WJEC A-Level Chemistry Revision
Explore WJEC A-Level Chemistry revision topics. Check the official specification for complete requirements for your exam year and route.
Course details
WJEC · A-Level Chemistry1410QSExam years · 2027
Assessment AS Units 1-2: 90 minutes each, 20% each of the full A level. A2 Units 3-4: 105 minutes each, 25% each. Unit 5: separate practical examination, 10%, not an NEA or endorsement; ask your centre for the 2027 topic notice.
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Reviewed course guidance
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Topic group
AS Unit 1: the language of chemistry, structure and matter
- Core
Scope: Core · Unit 1 §1.1
Open revision guidance
What to know
- Write neutral ionic formulae by balancing total positive and negative charge; use brackets when more than one polyatomic ion is needed. Common ions include NH₄⁺, OH⁻, NO₃⁻, HCO₃⁻, CO₃²⁻, SO₄²⁻ and PO₄³⁻; examples include NaCl, CaCO₃, (NH₄)₂SO₄ and Mg(NO₃)₂.
- An oxidation number is a bookkeeping charge. Free elements have oxidation number 0; a monatomic ion has its ionic charge; the sum is 0 in a neutral compound and equals the charge in a polyatomic ion. Oxygen is usually −2 and hydrogen +1, with known peroxide and metal-hydride exceptions.
- Balance an equation by changing coefficients, never formula subscripts. Ionic equations retain only species that change; spectator ions cancel. State symbols are (s), (l), (g) and (aq).
- For SO₄²⁻, four oxygens contribute −8, so sulfur is +6 to give the ion's −2 total. In H₂O₂ each O is −1; in NaH hydrogen is −1. Oxidation numbers are not the same as an atom's measured partial charge.
- For acid-carbonate reaction, CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l). Count oxygen and hydrogen as well as the metal, and check the total +2 charge on both sides.
What to do in an assessment
- For aluminium sulfate, combine Al³⁺ and SO₄²⁻ in the smallest whole-number ratio: Al₂(SO₄)₃. Check the net charge before using the formula.
- Balance atoms and charge in ionic equations. For example, Ag⁺(aq) + Cl⁻(aq) → AgCl(s); add state symbols from the conditions rather than guessing them.
In plain language
A formula is the smallest particle ratio that has the right charge. An equation is balanced only when both atoms and charge are conserved.
Self-check
Write the formula of calcium nitrate and the net ionic equation for its precipitation with sodium carbonate.
Show the answer
Ca(NO₃)₂; Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s).
- Core
Scope: Core · Unit 1 §1.2
Open revision guidance
What to know
- α decay emits a helium nucleus, so A falls by 4 and Z by 2; β⁻ decay changes a neutron to a proton, so Z rises by 1 and A is unchanged. In positron emission or electron capture a proton becomes a neutron, so Z falls by 1. γ emission changes neither A nor Z.
- A half-life is the time for the number of undecayed nuclei or activity to halve. After n half-lives, N=N₀(½)ⁿ; decay is random for an individual nucleus but predictable for a large population.
- α is strongly ionising and weakly penetrating; β is intermediate; γ is weakly ionising and most penetrating. Electric/magnetic deflection depends on charge and mass. Ionising radiation can damage cells and DNA. Applications include Tc-99m medical imaging, radio-dating with carbon-14, industrial thickness gauges and isotopic tracers/analysis; justify benefit and control exposure with containment, shielding, time and distance.
- First ionisation energy removes one electron from each gaseous atom. Successive values rise overall; a large jump identifies removal from an inner shell. Across a period effective nuclear charge generally rises; down a group added shells and shielding lower attraction.
- Orbitals have quantised energies: s is spherical; p has two lobes. For Z=1–36 fill 1s, 2s, 2p, 3s, 3p, 4s, then 3d; account for the familiar chromium/copper arrangements when giving full configurations.
- An emission photon is released when an electron falls to a lower level; absorption raises it. Hydrogen has a line spectrum because only particular transitions are allowed; Lyman transitions end at n=1. The series limit corresponds to ionising ground-state hydrogen.
- Use E=hf and c=fλ. Frequency increases and wavelength decreases from infrared through visible to ultraviolet; photon energy increases with frequency.
- Nuclear equations conserve A and Z. Examples are ¹⁴₆C → ¹⁴₇N + ⁰₋₁e, ²²₁₁Na → ²²₁₀Ne + ⁰₊₁e, and ⁷₄Be + ⁰₋₁e → ⁷₃Li. Neutrinos or antineutrinos may also be shown; they have zero A and Z and do not alter this bookkeeping.
- In an electric field, positive alpha and beta-plus particles deflect towards the negative plate, beta-minus towards the positive plate, and gamma is undeflected. In a magnetic field the charged particles bend perpendicular to their motion and the field; opposite charges curve oppositely. Alpha is stopped by paper/skin, beta by a suitable thin metal sheet, while gamma is attenuated rather than wholly stopped by thick dense shielding. Internal alpha exposure can be particularly hazardous.
- For a sample initially giving 800 counts per second after background correction, a 3-hour half-life gives 400 after 3 hours, 200 after 6 and 100 after 9. Use corrected activity, not an uncorrected background-plus-source count.
- One orbital holds at most two electrons of opposite spin. An s subshell has one spherical orbital; a p subshell has three mutually perpendicular two-lobed orbitals; a d subshell has five orbitals. For the first 36 elements the usual filling sequence continues through 4s, 3d and then 4p, ending at krypton [Ar]3d¹⁰4s²4p⁶. Equal-energy orbitals fill singly before pairing. The familiar exceptions are Cr [Ar]3d⁵4s¹ and Cu [Ar]3d¹⁰4s¹.
- First molar ionisation energy refers to X(g) → X⁺(g) + e⁻ for one mole of gaseous atoms and is reported in kJ mol⁻¹. Successive ionisation removes electrons from the preceding gaseous ion. A large jump after the second value points to two outer-shell electrons; it is not the energy needed to remove two electrons in one step.
- Hydrogen's Lyman lines are transitions to n=1 in the ultraviolet. Their frequencies approach a convergence limit as the upper level approaches the free-electron state. Ionisation from ground-state hydrogen requires E = h f_limit per atom; multiply by Avogadro's constant and divide by 1000 for kJ mol⁻¹. For f_limit = 3.29 × 10¹⁵ s⁻¹, h = 6.63 × 10⁻³⁴ J s and Nₐ = 6.02 × 10²³ mol⁻¹, the result is approximately 1.31 × 10³ kJ mol⁻¹.
What to do in an assessment
- For a nuclear equation, conserve both nucleon number A and proton number Z to identify the daughter nuclide and emitted particle. For elapsed time, count half-lives and halve the remaining number/activity at each interval.
- Convert wavelength to metres before using f=c/λ, then calculate E=hf. State whether the result is per photon; multiply by Avogadro’s constant only if energy per mole is required.
- For isotope decay, use N=N₀(½)ⁿ with n=t/t½. For electronic structure, place the large successive-ionisation-energy jump after the last electron in the occupied outer shell.
In plain language
Radioactive and light spectra are evidence that nuclei and electrons change in specific, countable ways—not continuously.
Self-check
A hydrogen transition has λ=4.86×10⁻⁷ m. Calculate its frequency and energy per photon (c=3.00×10⁸ m s⁻¹; h=6.63×10⁻³⁴ J s).
Show the answer
f=c/λ=6.17×10¹⁴ s⁻¹; E=hf=4.09×10⁻¹⁹ J per photon (3 s.f.).
- Core
Scope: Core · Unit 1 §1.3
Open revision guidance
What to know
- Aᵣ is the weighted mean mass of an element’s isotopes relative to ¹²C/12; Mᵣ is the sum of Aᵣ values in a molecule or formula unit. For example, chlorine’s roughly 3:1 ³⁵Cl:³⁷Cl abundance gives Aᵣ≈35.5. In a mass spectrometer, a sample is ionised, ions are accelerated and separated by m/z, then detected; isotope peaks give abundance and fragments give structural clues. Chlorine gas has M⁺ peaks at 70, 72 and 74 in about 9:6:1 proportions.
- One mole contains Nₐ=6.02×10²³ entities. Use n=m/M, c=n/V, and convert cm³ to dm³ by dividing by 1000. The ideal-gas equation is pV=nRT with pressure in Pa, volume in m³ and T in K (R=8.31 J mol⁻¹ K⁻¹); use it to correct gas amount/volume when conditions differ. A molar gas volume changes with temperature and pressure.
- For formulae, convert each element’s mass or percentage to moles and divide by the smallest to get the empirical formula; use Mᵣ÷empirical-formula mass to find the molecular multiplier. For example CH₂O with Mᵣ=180 gives C₆H₁₂O₆. For reactions, use the balanced-equation mole ratio and identify the limiting reagent before calculating product.
- In titration, n=cV and coefficients give the reacting ratio. Solubility is concentration at saturation. Atom economy compares desired-product Mᵣ with total reactant Mᵣ; percentage yield compares actual and theoretical product mass.
- Percentage error is uncertainty ÷ measured value ×100. Report answers to a precision justified by the data; do not imply a precise answer from imprecise measurements.
- Relative isotopic mass refers to a particular isotope; relative molecular mass applies to molecules, and relative formula mass to a formula unit such as NaCl. These relative masses have no unit; molar mass has units g mol⁻¹.
- For 40.0% C, 6.7% H and 53.3% O, the mole values per 100 g are approximately 3.33, 6.7 and 3.33, giving 1:2:1 and empirical formula CH₂O. Molecular mass 180 then gives C₆H₁₂O₆. Do not turn a near-half-integer ratio into an integer by premature rounding.
- A saturated solution containing 20.0 g of solute per dm³ and molar mass 100 g mol⁻¹ has molar solubility 0.200 mol dm⁻³. For an ideal gas at 100 kPa and 298 K, molar volume RT/p = 0.0248 m³ mol⁻¹ = 24.8 dm³ mol⁻¹. Different temperatures or pressures require a different volume.
- Use stoichiometric coefficients in atom economy: desired-product formula mass times its coefficient divided by the sum of all reactant formula masses times their coefficients. For CaCO₃ → CaO + CO₂ with masses 100, 56 and 44, atom economy for CaO is 56%. If 10.0 g CaCO₃ gives 4.80 g CaO, theoretical mass is 5.60 g and yield is 85.7%.
- If each burette reading has uncertainty ±0.05 cm³, a difference of two readings has a maximum absolute uncertainty ±0.10 cm³. For a 25.00 cm³ titre this is 0.40%. Uncertainty estimates are not a measured error relative to a known true value.
What to do in an assessment
- For 25.0 cm³ of 0.100 mol dm⁻³ acid, n=0.100×0.0250=2.50×10⁻³ mol. Apply the equation coefficients before calculating the unknown amount.
- For 40.0% C, 6.7% H and 53.3% O: divide each by Aᵣ, then by the smallest to obtain the simplest ratio. Show the unrounded ratios before choosing integers.
- Check units at every step, keep guard digits, and round only the final value to the least precise justified significant figures.
In plain language
The mole turns particle counts into measurable masses and volumes; a balanced equation then tells you how those amounts react.
Self-check
What volume of 0.100 mol dm⁻³ NaOH neutralises 25.0 cm³ of 0.0800 mol dm⁻³ H₂SO₄?
Show the answer
H₂SO₄+2NaOH→Na₂SO₄+2H₂O. n(H₂SO₄)=0.0800×0.0250=0.00200 mol, so n(NaOH)=0.00400 mol and V=0.00400/0.100=0.0400 dm³=40.0 cm³.
- Core
Scope: Core · Unit 1 §§1.4–1.5
Open revision guidance
What to know
- Ionic bonding is electrostatic attraction between oppositely charged ions in a lattice. Covalent bonding is a shared pair attracted to both nuclei; a coordinate bond has both bonding electrons supplied by one atom. Many bonds are intermediate in ionic/covalent character.
- For example, a lone pair on NH₃ bonds to H⁺ to make NH₄⁺; the coordinate bond is still a shared pair after formation. In NaCl each ion has six nearest opposite ions; in CsCl each has eight.
- Electronegativity difference helps predict bond polarity, but molecular shape determines whether bond dipoles cancel. London forces arise from temporary dipoles; permanent dipole attractions and hydrogen bonding add stronger intermolecular attractions. These affect boiling temperature and solubility, not the intramolecular covalent bond.
- VSEPR places electron regions as far apart as possible. Simple ideal angles: linear 180°, trigonal planar 120°, tetrahedral 109.5°, octahedral 90°/180°; lone pairs repel more strongly and reduce adjacent bond angles.
- NaCl and CsCl are giant ionic lattices with different coordination; diamond is a 3-D tetrahedral covalent network, graphite has three bonds per carbon and delocalised electrons in layers, iodine is molecular, and ice is an open hydrogen-bonded molecular solid. Metals are positive ions in a delocalised “electron sea”.
- Structure explains properties: mobile ions conduct only when molten/dissolved; graphite and metals conduct as solids; giant covalent solids are hard/high-melting, while simple molecules melt at lower temperatures. Solubility depends on whether new solute–solvent attractions compensate for separating the original structure.
- Bond formation reflects attraction between positive nuclei and shared electrons balanced against nucleus-nucleus and electron-electron repulsions. Intermediate bond character means that real bonds need not fit a perfectly ionic or perfectly covalent model.
- Hydrogen bonding needs H bonded to a strongly electronegative atom such as O, N or F and a suitable lone pair on another molecule. It explains water's unusually high boiling temperature and ice's open, lower-density lattice. It is not the covalent O-H bond itself.
- CH₄ has four bonding regions and is tetrahedral (109.5°). NH₃ has three bonds and one lone pair, giving trigonal pyramidal shape (about 107°); H₂O has two bonds and two lone pairs and is bent (about 104.5°). CO₂ is linear (180°), BF₃ trigonal planar (120°), and SF₆ octahedral (90° between adjacent bonds). Treat a multiple bond as one VSEPR region.
- Diamond is hard because covalent bonds extend throughout its three-dimensional structure; it has no mobile electrons and is insoluble. Graphite is soft between layers because the interlayer attractions are weak, but conducts through its delocalised electrons. Thus giant covalent does not mean every such material is hard or insulating.
- Iodine consists of I₂ molecules in a crystal with comparatively weak London attractions between them; it has much lower melting temperature than a giant covalent solid. Metals are malleable because layers can slide while non-directional metallic attraction persists. Ionic crystals are brittle because displacement can bring like charges alongside one another.
What to do in an assessment
- Draw a dot-and-cross/Lewis diagram, count bonding and lone-pair regions, predict shape/angle, then use electronegativity and vector symmetry to decide molecular polarity.
- Compare structure, particle attractions and mobile charge before predicting melting, hardness, solubility or conductivity; state the phase in which conduction is possible.
In plain language
Bonding describes what holds particles together; the whole arrangement explains how a material behaves.
Self-check
Why does solid NaCl not conduct, whereas molten NaCl does, and why is CO₂ non-polar?
Show the answer
In solid NaCl the ions are fixed in the lattice; on melting the ions can move and carry charge. CO₂ is linear, so equal C=O bond dipoles cancel.
View diagrams
Sodium chloride: ionic transferBefore transfer, sodium is 2,8,1 and chlorine is 2,8,7. The sodium outer electron transfers to chlorine. Separate bracketed ions result: Na⁺ has configuration 2,8 and Cl⁻ has 2,8,8. Dots identify sodium-origin electrons; crosses identify chlorine-origin electrons. This is electron transfer, not a shared covalent pair.
Hydrogen · H₂Two hydrogen atoms contribute one electron each to one shared pair. The dot comes from the left H atom and the cross from the right H atom; each hydrogen now has a full first shell.
Water · H₂OOxygen shares one electron with each H, making two shared pairs. Its four remaining outer-shell electrons make two lone pairs.
Tetrahedral methaneMethane has four C-H bonds arranged tetrahedrally. Two bonds are drawn in the plane, a solid wedge points toward the viewer, and a hashed wedge points away. The ideal three-dimensional H-C-H bond angle is 109.5°; the perspective sketch is not a planar or scaled projection. Methane is not chiral.
In dot-and-cross diagrams, a dot is an electron from the atom named in the explanation and a cross is from the other atom. The symbols show electron origin, not a charge. Shared pairs belong between bonded atoms; lone pairs stay on one atom.
The dot-and-cross and ionic shell sketches show outer-shell electrons only; inner-shell electrons are omitted from those sketches. Full sodium and chlorine configurations are printed beside the ions. Displayed structural formulae instead draw every atom, including each hydrogen, and every covalent bond.
Compare electron transfer with covalent sharing, then use the methane model to check tetrahedral geometry. - Core
Scope: Core · Unit 1 §1.6
Open revision guidance
What to know
- The Periodic Table is ordered by atomic number. The outer-electron configuration identifies s-, p- and d-block position. Oxidation and reduction are electron loss and gain (OIL RIG); track oxidation numbers to identify both half-changes.
- Across a period, ionisation energy and electronegativity generally rise as effective nuclear charge rises; down a group added shells and shielding usually lower attraction. Melting temperature often peaks near the middle of a period then falls, but the pattern changes with metallic, giant-covalent or molecular structure; down groups compare actual bonding/structure rather than assume one universal trend.
