A Level Chemistry (A2) equations and key facts

Every chapter of A Level Chemistry (A2) on one page: the 262 equations, definitions and facts to remember, in syllabus order. Use it for a last look before a test, then check yourself.

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Chemical energetics

Lattice energy and Born-Haber cycles

Lattice energy, ΔH_latt: gaseous ions → one mole of solid, e.g. Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s). Always negative.
Enthalpy change of atomisation, ΔH_at: one mole of gaseous atoms formed from the element in its standard state, e.g. ½Cl₂(g) → Cl(g). Always positive.
First electron affinity, EA₁: X(g) + e⁻ → X⁻(g), per mole. Usually exothermic.
Second electron affinity is endothermic: the electron is added to a negative ion, which repels it.
ΔH_latt = ΔH_f − (sum of all atomisation, ionisation and electron affinity steps)
Electron affinity becomes less exothermic down Groups 16 and 17 (more shells, more shielding, weaker attraction). Exceptions: O and F are less exothermic than S and Cl, because the atoms are so small that the added electron is repelled strongly by the electrons already in the outer shell.
Higher ionic charge or smaller ionic radius → more exothermic lattice energy.

Enthalpies of solution and hydration

Enthalpy change of hydration, ΔH_hyd: one mole of gaseous ions dissolves in water to form a very dilute solution, e.g. Na⁺(g) + aq → Na⁺(aq). Always negative.
Enthalpy change of solution, ΔH_sol: one mole of solid dissolves in water to form a very dilute solution, e.g. NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq).
ΔH_sol = ΣΔH_hyd − ΔH_latt (with ΔH_latt as the negative value, gaseous ions → solid)
Count every ion: for MgCl₂, ΣΔH_hyd = ΔH_hyd(Mg²⁺) + 2 × ΔH_hyd(Cl⁻).
Smaller ionic radius or higher ionic charge → more exothermic ΔH_hyd.
ΔH_sol is negative when the hydration enthalpies together are more exothermic than the lattice energy.

Entropy change, ΔS

Entropy order for one substance: solid < liquid < gas.
ΔS^⦵ = ΣS^⦵(products) − ΣS^⦵(reactants)
Multiply each S^⦵ by the number of moles in the balanced equation.
Units of S and ΔS: J K⁻¹ mol⁻¹ (joules, not kilojoules).
More moles of gas in the products → ΔS positive; fewer moles of gas → ΔS negative.
Raising the temperature increases entropy: the particles have more energy and more ways to share it.
Standard entropies of elements are not zero (unlike enthalpies of formation).

Gibbs free energy change, ΔG

Gibbs equation: ΔG^⦵ = ΔH^⦵ − TΔS^⦵, with T in kelvin.
Convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ (÷ 1000) before using it with ΔH in kJ mol⁻¹.
ΔG ≤ 0: feasible. ΔG > 0: not feasible.
ΔH negative, ΔS positive: feasible at all temperatures. ΔH positive, ΔS negative: never feasible.
ΔH negative, ΔS negative: feasible only at low temperatures. ΔH positive, ΔS positive: feasible only at high temperatures.
The temperature at which feasibility changes (ΔG = 0): T = ΔH ÷ ΔS, with both in the same energy unit.

Electrochemistry

Electrolysis

Cathode: the species with the more positive E^⦵ is reduced first. In aqueous solution, reactive metals (K, Na, Mg, Al) give hydrogen; Cu and Ag are deposited.
Anode: concentrated aqueous halide gives the halogen; very dilute halide, sulfate or nitrate solutions give oxygen.
charge = current × time (Q = It), with Q in coulombs, I in amperes and t in seconds
Faraday constant = Avogadro constant × charge on one electron (F = Le); F = 96 500 C mol⁻¹
moles of electrons = Q ÷ F; then use the half-equation, e.g. Cu²⁺ + 2e⁻ → Cu needs 2 mol of electrons per mole of copper.
Gases: 2H⁺ + 2e⁻ → H₂; 2Cl⁻ → Cl₂ + 2e⁻; 4OH⁻ → O₂ + 2H₂O + 4e⁻ (4 mol of electrons per mole of O₂).
To find L by electrolysis: measure current, time and the mass gained by a copper cathode, find the charge per mole of electrons (F), then L = F ÷ e.

