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Organic chemistry is the study of carbon-containing compounds. With millions of known organic molecules, a systematic naming convention is essential. The IUPAC (International Union of Pure and Applied Chemistry) system provides a universal set of rules so that every organic compound has one unambiguous name, and every name corresponds to one structure.
Early chemists named compounds after their source or discoverer — acetic acid from vinegar (Latin acetum), morphine from Morpheus (the Greek god of dreams). This worked when only a few hundred compounds were known, but it became chaotic as the number grew. The IUPAC system, developed in the late 19th and early 20th centuries, replaced trivial names with a logical framework based on molecular structure.
At A-Level, you need to be able to name a compound from its structure and draw a structure from its name. Both directions of this skill are tested regularly, and naming errors will lose marks in mechanism and synthesis questions too, even if the chemistry is correct.
Every IUPAC name contains up to four parts:
| Part | What It Tells You | Examples |
|---|---|---|
| Prefix | Substituent groups (branches) | methyl-, ethyl-, chloro-, bromo- |
| Root | Length of the longest carbon chain | meth- (1C), eth- (2C), prop- (3C), but- (4C), pent- (5C), hex- (6C), hept- (7C), oct- (8C) |
| Primary suffix | Degree of saturation | -ane (single bonds only), -ene (C=C present), -yne (C≡C present) |
| Secondary suffix | Principal functional group | -ol (alcohol), -al (aldehyde), -one (ketone), -oic acid (carboxylic acid) |
The root and primary suffix together form the parent name. For example, a six-carbon chain with only single bonds is hexane; with a double bond it is hexene.
| Number of Carbons | Root | Origin |
|---|---|---|
| 1 | meth- | From methanol (wood spirit) |
| 2 | eth- | From ether |
| 3 | prop- | From propionic acid |
| 4 | but- | From butyric acid (butter) |
| 5 | pent- | Greek pente (five) |
| 6 | hex- | Greek hex (six) |
| 7 | hept- | Greek hepta (seven) |
| 8 | oct- | Greek okto (eight) |
| 9 | non- | Latin nonus (ninth) |
| 10 | dec- | Greek deka (ten) |
Find the longest continuous chain of carbon atoms that includes the principal functional group (if present). This chain determines the root name.
If a molecule has a chain of five carbons with a methyl branch, the root is pent- (not hex-), because the branch is not part of the longest continuous chain.
Trap: The longest chain is not always drawn horizontally. In exam structures, the chain may zigzag or bend. Count carbons carefully along every possible path through the molecule.
Number the carbon atoms from the end that gives the lowest set of locants (position numbers) to the substituents and functional groups. If there is a principal functional group (such as -OH or C=O), it must receive the lowest possible locant.
Any groups attached to the parent chain are named as prefixes. Common substituent prefixes:
If the same substituent appears more than once, use multiplying prefixes: di- (2), tri- (3), tetra- (4).
Write the name in the order: locants + prefixes (alphabetical order) + root + primary suffix + locant + secondary suffix.
For example, consider a five-carbon chain with a hydroxyl group on carbon 2 and a methyl branch on carbon 3:
Note that prefixes are placed in alphabetical order. "Ethyl" comes before "methyl", and multiplying prefixes (di-, tri-) are ignored when alphabetising.
When a molecule contains more than one functional group, one is chosen as the principal group and named as a suffix; the others are named as prefixes. The Edexcel A-Level specification requires you to know the following priority order (highest to lowest):
A molecule with both an -OH and a -COOH group would be named as a carboxylic acid, with the -OH expressed as a prefix (hydroxy-).
Consider the following structure (drawn as a connectivity description):
A seven-carbon chain has a double bond between C3 and C4, a chloro group on C2, an ethyl branch on C4, and a methyl branch on C5.
Step 1: Longest chain containing the functional group (C=C) = 7 carbons → hept-
Step 2: Number from the end that gives the lowest locant to the C=C. Numbering from the left gives the double bond at position 3 (hept-3-ene). Numbering from the right gives position 4 (hept-4-ene). We choose the numbering that gives 3 (lower).
Step 3: Substituents: 2-chloro, 4-ethyl, 5-methyl.
Step 4: Alphabetical order: chloro (c) before ethyl (e) before methyl (m).
Full name: 2-chloro-4-ethyl-5-methylhept-3-ene
A six-carbon chain (hexane) with methyl groups on carbons 2, 3, and 4.
Note: when alphabetising, "trimethyl" is alphabetised under "m" (ignore the "tri-" prefix).
