 9: Rate equations

Rate equation: Rate = k[A]^m [B]^n, an experimentally determined relationship.
Order of reaction with respect to a reactant: The power to which its concentration is raised in the rate equation. The overall order is the sum of the orders.
Rate-determining step: The slowest step in a mechanism; it limits the overall rate.
Units of k for overall orders 1, 2 and 3: First: s⁻¹. Second: mol⁻¹ dm³ s⁻¹. Third: mol⁻² dm⁶ s⁻¹.
Rate = k[A][B]². [A] is halved and [B] is tripled. What happens to the rate?: ½ × 3² = 4.5, so the rate increases by a factor of 4.5.
Shapes of rate against [A] for orders 0, 1 and 2: Order 0: horizontal line. Order 1: straight line through the origin. Order 2: upward curve through the origin.
The Arrhenius equation and its straight-line form: k = A e^(−Ea/RT). ln k = −Ea/(RT) + ln A: plot ln k against 1/T; gradient = −Ea/R, intercept = ln A.
What can change the value of k?: Temperature (k rises as T rises) or adding a catalyst. Never a change in concentration.
What do the orders tell you about the rate-determining step?: Species in the rate equation take part in the rate-determining step (or a step before it). A zero-order reactant reacts after it.

A Level Topic 10: Equilibrium constant Kp for homogeneous systems

Mole fraction: x(A) = moles of A ÷ total moles of gas in the mixture. No units; all mole fractions add up to 1.
Partial pressure: p(A) = x(A) × total pressure: the pressure gas A would exert if it alone filled the container. Partial pressures add up to the total pressure.
Kp expression for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), with units in kPa: Kp = p(NH₃)² ÷ (p(N₂) × p(H₂)³). Units: kPa⁻².
Units of Kp for N₂O₄(g) ⇌ 2NO₂(g) with pressures in kPa: kPa (Δn = 2 − 1 = +1).
Effect of raising total pressure on Kp and on the position: Kp does not change at constant temperature. The position shifts to the side with fewer moles of gas.
Forward reaction endothermic, temperature raised: effect on Kp: The equilibrium shifts right, so Kp increases.
0.900 mol A, 0.300 mol B, 0.600 mol C at a total pressure of 180 kPa. Partial pressures?: Total 1.800 mol. p(A) = 0.500 × 180 = 90.0 kPa; p(B) = 30.0 kPa; p(C) = 60.0 kPa. They sum to 180 kPa.
Classic trap when finding the total moles of gas for mole fractions: Use equilibrium moles, not initial moles, and include products that started at zero.

A Level Topic 11: Electrode potentials and electrochemical cells

Standard electrode potential, E⦵: The EMF of a half-cell measured against the standard hydrogen electrode at 298 K, 100 kPa and 1.00 mol dm⁻³ solutions of ions.
Salt bridge: Usually filter paper soaked in saturated KNO₃(aq). Ions move through it to balance charge. It does not carry electrons.
The standard hydrogen electrode: Platinum electrode; H₂ gas at 100 kPa; 1.00 mol dm⁻³ H⁺(aq); 298 K. E⦵ = 0.00 V by definition. Pt(s) | H₂(g) | H⁺(aq)
Conventional cell representation: what goes on the left and the right?: Left: the negative electrode, written as an oxidation, e.g. Zn(s) | Zn²⁺(aq). Right: the positive electrode, written as a reduction, e.g. Ag⁺(aq) | Ag(s). || is the salt bridge.
EMF of a cell from electrode potentials: EMF = E⦵(right) − E⦵(left) = E⦵(positive) − E⦵(negative). Do not multiply E⦵ when a half-equation is scaled.
EMF of Zn | Zn²⁺ || Cu²⁺ | Cu (E⦵ Zn²⁺/Zn −0.76 V, Cu²⁺/Cu +0.34 V): +0.34 − (−0.76) = +1.10 V
How do you decide whether a redox reaction is feasible from E⦵ values?: The more positive half-equation goes forwards (reduction) and the other goes backwards. It is feasible if the oxidising agent’s E⦵ is more positive than the reducing agent’s.
Raising [Cu²⁺] in a Cu²⁺/Cu half-cell: effect on its electrode potential: The equilibrium Cu²⁺ + 2e⁻ ⇌ Cu shifts right, so E becomes more positive.
Alkaline hydrogen–oxygen fuel cell: electrode half-equations: Positive: O₂ + 2H₂O + 4e⁻ → 4OH⁻ Negative: H₂ + 2OH⁻ → 2H₂O + 2e⁻ Overall 2H₂ + O₂ → 2H₂O; EMF +1.23 V.
Required practical 8: why a high-resistance voltmeter?: So that negligible current flows and the reading is the EMF.

