Master the naming rules for all eight organic functional group classes — alcohols, aldehydes, ketones, carboxylic acids, esters, amines, amides and haloalkanes — and learn to classify and draw structural isomers systematically.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Ethanol (the alcohol in wine and beer) and dimethyl ether (an industrial solvent and propellant) have exactly the same molecular formula: C₂H₆O. Same atoms, same count — yet one is metabolised by your liver at a predictable rate and the other causes anaesthesia and is used in refrigeration.
Before you read on, try to draw two different structural arrangements using 2 carbons, 6 hydrogens, and 1 oxygen. Write down how you would decide which arrangement is an “alcohol” and which is not. You will return to this at the end of the lesson.
Core Content
Hydroxyl group · -ol suffix · 1°/2°/3° based on C–OH neighbours
Alcohols all contain the -OH group, but WHERE on the carbon skeleton it sits — and how many other carbons surround it — determines both the name and the entire chemistry of that alcohol.
The hydroxyl group (–OH) bonded to a carbon is the defining feature of an alcohol. The suffix is -ol, replacing the final -e of the parent alkane name. Number the chain from the end closer to the –OH group to give it the lowest possible locant.
Alcohols are classified by the number of carbon atoms directly bonded to the carbon bearing the –OH group:
Carbonyl group position · -al / -one / -oic acid · terminal vs internal
Aldehydes, ketones, and carboxylic acids all contain the carbonyl group (C=O) — the position of that carbonyl and what else is bonded to it is the only thing separating three completely different functional group classes.
Aldehyde (–CHO): The carbonyl carbon is terminal and also has a hydrogen directly bonded to it. Suffix: -al. The aldehyde carbon is always C1 — a locant is never written. Examples: methanal (HCHO), ethanal (CH₃CHO), propanal (CH₃CH₂CHO), butanal.
Ketone (R–CO–R′): The carbonyl carbon is internal — bonded to two other carbons. Suffix: -one. A locant is required for chains of five or more carbons. For C4, the only possible internal position is C2, so “butan-2-one” is correct (also written as butanone). Examples: butan-2-one, pentan-2-one, pentan-3-one. The carbonyl cannot be at C1 — that would make it an aldehyde.
Carboxylic acid (–COOH): The carbonyl carbon bears both a C=O and an –OH on the same carbon, always terminal. Suffix: -oic acid. Always C1 — no locant. Examples: methanoic acid (HCOOH), ethanoic acid (CH₃COOH), propanoic acid, butanoic acid.
| Group | Carbonyl position | Additional feature | Suffix | Locant? |
|---|---|---|---|---|
| Aldehyde | Terminal (C1) | H on carbonyl C | -al | Never |
| Ketone | Internal | Two C neighbours | -one | Yes (C5+) |
| Carboxylic acid | Terminal (C1) | –OH on carbonyl C | -oic acid | Never |
Alkyl alkanoate · lone pair basicity · halogen as prefix
Esters smell like fruit, amines smell like fish, amides form the backbone of every protein in your body, and haloalkanes are the building blocks for most organic synthesis reactions — each has a structural signature you can read instantly once you know the pattern.
Esters (R–COO–R′): Named as alkyl alkanoate — the alkyl group (from the alcohol, the –OR′ part) comes FIRST, the alkanoate (from the acid, the RCOO– part) comes second.
Amines (R–NH₂): Suffix -amine. Classified by alkyl groups attached to N: primary (1°) = one alkyl (R–NH₂); secondary (2°) = two alkyls (R–NH–R′); tertiary (3°) = three alkyls. Amines are weak bases — the lone pair on N accepts H⁺ from acids.
Amides (R–CONH₂): Suffix -amide. The carbonyl carbon is always C1 — no locant. Examples: ethanamide (CH₃CONH₂), propanamide. Amides are neutral — the nitrogen lone pair is delocalised into the adjacent C=O and is not available for proton acceptance.