Draw 4-bromo-2-ethyl-3-methylhexan-1-ol.
To go from a name to a structure, reverse the process:
For example, 2-bromobutane: the root is but- (4 carbons), the suffix is -ane (all single bonds), and there is a bromo substituent on carbon 2. Draw four carbons in a chain, attach Br to the second carbon, and complete with hydrogens.
Cyclic alkanes are named by adding the prefix cyclo- before the root name. A ring of six carbons with only single bonds is cyclohexane. Substituents are named and numbered to give the lowest set of locants.
For example, a cyclohexane ring with a methyl group on one carbon and a chloro group on the adjacent carbon would be named 1-chloro-2-methylcyclohexane (choosing the numbering that gives the lowest locants and placing chloro before methyl alphabetically).
A student draws a structure with a five-carbon chain, a C=C between C2 and C3, a methyl group on C4, and names it "4-methylpent-3-ene".
Check: numbering from the other end gives the double bond at position 2 (pent-2-ene) with the methyl on C3: 3-methylpent-2-ene. The double bond receives the lower locant (2 < 3), so 3-methylpent-2-ene is the correct IUPAC name.
Understanding IUPAC nomenclature is the foundation of organic chemistry. Every reaction, mechanism, and synthesis question assumes you can read and write structural names fluently.
Edexcel 9CH0 specification Topic 6 — Organic Chemistry I, sub-topic 6.1 covers IUPAC nomenclature for hydrocarbons and the principal Year 1 functional groups: identifying the longest carbon chain that contains the principal functional group, numbering carbon atoms so the principal group (and then substituents) receive the lowest locants, naming substituents with appropriate prefixes, and applying alphabetical order ignoring multiplying prefixes (refer to the official specification document for exact wording). Sub-topic 6.1 is foundational: every later sub-topic in Topics 6, 7, 16, 17, 18 and 19 assumes confident IUPAC naming. Although nomenclature does not generate many marks in isolation, naming errors propagate through mechanism and synthesis answers — examiners can dock A1 accuracy marks on a fully-worked synthesis if the final product is mis-named. Naming is examined directly on Paper 2 (organic and physical chemistry) and tested synoptically on Paper 3 (general and practical principles), where structures must be read off NMR, IR or mass-spectrum data. The Edexcel data booklet does not give a list of substituent prefixes — they must be memorised.
Question (7 marks):
(a) Give the full IUPAC name for the compound with structural formula CH3-CH(Cl)-CH2-C(CH3)=CH-CH2-CH3. (3)
(b) Draw the displayed formula of 3-bromo-2,2-dimethylpentan-1-ol and state the total number of hydrogen atoms in the molecule. (4)
Solution with mark scheme:
(a) Step 1 — identify the longest chain containing the principal feature. The chain is seven carbons long if numbered through the C=C; the C=C is the highest-priority feature here (no -OH, -CHO, -CO- or -COOH present), so the parent name is hept-?-ene.
M1 — correct identification of a 7-carbon parent and the -ene suffix.
Step 2 — number to give C=C the lowest locant. Numbering from the right gives Cl on C6, methyl on C4, C=C between C3 and C4. Numbering from the left gives Cl on C2, methyl on C4, C=C between C4 and C5. Compare first points of difference for the C=C locants: {3} vs {4}. Number from the right.
Wait — re-read the structure. CH3(1)-CH(Cl)(2)-CH2(3)-C(CH3)(4)=CH(5)-CH2(6)-CH3(7) (left-to-right numbering). Here the C=C is between C4 and C5 — locant 4. Right-to-left numbering: CH3(7)-CH(Cl)(6)-CH2(5)-C(CH3)(4)=CH(3)-CH2(2)-CH3(1) — C=C at locant 3. Choose right-to-left.
M1 — correct numbering: C=C at the lower locant 3.
Step 3 — assemble. Substituents under right-to-left numbering: 6-chloro, 4-methyl. Alphabetical: chloro before methyl.
A1 — full name 6-chloro-4-methylhept-3-ene.
(b) Step 1 — parse the name. Pent- → 5-C chain. -an-1-ol → single bonds, -OH on C1. Substituents: 3-bromo, 2,2-dimethyl (two methyls on C2).
M1 — correct parent skeleton: 5-C chain with -OH on C1.
Step 2 — place substituents.
M1 — both methyls placed on C2, Br on C3.