A Level Topic 12: Acids and bases

Brønsted–Lowry acid and base: An acid is a proton donor; a base is a proton acceptor.
pH: −log₁₀[H⁺].
Kw: expression, value at 298 K and units: Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K.
Pure water at a higher temperature has a pH below 7. Is it acidic?: No, it is still neutral, because [H⁺] = [OH⁻]. Kw increases with temperature because the dissociation of water is endothermic.
pH of 0.0100 mol dm⁻³ NaOH at 298 K: [H⁺] = Kw ÷ [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 0.0100 = 1.00 × 10⁻¹² mol dm⁻³, so pH = 12.00.
pH of a weak acid: the two assumptions: [H⁺] = [A⁻], and [HA] at equilibrium equals the starting concentration (only slight dissociation). So [H⁺] = √(Ka × [HA]).
pH of a buffer made from HA and its salt: [H⁺] = Ka × [HA] ÷ [A⁻] (moles can be used, as the volume cancels), then pH = −log₁₀[H⁺]. At half-neutralisation pH = pKa.
Buffer: A solution that keeps an approximately constant pH despite dilution or the addition of small amounts of acid or base.
How does an acidic buffer (CH₃COOH / CH₃COO⁻) resist added acid and added alkali?: Added H⁺: H⁺ + A⁻ → HA. Added OH⁻: OH⁻ + HA → A⁻ + H₂O. The large reservoirs of HA and A⁻ mean the ratio [HA]/[A⁻] barely changes.
Indicator for a weak acid–strong base titration, and why: Phenolphthalein (range 8.3–10.0), because the vertical section and the equivalence point are in the alkaline range.
Ka vs pKa: which way means a stronger acid?: Larger Ka means stronger acid; larger pKa means weaker acid.

AS Topic 13: Periodicity

How is an element’s block (s, p, d or f) decided?: By the sub-shell that holds its highest-energy electron.
Which block are helium and gallium in?: Helium (1s²): s block. Gallium: p block, because its last electron goes into 4p, even though it contains 3d¹⁰.
Why does atomic radius decrease from Na to Cl?: Nuclear charge increases, electrons are added to the same shell so shielding is similar, and the stronger attraction pulls the outer electrons closer.
Why does silicon have the highest melting point in Period 3?: It is macromolecular: many strong covalent bonds must be broken to melt it.
Melting points of P₄, S₈ and Cl₂: order and reason: S₈ > P₄ > Cl₂. Only van der Waals forces are overcome; more electrons per molecule (S₈ 128, P₄ 60, Cl₂ 34) give stronger forces.
Why does the melting point rise from Na to Al?: Ion charge goes +1, +2, +3, delocalised electrons per atom go 1, 2, 3 and the ions get smaller, so metallic bonding gets stronger.
Trend in first ionisation energy across Period 3: A general increase from Na (lowest) to Ar (highest), with small drops at Al and at S.

AS Topic 14: Group 2, the alkaline earth metals

Trends down Group 2 (Mg to Ba): atomic radius and first ionisation energy: Atomic radius increases (one more shell each time). First ionisation energy decreases: the outer electron is further away and more shielded.
Why does the melting point generally decrease from Ca to Ba?: Each ion is 2+, but the ions get larger, so the attraction between the ions and the delocalised electrons is weaker.
Magnesium with cold water and with steam: Cold water: very slow, Mg + 2H₂O → Mg(OH)₂ + H₂. Steam: burns with a bright white light, Mg + H₂O → MgO + H₂.
Solubility trends of Group 2 hydroxides and sulfates: Hydroxides: solubility increases down the group (Mg(OH)₂ sparingly soluble). Sulfates: solubility decreases down the group (BaSO₄ insoluble).
Test for sulfate ions, and why the acid is added: Add dilute HCl (or HNO₃), then BaCl₂(aq): a white precipitate of BaSO₄. The acid removes carbonate (and sulfite) ions, which would also give a white precipitate. Never use H₂SO₄.
Role of magnesium in titanium extraction: TiCl₄ + 2Mg → Ti + 2MgCl₂. Magnesium is oxidised, so it is the reducing agent.
Uses of Mg(OH)₂, Ca(OH)₂, CaO/CaCO₃ and BaSO₄: Mg(OH)₂: antacid. Ca(OH)₂: raises the pH of acidic soil. CaO or CaCO₃: removes SO₂ from flue gases. BaSO₄: barium meal for X-rays (insoluble, so toxic Ba²⁺ is not absorbed).

AS Topic 15: Group 7(17), the halogens

Why does the boiling point of the halogens increase down the group?: More electrons per molecule (Cl₂ 34, Br₂ 70, I₂ 106), so stronger van der Waals forces between molecules need more energy to overcome. The X–X bond does not break.
Trends down Group 7 in the oxidising ability of X₂ and the reducing ability of X⁻: Oxidising ability of the halogens decreases; reducing ability of the halide ions increases.
Chlorine water added to potassium iodide solution: result and ionic equation: Brown solution. Cl₂ + 2I⁻ → 2Cl⁻ + I₂
Concentrated H₂SO₄ with NaCl and with NaBr: observations: NaCl: steamy white fumes of HCl only (acid–base, not redox). NaBr: steamy fumes of HBr, then orange-brown fumes of Br₂, and SO₂ forms (S +6 → +4).
Reduction products of H₂SO₄ with sodium iodide, and their observations: SO₂ (S +4); S, a yellow solid (S 0); H₂S, a bad-egg smell (S −2). I₂ appears as a black solid or purple vapour.
Silver nitrate test: precipitate colours and ammonia solubility: Cl⁻: white, dissolves in dilute NH₃. Br⁻: cream, dissolves only in concentrated NH₃. I⁻: yellow, insoluble in concentrated NH₃. Acidify with dilute HNO₃ first.
Chlorine with cold, dilute NaOH: equation and type of reaction: Cl₂ + 2NaOH → NaCl + NaClO + H₂O. Disproportionation: Cl goes from 0 to −1 and to +1. The NaClO solution is bleach.
Why is chlorine added to drinking water despite being toxic?: It kills bacteria. The doses are controlled and the health benefits outweigh the toxic effects.

