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AS Topic 1: Unit 1 – Structure, Bonding and Introduction to Organic Chemistry

Empirical formula vs molecular formula: Empirical: the lowest whole-number ratio of the atoms present. Molecular: the real number of atoms of each element in one molecule (C₂H₄ has empirical formula CH₂).
Relative molecular mass vs relative formula mass: Both are sums of Ar values and have no units. Relative molecular mass is for simple molecular substances; relative formula mass is for compounds with giant structures, such as NaCl and SiO₂.
The ideal gas equation, with the units it needs: pV = nRT, with p in Pa, V in m³, T in K and R = 8.31 J mol⁻¹ K⁻¹. Convert kPa × 1000, cm³ × 10⁻⁶, °C + 273.
1.06 g of sodium carbonate (M = 106.0 g mol⁻¹) is dissolved to make 100 cm³. Concentration?: n = 1.06 ÷ 106.0 = 0.0100 mol. c = 0.0100 ÷ 0.100 = 0.100 mol dm⁻³ (10.6 g dm⁻³).
Atom economy (as defined in this specification): Molar mass of the desired product ÷ sum of the molar masses of all products × 100%. It depends only on the equation; an addition reaction with one product has 100%.
Ionic equation for an acid reacting with a carbonate, and what you see: CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l). Fizzing; the gas turns limewater cloudy.
Core Practical 1 (molar volume of a gas): two reasons the result is often low: Any two: gas escapes before the bung is fitted; gas dissolves in water if collected over water; an oxide layer on the metal; leaks at joints.
Electrons in a positive ion and in a negative ion: A positive ion has fewer electrons than protons; a negative ion has more electrons than protons. Neutrons = mass number − atomic number.
Why do isotopes of an element have the same chemical properties?: They have the same number of protons and so the same electron arrangement, which governs chemistry. They differ only in neutrons, so physical properties such as mass differ.
Relative atomic mass from a mass spectrum: chlorine is 75% ³⁵Cl and 25% ³⁷Cl: Weighted mean: (35 × 75 + 37 × 25) ÷ 100 = (2625 + 925) ÷ 100 = 35.5
How many molecular ion peaks does Cl₂ give, and at which m/z?: Three: ³⁵Cl–³⁵Cl at 70, ³⁵Cl–³⁷Cl at 72, ³⁷Cl–³⁷Cl at 74. Their heights follow the probability of each combination, so they are not equal.
First ionisation energy: The energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions: X(g) → X⁺(g) + e⁻.
What do successive ionisation energies show? Use magnesium as the example.: A large jump shows an electron being removed from a shell closer to the nucleus. For Mg the jump comes at the third electron: two electrons in the outer shell (shells 2, 8, 2).
Orbitals per sub-shell and electrons each holds: s: 1 orbital, 2 electrons. p: 3 orbitals, 6 electrons. d: 5 orbitals, 10 electrons. Each orbital holds two electrons of opposite spin; orbitals in a sub-shell fill singly before pairing.
Ionic bonding: Strong net electrostatic attraction between oppositely charged ions in a giant lattice.
Covalent bond and dative covalent bond: Covalent: strong electrostatic attraction between two nuclei and the shared pair of electrons between them. Dative: a covalent bond where one atom supplies both electrons of the shared pair (e.g. N to H⁺ in NH₄⁺).
What gives an ionic compound covalent character?: Polarisation: a small, highly charged cation (high polarising power) distorts the electron cloud of a large, highly charged anion (high polarisability). Covalent character: AlF₃ < AlCl₃ < AlI₃.
Why do the radii of N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺ decrease in that order?: All have 10 electrons in the same arrangement, but the number of protons rises from 7 to 13, so the electrons are pulled in more strongly.
Shapes and bond angles of CH₄, NH₃ and H₂O: CH₄: 4 bonding pairs, tetrahedral, 109.5°. NH₃: 3 bonding pairs + 1 lone pair, pyramidal, about 107°. H₂O: 2 bonding pairs + 2 lone pairs, bent, about 104.5°. Lone pairs repel more than bonding pairs.
Shapes of gaseous PCl₅ and SF₆: PCl₅(g): 5 bonding pairs, trigonal bipyramidal, 120° and 90°. SF₆: 6 bonding pairs, octahedral, 90°.
Why is CO₂ non-polar but H₂O polar?: Both have polar bonds. CO₂ is linear and symmetrical, so the bond dipoles cancel. H₂O is bent, so they do not cancel and there is a net dipole.
Why do metals conduct electricity?: When a potential difference is applied, the mobile delocalised electrons drift through the lattice, carrying charge.
Why are metals malleable?: Layers of ions slide while the electron sea keeps holding them, since the bonds are not between fixed pairs of atoms.
Hazard vs risk: The property versus the likelihood of harm. A hazard is something with the potential to cause harm; risk is the likelihood that it causes harm under the conditions of use.
