Organic chemistry is a language exam disguised as a science exam. The molecules are simple; the marks are lost in the words. Candidates write CnH2n+2 for an alkene, say "saturated means full", tell the examiner bromine water "turns clear" (it decolourises — a different thing), give ethene a substitution reaction and ethane an addition, run cracking backwards, ferment glucose at 300 °C, draw poly(ethene) with its C=C still intact, name an ester backwards, and confuse isomers with homologues. Every one of these is a named, repeat offender in examiners' reports. This guide takes the twelve worst, explains the precise wording the mark scheme wants, and then drills it until the language is automatic.
Twelve traps that cost students marks on Topic 11 questions. Every one of them appears on challenge papers regularly.
Six challenging questions broken down step by step. Try each step yourself before revealing the next.
For n = 14: CnH2n predicts H = 28; CnH2n+2 predicts H = 30. So X (C₁₄H₂₈) fits CnH2n → an alkene, and Y (C₁₄H₃₀) fits CnH2n+2 → an alkane. The reasoning — the fit to the general formula — is the mark, not the bare family name. This two-second arithmetic works for any hydrocarbon the paper throws at you.
A homologous series is a family of compounds with: (1) the same general formula; (2) the same functional group, and therefore similar chemical properties; (3) each member differing from the next by CH₂, giving a trend in physical properties (boiling point rises along the series). Any three of those clauses score; "they are similar compounds" scores nothing because it names no shared feature.
X, the alkene, is unsaturated: its C=C double bond undergoes an addition reaction with bromine, so the orange solution is decolourised. Y, a saturated alkane, does not react (in the absence of UV light), so the bromine water stays orange. Say the result for both compounds — a distinguishing test needs both outcomes stated.
Y (C₁₄H₃₀) is a much larger molecule than propane (C₃H₈): the attractive forces between larger molecules are stronger, so more energy is needed to separate them and the boiling point is higher. This is the trend the homologous-series definition promised — and it is the entire logic of fractional distillation of petroleum: bigger molecules, higher boiling point, lower down the column.
Three recurring errors: assigning X to the alkanes because "it has lots of hydrogens" (the test is the formula fit, not the impression of plenty); defining a homologous series as "compounds with similar formulas" (no credit — name the shared general formula and functional group); and explaining the boiling-point difference with "bigger molecules have stronger bonds" (the covalent bonds are irrelevant — it is the attractions between molecules that strengthen).
Ethane is saturated — single bonds only — so bromine cannot add; it can only replace a hydrogen atom. This is substitution, it requires ultraviolet light (a photochemical reaction — the light supplies the energy to start it), and it produces two products: C₂H₆ + Br₂ → C₂H₅Br + HBr. In the dark, nothing happens — a fact the examiner loves to test.
Ethene's C=C double bond opens, and both bromine atoms add across it: C₂H₄ + Br₂ → C₂H₄Br₂ (1,2-dibromoethane). This is addition: it happens rapidly at room temperature, no light required, and gives one product only — nothing is displaced, so nothing else is made. The double bond is the reactive site; that is what "functional group" means in practice.
In substitution, the bromine molecule splits its atoms between two destinations: one Br goes onto the carbon chain, and the displaced H leaves with the other Br as HBr — two products by necessity. In addition, both bromine atoms end up on the same molecule (one on each carbon of the opened double bond) — nothing leaves, one product. This atom-bookkeeping is the fastest way to check you wrote the right reaction type.
Shake each gas with aqueous bromine. Ethene: the orange solution is decolourised (orange → colourless) — rapidly, at room temperature, by addition. Ethane: the solution remains orange (no reaction without UV, and even under UV substitution is slow). Both observations are needed for a "distinguish" question, and the word is decolourised or colourless — never "clear", which means transparent and describes both solutions.
Mark schemes routinely allocate a whole mark to ultraviolet light for the alkane reaction — and a common error is attaching UV to the alkene reaction instead, which actively signals confusion. If a question says a mixture of ethane and bromine "showed no change", the expected explanation is: no UV light was present, and ethane cannot react by addition because it has no C=C.
Catalytic cracking breaks large, less useful alkanes from the heavy fractions into smaller, more useful molecules — petrol-sized alkanes plus alkenes (and sometimes hydrogen). Direction check: cracking always goes big → small; if your sentence has small molecules joining up, you are describing polymerisation, its exact opposite.
The vapour of the heavy fraction is passed over a hot catalyst (around 500–600 °C, silica/alumina) — the process is a thermal decomposition. "High temperature and a catalyst" earns the condition marks. No pressure figure, no acid, no yeast — keep the ethanol-route conditions from wandering in here.
Write the totals first: 12 carbons, 26 hydrogens. Octane takes 8 C and 18 H, leaving 4 C and 8 H: the other product is C₄H₈ — butene (an alkene: 8 = 2 × 4). The equation: C₁₂H₂₆ → C₈H₁₈ + C₄H₈. Every cracking equation is this same subtraction; the only skill is refusing to guess before counting.
An alkane CnH2n+2 has exactly two "spare" hydrogens beyond 2-per-carbon. Split it into two pieces and each piece would need those two spares to be an alkane — but there is only one pair to give. So one fragment must make do with H = 2n — a C=C double bond forms and that product is an alkene. This shortage argument, or simply "there are not enough hydrogen atoms for both products to be alkanes", earns both marks.
Bubble the gas through aqueous bromine: it is decolourised, orange → colourless — confirming a C=C double bond. (An alkane by-product would leave it orange.) Same test, same wording discipline as always: colour before, colour after, and the verb is decolourise.
