What are the basic rules for naming organic compounds according to IUPAC nomenclature?

The IUPAC nomenclature for naming organic compounds involves identifying the longest carbon chain as the parent hydrocarbon, numbering the chain to give the lowest numbers to substituents, and naming substituents as prefixes. Functional groups are identified and named according to their priority, and multiple identical groups are indicated with prefixes like di-, tri-, etc. This systematic approach ensures consistency and clarity in naming organic compounds, which is crucial for Cambridge IGCSE Chemistry students.

How do you name alkanes in organic chemistry?

Alkanes are named based on the number of carbon atoms in the longest continuous chain. The names end with the suffix '-ane'. For example, methane (CH4) has one carbon, ethane (C2H6) has two carbons, and so on. If there are branches or substituents, they are named as prefixes with their position indicated by numbers. This foundational knowledge is essential for students studying organic chemistry at the Cambridge IGCSE level.

What is the significance of functional groups in naming organic compounds?

Functional groups are specific groups of atoms within molecules that determine the chemical properties of those molecules. In IUPAC nomenclature, functional groups are prioritized and often determine the suffix or prefix of the compound's name. For example, alcohols have the suffix '-ol', and carboxylic acids have the suffix '-oic acid'. Understanding functional groups is crucial for Cambridge IGCSE Chemistry students as it affects both the naming and reactivity of organic compounds.

How do you differentiate between isomers when naming organic compounds?

Isomers are compounds with the same molecular formula but different structural arrangements. In naming isomers, the structure is analyzed to determine the longest carbon chain, the position of substituents, and the type of functional groups. The name reflects these differences, such as using numbers to indicate the position of branches or functional groups. This precise naming is important for students learning organic chemistry in the Cambridge IGCSE curriculum to distinguish between different isomers.

Why is it important to learn the IUPAC naming system in organic chemistry?

The IUPAC naming system provides a standardized method for naming organic compounds, ensuring clear communication and understanding among chemists worldwide. For Cambridge IGCSE Chemistry students, mastering this system is vital as it forms the basis for identifying and discussing organic compounds in both academic and professional settings. It also aids in understanding the structure and properties of compounds, which is essential for further studies in chemistry.

Why is "Mixing up the stems (e.g. saying 'but' means 5 carbons)." a common mistake in Naming Organic Compounds?

Why it happens: Off-by-one slip when reciting the order from memory. How to avoid it: Learn the order in sequence: meth (1), eth (2), prop (3), but (4). Source: 0620/42 — recurring

Why is "Naming an alcohol with the '-ane' ending (e.g. 'propane' for $-\text{OH}$)." a common mistake in Naming Organic Compounds?

Why it happens: Forgetting to change the suffix when a functional group is present. How to avoid it: If -OH is present the suffix becomes '-ol' (propan-1-ol); if -COOH is present it becomes '-oic acid' (propanoic acid).

Why is "Forgetting the position number when the functional group can sit on more than one carbon." a common mistake in Naming Organic Compounds?

Why it happens: Rushing and giving only the stem + suffix. How to avoid it: Add a number whenever the group could be in more than one place — propan-1-ol vs propan-2-ol; but-1-ene vs but-2-ene.

Why is "Not counting the carbon inside $-\text{COOH}$ as part of the chain." a common mistake in Naming Organic Compounds?

Why it happens: The carboxyl carbon is easy to overlook when counting. How to avoid it: Always include the -COOH carbon: CH_3COOH has 2 carbons → ethanoic acid, not methanoic acid.

Cambridge IGCSE Chemistry (0620)

Naming Organic Compounds

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Short Notes - Naming Organic Compounds

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Detailed Study Notes

Detailed notes on Organic Chemistry for Cambridge IGCSE Chemistry, covering key concepts, explanations, examples, and exam-focused revision points.

Naming Organic Compounds — Cambridge IGCSE 0620 Chemistry Extended (2026)

IUPAC nomenclature for hydrocarbons and their derivatives. Stem (number of carbons) + suffix (functional group). Core skill for the rest of organic chemistry.

At a glance

Stem based on number of carbons: meth (1), eth (2), prop (3), but (4), pent (5), hex (6).
Suffix based on functional group: -ane (alkane), -ene (alkene), -ol (alcohol), -oic acid (carboxylic acid).
Position number added when needed: e.g. but-1-ene vs but-2-ene.
Lowest possible locants — number from the end nearest the functional group.
Branched: name longest chain as parent; substituents named as prefixes (methyl, ethyl).

What you’ll learn

Mapped to the Cambridge IGCSE 0620 syllabus (2026-2028).