Homolytic vs heterolytic fission: Radicals versus ions; half-arrows versus full arrows.
Homologous series: Members share a functional group and a general formula; neighbouring members differ by CH₂.
Structural isomers of C₅H₁₂: Three: pentane, 2-methylbutane and 2,2-dimethylpropane.
Free radical substitution of an alkane with chlorine: conditions and the three stages: UV light. Initiation: Cl₂ → 2Cl•. Propagation: Cl• + RH → R• + HCl; R• + Cl₂ → RCl + Cl•. Termination: two radicals join (e.g. Cl• + R• → RCl).
Why is free radical substitution of limited use in synthesis?: The product can be substituted further, and termination gives side products, so a mixture forms with a low yield of any one compound.
Cracking vs reforming: Cracking shortens chains and makes alkenes; reforming keeps the carbon count and changes shape (to branched or cyclic molecules).
Why is carbon monoxide toxic, and why is bioethanol not fully carbon-neutral?: CO binds to haemoglobin more strongly than O₂, reducing oxygen transport. Bioethanol: the CO₂ released was taken in by photosynthesis, but farming, processing and transport add CO₂.
σ bond vs π bond: σ: head-on overlap of orbitals on the axis between the two nuclei. π: sideways overlap of p orbitals, with electron density above and below the plane. The π bond is weaker.
Two conditions for E/Z (geometric) isomerism: A C=C, which has restricted rotation, and two different atoms or groups on each carbon of the C=C.
How are E and Z assigned?: On each C=C carbon, the attached atom with the higher atomic number has priority. Higher-priority groups on the same side = Z; on opposite sides = E.
Conditions for hydrogenation and hydration of an alkene: Hydrogenation: H₂, nickel catalyst, about 150 °C → alkane. Hydration: steam, phosphoric acid catalyst, high temperature and pressure → alcohol.
Propene + HBr: major product and why: 2-bromopropane. It forms via the secondary carbocation CH₃CH⁺CH₃, which is more stable than the primary one because more alkyl groups release electron density onto C⁺.
What evidence supports a two-step (carbocation) mechanism for Br₂ adding to ethene?: In sodium chloride solution, BrCH₂CH₂Cl forms as well as BrCH₂CH₂Br: a carbocation forms first and is attacked by any available nucleophile. No ClCH₂CH₂Cl forms.
Repeat unit of poly(chloroethene), PVC, and a problem with incinerating it: –[CH₂–CHCl]ₙ– from CH₂=CHCl. Incineration releases HCl; it is removed by scrubbers containing a base such as calcium hydroxide.

AS Topic 2: Unit 2 – Energetics, Group Chemistry, Halogenoalkanes and Alcohols

Standard conditions (energetics): 100 kPa and a specified temperature, usually 298 K. Substances in their standard states. Symbol ⦵.
Exothermic vs endothermic (sign of ΔH): Exothermic: ΔH negative; surroundings get hotter. Endothermic: ΔH positive; surroundings get colder.
Standard enthalpy change of formation: The enthalpy change when 1 mol of a compound is formed from its elements, under standard conditions with all substances in their standard states, e.g. Na(s) + ½Cl₂(g) → NaCl(s). It is zero for an element.
Standard enthalpy change of neutralisation: The enthalpy change when an acid and a base react to form 1 mol of water, under standard conditions with all substances in their standard states.
25.0 cm³ of solution warms by 4.4 °C when 0.00500 mol of a metal reacts (c = 4.18 J g⁻¹ °C⁻¹). ΔH?: Energy = 25.0 × 4.18 × 4.4 = 459.8 J = 0.4598 kJ. ΔH = −0.4598 ÷ 0.00500 = −92 kJ mol⁻¹ (negative because the temperature rose).
Hess's law, and ΔrH from formation data: The overall enthalpy change is the same whatever route is taken, if the start and end conditions match. ΔrH = ΣΔfH(products) − ΣΔfH(reactants).
Bond enthalpy vs mean bond enthalpy: One specific bond in one molecule vs an average over many.
Why is a measured enthalpy change of combustion usually less exothermic than the data value?: Heat is lost to the air and the can, combustion is incomplete, fuel evaporates, and conditions are not standard.
London force: An attraction between an instantaneous dipole in one molecule and the dipole it induces in a neighbouring molecule.
Hydrogen bond: An attraction between a δ+ hydrogen atom bonded to N, O or F and a lone pair on an N, O or F atom of another molecule. The X–H···Y arrangement is close to linear.
Why does water boil higher than HF and NH₃?: Water can form the most hydrogen bonds per molecule (2 H and 2 lone pairs). NH₃ has only one lone pair; HF has only one H. It is the number of hydrogen bonds, not the strength of each.
Why is ice less dense than liquid water?: In ice each molecule is hydrogen bonded to four others in a tetrahedral arrangement, an open structure with empty space. On melting some hydrogen bonds break and the molecules move closer.
