Isomerism: structural isomerism and stereoisomerism

1. Overview

Isomers are compounds with the same molecular formula but a different arrangement of atoms. In structural isomers the atoms are bonded together in a different order; in stereoisomers they are bonded in the same order but arranged differently in space. This section covers both families, how to spot each kind in a given structure (rings included) and how to find every isomer of a molecular formula, skills you will use in almost every later organic topic.

Key Definitions

• Structural isomers: compounds with the same molecular formula but different structural formulae.
• Chain isomerism: structural isomerism in which the isomers have different carbon skeletons, for example a straight chain and a branched chain.
• Positional isomerism: structural isomerism in which the isomers have the same carbon skeleton and the same functional group, but the functional group or substituent is on a different carbon atom.
• Functional group isomerism: structural isomerism in which the isomers have different functional groups, so they belong to different homologous series.
• Stereoisomers: compounds with the same molecular formula and the same structural formula, but a different arrangement of their atoms in space.
• Geometrical (cis/trans) isomerism: stereoisomerism caused by restricted rotation about a C=C bond (or in a ring), when each of the two carbon atoms involved carries two different atoms or groups; groups on the same side are cis, on opposite sides trans.
• Chiral centre: a carbon atom bonded to four different atoms or groups of atoms.
• Optical isomers (enantiomers): two stereoisomers that are mirror images of each other and are non-superimposable: neither can be turned in space to fit exactly onto the other.

Content

Two families of isomers

Family	What differs	Types
Structural	the structural formula	chain, positional, functional group
Stereo	only the arrangement in space	geometrical (cis/trans), optical

Structural isomers have different names. Stereoisomers share a name, apart from a prefix such as cis- or trans-. Two drawings with the same name and no cis/trans or mirror-image difference are the same compound: bending a chain on paper does not make a new isomer.

Structural isomerism

Chain isomerism. The carbon skeleton differs: straight against branched, or branches in different places. Butane, CH₃CH₂CH₂CH₃, and methylpropane, CH₃CH(CH₃)CH₃, are both C₄H₁₀. An open chain needs at least four carbon atoms before it can branch.

Positional isomerism. Same skeleton, same functional group, but the group (or a substituent such as a halogen atom) is on a different carbon. Propan-1-ol, CH₃CH₂CH₂OH, and propan-2-ol, CH₃CH(OH)CH₃, are positional isomers; so are but-1-ene, CH₂=CHCH₂CH₃, and but-2-ene, CH₃CH=CHCH₃. Number from the end that gives the lower number: CH₃CH₂CH₂CH₂Cl is 1-chlorobutane, and "4-chlorobutane" is the same molecule, not another isomer.

Functional group isomerism. The functional groups differ, so the isomers belong to different homologous series and react differently: warm acidified potassium dichromate(VI) oxidises propanal but not propanone.

Formula	Isomer 1	Isomer 2
C₃H₆O	propanal, CH₃CH₂CHO (aldehyde)	propanone, CH₃COCH₃ (ketone)
C₄H₈O₂	butanoic acid, CH₃CH₂CH₂COOH	ethyl ethanoate, CH₃COOCH₂CH₃ (ester)
C₂H₆O	ethanol, CH₃CH₂OH (alcohol)	methoxymethane, CH₃OCH₃ (ether)

You do not need to name ethers (R-O-R′), but a question may show you one.

Ring or C=C? A ring uses up two H atoms, just as a C=C bond does, so propene, CH₂=CHCH₃, and cyclopropane, a ring of three CH₂ groups, are both C₃H₆. Only propene has the C=C functional group, and only cyclopropane has a ring. They are in different homologous series, so they are usually called functional group isomers; if asked to explain, give both differences.

Worked example 1: naming the type of isomerism

Question: State the type of isomerism shown by each pair.