Haloalkanes (R–X): Halogen as a prefix — fluoro-, chloro-, bromo-, iodo- — with a locant. Examples: 1-chlorobutane, 2-bromobutane. Classification as 1°/2°/3° follows the same logic as alcohols (count carbon neighbours of the C bonded to X).
| Group | Structure | Naming pattern | Key feature |
|---|---|---|---|
| Ester | R–COO–R′ | Alkyl alkanoate (alkyl first) | Fruit-like odour; formed from acid + alcohol |
| Amine | R–NH₂ | [chain]-amine | Weak base; lone pair on N available |
| Amide | R–CONH₂ | [chain]-amide | Neutral; lone pair delocalised into C=O |
| Haloalkane | R–X | [halo]-[alkane] + locant | Polar C–X bond; 1°/2°/3° applies |
Same molecular formula · different arrangement · three isomer types
Structural isomers prove that molecular formula alone tells you almost nothing useful about a compound — the arrangement of atoms changes everything about physical properties, chemical reactivity, and biological activity.
Structural isomers are compounds with the same molecular formula but different structural arrangements of atoms. There are three types to distinguish:
Chain isomers: Same functional group and molecular formula, but a different carbon skeleton (branching pattern). Example: butane vs 2-methylpropane (both C₄H₁₀). The functional group (alkane) is identical; only the chain architecture changes. Different surface areas produce different boiling points.
Position isomers: Same functional group, same carbon skeleton, but the functional group is at a different position on the chain. Example: propan-1-ol vs propan-2-ol (both C₃H₈O). Both are alcohols with a 3-carbon chain, but –OH at C1 (primary) vs C2 (secondary). This matters chemically: propan-1-ol (primary) is oxidised to propanal then to propanoic acid; propan-2-ol (secondary) is oxidised only to propanone.
Functional group isomers: Same molecular formula but entirely different functional groups. Example: ethanol (C₂H₆O, alcohol) vs methoxymethane (dimethyl ether, C₂H₆O). Another pair: propanal vs propanone (both C₃H₆O, but aldehyde vs ketone).
Because the –OH group in ethanol can form hydrogen bonds (donating H to an electronegative acceptor) but the C–O–C ether linkage cannot donate H-bonds, ethanol (bp 78°C) has a dramatically higher boiling point than dimethyl ether (bp −24°C) despite both having the molecular formula C₂H₆O. Structure dictates intermolecular forces, which determine physical properties.
Coming up: In Lesson 3 you will see how chain length and functional group together determine physical properties such as boiling point — the first systematic comparison across homologous series.
Worked Examples
GIVEN: Five structural formulas. FIND: IUPAC name for each. METHOD: (1) Identify functional group from structural signature. (2) Apply the suffix/prefix rule. (3) Number the chain for lowest locant.
(a) CH₃CH₂CH₂OH: Terminal –OH → alcohol. 3C chain → prop-. Suffix -ol. –OH at C1 (numbered from the end closest to –OH). ANSWER: propan-1-ol (primary alcohol — C–OH bonded to 1 other carbon).
(b) CH₃CH₂CHO: Terminal C=O with H on same carbon → aldehyde. 3C chain → prop-. Suffix -al. Always C1. ANSWER: propanal.
(c) CH₃COCH₃: Internal C=O flanked by two carbons → ketone. 3C chain → prop-. Suffix -one. C=O at C2 (the only possible internal position in a 3C chain). ANSWER: propanone (locant omitted as unambiguous for 3C).
(d) CH₃CH₂COOCH₃: –COO– linkage → ester. Named as alkyl alkanoate. RCOO– part: CH₃CH₂COO– = propanoate (3C acid). –OR′ part: –OCH₃ = methyl. Alkyl first. ANSWER: methyl propanoate.
(e) CH₃CH₂NH₂: –NH₂ on terminal carbon → primary amine. 2C chain → eth-. Suffix -amine. N bonded to one alkyl group only. ANSWER: ethanamine (primary amine).
GIVEN: Molecular formula C₄H₈O, carbonyl group (C=O) required. FIND: All structural isomers. METHOD: Consider (a) aldehyde structures (terminal C=O + H), then (b) ketone structures (internal C=O). Vary the carbon skeleton for each.
