Organic structure and nomenclature: WACE Year 12 Chemistry

Syllabus dot point

“Identify functional groups and apply IUPAC rules to name and draw structural formulas for the main families of organic compounds”

8 min readUpdated 2 Jun 2026

Quick answer

Organic compounds are grouped into homologous series by their functional group (alkane, alkene, haloalkane, alcohol, aldehyde, ketone, carboxylic acid, ester, amine). IUPAC naming builds a name from the longest carbon chain that contains the main functional group, a suffix for that group, and numbered prefixes for branches and substituents.

What this dot point is asking

Organic chemistry is the chemistry of carbon compounds. Carbon forms four covalent bonds and chains, branches and rings, so the number of possible structures is enormous. To bring order, compounds are grouped into homologous series: families with the same functional group, the same general formula, and a $\text{CH}_2$ difference between successive members, showing a gradual trend in physical properties.

The functional groups you must know

Series	Functional group	Suffix or class	Example
Alkane	C-C single bonds	-ane	propane
Alkene	C=C double bond	-ene	propene
Haloalkane	C-X (F, Cl, Br, I)	halo- prefix	2-chloropropane
Alcohol	C-OH	-ol	propan-1-ol
Aldehyde	terminal -CHO	-al	propanal
Ketone	internal C=O	-one	propan-2-one
Carboxylic acid	-COOH	-oic acid	propanoic acid
Ester	-COO-	-yl ...-oate	methyl propanoate
Amine	C-NH2	-amine	propan-1-amine

Functional group

A specific atom or group of atoms within a molecule that gives the molecule its characteristic chemical reactions. Members of the same homologous series share a functional group and react similarly.

IUPAC naming rules

A systematic name is built in a fixed order.

Identify the principal functional group and choose the longest continuous carbon chain that contains it. This chain gives the stem (meth-, eth-, prop-, but-, pent-, hex-) and the suffix.
Number the chain so the principal functional group (or, for alkanes, the first branch point) gets the lowest possible locant.
Name substituents (branches like methyl, ethyl; halogens like chloro, bromo) as prefixes, each with its locant.
Order the prefixes alphabetically, use di-, tri-, tetra- for repeats (ignored when alphabetising), put hyphens between numbers and letters and commas between numbers.

Worked names

$\text{CH}_3\text{-CH(OH)-CH}_2\text{-CH}_3$: Principal group is -OH (alcohol, suffix -ol). Longest chain containing it is four carbons (butan-). Number for the lowest locant: the OH is on C2. Name: butan-2-ol.
$(\text{CH}_3)_2\text{CH-CH}_2\text{-COOH}$: Principal group is -COOH (carboxylic acid), so its carbon is C1. The longest chain containing it has four carbons (butanoic acid) with a methyl on C3. Name: 3-methylbutanoic acid.
$\text{CH}_3\text{-CH}_2\text{-COO-CH}_3$: This is an ester. The alkyl group from the alcohol comes first (methyl), the acyl part from the acid takes the -oate suffix (propanoate). Name: methyl propanoate.

Naming esters

An ester is named alcohol-part first as the alkyl group, then the acid-part as the -oate. Methyl propanoate comes from methanol and propanoic acid, not the other way round.

Representing structures

You should be able to draw structures three ways: a full structural formula (every bond shown), a condensed structural formula (e.g. $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ , branches in brackets), and a skeletal formula (carbons at line vertices and ends, carbon-hydrogens implied, heteroatoms drawn). The examination usually accepts any clear unambiguous structure unless one type is specified.

Naming a branched alcohol

Question

Give the IUPAC name of $\text{CH}_3\text{-CH}_2\text{-CH(CH}_3\text{)-CH}_2\text{-OH}$ .

Solution

The principal group is -OH, so the chain must contain it and the suffix is -ol. The longest chain containing the carbon bearing OH is four carbons (butan-). Number from the OH end so it gets the lowest locant: OH on C1, methyl on C2.

Name: 2-methylbutan-1-ol.

When answering nomenclature questions in the WACE examination, always pick the longest chain that includes the main functional group (not simply the longest chain), and number for the lowest locant on the principal group.

Numbering from the wrong end

A common error is numbering the chain to give a branch the lowest locant when the rule is to give the principal functional group the lowest locant first. Another frequent slip is choosing the longest chain that does not contain the functional group. A third is writing ester names backwards, swapping the alcohol-derived alkyl group with the acid-derived -oate part.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style6 marks

Give the IUPAC name for each of the following, showing how you applied the naming rules: (a)

(\text{CH}_3)_2\text{CHCH}_2\text{CH}_2\text{OH}

; (b)

\text{CH}_3\text{CH}_2\text{CHClCOOH}

; (c) the ester formed from butanoic acid and ethanol.

Show worked answer →

A 6 mark question rewards each name with correct chain choice and numbering.

(a) The principal group is $\text{-OH}$ (suffix $\text{-ol}$ ). The longest chain containing the $\text{OH}$ carbon is four carbons (butan-), with a methyl branch. Numbering from the $\text{OH}$ end gives $\text{OH}$ on C1 and methyl on C3: 3-methylbutan-1-ol.

(b) The principal group is $\text{-COOH}$ , so its carbon is C1 (suffix $\text{-oic acid}$ ). The chain is four carbons (butanoic acid) with a chloro substituent on C2: 2-chlorobutanoic acid.

(c) An ester is named alkyl (from the alcohol) then $\text{-oate}$ (from the acid). Ethanol gives ethyl and butanoic acid gives butanoate: ethyl butanoate.

Markers reward the correct longest chain containing the principal group, lowest-locant numbering, and the alcohol-first ester naming convention.

WACE 2023-style5 marks

A student names a compound 3-methylbutan-3-ol. (a) Explain why this name is incorrect. (b) Draw the structure the student probably intended and give its correct IUPAC name. (c) State the general rule about numbering that the student broke.

Show worked answer →

A 5 mark answer needs the error, the corrected name, and the rule.

(a) The name is incorrect because the chain has not been numbered to give the principal functional group ( $\text{-OH}$ ) the lowest possible locant. On a four-carbon chain, numbering from the other end places the $\text{OH}$ on C2, not C3, so "butan-3-ol" should be "butan-2-ol".

(b) The intended structure is $(\text{CH}_3)_2\text{C(OH)CH}_2\text{CH}_3$ , with the $\text{OH}$ and a methyl on the same carbon of a butane chain. Numbering for the lowest $\text{OH}$ locant gives the $\text{OH}$ and methyl on C2: the correct name is 2-methylbutan-2-ol.

(c) The rule broken is: number the parent chain so that the principal functional group receives the lowest possible locant (the functional group takes priority over branches when choosing the direction of numbering).

Markers reward identifying the high-locant error, the corrected name 2-methylbutan-2-ol, and the lowest-locant-for-the-principal-group rule.

Practise this

Sources & how we know this

Chemistry ATAR course: syllabus and support materials — SCSA
Chemistry: past ATAR course examinations — SCSA