Organic chemistry

What the exam asks

Expect a structure, formula or spectrum in the stem and a question that requires reading it rather than recalling it. The commonest items are: count degrees of unsaturation from a formula, decide whether two structures are isomers or the same compound written differently, rank two or three compounds by acidity and justify it, identify a functional group from an IR peak or a proton count, and say which nitrogen in a molecule is the basic one. The trap that catches most candidates is ranking acidity by looking at the acid instead of the anion. Nothing about an O-H bond tells you whether it will ionise; what tells you is whether the resulting negative charge has somewhere to go. Draw the conjugate base, ask how many atoms share the charge and how electronegative they are, and the ranking falls out. The second trap is a degrees of unsaturation count that forgets rings exist, which turns every cyclic answer into a wrong one.

Organic chemistry at this level is a language problem before it is a chemistry problem. A molecule is a skeleton with functional groups hung off it, and nearly everything a question asks turns on recognising the group and knowing what that group makes the molecule do. The skeleton is scenery. The group is the plot.

The exam does not test nomenclature for its own sake and it does not expect a memorised reagent table here. It gives you a structure, or a formula, or a spectrum, and asks you to extract something from it: how many rings and double bonds the formula demands, whether two drawings are the same compound or different ones, which of two acids gives up its proton more readily, which functional group a peak belongs to. All of those are inference from a rule, which is why the rules are worth more than the examples.

Two ideas carry most of the weight. Degrees of unsaturation converts a molecular formula into a count of rings plus pi bonds, which narrows a structure question before you have drawn anything. And acidity is never about the acid: it is about the stability of what is left behind after the proton goes. Resonance, electronegativity, induction and hybridisation are four ways of asking the same question, which is how well the anion can cope with its charge.

What to hold

• A functional group is the reactive part of a molecule, and the carbon skeleton mostly modifies what that group does rather than doing anything itself.
• Degrees of unsaturation from a molecular formula is (2C+2+N-H-X)/2, where X counts halogens and oxygen is ignored entirely.
• Each degree of unsaturation is one ring or one pi bond, so a benzene ring accounts for four: three double bonds and the ring itself.
• Constitutional isomers have the same formula but a different connectivity; stereoisomers have identical connectivity and differ only in arrangement in space.
• Acidity is decided by the stability of the conjugate base, so the question to ask is always what the anion looks like after the proton leaves.
• A negative charge spread over several atoms by resonance is more stable than one stuck on a single atom, which is why a carboxylic acid with a pKa near 5 beats an alcohol near 16.
• A phenol, with a pKa near 10, sits between them: its anion is delocalised into the ring, but onto carbons rather than onto a second electronegative oxygen.
• An electron-withdrawing group near an acidic site pulls charge away from the anion and stabilises it, raising acidity, and the effect fades quickly as the group is moved further away.
• Charge is more comfortable on a more electronegative atom, which is why an O-H is far more acidic than the N-H or C-H of a comparable compound.
• More s character holds a lone pair closer to the nucleus and stabilises it, which is why a terminal alkyne C-H is measurably acidic while an alkane C-H is not.
• An amine is basic because its nitrogen lone pair is available; an amide nitrogen is not basic, because that lone pair is delocalised into the neighbouring carbonyl.
• In an infrared spectrum, a broad absorption near 3300 suggests O-H and a strong sharp one near 1700 suggests C=O. Those two peaks resolve most first-year IR questions.
• In a proton NMR spectrum, the number of signals counts chemically distinct environments, integration gives the relative number of protons, and splitting into n + 1 peaks counts n neighbouring protons.
• In a mass spectrum, a large peak two units above the molecular ion signals a halogen: about one third the height for chlorine, roughly equal height for bromine.