Standard electrode potentials, cell potentials and the Nernst equation

Standard conditions: 298 K, solutions of 1.00 mol dm⁻³, gases at 101 kPa.
Standard hydrogen electrode: H₂(g) at 101 kPa over platinum in 1.00 mol dm⁻³ H⁺(aq); its E^⦵ is 0.00 V.
Two ions of the same element (e.g. Fe³⁺/Fe²⁺): use a platinum electrode in a solution with both ions at 1.00 mol dm⁻³.
E^⦵_cell = E^⦵(more positive half-cell) − E^⦵(more negative half-cell)
A reaction is feasible if E^⦵_cell is positive: the oxidising agent must come from the half-equation with the more positive E^⦵.
Nernst equation at 298 K: E = E^⦵ + (0.059 ÷ z) × log([oxidised species] ÷ [reduced species]); a solid metal is left out of the ratio.
ΔG^⦵ = −nE^⦵_cellF, giving joules per mole; n = moles of electrons transferred.

Equilibria

Acids and bases

pH = −log[H⁺]; [H⁺] = 10^−pH; pK_a = −log K_a
K_w = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K; for a strong alkali, [H⁺] = K_w ÷ [OH⁻]
K_a = [H⁺][A⁻] ÷ [HA]; for a weak acid alone, [H⁺] = √(K_a × [HA])
Buffer: [H⁺] = K_a × [HA] ÷ [A⁻], or pH = pK_a + log([A⁻] ÷ [HA])
Buffer action: A⁻ + H⁺ → HA and HA + OH⁻ → A⁻ + H₂O. In blood: HCO₃⁻ + H⁺ ⇌ H₂CO₃.
K_sp is the product of the ion concentrations in a saturated solution, each raised to the power of its number in the formula, e.g. K_sp(PbCl₂) = [Pb²⁺][Cl⁻]²
Common ion effect: adding an ion already in the equilibrium makes the salt less soluble.

Partition coefficients

K_pc = [solute in organic solvent] ÷ [solute in water], at equilibrium. Check which way round the question defines it.
Use concentrations, not masses: concentration = mass ÷ volume of that layer.
The solute must be in the same physical and molecular state in both solvents.
K_pc has no units and changes only with temperature.
Non-polar solutes (e.g. I₂) have large K_pc values between an organic solvent and water; polar solutes have small ones.
Extracting with several small portions of solvent removes more solute than one large portion of the same total volume.

Reaction kinetics

Simple rate equations, orders of reaction and rate constants

rate = k[A]^m[B]ⁿ; overall order = m + n
Units of k: overall order 1 → s⁻¹; order 2 → dm³ mol⁻¹ s⁻¹; order 3 → dm⁶ mol⁻² s⁻¹
Concentration–time graph: zero order is a straight line going down; first order is a curve with a constant half-life.
Rate–concentration graph: zero order is a horizontal line; first order is a straight line through the origin; second order is an upward curve.
First order: half-life is constant and does not depend on concentration; k = 0.693 ÷ t_½
An intermediate is made in one step and used up in a later step; a catalyst is used in one step and re-formed later.
Raising the temperature increases k, so the rate increases.

Homogeneous and heterogeneous catalysts

Heterogeneous sequence: adsorption of reactants → bonds weakened → reaction → desorption of products.
Haber process: solid iron with N₂(g) and H₂(g). Adsorption weakens the N≡N and H−H bonds.
Catalytic converter: Pt, Pd and Rh on a ceramic honeycomb; 2NO + 2CO → N₂ + 2CO₂.
Adsorption must be strong enough to weaken bonds but weak enough for the products to leave the surface.
Homogeneous, in the atmosphere: SO₂ + NO₂ → SO₃ + NO, then NO + ½O₂ → NO₂.
Homogeneous, in solution: S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺, then 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂.
The uncatalysed S₂O₈²⁻/I⁻ reaction is slow because two negative ions repel each other.

Group 2

Trends in the properties of the Group 2 metals, magnesium to barium

Carbonates: MCO₃ → MO + CO₂. Nitrates: 2M(NO₃)₂ → 2MO + 4NO₂ + O₂.
Thermal stability increases down the group: MgCO₃ decomposes most easily, BaCO₃ least easily.
Smaller cation → higher charge density → more polarisation of the anion → lower decomposition temperature.
Hydroxides: solubility increases down the group; Mg(OH)₂ is the least soluble, Ba(OH)₂ the most (highest pH).
Sulfates: solubility decreases down the group; MgSO₄ is soluble, BaSO₄ is insoluble.
Sulfates: SO₄²⁻ is large, so lattice energy changes little, while ΔH_hyd of the cation becomes much less exothermic. ΔH_sol becomes more endothermic.
Hydroxides: OH⁻ is small, so lattice energy becomes less exothermic faster than ΔH_hyd does. ΔH_sol becomes more exothermic.