A Level Topic 16: Properties of Period 3 elements and their oxides

Amphoteric oxide: Reacts with both acids and bases (Al₂O₃).
Sodium vs magnesium with cold water: Na: vigorous, 2Na + 2H₂O → 2NaOH + H₂, pH 13–14. Mg: very slow, Mg + 2H₂O → Mg(OH)₂ + H₂, pH 9–10.
Equations: P₄O₁₀, SO₂ and SO₃ with water: P₄O₁₀ + 6H₂O → 4H₃PO₄ SO₂ + H₂O → H₂SO₃ SO₃ + H₂O → H₂SO₄
Why do Al₂O₃ and SiO₂ give pH 7 with water?: Both are insoluble: their giant structures are too strongly bonded to dissolve. They are not neutral oxides: Al₂O₃ is amphoteric and SiO₂ is acidic.
Why does MgO have a higher melting point than Na₂O?: Both are giant ionic, but Mg²⁺ has a higher charge and is smaller than Na⁺, so the attraction to O²⁻ is stronger.
Why does P₄O₁₀ have a low melting point?: It is simple molecular: only van der Waals forces between molecules are overcome, not covalent bonds.
Al₂O₃ with an acid and with sodium hydroxide: Al₂O₃ + 6H⁺ → 2Al³⁺ + 3H₂O Al₂O₃ + 2OH⁻ + 3H₂O → 2Al(OH)₄⁻
SiO₂ with sodium hydroxide: Only with hot, concentrated NaOH: SiO₂ + 2OH⁻ → SiO₃²⁻ + H₂O

A Level Topic 17: Transition metals

What gives transition metals (Ti–Cu) their characteristic properties, and what are the four properties?: An incomplete d sub-level in atoms or ions. Properties: complex formation, coloured ions, variable oxidation state and catalytic activity.
Ligand: A molecule or ion that forms a co-ordinate bond with a transition metal by donating a pair of electrons.
Co-ordination number: The number of co-ordinate bonds to the central metal atom or ion.
Why does replacing H₂O by Cl⁻ change the co-ordination number from 6 to 4?: Cl⁻ is larger than H₂O, so fewer fit round the metal: e.g. [Co(H₂O)₆]²⁺ + 4Cl⁻ → [CoCl₄]²⁻ + 6H₂O (octahedral to tetrahedral).
The chelate effect: Multidentate ligands replace monodentate ones. ΔH is small (similar bonds broken and made) but the number of particles increases, so ΔS is positive and ΔG = ΔH − TΔS is negative.
Why is carbon monoxide toxic?: CO replaces the O₂ that is co-ordinately bonded to Fe(II) in haem, so oxygen cannot be transported in the blood.
Why are transition metal complexes coloured?: Some visible wavelengths are absorbed, moving d electrons from the ground state to an excited state (ΔE = hν = hc/λ). The colour seen is the light transmitted or reflected.
Colours of vanadium oxidation states when vanadate(V) is reduced by zinc in acid: VO₂⁺ (+5) yellow → VO²⁺ (+4) blue → V³⁺ (+3) green → V²⁺ (+2) violet.
Manganate(VII) titrations: mole ratio with Fe²⁺ and with ethanedioate: MnO₄⁻ : Fe²⁺ = 1 : 5. MnO₄⁻ : C₂O₄²⁻ = 2 : 5 (the mixture is warmed). The end point is the first permanent pale pink.
Heterogeneous vs homogeneous catalyst, with an example of each: Heterogeneous: a different phase; works at surface active sites (V₂O₅ in the Contact process, Fe in the Haber process). Homogeneous: the same phase; works through an intermediate (Fe²⁺ in S₂O₈²⁻ + I⁻).
Autocatalyst: A product of the reaction that catalyses it, e.g. Mn²⁺ in the reaction of MnO₄⁻ with C₂O₄²⁻.