13.1 — Recognise and use IUPAC names for the first six members of homologous series.
13.1 — Identify the functional group from a name or structure.

Carbon-count stems

Memorise meth, eth, prop, but, pent, hex for 1, 2, 3, 4, 5, 6 carbons.

Stems by number of carbons.

Carbons	Stem
1	meth-
2	eth-
3	prop-
4	but-
5	pent-
6	hex-

Cambridge tip. Cambridge IGCSE only typically tests up to 6 carbons. Beyond: hept (7), oct (8), non (9), dec (10) — useful but rarely needed.

Worked qualitative. A compound has 4 carbons. Stem: but-.

1 → meth, 2 → eth, 3 → prop, 4 → but.
5 → pent, 6 → hex.
Memorise: My Easy Pet Bunny Penny Hops.

Suffixes for functional groups

-ane: alkane. -ene: alkene. -ol: alcohol. -oic acid: carboxylic acid.

Common suffixes.

Family	Suffix	Functional group
Alkanes	-ane	C-C single bonds only
Alkenes	-ene	C=C double bond
Alcohols	-ol	-OH
Carboxylic acids	-oic acid	-COOH

Examples.

methANE: CH₄ — 1 C, alkane.
ethENE: C₂H₄ — 2 C, alkene.
propAN-1-OL: CH₃CH₂CH₂OH — 3 C, alcohol on C1.
ethanOIC ACID: CH₃COOH — 2 C, carboxylic acid.

Worked. Name: $\text{CH}_{3}\text{CH}_{2}\text{CH}_{3}$ .

3 carbons: prop-.
All single bonds: -ane.
Name: propane.

Worked. Name: $\text{CH}_{3}\text{CH}=\text{CHCH}_{3}$ .

4 carbons: but-.
Double bond: -ene.
Position of double bond: between C2 and C3.
Name: but-2-ene.

Every IUPAC name = stem (carbon count) + locant (position) + suffix (functional group).

-ane: single bonds.
-ene: double bond present.
-ol: OH group.
-oic acid: COOH group.
Combine stem + suffix.

Position numbers — keep it lowest

Number carbons from the end nearest the functional group, giving the lowest possible position number.

Why position numbers? Some structures have multiple possible places for the functional group. Numbers tell you exactly where.

Worked. $\text{CH}_{2}=\text{CHCH}_{2}\text{CH}_{3}$ — 4 carbons, double bond on the leftmost.

Number from LEFT: 1-2 double bond → "but-1-ene".
Number from RIGHT: 3-4 double bond → "but-3-ene".
Choose the lowest position number: but-1-ene.

Number from whichever end gives the lower locant — here the left, so it is but-1-ene.

Worked. $\text{CH}_{3}\text{CH(OH)CH}_{3}$ — propan-2-ol (OH on the middle carbon).

Numbering doesn't matter here as left and right give the same answer (2).

Tip. When more than one functional group / substituent is present, number to give the LOWEST SUM of locants.

Cambridge tip. Most IGCSE questions are simple enough that lowest-locant rule applies straightforwardly. Branched-chain naming is rare at Cambridge IGCSE — but present in other syllabuses, so worth knowing the principle.

Number from end nearest the functional group.
Lowest position number wins.
Helps differentiate but-1-ene from but-2-ene, etc.

Quick recap

Stems: meth (1), eth (2), prop (3), but (4), pent (5), hex (6).
Suffixes: -ane, -ene, -ol, -oic acid.
Combine: stem + suffix + position number when needed.
Number from end nearest functional group.
Lowest locant rule.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

IUPAC nomenclature — systematic naming of organic compounds.
Functional group — atom or group of atoms responsible for the characteristic chemistry of a class.
Locant — position number indicating where a substituent or functional group is on the carbon chain.

How it’s examined

Naming appears every Paper 2 (2-3 marks: name a structure, draw a structure from a name) and most Paper 4s (3-4 marks). Examiner reports flag students missing the position number (e.g. writing "butene" instead of "but-1-ene" when the position matters).

Sources: Cambridge IGCSE Chemistry 0620 syllabus 2026-2028 (13.1); 0620/22 May/Jun 2024 — Q19 (organic naming); 0620 Examiner Reports 2022-2024. Last reviewed 2026-05-07.

Take this whole topic with you

Step-by-step worked examples — Naming Organic Compounds

Step-by-step solutions to past-paper-style questions on naming organic compounds, written exactly the way a tutor would explain them at the board.