- Group 1 and Group 2 metals become more reactive down their groups. Group 2 metals form oxides with oxygen and hydroxides plus hydrogen with water; Mg reacts slowly with cold water but with steam gives MgO and H₂. Down Group 2, hydroxide solubility and carbonate/hydroxide thermal stability increase, while sulfate solubility decreases. Group 1 and 2 oxides/hydroxides are basic.
- Mg²⁺, Ca²⁺ and Ba²⁺ give white carbonate precipitates under suitable conditions. With OH⁻, Mg(OH)₂ is a white precipitate, Ca(OH)₂ is less readily precipitated because it is more soluble, and dilute Ba²⁺ normally gives no precipitate because Ba(OH)₂ is relatively soluble. Sulfate gives no ordinary precipitate with Mg²⁺, limited precipitation of sparingly soluble CaSO₄ and insoluble white BaSO₄. State concentrations and interpret the whole pattern.
- Flame-test colours: Li crimson, Na yellow, K lilac, Ca orange-red, Sr crimson, Ba apple-green; Mg has no characteristic flame colour. Halogens become less volatile down Group 7 (gas → liquid → solid); their oxidising power/reactivity also decreases down the group.
- Halogens react with metals to form halide salts. A more reactive halogen displaces a less reactive halide: Cl₂ oxidises Br⁻/I⁻; Br₂ oxidises I⁻. Ag⁺ gives AgCl white, AgBr cream and AgI yellow; dilute ammonia dissolves AgCl, while AgBr and AgI remain in dilute ammonia. For example, Cl₂+2I⁻→2Cl⁻+I₂ transfers electrons from iodide to chlorine.
- Redox applies across s-, p- and d-block chemistry: Fe²⁺→Fe³⁺+e⁻ is oxidation, while the matching electron gain by another species is reduction. Use electrons and oxidation numbers together to check the direction.
- Chlorine disinfects water; fluoride ions are used for dental health. Evaluate benefits alongside exposure, environmental, ethical and consent concerns rather than treating a policy judgement as a chemical fact.
- Across Period 3, metallic bonding strengthens from Na to Al, giant covalent Si has a high melting temperature, then molecular P₄, S₈ and Cl₂ and atomic Ar have much lower values governed mainly by intermolecular attractions. Down Group 1, increased ion size generally weakens metallic attraction and lowers melting temperature; do not apply that trend to every other group.
- Group 2 reactions include 2Mg + O₂ → 2MgO, Mg + H₂O(g) → MgO + H₂, and Ca + 2H₂O(l) → Ca(OH)₂ + H₂. Carbonates decompose as MCO₃ → MO + CO₂; hydroxides as M(OH)₂ → MO + H₂O. Larger cations down the group polarise the anions less, increasing thermal stability.
- For a gravimetric carbonate determination, precipitate quantitatively, wash away soluble salts and dry to constant mass. A 0.197 g BaCO₃ precipitate with M = 197 g mol⁻¹ represents 0.00100 mol Ba²⁺ in the tested aliquot. Scale to the original flask only after this mole calculation.
- To prepare a soluble salt with an insoluble base/carbonate, add excess solid to warm acid, filter, gently concentrate the filtrate, cool to crystallise, then filter, rinse and dry. To make an insoluble salt, mix appropriate soluble salts, collect the precipitate, wash and dry. The reaction and isolation method are different choices.
What to do in an assessment
- Write ionic equations for displacement and precipitation, balance charge, then use oxidation numbers to show which species is oxidised/reduced.
- For gravimetry, precipitate quantitatively, filter, wash, dry to constant mass and weigh. Convert mass to moles and use the balanced equation; avoid losing fine precipitate and include balance/filter uncertainties.
- For unknowns, use small labelled samples and a decision table: flame test, selected cation/anion reagents, observations and confirmatory steps. Wear eye protection; handle corrosive alkalis, silver salts and hot apparatus carefully; collect metal-containing waste.
In plain language
Elements in a group share outer-electron patterns, so their chemistry changes in orderly ways that can be tested by reactions and precipitates.
Self-check
Among Cl₂, Br₂ and I₂, a solution oxidises I⁻ but not Br⁻ under comparable displacement-test conditions. Which halogen is consistent with these observations?
Show the answer
Among Cl₂, Br₂ and I₂, it is Br₂. Bromine oxidises iodide but cannot displace itself from bromide; an inability to displace bromide is consistent with being bromine itself.
- Core
Scope: Core · Unit 1 §1.7(a–c)
Open revision guidance
What to know
- In a closed system, a reversible reaction reaches dynamic equilibrium when forward and reverse rates are equal; concentrations remain constant but both reactions continue.
- Le Chatelier: a concentration change is opposed by a shift that consumes the added species; higher pressure favours the side with fewer gas molecules; temperature changes favour the endothermic direction when heat is added. A catalyst changes the time to equilibrium, not its position or Kc.
- For aA+bB ⇌ cC+dD, Kc=[C]^c[D]^d/[A]^a[B]^b, using equilibrium concentrations and omitting pure solids/liquids. The value depends on temperature; substitute concentrations with powers and keep the stated units/conventions.
What to do in an assessment
- Write the balanced equation first, construct an ICE table, use stoichiometric changes and substitute equilibrium concentrations into Kc.
- When predicting a shift, state the imposed change, the opposing response and the final direction. For temperature, identify the endothermic direction explicitly.
In plain language
At equilibrium both directions still happen; their equal rates make the measured composition steady.
Self-check
For H₂(g)+I₂(g)⇌2HI(g), [H₂]=0.20, [I₂]=0.20 and [HI]=0.80 mol dm⁻³. Calculate Kc.
Show the answer
Kc=[HI]²/([H₂][I₂])=0.80²/(0.20×0.20)=16.
- Core
Scope: Core · Unit 1 §1.7(d–g)
Open revision guidance
What to know
- A Brønsted–Lowry acid donates H⁺; a base accepts H⁺. Strong acids dissociate almost completely in water; weak acids dissociate only partially. Strength is not the same as concentration.
- pH=−log₁₀[H⁺] and [H⁺]=10⁻ᵖᴴ mol dm⁻³. A one-unit pH decrease means tenfold greater hydrogen-ion concentration.
- Acid–base titration uses a measured titre and balanced stoichiometry. An indicator changes colour over a transition range; choose one whose range contains the steep pH change. Concordant titres improve precision; they do not remove calibration or endpoint bias.
- To prepare a soluble salt by titration when both reactants are soluble, first determine the exact neutralising volumes using an indicator. Repeat those measured volumes without indicator, gently concentrate the neutral solution, cool to crystallise, then filter, wash with a little cold water and dry. This avoids both excess soluble reactant and indicator contamination.
- A suitable primary standard has known purity and composition and can be weighed accurately. Dissolve a weighed mass of anhydrous Na₂CO₃, quantitatively transfer to a volumetric flask, make to the mark and mix. Titrate an aliquot with the acid using a suitable endpoint; Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ gives the 1:2 ratio.
- For back titration, 1.00 g of limestone reacts with 50.0 cm³ of 0.500 mol dm⁻³ HCl; unused acid needs 10.0 cm³ of 0.500 mol dm⁻³ NaOH. Initial HCl is 0.0250 mol and unused HCl 0.00500 mol, so carbonate consumed 0.0200 mol H⁺ and contains 0.0100 mol CaCO₃, or 1.00 g for M = 100 g mol⁻¹. This example is 100% CaCO₃ within the stated assumptions.
- For matched mixture aliquots requiring 10.0 cm³ of 0.100 mol dm⁻³ HCl to phenolphthalein and 15.0 cm³ to methyl orange, carbonate moles are 0.00150−0.00100 = 0.000500 and hydroxide moles are 2×0.00100−0.00150 = 0.000500. The second result is the total acid to its endpoint, not an extra 15.0 cm³ after the first endpoint.
What to do in an assessment
- Rinse a burette with titrant and pipette with analyte, remove air bubbles, read the meniscus consistently, perform a rough run then repeat to concordance; record initial/final readings and uncertainty. Standardise an acid by weighing/dissolving a suitable primary standard, pipetting a known aliquot, titrating to endpoint and calculating the acid concentration.
- To prepare a soluble salt from an acid and insoluble base/carbonate, warm the acid, add excess solid, filter off excess, concentrate the filtrate and crystallise, then rinse/dry crystals. For a back titration of limestone, react a measured excess of standard acid, stop when effervescence ceases, then titrate unused acid; the difference gives carbonate amount.
- For a NaOH/Na₂CO₃ mixture double titration, use matched aliquots: to the phenolphthalein endpoint n(H⁺)=n(OH⁻)+n(CO₃²⁻); to the methyl-orange endpoint n(H⁺)=n(OH⁻)+2n(CO₃²⁻). Subtract the first result from the second to find carbonate, then find hydroxide. Calculate n=cV in dm³ and convert to concentration or mass.
- Explain why a weak acid has a higher pH than a strong acid of equal concentration; select an indicator whose transition range lies within the steep pH change at the endpoint.
In plain language
Acid strength tells how much an acid ionises; pH tells the hydrogen-ion concentration that results.
Self-check
Find [H⁺] when pH=3.00, and distinguish a 0.10 mol dm⁻³ strong acid from a 0.10 mol dm⁻³ weak acid.
Show the answer
[H⁺]=10⁻³·⁰⁰=1.00×10⁻³ mol dm⁻³. The strong acid dissociates nearly completely; the weak acid only partially, so its [H⁺] and pH differ despite equal analytical concentration.
Topic group
AS Unit 2: energy, rate and chemistry of carbon compounds
- Core
Scope: Core · Unit 2 §2.1
Open revision guidance
What to know
- Enthalpy change is heat transferred at constant pressure. ΔHc is for complete combustion of one mole; ΔfH° is formation of one mole from elements in their standard states. Exothermic ΔH is negative; endothermic ΔH is positive.
- Hess’s law follows from enthalpy being a state function: add/reverse/scale equations and their ΔH values to obtain the target reaction. Average bond enthalpy estimates ΔH≈Σ(bonds broken)−Σ(bonds formed); it is approximate because bond enthalpies are mean gas-phase values.
- Calorimetry uses q=mcΔT and ΔH=−q/n for the reacting amount. State the mass/volume assumption and use the balanced equation to identify n. Heat loss, incomplete combustion, solution heat capacity, thermometer resolution and transfer delay bias results.
- Standard conditions use defined standard states, pressure 100 kPa and a stated temperature, commonly 298 K; solutions are conventionally taken at 1 mol dm⁻³. Formation must give exactly one mole of compound: ½H₂(g) + ½Cl₂(g) → HCl(g). An element in its standard state has standard formation enthalpy zero.
- With formation data, ΔHreaction = ΣνΔfH°(products) − ΣνΔfH°(reactants). With combustion data for paths to common final products, subtract product combustion values from reactant combustion values with their coefficients. Average bond estimates instead use gas-phase bonds broken minus bonds formed.
- For H₂ + Cl₂ → 2HCl, using mean bond enthalpies H-H 436, Cl-Cl 242 and H-Cl 431 kJ mol⁻¹ gives ΔH ≈ 436+242−2×431 = −184 kJ mol⁻¹ for the written reaction. Reversing it changes the sign; halving it gives the value per mole of HCl formed.
- For a solution calorimeter, insulate the cup, record temperatures before and after mixing, and use a temperature-time plot/extrapolation where the method requires a cooling correction. For the indirect MgO/CO₂ reaction, matched acid routes must use amounts and final conditions that support a valid Hess comparison.
What to do in an assessment
- For an indirect Hess cycle, write every equation in the direction needed, reverse the sign when reversing an equation, multiply ΔH when scaling it, then add and cancel species. For MgO(s)+CO₂(g)→MgCO₃(s), measure both MgO+acid and MgCO₃+acid temperature changes; because both reach the same aqueous products, ΔHformation = ΔH(MgO+acid)−ΔH(MgCO₃+acid).
- For combustion, record initial/final water temperature and fuel mass; calculate q of water, divide by fuel moles, state the sign for the reaction and evaluate heat loss. Use a screen and consistent burner distance; keep the burner cap on between weighings and avoid inhaling combustion products.
In plain language
A Hess cycle is energy bookkeeping: the route may change, but the total enthalpy change between the same start and finish does not.
Self-check
A reaction warms 100.0 g of water by 5.00 °C. Take c=4.18 J g⁻¹ K⁻¹ and 0.0100 mol reacting. Find ΔH assuming no heat loss.
Show the answer
q=100.0×4.18×5.00=2090 J; ΔH=−2.090 kJ/0.0100 mol=−209 kJ mol⁻¹.
View diagrams
Exothermic energy profileQualitative exothermic profile: the products lie at a lower energy level than the reactants, so energy is released. Activation energy Eₐ is measured from the reactant level to the peak; net energy change ΔH runs from reactants to products and is negative. The horizontal axis is reaction progress, not time. No numerical energies are implied.
Endothermic energy profileQualitative endothermic profile: the products lie at a higher energy level than the reactants, so energy is absorbed. Activation energy Eₐ is measured from the reactant level to the peak; net energy change ΔH runs from reactants to products and is positive. The horizontal axis is reaction progress, not time. No numerical energies are implied.
Use the labelled model with the explanation; arrows represent electron-pair movement or the stated physical flow, not interchangeable processes. - Core
Scope: Core · Unit 2 §2.2
Open revision guidance
What to know
- Rate is change in concentration or amount per time. Collision theory requires collisions with sufficient energy and suitable orientation. Higher concentration/pressure or surface area usually increases successful collisions; higher temperature increases both collision frequency and the fraction with E≥Ea.
- An energy profile plots energy through reaction progress: activation energy is the reactant-to-peak gap and ΔH is product minus reactant energy. A Boltzmann distribution is a spread of molecular energies; raising temperature broadens/lowers the curve and increases the area above Ea. A catalyst is selective, is regenerated overall and provides a lower-Ea route without changing ΔH or the equilibrium constant.
- Initial-rate data can establish the relationship between reactant concentration and rate by changing one concentration at a time. Colorimetry follows a coloured species; gas volume/mass loss, a disappearing precipitate cross and an iodine clock give other time-dependent measures.
- For the iodine clock, the initial thiosulfate removes iodine as it forms: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. Once a fixed thiosulfate amount is consumed, the first persistent iodine-starch blue colour marks the endpoint. Keep that amount, total volume and temperature fixed when using 1/time as a comparative rate.
- For a precipitation method such as thiosulfate plus acid, record the time to a defined visual obscuring endpoint with constant depth and viewing conditions; evaluate its subjective threshold and toxic sulfur dioxide risk. A continuous gas-collection method instead gives many volume-time points and allows tangent gradients.
What to do in an assessment
- Calculate an average rate as Δquantity/Δt; obtain an instantaneous rate from a tangent to a concentration–time curve. Keep units (e.g. mol dm⁻³ s⁻¹ or cm³ s⁻¹) consistent. Compare initial rates while varying one reactant concentration at a time and holding other conditions fixed; a doubling that doubles rate indicates first-order dependence over that range.
- For colorimetry, blank/zero the instrument with a suitable reference, use the same wavelength and cuvette orientation, and record absorbance at fixed intervals. For gas collection, seal the apparatus, start timing consistently and record volume at fixed intervals; for a clock, define the same visible endpoint each run. Repeat, identify anomalies and compare gradients with uncertainty.
- Control temperature, volumes, concentrations, surface area, mixing and timing. Protect eyes/skin from reactants; assess gas pressure/leaks and dispose of reaction mixtures safely.
In plain language
A reaction is faster when more collisions succeed each second; a catalyst makes success possible at lower energy.
Self-check
A concentration falls from 0.80 to 0.50 mol dm⁻³ in 30 s. Find the average disappearance rate.
Show the answer
(0.80−0.50)/30=0.010 mol dm⁻³ s⁻¹.
View diagrams
Maxwell-Boltzmann energy distribution at two temperaturesThe normalised density follows f(E,T) = 2/√π · √E/T^(3/2) · exp(−E/T). The higher-temperature curve has a lower, right-shifted peak. Eₐ is unchanged by temperature; a catalyst lowers the threshold without changing either distribution. Shading marks the higher-temperature population at or above Eₐ. The full, untruncated distributions each have unit area; the sampled finite SVG areas are not asserted to be exactly equal.
Use the labelled model with the explanation; arrows represent electron-pair movement or the stated physical flow, not interchangeable processes. - Core
Scope: Core · Unit 2 §2.3
Open revision guidance
What to know
- Chemical synthesis and energy production have social, economic and environmental costs and benefits: useful products, employment and reliable energy must be weighed against resource use, emissions, waste, ecosystem effects and health impacts.
- Green chemistry improves sustainability across development: prevent waste, improve atom economy and yield, use safer/renewable feedstocks and solvents, reduce energy demand, and prefer selective catalysts. It is an evaluative framework, not a guarantee that one process has no impact.
What to do in an assessment
- Compare alternatives using a stated system boundary and evidence: energy input, carbon/other emissions, waste, toxicity, resource lifetime, cost and who receives benefits or risks.
- Support a recommendation with trade-offs and data; distinguish measured environmental effects from assumptions and do not equate “renewable” with “impact-free”.
In plain language
Greener chemistry tries to make the useful product while using fewer resources and creating less harmful waste.
Self-check
Why is higher atom economy useful but not, by itself, proof that a process is sustainable?
Show the answer
It places more reactant atoms in the desired product, reducing potential by-products; it says nothing by itself about energy, toxicity, solvent, emissions, yield, feedstock or lifecycle impact.