A Level Topic 18: Reactions of ions in aqueous solution

Metal-aqua ion: A metal ion with six water ligands, each bonded by a co-ordinate bond from oxygen, for example [Cu(H₂O)₆]²⁺.
Why are [M(H₂O)₆]³⁺ solutions more acidic than [M(H₂O)₆]²⁺ solutions?: M³⁺ has a higher charge and smaller radius, so a higher charge/size ratio. It is more polarising, weakening the O–H bonds in the water ligands, so H⁺ is released more easily.
Fe²⁺, Cu²⁺, Fe³⁺ and Al³⁺ with a little NaOH(aq): precipitate colours: Fe²⁺: green. Cu²⁺: pale blue. Fe³⁺: brown. Al³⁺: white.
Which hydroxide precipitate dissolves in excess NaOH, and which in excess NH₃?: Excess NaOH: aluminium hydroxide only (amphoteric), giving a colourless solution of [Al(H₂O)₂(OH)₄]⁻. Excess NH₃: copper hydroxide only, giving deep blue [Cu(NH₃)₄(H₂O)₂]²⁺.
Sodium carbonate with 2+ aqua ions vs 3+ aqua ions: 2+ (Fe²⁺, Cu²⁺): a metal carbonate precipitates (FeCO₃, CuCO₃). 3+ (Fe³⁺, Al³⁺): the hydroxide precipitates and CO₂ is given off (effervescence).
Equation: [Fe(H₂O)₆]³⁺ with ammonia (a few drops): [Fe(H₂O)₆]³⁺ + 3NH₃ → [Fe(H₂O)₃(OH)₃] + 3NH₄⁺ (brown precipitate).
Why does an iron(II) hydroxide precipitate darken on standing in air?: Iron(II) is oxidised to iron(III) by oxygen in the air, so the green precipitate turns brown.
Why do the neutral hydroxides precipitate?: They carry no charge (e.g. [Cu(H₂O)₄(OH)₂]), so they are not stabilised by water and come out of solution.

AS Topic 19: Introduction to organic chemistry

Homologous series: four features: Same general formula; successive members differ by CH₂; same functional group, so similar chemical properties; a gradual trend in physical properties.
Functional group: An atom or group of atoms responsible for the typical reactions of a compound.
Name CH₃COCH₂CH₂CH₃ and CH₂=CHCH(OH)CH₃: Pentan-2-one. But-3-en-2-ol (–OH takes the suffix and the lower number).
Structural isomers vs stereoisomers: Structural isomers: same molecular formula, different structural formula. Stereoisomers: same structural formula, different arrangement of atoms in space.
The three types of structural isomer, with examples: Chain: butane / methylpropane. Position: 1-chloropropane / 2-chloropropane. Functional group: hex-1-ene / cyclohexane.
Why do E–Z isomers exist, and when are they possible?: Restricted rotation about the planar C=C bond. Each carbon of the C=C must carry two different groups.
How do you assign E or Z?: Rank the two groups on each C=C carbon by atomic number (CIP rules). Higher-priority groups on the same side: Z. On opposite sides: E.
Curly arrow: Shows the movement of an electron pair. It starts at a lone pair or a covalent bond and ends where the new bond forms or at the atom that keeps the pair.
E/Z vs cis/trans: The specification uses E and Z with CIP priorities. Z is not always the "both H on one side" form — apply the rules each time.

AS Topic 20: Alkanes

How does fractional distillation of crude oil separate it?: The vapour rises through a column that is hot at the bottom and cool at the top. Each hydrocarbon condenses where the temperature falls below its boiling point; short chains collect near the top.
Why do longer-chain alkanes have higher boiling points?: More electrons and more surface contact, so stronger van der Waals forces between molecules, which need more energy to overcome.
Thermal vs catalytic cracking: conditions and main products: Thermal: high pressure, high temperature; high percentage of alkenes. Catalytic: slight pressure, high temperature, zeolite catalyst; motor fuels and aromatic hydrocarbons.
Economic reasons for cracking: Distillation gives a surplus of long-chain fractions; demand is higher for short-chain fractions such as petrol, which are worth more. Cracking also makes alkenes for polymers and other chemicals.
Where do NO and SO₂ from engines and power stations come from?: NO: nitrogen and oxygen from the air reacting at high engine temperature (N₂ + O₂ → 2NO). SO₂: sulfur impurities in the fuel.
What does a catalytic converter do? Give one equation.: Converts CO, NOₓ and unburned hydrocarbons into CO₂, N₂ and H₂O. For example 2CO + 2NO → 2CO₂ + N₂.
Chlorination of methane: conditions and the two propagation steps: UV light; free-radical substitution. Cl• + CH₄ → •CH₃ + HCl •CH₃ + Cl₂ → CH₃Cl + Cl•
Initiation and termination steps in the chlorination of methane: Initiation: Cl₂ → 2Cl• Termination (any two radicals): Cl• + Cl• → Cl₂; •CH₃ + Cl• → CH₃Cl; •CH₃ + •CH₃ → C₂H₆