1How a name is built (stem + suffix)
Getting started• stem, suffix
Question
Explain how the name 'propan-1-ol' tells you the structure of the compound. Break the name into its parts.
Step-by-step solution
1. Step 1
  The stem 'prop' tells you the longest carbon chain has 3 carbon atoms (meth- 1, eth- 2, prop- 3, but- 4).
2. Step 2
  The suffix '-ol' tells you the functional group is an alcohol ( $-\text{OH}$ ), so it belongs to the alcohol homologous series.
3. Step 3
  The position number '1' tells you the $-\text{OH}$ is on carbon 1 (the end carbon).
4. Step 4
  Putting it together gives $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{OH}$ .
Answer
prop = 3 carbons; -ol = alcohol ( $-\text{OH}$ ); 1 = $-\text{OH}$ on carbon 1. Structure: $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{OH}$ .
Examiner tip
Always read a name in three parts: stem (how many C), suffix (which family), number (where the group is).
2Stems for 1-4 carbons
Getting started• stem
Question
Give the stem used for organic compounds with 1, 2, 3 and 4 carbon atoms in the longest chain.
Step-by-step solution
1. Step 1
  1 carbon → meth.
2. Step 2
  2 carbons → eth.
3. Step 3
  3 carbons → prop.
4. Step 4
  4 carbons → but.
Answer
1: meth, 2: eth, 3: prop, 4: but.
Examiner tip
On the 0620 syllabus you only need straight chains up to 4 carbons. Learn the order 'meth-eth-prop-but'.
3Name a carboxylic acid from its formula
Building confidence• Adapted from 0620/42 May/Jun 2024 Q14• name, carboxylic acid
Question
Name the compound $\text{CH}_{3}\text{CH}_{2}\text{COOH}$ .
Step-by-step solution
1. Step 1
  Count the carbons in the longest chain — including the carbon in $-\text{COOH}$ . There are 3 carbons → stem 'prop'.
2. Step 2
  The $-\text{COOH}$ group is the carboxylic acid functional group → suffix '-oic acid'.
3. Step 3
  Combine the stem and suffix: prop + anoic acid → propanoic acid. No position number is needed because $-\text{COOH}$ is always on an end carbon (carbon 1).
Answer
Propanoic acid.
Examiner tip
Remember to count the carbon inside $-\text{COOH}$ as part of the chain — a common slip is calling this 'ethanoic acid'.
4Draw a structure from a name
Building confidence• structure, displayed formula
Question
Give the structural formula of but-2-ene and explain what each part of the name tells you.
Step-by-step solution
1. Step 1
  'but' → 4 carbons in the chain.
2. Step 2
  '-ene' → an alkene, so there is one $\text{C}=\text{C}$ double bond.
3. Step 3
  '2' → the double bond starts on carbon 2 (between C2 and C3).
4. Step 4
  Build the chain C1–C2=C3–C4 and add hydrogens: $\text{CH}_{3}\text{CH}=\text{CHCH}_{3}$ .
Answer
But-2-ene = $\text{CH}_{3}\text{CH}=\text{CHCH}_{3}$ (4 carbons, C=C between carbons 2 and 3).
Examiner tip
For but-1-ene the double bond is on carbon 1 ( $\text{CH}_{2}=\text{CHCH}_{2}\text{CH}_{3}$ ) — these two are structural isomers.
5Structural isomers of C₄H₁₀
Stretch• isomers, structure
Question
Draw and name the two structural isomers with the molecular formula $\text{C}_{4}\text{H}_{10}$ , and state what makes them isomers.
Step-by-step solution
1. Step 1
  Isomer 1 — a straight chain of 4 carbons: $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH}_{3}$ = butane.
2. Step 2
  Isomer 2 — a branched chain: a 3-carbon chain with a $\text{CH}_{3}$ branch on the middle carbon, $\text{CH}_{3}\text{CH(CH}_{3}\text{)CH}_{3}$ = methylpropane (2-methylpropane).
3. Step 3
  Both have the same molecular formula $\text{C}_{4}\text{H}_{10}$ but a different arrangement of atoms (one straight, one branched).
4. Step 4
  Same molecular formula + different structural formula = structural isomers.
Answer
Butane $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH}_{3}$ and methylpropane $\text{CH}_{3}\text{CH(CH}_{3}\text{)CH}_{3}$ — both $\text{C}_{4}\text{H}_{10}$ , so they are structural isomers.
Examiner tip
Structural isomers must have the SAME molecular formula. Check by counting atoms in each: both come to 4 C and 10 H.
6Identify the homologous series
Stretch• Paper 4 (Extended) structured style• homologous series, classify
Question
For each compound, state which homologous series it belongs to and justify your answer: (a) $\text{C}_{2}\text{H}_{4}$ , (b) $\text{CH}_{3}\text{OH}$ , (c) $\text{HCOOH}$ , (d) $\text{C}_{2}\text{H}_{6}$ .
Step-by-step solution
1. Step 1
  (a) $\text{C}_{2}\text{H}_{4}$ fits $\text{C}_{n}\text{H}_{2n}$ and has a C=C → alkene (ethene).
2. Step 2
  (b) $\text{CH}_{3}\text{OH}$ contains an $-\text{OH}$ group → alcohol (methanol).
3. Step 3
  (c) $\text{HCOOH}$ contains a $-\text{COOH}$ group → carboxylic acid (methanoic acid).
4. Step 4
  (d) $\text{C}_{2}\text{H}_{6}$ fits $\text{C}_{n}\text{H}_{2n+2}$ with single bonds only → alkane (ethane).
Answer
(a) alkene, (b) alcohol, (c) carboxylic acid, (d) alkane — identified from the functional group / general formula.
Examiner tip
Spot the functional group first: C=C → alkene; $-\text{OH}$ → alcohol; $-\text{COOH}$ → carboxylic acid; none → alkane.

