What this chapter covers13

Organic chemistry looks, at first, like an enormous amount of memorising. It is not. Almost everything in the rest of this course follows from four small ideas introduced here: that a molecule can be written down in several different ways and you must be fluent in all of them; that the reactions of a compound are decided by one small part of it; that a carbon atom takes one of three shapes depending on how many things are attached to it; and that two molecules can share a formula and still be different compounds. Get these right and the later topics become a short list of reactions attached to ideas you already have. Get them wrong and every later topic feels like a new subject.

Read the verbs, because they set the price of each statement. 13.1.4 says interpret and use — you have to go both ways, from a structure to a formula and from a formula to a structure. 13.1.5 says understand and use, which means producing a name you have never seen for a molecule you have never seen, not recognising one from a list. 13.3.2 says describe and explain: a bond angle on its own is half an answer. And 13.4.6, which asks you to deduce every isomer of a molecular formula, is the one statement on this page that is pure method — there is nothing to learn, only a procedure to carry out without missing anything.

Hydrocarbons, alkanes and the idea of a functional group13.1.1–13.1.3

Three statements, and the first two are one sentence each.

The word that earns the mark in the first definition is only. Ethanol, CH₃CH₂OH, contains carbon and hydrogen; it is not a hydrocarbon, because it also contains oxygen. Poly(ethene) is a hydrocarbon. Chloromethane is not. It is a narrow definition and examiners test its edges.

The second statement is more interesting than it looks. Saying an alkane has no functional group is a claim about its chemistry: there is no part of the molecule that a reagent can pick out. Every bond in an alkane is either C–C or C–H, both strong and both very nearly non-polar, so there is nowhere for a polar reagent to attack and nothing for a nucleophile to find. That is the whole explanation of why alkanes are so unreactive, and it is examined directly in 14.1.5.

The classes of compound you are given13.1.3

The syllabus supplies a table of functional groups in the exam, so the job is not to memorise it but to recognise a group on sight and know what it does. The eight classes below are the ones this course uses most; esters and nitriles are in the table too, with their chemistry in later topics.

Five ways to write the same molecule13.1.4

Statement 13.1.4 is one of the most important in the whole syllabus, and it is worth being blunt about why: every organic question you will ever answer begins by turning something written on paper into a molecule in your head. If you cannot do that step reliably, nothing that follows can be attempted. There are five ways of writing a molecule down, and each throws away a different amount of information.

Reading a skeletal formula

Skeletal formulae appear constantly in Cambridge papers, and they are the form students most often misread. Three rules cover every case:

• Every corner and every free end of the line is a carbon atom. A line of five segments therefore has six carbons, not five.
• Every carbon has four bonds. Whatever is not drawn is hydrogen, so count the lines meeting that corner and add hydrogens until you reach four.
• Anything that is not carbon or hydrogen is written in — O, N, Cl, and the hydrogens attached directly to them.

General formulae and homologous series13.1.4, 13.2.1(a)

The general formula is the part that does real work. C_nH_2n+2 is not a piece of trivia about alkanes — it is a statement that a saturated chain of n carbons has exactly 2n+2 hydrogens, which lets you check any structure you have drawn in two seconds. If your "C₅H₁₂" isomer comes out with 11 hydrogens, you have left a carbon with three bonds somewhere.

Molecular and empirical formulae from a structure13.1.6

This statement asks for one skill in two directions. Going from a structure to a molecular formula is careful counting; going from a molecular formula to an empirical formula is arithmetic.

To get the molecular formula: count the atoms. From a structural formula it is bookkeeping — CH₃(CH₂)₂COOH is 1 + 2 + 1 = 4 carbons, 3 + 4 + 1 = 8 hydrogens, and 2 oxygens, so C₄H₈O₂. From a skeletal formula it is the three rules above. From a displayed formula it is simply slow and careful reading, and the only real danger is missing something.

To get the empirical formula: divide every subscript by their highest common factor. C₂H₄O₂ → CH₂O. C₆H₆ → CH. C₄H₁₀ → C₂H₅. Very often the factor is 1 and the two formulae are identical, which is a correct answer, not a failure to simplify.

