UNIT 2 CHEMISTRY · PEARSON AAQ · E1.3
IUPAC Naming of Organic Compounds
FREE CHEMISTRY REVISION · TOPIC 52
IUPAC Naming of Organic Compounds
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.3
What you will learn
Construct and interpret systematic names for alkanes, alkenes, halogenoalkanes and alcohols containing up to six carbon atoms in total. Use the structure to choose the parent chain, identify the defining group, number it and locate substituents. Names must recover the original connections; merely counting all carbons is not enough.
1. The components of a name
| Parent carbons | Stem | Unbranched alkane |
|---|---|---|
| 1 | meth- | methane |
| 2 | eth- | ethane |
| 3 | prop- | propane |
| 4 | but- | butane |
| 5 | pent- | pentane |
| 6 | hex- | hexane |
| Family | Feature | Naming pattern | Example |
|---|---|---|---|
| Alkane | Saturated C/H chain | stem + ane | 2-methylpropane |
| Alkene | C=C | stem + position + ene | but-1-ene |
| Halogenoalkane | C–F, C–Cl, C–Br or C–I | locanted halo prefix + alkane parent | 2-bromopropane |
| Alcohol | Alcohol –OH | alkan + position + ol | butan-2-ol |
A locant is a position number. An alkyl substituent is a branch obtained by removing one H from an alkane: CH₃– is methyl and CH₃CH₂– is ethyl. Halogen prefixes are fluoro-, chloro-, bromo- and iodo-. The number identifies the parent carbon to which the branch or halogen is bonded.
2. Choose the parent and the defining feature
- For these simple alkanes and halogenoalkanes, choose the longest continuous carbon chain. Follow bonds through bends; do not jump across a gap or count a branch twice.
- For the alkene examples here, select a parent chain containing the C=C and then the longest suitable chain. The worked molecules contain one C=C.
- For an alcohol, the parent must include the carbon bearing the principal –OH group. Choose the longest suitable chain that includes that carbon.
- The alcohol is the principal characteristic group in the alcohol examples and supplies the -ol suffix; methyl and halogens are prefixes. Do not let a prefix override the OH position.
- If several equally long routes exist, choose a valid route that accounts for the defining group and gives the appropriate substituent numbering. These simple examples avoid advanced tied-parent cases.
3. Number with a reason
Compare numbering from both ends. For the alkene examples, give C=C the lower locant before a methyl branch. For alcohol examples, give the –OH suffix group its lower locant before prefix substituents. For simple substituted alkanes without a suffix group or multiple bond, compare the ordered substituent locants at their first point of difference. Do not add all locants and choose the smaller total as a general rule.
When equivalent numbering gives a complete locant tie, assign the lower locant to the prefix cited first alphabetically where needed, as in 1-bromo-3-chloropropane. List different simple prefixes alphabetically: bromo before chloro before methyl. Ignore multiplicative di- and tri- when alphabetising simple substituent prefixes.
Use commas between numbers and hyphens between a number and a word: 2,3-dimethylbutane. Write one continuous compound name without spaces. Repeated substituents need every attachment locant: 2,2-dimethylbutane has two methyl groups on C2; it is different from 2,3-dimethylbutane. Both contain six C in total.
4. Ten worked naming examples
Worked naming 1 — butane
Given structure: CH₃CH₂CH₂CH₃
Four connected carbons in one unbranched saturated chain: but- + -ane. No position number is needed.
Check: C4H10; all connections must remain as written.
Worked naming 2 — 2-methylpropane
Given structure: CH₃CH(CH₃)CH₃
The longest continuous chain has three C, not four. The remaining CH₃ is a methyl substituent on C2. The whole molecule still has four carbons.
Check: C4H10; all connections must remain as written.
Worked naming 3 — 2-methylpentane
Given structure: CH₃CH(CH₃)CH₂CH₂CH₃
Choose a five-carbon chain. Number from the nearer branch end: methyl gets 2 rather than 4. Six total carbons does not imply a six-carbon parent chain.
Check: C6H14; all connections must remain as written.
Worked naming 4 — 2,3-dimethylbutane
Given structure: CH₃CH(CH₃)CH(CH₃)CH₃
Longest chain: four C. Two methyl substituents attach at C2 and C3. Use di- and both locants, separated by a comma.
Check: C6H14; all connections must remain as written.
Worked naming 5 — but-1-ene
Given structure: CH₂=CHCH₂CH₃
Choose the four-carbon chain containing C=C. Number so the double bond starts at 1 rather than 3. The alkene locant identifies the first carbon of that double bond.
Check: C4H8; all connections must remain as written.
Worked naming 6 — 3-methylpent-1-ene
Given structure: CH₂=CHCH(CH₃)CH₂CH₃
Choose the five-carbon chain including C=C. Give the double bond its lower locant 1, then place methyl at 3. There are six carbons altogether.