Why does 2,2-dimethylpropane boil lower than pentane?: Same number of electrons, but the branched molecule is compact with less surface contact, so the London forces are weaker.
Why does boiling temperature rise from HCl to HI?: More electrons, so stronger London forces, which outweigh the falling permanent dipole.
Why do halogenoalkanes not dissolve well in water?: They are polar but cannot hydrogen bond with water, so the energy needed to break water’s hydrogen bonds is not repaid.
Oxidation number rules: the exceptions for hydrogen and oxygen: H is +1, but −1 in metal hydrides (NaH, CaH₂). O is −2, but −1 in peroxides (H₂O₂, Na₂O₂) and positive with F.
Oxidation number of Se in H₂SeO₄: 2(+1) + x + 4(−2) = 0, so x = +6.
Oxidising agent versus oxidised: The oxidising agent is the species that gets reduced.
Disproportionation: Atoms of one element, all in the same species, are both oxidised and reduced. Example: Cl₂ + 2NaOH → NaCl + NaClO + H₂O (Cl 0 → −1 and 0 → +1).
Why does first ionisation energy fall down Groups 1 and 2?: An extra shell each time places the outer electron further out, with more inner-shell shielding; these outweigh the larger nuclear charge, so the attraction is weaker.
Solubility trends down Group 2 for hydroxides and sulfates: Hydroxides become more soluble (Mg(OH)₂ least, Ba(OH)₂ most). Sulfates become less soluble (MgSO₄ most, BaSO₄ least).
Why do Group 2 carbonates become more thermally stable down the group?: The cation gets larger, so it is less polarising: it distorts the carbonate ion’s electron cloud less and weakens the C–O bonds less. Order: MgCO₃ < CaCO₃ < SrCO₃ < BaCO₃.
What does heating sodium nitrate give, and what does a Group 2 nitrate give?: NaNO₃ (and KNO₃): the nitrite plus O₂ only. Group 2 nitrates (and LiNO₃): the oxide + NO₂ + O₂.
Chlorine with cold, dilute NaOH vs hot NaOH: Cold: Cl₂ + 2NaOH → NaCl + NaClO + H₂O (chlorate(I), +1). Hot: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O (chlorate(V), +5).
Solid NaCl, NaBr and NaI with concentrated sulfuric acid: what forms?: NaCl: HCl only (steamy fumes); not redox. NaBr: HBr, then Br₂ + SO₂ (S +6 → +4). NaI: HI, then I₂ with SO₂, S and H₂S. Reducing power HCl < HBr < HI.
Silver halide precipitates and their behaviour with ammonia: AgCl white, dissolves in dilute NH₃. AgBr cream, dissolves in concentrated NH₃ only. AgI yellow, does not dissolve.
Colour of iodine in water and in cyclohexane: Brown in water; violet in cyclohexane (a non-polar solvent). Bromine is orange to red-brown in cyclohexane.
Activation energy: The minimum energy colliding particles need for reaction to occur.
Maxwell-Boltzmann distribution: shape, and the effect of a higher temperature: Starts at zero, one asymmetric peak, long tail that never touches the energy axis; area = number of molecules. Higher T: peak lower and further right, same area, much more area beyond Ea.
Main reason a higher temperature increases the rate: A greater proportion of collisions have energy ≥ Ea, so successful collisions are more frequent. Collision frequency rises only slightly.
How does a catalyst appear on a Maxwell-Boltzmann diagram and on a reaction profile?: Maxwell-Boltzmann: the curve is unchanged; only the Ea mark moves left. Profile: a lower route, usually two humps with an intermediate in the dip; ΔH unchanged.
Dynamic equilibrium: the two statements needed: Forward and backward rates are equal, and the concentrations of reactants and products do not change. It needs a closed system.
Effect of a catalyst on the position of equilibrium: None: it speeds up the forward and backward reactions equally, so equilibrium is reached sooner but the yield is unchanged.
Run A takes 50 s and run B takes 20 s. How many times faster is B?: Rate ∝ 1/t: (1/20) ÷ (1/50) = 2.5 times faster.
Nucleophile: Donates a lone pair of electrons to form a new covalent bond.
Halogenoalkane with KOH: aqueous vs ethanolic: KOH(aq), heat under reflux: OH⁻ acts as a nucleophile; substitution gives an alcohol. KOH in ethanol, heat under reflux: OH⁻ acts as a base; elimination gives an alkene.
Halogenoalkane + KCN in ethanol: product: A nitrile with one more carbon atom (bromoethane → propanenitrile, CH₃CH₂CN). Heat under reflux.
Why does hydrolysis rate follow R–I > R–Br > R–Cl?: Bond enthalpy, not polarity: C–I is the weakest bond, so it breaks most easily. C–Cl is the most polar but reacts slowest.