AS Topic 21: Halogenoalkanes

Why do nucleophiles attack halogenoalkanes?: The halogen is more electronegative than carbon, so the C–X bond is polar: C is δ+ and attracts the lone pair of a nucleophile.
Nucleophilic substitution: reagents and conditions for OH⁻, CN⁻ and NH₃: OH⁻: NaOH or KOH(aq), warm → alcohol. CN⁻: KCN in aqueous ethanol, reflux → nitrile (one more carbon). NH₃: excess NH₃ in ethanol, heated in a sealed container → primary amine.
Order of reactivity of halogenoalkanes, and the reason: Iodo > bromo > chloro > fluoro. The C–X bond enthalpy decreases from C–F to C–I, and the weaker bond breaks faster. Polarity predicts the opposite order.
Why is excess ammonia used to make a primary amine?: The amine product has a lone pair and can itself act as a nucleophile; excess NH₃ reduces further substitution.
Substitution vs elimination with KOH: conditions that favour each: Substitution (alcohol): aqueous, warm; OH⁻ acts as a nucleophile. Elimination (alkene): ethanol, hot; OH⁻ acts as a base.
Name the product of 1-bromopropane with KCN: Butanenitrile, CH₃CH₂CH₂CN (the C of C≡N is counted).
How do chlorine radicals destroy ozone? Give the equations.: UV breaks C–Cl in CFCs, releasing Cl•. Cl• + O₃ → ClO• + O₂ ClO• + O₃ → 2O₂ + Cl• Cl• is regenerated, so it acts as a catalyst.
Hydrolysis with aqueous silver nitrate: which halogenoalkane gives a precipitate first?: The iodoalkane (yellow AgI), then bromo (cream AgBr), then chloro (white AgCl), following C–X bond enthalpy.

AS Topic 22: Alkenes

Why do alkenes react with electrophiles?: The C=C bond has two shared pairs between the same two carbons, so it is a centre of high electron density that attracts electron-deficient species.
Electrophile: An electron pair acceptor, e.g. H⁺, Hδ+ in HBr, or Brδ+ in Br₂.
Test for a C=C bond: Shake with bromine water at room temperature: orange → colourless for an alkene. An alkane stays orange.
Carbocation stability order, and why: Tertiary > secondary > primary. More alkyl groups release electron density and spread the positive charge.
Major product of 2-methylpropene + HBr: 2-bromo-2-methylpropane, formed via the more stable tertiary carbocation (CH₃)₃C⁺.
Product of ethene with cold concentrated sulfuric acid: Ethyl hydrogensulfate, CH₃CH₂OSO₂OH. The hydrogensulfate bonds to carbon through its negative O atom.
Why are polyalkenes unreactive?: The chains are saturated, with strong, non-polar C–C and C–H bonds and no C=C for electrophiles to attack.
How does a plasticiser make PVC flexible?: Its small molecules get between the polymer chains, push them apart and weaken the intermolecular forces, so the chains slide more easily.

AS Topic 23: Alcohols

Hydration of ethene: equation and conditions: CH₂=CH₂ + H₂O → CH₃CH₂OH. Steam, about 300 °C, high pressure, acid catalyst such as phosphoric acid on a solid support.
Fermentation of glucose: equation and conditions: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Yeast enzymes, around 35 °C, anaerobic, aqueous.
Fermentation temperature: Too cold → slow; too hot → yeast enzymes denature. About 35 °C is a compromise.
Why must fermentation be done without air?: Oxygen would oxidise ethanol to ethanoic acid, and yeast would respire aerobically to CO₂ and water.
Why is bioethanol only "carbon neutral" on paper?: 6 CO₂ are absorbed in photosynthesis and 6 released by fermentation (2) and combustion (4). But fossil fuels used for farm machinery, fertiliser, distillation and transport release more CO₂.
Primary, secondary and tertiary alcohols: The carbon bonded to –OH is bonded to one, two or three other carbons respectively (methanol counts as primary).
Oxidising agent for alcohols and its colour change: Acidified potassium dichromate(VI), K₂Cr₂O₇ / H₂SO₄: orange → green (Cr₂O₇²⁻ → Cr³⁺).
Primary alcohol: how do you get the aldehyde rather than the carboxylic acid?: Aldehyde: distil it off as it forms (it has no hydrogen bonding, so it boils lowest). Carboxylic acid: heat under reflux with excess dichromate.
Products of oxidising a secondary and a tertiary alcohol: Secondary: a ketone. Tertiary: no reaction (stays orange), because there is no H on the carbon bonded to –OH.
Equation for the oxidation of propan-1-ol to propanal: CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O
Tollens’ and Fehling’s results with an aldehyde: Tollens’: silver mirror. Fehling’s: blue → red precipitate. A ketone gives no change with either. Warm in a water bath.
Dehydration of an alcohol: reagent and conditions: Heat with concentrated sulfuric acid or concentrated phosphoric acid. An alkene and water form.
Outline the three stages of the mechanism for alcohol elimination: A lone pair on O takes H⁺, forming R–OH₂⁺; the C–O bond breaks and H₂O leaves, giving a carbocation; electrons from an adjacent C–H bond form C=C and H⁺ is released.
Which alkenes form when 2-methylbutan-2-ol is dehydrated?: 2-methylbut-1-ene and 2-methylbut-2-ene. Neither shows E–Z isomerism.
Why is making ethene from ethanol useful?: Ethanol from fermentation → ethene → poly(ethene), so addition polymers can be made without crude oil.
Required practical 5: purifying cyclohexene from the crude distillate: Shake with sodium carbonate solution in a separating funnel (release CO₂), discard the aqueous layer, wash with water, dry with anhydrous CaCl₂ or MgSO₄, then redistil.