Aldehyde structures (C=O at C1 with H on carbonyl C):
Straight-chain: CH₃CH₂CH₂CHO = butanal ✔ C₄H₈O ✔
Branched: (CH₃)₂CHCHO = 2-methylpropanal ✔ C₄H₈O ✔
Ketone structures (internal C=O): A 4C ketone has C=O at C2. C=O at C3 in a 4C chain is the same compound numbered from the other end.
CH₃COCH₂CH₃ = butan-2-one ✔ C₄H₈O ✔
ANSWER: Three isomers. (1) Butanal — straight-chain aldehyde. (2) 2-methylpropanal — branched aldehyde (chain isomer of butanal). (3) Butan-2-one — ketone (functional group isomer of both aldehyde isomers — same C₄H₈O formula, different functional group).
Check the molecular formula claim. Propan-1-ol (CH₃CH₂CH₂OH) → C₃H₈O ✔. Propan-2-ol (CH₃CHOHCH₃) → C₃H₈O ✔. The student is correct that both have the formula C₃H₈O and they are structural isomers. [1 mark]
Evaluate the isomer type classification. The student calls them “functional group isomers” — this is WRONG. Functional group isomers have different functional groups. Both propan-1-ol and propan-2-ol contain the –OH (hydroxyl) group — the functional group is identical in both compounds. [1 mark]
State the correct isomer type. These are position isomers: same molecular formula (C₃H₈O), same functional group (–OH, alcohol), same 3-carbon skeleton, but –OH is at C1 in propan-1-ol and at C2 in propan-2-ol. Only the position of the group changes. [1 mark]
Extend with chemical consequences. Propan-1-ol has a primary C–OH (bonded to 1 other carbon); propan-2-ol has a secondary C–OH (bonded to 2 carbons). Primary alcohols are oxidised to aldehydes then carboxylic acids under mild oxidising conditions; secondary alcohols are oxidised only to ketones — oxidation stops at that stage. [1 mark]
ANSWER SUMMARY: The molecular formula claim is correct (both C₃H₈O, both structural isomers) ✔. The isomer type classification is wrong ✗ — the correct type is position isomers, not functional group isomers. The functional group (–OH) is the same in both compounds; only its position on the 3C chain differs. The position difference means propan-1-ol is primary (oxidised → aldehyde → carboxylic acid) while propan-2-ol is secondary (oxidised → ketone only). [1 mark for structured evaluation]
Activities
CH₃CH₂OH
CH₃CH₂CHO
CH₃COCH₂CH₃
CH₃CH₂COOH
CH₃COOCH₂CH₂CH₃
CH₃CH₂CH₂NH₂
1. Butane (CH₃CH₂CH₂CH₃) vs 2-methylpropane (CH₃CH(CH₃)CH₃)
2. Butan-1-ol (CH₃CH₂CH₂CH₂OH) vs Butan-2-ol (CH₃CH₂CHOHCH₃)
3. Butanal (CH₃CH₂CH₂CHO) vs Butan-2-one (CH₃COCH₂CH₃)
Check Your Understanding
Click an option to check your answer. Aim for all 5 before looking at the answers accordion.
1. Which compound is the ester formed from propan-1-ol and ethanoic acid?
2. A compound has the structure CH₃CH₂CH(NH₂)CH₃. What is the IUPAC name and amine classification?
3. Propanal and propanone both have C₃H₆O. What type of isomers are they?
4. Which statement correctly identifies the type of isomers formed by butan-1-ol and butan-2-ol?
5. A student is given an unknown compound with molecular formula C₃H₆O. Which of the following correctly lists ALL possible structural isomers of this formula that contain only one functional group?