Oxidising a primary alcohol to an aldehyde vs to a carboxylic acid: Both: K₂Cr₂O₇ / dilute H₂SO₄, orange → green. Aldehyde: distil off as it forms. Acid: excess oxidising agent, heat under reflux. A tertiary alcohol does not react (stays orange).
Test for an –OH group with PCl₅: Steamy fumes of HCl, using a dry sample: ROH + PCl₅ → RCl + POCl₃ + HCl. Positive for alcohols, carboxylic acids and water.
Why is 50% (not fully concentrated) sulfuric acid used with KBr to brominate an alcohol?: Fully concentrated sulfuric acid oxidises the HBr to bromine, wasting the HBr.
Molecular ion peak: Found at the largest m/z value, if you ignore the tiny M+1 peak from carbon-13. Its m/z gives Mr.
Mass spectrum M⁺ patterns for one Cl and one Br: One Cl: two peaks 2 apart, heights about 3 : 1. One Br: two peaks 2 apart, heights about 1 : 1.
IR: how do the O–H absorptions of an alcohol and a carboxylic acid differ?: Alcohol: broad, about 3550–3230 cm⁻¹. Acid: very broad, about 3300–2500 cm⁻¹, always with a strong C=O near 1700 cm⁻¹. Hydrogen bonding makes O–H bands broad.

AS Topic 3: Unit 3 – Practical Skills in Chemistry I

Accuracy vs precision: Tight repeats can all be wrong if a systematic error is present. Accurate: close to the true value. Precise: repeat results close to each other.
Random error vs systematic error: Random: unpredictable scatter, reduced by repeating and averaging. Systematic: the same shift in every reading, not removed by repeats.
Percentage uncertainty of a titre or a temperature change: Double the single-reading uncertainty (two readings are subtracted), then divide by the value and × 100. Add percentage uncertainties for the overall result.
Reflux (set-up): Vertical condenser, open at the top, anti-bumping granules, water bath or electric heater for flammable liquids.
Distillation (set-up): Thermometer bulb at the side arm; water into the condenser at the bottom; collect over a narrow range.
Using a separating funnel: Stopper, invert, vent through the tap; let layers settle; drain the lower layer.
Drying an organic liquid: Add anhydrous salt until some moves freely and the liquid is clear; decant or filter.
Mass by difference vs taring: Weighing by difference accounts for solid left in the weighing bottle.
Test for the ammonium ion: Warm with NaOH(aq): ammonia is given off, which turns damp red litmus blue.
Flame test colours for Ca²⁺, Sr²⁺ and Ba²⁺: Ca²⁺ orange-red, Sr²⁺ red, Ba²⁺ pale green.

A Level Topic 4: Unit 4 – Rates, Equilibria and Further Organic Chemistry

Rate equation and order of reaction: rate = k[A]ᵐ[B]ⁿ, with orders found by experiment (usually 0, 1 or 2). Order with respect to A is m; overall order is m + n.
Units of k for zero, first and second order overall: Zero: mol dm⁻³ s⁻¹. First: s⁻¹. Second: mol⁻¹ dm³ s⁻¹.
rate = k[A][B]; rate 3.0 × 10⁻⁴ mol dm⁻³ s⁻¹, [A] = 0.20, [B] = 0.050 mol dm⁻³. k?: k = 3.0 × 10⁻⁴ ÷ (0.20 × 0.050) = 0.030 mol⁻¹ dm³ s⁻¹
Concentration-time and rate-concentration graphs for zero and first order: Zero order: straight line with negative gradient; rate-concentration horizontal. First order: curve with a constant half-life; rate-concentration a straight line through the origin.
SN1 vs SN2 for halogenoalkane hydrolysis: SN1 (tertiary): rate = k[RX]; slow ionisation to a carbocation, then fast attack by OH⁻. SN2 (primary): rate = k[RX][OH⁻]; one step via a transition state, OH⁻ attacking opposite the halogen.
Intermediate vs catalyst in a mechanism: Intermediate: made in one step, used in a later one. Catalyst: used in one step, regenerated later.
Finding Ea from an Arrhenius plot: ln k = −(Ea/R)(1/T) + ln A. Plot ln k against 1/T (T in K); gradient = −Ea/R, so Ea = −gradient × 8.31 J mol⁻¹.
Iodination of propanone (Core Practical 9a): what does zero order in iodine tell you?: Iodine is not in the rate-determining step; it reacts after it. rate = k[CH₃COCH₃][H⁺]; H⁺ is a catalyst (regenerated).
How does a solid (heterogeneous) catalyst work?: Reactant gases adsorb weakly onto active sites; their bonds weaken and they meet in a good orientation, so Ea falls. Products desorb and free the sites.
Entropy, and the order of entropy for the three states: A measure of disorder: the random dispersal of molecules and of energy quanta between them. S(solid) < S(liquid) < S(gas); a perfect crystal at 0 K has zero entropy.
