How do we name organic molecules and account for their isomers?
Apply IUPAC nomenclature and identify structural and stereo isomers of organic compounds.
Systematic IUPAC naming of organic compounds, structural isomers, chain branching, and cis-trans (geometric) isomerism.
Reviewed by: AI editorial process; not yet individually human-reviewed
Have a quick question? Jump to the Q&A page
What this dot point is asking
You must name organic compounds systematically and identify their isomers.
IUPAC naming rules
Build the name in three parts.
- Stem: the number of carbons in the longest chain that contains the main functional group (meth, eth, prop, but, pent, hex and so on).
- Suffix: the main functional group (-ane, -ene, -ol, -al, -one, -oic acid, -oate, -amine, -amide).
- Prefixes: branches and other substituents, each with a locant (number) showing its position.
Number the chain from the end that gives the main functional group the lowest locant. List substituents alphabetically, and use di, tri and tetra for repeated groups.
Structural isomerism
For example, exists as butane (straight chain) and 2-methylpropane (branched). Chain branching lowers the boiling point because branched molecules pack less closely, so dispersion forces are weaker.
Cis-trans (geometric) isomerism
A carbon-carbon double bond cannot rotate, so groups attached to it are fixed in position. When each double-bond carbon carries two different groups, two arrangements exist: cis (matching groups on the same side) and trans (on opposite sides). These geometric isomers can have different physical properties such as melting point.
In the exam, identify and number the longest chain through the main functional group, name substituents alphabetically with locants, and check for both structural and geometric isomers when asked for isomers of a formula.
Exam-style practice questions
Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2024 TASC4 marksWrite an equation for the synthesis of chloropropane from propene gas if just one molecule of chlorine gas is involved, and identify the main two isomers of chloropropane that would be produced.Show worked answer →
Propene (CH3CH=CH2) undergoes addition with HCl, or here addition of a single chlorine across the double bond. For the addition of HCl the equation is CH3CH=CH2 + HCl -> C3H7Cl. (If adding Cl2 across the double bond the product would be 1,2-dichloropropane; the two monochloro isomers below arise from addition of H-Cl following Markovnikov and anti-Markovnikov outcomes.) (2 marks)
The two structural isomers of chloropropane are: 1-chloropropane, CH3CH2CH2Cl (chlorine on an end carbon), and 2-chloropropane, CH3CHClCH3 (chlorine on the middle carbon). They have the same molecular formula C3H7Cl but differ in the position of the chlorine, so they are positional structural isomers. (2 marks)
2022 TASC4 marksTwo isomers of C4H8, Isomer A and Isomer B, both immediately decolourise bromine. When reacted with steam (H+ catalyst), Isomer A gives two products that on oxidation give an aldehyde and a ketone respectively, while Isomer B gives a single product (Compound C). Draw and name Isomer A, and explain what the bromine and steam results reveal.Show worked answer →
Both isomers decolourise bromine immediately, which shows both contain a carbon-carbon double bond (they are alkenes, undergoing addition). (1 mark)
Isomer A gives two different products with steam because the double bond is not symmetrical, so water can add either way (Markovnikov and anti-Markovnikov), giving a primary alcohol (oxidised to an aldehyde) and a secondary alcohol (oxidised to a ketone). This identifies Isomer A as but-1-ene, CH2=CHCH2CH3. (2 marks)
Name and structure of Isomer A: but-1-ene, CH2=CH-CH2-CH3. Isomer B gives only one hydration product, consistent with the symmetrical but-2-ene (CH3CH=CHCH3), where addition either way gives the same secondary alcohol (butan-2-ol). (1 mark)