IB Chemistry HL Topic 3 — Classification of Matter Paper 1 & 2 Core idea ~12 min read

Structural Isomerism

Two compounds can be built from exactly the same atoms and still be completely different substances. One might be a gas, the other a liquid; one might react and the other refuse to. All that changed was the order the atoms were joined up in. Counting those possibilities is a standard exam question, and it is much easier with a system.

📘 What you need to know

The three types

Every structural isomerism question is one of these three, and you can usually tell which by comparing the two structures side by side.

Three ways to rearrange the same atoms In each row the molecular formula on the left equals the one on the right BRANCHED CHAIN both C₄H₁₀ CH₃CH₂CH₂CH₃ vs CH₃CH(CH₃)CH₃ butane, chain of 4 2-methylpropane, chain of 3 POSITIONAL both C₄H₁₀O CH₃CH₂CH₂CH₂OH vs CH₃CH₂CH(OH)CH₃ butan-1-ol, OH on carbon 1 butan-2-ol, OH on carbon 2 FUNCTIONAL GROUP both C₄H₁₀O CH₃CH₂CH₂CH₂OH vs CH₃CH₂OCH₂CH₃ butan-1-ol, an alcohol ethoxyethane, an etherPositional isomers share a class; functional group isomers do not.
The bottom two rows both use C4H10O, which shows the difference clearly. Butan-1-ol and butan-2-ol are both alcohols and behave similarly. Ethoxyethane is an ether and behaves nothing like either of them.

Branched chain isomerism

The longest chain gets shorter and the leftover carbons hang off it as branches. You need at least four carbons before this is possible — with three you can shuffle atoms all you like and still only get propane.

Positional isomerism

Same chain, same functional group, but the group has moved. Butan-1-ol and butan-2-ol are the standard pair. Because the class is unchanged, positional isomers have similar chemical properties but slightly different physical ones.

Functional group isomerism

The atoms reorganise so thoroughly that you end up in a different class. These isomers have very different chemical properties, because the reactive group is not the same. Three pairs of classes are worth memorising:

The alkene and cycloalkane pair catches people out because there is no oxygen to spot. C3H6 can be propene, with a double bond, or cyclopropane, a three-carbon ring. Closing a ring costs you two hydrogens, exactly like making a double bond does.

Primary, secondary and tertiary

Once you can spot positional isomers, you need the vocabulary for describing them. For an alcohol or a halogenoalkane, look at the carbon that carries the group and count how many other carbon atoms are attached to it.

Count the carbons on the carbon that holds the group Not the carbons in the molecule — only the ones touching that one carbon PRIMARY C OH CH₃ H 1 carbon attached e.g. propan-1-ol SECONDARY C OH CH₃ CH₃ 2 carbons attached e.g. propan-2-ol TERTIARY C OH CH₃ CH₃ CH₃ 3 carbons attachedMethanol counts as primary even though it has no other carbon at all. The tertiary example is 2-methylpropan-2-ol, C₄H₁₀O.
This classification decides how the molecule reacts later on. Primary alcohols oxidise all the way to carboxylic acids, secondary alcohols stop at ketones, and tertiary alcohols resist oxidation because there is no hydrogen left on that carbon to remove.
Amines use a different rule. For an amine you count the carbons attached to the nitrogen, not to a carbon. So CH3NH2 is a primary amine because the nitrogen has one carbon on it, and (CH3)3N is tertiary because the nitrogen has three. Mixing this up with the alcohol rule is a classic mistake.

The trap: same molecule, different drawing

Molecules are three-dimensional and free to rotate, so the same compound can be drawn in dozens of ways that look different on paper. Bending a chain through 90 degrees, flipping it left to right, or drawing a branch pointing down instead of up all produce the same molecule.

The test that never fails Name both structures. Same name means the same compound.
Different names means they really are isomers.
This is why the naming page came first. Counting isomers by eye leads to double-counting; counting them by name does not. If you end up with two structures you both want to call 2-methylbutane, you have drawn the same thing twice.