ΔSsurroundings and ΔStotal: ΔSsurroundings = −ΔH ÷ T (ΔH in J mol⁻¹, T in K). ΔStotal = ΔSsystem + ΔSsurroundings. A change is feasible when ΔStotal > 0.
ΔH = +52 kJ mol⁻¹, ΔSsystem = +130 J mol⁻¹ K⁻¹. When is the reaction feasible?: Changeover T = ΔH ÷ ΔSsystem = 52 000 ÷ 130 = 400 K. Both positive, so feasible above 400 K.
Thermodynamic vs kinetic stability: Thermodynamically stable: ΔStotal for the change is negative, so it is not feasible. Kinetically stable: feasible, but the activation energy is so high the rate is negligible (petrol in air).
Lattice energy (as defined in this specification): The enthalpy change when gaseous ions come together to form 1 mol of ionic solid, e.g. Ca²⁺(g) + 2I⁻(g) → CaI₂(s). Always negative.
First electron affinity, and the sign of the second: First: the enthalpy change when 1 mol of gaseous atoms each gain one electron to form 1 mol of gaseous 1− ions; usually negative. The second electron affinity is always positive.
Born-Haber: ΔfH(LiCl) −409; ΔatH(Li) +159; IE(Li) +520; ΔatH(Cl) +122; EA(Cl) −349 kJ mol⁻¹. Lattice energy?: Steps to gaseous ions: 159 + 520 + 122 − 349 = +452. Lattice energy = −409 − 452 = −861 kJ mol⁻¹.
Enthalpy change of solution from lattice and hydration enthalpies: ΔsolH = −(lattice energy) + Σ ΔhydH(ions). For CaI₂: 2074 + (−1616) + 2(−297) = −136 kJ mol⁻¹.
What does an experimental lattice energy much more exothermic than the theoretical value show?: Partial covalent bonding: the cation polarises the anion, drawing electron density between the nuclei. Biggest for a small, highly charged cation with a large anion.
Homogeneous vs heterogeneous equilibrium: Homogeneous: every species in one phase. Heterogeneous: species in more than one phase. Solids never appear in K expressions.
Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), with units: Kc = [NH₃]² ÷ ([N₂][H₂]³). Units dm⁶ mol⁻².
Mole fraction and partial pressure: x(A) = moles of A ÷ total moles of gas. p(A) = x(A) × total pressure. The partial pressures add up to the total pressure.
C₂H₄ + H₂ ⇌ C₂H₆: p(C₂H₄) 0.40, p(H₂) 0.25, p(C₂H₆) 1.10 atm. Kp?: Kp = 1.10 ÷ (0.40 × 0.25) = 11.0 atm⁻¹
Which changes alter the value of K?: Only temperature. Concentration and pressure move the position of equilibrium; a catalyst changes neither.
Effect of raising the temperature on K for an exothermic forward reaction: K gets smaller, and the equilibrium moves left so the composition matches the new K. (K changes first.)
Link between ΔStotal and K: ΔStotal = R ln K (R = 8.31 J K⁻¹ mol⁻¹). A large positive ΔStotal means a very large K.
Brønsted–Lowry acid and base, and a conjugate pair: Acid: proton (H⁺) donor. Base: proton acceptor. A conjugate pair is two species that differ by one H⁺, e.g. HSO₄⁻/SO₄²⁻.
Kw vs pKw: Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ and pKw = 14.00, both at 298 K only.
Ka vs pKa: Larger Ka means stronger acid; smaller pKa means stronger acid.
pH of 0.110 mol dm⁻³ weak acid with pKa 4.13: Ka = 10⁻⁴·¹³ = 7.41 × 10⁻⁵. [H⁺] = √(7.41 × 10⁻⁵ × 0.110) = 2.86 × 10⁻³. pH = 2.54
pH of 0.0215 mol dm⁻³ KOH at 298 K: [H⁺] = Kw ÷ [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 0.0215 = 4.65 × 10⁻¹³. pH = 12.33
Buffer solution: A solution whose pH changes very little when a small amount of acid or alkali is added to it.
How does an acidic buffer (HA and A⁻) resist added acid and added alkali?: Added H⁺: H⁺ + A⁻ → HA. Added OH⁻: OH⁻ + HA → A⁻ + H₂O. HA and A⁻ are present in large amounts, so the ratio [HA]/[A⁻] and so [H⁺] change only slightly.
Half-neutralisation point: Half the equivalence volume; here [HA] = [A⁻] and pH = pKa.
Choosing an indicator for weak acid–strong base and strong acid–weak base titrations: Its whole colour-change range must lie in the steep section. Weak acid + strong base: phenolphthalein (about 8.2–10.0). Strong acid + weak base: methyl orange (about 3.1–4.4). Weak + weak: no sharp change; use a pH meter.