Model Answers — Naming Organic Compounds

High-scoring sample answers for naming organic compounds on the Cambridge IGCSE 0620 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 4 short-answer style1 mark
State the number of carbon atoms in the longest chain of a compound whose name begins with the stem 'but-'. (1 mark)
Model answer
Four (4) carbon atoms.
Why this scores
One mark for '4'. Stems: meth- 1, eth- 2, prop- 3, but- 4.
Question 2
Paper 4 short-answer style2 marks
State the homologous series indicated by each of these suffixes: (i) -ene and (ii) -oic acid. (2 marks)
Model answer
(i) '-ene' indicates an alkene (a compound with a $\text{C}=\text{C}$ double bond). (ii) '-oic acid' indicates a carboxylic acid (a compound with the $-\text{COOH}$ group).
Why this scores
One mark each. The suffix always names the functional group / homologous series.
Question 3
Paper 4 structured style3 marks
Name the compound $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{OH}$ and explain how the name is built up. (3 marks)
Model answer
The compound is propan-1-ol. There are three carbon atoms in the chain, so the stem is 'prop'. It contains an $-\text{OH}$ group, so it is an alcohol and takes the suffix '-ol'. The $-\text{OH}$ is on carbon 1, giving the position number, so the full name is propan-1-ol.
Why this scores
Three marks: (1) correct name propan-1-ol; (2) stem = 3 carbons; (3) -ol = alcohol / $-\text{OH}$ group. The position number secures the full name.
Question 4
Paper 4 structured style4 marks
Define the term structural isomers and give one example, naming both compounds. (4 marks)
Model answer
Structural isomers are compounds that have the same molecular formula but different structural formulae (a different arrangement of their atoms). For example, butane ( $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH}_{3}$ ) and methylpropane ( $\text{CH}_{3}\text{CH(CH}_{3}\text{)CH}_{3}$ ) both have the molecular formula $\text{C}_{4}\text{H}_{10}$ , but one is a straight chain and the other is branched.
Why this scores
Four marks: (1) same molecular formula; (2) different structural formula / arrangement; (3) a valid pair with the same molecular formula; (4) correctly named (butane and methylpropane).
Question 5
Paper 4 (Extended) structured style5 marks
A compound has the structural formula $\text{CH}_{3}\text{COOH}$ . Name the compound, state its homologous series, and give its general functional group. (5 marks)
Model answer
The compound is ethanoic acid. Counting the carbons (including the one in $-\text{COOH}$ ) gives two carbons, so the stem is 'eth'. The $-\text{COOH}$ group means it is a carboxylic acid, taking the suffix '-oic acid', so the name is ethanoic acid. It belongs to the carboxylic acid homologous series, and its functional group is the carboxyl group, $-\text{COOH}$ .
Why this scores
Five marks: (1) name ethanoic acid; (2) 2 carbons / stem 'eth'; (3) -oic acid suffix; (4) carboxylic acid series; (5) functional group $-\text{COOH}$ . Counting the carboxyl carbon is the key step.
Question 6
Paper 4 (Extended) structured style6 marks
But-1-ene and but-2-ene both have the molecular formula $\text{C}_{4}\text{H}_{8}$ . Draw the structural formula of each, explain why they are structural isomers, and explain how their names show the difference between them. (6 marks)
Model answer
But-1-ene has the structure $\text{CH}_{2}=\text{CHCH}_{2}\text{CH}_{3}$ , with the $\text{C}=\text{C}$ double bond between carbons 1 and 2. But-2-ene has the structure $\text{CH}_{3}\text{CH}=\text{CHCH}_{3}$ , with the double bond between carbons 2 and 3. Both compounds have the same molecular formula, $\text{C}_{4}\text{H}_{8}$ , but a different structural formula because the double bond is in a different position, so they are structural isomers. The stem 'but' shows both have four carbons and the suffix '-ene' shows both are alkenes; the difference is shown by the position number — '1' means the double bond starts on carbon 1, while '2' means it starts on carbon 2.
Why this scores
Six marks: (1) correct structure of but-1-ene; (2) correct structure of but-2-ene; (3) same molecular formula $\text{C}_{4}\text{H}_{8}$ ; (4) different position of the C=C / different structural formula; (5) therefore structural isomers; (6) position number distinguishes the names.

