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Spec Mapping — OCR H432 Module 4.1.1 — Basic concepts of organic chemistry, covering empirical, molecular, general, structural, displayed and skeletal formulae; the homologous-series concept (shared functional group, common general formula, gradual physical-property trends, identical chemistry); the saturated/unsaturated distinction; and the aliphatic/alicyclic/aromatic classification (refer to the official OCR H432 specification document for exact wording).
Organic chemists use several distinct types of formula to represent the same molecule, each emphasising a different aspect of its structure: the atom count alone, the simplest whole-number ratio, the connectivity pattern, every individual bond, or just the carbon skeleton with heteroatoms picked out. Each format has its place. The homologous-series concept then groups related compounds into families that share a general formula and a functional group, so that one piece of chemistry covers thousands of compounds. This lesson formalises both ideas — the five formula types and the homologous-series machinery — and shows how they connect. Mastering this material is what lets you decode any organic-chemistry question on the OCR H432 paper from a single line of structural notation, and what lets you predict the physical properties of a compound you have never seen before from the chain length and functional group alone.
Key Definition: a homologous series is a family of organic compounds with the same functional group and the same general formula (CnH2n+2 for alkanes, CnH2n for alkenes, CnH2n+1OH for alcohols, etc.), in which successive members differ by an additional −CH2− unit, share identical chemistry but show gradually changing physical properties (boiling point, viscosity, volatility, solubility).
The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound.
Key Definition — Empirical formula: The simplest whole-number ratio of atoms of each element present in a compound.
Example: glucose has a molecular formula C₆H₁₂O₆ but an empirical formula CH₂O.
The molecular formula gives the actual number of atoms of each element present in one molecule.
Key Definition — Molecular formula: The actual number of atoms of each element in a molecule.
Example: butane is C₄H₁₀; the empirical formula is C₂H₅.
The general formula expresses the ratio of atoms for any member of a homologous series using n for the number of carbons.
| Homologous series | General formula |
|---|---|
| Alkane | CₙH₂ₙ₊₂ |
| Alkene (one C=C) | CₙH₂ₙ |
| Cycloalkane | CₙH₂ₙ |
| Alcohol (saturated, one OH) | CₙH₂ₙ₊₁OH |
| Haloalkane (one halogen) | CₙH₂ₙ₊₁X |
| Carboxylic acid | CₙH₂ₙO₂ |
| Ketone / aldehyde | CₙH₂ₙO |
A structural formula shows the minimum detail needed to unambiguously identify the molecule — it shows which atoms are joined to which, without drawing every bond.
Examples:
A displayed formula shows every bond and every atom explicitly. You draw every C–H and every C–C single bond individually.
Ethanol displayed (every C–H, C–C, C–O and O–H bond drawn as a separate line):
A skeletal formula is the most compact representation. Carbon atoms are implied at every vertex and line end; hydrogen atoms on carbons are implied (each carbon is assumed to have enough H atoms to complete four bonds). Functional groups and heteroatoms are shown explicitly.
(Four zig-zag carbons with –OH on the terminal carbon.)
graph LR
A["Molecular formula<br/>C2H6O"] --> B["Empirical formula<br/>C2H6O cannot simplify"]
A --> C["Structural formula<br/>CH3CH2OH"]
A --> D["Displayed formula<br/>all bonds shown"]
A --> E["Skeletal formula<br/>vertices imply C"]
OCR examiners use precise verbs. "State the empirical formula" expects a simplest ratio with no further detail. "Draw the displayed formula" demands every C–H bond drawn explicitly — a structural or skeletal answer loses marks. "Draw the skeletal formula" allows the compact form but every heteroatom must still appear. When the question says simply "give the formula", any unambiguous representation is acceptable. Reading the verb is half the mark.