How is blood kept close to pH 7.4?: Mainly the H₂CO₃ ⇌ H⁺ + HCO₃⁻ buffer. Extra acid: H⁺ + HCO₃⁻ → H₂CO₃, removed as CO₂ through the lungs.
Chiral centre, and optical isomers: A chiral centre is a carbon atom carrying four different atoms or groups. Optical isomers (enantiomers) are non-superimposable mirror images.
Optical activity, and a racemic mixture: A single optical isomer rotates the plane of polarisation of plane-polarised monochromatic light; enantiomers rotate it by equal angles in opposite directions. A racemic mixture (equal amounts of both) shows no net rotation.
How optical activity distinguishes SN1 from SN2: SN1: planar carbocation attacked equally from either face → racemic product, zero rotation. SN2: one-sided attack with inversion → a single enantiomer, non-zero rotation.
Tests that distinguish an aldehyde from a ketone: Warm with Tollens’ reagent: aldehyde gives a silver mirror. Warm with Fehling’s or Benedict’s: aldehyde gives a red precipitate (Cu₂O); a ketone stays blue. 2,4-DNPH gives a yellow-orange precipitate with both.
Adding HCN to a carbonyl: mechanism type, why KCN is added, and the stereochemistry: Nucleophilic addition. HCN is a weak acid, so KCN supplies enough CN⁻. CN⁻ attacks the planar C=O equally from both sides, so a chiral product forms as a racemic mixture.
Reducing aldehydes, ketones and carboxylic acids: LiAlH₄ in dry ether. Aldehyde → primary alcohol (2[H]); ketone → secondary alcohol (2[H]); carboxylic acid → primary alcohol (4[H]).
Iodoform test: reagents, result and what it shows: Iodine with NaOH(aq), warm: pale yellow precipitate of CHI₃. Shows a CH₃CO– group (ethanal and methyl ketones), or CH₃CH(OH)–.
Acyl chloride reactions with water, alcohols and ammonia: Water → carboxylic acid + HCl. Alcohol → ester + HCl. Concentrated ammonia (2 mol) → amide + NH₄Cl. Fast at room temperature and not reversible.
Ester hydrolysis: acidic vs alkaline: Both heated under reflux. Acidic (dilute HCl or H₂SO₄): carboxylic acid + alcohol; reversible. Alkaline (NaOH(aq)): carboxylate salt + alcohol; goes to completion.
Terylene: Ethane-1,2-diol + benzene-1,4-dicarboxylic acid. Repeat unit –OC–C₆H₄–COO–CH₂CH₂–O–.
¹³C NMR: what does the number of peaks tell you?: The number of different carbon environments, not the number of carbons. Symmetry reduces the count.
¹H NMR: what do peak area and splitting tell you?: Relative area: the number of H atoms in that environment. Splitting (n + 1 rule): n non-equivalent H atoms on adjacent carbons give n + 1 lines. Equivalent H do not split each other; O–H is usually a singlet.
HPLC vs GC: Both are column chromatography that separate by retention time. HPLC: liquid mobile phase under high pressure; solid particles packed in the column. GC: unreactive carrier gas; liquid stationary phase on a solid support. Linked to MS for identification.

A Level Topic 5: Unit 5 – Transition Metals and Organic Nitrogen Chemistry

Standard electrode potential, E⦵: The voltage of a half-cell measured against a standard hydrogen electrode, both under standard conditions: 298 K, 100 kPa for any gas, 1.00 mol dm⁻³ for each ion.
Standard hydrogen electrode: Platinum coated with platinum black, H₂ at 100 kPa, H⁺(aq) at 1.00 mol dm⁻³, 298 K. E⦵ = 0.00 V by definition.
Al³⁺/Al −1.66 V and Cu²⁺/Cu +0.34 V. Cell diagram and E⦵cell: Al(s) | Al³⁺(aq) || Cu²⁺(aq) | Cu(s). E⦵cell = E⦵(right) − E⦵(left) = +0.34 − (−1.66) = +2.00 V. Overall 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu.
Why a salt bridge of potassium nitrate, and a high-resistance voltmeter?: The salt bridge completes the circuit by ion movement without mixing the solutions; nitrate forms no precipitates. The high-resistance voltmeter draws almost no current, so it reads the true emf.
Limitations of using E⦵ to predict whether a reaction happens: E⦵ predicts thermodynamics, not rate: a feasible reaction can be too slow if its activation energy is large. Non-standard conditions change E and can reverse a prediction when E⦵cell is close to zero.
Iron(II) with manganate(VII): ionic equation, ratio and end point: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. Fe²⁺ : MnO₄⁻ = 5 : 1. Self-indicating: first permanent pale pink. Use dilute sulfuric acid (HCl would be oxidised to chlorine).