- Core
Scope: Core · Unit 2 §2.4
Open revision guidance
What to know
- A molecular formula counts atoms; a displayed formula shows every bond; a shortened/condensed formula groups atoms; a skeletal formula omits carbon symbols and C–H bonds at line ends/vertices. Each representation must encode the same connectivity and valency.
- Choose the longest chain containing the principal functional group/multiple bond, number it to give that group/bond the lowest locant, and state positions for branches/substituents. Use alkane -ane, alkene -ene, alcohol -ol, carboxylic acid -oic acid; halogeno substituents use fluoro-, chloro-, bromo-, iodo-. List different substituent prefixes alphabetically and use di-/tri- for repeats.
- Structural isomers share a molecular formula but differ in connectivity. Chain length and functional groups affect intermolecular forces, so melting/boiling temperature and water solubility change.
- An electrophile accepts an electron pair; a nucleophile donates one; a radical has an unpaired electron. Homolytic fission gives one electron to each fragment; heterolytic fission gives both to one fragment.
- For a homologous series, longer chains usually have stronger London forces and higher boiling temperatures; increased branching often lowers boiling temperature at a fixed formula by reducing effective contact. Melting also depends strongly on packing, so do not assume it follows a universal branching rule.
- Alcohols and acids can hydrogen-bond with water, but increasing a non-polar hydrocarbon chain usually reduces aqueous solubility. Butan-1-ol and butan-2-ol are position isomers; butane and 2-methylpropane are chain isomers. Compare actual atom connectivity, not the orientation of a drawing.
What to do in an assessment
- Convert between displayed, shortened and skeletal formulae by counting carbons, hydrogens, bonds and charge. Number the chain before adding substituent positions and functional-group suffixes.
- Annotate a mechanism with curved arrows starting at an electron pair or bond; mark δ+/δ− or radicals and show all products, charges and regenerated catalyst if present.
In plain language
Organic names and formula drawings are different ways of recording which atoms are connected; mechanisms track the movement of electron pairs.
Self-check
Give a shortened formula and IUPAC name for CH₃CH₂CH(OH)CH₃.
Show the answer
The shortened formula is CH₃CH₂CH(OH)CH₃ and the name is butan-2-ol.
View diagrams
Curly arrows show electron-pair movementA full-headed curly arrow represents movement of an electron pair. Its tail starts at the electron-pair source (a lone pair or a covalent bond) and its head points to where that pair is accepted. Here the example is hydroxide attacking the carbon bonded to bromine in bromoethane; the second arrow starts at the C-Br bond.
Tetrahedral methaneMethane has four C-H bonds arranged tetrahedrally. Two bonds are drawn in the plane, a solid wedge points toward the viewer, and a hashed wedge points away. The ideal three-dimensional H-C-H bond angle is 109.5°; the perspective sketch is not a planar or scaled projection. Methane is not chiral.
Use the labelled model with the explanation; arrows represent electron-pair movement or the stated physical flow, not interchangeable processes. - Core
Scope: Core · Unit 2 §2.5
Open revision guidance
What to know
- Alkanes contain C–C/C–H σ bonds. Complete combustion gives CO₂ and H₂O; limited oxygen can give CO and/or carbon/soot. Fossil fuels supply energy/feedstock but release greenhouse and acidic gases and are finite.
- Photochemical radical substitution: initiation splits Cl₂ homolytically under UV (Cl₂→2Cl•); propagation includes Cl•+CH₄→HCl+CH₃• and CH₃•+Cl₂→CH₃Cl+Cl•; termination combines radicals. Further substitutions at other C–H bonds can occur.
- An alkene C=C contains one σ and one π bond; the π region is electron-rich and reacts more readily than an alkane. Restricted rotation permits E–Z isomers when each alkene carbon has two different substituents.
- Electrophilic addition uses the electron-rich pi bond to attack an electrophile. With HBr, the H-Br pair moves to Br as the alkene acquires H, then Br⁻ attacks the carbocation. With Br₂, ethene forms 1,2-dibromoethane; a cyclic bromonium ion is the more realistic intermediate, followed by bromide attack. A simplified school carbocation drawing is a model, not proof of the actual intermediate.
- Bromine water and potassium manganate(VII) are unsaturation tests. Catalytic hydrogenation uses H₂, a nickel catalyst and heating to convert alkenes to alkanes; hydrogenation is also used to harden unsaturated oils. Addition polymerisation opens C=C bonds to make saturated carbon-chain polymers used in important plastics/fibres.
- Sigma bonding is head-on orbital overlap along the internuclear axis; an alkene pi bond is sideways overlap above and below it. Rotation would disrupt the pi overlap. In HBr addition to propene, forming the secondary carbocation CH₃CH⁺CH₃ is favoured over the primary alternative, leading mainly to 2-bromopropane.
- Complete methane combustion is CH₄ + 2O₂ → CO₂ + 2H₂O. Incomplete combustion can give toxic CO; sulfur impurities form acidic SO₂, and high-temperature air combustion can form nitrogen oxides. Distinguish greenhouse effects from direct toxicity and acidification.
- A termination example is CH₃• + Cl• → CH₃Cl or 2CH₃• → C₂H₆: no radical remains. For ethene polymerisation, nCH₂=CH₂ → [-CH₂-CH₂-]ₙ. Substituents stay attached to their original chain carbon; poly(chloroethene) has [-CH₂-CHCl-]ₙ rather than losing chlorine.
- Bromine water decolourises with unsaturation under the ordinary test conditions. Cold dilute manganate(VII) also changes colour: acidified reagent loses purple as Mn²⁺ forms, while neutral/alkaline conditions can give brown MnO₂. State the medium rather than promise one observation for every permanganate test.
What to do in an assessment
- Write initiation, propagation and termination separately and ensure each propagation step regenerates a radical. Use UV for photochemical initiation and state that substitution can continue at other C–H bonds.
- For Br₂ addition, show the pi-pair attack on polarised Br₂ and Br-Br bond fission; a bromonium intermediate is followed by Br⁻ attack to form CH₂BrCH₂Br. For HBr addition, use the carbocation route and compare carbocation stability before choosing orientation.
- Draw curly arrows from electron sources; for E–Z, assign priority on each alkene carbon and compare the higher-priority groups. Show the repeat unit with continuation bonds.
- Test a small sample: bromine water decolourises; acidified manganate(VII) loses purple colour. Use eye protection and a fume cupboard for volatile halogens/flammable hydrocarbons; avoid inhalation and ignition sources.
In plain language
Alkanes tend to substitute; the electron-rich double bond in alkenes tends to add atoms across it.
Self-check
What is the major product of normal addition of HBr to propene, and why?
Show the answer
2-bromopropane. Proton addition forms the more stable secondary carbocation, then Br⁻ attacks it.
View diagrams
Propene + hydrogen bromide: electrophilic additionElectrophilic addition occurs in two steps. The alkene π pair attacks the δ+ hydrogen of polarised Hδ+-Brδ− while the H-Br bond pair moves to bromine, giving a secondary propan-2-yl carbocation and Br−. A bromide lone pair then forms the C-Br bond. The neutral product is 2-bromopropane. The displayed atom balance is C₃H₆ + HBr → C₃H₇Br; no concerted step is implied.
Use the labelled model with the explanation; arrows represent electron-pair movement or the stated physical flow, not interchangeable processes. - Core
Scope: Core · Unit 2 §2.6
Open revision guidance
What to know
- Aqueous OH⁻ substitutes halide in a primary halogenoalkane by an SN2 pathway: the nucleophile attacks as the C–X bond breaks in one step. Ethanolic base and heat favour elimination; for example 1-bromopropane forms propene by loss of HBr.
- C–X polarity makes carbon partially positive; bond enthalpy and steric access affect substitution. Down the halogen group the C–X bond becomes weaker, so hydrolysis generally becomes faster despite the changing polarity.
- Hydrolysis releases X⁻; Ag⁺ gives AgCl white, AgBr cream and AgI yellow. In CFCs C–Cl is weaker than C–H and C–F; UV in the upper atmosphere can break C–Cl to release chlorine radicals that catalyse ozone loss. Compare bond strengths with exposure/atmospheric chemistry when assessing impact.
- Halogenoalkanes have been used as solvents, anaesthetics and refrigerants; toxicity and environmental effects require controlled, regulated use.
- Elimination with hot ethanolic hydroxide removes H from a neighbouring carbon as halide leaves: CH₃CH₂CH₂Br + OH⁻ → CH₃CH=CH₂ + Br⁻ + H₂O. Aqueous hydroxide instead favours substitution to propan-1-ol; solvent, temperature and substrate influence the product balance.
- For reflux, clamp the flask and vertical condenser, circulate cooling water in at the lower end and out at the top, and keep the system open to a safe outlet rather than sealed against pressure. Add anti-bumping granules before heating. In a hydrolysis-rate comparison, use the same solvent, temperature and concentration and compare the time for a defined silver-halide precipitate.
- A simplified CFC ozone-loss cycle is Cl• + O₃ → ClO• + O₂ followed by ClO• + O → Cl• + O₂. The chlorine radical is regenerated, so many ozone molecules can be destroyed by one radical. A strong C-F bond does not prevent a weaker C-Cl bond in the same molecule from being broken by high-energy UV.
What to do in an assessment
- Draw the SN2 arrow from OH⁻ to the carbon attached to X and a simultaneous arrow from C–X to X; show the alcohol and halide product.
- For the specified substitution, heat measured 1-bromobutane with aqueous NaOH under reflux: a vertical water condenser returns vapour to the flask. Use anti-bumping granules, secure glassware, flowing condenser water and an electric heater where flammable solvent is present.
- For the halide test, hydrolyse a small sample, acidify as specified, add AgNO₃ and record precipitate colour/solubility. Avoid skin contact and collect silver/halogen waste separately.
In plain language
The carbon–halogen bond lets a nucleophile replace the halogen; controlled heating speeds the reaction without losing volatile reactants.
Self-check
What does aqueous OH⁻ do to 1-bromobutane, and what is the reaction type?
Show the answer
It replaces Br to form butan-1-ol and Br⁻ by nucleophilic substitution (SN2 for a primary halogenoalkane).
View diagrams
Hydroxide + bromoethane: one-step SN2 substitutionHydroxide uses an oxygen lone pair to form a bond to the primary carbon of bromoethane while the C-Br bond pair leaves for bromine. Both electron-pair changes occur in one SN2 step: ethanol and Br− form, with no carbocation intermediate.
Use the labelled model with the explanation; arrows represent electron-pair movement or the stated physical flow, not interchangeable processes. - Core
Scope: Core · Unit 2 §2.7
Open revision guidance
What to know
- Industrial hydration reacts ethene with steam over an acid catalyst at elevated temperature and pressure to make ethanol. Fermentation uses sugar/yeast warm without oxygen and is followed by distillation; compare renewable feedstock with land use, yield and process energy.
- Alcohols are primary/secondary/tertiary according to the number of carbon groups attached to the carbon bearing OH. Heating with a dehydrating acid removes H₂O to form an alkene. Oxidation gives a primary alcohol → aldehyde (distil) or carboxylic acid (reflux), secondary → ketone; tertiary alcohols resist oxidation under these conditions.
- Acidified dichromate(VI) oxidises primary/secondary alcohols (orange to green). Carboxylic acids effervesce with hydrogencarbonate as CO₂ forms; acids also react with bases to make salts/water and with carbonates to make salt/water/CO₂.
- A carboxylic acid plus an alcohol forms an ester and water in a reversible, acid-catalysed esterification. Simple distillation separates liquids using boiling-point differences; reflux retains volatile reactants during heating.
- Hydration is CH₂=CH₂ + H₂O ⇌ CH₃CH₂OH, commonly using a phosphoric-acid catalyst at about 300 °C and high pressure. Fermentation is C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ with yeast under warm oxygen-limited conditions, commonly about 30-40 °C; ethanol concentration eventually inhibits the organisms. Separation and lifecycle energy affect the comparison.
- For alcohol dehydration, hot concentrated acid or a suitable heated solid catalyst gives alkene plus water. Ethanol gives ethene. Ethanoic acid and ethanol give ethyl ethanoate: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O.
- With hydrogencarbonate: CH₃COOH + NaHCO₃ → CH₃COONa + CO₂ + H₂O; with carbonate two acid molecules are required. Test the evolved gas with limewater. For simple distillation, collect a stated boiling fraction with the thermometer bulb at the vapour outlet and keep the receiving system unsealed.
What to do in an assessment
- To preserve an aldehyde, distil it from the oxidation mixture; to make a carboxylic acid, reflux the primary alcohol with oxidant. Use a water bath/electric heater, never a naked flame near ethanol.
- For ester preparation, measure reagents, add acid catalyst carefully, heat under reflux, then distil the product; use a separating funnel if an immiscible layer forms. Do not identify by deliberately inhaling vapour.
- Dichromate(VI) is toxic/oxidising; use eye protection, small quantities and designated waste. Acids are corrosive; control splashes and assess hot glassware/pressure.
In plain language
Changing the conditions directs an alcohol towards an alkene, aldehyde, acid or ketone; an acid and alcohol can join to make an ester.
Self-check
What product is expected when a secondary alcohol is oxidised, and how can an ester be made from a carboxylic acid?
Show the answer
A ketone. Warm the acid and an alcohol with an acid catalyst; esterification produces ester and water and is reversible.
- Core
Scope: Core · Unit 2 §2.8
Open revision guidance
What to know
- A mass spectrum plots relative abundance against m/z. The molecular-ion peak can indicate Mᵣ; fragment ions give structural clues. Compare candidate formulae with exact/relative mass and isotope patterns rather than treating one peak as a full structure.
- IR absorption occurs when radiation matches a bond vibration that changes dipole moment. The spectrum’s functional-group region and fingerprint region are compared with reference data; broad O–H and strong C=O absorptions are useful indicators, not standalone proof.
- ¹³C NMR low-resolution spectra show chemically distinct carbon environments; the number of signals constrains symmetry. Low-resolution ¹H NMR gives proton environments and relative signal areas; combine spectra with formula and chemical tests to choose a consistent structure.
- A molecular ion is the ionised intact molecule, often M⁺•, not necessarily the most abundant peak. The base peak is the tallest, defined as 100% relative abundance. For ethanol, M⁺ can occur at m/z 46 and a CH₂OH⁺ fragment at 31; fragment and isotope patterns must fit the same candidate.
- Using typical reference ranges, alcohol O-H is broad around 3200-3600 cm⁻¹, carboxylic-acid O-H extends roughly 2500-3300 cm⁻¹, and C=O is strong around 1650-1750 cm⁻¹ depending on the functional group. Compare the fingerprint region with a reference spectrum; a single broad absorption is not an entire identification.
- Propanone has two ¹³C environments: two equivalent methyl carbons give one signal and the carbonyl carbon another. Ethanol has two carbon environments and three low-resolution proton environments, CH₃, CH₂ and OH, with nominal integrals 3:2:1 when exchange and measurement permit. Unit 4 adds high-resolution proton splitting.
What to do in an assessment
- Calculate m/z from ion charge, identify the molecular ion/isotope pattern, then match plausible fragments to bonds that could break.
- Count distinct ¹³C and ¹H environments, use relative proton areas where supplied, identify diagnostic IR absorptions, and reject candidates that fail any one independent constraint.
In plain language
Each instrument supplies a different clue: mass and fragments, bond types, or chemically distinct atoms. A structure must fit all of them.
Self-check
A compound has three ¹³C NMR signals. What can you infer, and what can you not infer from this alone?
Show the answer
It has three chemically distinct carbon environments in the measured spectrum; this does not uniquely determine the connectivity or prove a structure without the molecular formula and other evidence.
Topic group
A2 Unit 3: physical and inorganic chemistry
- Core
Scope: Core · Unit 3 §3.1(a, b, c, d, e, f); Unit 3 §3.1 specified practical 1
Open revision guidance
What to know
- Oxidation releases electrons; reduction accepts them. Write Zn(s) → Zn²⁺(aq) + 2e⁻ and Cu²⁺(aq) + 2e⁻ → Cu(s). Combining them gives Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), conserving atoms and charge.
- A half-cell separates a redox couple from another couple. A metal electrode contacts its ion solution; a couple such as Fe³⁺/Fe²⁺ needs an inert platinum electrode because both forms are dissolved. A salt bridge completes the ionic circuit without freely mixing the solutions. Electrons flow through the external wire, not through the salt bridge.
- Cell notation records phases with | and the salt bridge with ||: Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s). Write the oxidation half-cell on the left for this operating cell. For the iron couple use Pt(s)|Fe²⁺(aq),Fe³⁺(aq).
- A standard electrode potential E° is the reduction potential measured relative to the standard hydrogen electrode, assigned 0.00 V. The hydrogen electrode uses an inert platinum surface, hydrogen at 100 kPa, unit hydrogen-ion activity (approximated by 1 mol dm⁻³ solution) and a stated temperature, usually 298 K. These are comparison conditions, not an absolute potential measurement.
- For a standard cell, E°cell = E°reduction at cathode − E°reduction at anode. With Cu²⁺/Cu = +0.34 V and Zn²⁺/Zn = −0.76 V, E°cell = +1.10 V. Do not multiply E° by a stoichiometric coefficient. A positive value supports thermodynamic feasibility in the stated direction, but not a fast observable reaction; kinetic barriers and non-standard concentrations matter.