Chemistry of transition elements

General physical and chemical properties of the transition elements, titanium to copper

Transition element: a d-block element that forms at least one stable ion with incomplete d orbitals.
Zn²⁺ is [Ar] 3d¹⁰ (full) and Sc³⁺ is [Ar] 3d⁰ (empty), so zinc and scandium are not transition elements.
Atoms: Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹. Ions lose 4s electrons first, e.g. Fe²⁺ is [Ar] 3d⁶.
3d_xy orbital: four lobes in the xy plane, pointing between the x and y axes.
3d_z² orbital: two lobes along the z-axis with a ring of electron density around the middle in the xy plane.
Catalysts because they have more than one stable oxidation state and vacant d orbitals that can accept lone pairs (dative bonds).
Complex ions form because vacant d orbitals of suitable energy accept lone pairs from ligands.

General characteristic chemical properties of the transition elements, titanium to copper

Monodentate (one bond): H₂O, NH₃, Cl⁻, CN⁻. Bidentate (two): 1,2-diaminoethane (en), C₂O₄²⁻. Polydentate: EDTA⁴⁻ forms six bonds.
Shapes: octahedral (6 bonds, 90°); tetrahedral (4, 109.5°); square planar (4, 90°); linear (2, 180°).
Charge on a complex = charge on the metal ion + total charge of the ligands.
Copper(II): [Cu(H₂O)₆]²⁺ blue; with OH⁻ or a little NH₃, pale blue precipitate of Cu(OH)₂; excess NH₃ gives deep blue [Cu(NH₃)₄(H₂O)₂]²⁺; concentrated HCl gives yellow [CuCl₄]²⁻.
Cobalt(II): [Co(H₂O)₆]²⁺ pink; with OH⁻, blue precipitate of Co(OH)₂; excess NH₃ gives [Co(NH₃)₆]²⁺; concentrated HCl gives blue [CoCl₄]²⁻.
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O (1 : 5); 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O (2 : 5)
2Cu²⁺ + 4I⁻ → 2CuI + I₂, then I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻; so moles of Cu²⁺ = moles of S₂O₃²⁻.

Colour of complexes

Degenerate orbitals have the same energy; non-degenerate orbitals have different energies.
Octahedral complex: two d orbitals higher (3d_z², 3d_x²−y²), three lower. Tetrahedral complex: three higher, two lower.
Energy gap = Planck constant × frequency absorbed (ΔE = hf): a larger ΔE means light of higher frequency (shorter wavelength) is absorbed.
Colour seen = complementary colour of the light absorbed, e.g. orange-red absorbed, blue seen.
Copper(II): [Cu(H₂O)₆]²⁺ blue; Cu(OH)₂(H₂O)₄ pale blue precipitate; [Cu(NH₃)₄(H₂O)₂]²⁺ deep blue; [CuCl₄]²⁻ yellow (the solution often looks yellow-green).
Cobalt(II): [Co(H₂O)₆]²⁺ pink; Co(OH)₂(H₂O)₄ blue precipitate; [Co(NH₃)₆]²⁺ yellow-brown; [CoCl₄]²⁻ blue.
No colour if the d orbitals are empty (3d⁰) or full (3d¹⁰): no d electron can be promoted.

Stereoisomerism in transition element complexes

Square planar [MA₂B₂]: cis and trans isomers. Tetrahedral [MA₂B₂]: one form only.
Octahedral [MA₄B₂]: two isomers, cis (B ligands at 90°) and trans (B ligands at 180°).
[Ni(en)₃]²⁺: two optical isomers, no cis/trans forms.
[Ni(en)₂(H₂O)₂]²⁺: three stereoisomers. The trans form has a plane of symmetry; the cis form exists as two optical isomers.
Trans isomers: the dipoles are opposite and cancel, so the complex is non-polar. Cis isomers: the dipoles do not cancel, so the complex is polar.
A bidentate ligand can only join two positions that are next to each other (90° apart).