Naming: the four decisions13.1.5

Systematic naming is not a list to learn. It is four decisions taken in a fixed order, and once you take them in that order the name assembles itself. Every name has the same shape:

(locants)-(substituents)  +  stem  +  (un)saturation  +  suffix

Naming with a functional group13.1.5

Once there is a functional group, decision 1 takes over from decision 3 — the group gets the lowest locant whatever that does to the substituents.

Breaking bonds: homolytic and heterolytic fission13.2.1(c)–(d)

A covalent bond is a shared pair of electrons. When it breaks, the pair has to go somewhere, and there are exactly two possibilities: one electron to each atom, or both to one atom. Everything about organic mechanisms follows from which of the two happens.

Which one happens is decided by the bond. A non-polar bond — Cl–Cl, Br–Br, C–H — has no reason to give both electrons to one end, so when it is supplied with enough energy by ultraviolet light or by heat it breaks homolytically. A polar bond — C–Br, C–O, C=O — is already pulling electrons towards one atom, and in a polar solvent that pull can be completed: it breaks heterolytically.

The mechanism itself is carried out in topic 14.1, where alkanes react with chlorine in ultraviolet light. What matters here is the vocabulary, because 13.2.1(d) names all four terms and they are examined as definitions in their own right.

Nucleophiles and electrophiles13.2.1(e)

Naming the reaction13.2.1(f)

Five words, and each one describes what happened to the molecules rather than what the reagent was. You can work out which applies by counting.

Oxidation and reduction, [O] and [H]13.2.1(g)

Oxidation numbers work perfectly well in organic chemistry, but they are laborious and rarely asked for. In practice the working definitions are:

• Oxidation — oxygen gained, or hydrogen lost.
• Reduction — hydrogen gained, or oxygen lost.

So the sequence primary alcohol → aldehyde → carboxylic acid is two successive oxidations: the first removes two hydrogens, the second adds an oxygen.

The four mechanisms, and the curly arrow13.2.2

Statement 13.2.2 names four mechanisms. Each is the meeting of a reagent type with a functional group type, and the name is built from the two halves: who attacks, then what happens to the molecule.

Straight-chained, branched and cyclic13.3.1

The shortest statement in the topic, and worth two minutes. Three words describe the shape of a carbon skeleton:

• Straight-chained — every carbon is joined to at most two others, so the skeleton is one unbranched line. Butane is straight-chained. The name is a convention, not a description: the real molecule is a zig-zag at 109.5°, and it coils freely in solution.
• Branched — at least one carbon is joined to three or four other carbons, so a side chain hangs off the main one. 2-methylbutane is branched.
• Cyclic — the carbons form a closed ring. Cyclohexane is cyclic. A ring costs two hydrogens compared with the open chain, which is why cyclohexane is C₆H₁₂ and not C₆H₁₄.

The word aliphatic covers all three: it means a compound that is not aromatic, and every molecule on this page is aliphatic. Benzene rings are topic 25.

sp³, sp² and sp: shape and bond angle13.3.2

A carbon atom has four outer electrons and makes four bonds. What changes from molecule to molecule is how many of those bonds are σ bonds — and that number, on its own, fixes the shape around that carbon.

The reasoning behind the names is that the 2s orbital and some of the 2p orbitals mix to give a new set of equivalent orbitals pointing in the right directions. Mix the 2s with all three 2p orbitals and you get four sp³ orbitals pointing at the corners of a tetrahedron. Mix it with two of them and you get three sp² orbitals in a plane at 120°, leaving one p orbital untouched. Mix it with one and you get two sp orbitals pointing in opposite directions, leaving two p orbitals untouched. The leftover p orbitals are the ones that make π bonds — which is why the sums always work out.

σ and π bonds13.3.3

Two consequences run through the whole of organic chemistry:

• Every bond contains exactly one σ. A single bond is 1σ. A double bond is 1σ + 1π. A triple bond is 1σ + 2π. So counting σ bonds is just counting the lines in a displayed formula; counting π bonds is counting the extra lines.
• The π electrons are the reactive ones. They stick out above and below the molecule, they are held less tightly, and they are the reason an alkene attracts electrophiles while an alkane attracts nothing. Topic 14.2 is, in a sense, one long consequence of this sentence.