• (a) hexane, CH₃CH₂CH₂CH₂CH₂CH₃, and 2-methylpentane, CH₃CH(CH₃)CH₂CH₂CH₃
• (b) hexan-2-one, CH₃COCH₂CH₂CH₂CH₃, and hexan-3-one, CH₃CH₂COCH₂CH₂CH₃
• (c) propanoic acid, CH₃CH₂COOH, and methyl ethanoate, CH₃COOCH₃
• (d) the two forms of hex-3-ene, CH₃CH₂CH=CHCH₂CH₃, with the ethyl groups on the same side or on opposite sides of the C=C bond
• (e) the two mirror-image forms of pentan-2-ol, CH₃CH(OH)CH₂CH₂CH₃

Ask what differs between the two in each pair.

• (a) A straight chain against a branched one: the skeleton.
• (b) Same chain and same ketone group, but the C=O is on carbon 2 in one and carbon 3 in the other.
• (c) A carboxylic acid against an ester.
• (d) Nothing in the structural formula. Rotation about the C=C is restricted, and each of its carbons carries H and CH₂CH₃.
• (e) Nothing in the structural formula. Carbon 2 is bonded to H, OH, CH₃ and CH₂CH₂CH₃, so it is a chiral centre.

Answer: (a) chain; (b) positional; (c) functional group; (d) geometrical (cis/trans); (e) optical.

Stereoisomerism

Stereoisomers share a structural formula, so you need a three-dimensional picture to tell them apart. Geometrical (cis/trans) isomerism comes from restricted rotation about a C=C bond or within a ring; optical isomerism comes from a chiral centre.

E/Z names (optional). You may see Z (same side) and E (opposite sides) instead of cis and trans; they rank the groups on each carbon of the C=C by set rules. When each carbon of the C=C carries one H atom, Z is cis and E is trans. You do not need E/Z: cis and trans are enough.

Geometrical (cis/trans) isomerism in alkenes

The two carbons of a C=C bond and the four atoms bonded to them lie in one plane, with bond angles of about 120°. In but-2-ene, CH₃CH=CHCH₃, each carbon of the C=C carries one CH₃ and one H. In cis-but-2-ene the two CH₃ groups are on the same side of the C=C bond; in trans-but-2-ene they are on opposite sides, diagonally across it.

An alkene shows cis/trans isomerism only when each carbon of the C=C is bonded to two different atoms or groups. If either carbon carries two identical groups, swapping them changes nothing: but-1-ene, CH₂=CHCH₂CH₃, has no cis/trans isomers, because carbon 1 carries two H atoms.

Why rotation about a C=C bond is restricted. A C=C bond is one σ bond plus one π bond.

• The σ bond comes from end-on overlap along the line between the nuclei. It is symmetrical around that line, so rotation does not weaken it. That is why groups joined by a C-C single bond rotate freely, and CH₂ClCH₂Cl has no cis/trans isomers.
• The π bond comes from sideways overlap of two parallel p orbitals, above and below the plane of the molecule. Twisting one carbon through 90° would leave the p orbitals at right angles with no overlap, so rotation would break the π bond.
• Taking the σ part of C=C to be like a C-C single bond, Data Booklet average bond energies give a rough size for the π bond (as in topic 3.4): $E(\text{C=C}) - E(\text{C-C}) = 610 - 350 = 260$ kJ mol⁻¹. Almost no collisions at room temperature have this much energy, so rotation is restricted.

So cis and trans isomers do not turn into each other on their own at room temperature: they are separate compounds with different physical properties. For example, cis-1,2-dichloroethene, CHCl=CHCl, is polar, while in the trans isomer the C-Cl dipoles cancel, so it is non-polar and boils lower.

Cis/trans isomerism in ring compounds

Ring bonds cannot rotate freely either: turning one ring carbon right round would break the ring.

• Groups on two ring carbons. If two ring carbons each carry two different groups (such as a substituent and an H atom), the substituents can point to the same face of the ring (cis) or to opposite faces (trans). A six-membered ring is not flat, but each ring carbon still has one bond towards each face.
• A C=C bond inside a small ring. In rings of three to six carbon atoms, the ring must continue from both carbons of the C=C on the same side, because a trans arrangement could not close so small a ring. So a C=C inside such a ring gives no cis/trans isomers. A C=C in a side chain is judged like any other alkene.

