Hydrocarbons and IUPAC nomenclature (SACE Stage 2 Chemistry)

Syllabus dot point

“Classify hydrocarbons as alkanes, alkenes and alkynes, and apply IUPAC nomenclature to name straight-chain and branched compounds.”

9 min readUpdated 2 Jun 2026

Quick answer

Hydrocarbons contain only carbon and hydrogen. Alkanes ( $\text{C}_n\text{H}_{2n+2}$ ) are saturated with single bonds; alkenes ( $\text{C}_n\text{H}_{2n}$ ) contain a $\text{C}=\text{C}$ double bond; alkynes ( $\text{C}_n\text{H}_{2n-2}$ ) contain a $\text{C}\equiv\text{C}$ triple bond. IUPAC names are built from the longest carbon chain (the stem), a suffix for the family (-ane, -ene, -yne), and prefixes with locant numbers for branches, choosing numbering that gives the lowest locants.

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What this dot point is asking
Lead worked calculation
Classifying hydrocarbons
The IUPAC naming rules
Worked naming logic
Drawing from a name
Why it matters
Exam-style questions

What this dot point is asking

SACE expects you to classify hydrocarbons, recall the general formulae, and name and draw straight-chain and branched compounds using the IUPAC rules.

Lead worked calculation

Naming a branched alkene step by step

Name the compound $\text{CH}_2=\text{CHCH}_2\text{CH}(\text{CH}_3)\text{CH}_3$ .

Find the longest chain containing the double bond

The longest continuous chain is $5$ carbons, so the stem is pent and, with a double bond, the suffix is -ene.

Number to give the lowest locants

Number from the end nearest the double bond. The double bond starts at $\text{C-1}$ , so it is pent-1-ene. A methyl branch sits on $\text{C-4}$ .

Assemble the name

Combine the branch (with its locant) and the chain: 4-methylpent-1-ene. The double bond gets the lowest possible number, which fixes the direction of numbering before the branch is located.

Classifying hydrocarbons

Saturated and unsaturated

A saturated hydrocarbon (alkane) has only single $\text{C-C}$ bonds, so each carbon holds the maximum number of hydrogen atoms. An unsaturated hydrocarbon contains at least one double ( $\text{C}=\text{C}$ , alkene) or triple ( $\text{C}\equiv\text{C}$ , alkyne) bond, so it has fewer hydrogen atoms and can undergo addition reactions.

Family	Bond	General formula	Example
Alkane	single	$\text{C}_n\text{H}_{2n+2}$	ethane $\text{C}_2\text{H}_6$
Alkene	one $\text{C}=\text{C}$	$\text{C}_n\text{H}_{2n}$	ethene $\text{C}_2\text{H}_4$
Alkyne	one $\text{C}\equiv\text{C}$	$\text{C}_n\text{H}_{2n-2}$	ethyne $\text{C}_2\text{H}_2$

The IUPAC naming rules

Find the longest continuous carbon chain that contains any double or triple bond; this gives the stem (meth-, eth-, prop-, but-, pent-, hex-, ...).
Add the suffix for the family: -ane (alkane), -ene (alkene), -yne (alkyne).
Number the chain from the end that gives the lowest locant to the double/triple bond first, then to branches.
Name and locate branches as prefixes (methyl, ethyl, ...) with their locant numbers, listed alphabetically.
Use di-, tri-, tetra- for repeated identical branches, and separate numbers with commas and numbers from words with hyphens.

Worked naming logic

For $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3$ : the longest chain is $5$ carbons (pentane); a methyl branch is on the second carbon when numbered from the nearer end, giving 2-methylpentane. Always check both numbering directions and keep the one giving lower locants.

Lowest-locant priority order

When numbering a chain, give the lowest locant first to the principal functional group or multiple bond, then to substituents. For a simple branched alkane with no double bond, number from whichever end gives the substituents the lowest set of numbers. This single rule resolves most naming ambiguities.

Drawing from a name

Reverse the process: draw the stem chain, place the multiple bond at its locant, attach branches at their locants, then fill remaining bonds with hydrogen so every carbon has four bonds. Checking that every carbon has exactly four bonds is the quickest way to catch a drawing error.

Not choosing the genuinely longest chain

The classic SACE error is naming from a chain that is not the longest, or numbering from the wrong end. Always trace every path through the molecule to find the true longest chain (it may bend), then number to give the double/triple bond, and then branches, the lowest locants. A second slip is forgetting the locant for the multiple bond (e.g. writing "pentene" instead of "pent-2-ene") and listing branches out of alphabetical order.

Why it matters

Systematic naming gives every organic compound a single unambiguous name worldwide, essential for communicating structures, predicting reactions, and reading the chemistry of fuels, plastics and biological molecules that follow in this topic.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2022-style4 marks

Give the systematic IUPAC name for each compound: (a)

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

; (b)

\text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3

; (c)

\text{CH}_3\text{C(CH}_3)_2\text{CH}_2\text{CH}_3

. Justify the numbering used in (b).

Show worked answer →

(a) Longest chain is $5$ carbons (pentane) with a methyl branch on $\text{C-3}$ : 3-methylpentane. (1 mark)

(b) Longest chain is $5$ carbons with a double bond. Numbering from the end nearest the double bond gives the lowest locant: the double bond is between $\text{C-2}$ and $\text{C-3}$ , so it is pent-2-ene. Numbering from the other end would give pent-3-ene (higher), so the first numbering is correct. (2 marks)

SACE 2020-style4 marks

Propane (

\text{C}_3\text{H}_8

) undergoes complete combustion. (a) Write the balanced equation. (b) Calculate the volume of oxygen at standard laboratory conditions (

V_m = 24.5\ \text{L mol}^{-1}

) needed to completely burn

11.0\ \text{g}

of propane. (

M(\text{C}_3\text{H}_8) = 44.0\ \text{g mol}^{-1}

Show worked answer →

(a) $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$ . (1 mark)

(b) $n(\text{C}_3\text{H}_8) = \dfrac{11.0}{44.0} = 0.250\ \text{mol}$ . From the $1:5$ ratio, $n(\text{O}_2) = 5 \times 0.250 = 1.25\ \text{mol}$ . (2 marks)

$V(\text{O}_2) = nV_m = 1.25 \times 24.5 = 30.6\ \text{L}$ . (1 mark)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board