Short Answer
6. For each compound below, state the functional group class and give the correct IUPAC name: (a) CH₃CH₂CH₂COOH, (b) CH₃CONH₂, (c) CH₃CH(CH₃)OH, (d) CH₃CH₂COOCH₂CH₃ 4 MARKS
7. Draw and name all structural isomers of C₃H₆O. For each, identify the isomer type relative to the other and explain the key structural difference. 4 MARKS
8. A pharmacologist is investigating two compounds with the molecular formula C₄H₉NO. Compound A reacts with hydrochloric acid to form a salt. Compound B does not react with HCl. (a) Identify the functional groups that could give this difference in reactivity. (b) Name and draw one possible structure for each compound. (c) Identify the type of isomers that Compound A and Compound B represent, and justify your answer. 5 MARKS
1. CH₃CH₂OH: Alcohol (–OH on terminal carbon). 2C chain. Name: ethanol. Primary alcohol (C–OH bonded to 1 carbon). No locant needed for C2.
2. CH₃CH₂CHO: Aldehyde (terminal C=O with H on same carbon). 3C chain. Suffix -al. Name: propanal. Always at C1; no locant written.
3. CH₃COCH₂CH₃: Ketone (internal C=O flanked by two carbons). 4C chain. Suffix -one. C=O at C2. Name: butan-2-one (also accepted: butanone).
4. CH₃CH₂COOH: Carboxylic acid (terminal –COOH). 3C chain. Suffix -oic acid. Name: propanoic acid. Always C1; no locant.
5. CH₃COOCH₂CH₂CH₃: Ester (–COO–). Alkanoate part: CH₃COO– = ethanoate. Alkyl part: –OCH₂CH₂CH₃ = propyl. Name: propyl ethanoate.
6. CH₃CH₂CH₂NH₂: Amine (–NH₂). 3C chain. Suffix -amine. –NH₂ at C1. Name: propan-1-amine (or propylamine). Primary amine — N bonded to one alkyl group. Basic because the lone pair on N is available to accept H⁺.
1. Butane vs 2-methylpropane: Both C₄H₁₀. Chain isomers — same functional group class (alkane), same molecular formula, but different carbon skeleton: butane has a straight 4-carbon chain; 2-methylpropane has a branched 3-carbon chain with a methyl group at C2.
2. Butan-1-ol vs Butan-2-ol: Both C₄H₁₀O. Position isomers — same formula, same functional group (–OH, alcohol), same 4-carbon skeleton, but –OH is at C1 (primary alcohol) in butan-1-ol and at C2 (secondary alcohol) in butan-2-ol.
3. Butanal vs Butan-2-one: Both C₄H₈O. Functional group isomers — same formula, but different functional groups. Butanal has a terminal C=O with H (aldehyde, suffix -al); butan-2-one has an internal C=O between two carbons (ketone, suffix -one).
1. B — Propyl ethanoate. Ester from propan-1-ol + ethanoic acid. Alkyl part (from propan-1-ol) = propyl. Alkanoate part (from ethanoic acid) = ethanoate. Alkyl comes first → propyl ethanoate. Option A (ethyl propanoate) would come from ethanol + propanoic acid. Option D misspells the suffix (-propanate is not correct IUPAC).
2. C — Butan-2-amine; primary. CH₃CH₂CH(NH₂)CH₃: longest chain = 4C (butane). –NH₂ at C2 from the end that gives the lower locant (C2 vs C3 from the other end). Name: butan-2-amine. Primary amine because N is bonded to only one alkyl group (–NH₂ on C2; N has H₂ + one carbon neighbour). Option A has correct name but wrong classification; secondary amine requires two alkyl groups on N.
3. B — Functional group isomers. Propanal has a terminal C=O (aldehyde); propanone has an internal C=O (ketone). These are entirely different functional groups — the hallmark of functional group isomers. Option A incorrectly describes position isomers (same functional group, different position) — but here the group itself differs.
4. C — Position isomers. Butan-1-ol and butan-2-ol: both C₄H₁₀O ✔, same –OH group ✔, same 4-carbon skeleton ✔, –OH at C1 vs C2. This matches the definition of position isomers exactly. Options A and D incorrectly treat 1° vs 2° as different functional group classes — they are classifications of the same class (alcohol), not different classes.