Worked example 2: spotting cis/trans isomerism

Question: Which of these show geometrical (cis/trans) isomerism? Give a reason for each.

• A: CH₃CH₂CH=CHCl
• B: CH₂=CClCH₂CH₃
• C: (CH₃CH₂)₂C=CHCH₃
• D: HOCH₂CH=CHCH₂OH
• E: 1,2-dimethylcyclopentene: a five-membered carbon ring containing the C=C bond, with a CH₃ group on each carbon of the C=C

List the two groups on each carbon of the C=C, then apply the conditions.

First C=C carbon	Second C=C carbon	Cis/trans?
A: H, Cl	H, CH₂CH₃	yes
B: H, H	Cl, CH₂CH₃	no: two H on one carbon
C: CH₂CH₃, CH₂CH₃	H, CH₃	no: two ethyl groups on one carbon
D: H, CH₂OH	H, CH₂OH	yes
E: CH₃, ring	CH₃, ring	no: C=C inside a small ring

In D, the same pair on both carbons does not matter: the two groups on each carbon differ. E meets that condition, but its ring bonds must be cis, so only one arrangement exists.

Answer: A and D show cis/trans isomerism; B, C and E do not.

Chiral centres and optical isomers

A chiral centre is a carbon atom bonded to four different atoms or groups. Its four bonds point to the corners of a tetrahedron (about 109.5°), so four different groups can be arranged around it in exactly two ways that are mirror images. Like your left and right hands, the two cannot be turned so that one fits exactly on the other: they are non-superimposable. These two forms are optical isomers (enantiomers), one kind of stereoisomer.

• Butan-2-ol, CH₃CH₂C*H(OH)CH₃: carbon 2 is bonded to H, OH, CH₃ and CH₂CH₃, so it is a chiral centre (marked with an asterisk), and butan-2-ol has two optical isomers.
• Propan-2-ol, CH₃CH(OH)CH₃: carbon 2 carries two CH₃ groups. Its mirror image can be turned to fit the original, so it has no optical isomers.

Compare whole groups: CH₃ and CH₂CH₃ differ even though both join through a carbon atom. How the two isomers differ in properties (their effect on plane-polarised light) is A Level work, in topic 29.4.

Drawing a pair of optical isomers. A flat cross with 90° angles does not show which isomer you mean. Instead:

1. Draw the chiral carbon with two plain bonds (in the plane of the paper), one solid wedge (towards you) and one hashed wedge (away from you).
2. Put one group on each bond, joined through the atom that is actually bonded: write HO-C, so that the line meets the O, not the H.
3. Draw a dashed vertical mirror line and then the reflection: each group keeps its type of bond, but left and right swap.

For butan-2-ol, one isomer could have CH₃ up and CH₂CH₃ down on plain bonds, OH on a wedge to the left and H on a hashed bond to the right; its mirror image has the OH wedge on the right and the H hashed bond on the left. With one chiral centre, swapping any two groups on one drawing also gives the other isomer.

More than one chiral centre. Each chiral centre can take either of its two arrangements, so $n$ chiral centres give up to $2^n$ stereoisomers, and each C=C bond that shows cis/trans isomerism doubles the count again. In a symmetrical molecule some arrangements turn out to be identical, so $2^n$ is a maximum.

Finding chiral centres in a structure

• Rule out every carbon in a double or triple bond (fewer than four groups attached) and every CH₃ or CH₂ group (identical H atoms).
• For each carbon left, write out its four groups in full. In a skeletal formula, add the hidden H atoms first: a carbon with three single bonds drawn carries one H atom.
• In a ring, walk round the ring both ways from the carbon. If the two routes meet different groups, or meet the same group after a different number of steps, they count as two different groups.

