SACE Board SACE 2022-style practice: Give the systematic IUPAC name for each compound: (a) CH_3CH_2CH(CH_3)CH_2CH_3; (b) CH_3CH=CHCH_2CH_3; (c) CH_3C(CH_3)_2CH_2CH_3. Justify the numbering used in (b).

(a) Longest chain is 5 carbons (pentane) with a methyl branch on 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 C-2 and 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) (c) Longest chain is 4 carbons (butane) with two methyl groups on C-2: **2,2-dimethylbutane**. (1 mark)

§-Syllabus dot point

SAChemistrySyllabus dot point

How are hydrocarbons classified and named systematically?

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

Classifying alkanes, alkenes and alkynes, applying IUPAC rules to name and draw straight-chain and branched hydrocarbons, general formulae, and worked SACE-style naming and combustion-stoichiometry examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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

Classifying hydrocarbons

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.

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.

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 practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20224 marksGive 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 20204 marksPropane (

\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)