5. D — Propanal and propanone. C₃H₆O: propanal (CH₃CH₂CHO, aldehyde) and propanone (CH₃COCH₃, ketone) both satisfy the formula. Propan-1-ol and propan-2-ol are C₃H₈O (two extra H) — a different molecular formula — so they are not C₃H₆O isomers. Options A and B each list only one of the two correct isomers.
Q6 (4 marks — 1 mark each):
(a) CH₃CH₂CH₂COOH: Carboxylic acid (–COOH at terminal carbon). 4C chain. Name: butanoic acid [1].
(b) CH₃CONH₂: Amide (–CONH₂). 2C chain. Suffix -amide. Name: ethanamide [1].
(c) CH₃CH(CH₃)OH: The carbon with –OH: CH₃–CH(OH)–CH₃ → 3-carbon chain with –OH at C2 → propan-2-ol. Secondary alcohol (C–OH bonded to 2 other carbons). Name: propan-2-ol [1].
(d) CH₃CH₂COOCH₂CH₃: Ester. Alkanoate part: CH₃CH₂COO– = propanoate (3C acid). Alkyl part: –OCH₂CH₃ = ethyl. Name: ethyl propanoate [1].
Q7 (4 marks): C₃H₆O has two structural isomers containing a single C=O functional group:
Isomer 1: Propanal (CH₃CH₂CHO) — aldehyde. Terminal C=O with H on carbonyl C [1].
Isomer 2: Propanone (CH₃COCH₃) — ketone. Internal C=O flanked by two methyl groups [1].
These are functional group isomers [1]: same molecular formula C₃H₆O, but different functional groups. Key structural difference: in propanal, C=O is at C1 with an H (aldehyde); in propanone, C=O is at C2 between two carbon groups (ketone) [1].
Q8 (5 marks):
(a) Compound A reacts with HCl → must contain an amine (–NH₂): the undelocalised lone pair on N accepts H⁺ from HCl, forming an ammonium salt (R–NH₃⁺Cl⁻) [1]. Compound B does not react with HCl → must contain an amide (–CONH₂): the N lone pair is delocalised into the adjacent C=O by resonance and cannot accept a proton — amides are neutral [1].
(b) Compound A (amine): 4-aminobutan-2-one (CH₃COCH₂CH₂NH₂) contains a primary amine and a ketone, formula C₄H₉NO ✔ [1]. Compound B (amide): butanamide (CH₃CH₂CH₂CONH₂) contains an amide at C1, formula C₄H₉NO ✔ [1].
(c) Compound A and Compound B are functional group isomers [1]: same molecular formula (C₄H₉NO) but completely different functional groups — Compound A contains an amine (–NH₂) and Compound B contains an amide (–CONH₂). Because the functional groups differ (not just their positions), these are functional group isomers, not position isomers.
Review Drills
What is the IUPAC name for the compound formed when ethanol reacts with propanoic acid to form an ester?
How do you distinguish between a 1°, 2°, and 3° alcohol? Give one example of each.
Why are amines weak bases but amides are neutral? Use lone pair location to explain.
Classify these three isomer pairs: (A) butane / 2-methylpropane, (B) propan-1-ol / propan-2-ol, (C) propanal / propanone.
What structural feature must be present to confirm a compound is an aldehyde rather than a ketone? How does this affect the IUPAC suffix?
A student writes the name “ethanoate methyl” for an ester. What is wrong and what is the correct name?
Back at the start, you were asked why ethanol gets metabolised by your liver while dimethyl ether is used as an anaesthetic propellant. Now you know: the –OH group in ethanol can be oxidised by alcohol dehydrogenase in the liver — the enzyme recognises and acts on the hydroxyl group. The C–O–C ether linkage in dimethyl ether is chemically inert under physiological conditions: no enzyme can oxidise it, so it passes through the body without being broken down. The difference comes down entirely to the functional group. Same molecular formula (C₂H₆O), different connectivity, completely different biochemistry. These are functional group isomers: ethanol has O bonded to C and H (C–O–H = hydroxyl/alcohol), while dimethyl ether has O bonded to two carbons (C–O–C = ether).
Tick when you’ve finished all activities and checked your answers.
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