Iodine–thiosulfate titration: equation, ratio and indicator: 2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻. S₂O₃²⁻ : I₂ = 2 : 1. Starch added near the end: blue-black to colourless.
Hydrogen fuel cell electrode equations with an alkaline electrolyte: Negative: H₂ + 2OH⁻ → 2H₂O + 2e⁻. Positive: O₂ + 2H₂O + 4e⁻ → 4OH⁻. Overall 2H₂ + O₂ → 2H₂O.
Transition metal, and why scandium and zinc are not: A d-block element with at least one stable ion with incompletely filled d-orbitals. Sc forms only Sc³⁺ (d⁰) and Zn only Zn²⁺ (d¹⁰).
Electron configurations of Cr, Cu and Fe³⁺: Cr: [Ar] 3d⁵ 4s¹. Cu: [Ar] 3d¹⁰ 4s¹. Fe³⁺: [Ar] 3d⁵ (4s electrons are removed first).
Ligand, and coordination number: Ligand: a molecule or ion that bonds to a metal ion by donating a lone pair (a dative bond). Coordination number: the total number of dative bonds to the metal ion.
Shapes of [Fe(H₂O)₆]³⁺, [CoCl₄]²⁻ and cis-platin: Octahedral (90°), coordination number 6. Tetrahedral (109.5°), 4: large Cl⁻ ligands. Square planar (90°), 4: [Pt(NH₃)₂Cl₂] with the two Cl on the same side.
Why are transition metal complexes coloured?: Ligands split the 3d orbitals into two levels. A d electron absorbs visible light of frequency f where ΔE = hf. The complementary colour is transmitted and seen. d⁰ and d¹⁰ ions are colourless.
Why is carbon monoxide toxic (haemoglobin)?: CO replaces O₂ on the Fe²⁺ by ligand exchange and binds far more strongly, so the change is not easily reversed and less oxygen is carried.
Colours of vanadium in oxidation states +5, +4, +3, +2: VO₂⁺ (+5) yellow; VO²⁺ (+4) blue; V³⁺ (+3) green; V²⁺ (+2) violet.
Cr³⁺(aq) with NaOH(aq) and with NH₃(aq), then excess: Grey-green precipitate with both. Excess NaOH: dark green solution (amphoteric). Excess NH₃: purple solution (ligand exchange).
Why is an EDTA⁴⁻ complex more stable than the aqua complex it replaces?: [M(H₂O)₆]ⁿ⁺ + EDTA⁴⁻ → 7 particles from 2, so ΔSsystem is large and positive. ΔH ≈ 0, so ΔStotal is positive and K is large (the chelate effect).
Heterogeneous vs homogeneous catalysts: Heterogeneous: different phase; reaction at the surface (V₂O₅ in the Contact process). Homogeneous: same phase; goes via an intermediate species (Fe²⁺ with S₂O₈²⁻ and I⁻).
Catalyst vs autocatalyst: An autocatalyst forms during the reaction and then catalyses it (Mn²⁺ in the manganate(VII)–ethanedioate reaction).
Three lines of evidence against the Kekulé structure of benzene: Thermochemical: benzene is more stable than predicted for three C=C. X-ray: all six C–C bonds equal, about 0.140 nm. IR: no alkene C=C band near 1650 cm⁻¹.
The delocalised model of benzene: Each carbon forms three σ bonds at 120° and has one electron in a p-orbital perpendicular to the ring. The p-orbitals overlap sideways all round, so six π-electrons are delocalised above and below the ring.
Why does benzene need a halogen carrier to react with bromine, unlike cyclohexene?: Its π-electrons are delocalised, so electron density between any two carbons is too low to polarise Br₂. FeBr₃ makes Br⁺. It substitutes rather than adds, keeping the stable delocalised ring.
Nitration of benzene: reagents, conditions and electrophile: Concentrated HNO₃ + concentrated H₂SO₄, at or below about 55 °C. Electrophile NO₂⁺: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. Product nitrobenzene.
Friedel-Crafts acylation: reagents and product: An acyl chloride with anhydrous AlCl₃, warm under reflux. RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻; product an aromatic ketone C₆H₅COR + HCl.
Phenol with bromine water: observation, product and reason: Orange decolourised and a white precipitate of 2,4,6-tribromophenol (C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr). An O lone pair is partly delocalised into the ring, raising its electron density.
Order of base strength: ammonia, butylamine, phenylamine: Butylamine > ammonia > phenylamine. The alkyl group releases electron density onto N; in phenylamine the N lone pair is delocalised into the ring and is less available.
Making phenylamine from nitrobenzene: Heat under reflux with tin and concentrated HCl, then add NaOH(aq) until alkaline to free the amine from C₆H₅NH₃⁺. ArNO₂ + 6[H] → ArNH₂ + 2H₂O.