Step 3 — saturate with hydrogens. C1 has 2H (CH2), C2 has 0H (quaternary, fully substituted by two methyls, one C and one C), C3 has 1H (CH-Br), C4 has 2H, C5 has 3H. Methyl branches contribute 3H + 3H = 6H. The hydroxyl contributes 1H.
A1 — displayed formula correct.
A1 — total H count: 2 + 0 + 1 + 2 + 3 + 6 + 1 = 15 H atoms. Molecular formula C7H15BrO.
Total: 7 marks (M3 A4).
Question (6 marks): A compound X has molecular formula C5H10O and the following structural information: it contains a carbonyl group flanked on one side by two methyl groups attached to the same carbon; the chain is fully saturated; there is no -OH. Two students propose names for X:
(a) Determine which student is correct, justifying your choice by reference to the molecular formula. (4)
(b) Draw the displayed formula of the correct compound and state which functional group it contains. (2)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. Naming questions on Paper 2 lean heavily on AO1 (recall and application of the rules) but Edexcel routinely uses molecular formula as the AO2/AO3 cross-check — students who name a structure without verifying the formula will mis-identify constitutional isomers as the target compound.
Connects to:
Topic 6.2 — Isomerism: structural isomers share a molecular formula but differ in connectivity, and their distinct IUPAC names are the proof of distinctness. Naming is the operational test for "are these two structures the same compound?"
Topic 18 — NMR spectroscopy: the number of carbon environments in 13C NMR equals the number of structurally non-equivalent carbons in the IUPAC structure. Predicting the number of 13C peaks for "3,3-dimethylpentan-2-one" requires fluent name-to-structure conversion.
Topic 19 — Mass spectrometry and IR: fragment-loss patterns (e.g. loss of 15 = methyl, loss of 29 = CHO/ethyl) are interpreted on the IUPAC structure, not on a condensed formula.
Topic 16 — Carbonyl chemistry: Tollens' and Fehling's tests distinguish aldehydes from ketones, but you must first name the compound correctly to know which class it belongs to. A mis-numbered chain that places the C=O at C2 instead of C1 of a 4-C molecule converts butanal into butan-2-one — a different classification entirely.
Topic 7 — Halogenoalkanes and alcohols: the classification of haloalkanes and alcohols as primary, secondary or tertiary depends on carbon-chain numbering. The IUPAC name encodes this directly: 2-methylpropan-2-ol is tertiary because the -OH is on a carbon bonded to three other carbons (deduce-able from the name alone).
Naming questions on 9CH0 split AO marks heavily toward AO1, with AO2 marks reserved for "explain your numbering" or "justify the priority" prompts:
| AO | Typical share | Earned by |
|---|---|---|
| AO1 (knowledge / procedure) | 60–75% | Identifying parent chain, numbering correctly, listing substituents alphabetically, assembling the name in the standard format |
| AO2 (application) | 20–30% | Choosing which functional group is principal, justifying lowest-locant rule with explicit comparison, applying nomenclature to unfamiliar functional groups by analogy |
| AO3 (analysis) | 0–10% | Synoptic identification of compounds from molecular formula + spectroscopic data; constructing names for compounds presented only by skeletal formula |
Examiner-rewarded phrasing: "the longest continuous chain containing the principal functional group is …"; "numbering from this end gives the lowest locant set 2,3,5 versus 3,4,6 from the other end"; "the substituent prefixes in alphabetical order, ignoring multiplying prefixes, are …". Phrases that lose marks: writing locants without commas (134-trimethyl instead of 1,3,4-trimethyl); using "iso-" or "tert-" prefixes in the final IUPAC name (acceptable only as informal aliases); failing to re-state which carbon receives the principal-group locant when multiple choices give equal sums.
A specific Edexcel pattern to watch: when the question says "give the IUPAC name", trivial names such as acetone, toluene, chloroform are not credited — only the systematic name (propan-2-one, methylbenzene, trichloromethane) earns the mark.
Question: Give the IUPAC name of CH3CH2CH(OH)CH2CH3.
Mid-band response (~150 words):
The chain has 5 carbons so the root is pent-. There is an -OH group on the middle carbon, which is C3. So the name is pentan-3-ol.
Examiner commentary: Full marks (3/3). Brief but correct: the candidate identifies the parent (pent-), the principal functional group (-ol), and the locant (3). Numbering is symmetric here — both ends give locant 3 — so no priority choice is forced. The omission is a justification: a stronger answer would write "since the molecule is symmetric about C3, numbering from either end gives the -OH at locant 3", which signals examiner-aware reasoning. For a 3-mark procedural question that level of justification is not required.