AS Topic 24: Organic analysis

Required practical 6: test for a carboxylic acid: Add Na₂CO₃ or NaHCO₃: effervescence; the CO₂ turns limewater cloudy.
How does acidified dichromate distinguish alcohols?: Primary and secondary alcohols turn it from orange to green. A tertiary alcohol leaves it orange.
In a test sequence, why test for an aldehyde before using dichromate?: Aldehydes are also oxidised by dichromate (orange → green), so an aldehyde would be mistaken for an alcohol.
How does high-resolution mass spectrometry find a molecular formula?: Molecules with the same whole-number Mr have different precise masses. The m/z of the molecular ion to four decimal places matches only one formula.
IR absorptions: alcohol O–H vs carboxylic acid O–H: Alcohol O–H: 3230–3550 cm⁻¹, broad. Acid O–H: 2500–3000 cm⁻¹, very broad.
IR absorption range for C=O: 1680–1750 cm⁻¹, strong and sharp.
Fingerprint region: The region below about 1500 cm⁻¹, with a pattern unique to each molecule. Comparing it with a known spectrum identifies the compound.
Why do CO₂, methane and water vapour contribute to global warming?: Their bonds (C=O, C–H, O–H) absorb infrared radiation emitted by the Earth’s surface, and they re-emit energy, some back towards the surface.

A Level Topic 25: Optical isomerism

Chiral centre (asymmetric carbon): A carbon atom bonded to four different atoms or groups.
Enantiomers: Non-superimposable mirror images that differ in their effect on plane-polarised light.
Racemic mixture (racemate): A mixture of equal amounts of the two enantiomers. It is optically inactive.
Why is a racemate optically inactive?: Each enantiomer rotates the plane of plane-polarised light by the same angle in opposite directions, and there are equal amounts of each, so the effects cancel.
Why does a reaction at a planar group give a racemate?: The planar group (C=O or a carbocation) can be attacked from either side with equal probability, giving equal amounts of the two enantiomers.
Which carbon in butan-2-ol is chiral, and what are its groups?: C2: it carries H, OH, CH₃ and C₂H₅.
Optically active vs racemate: A single enantiomer is optically active; an equal mixture is not, even though every molecule in it is chiral.

A Level Topic 26: Aldehydes and ketones

Reducing agent for aldehydes and ketones, and the products: NaBH₄ in aqueous solution. Aldehyde → primary alcohol; ketone → secondary alcohol. Nucleophilic addition.
Equation for reducing propanal with NaBH₄: CH₃CH₂CHO + 2[H] → CH₃CH₂CH₂OH (propan-1-ol)
Nucleophile in NaBH₄ reduction, and the mechanism outline: H⁻. It attacks the δ+ carbonyl carbon, the C=O π electrons move to O giving an alkoxide, then O⁻ is protonated by water (or H⁺).
Product of ethanal with KCN followed by dilute acid: 2-hydroxypropanenitrile, CH₃CH(OH)CN, by nucleophilic addition of CN⁻. It forms as a racemic mixture.
Hazards of KCN, and why KCN is used rather than HCN: KCN is highly toxic and acid releases toxic, volatile HCN: use a fume cupboard and gloves. HCN is hard to handle and, as a weak acid, gives few CN⁻ ions; KCN gives plenty.
Which carbonyl compounds give a racemic mixture with KCN?: Aldehydes (except methanal) and unsymmetrical ketones. Methanal and symmetrical ketones such as propanone give no chiral centre.
Why are ketones not oxidised by Tollens’ or Fehling’s?: The C=O carbon has no H, so oxidation would need a C–C bond to break.
Mole ratios: aldehyde to silver (Tollens’) and aldehyde to Cu₂O (Fehling’s): 1 mol aldehyde → 2 mol Ag. 1 mol aldehyde → 1 mol Cu₂O.

A Level Topic 27: Carboxylic acids and derivatives

Name the ester CH₃CH₂CH₂COOCH₃: Methyl butanoate (from butanoic acid and methanol). The alcohol part is named first.
Esterification: reagents and conditions: Carboxylic acid + alcohol, warm, with a concentrated H₂SO₄ catalyst. Reversible: CH₃COOH + CH₃OH ⇌ CH₃COOCH₃ + H₂O
Acid vs alkaline hydrolysis of an ester: Acid (dilute acid, reflux): reversible, gives the carboxylic acid. Alkaline (NaOH(aq), reflux): goes to completion, gives the carboxylate salt.
Soap and biodiesel from fats and oils: Soap: fat + 3NaOH → 3 sodium salts of long-chain acids + glycerol. Biodiesel: oil + 3CH₃OH (catalyst) → 3 methyl esters + glycerol.
Four uses of esters: Solvents, plasticisers, perfumes and food flavourings.
Products of ethanoyl chloride with water, ethanol, ammonia and methylamine: Water: ethanoic acid. Ethanol: ethyl ethanoate. Ammonia: ethanamide. Methylamine: N-methylethanamide. HCl is the by-product (as a salt with excess NH₃ or amine).
Name the mechanism for acyl chlorides with nucleophiles: Nucleophilic addition–elimination.
Why is ethanoic anhydride used rather than ethanoyl chloride to make aspirin?: Its by-product is ethanoic acid, not corrosive HCl gas; it reacts less vigorously so it is easier to control; it is hydrolysed less readily, and is cheaper.
Required practical 10: how do you recrystallise an impure organic solid?: Dissolve in the minimum volume of hot solvent, hot-filter if needed, cool to crystallise, filter under reduced pressure, wash with a little cold solvent and dry.
Fats and oils: Esters of glycerol with three long-chain carboxylic acids.