Stability constants, Kstab

K_stab = [complex] ÷ ([aqueous metal ion] × [ligand]ⁿ), where n is the number of ligands that go in.
Example: for Cu²⁺(aq) + 4Cl⁻(aq) ⇌ [CuCl₄]²⁻(aq), K_stab = [[CuCl₄]²⁻] ÷ ([Cu²⁺][Cl⁻]⁴).
Never include [H₂O] in the expression.
Units depend on n: one ligand dm³ mol⁻¹; two ligands dm⁶ mol⁻²; four ligands dm¹² mol⁻⁴.
Larger K_stab (or larger log K_stab) = more stable complex.
Ligand exchange goes towards the complex with the larger K_stab.

An introduction to A Level organic chemistry

Formulas, functional groups and the naming of organic compounds

Acyl chloride, RCOCl: name ends in -oyl chloride, e.g. CH₃COCl is ethanoyl chloride. The COCl carbon is carbon 1.
Amide, RCONH₂: name ends in -amide. A group on the nitrogen is shown with N-, e.g. CH₃CONHCH₃ is N-methylethanamide.
Ester, RCOOR′: the alkyl group R′ (from the alcohol) comes first, then the -oate part (from the acid), e.g. HCOOCH₂CH₃ is ethyl methanoate.
Ring positions: carbon 1 has the main group (–OH in phenol, –COOH in benzoic acid, –NH₂ in phenylamine); position 2 is next to it, 3 is one further on, 4 is directly opposite.
Each group on a benzene ring replaces one H: C₆H₅– with one group, C₆H₄ with two, C₆H₃ with three.
Amino acids are named as amino-substituted acids, e.g. H₂NCH₂COOH is 2-aminoethanoic acid.
General formula of a homologous series: alkyl group + functional group, e.g. acyl chlorides C_nH_2n+1COCl.

Characteristic organic reactions

Electrophile: electron-pair acceptor. Nucleophile: electron-pair donor.
Electrophilic substitution: benzene ring + electrophile; one ring H is replaced, e.g. C₆H₆ + Br₂ → C₆H₅Br + HBr (AlBr₃ catalyst).
Ring reactions that are electrophilic substitution: halogenation with a catalyst, nitration, Friedel–Crafts alkylation and acylation.
Addition–elimination: acyl chloride + nucleophile, e.g. CH₃COCl + H₂O → CH₃COOH + HCl.
Alkenes give electrophilic addition; arenes give electrophilic substitution.
Halogen + ultraviolet light on an alkyl side chain is free-radical substitution, not electrophilic substitution.

Shapes of aromatic organic molecules; σ and π bonds

Each ring carbon: sp², three σ bonds, bond angles 120°, planar.
σ bond: end-on overlap of orbitals. π bond: sideways overlap of p orbitals.
Benzene has 12 σ bonds (6 C–C and 6 C–H) and 6 delocalised π electrons.
All C–C bonds in benzene are equal in length, between a C–C single bond and a C=C double bond.
The ring must be flat so that the p orbitals are parallel and can overlap.
Atoms joined directly to the ring lie in the plane of the ring. An alkyl carbon such as –CH₃ is sp³ (109.5°), so its H atoms are not all in the plane.

Isomerism: optical

Optically active: the substance rotates the plane of plane polarised light.
One enantiomer rotates the plane clockwise (+), the other anticlockwise (−), by the same angle at the same concentration.
Racemic mixture: 50% of each enantiomer; optically inactive because the two rotations cancel.
Making a chiral compound from non-chiral reagents with no chiral catalyst gives a racemic mixture.
Enantiomers cannot be separated by distillation or simple crystallisation: their physical properties are identical.
Drug synthesis: separate the racemic mixture into pure enantiomers (difficult and costly), or use a chiral catalyst to make only one enantiomer.
A single enantiomer drug can mean a lower dose and fewer side effects.

Hydrocarbons

Arenes

Halogenation of the ring: Cl₂ or Br₂ with AlCl₃ or AlBr₃, e.g. C₆H₆ + Br₂ → C₆H₅Br + HBr. The catalyst makes the electrophile: Br₂ + AlBr₃ → Br⁺ + AlBr₄⁻.
Nitration: concentrated HNO₃ and concentrated H₂SO₄, 25 °C to 60 °C. HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O.
Friedel–Crafts, AlCl₃ and heat: CH₃Cl gives methylbenzene (alkylation); CH₃COCl gives phenylethanone, C₆H₅COCH₃ (acylation). HCl also forms.
Side-chain oxidation: hot alkaline KMnO₄, then dilute acid, turns any alkyl side chain into –COOH (benzoic acid).
Hydrogenation: H₂ with Pt or Ni, heat, turns the benzene ring into a cyclohexane ring (3 mol of H₂ per ring).
Halogen + ultraviolet light, no catalyst: substitution in the side chain, e.g. C₆H₅CH₂Cl.
Directing effects: –NH₂, –OH and –R (alkyl) direct to positions 2, 4 and 6; –NO₂, –COOH and –COR direct to positions 3 and 5.

