Pick a bottle from the shelf, pick a test from the rack, and the tube responds in real time. Every observation is written the way it should appear in an answer: colour before, colour after, and what the change proves. "It changes colour" earns nothing on its own.
The bench is idle. Pick a sample from the shelf and a reagent from the rack.
A full-marks observation has three parts: what you saw before, what you saw after, and the speed or conditions if they matter. "Bromine water changed" is not an observation. "The orange-brown bromine water was decolourised immediately, in the dark" is.
Every bottle on the shelf, unlabelled. Run a test, pick the result you got, and watch the field cut down. This is how a distinguishing question actually works — you are not looking for a reaction, you are looking for a split. A test that everything passes tells you nothing.
Watch the split bar on each test. A test where every remaining bottle gives the same result is marked no split: running it would use up a tube and tell you nothing. Choosing the test that cuts the field hardest is the skill being assessed, not remembering that bromine water goes colourless.
Tap an arrow for reagents, conditions and the balanced equation. Tap a box to light up everything that family does. Filter by reaction type to see the pattern: haloalkanes are the hub — almost every route in Level 2 goes through one.
Tap any arrow on the map.
Reagents are written in the forms the standard uses: H₂/Pt for hydrogenation, H₂O/H⁺ for hydration, KOH(aq) for hydrolysis. Other chemistry is equally valid and appears in textbooks — nickel instead of platinum, steam over H₃PO₄ instead of acid then water, NaOH(aq) instead of KOH(aq) — but answers are marked against the forms above, so those are what is shown on the arrows.
Esterification, hydrolysis and amide formation are not here on purpose: they belong to Level 3. Switch the level to L3 in the header and the map grows to the whole of AS91391.
At Level 3 the map carries the derivatives as well: aldehydes and ketones, esters, acyl chlorides, amides and condensation polymers. Reduction and hydrolysis join the reaction types. Everything here is inside AS91391.
Exam questions rarely ask for one reaction. They ask you to get from an alkane to an amine, or an alkene to a carboxylic acid, and expect every reagent along the way. Pick a starting family and a target and the shortest routes are worked out from the map — including the ones that turn out to be impossible, which is worth seeing too.
Only the shortest routes are shown. Longer ones exist — you could always go alkene → haloalkane → alcohol instead of hydrating the alkene directly — and they are not wrong, just less efficient. If a question limits your reagents, count the steps yourself from the map.
Adding H–Br or H–OH across an unsymmetrical double bond gives two possible products, and they are not equally likely. Choose an alkene and a reagent to see both carbocations, why one wins, and what to call each product.
The same C=C opens up, but this time it joins to the next monomer instead of a small molecule. Nothing is lost, so the empirical formula of the polymer matches the monomer. Pick a monomer to build the repeating unit.
Two different monomers, each carrying two functional groups, joined end to end — and a small molecule thrown out at every join. That lost molecule is the entire difference from addition polymerisation, and it is why these polymers can be hydrolysed back apart while poly(ethene) cannot.
Diol + dicarboxylic acid
The –OH of one monomer and the –COOH of the next form an ester link, losing water. Terylene and PET bottles are made this way.
Diamine + dicarboxylic acid
The –NH2 and the –COOH form an amide link, again losing water. This is nylon; the same link, made by proteins from amino acids, is a peptide bond.
Use a diacyl chloride in place of the diacid and the reaction runs cold and to completion, losing HCl instead of water — the classroom nylon rope trick.
Acidified dichromate turns orange to green when it oxidises something. Whether it does, and where it stops, depends entirely on how many hydrogen atoms sit on the carbon carrying the –OH. Pick a class of alcohol and step through.
Once you have a carbonyl compound you can say which kind it is. 2,4-DNPH gives an orange precipitate with an aldehyde or a ketone, so it finds the C=O and stops there. Tollens’ (silver mirror) and Fehling’s (brick-red precipitate) go further: only the aldehyde is easily enough oxidised to reduce them.
The ladder also runs downwards. NaBH₄ reduces an aldehyde back to a primary alcohol and a ketone back to a secondary one — it cannot go past the alcohol, for the same reason oxidation cannot go past the ketone.