A Level Topic 28: Aromatic chemistry

Bonding in benzene: A planar ring of six carbons, bond angles 120°. Each C has one p electron; the p orbitals overlap to form a ring of six delocalised electrons above and below the plane. All C–C bonds are the same length, between single and double.
How does hydrogenation data show benzene’s extra stability?: Predicted ΔH for cyclohexa-1,3,5-triene = 3 × ΔH(cyclohexene). Benzene’s measured value is less exothermic, so benzene is more stable, because of delocalisation.
Why does benzene undergo substitution rather than addition?: Addition would break the delocalised ring and lose its extra stability. Substitution keeps the delocalised ring.
Nitration of benzene: reagents and electrophile: Concentrated HNO₃ and concentrated H₂SO₄. Electrophile NO₂⁺: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻
Friedel–Crafts acylation: reagents and electrophile: An acyl chloride with an AlCl₃ catalyst (dry). Electrophile RCO⁺: RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻. The product is a phenyl ketone.
Why is nitration of benzene important?: It is a step in the manufacture of explosives, and reducing nitrobenzene gives aromatic amines used to make dyes.
Product of benzene with propanoyl chloride and AlCl₃: 1-phenylpropan-1-one, C₆H₅COCH₂CH₃, plus HCl.

A Level Topic 29: Amines

Primary / secondary / tertiary amine: One / two / three carbons bonded to N (RNH₂, R₂NH, R₃N).
Quaternary ammonium ion: Four carbons bonded to N, R₄N⁺. Its salts are R₄N⁺X⁻.
Two ways to make a primary aliphatic amine: Halogenoalkane + excess NH₃ in ethanol, heated in a sealed container (gives a mixture by further substitution). Reduction of a nitrile, e.g. LiAlH₄ or H₂ with a Ni catalyst (primary amine only).
How is phenylamine made from nitrobenzene?: Reduction, e.g. Sn and concentrated HCl under reflux, then NaOH. C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O
Base strength order: methylamine, ammonia, phenylamine: Methylamine > ammonia > phenylamine.
Explain the base strength order of primary aliphatic amine > NH₃ > phenylamine: A base accepts H⁺ with the N lone pair. Alkyl groups release electron density, so N is more electron-rich. In phenylamine the lone pair is delocalised into the ring, so it is less available.
Cationic surfactant: A surfactant whose ionic head group is positive, e.g. a quaternary ammonium salt with one long hydrocarbon chain.
Product of ethanoyl chloride with propylamine: N-propylethanamide, CH₃CONHCH₂CH₂CH₃ (plus propylammonium chloride with excess amine).

A Level Topic 30: Polymers

Condensation polymerisation: Monomers with two functional groups join, losing a small molecule (usually H₂O) at each link.
Monomers of Terylene: Benzene-1,4-dicarboxylic acid and ethane-1,2-diol. It is a polyester.
Monomers of nylon 6,6 and of Kevlar: Nylon 6,6: hexanedioic acid and hexane-1,6-diamine. Kevlar: benzene-1,4-dicarboxylic acid and benzene-1,4-diamine.
Forces between polymer chains: polyester vs polyamide: Polyester: van der Waals and permanent dipole–dipole (no H on N or O). Polyamide: van der Waals and hydrogen bonds between N–H on one chain and the C=O oxygen of another.
Why are polyesters and polyamides biodegradable but polyalkenes not?: The ester and amide links have a polar C=O with a δ+ carbon that nucleophiles (H₂O, OH⁻) attack, so they can be hydrolysed. Polyalkenes have only non-polar C–C and C–H bonds and are inert.
Mr of the repeating unit of nylon 6,6 (monomer Mr values 146.0 and 116.0): 146.0 + 116.0 − 2 × 18.0 = 226.0
One advantage and one disadvantage of incinerating waste polymers: Advantage: less volume, and the energy can generate electricity. Disadvantage: CO₂ is released, and PVC gives toxic HCl that must be removed.
Products of acid hydrolysis of a polyamide: The dicarboxylic acid and the ammonium salt of the diamine (not the free amine).