Halogen compounds

C₆H₆ + Cl₂ → C₆H₅Cl + HCl, with AlCl₃ catalyst.
Catalyst role: Cl₂ + AlCl₃ → Cl⁺ + AlCl₄⁻; AlCl₃ accepts a lone pair from chlorine.
Methylbenzene + Cl₂ with AlCl₃ gives 2-chloromethylbenzene and 4-chloromethylbenzene.
Chloroethane: CH₃CH₂Cl + NaOH → CH₃CH₂OH + NaCl (nucleophilic substitution, heat). Chlorobenzene: no reaction under these conditions.
C–Cl bond in chlorobenzene: shorter and stronger than in chloroethane (partial double-bond character).
Test: warm with NaOH(aq), acidify with dilute HNO₃, add AgNO₃(aq). A white precipitate of AgCl forms from a chloroalkane only.
A halogen on a side chain (e.g. C₆H₅CH₂Cl) behaves like a halogenoalkane and is hydrolysed.

Hydroxy compounds

Alcohols

alcohol + acyl chloride → ester + hydrogen chloride (ROH + R′COCl → R′COOR + HCl).
CH₃COCl + CH₃CH₂OH → CH₃COOCH₂CH₃ + HCl.
Conditions: room temperature, no catalyst, dry reagents; the reaction is vigorous and not reversible.
Mole ratio: one –OH group reacts with one acyl chloride molecule and gives one HCl.
Naming the ester: alkyl part from the alcohol, -oate part from the acyl chloride.
Water must be absent: CH₃COCl + H₂O → CH₃COOH + HCl.

Phenol

Making phenol: phenylamine + NaNO₂ and dilute HCl below 10 °C gives the diazonium salt; warming it with water gives phenol and N₂. C₆H₅N₂⁺ + H₂O → C₆H₅OH + N₂ + H⁺.
With alkali: C₆H₅OH + NaOH → C₆H₅O⁻Na⁺ + H₂O. With sodium: 2C₆H₅OH + 2Na → 2C₆H₅O⁻Na⁺ + H₂.
Phenol is too weak an acid to release CO₂ from NaHCO₃(aq).
Nitration: dilute HNO₃(aq) at room temperature gives 2-nitrophenol and 4-nitrophenol.
Bromination: Br₂(aq) is decolourised and a white precipitate forms. C₆H₅OH + 3Br₂ → C₆H₂Br₃OH (2,4,6-tribromophenol) + 3HBr.
Coupling: phenol in NaOH(aq) + diazonium salt, below 10 °C, gives a coloured azo compound (–N=N–), joined at the 4-position.
Acid strength: ethanol < water < phenol. The –OH group directs to positions 2, 4 and 6.

Carboxylic acids and derivatives

Carboxylic acids

C₆H₅CH₃ + 3[O] → C₆H₅COOH + H₂O (hot alkaline KMnO₄, then dilute acid to turn benzoate ions into benzoic acid).
3RCOOH + PCl₃ → 3RCOCl + H₃PO₃ (heat).
RCOOH + PCl₅ → RCOCl + POCl₃ + HCl; RCOOH + SOCl₂ → RCOCl + SO₂ + HCl.
HCOOH + [O] → CO₂ + H₂O, with Fehling's reagent, Tollens' reagent, acidified KMnO₄ or acidified K₂Cr₂O₇.
HOOCCOOH + [O] → 2CO₂ + H₂O, with warm acidified KMnO₄.
Acid strength: alcohol < phenol < carboxylic acid. Only carboxylic acids release CO₂ from NaHCO₃(aq).
More Cl atoms, and Cl atoms closer to –COOH, give a stronger acid (lower pK_a): CH₃COOH < ClCH₂COOH < Cl₂CHCOOH < Cl₃CCOOH.