- In an alkaline hydrogen fuel cell, the anode oxidises hydrogen: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻; the cathode reduces oxygen: O₂ + 2H₂O + 4e⁻ → 4OH⁻. Overall 2H₂ + O₂ → 2H₂O. Reactants are supplied continuously. Water is the point-of-use product, but hydrogen manufacture, electricity source, storage, transport and catalyst cost affect the overall environmental and economic benefit.
What to do in an assessment
- Construct the specified cell using cleaned metal electrodes, suitable measured ion solutions, a nitrate salt bridge and a high-resistance voltmeter. Record temperature, concentrations, electrode identity, polarity and voltage. Avoid a reactive bridge ion and prevent an electrical short.
- Identify the lower reduction potential as the oxidation half-cell, reverse its half-equation, then balance electrons before adding. The zinc electrode is the negative anode and copper the positive cathode in this galvanic cell; nitrate migrates towards the zinc side and potassium towards the copper side in the illustrated nitrate bridge.
- For Fe³⁺/Fe²⁺, include both ions and the inert electrode. Compare observed voltage with the standard prediction and explain concentration, temperature, contamination and contact resistance rather than forcing the measurement to match.
In plain language
Separate electron transfer into two half-cells to obtain a measurable voltage.
Self-check
An oxidation half-cell has E° = −0.44 V and a reduction half-cell E° = +0.80 V. What is E°cell, and does it guarantee a rapid reaction?
Show the answer
E°cell = 0.80 − (−0.44) = +1.24 V. The direction is thermodynamically feasible under standard conditions; activation barriers can still make it slow.
View diagrams
Zn/Cu electrochemical cell · standard conditionsZinc is oxidised at the negative anode; copper(II) ions are reduced at the positive cathode. The KNO₃ bridge maintains ionic balance: NO₃⁻ moves toward the anode and K⁺ toward the cathode. Electrons travel only through the external circuit.
Use the labelled model with the explanation; arrows represent electron-pair movement or the stated physical flow, not interchangeable processes. - Core
Scope: Core · Unit 3 §3.2(a, b, c, d); Unit 3 §3.2 specified practical 1, 2
Open revision guidance
What to know
- In acidic solution, balance O with H₂O, H with H⁺ and charge with electrons. Required reduction half-equations are Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O and MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Thiosulfate oxidation is 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻.
- To titrate Fe²⁺ with permanganate, combine Fe²⁺ → Fe³⁺ + e⁻ five times with the permanganate equation: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. Acidify with suitable dilute sulfuric acid, not hydrochloric acid, which can itself be oxidised. The first persistent pale pink excess of MnO₄⁻ is the endpoint; no separate indicator is needed.
- Iodide reduces Cu²⁺ while CuI precipitates: 2Cu²⁺ + 4I⁻ → 2CuI(s) + I₂. Titrate the released iodine using I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. Together these equations show n(Cu²⁺) = n(S₂O₃²⁻). Precipitation helps drive the copper reaction; it is not Cu metal formation.
- Use excess iodide in the measured copper aliquot. Titrate the iodine promptly with standard thiosulfate; add starch only near the endpoint when the solution is pale, then continue dropwise until the blue iodine-starch colour disappears. Treat dilution and aliquot factors separately from the reacting ratio.
- For 25.00 cm³ of 0.0200 mol dm⁻³ MnO₄⁻, n = 5.00 × 10⁻⁴ mol. This reacts with 2.50 × 10⁻³ mol Fe²⁺. If the iron aliquot is 25.0 cm³, c(Fe²⁺) = 0.100 mol dm⁻³. The same volume of 0.0200 mol dm⁻³ thiosulfate instead corresponds to 5.00 × 10⁻⁴ mol copper ions, not five times this amount.
What to do in an assessment
- Prepare or standardise the titrant as required by the supplied method. Rinse apparatus with the appropriate solution, pipette an aliquot, record both burette readings, run a rough titration then obtain justified concordant titres. Use a white tile and mix continuously near the endpoint.
- For copper in a salt, weigh and quantitatively dissolve the sample, make up a known volumetric flask, titrate an aliquot and scale its moles to the full flask. For a hydrate, compare the measured copper fraction with the candidate formula mass.
- Control oxidant and acid hazards, protect eyes, and collect copper/chromium-containing waste. Method preparation does not certify that the learner has performed the separately assessed practical examination.
In plain language
Balance the electrons first; use the resulting mole ratio in the titration.
Self-check
20.00 cm³ of 0.100 mol dm⁻³ thiosulfate titrates iodine released from a copper aliquot. How many moles and what mass of copper does the aliquot contain? Take M(Cu) = 63.5 g mol⁻¹.
Show the answer
n(S₂O₃²⁻) = 0.100 × 0.02000 = 0.00200 mol. n(Cu²⁺) is also 0.00200 mol, so m(Cu) = 0.00200 × 63.5 = 0.127 g.
- Core
Scope: Core · Unit 3 §3.3(a, b, c, d, e)
Open revision guidance
What to know
- An amphoteric oxide or hydroxide reacts with both acids and bases. Al(OH)₃ + 3H⁺ → Al³⁺ + 3H₂O; with excess hydroxide Al(OH)₃ + OH⁻ → [Al(OH)₄]⁻. Similarly Pb(OH)₂ + 2H⁺ → Pb²⁺ + 2H₂O and Pb(OH)₂ + 2OH⁻ → [Pb(OH)₄]²⁻. Lead can be passivated by an insoluble salt, so an apparent absence of reaction needs interpretation.
- The metals also dissolve in suitable acids and strong alkali. For aluminium, 2Al + 6H⁺ → 2Al³⁺ + 3H₂ and 2Al + 2OH⁻ + 6H₂O → 2[Al(OH)₄]⁻ + 3H₂. The simple non-oxidising acid equation for lead is Pb + 2H⁺ → Pb²⁺ + H₂, although insoluble coatings can inhibit it. In alkali, Pb + 2OH⁻ + 2H₂O → [Pb(OH)₄]²⁻ + H₂. An oxide coating and conditions affect the observed rate.
- The inert pair effect is the increasing reluctance of the outer ns² pair to participate in bonding down the p-block. Lower oxidation states become more stable: Group 3 has +1 increasingly favoured relative to +3, Group 4 has +2 relative to +4, and Group 5 has +3 relative to +5. These are the specification's traditional group labels (modern Groups 13, 14 and 15).
- The octet rule is not universal. BF₃ has only six electrons around boron; PCl₅ has ten around phosphorus, SF₆ twelve around sulfur, and ClF₃ ten around chlorine when its three bonds and two lone pairs are counted. Avoid applying a simple second-period octet rule indiscriminately to later-period elements.
- In Al₂Cl₆, two chloride atoms bridge the aluminium centres, with lone-pair donation into the electron-deficient centres; each Al has four neighbouring Cl atoms. In H₃N→BF₃, nitrogen donates its lone pair to boron to form a coordinate bond; BF₃ accepts the pair rather than supplying it.
- Hexagonal boron nitride has planar covalent layers with weaker attractions between layers, giving lubricating behaviour; unlike graphite it lacks freely conducting delocalised electrons. Cubic boron nitride is a three-dimensional tetrahedral network, extremely hard and used in cutting tools. Both have strong B-N bonding and high thermal stability.
What to do in an assessment
- Test the amphoteric claim with separate fresh samples in acid and in excess alkali. Record initial precipitation and subsequent dissolution; do not confuse a metal reaction producing hydrogen with hydroxide dissolution.
- Draw the Al₂Cl₆ dimer with two bridging chlorides, then mark donor and acceptor. For NH₃BF₃, start the donation arrow at nitrogen's lone pair and end at boron. Check valency and total charge.
- Use network dimension, interlayer forces and mobile charge carriers to compare the two BN structures rather than assuming that every graphite-like solid conducts electricity.
In plain language
Electron availability and structure explain why p-block chemistry goes beyond the octet rule.
Self-check
Why does Al(OH)₃ dissolve in both acid and excess hydroxide, and what lower oxidation state becomes more stable down Group 4?
Show the answer
It is amphoteric: acid forms Al³⁺ and water; hydroxide forms [Al(OH)₄]⁻. The +2 state becomes more stable relative to +4 down Group 4 because of the inert pair effect.
- Core
Scope: Core · Unit 3 §3.3(f, g, h, i)
Open revision guidance
What to know
- Carbon(II) in CO is readily oxidised to carbon(IV), so CO reduces some metal oxides: CuO + CO → Cu + CO₂. Down Group 4 the +2 state becomes more stable. Lead(IV) is therefore an oxidant: PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O. Lead is reduced from +4 to +2 while chloride is oxidised to chlorine.
- CO₂ is a linear simple covalent molecule, a gas at room conditions; weak intermolecular attractions give low boiling/sublimation temperatures. It is an acidic oxide: CO₂ + 2OH⁻ → CO₃²⁻ + H₂O. PbO is a high-melting extended solid with substantial ionic bonding and is amphoteric: PbO + 2H⁺ → Pb²⁺ + H₂O and PbO + 2OH⁻ + H₂O → [Pb(OH)₄]²⁻.
- CCl₄ and SiCl₄ are covalent molecular liquids, whereas PbCl₂ is predominantly ionic and a sparingly soluble solid. CCl₄ does not hydrolyse readily under the usual conditions. SiCl₄ reacts vigorously with water: SiCl₄ + 2H₂O → SiO₂ + 4HCl; hydrated silica may be produced in practice. The differing accessibility of a reaction pathway matters, not just whether chlorine is present.
- Adding OH⁻ to Pb²⁺ gives a white Pb(OH)₂ precipitate, which dissolves in excess hydroxide as [Pb(OH)₄]²⁻. Cl⁻ gives white PbCl₂ and I⁻ gives yellow PbI₂: Pb²⁺ + 2X⁻ → PbX₂(s). PbCl₂ and PbI₂ are more soluble in hot water; cooling a hot PbI₂ solution can give yellow crystalline plates.
- All lead compounds are toxic. Chlorine, concentrated hydrochloric acid and silicon tetrachloride require suitable controlled laboratory procedures and ventilation; these equations are preparation for supervised work, not home experiments.
What to do in an assessment
- Track carbon and lead oxidation numbers separately, then identify the species supplying and accepting electrons. A reducing agent itself becomes oxidised.
- Compare chloride bonding and water reaction independently. Dissolving an ionic chloride is not the same process as hydrolysing a molecular chloride.
- For an unknown lead solution use separate small aliquots for hydroxide, chloride and iodide, noting both colour and changes in excess reagent.
In plain language
The stable oxidation state and bonding change down Group 4.
Self-check
Balance the reaction of PbO₂ with concentrated HCl and identify the oxidising agent.
Show the answer
PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O. PbO₂ is the oxidising agent: Pb(IV) gains two electrons to become Pb(II); two chloride ions lose two electrons to form Cl₂.
- Core
Scope: Core · Unit 3 §3.3(j, k, l)
Open revision guidance
What to know
- Disproportionation oxidises and reduces the same starting element. In cold dilute NaOH, Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O: chlorine changes from 0 to −1 and +1. In hot concentrated alkali, 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O, giving −1 and +5. Heating can also convert chlorate(I) to chlorate(V): 3ClO⁻ → 2Cl⁻ + ClO₃⁻.
- Chlorine in water produces HCl and HClO in an equilibrium; HClO/ClO⁻ oxidises coloured substances and biological molecules, giving bleaching and bactericidal effects. Oxidising power is the chemical basis, but dose, contact time, pH and harmful by-products matter for water treatment.
- NaCl with concentrated H₂SO₄ gives HCl fumes: NaCl + H₂SO₄ → NaHSO₄ + HCl. Chloride is not a strong enough reducing agent to reduce the acid significantly under these conditions; this is an acid-base displacement, not a redox step.
- NaBr first gives HBr by the analogous acid-base reaction. HBr then reduces sulfuric acid to SO₂: 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O. Brown bromine vapour is produced and sulfur is reduced from +6 to +4.
- Iodide is a stronger reducing agent. HI formed initially can reduce H₂SO₄ to SO₂, sulfur or H₂S: 2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O; 6HI + H₂SO₄ → 3I₂ + S + 4H₂O; 8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O. Expect iodine and, depending on conditions, yellow sulfur and toxic gases. Do not identify a gas by deliberately smelling it.
What to do in an assessment
- Assign oxidation numbers to every chlorine or sulfur-containing product. Match the number of electrons lost and gained before deciding whether an equation is disproportionation.
- For the halide/sulfuric acid reactions, separate the initial proton-transfer stage from subsequent redox. Use the observed products and balanced equations rather than one vague statement that all halides react alike.
In plain language
Halogens can undergo disproportionation, while halide reducing power increases down the group.
Self-check
Why is Cl₂ reacting with cold OH⁻ a disproportionation reaction, and what is the chlorate(V) oxidation number?
Show the answer
Chlorine starts at 0 and forms both Cl⁻ (−1) and ClO⁻ (+1), so it is reduced and oxidised. In ClO₃⁻, chlorine is +5.
- Core
Scope: Core · Unit 3 §3.4(a, b, c, d, e, f, j)
Open revision guidance
What to know
- The 3d and 4s levels are close in energy, allowing several oxidation states. Remove 4s electrons before 3d electrons when writing ions. Important states here are Cr(III, VI), Mn(II, IV, VII), Fe(II, III), Co(II, III) and Cu(I, II). For example Fe is [Ar]3d⁶4s², Fe²⁺ is [Ar]3d⁶ and Fe³⁺ is [Ar]3d⁵.
- A ligand donates a lone pair into a coordinate bond with a metal ion. Coordination number counts donor atoms, not necessarily ligand molecules. Six monodentate ligands give an octahedral complex with adjacent 90° angles; four can give a tetrahedral complex with approximately 109.5° angles. Do not assume that every four-coordinate complex is tetrahedral.
- Ligands split the five d orbitals into energy groups; absorption promotes an electron between them when hf matches the gap. The light not absorbed determines the observed complementary colour. Ligand identity and oxidation state change the splitting. A colourless d⁰ or d¹⁰ ion lacks this particular d-d transition; other transition-metal species can still be coloured by charge-transfer processes.
- Recognise the required aqueous colours: chromate(VI) CrO₄²⁻ yellow; dichromate(VI) Cr₂O₇²⁻ orange; permanganate(VII) MnO₄⁻ purple; Fe²⁺ pale green; Fe³⁺ yellow/brown in usual aqueous solutions; Cu²⁺ blue; Co²⁺ pink. Cr³⁺ solutions are often green, while the pure hexaaqua chromium(III) ion is violet; its ligands and preparation affect the observation.
- [Cu(H₂O)₆]²⁺ is blue and approximately octahedral. Excess ammonia forms deep-blue [Cu(NH₃)₄(H₂O)₂]²⁺, still approximately octahedral. High chloride concentration forms tetrahedral [CuCl₄]²⁻, yellow; mixtures with the blue aquo complex often look green. [Co(H₂O)₆]²⁺ is pink and octahedral, while [CoCl₄]²⁻ is blue and tetrahedral.
- Ligand exchange can change both colour and coordination number: [Co(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CoCl₄]²⁻ + 6H₂O. Adding concentrated HCl increases chloride availability; dilution with water favours the aquo complex. Changing coordination from six to four is not changing cobalt's +2 oxidation state.
- With OH⁻, Cr³⁺ gives grey-green Cr(OH)₃, soluble in excess as a hydroxo complex such as [Cr(OH)₆]³⁻. Fe²⁺ gives green Fe(OH)₂, which browns on standing as air oxidises it. Fe³⁺ gives red-brown Fe(OH)₃; Cu²⁺ gives blue Cu(OH)₂. The iron and copper hydroxides do not dissolve in excess aqueous NaOH under the usual test conditions.
What to do in an assessment
- Write ligand-exchange equations with ligands, water and overall charges balanced. Draw an octahedron with four equatorial and two axial donor positions; distinguish its 90° adjacent angles from tetrahedral 109.5° geometry.
- Use fresh labelled aliquots for OH⁻ and ligand tests. Record initial precipitates, the result in excess and changes on standing; concentrated HCl is corrosive and the metal-containing waste requires collection.
- Explain colour using energy splitting and absorption before naming the observed colour. Oxidation state alone cannot predict the colour of every complex.
In plain language
Ligands change bonding, shape and colour without necessarily changing oxidation state.
Self-check
Pink [Co(H₂O)₆]²⁺ becomes blue [CoCl₄]²⁻. What changes and what stays the same?
Show the answer
Water ligands are replaced by chloride, coordination falls from six to four, and geometry changes from octahedral to tetrahedral. Cobalt remains in oxidation state +2.
- Core
Scope: Core · Unit 3 §3.4(g, h, i)
Open revision guidance
What to know
- A heterogeneous catalyst is in a different phase from the reacting material. Reactants adsorb onto surface sites, bonds are weakened or brought into favourable positions, reaction occurs by a lower-activation-energy route and products desorb. Surface poisoning or reduced available area can decrease activity.
- Nickel catalyses hydrogenation of alkenes: CH₂=CH₂ + H₂ → CH₃CH₃. Iron catalyses the Haber reaction N₂ + 3H₂ ⇌ 2NH₃. Both illustrate surface catalysis and regeneration of the catalyst, not a change to the equilibrium constant.