Worked examples

WORKED EXAMPLE

How many structural isomers does C4H10 have?

Step 1: check which types of isomerism are possible No functional group, so no functional group isomerism and nothing to reposition. That leaves branched chain isomerism only. Step 2: start with the longest possible chain chain of 4: CH3CH2CH2CH3 = butane Step 3: shorten the chain by one and branch the spare carbon chain of 3 with a methyl on carbon 2: CH3CH(CH3)CH3 = 2-methylpropane You cannot put the methyl on carbon 1 or 3 — that just makes butane again, which the name check confirms. Step 4: shorten again A chain of 2 cannot hold two extra carbons as separate branches without becoming a longer chain, so we are finished. 2 isomers: butane and 2-methylpropane Four carbons is the smallest alkane that has any isomers at all. C1, C2 and C3 have exactly one each.
WORKED EXAMPLE

Deduce all the structural isomers of C4H9Cl and classify each one as primary, secondary or tertiary.

Step 1: sort out the possible carbon skeletons first Four carbons gives two skeletons: a straight chain of 4, and a chain of 3 with a methyl branch. Do the chlorine positions on each skeleton in turn. Step 2: put the Cl on the straight chain Cl on carbon 1: 1-chlorobutane, primary Cl on carbon 2: 2-chlorobutane, secondary Carbon 3 is the same as carbon 2 counted from the other end, and carbon 4 is the same as carbon 1. Both would repeat a name we already have. Step 3: put the Cl on the branched skeleton Cl on an end CH3: 1-chloro-2-methylpropane, primary Cl on the central carbon: 2-chloro-2-methylpropane, tertiary Step 4: count them 4 isomers — two primary, one secondary, one tertiary Working skeleton by skeleton stops you missing any. Doing it at random almost always loses one of the branched pair.
WORKED EXAMPLE

Name three compounds with the molecular formula C3H6O and state the type of isomerism between them.

Step 1: see which general formula it fits C3H6O fits CnH2nO, so aldehydes and ketones are both in play Step 2: write those two CH3CH2CHO = propanal, an aldehyde CH3COCH3 = propanone, a ketone Step 3: look for a third option A C=C plus an OH also uses up two hydrogens, and that fits too. CH2=CHCH2OH = prop-2-en-1-ol, an unsaturated alcohol propanal, propanone and prop-2-en-1-ol — functional group isomers Three different classes from one formula. Their chemical properties are completely different, which is the hallmark of functional group isomerism.
WORKED EXAMPLE

Classify each of these as primary, secondary or tertiary: butan-2-ol, 2-methylpropan-2-ol, and the amine (CH3)2NH.

Step 1: butan-2-ol — find the carbon with the OH That is carbon 2. It has carbon 1 on one side and carbon 3 on the other. 2 carbons attached, so secondary Step 2: 2-methylpropan-2-ol — same question Carbon 2 carries the OH, and it is joined to carbon 1, carbon 3 and the methyl branch. 3 carbons attached, so tertiary Step 3: the amine — switch rules and count on the nitrogen The nitrogen has two methyl groups and one hydrogen. 2 carbons on the N, so secondary secondary, tertiary, secondary Note that (CH3)2NH has no branched carbon at all, yet it is still a secondary amine. The rule really does move to the nitrogen.

Comparing the three types at a glance

TypeWhat changesChemical propertiesExample pair
branched chainlength of the longest chainsimilarpentane and 2-methylbutane
positionalwhich carbon carries the groupsimilarpropan-1-ol and propan-2-ol
functional groupthe class of compoundvery differentpropanal and propanone

💡 Exam tip

⚠ Common mix-up

Up next: Cis-Trans Isomers — so far the atoms have been joined in a different order. Next we look at isomers where the connections are identical and only the shape in space differs.

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