Esters

CH₃COCl + CH₃CH₂OH → CH₃COOCH₂CH₃ (ethyl ethanoate) + HCl.
C₆H₅COCl + C₆H₅OH → C₆H₅COOC₆H₅ (phenyl benzoate) + HCl.
Conditions: room temperature, no catalyst; the reaction goes to completion.
HCl gas is given off as steamy, acidic fumes, so use a fume cupboard.
Phenol + carboxylic acid: very poor yield of ester; use the acyl chloride.
Name: first word from the alcohol or phenol (ethyl, phenyl), second word from the acyl chloride (ethanoate, benzoate).
1 mol of –OH compound gives 1 mol of ester and 1 mol of HCl.

Acyl chlorides

CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl
CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl
3CH₃COOH + PCl₃ → 3CH₃COCl + H₃PO₃ (heat)
With water: RCOCl + H₂O → RCOOH + HCl (steamy fumes)
With an alcohol or phenol: RCOCl + R'OH → RCOOR' + HCl (an ester)
With ammonia: RCONH₂; with a primary amine R'NH₂: RCONHR' (amides)
Ease of hydrolysis: acyl chloride (cold water) > alkyl chloride (needs heating with NaOH(aq)) > aryl chloride (not hydrolysed, because a lone pair on Cl overlaps with the ring and strengthens the C–Cl bond)

Nitrogen compounds

Primary and secondary amines

CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br (ethanol, heat, under pressure)
Halogenoalkane + primary amine (ethanol, sealed tube, heat) → secondary amine
Amide to amine: CH₃CONH₂ + 4[H] → CH₃CH₂NH₂ + H₂O (LiAlH₄)
Nitrile to amine: CH₃CN + 4[H] → CH₃CH₂NH₂ (LiAlH₄, or H₂ with Ni)
Amine + acyl chloride at room temperature → amide + HCl
In water: RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻; with acid: RNH₂ + HCl → RNH₃⁺Cl⁻
Base strength: ammonia < ethylamine < diethylamine

Phenylamine and azo compounds

C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O (hot Sn and concentrated HCl, then NaOH(aq))
C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr: bromine water is decolourised and a white precipitate of 2,4,6-tribromophenylamine forms
C₆H₅NH₂ + HNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + 2H₂O (NaNO₂ and dilute HCl, below 10 °C)
Warming the diazonium salt with water: C₆H₅N₂⁺ + H₂O → C₆H₅OH + N₂ + H⁺
Coupling: C₆H₅N₂⁺Cl⁻ + C₆H₅OH + NaOH → C₆H₅N=NC₆H₄OH + NaCl + H₂O (cold, alkaline)
Base strength: phenylamine < ammonia < ethylamine

Amides

CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl (room temperature)
Acyl chloride + primary amine: RCOCl + R'NH₂ → RCONHR' + HCl
Acid hydrolysis: CH₃CONH₂ + H₂O + HCl → CH₃COOH + NH₄Cl (heat)
Alkaline hydrolysis: CH₃CONH₂ + NaOH → CH₃COONa + NH₃ (heat)
Reduction: RCONH₂ + 4[H] → RCH₂NH₂ + H₂O (LiAlH₄)
Base strength: amide (neutral) < phenylamine < ammonia < alkyl amine

Amino acids

Zwitterion: ⁺H₃NCHRCOO⁻; ionic attractions between zwitterions give solid amino acids high melting points
pH below the isoelectric point: ⁺H₃NCHRCOOH, positive, moves to the negative electrode
pH above the isoelectric point: H₂NCHRCOO⁻, negative, moves to the positive electrode
pH equal to the isoelectric point: no overall charge, stays near the start
Smaller ions and ions with a larger charge move further in the same time
M_r of a peptide = sum of the M_r of the amino acids − 18 for each peptide bond
Two different amino acids A and B can give four dipeptides: A–A, B–B, A–B and B–A

Polymerisation

Condensation polymerisation

Diol + dicarboxylic acid → polyester + H₂O; diol + dioyl chloride → polyester + HCl
Diamine + dicarboxylic acid → polyamide + H₂O; diamine + dioyl chloride → polyamide + HCl
Two monomers: the repeat unit contains one of each monomer, for example –O–R–O–CO–R'–CO–
One monomer (HO–R–COOH or H₂N–R–COOH): the repeat unit is –O–R–CO– or –NH–R–CO–
M_r of repeat unit from a diol or diamine with a dicarboxylic acid = sum of both M_r − 2 × 18
With a dioyl chloride: sum of both M_r − 2 × 36.5
One monomer with both groups: M_r of repeat unit = M_r of monomer − 18