- Vanadium(V) oxide catalyses the contact process, 2SO₂ + O₂ ⇌ 2SO₃. A simplified redox cycle is V₂O₅ + SO₂ → V₂O₄ + SO₃, followed by 2V₂O₄ + O₂ → 2V₂O₅. Manganese(IV) oxide catalyses 2H₂O₂ → 2H₂O + O₂.
- A homogeneous catalyst shares the reactant phase. Variable oxidation states allow an intermediate electron-transfer route. For example iron ions can catalyse persulfate oxidation of iodide: S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺, then 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. Adding the steps cancels the iron species.
- A catalyst supplies a different mechanism, often with several lower barriers. It changes forward and reverse rates but not ΔH, ΔG or the equilibrium composition at a fixed temperature. Industrial choice also involves selectivity, lifetime, cost and safe handling.
What to do in an assessment
- Classify by the physical phases, not by whether the catalyst is a metal. For a redox cycle, add the intermediate equations to show regeneration and recover the net reaction.
- Link each named catalyst to its required reaction. Explain adsorption and desorption for a surface catalyst, and intermediate oxidation-state changes for a homogeneous example.
In plain language
Surface sites and variable oxidation states offer alternative reaction routes.
Self-check
Name the catalysts for the Haber and contact processes, and explain why neither increases the equilibrium yield at unchanged temperature.
Show the answer
Iron is used for Haber; vanadium(V) oxide for contact. They lower activation barriers and accelerate both directions, so equilibrium is reached faster without changing its position.
- Core
Scope: Core · Unit 3 §3.5(a, b, c, d, e); Unit 3 §3.5 specified practical 1
Open revision guidance
What to know
- For rate = k[A]ᵐ[B]ⁿ, m and n are experimental orders and the overall order is m+n. They cannot generally be read from the overall balanced equation. Compare experiments in which only one concentration changes: doubling it leaves rate unchanged for zero order, doubles it for first order and quadruples it for second order.
- If doubling [A] doubles rate and doubling [B] quadruples rate, rate = k[A][B]². With rate 8.00 × 10⁻⁴ mol dm⁻³ s⁻¹, [A] = 0.100 and [B] = 0.200 mol dm⁻³, k = 0.200 dm⁶ mol⁻² s⁻¹. Derive units from rate divided by the concentration factors; the units depend on overall order.
- Sampling and quenching means withdrawing an aliquot at a known time, rapidly stopping or greatly slowing its reaction, then measuring its composition. A suitable quench may use rapid cooling or chemical removal of a reactant, but must not destroy the analyte being measured. Correct for any sampling or dilution effects.
- An initial rate is the tangent gradient at t=0 on a concentration-time curve. A zero-order concentration-time graph is linear with slope −k; for a first-order reaction ln[A] against t is linear with slope −k and equal successive half-lives. A clock's fixed endpoint gives a measure proportional to 1/t only when the amount needed to reach that endpoint is fixed and the concentration change is small.
- The rate-determining step is the slow step in a proposed mechanism. A candidate mechanism must sum to the overall equation and be consistent with the measured rate law. An elementary slow collision between A and B supports rate proportional to [A][B]; a slow unimolecular step supports first order in that intermediate/reactant. Rate data constrain a mechanism but do not uniquely prove it.
- For the specified acidic peroxide/iodide reaction, H₂O₂ + 2I⁻ + 2H⁺ → I₂ + 2H₂O. Follow iodine appearance, for example with a controlled clock system, or sample and titrate quenched aliquots. Vary one initial concentration at a time, compensate volumes with water and hold temperature and acidity constant when testing iodide or peroxide order.
What to do in an assessment
- Write each rate ratio with all concentration ratios. Solve the order from the isolated comparison, then substitute one complete experiment to find k and check it against another.
- For the peroxide/iodide investigation, state the defined endpoint and the fixed quantity it represents. Use repeats, consistent mixing and a thermostated bath; do not treat a smaller time as a rate with molar units without conversion.
- Draw a large tangent and calculate Δconcentration/Δtime from two points on that tangent, not from two arbitrary points on the curve.
In plain language
Measured rates tell you how concentrations influence the pathway.
Self-check
Doubling A makes the rate four times larger; doubling B makes no difference. Write the rate equation and the units of k when rate is in mol dm⁻³ s⁻¹.
Show the answer
rate = k[A]²[B]⁰ = k[A]². Overall order is two and k has units dm³ mol⁻¹ s⁻¹.
- Core
Scope: Core · Unit 3 §3.5(f)
Open revision guidance
What to know
- The Arrhenius equation is k = A exp(−Eₐ/RT). A is the frequency factor, incorporating encounter frequency and orientation; Eₐ is the activation energy. Use T in kelvin and Eₐ in J mol⁻¹ when R = 8.31 J mol⁻¹ K⁻¹.
- Taking natural logarithms gives ln k = ln A − Eₐ/(RT). A plot of ln k against 1/T is a straight line: gradient = −Eₐ/R and intercept = ln A. Therefore Eₐ = −R × gradient and A = exp(intercept). A has the same units as k.
- If gradient = −6.00 × 10³ K, Eₐ = 8.31 × 6.00 × 10³ = 49.9 kJ mol⁻¹. If intercept = 12.0, A = e¹² = 1.63 × 10⁵ in the units of k. Do not use a base-10 logarithm in the natural-log form without the 2.303 conversion factor.
- At two temperatures, ln(k₂/k₁) = Eₐ/R × (1/T₁ − 1/T₂). For Eₐ = 50.0 kJ mol⁻¹, T₁ = 300 K and T₂ = 310 K, ln(k₂/k₁) ≈ 0.647, so k₂/k₁ ≈ 1.91. Temperature can increase the rate constant substantially without altering the measured concentration.
- A catalyst changes the reaction pathway and effective activation barrier. It commonly gives a larger k at the same temperature; it does not merely heat the sample or increase the reactants' molecular energy.
What to do in an assessment
- Convert all temperatures to K, compute reciprocal temperatures and natural logarithms, label axes with the underlying quantities and units, and obtain a best-fit gradient and intercept.
- Estimate uncertainty from data scatter or plausible gradient limits. Do not round 1/T values so early that distinct measurements become identical.
In plain language
A reciprocal-temperature plot separates activation energy from the frequency factor.
Self-check
An ln k against 1/T plot has gradient −4.00 × 10³ K. Find Eₐ using R = 8.31 J mol⁻¹ K⁻¹.
Show the answer
Eₐ = −R × gradient = 3.324 × 10⁴ J mol⁻¹ = 33.2 kJ mol⁻¹ to three significant figures.
- Core
Scope: Core · Unit 3 §3.6(a, b, c, d)
Open revision guidance
What to know
- Atomisation forms one mole of gaseous atoms from an element in its standard state: Na(s) → Na(g), or ½Cl₂(g) → Cl(g). Lattice formation forms one mole of ionic solid from gaseous ions and is exothermic; lattice breaking reverses it and is endothermic. Always state which sign convention is used.
- Hydration transfers gaseous ions into aqueous solution, for example Na⁺(g) → Na⁺(aq), and is normally exothermic. Enthalpy of solution refers to dissolving one mole of solid in sufficient water under the stated conditions. ΔHsolution = ΔHlattice breaking + ΣΔHhydration.
- If lattice breaking costs +770 kJ mol⁻¹ and hydration releases −760 kJ mol⁻¹, ΔHsolution = +10 kJ mol⁻¹. The solution cools when dissolution draws heat from its surroundings. Endothermic solution can nevertheless occur because entropy also affects feasibility; ΔH alone does not decide solubility.
- A Born-Haber cycle connects elemental standard states to solid formation through gaseous atoms and ions. For NaCl: atomise sodium, atomise half a mole of chlorine molecules, ionise sodium, add an electron to chlorine (electron affinity), then form the lattice. Sum the signed steps to ΔfH°. Multiply ionisation/electron-affinity steps by the stoichiometric number of ions when necessary.
- Using ΔfH°(NaCl) = −411, Na atomisation +108, ½Cl₂ atomisation +121, Na first ionisation +496 and Cl electron affinity −349 kJ mol⁻¹ gives lattice formation = −411 − (108 + 121 + 496 − 349) = −787 kJ mol⁻¹.
- A more negative formation enthalpy suggests greater stability relative to the elements enthalpically, but cannot alone establish the preferred reaction against every competing product or at every temperature. Consider the chosen reference reaction and Gibbs energy.
What to do in an assessment
- Write the gaseous atoms and ions explicitly in every cycle stage. Use half a mole of a diatomic molecule when forming one mole of its atoms, and include any second ionisation step for a divalent ion.
- For dissolution, reverse a quoted lattice formation value before adding hydration values. Keep a complete signed equation before isolating the unknown.
In plain language
A signed energy cycle distinguishes breaking a lattice from hydrating its ions.
Self-check
An ionic solid has lattice formation −800 kJ mol⁻¹ and total hydration −835 kJ mol⁻¹. Find ΔHsolution.
Show the answer
Lattice breaking is +800 kJ mol⁻¹, so ΔHsolution = 800 − 835 = −35 kJ mol⁻¹: dissolution is exothermic.
- Core
Scope: Core · Unit 3 §3.7(a, b, c, d, e)
Open revision guidance
What to know
- Entropy S describes the distribution of energy and freedom available to particles. Other factors being equal, gas has more possible arrangements than liquid, and liquid more than solid. More gaseous particles often increases entropy, but use actual absolute entropies where data are supplied.
- For a reaction, ΔS = ΣνS(products) − ΣνS(reactants), including the equation coefficients. S and ΔS are commonly quoted in J mol⁻¹ K⁻¹. Spontaneous natural changes increase total entropy of system plus surroundings towards equilibrium; the system alone need not increase in entropy.
- Gibbs energy combines enthalpy and entropy: ΔG = ΔH − TΔS. At constant temperature and pressure, negative ΔG makes the stated direction thermodynamically feasible; positive ΔG favours the reverse, and ΔG = 0 describes equilibrium. Feasibility does not guarantee a fast rate.
- Convert ΔS to kJ mol⁻¹ K⁻¹ if ΔH is in kJ mol⁻¹. With ΔH = +20.0 kJ mol⁻¹ and ΔS = +100 J mol⁻¹ K⁻¹, at 298 K ΔG = 20.0 − 298 × 0.100 = −9.80 kJ mol⁻¹. An endothermic reaction can occur because the entropy contribution outweighs the enthalpy cost.
- If ΔH and ΔS are both positive, feasibility is favoured above T = ΔH/ΔS; both negative favours lower temperatures. Negative ΔH with positive ΔS is favourable at all temperatures in this simple constant-value model, and the reverse sign combination is unfavourable. Values may themselves vary with temperature, so a calculated threshold is an approximation.
What to do in an assessment
- Use a products-minus-reactants table with coefficients, then convert units before substituting T in kelvin. Check whether your sign prediction agrees with the number of gas particles and phase changes.
- Separate a thermodynamic prediction from evidence about activation energy, rate and equilibrium composition.
In plain language
A reaction's enthalpy and entropy together determine its thermodynamic direction.
Self-check
ΔH = +30.0 kJ mol⁻¹ and ΔS = +120 J mol⁻¹ K⁻¹. Above what approximate temperature is the reaction feasible?
Show the answer
Set ΔG = 0: T = 30.0/0.120 = 250 K. With constant ΔH and ΔS, ΔG is negative above 250 K; this says nothing by itself about reaction speed.
- Core
Scope: Core · Unit 3 §3.8(a, b, c); Unit 3 §3.8 specified practical 1
Open revision guidance
What to know
- Kc uses equilibrium concentrations; Kp uses gas partial pressures. For N₂ + 3H₂ ⇌ 2NH₃, Kp = p(NH₃)²/[p(N₂)p(H₂)³]. If pressures are in kPa, its units are kPa⁻². Pure solids and pure liquids do not appear; ensure that the expression matches the chosen equation.
- A partial pressure is mole fraction × total pressure. With equilibrium mole fractions 0.20 N₂, 0.60 H₂ and 0.20 NH₃ at 100 kPa, pressures are 20, 60 and 20 kPa. Kp = 20²/(20 × 60³) = 9.26 × 10⁻⁵ kPa⁻². Use equilibrium amounts, not starting amounts.
- For A + B ⇌ C + D starting with 1.00 mol each of A and B and no products in a fixed volume, let x react. Equilibrium amounts are 1−x, 1−x, x and x; equal powers make volume cancel. If Kc = 4.00, x²/(1−x)² = 4, so x/(1−x) = 2 and x = 0.667 mol. Reject roots giving negative amounts.
- Large equilibrium constants indicate a product-favoured position for the stated reaction; small ones favour reactants. A numerical value changes when an equation is reversed or scaled. A constant describes position, not the speed of reaching it.
- Temperature changes K. Heating decreases K for an exothermic forward reaction and increases K for an endothermic one. Concentration, pressure and a catalyst can change amounts or the time taken to equilibrate, but do not change the value of K at a fixed temperature.
- The specified ester equilibrium is CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. In the homogeneous mixture, use the appropriate concentration expression for all four species rather than automatically omitting water that is a changing reactant/product rather than a pure solvent phase. Establish equilibrium at controlled temperature and determine remaining acid by titration of a measured sample, accounting for any acid catalyst separately.
What to do in an assessment
- Make an initial-change-equilibrium table and use the equation coefficients for every change. Convert equilibrium moles into concentrations or partial pressures before substituting.
- For the ester practical, measure starting quantities, retain volatile components with a suitable closed or reflux arrangement from the approved method, allow equilibration, and titrate samples promptly. Use a suitable blank/correction for the catalyst and compare repeated measurements at the same temperature.
In plain language
The equation and equilibrium amounts determine the constant and its units.
Self-check
For H₂(g) + I₂(g) ⇌ 2HI(g), partial pressures are 10.0, 10.0 and 40.0 kPa respectively. Find Kp and its units.
Show the answer
Kp = 40.0²/(10.0 × 10.0) = 16.0. The pressure powers cancel, so this Kp is dimensionless.
- Core
Scope: Core · Unit 3 §3.9(a, b, c, d, h)
Open revision guidance
What to know
- A Brønsted-Lowry acid donates a proton and a base accepts it. Conjugate partners differ by one H⁺: HA/A⁻ and BH⁺/B. Strong and weak describe extent of ionisation, not how concentrated a solution is.
- For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA] and pKa = −log₁₀Ka. Larger Ka and smaller pKa indicate a stronger acid in comparable conditions. For a weak base, proton acceptance from water is incomplete; the conjugate acid and base equilibria are linked.
- Kw = [H⁺][OH⁻]. At 298 K take Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶, so pH + pOH = 14.00. Kw changes with temperature; neutral means equal [H⁺] and [OH⁻], not invariably pH 7 at every temperature.
- A 0.0100 mol dm⁻³ fully dissociated monoprotic acid has pH 2.00. A 0.0200 mol dm⁻³ NaOH solution has [H⁺] = Kw/[OH⁻] = 5.00 × 10⁻¹³ mol dm⁻³ and pH 12.30 at 298 K. Account for the number of hydroxide ions from each formula unit of a strong base and for dilution after mixing.
- For a weak monoprotic acid of concentration c, let x = [H⁺] from dissociation. Ka = x²/(c−x). If x is much smaller than c, x ≈ √(Ka c). With Ka = 1.80 × 10⁻⁵ mol dm⁻³ and c = 0.100 mol dm⁻³, x = 1.34 × 10⁻³ and pH = 2.87; x/c is about 1.34%, so the approximation is justified. Otherwise solve the quadratic and reject a negative root.
- Salt hydrolysis explains non-neutral solutions. NaCl, from strong acid and strong base, has no significant hydrolysis. NH₄Cl contains the weak-base conjugate acid: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, so it is acidic. Sodium ethanoate contains the weak-acid conjugate base: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, so it is alkaline.
What to do in an assessment
- For a mixture first calculate and neutralise moles using stoichiometry, then find the concentration of the remaining strong acid/base or the resulting weak-acid system. Do not average two pH values.
- State temperature, constant units and approximation. Check the calculated dissociation fraction before treating [HA] as unchanged.
In plain language
Start from the equilibria and remaining moles, not from the word acid or base alone.
Self-check
At 298 K, find the pH of 0.00100 mol dm⁻³ NaOH using Kw = 1.00 × 10⁻¹⁴.
Show the answer
[H⁺] = 1.00 × 10⁻¹¹ mol dm⁻³, so pH = 11.00.
- Core
Scope: Core · Unit 3 §3.9(e, f, g, i); Unit 3 §3.9 specified practical 1
Open revision guidance
What to know
- An acidic buffer contains a weak acid HA and its conjugate base A⁻. Added H⁺ is removed by A⁻ + H⁺ → HA; added OH⁻ is removed by HA + OH⁻ → A⁻ + H₂O. Both components must remain available. A basic buffer such as NH₃/NH₄⁺ works through the analogous conjugate pair.
- From Ka, [H⁺] ≈ Ka[HA]/[A⁻], giving pH = pKa + log₁₀([A⁻]/[HA]). With equal component concentrations, pH ≈ pKa. If pKa = 4.76 and [A⁻]/[HA] = 2.00, pH = 5.06. Volume cancels from a ratio of moles only when both species occupy the same final solution.
- For added strong acid or base, first update buffer moles by complete neutralisation, then apply the equilibrium approximation. From 0.0100 mol HA and 0.0100 mol A⁻, adding 0.00200 mol H⁺ leaves 0.0120 mol HA and 0.00800 mol A⁻: pH = 4.76 + log₁₀(0.00800/0.0120) = 4.58. Buffer capacity is finite and falls when either component is exhausted.