Worked example 3: chiral centres and the number of stereoisomers

Question: Identify any chiral centres and state the total number of stereoisomers.

• (a) 2-chlorohex-4-en-3-ol, CH₃CHClCH(OH)CH=CHCH₃
• (b) 2-methylcyclohexan-1-ol: a six-membered carbon ring with OH and H on carbon 1, and CH₃ and H on carbon 2, the next carbon
• (c) 4-methylcyclohexan-1-ol: the same ring, but with the CH₃ on carbon 4, directly across the ring from carbon 1

(a) Carbon 2 is bonded to H, Cl, CH₃ and CH(OH)CH=CHCH₃. Carbon 3 is bonded to H, OH, CHClCH₃ and CH=CHCH₃. Both are chiral centres: CH₃C*HClC*H(OH)CH=CHCH₃. The C=C carbons carry H and CH(OH)CHClCH₃, and H and CH₃: two different groups on each, so it shows cis/trans isomerism. Count: $2 \times 2 \times 2 = 2^3 = 8$.

Answer: two chiral centres (carbons 2 and 3); 8 stereoisomers.

(b) Carbon 1's two ring routes differ: one reaches the CH(CH₃) carbon at once, the other passes four CH₂ groups first. So carbon 1 is a chiral centre. Carbon 2 (H, CH₃, the CH(OH) carbon and a CH₂ group) is one too. Count: $2^2 = 4$: a cis pair (OH and CH₃ on the same face, as two mirror-image forms) and a trans pair. Here cis/trans only describes how the two chiral centres are arranged, so do not multiply by 2 again.

Answer: two chiral centres (carbons 1 and 2); 4 stereoisomers.

(c) From carbon 1, both ring routes pass two CH₂ groups and reach carbon 4, so they are identical: carbon 1 is not a chiral centre, and neither is carbon 4. But both carry two different groups and the ring stops rotation, so OH and CH₃ can be on the same face (cis) or on opposite faces (trans).

Answer: no chiral centres; 2 stereoisomers (cis and trans).

Deducing all the isomers of a molecular formula

Work in a fixed order, so that nothing is missed or counted twice.

1. See what the formula allows. Compare it with the alkane $\text{C}_n\text{H}_{2n+2}$, counting each halogen atom as an H atom and ignoring oxygen. Every 2 H atoms short of the alkane means one C=C bond, one C=O bond or one ring.
2. List the carbon skeletons: the longest straight chain first, then one carbon shorter with a branch, and so on. Include rings if the formula allows them and the question does not rule them out.
3. Place the functional group or substituent in every different position on each skeleton. Name each structure: two with the same name are the same isomer.
4. Check every structure for stereoisomers: a C=C bond with two different groups on each carbon, and chiral centres. Each one doubles the number of forms, unless the molecule is symmetrical.
5. Count what was asked. "Structural isomers" stops at step 3; "isomers", with stereoisomerism considered, includes step 4.

Worked example 4: all the isomers of C₃H₅Cl

Question: Deduce and name all the isomers, including stereoisomers, with molecular formula C₃H₅Cl.

Step 1: what the formula allows. Counting Cl as H, C₃H₅Cl is like C₃H₆. Propane is C₃H₈, so 2 H atoms are missing: one C=C bond or one ring.

Step 2: skeletons. Three carbons can form only the chain C=C-C or a three-membered ring.

Step 3: place the Cl atom. On the chain: CHCl=CHCH₃ (1-chloropropene), CH₂=CClCH₃ (2-chloropropene) and CH₂=CHCH₂Cl (3-chloropropene). On the ring every carbon is equivalent, giving one more: chlorocyclopropane. That is 4 structural isomers.

Step 4: stereoisomers. In 1-chloropropene, carbon 1 carries H and Cl and carbon 2 carries H and CH₃, so it has cis and trans forms. In 2-chloropropene and 3-chloropropene, carbon 1 carries two H atoms: no cis/trans. There are no chiral centres: the C=C carbons have only three groups, every CH₃ and CH₂ carbon (including CH₂Cl and the ring CH₂ groups) has identical H atoms, and the CHCl carbon of the ring has two identical ring routes.