Predicting the type of polymerisation

C=C in the monomer → addition polymer; nothing else is formed
Two functional groups per monomer (two –OH, two –NH₂, two –COOH, two –COCl, or one of each kind that react together) → condensation polymer plus H₂O or HCl
Backbone of carbon atoms only → addition
–CO–O– or –CO–NH– in the backbone → condensation (polyester or polyamide)
An ester or amide group in a side chain does not make a condensation polymer
A monomer with only one reactive group, such as ethanol or ethanoic acid, cannot form a condensation polymer

Degradable polymers

Poly(alkenes): non-polar, no bonds that can be hydrolysed, not biodegradable
Photodegradable: broken down by light (ultraviolet); it does not happen when the polymer is buried
Polyester + acid, heat → diol + dicarboxylic acid
Polyester + NaOH(aq), heat → diol + sodium salt of the dicarboxylic acid
Polyamide + acid, heat → dicarboxylic acid + salt of the diamine (–NH₃⁺ groups)
Polyamide + NaOH(aq), heat → sodium salt of the dicarboxylic acid + diamine
Hydrolysis of one ester or amide link uses one H₂O, or one NaOH in alkali

Organic synthesis

2,4-dinitrophenylhydrazine: orange precipitate with aldehydes and ketones
Tollens' reagent (silver mirror) or Fehling's solution (red precipitate): aldehydes only
Alkaline aqueous iodine: yellow precipitate of CHI₃ with CH₃CO– or CH₃CH(OH)– groups
Na₂CO₃ or NaHCO₃: CO₂ with carboxylic acids only, not with phenols or alcohols
NaOH(aq) reacts with phenols and carboxylic acids; sodium metal reacts with every –OH group (alcohol, phenol, acid) to give H₂
Br₂(aq): decolourised by C=C; white precipitate with phenol or phenylamine
One more carbon: halogenoalkane + KCN in ethanol, heat → nitrile; then hydrolyse to a carboxylic acid or reduce to an amine

Analytical techniques

Thin-layer chromatography

R_f = distance moved by the spot ÷ distance moved by the solvent front, both measured from the baseline
R_f is between 0 and 1 and has no units
Stronger attraction to the stationary phase → smaller R_f
More soluble in the mobile phase → larger R_f
A more polar solvent increases the R_f of polar compounds on a polar plate
Identify a spot by comparing its R_f with a known compound run on the same plate in the same solvent
Draw the baseline in pencil and keep the solvent level below it at the start

Gas / liquid chromatography

Mobile phase: unreactive carrier gas; stationary phase: high boiling point non-polar liquid on a solid support
Stronger attraction to (more soluble in) the stationary phase → longer retention time
On a non-polar column, similar compounds of higher boiling point have longer retention times
A higher column temperature shortens all the retention times
percentage of a component = (area of its peak ÷ total area of all peaks) × 100
Area of a triangular peak = ½ × base × height
Identify a peak by comparing its retention time with a pure sample under the same conditions

Carbon-13 NMR spectroscopy

Number of peaks = number of different carbon environments (not the number of carbon atoms)
Look for symmetry: propanone, CH₃COCH₃, has 3 carbons but only 2 peaks
Alkyl carbon: about δ 0–50
C–O (alcohols, esters): about δ 50–70
C=C and benzene ring carbons: about δ 110–160
C=O in acids and esters: about δ 160–185
C=O in aldehydes and ketones: about δ 190–220

Proton (1H) NMR spectroscopy

n + 1 rule: a proton with n equivalent protons on adjacent carbon atoms gives n + 1 lines
0 neighbours: singlet; 1: doublet; 2: triplet; 3: quartet
Equivalent protons do not split each other; O–H and N–H protons usually give a singlet and do not split their neighbours
TMS, Si(CH₃)₄: 12 equivalent protons give one sharp peak at δ = 0; it is unreactive and easily removed
Approximate δ: alkyl H 0.9–1.7; H–C–C=O 2.2–3.0; H–C–aryl 2.3–3.0; H–C–O 3.2–4.0
Approximate δ: aryl H 6.0–9.0; aldehyde H 9.3–10.5; –COOH 9.0–13.0
A peak that disappears with D₂O is an O–H or N–H proton