- Buffers stabilise pH for enzyme activity, blood chemistry and industrial preparations or fermentation. A buffer resists small changes; it does not keep pH exactly fixed after arbitrary addition or eliminate the need to monitor a process.
- For strong acid/strong base titration, the initial pH is low, there is a large steep change and equivalence is approximately pH 7 at 298 K. Weak acid/strong base starts at higher pH, has a buffer region and equivalence above 7; halfway to equivalence pH ≈ pKa. Strong acid/weak base has equivalence below 7 and a smaller jump. Weak acid/weak base has no large sharp vertical interval and the equivalence pH depends on both strengths.
- Choose an indicator whose transition range lies within the steep interval: phenolphthalein suits many weak-acid/strong-base titrations and methyl orange suits many strong-acid/weak-base titrations. For weak acid/weak base, a common indicator is unreliable; use a measured pH curve and suitable analysis rather than inventing a sharp endpoint.
- For the specified pH-probe titration, calibrate with suitable standard buffers, rinse the probe between solutions and immerse it consistently. Add measured titrant increments, allow readings to settle and use smaller increments near the changing region. Plot pH against cumulative volume and compare the actual curve with the acid/base strengths.
What to do in an assessment
- Sketch all four curves with labelled axes, starting pH, buffer region where appropriate and equivalence point. Distinguish an indicator's observed endpoint from the stoichiometric equivalence point.
- For buffer calculations, show the neutralisation table before substituting Ka or pKa. For a basic buffer use the conjugate acid constant or link pOH and pH through Kw at the stated temperature.
- Use a data logger if available, record calibration and temperature, and explain probe drift, mixing, additions and uncertainties in the chosen equivalence volume.
In plain language
Conjugate pairs resist pH change; titration curves show where an endpoint can be measured.
Self-check
A buffer has pKa = 4.76, 0.0200 mol HA and 0.0100 mol A⁻ in the same solution. Find its approximate pH.
Show the answer
pH = 4.76 + log₁₀(0.0100/0.0200) = 4.46. The concentration ratio is the same as the mole ratio here.
Topic group
A2 Unit 4: organic chemistry and analysis
- Core
Scope: Core · Unit 4 §4.1(a, b, c, d)
Open revision guidance
What to know
- Structural isomers have the same molecular formula but different connectivity. Stereoisomers have the same connectivity but different spatial arrangements. Restricted rotation about C=C produces E-Z isomerism; a tetrahedral carbon with four different attached groups can produce optical isomerism.
- For an alkene, each double-bond carbon must have two different groups. Assign priority separately at each carbon using atomic number (and the first point of difference further along a group). Higher-priority groups on the same side give Z, and on opposite sides E. A molecule with two identical groups on either double-bond carbon cannot have this E-Z pair.
- A chiral centre is commonly an asymmetric carbon bonded to four different groups. Butan-2-ol has H, OH, CH₃ and CH₂CH₃ at carbon 2. Its two mirror-image arrangements are not superimposable and are enantiomers. Use solid wedges towards the viewer and hashed wedges away; a flat four-bond cross without a depth convention cannot demonstrate the geometry.
- An optically active enantiomer rotates plane-polarised light. Its mirror partner rotates through the same angle in the opposite direction under the same conditions. The direction of rotation cannot be inferred merely from the drawing or an R/S label.
- A racemic mixture contains equal amounts of two enantiomers; their optical rotations cancel. Enzymes and other chiral biological receptors can distinguish the enantiomers, so equal formulae do not guarantee identical biological effects. Glycine has two H atoms on its alpha carbon and is not chiral.
What to do in an assessment
- List the four groups at each proposed stereocentre, checking complete connectivity rather than just the first attached atom. Draw its mirror image, keep every group attached to the same central carbon and test whether rotation, without bond breaking, superimposes the models.
- For but-2-ene compare the CH₃ groups on the two alkene carbons: same side is Z and opposite sides E. For a more complex example use priorities, not a memorised rule that the largest-looking group always wins.
In plain language
Connectivity can stay the same while three-dimensional arrangement changes.
Self-check
Does butan-2-ol show optical isomerism? Does propan-2-ol?
Show the answer
Butan-2-ol does: its OH-bearing carbon has four different groups. Propan-2-ol does not: that carbon has two identical CH₃ groups.
View diagrams
Optical isomerism · butan-2-olButan-2-ol has one chiral carbon attached to four different groups: OH, H, CH₃ and CH₂CH₃. The solid wedge points toward you; the hashed wedge points away. Optical isomerism is not E/Z isomerism.
Use the labelled model with the explanation; arrows represent electron-pair movement or the stated physical flow, not interchangeable processes. - Core
Scope: Core · Unit 4 §4.2(a, b, c, d)
Open revision guidance
What to know
- Benzene is a planar hexagon. Each carbon has three sigma bonds and a p orbital perpendicular to the plane. The six p orbitals overlap to delocalise six pi electrons above and below the ring; all six C-C bonds are equivalent and intermediate in length between typical single and double bonds.
- Delocalisation stabilises the ring. Benzene does not readily decolourise bromine water like an alkene because ordinary addition would destroy the aromatic electron system. Electrophilic substitution replaces a hydrogen while restoring aromaticity. An arene contains an aromatic ring; substituents can influence its reactivity.
- Nitration uses concentrated HNO₃/H₂SO₄ at controlled temperature, commonly around 50-55 °C for benzene. HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O generates the nitronium electrophile. A ring electron pair forms a C-N bond, giving a positively charged sigma intermediate that still has H on the attacked carbon. A base removes that proton; the C-H pair restores the delocalised system. Overall C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O.
- Halogenation requires activation, for example Br₂ with FeBr₃ or a suitable halogen carrier: C₆H₆ + Br₂ → C₆H₅Br + HBr. Activation polarises Br₂ and produces an electrophilic bromine species. Ring attack, a sigma intermediate and loss of H⁺ follow the same substitution pattern, regenerating the catalyst.
- Friedel-Crafts alkylation uses a halogenoalkane and anhydrous AlCl₃: C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl. The Lewis acid activates the C-Cl bond to supply the alkyl electrophile; ring attack and proton loss restore aromaticity. Do not confuse ring alkylation with the UV radical substitution of an alkane side chain.
- In chlorobenzene, a chlorine lone pair interacts with the aromatic pi system, giving the C-Cl bond some partial double-bond character; it is shorter/stronger and less easily hydrolysed than in a comparable chloroalkane. Do not simply treat the aryl chloride as a primary halogenoalkane undergoing the usual SN2 reaction.
- Benzene is carcinogenic; concentrated nitrating mixtures are corrosive and heat-producing. Mechanisms and conditions are for theoretical understanding and supervised authorised laboratory work, not home preparation.
What to do in an assessment
- Show the electrophile-generation equation before the ring mechanism. Start the attack arrow at a ring pi pair, retain H on the substituted carbon in the intermediate, show its positive charge, and start the restoring arrow at the C-H bond.
- Explain the evidence for delocalisation with equal bond lengths and greater stability than a localised three-double-bond model. A circle is a final delocalised-ring representation, not a licence to omit the intermediate's bonding.
In plain language
Aromatic substitution retains the stability of the delocalised ring.
Self-check
Why does benzene require a halogen carrier for bromination, and why is substitution preferred to addition?
Show the answer
The carrier activates/polarises Br₂ to provide a sufficiently strong electrophile. Substitution restores the aromatic pi system after proton loss; addition would remove its stabilising delocalisation.
View diagrams
Benzene nitration · electrophilic substitutionThe nitronium ion is the electrophile. In the σ complex, the attacked carbon keeps its H and the ring temporarily loses aromaticity; the positive charge is delocalised. Deprotonation restores aromaticity and regenerates H₂SO₄. The product circle denotes delocalisation, not the intermediate bond depiction.
Use the labelled model with the explanation; arrows represent electron-pair movement or the stated physical flow, not interchangeable processes. - Core
Scope: Core · Unit 4 §4.3(a, b, c, d)
Open revision guidance
What to know
- Aqueous hydroxide and reflux convert a suitable halogenoalkane into an alcohol by substitution: CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻. Reduction of an aldehyde gives a primary alcohol, RCHO + 2[H] → RCH₂OH; reduction of a ketone gives a secondary alcohol, RCOR′ + 2[H] → RCH(OH)R′, for example using NaBH₄.
- An alcohol reacts with a suitable hydrogen halide to replace OH by halogen: C₂H₅OH + HBr → C₂H₅Br + H₂O. Conditions depend on the halide and alcohol. Acid-catalysed reaction with a carboxylic acid is reversible esterification; reaction with ethanoyl chloride is more vigorous and produces an ester plus HCl: CH₃COCl + C₂H₅OH → CH₃COOC₂H₅ + HCl.
- Phenol has OH directly attached to a benzene ring, unlike benzyl alcohol C₆H₅CH₂OH. Phenol is a weak acid because its conjugate phenoxide ion is resonance-stabilised. C₆H₅OH + OH⁻ → C₆H₅O⁻ + H₂O; ordinary alcohols do not undergo this reaction appreciably with aqueous NaOH.
- Phenol reacts readily with bromine water without a halogen carrier: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr. Decolourisation and a white 2,4,6-tribromophenol precipitate distinguish this from benzene's usual bromine-water behaviour. Lone-pair interaction from oxygen increases ring electron density.
- Phenol forms phenyl ethanoate with ethanoyl chloride: CH₃COCl + C₆H₅OH → CH₃COOC₆H₅ + HCl. Aqueous iron(III) chloride can give a violet/purple complex with phenol; use it as a test alongside structure and other evidence, not a unique identification of every possible phenolic compound.
What to do in an assessment
- Check whether OH is attached directly to the ring before predicting acidity and bromination. Label products with the correct linkage: phenyl ethanoate contains CH₃CO-O-C₆H₅.
- Compare the two ester routes by reversibility, catalyst/conditions and by-product. Hydroxide favours substitution in an aqueous halogenoalkane preparation; hot ethanolic base instead can favour elimination.
- Phenol is toxic and corrosive; bromine and acid chlorides require small-scale controlled work, appropriate ventilation and waste handling.
In plain language
An OH group on an aromatic ring behaves differently from an alcohol's OH group.
Self-check
How can aqueous NaOH and bromine water distinguish phenol from benzene?
Show the answer
Phenol forms phenoxide with NaOH and decolourises bromine water with a white 2,4,6-tribromophenol precipitate. Benzene does not show these ordinary reactions without the relevant activating conditions.
- Core
Scope: Core · Unit 4 §4.4(a, b, c, e, f)
Open revision guidance
What to know
- Oxidation of a primary alcohol can give an aldehyde when the volatile product is distilled promptly; continued oxidation under reflux gives a carboxylic acid. A secondary alcohol gives a ketone. For ethanol: CH₃CH₂OH + [O] → CH₃CHO + H₂O; then CH₃CHO + [O] → CH₃COOH.
- Aldehydes are more easily oxidised than ketones under these mild tests. Fresh Tollens' reagent gives a silver deposit/mirror with an aldehyde as Ag⁺ is reduced. Fehling's reagent gives a brick-red Cu₂O precipitate with suitable aliphatic aldehydes on warming, while a simple ketone remains negative. Some aromatic aldehydes do not give the ordinary Fehling's result, so interpret the test conditions and candidate list.
- NaBH₄ reduces the C=O group: CH₃CHO + 2[H] → CH₃CH₂OH and CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃. This converts an aldehyde to a primary alcohol and a ketone to a secondary alcohol without assuming that every other functional group will be reduced.
- 2,4-dinitrophenylhydrazine gives a yellow/orange precipitate with aldehydes and ketones. The carbonyl compound forms a hydrazone derivative with loss of water. Purify the derivative and compare its melting temperature with reliable reference values to help identify the particular compound; the precipitate alone only establishes a compatible carbonyl group.
- The iodoform test uses iodine and alkali: a yellow CHI₃ precipitate indicates a methyl carbonyl CH₃CO- or a precursor oxidised to it, notably CH₃CH(OH)-. Ethanal and ethanol also give positive results. For a methyl ketone, RCOCH₃ + 3I₂ + 4OH⁻ → RCOO⁻ + CHI₃ + 3I⁻ + 3H₂O. A positive result is not a test for every alcohol or ketone.
- Use fresh Tollens' reagent, do not store it or allow residues to dry, and dispose of it promptly by the approved method because hazardous silver-containing residues can form. Do not handle dry 2,4-DNPH; use the supplied reagent under a risk-assessed laboratory method.
What to do in an assessment
- Use separate small aliquots so reagents from one test cannot alter another. Start with 2,4-DNPH to find a carbonyl, then distinguish suitable aldehyde/ketone candidates with oxidation tests and identify methyl-carbonyl candidates using iodoform.
- For a derivative melting-point identification, purify and dry the derivative, measure a narrow melting range and compare with an appropriate reference or a mixed melting point.
In plain language
Combine a carbonyl test with selective oxidation and structural evidence.
Self-check
Which of propanone, propanal and ethanol gives a positive 2,4-DNPH test, and which gives iodoform?
Show the answer
2,4-DNPH: propanone and propanal. Iodoform: propanone (CH₃CO-) and ethanol (oxidised to ethanal). Propanal lacks the required methyl-carbonyl group.
- Core
Scope: Core · Unit 4 §4.4(d)
Open revision guidance
What to know
- A carbonyl C=O bond is polar: oxygen is δ− and carbon δ+. A nucleophile attacks carbon, not oxygen. HCN addition produces a hydroxynitrile and forms a new C-C bond, increasing the carbon count by one.
- In the catalytic cyanide pathway, the carbon end of CN⁻ donates a lone pair to the carbonyl carbon while the C=O pi pair moves to oxygen. Ethanal produces CH₃CH(O⁻)CN. Oxygen's lone pair then acquires H from HCN and the H-CN bond pair returns to CN⁻, giving CH₃CH(OH)CN and regenerating cyanide.
- For propanone, CH₃COCH₃ + HCN → (CH₃)₂C(OH)CN. The product is not optically active at that central carbon because it has two identical methyl groups. Ethanal gives a new stereocentre; attack on the planar carbonyl from either face produces a racemic pair in an achiral environment.
- Use full curly arrows for electron pairs, not fishhook radical arrows. In the first step overall charge −1 passes from CN⁻ to oxygen; in protonation, the alkoxide plus neutral HCN gives neutral hydroxynitrile plus CN⁻. Cyanide and HCN are acutely toxic; this mechanism is theoretical preparation for strictly controlled authorised procedures.
What to do in an assessment
- Draw both simultaneous first-step arrows and show the negatively charged tetrahedral intermediate. Start the second attack arrow at the oxygen lone pair and include the bond-breaking arrow from H-CN to cyanide.
- Count carbons in the product and check that the old carbonyl carbon has four single bonds. Use stereochemistry to decide whether two different enantiomers are possible.
In plain language
A nucleophile adds across C=O and moves the electron pair onto oxygen.
Self-check
Give the shortened hydroxynitrile formula from ethanal and explain why both enantiomers form.
Show the answer
CH₃CH(OH)CN. The carbonyl is planar, allowing cyanide attack from either face; the new carbon has H, OH, CH₃ and CN groups, producing a racemic pair under achiral conditions.
View diagrams
HCN addition to ethanal · two-step mechanismCN⁻ attacks through carbon. The carbonyl π pair moves to oxygen, then O⁻ takes H⁺ from HCN while the H-C bond pair returns to carbon to regenerate CN⁻. The planar carbonyl creates a new stereogenic carbon, so attack from either face gives a racemic product.
Use the labelled model with the explanation; arrows represent electron-pair movement or the stated physical flow, not interchangeable processes. - Core
Scope: Core · Unit 4 §4.5(a, b, c, d, e)
Open revision guidance
What to know
- For the usual comparison, acid strength decreases from a carboxylic acid to phenol to water to ethanol. Carboxylate disperses its negative charge over two oxygens; phenoxide also has resonance stabilisation; an alkoxide has more localised charge. Alcohol acidity varies with structure, so use the stated compounds rather than a universal numerical pKa for all alcohols.
- A carboxylic acid reacts with aqueous carbonate/hydrogencarbonate to release CO₂ and with NaOH to give a salt. Phenol reacts with NaOH but does not normally release CO₂ from hydrogencarbonate. Ethanol does neither under these aqueous test conditions. All three can react with sufficiently reactive sodium metal, so that test alone does not distinguish their acid strengths.
- Oxidise a primary alcohol or aldehyde with suitable acidified dichromate(VI) under reflux to a carboxylic acid: RCH₂OH + 2[O] → RCOOH + H₂O; RCHO + [O] → RCOOH. A secondary alcohol does not simply produce an acid with the same carbon skeleton under these ordinary oxidation conditions.
- LiAlH₄ reduces a carboxylic acid to a primary alcohol: RCOOH + 4[H] → RCH₂OH + H₂O. Use a dry ether reaction medium and a controlled subsequent work-up in the approved supervised method; adding water directly to the active reducing agent is dangerous.
- An alkylbenzene with a benzylic H can be oxidised by hot permanganate to an aromatic carboxylic acid, followed by acidification. For methylbenzene, C₆H₅CH₃ + 3[O] → C₆H₅COOH + H₂O. The benzene ring is retained while the side-chain carbon becomes COOH.