Answer: 4 structural isomers and 5 isomers in all: cis-1-chloropropene, trans-1-chloropropene, 2-chloropropene, 3-chloropropene and chlorocyclopropane. The three open-chain compounds are positional isomers of each other; chlorocyclopropane has a ring in place of their C=C bond.

Key Equations

Maximum number of stereoisomers ($n$ = number of chiral centres + number of C=C bonds that show cis/trans isomerism + number of rings that show cis/trans isomerism without chiral centres, as in Worked example 3(c); a count, so no units): $$\text{stereoisomers} \le 2^n$$ Ring cis/trans forms that come from two chiral centres, as in Worked example 3(b), are already counted.

Rough size of the π bond in C=C (Data Booklet average bond energies; assumes the σ part of C=C is like a C-C single bond): $$E(\pi) \approx E(\text{C=C}) - E(\text{C-C}) = 260\ \text{kJ mol}^{-1}$$

Common Mistakes to Avoid

• Mistake: defining stereoisomers as having "the same molecular formula but a different structural formula". Fix: that defines structural isomers. Stereoisomers have the same structural formula and differ only in the arrangement of atoms in space: cis- and trans-but-2-ene are both CH₃CH=CHCH₃.
• Mistake: vague cis/trans reasons such as "groups on either side of the double bond" or "the double bond cannot move". Fix: say that the π bond restricts rotation about the C=C bond and that each carbon of the C=C carries two different groups. For "no cis/trans", name the carbon at fault: in CH₂=CClCH₂CH₃, carbon 1 carries two H atoms.
• Mistake: describing a chiral centre as "surrounded by four different molecules". Fix: write "a carbon atom bonded to four different atoms or groups"; in butan-2-ol these are H, OH, CH₃ and CH₂CH₃.
• Mistake: drawing optical isomers as flat crosses, or drawing the same isomer twice in two orientations. Fix: use plain, wedge and hashed bonds and reflect the first drawing in a mirror line, as for butan-2-ol above.
• Mistake: missing stereoisomers in a count, adding instead of multiplying, or counting ring cis/trans twice. Fix: check every structural isomer for a cis/trans C=C and for chiral centres (Worked example 4 has 5 isomers, not 4), then multiply: 2-chlorohex-4-en-3-ol has 2 × 2 × 2 = 8 stereoisomers, not 2 + 2 + 2 = 6. But 2-methylcyclohexan-1-ol has 4, not 8: its cis/trans forms come from its two chiral centres.
• Mistake: forgetting ring isomers, or giving cis/trans isomers to a C=C inside a ring. Fix: when a formula is 2 H atoms short of the alkane, list rings as well as alkenes (chlorocyclopropane in Worked example 4); a C=C inside a small ring, as in 1,2-dimethylcyclopentene, has no cis/trans isomers.
• Mistake: naming the wrong type of structural isomerism. Fix: check what actually changes. Ethanol and methoxymethane have different functional groups, so they are functional group isomers, not chain isomers: a two-carbon skeleton cannot branch.

Exam Tips

• Explain cis/trans isomerism with two points: rotation about the C=C would break the π bond (sideways overlap of p orbitals), so it is restricted; and each C=C carbon carries two different groups.
• Identify chiral centres by marking each with an asterisk on the given formula. Check every carbon with four single bonds, ring carbons included, after adding the hidden H atoms in a skeletal formula.
• Draw cis/trans isomers with angles of about 120° at the C=C and each group joined to the correct carbon; draw optical isomers in three dimensions on either side of a mirror line.
• Deduce or "how many": work through skeletons, then positions, then stereoisomers, name each structure, and check whether only structural isomers are wanted. When asked for the type, be specific: "positional isomerism", not just "structural isomerism".