A Level Topic 31: Amino acids, proteins and DNA

Zwitterion: An ion with a positive and a negative charge and no overall charge, H₃N⁺–CHR–COO⁻.
Form of an amino acid in acid and in alkaline solution: Acid: H₃N⁺–CHR–COOH (overall +1). Alkali: H₂N–CHR–COO⁻ (overall −1).
Primary, secondary and tertiary structure of a protein: what holds each?: Primary: the amino acid sequence, held by peptide links. Secondary: α-helix or β-pleated sheet, held by hydrogen bonds between peptide links. Tertiary: the 3D fold, held by hydrogen bonds and S–S bonds between side chains.
α-helix: A coil; hydrogen bonds between peptide links on different turns of the same chain.
Why does an enzyme act on only one enantiomer of a substrate?: The active site is stereospecific: its binding groups are in fixed 3D positions, and only one enantiomer can line up with all of them.
Base pairing in DNA and the number of hydrogen bonds: A pairs with T (2 hydrogen bonds); G pairs with C (3 hydrogen bonds).
How does cisplatin act as an anticancer drug?: Its Cl⁻ ligands are replaced as Pt bonds to nitrogen atoms on guanine in DNA. Binding two guanines distorts the DNA, so it cannot replicate and the cell cannot divide.
How are colourless amino acids located on a TLC plate?: Spray with ninhydrin (purple spots) or use UV light on a fluorescent plate.

A Level Topic 32: Organic synthesis

Percentage atom economy: (Mr of the desired product ÷ sum of Mr of all reactants) × 100.
Why do chemists aim for processes that need no solvent?: Solvents must be made, transported and separated from the product (often by energy-intensive distillation), become waste unless recycled, and are often flammable or toxic.
Why do chemists aim for fewer steps?: Product is lost at every step, so the overall yield falls; each step also needs its own reagents, separation, energy and time.
Overall yield of a three-step route with step yields 80%, 70% and 50%: 0.80 × 0.70 × 0.50 = 0.28, so 28%. Multiply the yields; never add or average them.
Which reactions in the course make a new C–C bond?: KCN with a halogenoalkane; KCN with an aldehyde or ketone; Friedel–Crafts acylation (and addition polymerisation).
Atom economy of an addition reaction: Exactly 100%: there is only one product. Substitution and elimination are always below 100%.
Route: 1-bromopropane to butylamine: KCN in aqueous ethanol, reflux → butanenitrile; then reduce (e.g. LiAlH₄, or H₂ with Ni) → butylamine.

A Level Topic 33: Nuclear magnetic resonance spectroscopy

Four reasons TMS is used as the standard: 12 equivalent H atoms give one strong single peak; its peak (δ = 0) is at lower δ than almost all other protons; it is inert; it is volatile, so it is easily removed.
Why are ¹H NMR samples dissolved in CDCl₃ or CCl₄?: Deuterium gives no signal in a ¹H spectrum, and CCl₄ has no H atoms, so the solvent does not swamp the sample’s peaks.
The n+1 rule: A ¹H signal is split into n + 1 lines by n H atoms on adjacent carbon atoms.
Splitting pattern of an ethyl group, CH₃CH₂–: A quartet (CH₂, 2H) and a triplet (CH₃, 3H).
What does integration of a ¹H spectrum give?: The relative number of H atoms in each environment, as a ratio. You need the molecular formula to get actual numbers.
How many peaks in the ¹³C NMR spectrum of methyl ethanoate, CH₃COOCH₃?: Three: the C=O carbon, the O–CH₃ carbon and the CH₃ on the C=O.
¹³C vs ¹H: ¹³C counts carbon environments and shows no splitting. ¹H counts hydrogen environments and gives integration and splitting.
Aldehyde vs ketone: Both show C=O near 190–220 in ¹³C. Only an aldehyde has a ¹H peak at δ 9–10.

A Level Topic 34: Chromatography

What does separation in chromatography depend on?: The balance between solubility in the moving phase and retention by the stationary phase.
Rf value: Distance moved by the spot ÷ distance moved by the solvent front, both measured from the baseline. No units; between 0 and 1.
Retention time (GC): The time from injection into the column until the component reaches the detector.
Stationary and moving phases in TLC, column chromatography and GC: TLC: solid on a plate; liquid solvent moving up. Column: solid packed in a column; solvent moving down. GC: solid, or liquid-coated solid, in a column; gas under pressure.
Spot centre 3.15 cm and solvent front 7.00 cm from the baseline. Rf?: 3.15 ÷ 7.00 = 0.45
TLC: why draw the baseline in pencil and keep the solvent below it?: Pencil graphite does not dissolve or separate, unlike ink. If the solvent were above the baseline, the samples would dissolve into the solvent pool.
What does GC-MS do?: Gas chromatography separates the components; each then passes into a mass spectrometer, which identifies it from its molecular ion (and precise m/z).
Why cover the TLC tank with a lid?: It saturates the air with solvent vapour, stopping solvent evaporating from the plate so the front rises evenly.

Flashcards by topic

Written for AQA A-level Chemistry (7405), syllabus 7405 (For teaching from September 2015).

Cards by Marlbridge Academic Team; 25 restate definitions from the 7405 revision notes. This deck was reviewed by Nouman Ahmed on 2026-11-05. Report an error.

All 7405 notes, practice and topics