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Conditions: aqueous glucose (sugar solution), yeast — whose enzymes are the catalyst — kept at 25–35 °C, in the absence of oxygen (anaerobic conditions). Four conditions, four half-marks in effect. Note the by-product: carbon dioxide — the same gas that makes bread rise and sparkling wine sparkle.
C₂H₄ + H₂O → C₂H₅OH — an addition reaction: steam adds across ethene's double bond. Conditions: 300 °C, 60 atm, phosphoric acid catalyst. Keep the two condition sets in separate mental boxes: the living route gets living conditions; the petrochemical route gets furnace-and-pressure conditions. Writing "yeast" here, or "300 °C" in fermentation, is an instant flag of confusion.
Advantages: it uses renewable raw materials (sugars from crops) and runs at low temperature with cheap equipment — low energy cost. Disadvantages: it is slow, runs as a batch process (stop, empty, clean, restart), and yields a dilute, impure ethanol solution that needs fractional distillation to concentrate. Hydration mirrors these: fast, continuous, essentially pure product — but the ethene comes from cracking petroleum, a finite resource, and the conditions are energy-expensive.
The yeast's enzymes are denatured at high temperature — the protein's shape is permanently destroyed, so the catalyst stops working and fermentation stops. The creditworthy word is denatured (or "enzymes destroyed / yeast killed by heat"). "The reaction gets too hot to work" is circular and scores nothing. The mirror question — why not run fermentation cold? — has the mirror answer: enzymes act too slowly at low temperature.
Fermentation yields perhaps 15% ethanol in water. The mixture is separated by fractional distillation: ethanol boils at 78 °C, water at 100 °C, so the ethanol vapour rises and is collected first. This links Topic 11 to the separation techniques of Topic 2 — and it is why the "disadvantage" of fermentation is not merely dilution but the extra energy and equipment the dilution demands.
Propene has a C=C, so it polymerises by addition: thousands of monomers link as each double bond opens, and the polymer is the only product — every atom of every monomer is in the chain. If a polymerisation releases water, it is not addition. That single sorting rule — anything lost or not? — classifies every polymer question you will meet.
The repeat unit of poly(propene): a two-carbon backbone joined by a single bond; one carbon carries H and CH₃, the other carries two H; continuation bonds pass through the brackets on both sides; subscript n. The checklist kills the three standard errors: C=C left intact (that is the monomer), no trailing bonds (chain looks sealed), and a three-carbon backbone (the CH₃ is a side group, not chain).
Nylon forms by condensation polymerisation: a dicarboxylic acid (COOH at both ends) reacts with a diamine (NH₂ at both ends). Each link forms an amide linkage, –CO–NH–, and releases one molecule of water — that elimination is what "condensation" means. Because each monomer has a reactive group at both ends, the chain grows in both directions indefinitely.
PET (Terylene) is a polyester: dicarboxylic acid + diol, linked by ester linkages –COO–. And the syllabus's favourite bridge to biology: proteins are natural polyamides — amino acids joined by the same –CO–NH– (peptide/amide) link as nylon. Identification drill: nitrogen in the linkage → amide → nylon or protein; two oxygens, no nitrogen → ester → PET.
Plastics are non-biodegradable: microbes cannot break them down, so they persist in landfill for centuries and accumulate in oceans, harming wildlife (entanglement, ingestion, microplastics in food chains). Burning them releases toxic gases — carbon monoxide from incomplete combustion, and, from chlorine-containing polymers like PVC, HCl. Name concrete problems; "bad for the environment" earns nothing.
Route 1: bacterial oxidation — leave ethanol (wine) exposed to air; bacteria oxidise it to ethanoic acid. This is how vinegar forms, and why opened wine sours. Route 2: heat ethanol with acidified potassium manganate(VII) — a strong oxidising agent; the purple colour is decolourised as the ethanol is oxidised to ethanoic acid. Either route: ethanol + oxygen (oxidation) → ethanoic acid.
A weak acid is one that is only partially dissociated (ionised) into its ions in aqueous solution: CH₃COOH ⇌ CH₃COO⁻ + H⁺ — note the equilibrium arrow. Contrast: a strong acid (HCl) is completely dissociated. Weak does not mean dilute, and strong does not mean concentrated — strength is about the fraction ionised, concentration about how much is dissolved. Consequence: ethanoic acid of the same concentration has a higher pH and reacts more slowly than hydrochloric acid.
Ethanoic acid does everything Topic 7 promised: with magnesium → magnesium ethanoate + hydrogen (effervescence, gas pops with lighted splint); with sodium carbonate → salt + water + CO₂ (fizzes, gas turns limewater milky); with sodium hydroxide → salt + water (neutralisation). Any one, with its observation, is the mark. The salts are ethanoates — e.g. sodium ethanoate, CH₃COONa.
Warm ethanoic acid + ethanol with a few drops of concentrated sulfuric acid (catalyst): the product is the ester ethyl ethanoate plus water. Name assembly: the alcohol contributes "ethyl" (-ol → -yl), the acid contributes "ethanoate" (-oic acid → -oate). CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. Esters are recognised in the flask by their sweet, fruity smell.
Esters are used as solvents (e.g. in nail varnish remover) and as artificial fruit flavourings and perfume ingredients — their volatility and fruity odours make them ideal. Exam packaging: this question type hands out marks in a fixed bundle — name of ester, catalyst, type of compound, use — so rehearse the bundle as one unit: ethyl ethanoate; concentrated H₂SO₄; an ester; flavourings/solvent.
Pairs of questions that look nearly identical but have different answers. Spot the key distinction.
Click each node to see how the subtopics connect.
Spot the error in each student's answer. Think before revealing.
Ten Cambridge-style challenge questions. Write your answer, then reveal the model answer with mark scheme and examiner's notes.