- Decarboxylation removes the carboxyl carbon as CO₂ or carbonate. Heating a sodium carboxylate with soda lime gives an alkane with one fewer carbon: CH₃COONa + NaOH → CH₄ + Na₂CO₃. Distinguish this from merely neutralising the acid with carbonate, which releases the reagent's carbonate carbon.
What to do in an assessment
- Compare the conjugate bases to explain relative acidity, then predict carbonate and alkali observations. Use separate aliquots and collect the identified gas rather than relying on a smell.
- For oxidation/reduction and decarboxylation, map every carbon from reactant to product. State reflux, oxidant or dry reducing conditions explicitly.
In plain language
The carboxyl group controls acidity and can be oxidised, reduced or removed.
Self-check
How does soda-lime decarboxylation of sodium propanoate differ from its reduction after conversion to propanoic acid?
Show the answer
Decarboxylation gives ethane, losing the carboxyl carbon. LiAlH₄ reduction of propanoic acid gives propan-1-ol, retaining all three carbons.
- Core
Scope: Core · Unit 4 §4.5(f, g, h, i, j)
Open revision guidance
What to know
- A carboxylic acid and alcohol form an ester plus water using an acid catalyst and heating: RCOOH + R′OH ⇌ RCOOR′ + H₂O. An acid chloride can be made using a suitable chlorinating reagent, for example CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl. Acid chlorides react with water to regenerate the acid and HCl; they are not stable in an aqueous preparation.
- Acid-catalysed ester hydrolysis is reversible. Alkaline hydrolysis produces carboxylate plus alcohol and is effectively driven to completion: CH₃COOC₂H₅ + OH⁻ → CH₃COO⁻ + C₂H₅OH. Acidify afterwards if the free carboxylic acid is wanted.
- An acid and ammonia form an ammonium carboxylate; heating it can give an amide plus water: CH₃COOH + NH₃ → CH₃COO⁻NH₄⁺, then CH₃COO⁻NH₄⁺ → CH₃CONH₂ + H₂O. Dehydration of a primary amide, for example with a suitable phosphorus(V) oxide reagent, gives a nitrile: CH₃CONH₂ → CH₃CN + H₂O.
- A halogenoalkane reacts with cyanide under suitable heated ethanolic substitution conditions: CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻. The cyanide carbon becomes part of the chain, so the product is propanenitrile. Carbonyl addition of HCN instead makes a hydroxynitrile.
- Nitrile hydrolysis gives a carboxylic acid in acid: RCN + 2H₂O + H⁺ → RCOOH + NH₄⁺. In alkali it gives a carboxylate: RCN + H₂O + OH⁻ → RCOO⁻ + NH₃. A primary amide hydrolyses similarly: RCONH₂ + H₂O + H⁺ → RCOOH + NH₄⁺, or RCONH₂ + OH⁻ → RCOO⁻ + NH₃. Heating/reflux and the appropriate medium are needed.
- LiAlH₄ reduction of a nitrile forms a primary amine: RCN + 4[H] → RCH₂NH₂. This retains the nitrile carbon in the chain. Acid chloride, cyanide and reducing-agent reactions require properly controlled supervised procedures and suitable waste arrangements.
What to do in an assessment
- Choose conditions by functional group: water hydrolyses an acid chloride rapidly, while esters, nitriles and amides normally require the relevant catalyst/medium and heating. Draw the linkage before naming the product.
- For a synthesis map, distinguish cyanide substitution (one additional carbon) from converting an existing acid carbon into an amide, nitrile or amine (the carbon remains).
In plain language
Track the acyl or nitrile carbon through each conversion.
Self-check
Starting from bromoethane, how can propanoic acid be obtained, and why does the chain grow?
Show the answer
Substitute Br with CN⁻ using suitable heated ethanolic conditions to form propanenitrile, then hydrolyse the nitrile under acidic reflux. Cyanide contributes the extra carbon, which becomes the carboxyl carbon.
- Core
Scope: Core · Unit 4 §4.6(a, b, c, d, e)
Open revision guidance
What to know
- A primary amine RNH₂ can be prepared from a halogenoalkane with excess ethanolic ammonia and heating in a suitable closed system: RBr + 2NH₃ → RNH₂ + NH₄Br. Excess ammonia reduces, but does not guarantee elimination of, further alkylation. LiAlH₄ reduction of RCN gives RCH₂NH₂ and retains the nitrile carbon.
- An aromatic amine is prepared by reducing a nitroarene, for example C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O. Tin/concentrated HCl and heating produce the protonated amine salt; subsequent alkali liberates phenylamine.
- An amine accepts H⁺ through its nitrogen lone pair: RNH₂ + H⁺ → RNH₃⁺. Alkyl groups generally increase electron availability compared with ammonia in the simple primary-amine comparison. In phenylamine, the lone pair interacts with the ring and is less available, making it a weaker base than ammonia or a typical primary aliphatic amine.
- Ethanoylation of a primary amine with ethanoyl chloride gives a substituted amide. CH₃COCl + 2C₂H₅NH₂ → CH₃CONHC₂H₅ + C₂H₅NH₃⁺Cl⁻ shows both acylation and removal of the HCl by a second amine molecule. The nitrogen lone pair in an amide is delocalised towards C=O; amides are much less basic than amines.
- Cold nitric(III) acid, HNO₂, is made in situ from nitrite and acid. A primary aliphatic amine gives mainly an alcohol and nitrogen: RNH₂ + HNO₂ → ROH + N₂ + H₂O; the transient aliphatic diazonium species is unstable. Phenylamine instead forms a benzenediazonium salt that can be retained in cold solution: C₆H₅NH₂ + HNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + 2H₂O. Keep aromatic diazotisation below about 5 °C and use the solution promptly; do not isolate or dry unstable diazonium salts.
What to do in an assessment
- Draw primary amine nitrogen with one carbon group and two H atoms. Do not classify an amine by the classification of the carbon atom to which nitrogen is attached.
- Separate the nitro reduction's acidic salt stage from alkaline liberation. For nitric(III) acid, compare aliphatic decomposition with aromatic diazonium formation rather than assuming both intermediates persist.
- Write both the organic product and the neutralised HCl product in an ethanoylation preparation.
In plain language
Nitrogen's lone pair controls amine basicity and its synthetic reactions.
Self-check
Why is phenylamine less basic than ethylamine, and what product type forms when a primary amine reacts with ethanoyl chloride?
Show the answer
The phenylamine lone pair is partly delocalised into the ring and less available to accept H⁺. Ethanoyl chloride gives an N-substituted amide, with HCl removed by base or additional amine.
- Core
Scope: Core · Unit 4 §4.6(f, g, h)
Open revision guidance
What to know
- A cold benzenediazonium solution couples with a suitably activated aromatic ring such as phenol (usually as phenoxide in alkaline conditions) or phenylamine under suitable conditions. Coupling commonly occurs para to the activating group if that position is available, forming an azo linkage Ar-N=N-Ar′.
- With phenol, the para product can be written C₆H₅-N=N-C₆H₄-OH, with OH para to the azo group. This is a substituted ring, not the impossible formula of an unsubstituted phenol retaining all five ring hydrogens after coupling.
- The -N=N- group is a chromophore within an extended conjugated electron system. Its electronic energy gap can match visible-light photon energies. The dye absorbs particular visible wavelengths, and the complementary light reaching the observer determines the perceived colour.
- Changing substituents changes electron distribution, energy gaps and absorbed wavelengths. A red-looking material does not necessarily absorb red light: it transmits or reflects red more strongly while absorbing other visible wavelengths. Relate the energy change to E = hf = hc/λ.
- Dye usefulness also depends on affinity for the fabric, chemical stability, safety and treatment of coloured waste; the presence of a chromophore alone does not establish a sustainable manufacturing process.
What to do in an assessment
- Generate and retain the diazonium compound only in cold solution using the approved supervised method. Identify the activated ring and the available coupling position, then replace its H by the azo-linked aryl group.
- Use an absorption spectrum to distinguish the absorbed wavelength from the observed complementary colour. Do not infer the spectrum from a colour name without supporting data.
In plain language
An extended conjugated system can absorb visible light.
Self-check
What is the azo chromophore, and why can changing a ring substituent change the dye's colour?
Show the answer
The azo group is -N=N-. Substituents alter the conjugated electron system and its energy gap, changing which visible wavelengths are absorbed and therefore the complementary colour observed.
- Core
Scope: Core · Unit 4 §4.7(a, b, c, d, e, f)
Open revision guidance
What to know
- An alpha-amino acid has the general formula H₂N-CH(R)-COOH, with amino and carboxyl groups attached to the same alpha carbon. The side chain R can be non-polar, polar, acidic or basic; glycine has R = H and alanine R = CH₃. Glycine is achiral; many other alpha-amino acids have a chiral alpha carbon.
- Amino acids are amphoteric. At low pH they are predominantly H₃N⁺-CH(R)-COOH; at high pH H₂N-CH(R)-COO⁻. In an appropriate intermediate pH range they exist largely as zwitterions H₃N⁺-CH(R)-COO⁻ with both charges but no net charge. This ionic character helps explain relatively high melting/decomposition temperatures and interaction with polar water; side chains affect solubility.
- Condensation joins one amino acid's COOH and another's NH₂, releasing H₂O and forming the peptide linkage -CO-NH-. Glycine plus alanine can give H₂N-CH₂-CO-NH-CH(CH₃)-COOH or the reverse sequence; they are distinct dipeptides. Preserve the free amino and carboxyl termini.
- Repeated condensation produces polypeptides; a linear chain of n residues contains n−1 peptide bonds and releases n−1 water molecules. Hydrolysis breaks peptide bonds. Proteins may contain one or more folded polypeptide chains.
- Primary structure is the amino-acid sequence. Secondary structure such as alpha helices and beta sheets is stabilised by hydrogen bonds between backbone C=O and N-H groups. Tertiary structure is overall folding, influenced by hydrogen bonding, ionic attractions, hydrophobic interactions and covalent disulfide links where cysteine residues occur.
- Proteins function as enzymes, structural materials, transporters and signalling molecules. An enzyme's active-site shape and chemistry depend on folding; substrate binding enables a lower-activation-energy pathway. Heat or unsuitable pH can disrupt folding and activity without necessarily breaking the primary peptide-bond sequence.
What to do in an assessment
- Draw each ionic amino-acid form with the appropriate nitrogen and oxygen charges. Use the pH relative to its acid/base behaviour; no universal pH makes every amino acid have the same net charge.
- Join COOH to NH₂, remove OH plus H as water, and count the peptide bonds and retained termini. Explain each protein-structure level using its specific bonding, not just the word shape.
In plain language
Two functional groups allow amino acids to form charged species and peptide chains.
Self-check
How many peptide bonds and water molecules are involved in making a linear tripeptide from three amino acids?
Show the answer
Two peptide bonds form and two water molecules are released. The tripeptide retains one amino terminus and one carboxyl terminus.
- Core
Scope: Core · Unit 4 §4.8(a, b, e); Unit 4 §4.8 specified practical 1, 2, 3, 4
Open revision guidance
What to know
- Plan an organic synthesis backwards from the desired functional group and carbon skeleton. For bromoethane → propanenitrile → propanoic acid, cyanide substitution adds one carbon and acidic hydrolysis converts that carbon to COOH. Give a reagent and conditions at every arrow and consider competing elimination or further reaction.
- For a liquid product, complete the reaction under suitable reflux, cool before transfer, and use a separating funnel when immiscible layers are present. Identify layers by the solvent density and a safe small water test if needed; do not assume that the organic layer is always on top. Stopper, mix gently, vent frequently away from people, allow separation and remove the stopper before draining.
- Wash the organic layer with a suitable reagent to remove impurities, manage any gas evolution, dry with an appropriate anhydrous drying agent, then decant/filter before distillation. Place the thermometer bulb at the condenser entrance to measure vapour temperature. Distillation separates by volatility, not by melting temperature.
- For a solid product, recrystallise using the minimum hot solvent in which product is soluble hot but much less soluble cold. Remove insoluble material by suitable hot filtration where needed, cool to crystallise, collect by reduced-pressure filtration, wash with a little cold solvent and dry. Excess solvent and warm washing reduce recovery.
- Measure the dried solid's melting range using a small well-packed capillary sample and slow heating near the expected temperature. A pure compound normally melts over a narrow range near its reference value; impurities commonly lower and broaden it. A matching melting range is supporting evidence, not unique proof of identity.
- A representative two-step solid preparation is nitrobenzene → phenylamine → N-phenylethanamide. First use the controlled Sn/HCl reduction and alkali liberation described in the amines lesson; isolate the amine by the authorised method. Then ethanoylate it: CH₃COCl + 2C₆H₅NH₂ → CH₃CONHC₆H₅ + C₆H₅NH₃Cl. Recrystallise, dry and measure the final solid's melting range. Record each actual stage and its yield; consecutive 75.0% and 80.0% yields give 60.0% overall, not 155%.
- For planning tests on a supplied unknown list, use functional groups to make a branching scheme: hydrogencarbonate for carboxylic acid, 2,4-DNPH for aldehyde/ketone, a suitable oxidation test to distinguish these, FeCl₃ or bromine water for phenol, and compatible spectra to resolve remaining candidates. Use fresh aliquots and confirmatory evidence; a negative test also helps eliminate candidates.
What to do in an assessment
- Before a specified one- or two-step preparation, record a balanced reaction, limiting reagent, theoretical yield, safe heating/separation method and chosen identity/purity checks. Do not count impurities as product or use wet mass to inflate yield.
- For a separating funnel, assess volatile solvent and gas pressure; for vacuum filtration, secure the correct flask and release vacuum safely. Use non-flame heating near flammable organic solvents.
- Reading and planning these methods does not certify performance of Unit 5. Follow the supplied supervised procedure and record what was actually carried out.
In plain language
A synthesis is complete only when its product is isolated and checked.
Self-check
Why does washing recrystallised product with warm solvent reduce yield, and why must it be dry before weighing?
Show the answer
Warm solvent dissolves more product and carries it away. Retained solvent adds mass and can affect melting behaviour, so a wet mass overstates the product yield.
- Core
Scope: Core · Unit 4 §4.8(c, d)
Open revision guidance
What to know
- Addition polymerisation joins unsaturated monomers by opening a C=C pi bond and forming a carbon-chain repeat unit; no small molecule is eliminated in the ideal addition step. Condensation polymerisation links monomers with two reactive functional groups and eliminates a small molecule such as water or HCl at each linkage.
- A diol and dicarboxylic acid form polyester ester links -CO-O-. Ethane-1,2-diol HO-CH₂CH₂-OH and terephthalic acid HOOC-C₆H₄-COOH give the repeat unit [-O-CH₂CH₂-O-CO-C₆H₄-CO-]ₙ. The two CO groups are para on the aromatic ring; continuation bonds join neighbouring units, not extra terminal H atoms inside each repeat.
- A diamine and dicarboxylic acid form polyamide -CO-NH- links. Hexane-1,6-diamine and hexanedioic acid give nylon-6,6 with repeat [-NH-(CH₂)₆-NH-CO-(CH₂)₄-CO-]ₙ. Acid chlorides can replace acids in appropriate preparations, releasing HCl rather than water.
- Identify monomers by cutting the ester or amide links and restoring OH, H or the appropriate precursor groups. Polyesters and polyamides can be hydrolysed at their functional links under suitable conditions; a typical polyalkene's carbon backbone does not contain those hydrolysable links.
- For a finite linear polymer, count the actual links and ends before stating the number of eliminated molecules. A repeat-unit drawing shows periodic connectivity, not the exact end-group structure of every real chain.
What to do in an assessment
- Draw both monomers with two functional groups, form the correct linkage, bracket a repeat unit and include continuation bonds through the brackets. Check atom conservation against the eliminated molecule.
- Compare polymer behaviour through its chemical linkage and intermolecular forces, then evaluate manufacture, durability and disposal rather than assuming hydrolysable automatically means harmless.
In plain language
The linkage shows how a polymer forms and how it can break down.
Self-check
What linkage distinguishes a polyester from a polyamide, and does ethene addition polymerisation eliminate water?
Show the answer
Polyester has -CO-O- links; polyamide has -CO-NH- links. Ethene addition polymerisation eliminates no water in the ideal polymer-forming step.
- Core
Scope: Core · Unit 4 §4.8(f)
Open revision guidance
What to know
- Combine molecular mass and formula, IR functional groups, ¹³C environments and ¹H environments before proposing a structure. High-resolution ¹H NMR adds splitting to the low-resolution information in Unit 2; chemical shift, relative integral and multiplicity are different measurements.
- For simple first-order cases, n equivalent neighbouring protons split a signal into n+1 lines: singlet, doublet, triplet or quartet. Equivalent protons do not normally split each other, and exchangeable OH/NH protons often behave differently. More complex coupling can occur, so use the stated simple conditions rather than an unrestricted n+1 rule.
- CH₃CH₂Br has two proton environments, integrals 3:2. CH₃ is split by two CH₂ neighbours into a triplet; CH₂ is split by three CH₃ neighbours into a quartet. Their shifts differ because the CH₂ is attached to Br. The ¹³C spectrum has two carbon environments.