Planar molecules13.3.4

A group of atoms is planar if all of them lie in one flat plane. The syllabus names ethene as the example, and ethene is the clean case: both carbons are sp², all six atoms are held in the same plane, and the π cloud sits above and below that plane.

The useful skill is deciding which part of a larger molecule is planar. The rule follows directly from the previous two sections: an sp² carbon and the three atoms bonded to it are coplanar, and so is any second sp² carbon joined to it by a double bond. An sp³ carbon forces nothing flat, because the single bonds it makes can rotate.

The map of isomerism13.4.1–13.4.2

Isomers are compounds with the same molecular formula but a different arrangement of atoms. That is the entry requirement, and everything after it is a question of what is different. The syllabus divides isomerism twice, and the whole of 13.4 fits on one small map.

Chain, positional and functional group isomerism13.4.1

The order in which to test a pair is the order the model below uses, because it is the order that cannot go wrong:

1. Do they have the same molecular formula? If not, they are not isomers at all, and nothing else matters.
2. Do they have the same functional group? If not, it is functional group isomerism.
3. Do they have the same carbon skeleton? If not, it is chain isomerism.
4. Otherwise the group has simply moved: positional isomerism.

Cis–trans isomerism13.4.3

Statement 13.4.3 asks for both the description and the explanation, and the explanation is the part worth marks: the origin is restricted rotation due to the presence of the π bond.

Rotation about a C–C single bond is free, because a σ bond is symmetrical about the line joining the nuclei — turning one end changes nothing about the overlap. Rotation about a C=C is not free, because the π bond depends on two p orbitals lying parallel to each other. Turning one end by 90° pulls them out of alignment and destroys the π bond entirely, which costs about 260 kJ mol⁻¹ — far more than the thermal energy available at room temperature. So the two ends are locked, and whatever is on one side stays on that side.

Finding cis–trans isomerism, including in rings13.4.5

13.4.5 says "including cyclic compounds", and the reason is that a ring restricts rotation just as a double bond does. The carbons of a ring cannot rotate relative to one another without breaking the ring, so a substituent is locked either above the ring or below it.

1,2-dichlorocyclohexane therefore exists in two forms: cis, with both chlorines on the same face of the ring, and trans, with one on each face. The test is unchanged — each of the two ring carbons carrying the groups must have two different things attached to it, which here means one Cl and one H each.

Chiral centres and optical isomers13.4.4

The word "non-superimposable" is doing the work. Your two hands are mirror images and you cannot lay one on the other palm-down and have the thumbs agree; a tetrahedral carbon with four different groups behaves the same way. Swap any two of the four groups and you have made the other isomer, not the same one turned round.

Enantiomers are almost impossible to tell apart by ordinary means: same melting point, same boiling point, same density, same reactions with ordinary reagents. They differ in exactly one respect that concerns this syllabus — they rotate the plane of plane-polarised light by equal amounts in opposite directions. One is labelled (+) and the other (−). A 50:50 mixture of the two rotates it not at all.

Finding the chiral centres in a structure13.4.5

Work through every carbon in turn and ask two questions. Does it have four single bonds? If it is part of a C=C or a C=O, it has only three σ bonds and cannot be chiral. Then: are the four things attached to it all different?

Deducing every isomer of a formula13.4.6

This statement is pure method, and chemguide's description of it — "a nightmare statement" — is honest. There is nothing to learn. The difficulty is entirely in being systematic enough not to miss one and not to draw the same one twice.

Self-test13

Thirty questions across the whole topic. Each one explains itself after you answer, and the explanation is the point — a question you got right for the wrong reason is worth reading too.

Definitions to learn13

These are the ones that get asked for word for word. Everything else on this page can be worked out; these cannot.

Data used on this page13

Bond energies used in the arguments on this page

The C=C figure is the one that carries an argument on this page: 610 against 350 for C–C means the second component of the double bond — the π bond — is worth about 260 kJ mol⁻¹, which is both why rotation is restricted and why the π bond, not the σ, is the one that breaks in an addition reaction.

Boiling points of the straight-chain alkanes

Relative atomic masses

H 1.0 · C 12.0 · N 14.0 · O 16.0 · F 19.0 · Cl 35.5 · Br 79.9 · I 126.9