Key Definitions and Keywords — Naming Organic Compounds

Definitions to memorise and the exact keywords mark schemes credit for naming organic compounds answers — sharpened from recent examiner reports for the 2026 0620 sitting.

Nomenclature (IUPAC name)
Examiner keyword
The systematic rules for naming organic compounds: a STEM gives the number of carbon atoms in the longest chain and a SUFFIX gives the homologous series / functional group.
Homologous series
Examiner keyword
A family of organic compounds with the same general formula and functional group, differing by $\text{CH}_{2}$ — e.g. alkanes (-ane), alkenes (-ene), alcohols (-ol), carboxylic acids (-oic acid).
Functional group
Examiner keyword
The specific atom or group of atoms responsible for the characteristic reactions of a homologous series, e.g. $\text{C}=\text{C}$ , $-\text{OH}$ or $-\text{COOH}$ .
Structural isomers
Examiner keyword
Compounds with the same molecular formula but different structural formulae (different arrangement of atoms), e.g. butane and methylpropane.
Molecular formula
A formula showing the actual number of atoms of each element in one molecule, e.g. $\text{C}_{4}\text{H}_{10}$ for both butane and methylpropane.

Common Mistakes and Misconceptions — Naming Organic Compounds

The traps other students keep falling into on naming organic compounds questions — taken from recent Cambridge IGCSE 0620 examiner reports and mark schemes — and how to avoid them.

✕Mixing up the stems (e.g. saying 'but' means 5 carbons).
0620/42 — recurring
Why it happens
Off-by-one slip when reciting the order from memory.
How to avoid it
Learn the order in sequence: meth (1), eth (2), prop (3), but (4).
✕Naming an alcohol with the '-ane' ending (e.g. 'propane' for $-\text{OH}$ ).
Why it happens
Forgetting to change the suffix when a functional group is present.
How to avoid it
If $-\text{OH}$ is present the suffix becomes '-ol' (propan-1-ol); if $-\text{COOH}$ is present it becomes '-oic acid' (propanoic acid).
✕Forgetting the position number when the functional group can sit on more than one carbon.
Why it happens
Rushing and giving only the stem + suffix.
How to avoid it
Add a number whenever the group could be in more than one place — propan-1-ol vs propan-2-ol; but-1-ene vs but-2-ene.
✕Not counting the carbon inside $-\text{COOH}$ as part of the chain.
Why it happens
The carboxyl carbon is easy to overlook when counting.
How to avoid it
Always include the $-\text{COOH}$ carbon: $\text{CH}_{3}\text{COOH}$ has 2 carbons → ethanoic acid, not methanoic acid.

Naming Organic Compounds — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Naming Organic Compounds

Short Notes - Naming Organic Compounds

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1How a name is built (stem + suffix)

2Stems for 1-4 carbons

3Name a carboxylic acid from its formula

4Draw a structure from a name

5Structural isomers of C₄H₁₀

6Identify the homologous series

Nomenclature (IUPAC name)

Homologous series

Functional group

Structural isomers

Molecular formula

✕Mixing up the stems (e.g. saying 'but' means 5 carbons).

✕Naming an alcohol with the '-ane' ending (e.g. 'propane' for −OH-\text{OH}−OH).

✕Forgetting the position number when the functional group can sit on more than one carbon.

✕Not counting the carbon inside −COOH-\text{COOH}−COOH as part of the chain.

Naming Organic Compounds — frequently asked questions

✕Naming an alcohol with the '-ane' ending (e.g. 'propane' for $-\text{OH}$ ).

✕Not counting the carbon inside $-\text{COOH}$ as part of the chain.