- An ester candidate with formula C₄H₈O₂, an IR C=O absorption around 1740 cm⁻¹, no broad acid O-H, and ¹H integrals 3:2:3 can be ethyl ethanoate CH₃COOCH₂CH₃. Approximate shifts near 2.0 (3H singlet), 4.1 (2H quartet) and 1.2 ppm (3H triplet) support that connectivity. The carbonyl-adjacent CH₃ has no neighbouring H across the carbonyl to give the ethyl pattern.
- Use a provided data sheet for exact permitted shift/absorption ranges; values vary with conditions. Fragment peaks such as a plausible acyl ion can support the mass-spectrum interpretation, but neither an absent molecular-ion peak nor one matching fragment alone is a complete structural proof.
What to do in an assessment
- Make a table of each observed signal: shift, integral, splitting and candidate environment. Link neighbouring groups only when the splitting is mutually consistent, then count all atoms and reject any structure inconsistent with IR or formula.
- Distinguish a 3:2 integral ratio from three or two peak lines. An integral measures relative proton numbers; splitting measures coupling under the stated assumptions.
In plain language
Independent spectra constrain the same structure in different ways.
Self-check
A simple CH₃CH₂- group gives integrals 3:2. Which signal is the quartet and why?
Show the answer
The 2H CH₂ signal is the quartet because it has three equivalent neighbouring CH₃ protons: n+1 = 4. The CH₃ signal is a triplet.
- Core
Scope: Core · Unit 4 §4.8(g); Unit 4 §4.8 specified practical 5
Open revision guidance
What to know
- Chromatography separates components through their different distributions or interactions between mobile and stationary phases. A component spending more time in the stationary phase travels more slowly; polarity and the actual phases determine the order.
- For paper or TLC, mark a pencil baseline, apply small concentrated spots above the solvent level, allow the solvent to rise in a suitable covered vessel, and mark the solvent front promptly. Rf = distance travelled by spot centre / distance travelled by solvent front, measured from the same baseline. Rf is dimensionless and normally between 0 and 1.
- For a spot travelling 3.0 cm and a front travelling 6.0 cm, Rf = 0.50. Compare with reference substances run under the same conditions. A matching Rf is supporting evidence, not unique identification; overlapping components may appear as one spot.
- Two-way paper chromatography first develops and dries the paper in one solvent, then rotates it through 90° and develops in a second solvent. Different selectivity can separate substances that overlapped in the first direction. Keep the original starting point and both solvent directions identifiable.
- Gas chromatography uses a carrier gas and an appropriate stationary phase; suitable volatile components elute with characteristic retention times under fixed conditions. HPLC pumps a liquid mobile phase through a packed column under high pressure. Compare retention times with standards under the same method, not across arbitrary different columns or solvents.
- Detector peak areas can estimate mixture composition when response factors or calibration are known. If corrected response areas are 30 and 70 in the same units, the corresponding proportions are 30% and 70%. Raw area percentages do not necessarily equal mass or mole percentages when detector responses differ.
What to do in an assessment
- For the specified two-way separation, use pencil rather than soluble ink, prevent the starting spots being submerged and dry between directions. Use a suitable visualisation method and handle solvents in accordance with their flammability/toxicity.
- Record both distances or retention times and method conditions. Distinguish identification by a reference from quantitative composition by a calibrated detector.
In plain language
Phase interactions separate a mixture; standards and calibration interpret the result.
Self-check
A TLC spot travels 2.4 cm while its solvent front travels 6.0 cm. Find Rf and explain why it does not uniquely identify the substance.
Show the answer
Rf = 2.4/6.0 = 0.40. Different substances can have the same Rf, and its value depends on the phases and conditions, so compare standards and other evidence.
Topic group
Practical preparation and mathematical skills
- Core
Scope: Core · Appendix A skills 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 16, 17
Open revision guidance
What to know
- Turn a chemical question into a testable prediction. For a rate investigation, identify the changed concentration as independent variable, measured initial rate as dependent variable, and temperature, total volume, mixing and other concentrations as controls. Select a measurement method that actually answers the question.
- Read the complete supplied method before starting and identify quantities, sequencing and safety-critical instructions. Choose apparatus by range and resolution; secure glassware and follow the approved supervised procedure rather than improvising hazardous substitutions.
- Distinguish a hazard from risk and a control: a corrosive acid is a hazard; splashing during transfer creates a risk; eye protection, a pipette filler, appropriate dilution and controlled handling reduce exposure. Assess hot glassware, flammable solvent, pressure, oxidants and toxic metal/cyanide species separately.
- Record the date, actual apparatus and reagents, deviations, raw measurements and qualitative observations such as precipitate colour or gas formation. Use quantity-and-unit table headings; do not retrospectively replace inconvenient readings with expected values.
- Repeat measurements and check for anomalies with a stated reason. A conclusion must use the observed pattern and its uncertainty; correlation alone cannot establish a cause, and a plausible model must remain open to conflicting evidence.
- Use software for graphing, regression or data logging with appropriate calibration and retained raw files. Explain the chosen fit and units; a spreadsheet's many decimal places do not improve the underlying measurement.
- Research with board specifications, data books and suitable scientific publications, checking author, publication date, relevance and limitations. Cite the exact source and page or URL for borrowed data or claims and distinguish them from your own observations. Use both appropriate printed and online evidence.
- Scientific understanding develops as observations challenge models, as with atomic structure and bonding. Peer scrutiny, reproducibility and transparent reporting help validate claims; they do not make a single unreplicated result unquestionable. Evaluate environmental, economic, human and ethical effects when recommending chemical technologies.
- Unit 5 is a separate practical examination worth 10%, not an NEA or a practical endorsement. It includes a 30-mark Experimental Task in one session of up to three hours, and a separate 30-mark, one-hour Practical Methods and Analysis paper with no hands-on work. Actual performance is assessed under controlled conditions; ticking a reading checklist does not certify competence.
- The specification requires annual Unit 5 themes to be shared by centres early in the spring term and posted on the board's qualification page. Ask your centre for the 2027 Instructions to Teachers/Exams Officers and current arrangements. The general guide does not substitute 2026 themes or assert an unverified 2027 topic list.
What to do in an assessment
- Write a short plan containing the scientific aim, variables, apparatus, quantities, risk controls, repeat strategy and analysis. Link each chosen step to the evidence needed, then compare the actual result with the prediction.
- When evaluating, separate random scatter from systematic bias and identify a realistic improvement with its expected effect. More repeats improve precision but do not remove a consistently miscalibrated instrument.
In plain language
Good evidence needs a suitable method, honest records and justified conclusions.
Self-check
All repeated titres are closely grouped but a burette consistently over-reads. Do repeats alone make the concentration accurate?
Show the answer
No. Close grouping indicates precision, while consistent calibration bias can make every result inaccurate. Calibration or an appropriate correction is needed.
- Core
Scope: Core · Appendix A skills 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37
Open revision guidance
What to know
- Measure mass using a tared suitable balance, time from a defined start, liquid volume using appropriate graduated/volumetric apparatus, gas volume with a leak-tested gas syringe or suitable collection arrangement, and temperature with a properly immersed sensor. Read liquid menisci at eye level and choose resolution appropriate to the expected quantity.
- Use a water bath, electric heater or sand bath suitable for the required temperature. Keep ignition sources away from flammable solvents, use controlled heating and clamps, and avoid pressure build-up. Reflux returns condensate to the reaction flask; distillation directs vapour to a separate receiver.
- A pH chart gives an approximate value by comparing indicator colour with its reference; coloured samples can interfere. A calibrated pH meter/probe provides more quantitative readings; use suitable standard buffers, rinse between samples, allow stabilisation and record temperature. A data logger records successive readings but still needs calibration and correct timing.
- For titration use a rinsed pipette and filler for the aliquot, a rinsed burette with a filled tip and no bubbles for titrant, and an indicator suitable for the curve's steep interval. Record both meniscus readings, mix, approach the endpoint dropwise and repeat to justified concordance. Indicators do not reliably locate weak-acid/weak-base equivalence.
- Make a standard solution by accurately weighing a suitable pure stable solute, dissolving it, quantitatively transferring and rinsing into a volumetric flask, making to the mark at suitable temperature and mixing thoroughly. Do not make a stated final volume merely by adding that volume of water to the solid.
- Use fresh aliquots for ion and functional-group tests and record reagent, conditions and actual observation. Gravity filtration with suitable fluted paper helps separate insoluble material; reduced-pressure filtration uses an appropriate side-arm flask, fitted funnel and wetted paper to collect crystals efficiently. Release vacuum safely before disassembly.
- Recrystallise a solid from minimum hot solvent, cool, filter, cold-wash and dry. For an immiscible liquid mixture use a separating funnel, identify the layers, vent pressure and remove the stopper before draining; dry the retained organic layer before distillation. Measure a dry solid's melting range with slow heating near the expected value.
- For paper/TLC chromatography use a pencil baseline, small spots above the solvent and a promptly marked front. Two-way development uses a second solvent after drying and a 90° rotation. For a cell, connect the two suitable electrodes through a high-resistance voltmeter and complete the ionic circuit with a non-reactive salt bridge.
- Use at least two rate approaches: a clock records time to a fixed chemical endpoint, whereas continuous gas collection or absorbance monitoring gives a quantity-time curve. Control temperature, concentrations and mixing; compare rates only after specifying the measured quantity and units.
- Handle corrosive, irritant, flammable and toxic materials with the assessed controls, appropriate ventilation and designated waste. Do not smell gases, mouth-pipette or seal a heated reflux system. Method knowledge supports written questions but reading does not certify performed practical skills.
What to do in an assessment
- Choose a technique by the physical and chemical problem: reflux to react without losing volatile reagents, distillation for a volatile product, filtration for an insoluble solid and a funnel for immiscible layers.
- Draw the apparatus with the measurement or collection position labelled. State the control that reduces each identified risk, and record uncertainty as well as the result.
In plain language
Choose, set up and interpret the apparatus rather than simply naming it.
Self-check
Why does a reflux condenser return liquid to the flask, and why is it unsuitable for collecting a distillate without changing the arrangement?
Show the answer
Its vertical arrangement condenses rising vapour back into the reaction flask. Distillation must direct condensed vapour into a separate receiver with a suitable vapour-temperature measurement.
- Core
Scope: Core · Appendix A skills 14, 15; Appendix C skills 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
Open revision guidance
What to know
- Convert before substituting: 250 cm³ = 0.250 dm³ = 2.50 × 10⁻⁴ m³. Gas calculations with SI R require Pa, m³ and K; molar concentrations commonly use dm³. Entropy in J mol⁻¹ K⁻¹ must be converted before combining with an enthalpy in kJ mol⁻¹.
- Standard form preserves significant figures: 0.0050 mol dm⁻³ = 5.0 × 10⁻³ mol dm⁻³, both with two significant figures. Keep unrounded intermediate values. For pH, decimal places reflect the precision of the concentration's significant figures rather than applying the same rounding rule blindly to the logarithm.
- Use balanced-equation ratios for formulae, reacting quantities, atom economy and yield. A 2:1 coefficient ratio means twice the moles, not necessarily twice the mass. A 3:1 mixture has fractions 3/4 and 1/4; a weighted mean uses these fractions, not an unweighted average.
- Select justified concordant titres and calculate their arithmetic mean: 24.80 and 24.90 cm³ give 24.85 cm³. A rough run or explained anomaly is not silently included or discarded. For isotope abundances 75% mass 35 and 25% mass 37, the weighted mean is 35.5.
- Estimate orders of magnitude and expected direction before evaluating: heating an exothermic equilibrium decreases K, so an increase suggests a conceptual or substitution error. Check a calculated titre against plausible reagent amounts and the apparatus range.
- Rearrange before inserting numbers: c = n/V becomes V = n/c; pV = nRT becomes n = pV/(RT). For rate = k[A][B]², k = rate/([A][B]²). Solve an equilibrium quadratic with a physically possible root; concentrations cannot be negative or consume more than the initial material.
- Use natural exponentials/logarithms for Arrhenius and base-10 logarithms for pH/pKa in their stated forms. Invert pH correctly as [H⁺] = 10⁻ᵖᴴ. The signs and log base matter over many orders of magnitude; log(10⁻⁵) is not +5.
- Interpret = as equality, < and > as comparisons, << and >> as much smaller/larger, ∝ as proportionality, and ~ as approximately. The equilibrium sign ⇌ represents a reversible system, not a statement that both concentrations are equal.
- For sums/differences, add maximum absolute uncertainties under the usual school worst-case convention; for products/quotients, add percentage uncertainties. Two burette uncertainties of ±0.05 cm³ give ±0.10 cm³ in a 25.00 cm³ titre, or 0.40%. A 2.00 g mass with ±0.01 g is 0.50%; combining these in a proportional concentration result gives a maximum 0.90% before any other contributions.
- Uncertainty is a range estimated from resolution, repeatability or other stated sources. Accuracy concerns agreement with a true/reference value; precision concerns spread. More displayed digits cannot justify reporting beyond the least reliable measurement.
What to do in an assessment
- Write the equation, rearrangement, units, numerical substitution and final justified rounding. Convert a percentage back to an absolute result uncertainty if requested.
- For a multi-stage titration, carry dilution and aliquot ratios explicitly and check the chemical ratio independently from the numerical arithmetic.
In plain language
Units and algebra make the calculation checkable; uncertainty limits its precision.
Self-check
Convert 0.0200 dm³ into cm³ and m³, preserving its three significant figures.
Show the answer
20.0 cm³ and 2.00 × 10⁻⁵ m³.
- Core
Scope: Core · Appendix A skills 13; Appendix C skills 14, 15, 16, 17, 18
Open revision guidance
What to know
- Translate a data table into a graph with the independent variable on the horizontal axis, dependent variable on the vertical axis, quantity-and-unit labels and a useful scale. Plot measured points accurately and use an appropriate smooth curve or best-fit line, not an unexplained join-the-dots construction.
- For a straight line y = mx + c, gradient m = Δy/Δx and c is the y-intercept. Choose widely separated points on the fitted line. With [A] = 0.100, 0.080 and 0.060 mol dm⁻³ at 0, 10 and 20 s, the straight-line slope is −0.00200 mol dm⁻³ s⁻¹; the positive disappearance rate is 0.00200.
- For a nonlinear concentration-time curve, draw a tangent at the requested time and obtain its gradient using two points on that tangent. An initial tangent through (0 s, 0.100 mol dm⁻³) and (10 s, 0.060 mol dm⁻³) gives disappearance rate 0.00400 mol dm⁻³ s⁻¹; those points need not both lie on the experimental curve.
- A transformed linear graph exposes parameters. For ln k against 1/T, slope = −Eₐ/R and intercept = ln A. For zero-order [A] against t, slope = −k; relate the algebraic form to the plotted coordinates before assigning units to a gradient.
- A calibration line connects measured response with concentration. If A = 2.50c with c in mol dm⁻³, an absorbance 0.300 gives c = 0.120 mol dm⁻³. Use the calibration's stated range and blank; extrapolating far beyond it is not automatically valid.
- Show measurement uncertainty where appropriate and assess scatter or plausible slope bounds. Distinguish a fitted trend from proof of a causal mechanism; control variables and independent evidence are still needed.
What to do in an assessment
- Calculate gradients with their units from the actual axes. For a decreasing concentration, report a negative concentration gradient but a positive disappearance rate unless the question explicitly asks otherwise.
- Explain why a tangent answers instantaneous rate while a secant over an interval answers average rate.
In plain language
The axes and line shape determine what a graph can tell you.
Self-check
A straight concentration-time line goes from 0.200 to 0.140 mol dm⁻³ over 30.0 s. What are its gradient and disappearance rate?
Show the answer
Gradient = (0.140−0.200)/30.0 = −0.00200 mol dm⁻³ s⁻¹; disappearance rate = +0.00200 mol dm⁻³ s⁻¹.
- Core
Scope: Core · Appendix C skills 19, 20, 21
Open revision guidance
What to know
- Use VSEPR electron regions to predict ideal geometry: two regions linear 180°, three trigonal planar 120°, four tetrahedral 109.5° and six octahedral 90° between adjacent positions. Lone pairs change the molecular shape and commonly compress bond angles, as in NH₃ and H₂O.
- Convert a three-dimensional arrangement into a two-dimensional drawing using ordinary bonds in the page, a solid wedge towards you and a hashed wedge away. A tetrahedron has four vertices, not four planar corners; an octahedron has six coordination positions, not eight.
- Symmetry can make atom environments equivalent in NMR or cancel equal bond dipoles, as in linear CO₂. Breaking symmetry can change the number of distinguishable environments. Do not confuse symmetry of a flat sketch with the actual molecule's three-dimensional symmetry.
- For E-Z, restricted rotation and different groups at each alkene carbon allow different relative arrangements; priorities determine E or Z. For optical isomerism, identify four different groups at a tetrahedral stereocentre and compare true non-superimposable mirror images. Rotating a molecule does not change its connectivity or turn one enantiomer into the other.
What to do in an assessment
- Draw and label shape, depth convention, relevant angles and all different substituents. Use a model or spatial reasoning to test superposition rather than deciding from two arbitrarily rotated flat drawings.
- Count chemically equivalent environments only after checking the actual connectivity and symmetry.
In plain language
A chemical drawing must encode real geometry, not just a recognisable formula.
Self-check
How many positions surround an octahedral centre, and why is a flat cross insufficient to prove optical isomerism?
Show the answer
Six positions surround an octahedral centre. A flat cross without a defined stereochemical convention does not show the tetrahedral depth needed to establish a mirror-image relationship.
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