Making an azo dye from phenylamine: Diazotisation: NaNO₂ + HCl below 10 °C → benzenediazonium chloride. Coupling: add to phenol in NaOH(aq) → a yellow-orange azo dye. Too warm and the diazonium ion decomposes to N₂ and phenol.
Amine + ethanoyl chloride: CH₃COCl + 2RNH₂ → CH₃CONHR + RNH₃⁺Cl⁻. An N-substituted amide forms fast at room temperature; two moles of amine are needed.
Nylon-6,6: monomers and small molecule lost: H₂N(CH₂)₆NH₂ + HOOC(CH₂)₄COOH (or the diacyl dichloride). Repeat unit –[NH(CH₂)₆NHCO(CH₂)₄CO]–; water (or HCl) is lost at each link.
Forms of 2-aminopropanoic acid in the solid, in acid and in alkali: Solid / neutral: zwitterion H₃N⁺CH(CH₃)COO⁻. Acid: H₃N⁺CH(CH₃)COOH. Alkali: H₂NCH(CH₃)COO⁻.
Why does poly(ethenol) dissolve in water, and one use?: Its many –OH groups hydrogen bond with water. Used for soluble hospital laundry bags and liquid-detergent capsules.
Grignard reagent: how it is made, and why everything must be dry: RX + Mg → RMgX in dry ether. Water reacts with RMgX to give the alkane RH, destroying the reagent.
Grignard reagent with CO₂, methanal, another aldehyde and a ketone: Then hydrolyse with dilute acid. CO₂ → carboxylic acid. Methanal → primary alcohol. Other aldehyde → secondary alcohol. Ketone → tertiary alcohol.
Recrystallisation: Dissolve the solid in the minimum volume of hot solvent, filter hot, cool to crystallise, filter, wash with a little cold solvent and dry.
Steam distillation is used for…: A volatile, water-immiscible compound that would decompose at its boiling point: it distils with steam below 100 °C.
Making aspirin (Core Practical 16): how is purity checked?: Measure the melting temperature: pure aspirin melts sharply at about 135–136 °C. A lower, broad range shows impurity (often unreacted 2-hydroxybenzoic acid) or water.
Reagents and conditions: halogenoalkane → amine, and nitrile → amine: Halogenoalkane → amine: excess ammonia in ethanol, heat in a sealed tube. Nitrile → amine: LiAlH₄ in dry ether.

A Level Topic 6: Unit 6 – Practical Skills in Chemistry II

Core Practical 9a: why are samples run into sodium hydrogencarbonate solution?: To quench the reaction: it removes the H⁺ catalyst, stopping the iodination so the iodine left can be titrated with thiosulfate.
Core Practical 9b (iodine clock): how is concentration kept proportional to volume?: Keep the total volume constant by making it up with water, changing only one solution’s volume at a time. Initial rate ∝ 1/t.
Core Practical 10: which graph gives the activation energy?: ln(1/t) (y) against 1/T in K⁻¹ (x). Ea = −gradient × 8.31, then divide by 1000 for kJ mol⁻¹.
Core Practical 11: finding Ka of a weak acid: Calibrate the pH meter with buffers; find the end-point volume V for one portion; add V/2 of alkali to an identical portion; its pH = pKa, so Ka = 10⁻ᵖᴴ.
Core Practical 12: preparing the half-cells: Clean the metals with emery paper; use a potassium nitrate salt bridge (no precipitates with metal ions); read the emf on a high-resistance voltmeter.
Iodate(V) releasing iodine: equation and ratio to thiosulfate: IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O, so IO₃⁻ : S₂O₃²⁻ = 1 : 6.
0.642 g KIO₃ (Mr 214.0) in 250.0 cm³; each 25.0 cm³ portion needs 22.50 cm³ thiosulfate. Its concentration?: KIO₃ per portion = (0.642 ÷ 214.0) ÷ 10 = 3.00 × 10⁻⁴ mol. Thiosulfate = 6 × 3.00 × 10⁻⁴ = 1.80 × 10⁻³ mol. c = 1.80 × 10⁻³ ÷ 0.02250 = 0.0800 mol dm⁻³.
Fe²⁺, Fe³⁺ and Zn²⁺ with NaOH(aq): Fe²⁺: green precipitate, browning in air. Fe³⁺: red-brown precipitate. Zn²⁺: white precipitate, dissolving in excess to a colourless solution.
A redox titration result differs from the accepted value by more than the total uncertainty. What does that suggest?: A systematic error that apparatus uncertainty cannot explain, such as air oxidising Fe²⁺ before titration (lowering every result).

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Written for Pearson Edexcel International Advanced Subsidiary/Advanced Level Chemistry (XCH11/YCH11), syllabus YCH11 (Issue 1, September 2017).

Cards by Marlbridge Academic Team; 26 restate definitions from the YCH11 revision notes. This deck was reviewed by Nouman Ahmed on 2026-11-05. Report an error.

All YCH11 notes, practice and topics