A homologous series is a family of organic compounds which:
| Name | Molecular formula | Boiling point / °C | State at room temp |
|---|---|---|---|
| Methane | CH₄ | −162 | Gas |
| Ethane | C₂H₆ | −89 | Gas |
| Propane | C₃H₈ | −42 | Gas |
| Butane | C₄H₁₀ | −1 | Gas |
| Pentane | C₅H₁₂ | 36 | Liquid |
| Hexane | C₆H₁₄ | 69 | Liquid |
| Octane | C₈H₁₈ | 126 | Liquid |
| Decane | C₁₀H₂₂ | 174 | Liquid |
Each successive alkane differs by CH₂ (adding about 14 g mol⁻¹). Boiling point rises steadily because larger molecules have more electrons, larger surface area, and therefore stronger London (dispersion) forces — more energy is required to overcome them.
As chain length increases within a series:
Two isomers with the same molecular formula can differ significantly in boiling point because of differences in surface area and therefore in London force strength:
| Isomer | B.p. / °C |
|---|---|
| Pentane (straight) | 36 |
| 2-methylbutane | 28 |
| 2,2-dimethylpropane | 10 |
More branched molecules are more spherical, have less surface contact between molecules, and therefore lower boiling points.
Adding one –CH₂– unit corresponds to adding 14 g mol⁻¹ to the molar mass. For alkanes this also adds approximately 20–30 °C to the boiling point (the increment shrinks as molecules get larger). The arithmetic of CnH2n+2 is more than a tidy mnemonic: it expresses the fact that an acyclic saturated hydrocarbon with n carbons has 4n valences from carbon (each C has 4) and n−1 of these are used in C–C bonds (counting bonds, not atoms). The remaining 4n−2(n−1)=2n+2 valences must be filled by hydrogen — hence the formula. Each ring or pi-bond costs two hydrogens, which is why cycloalkanes share CnH2n with alkenes (one ring, no double bond, vs no ring, one double bond — both single degrees of unsaturation).
Substitute n = 8 into CnH2n+2: C8H18. Yes — octane (or any of its 17 isomers) satisfies the formula. By contrast C₈H₁₆ has 2n hydrogens, so it must be a cycloalkane (e.g., cyclooctane), an alkene (e.g., oct-1-ene), or contain one ring/double bond — degree of unsaturation = 1. Whenever you meet an unfamiliar molecular formula, run the general-formula check first.
Within a homologous series the increment in boiling point per added –CH₂– can be approximated by a saturating function: each new methylene raises the b.p. by about 30 °C for n = 1→2 in alkanes, 25 °C for n = 2→3, 20 °C for n = 3→4 and so on, eventually plateauing at around 20 °C per methylene by the long-chain limit. The diminishing return is because fractional polarisability gain per added carbon decreases — adding the 20th methylene to a long chain is a smaller relative change in molecular size than adding the second methylene to ethane.
For alcohols the increments are slightly larger because hydrogen bonding still dominates intermolecular forces for short chains; by ~C8 the alcohols look more like alkanes plus a constant offset (~80 °C above the equivalent alkane) than like a hydrogen-bonded ensemble. OCR mark schemes accept "London forces increase with chain length / electron count" — but if you can quote the saturating trend in passing it tightens an A* answer.
A saturated compound has the maximum possible number of hydrogen atoms for a given number of carbons. Each additional degree of unsaturation (or ring) removes two hydrogens.
The chemical consequence is dramatic: saturated compounds (alkanes) only undergo a small number of reactions (combustion; free-radical substitution with halogens under UV light), whereas unsaturated compounds — by virtue of their reactive pi-electrons — undergo electrophilic addition with bromine, hydrogen halides, water (acid-catalysed), hydrogen (with a Ni or Pt catalyst), and ozonolysis. Sketching the saturation count for an unknown compound therefore narrows the universe of reactions to a manageable subset before any further data is examined. The 1832 Liebig–Wöhler tests on benzoyl chloride and the 1884 Baeyer test (cold dilute alkaline KMnO₄ decolourising) are early experimental probes of saturation that historically anchored the saturated/unsaturated distinction; today we use bromine-water decolourisation in PAG 7.
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