Check: C6H12; all connections must remain as written.
Worked naming 7 — 2-bromopropane
Given structure: CH₃CH(Br)CH₃
The parent is propane. Br replaces an H on C2, so add the prefix 2-bromo-. The ending remains -ane.
Check: C3H7Br; all connections must remain as written.
Worked naming 8 — 1-bromo-3-chloropropane
Given structure: BrCH₂CH₂CH₂Cl
Both directions give the locant set 1,3. Break this tie so the alphabetically first prefix bromo has the lower locant. Cite bromo before chloro.
Check: C3H6BrCl; all connections must remain as written.
Worked naming 9 — butan-2-ol
Given structure: CH₃CH(OH)CH₂CH₃
Choose a four-carbon chain containing the carbon bonded to OH. Number to put OH at 2 rather than 3. Replace the final e of butane with -2-ol.
Check: C4H10O; all connections must remain as written.
Worked naming 10 — 3-methylbutan-2-ol
Given structure: CH₃CH(OH)CH(CH₃)CH₃
Choose the four-carbon parent including the OH-bearing carbon. OH has suffix priority: number to give it 2, even though methyl then gets 3 rather than 2.
Check: C5H12O; all connections must remain as written.
5. Recognise halogens and alcohols
In a halogenoalkane, a halogen replaces H and makes one bond; do not add a halogen to an already complete four-bond carbon without removing H. In an alcohol, O makes a C–O bond and an O–H bond. Include that O–H hydrogen in the molecular formula.
The final e is removed before the vowel-starting -ol suffix: butane becomes butan-2-ol. Methanol and ethanol do not need a position number because the OH position is unambiguous. Propan-1-ol and propan-2-ol do need it. For alkenes, ethene and propene have no alternative double-bond positions, whereas but-1-ene and but-2-ene are different compounds.
6. Reverse the process: names to structures
| Name | Construction | Condensed result and check |
|---|---|---|
| hexane | Draw six C in one chain, single bonds; complete valencies | CH₃CH₂CH₂CH₂CH₂CH₃; C₆H₁₄ |
| 2,2-dimethylbutane | Draw four-C parent; put two CH₃ branches on C2 | CH₃C(CH₃)₂CH₂CH₃; C₆H₁₄. C2 has no H. |
| pent-2-ene | Five-C chain; C2=C3; complete H counts | CH₃CH=CHCH₂CH₃; C₅H₁₀ |
| 1-chlorobutane | Four-C chain; one terminal H replaced by Cl | ClCH₂CH₂CH₂CH₃; C₄H₉Cl |
| propan-1-ol | Three-C chain; C1–O–H | CH₃CH₂CH₂OH; C₃H₈O |
| 3-methylbutan-2-ol | Four-C parent; OH on C2 and CH₃ on C3 | CH₃CH(OH)CH(CH₃)CH₃; C₅H₁₂O |
After building a structure from a name, name your completed drawing afresh. If a longer valid parent appears or numbering gives a lower required locant, revisit the construction. An alkene name without E/Z can specify connectivity without specifying one spatial arrangement; that distinction is developed in E1.4.
Practice with answers
| Task | Answer with reason |
|---|---|
| Name CH₃CH₂CH(Cl)CH₃ | 2-chlorobutane: numbering from the right puts Cl at 2 rather than 3. |
| Name CH₃CH₂CH₂CH₂OH | butan-1-ol: start numbering at the OH end. |
| Draw 2-methylbutane | CH₃CH(CH₃)CH₂CH₃; five total C, C₅H₁₂. |
| Why is 4-methylpentane an incorrect choice here? | Number the same chain from the other end: the branch is at 2, giving 2-methylpentane. |
Common misconceptions
- Six total carbons always means hex-: a branch may leave a five- or four-carbon longest parent.
- Number from whichever end is drawn on the left: compare locants using the naming priorities.
- A methyl branch outranks OH: the alcohol suffix receives the lower locant.
- The double-bond locant refers to both carbon numbers: use its lower numbered carbon, e.g. pent-2-ene.
- Omit one repeated branch locant: 2,2- and 2,3- describe different attachments.
- Include di- when alphabetising: ignore simple multiplicative prefixes for that purpose.
- An alcohol has no H on oxygen: include O–H when completing formulae.
Examination tips
- Mark the parent chain and number it before writing the final name.
- State why the numbering is chosen: lower OH, C=C or substituent locants as appropriate.
- Check the complete molecule stays within the six-carbon scope of this lesson.
- Use modern locant placement, for example but-1-ene and butan-2-ol.
- For name-to-structure tasks, draw branches and functional groups before adding H.
- Finish with four bonds at C, two at O and one at H or a halogen.
Check your understanding
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