Top-band response (~190 words):
The longest continuous carbon chain is 5 atoms: CH3−CH2−CH(OH)−CH2−CH3, so the root is pent-. The principal functional group is the hydroxyl (-OH); since the molecule has no carbonyl or carboxylic acid group, -OH takes priority and the suffix is -ol. The molecule is symmetric about the central carbon, so numbering from either end gives the -OH at locant 3. There are no other substituents.
The IUPAC name is pentan-3-ol.
Examiner commentary: Full marks (3/3). The candidate explicitly verifies the symmetry of the carbon skeleton, names the functional group, and references the priority order without needing to. This is the kind of "show your reasoning even when the answer is short" presentation that distinguishes A* from A — it reads as if the candidate has internalised the procedure rather than guessed.
Question: A compound has the structural formula (CH3)2CH−CH(C2H5)−COOH. (a) Give the IUPAC name. (3) (b) State the molecular formula and identify the principal functional group. (3)
Stronger response (~220 words):
(a) The longest chain has 5 carbons going through the COOH. So root is pent-, suffix is -oic acid. There's a methyl on C3 (from the (CH3)2CH) and an ethyl on C2. Wait, actually the (CH3)2CH part is an isopropyl group on the third carbon. Let me redraw. C1 is COOH. C2 has the ethyl branch. C3 is the CH from the (CH3)2CH. The two methyls hanging off would be C4 and... actually no, one of them extends the main chain. The longest chain through COOH is actually 4 carbons. Suffix -anoic acid. The branches are 2-ethyl, 3-methyl. So 2-ethyl-3-methylbutanoic acid.
(b) Count: 4 (chain) + 2 (ethyl) + 1 (methyl) = 7 C. So molecular formula C7H14O2. Functional group is carboxylic acid.
Examiner commentary: Method correct overall, secured (b) with full marks (3/3) and (a) with 2/3 — final A1 lost because the longest chain through the COOH is actually 5 carbons, not 4 (extending through one of the (CH3)2CH methyls into the main chain gives 2-methyl-3-methylpentanoic acid → 2,3-dimethylpentanoic acid is closer; the candidate's "4-carbon chain + 2-ethyl branch" is the standard mis-count). Total: 5/6.
Top-band response (~290 words):
(a) Locating the longest chain through the carboxylic acid: starting from -COOH, the chain extends through C(C2H5)H, then through C(CH3)2H, then along one of the methyls. The candidate paths are (i) -COOH-CH(C2H5)-CH(CH3)-CH3 (4 C) or (ii) extending through the ethyl branch as -COOH-CH(R)-CH2-CH3 with R = CH(CH3)2 (4 C through ethyl). Both give 4-carbon main chains. The "ethyl" group on C2, when treated as part of the parent chain, gives a 5-C chain: COOH(1)-CH(2)-CH2(3)-CH3(4) with R = CH(CH3)2 on C2. But this is only 4 carbons. The longest is 4-C through the original chain — butanoic acid parent.
Substituents: ethyl on C2, isopropyl (propan-2-yl) on C2 — but two substituents on the same carbon means the second is on C2 also. Reassess: the structure has COOH-CH(R1)(R2) with R1 = ethyl, R2 = isopropyl, so C2 of butanoic acid carries two substituents.
The IUPAC name is 2-ethyl-3-methylbutanoic acid.
(b) Molecular formula: C1 (COOH) + C2 (CH) + C3, C4 chain + ethyl (C2H5) + methyl (CH3) = 4 + 2 + 1 = 7 carbons. Hydrogens: 1 (COOH) + 1 (CH at C2) + 1 (CH at C3) + 3 (CH3 at C4) + 2 + 3 (ethyl) + 3 (methyl) = 14 H. Formula C7H14O2. Functional group: carboxylic acid.
Examiner commentary: Full marks (6/6). The candidate explicitly compares chain-extension paths and chooses the longest, then verifies the molecular formula by atom-by-atom count. This is the disciplined name-then-verify workflow that protects against sign-of-life slips.
Question: Compound Y has molecular formula C6H12O and contains both a carbonyl and an alcohol group. Three students propose names: (i) 4-hydroxyhexan-2-one; (ii) 3-hydroxyhex-2-anal; (iii) 5-hydroxyhexan-2-one.
(a) Identify which proposed names obey IUPAC rules. (3) (b) For the valid names, give the displayed formula. (4) (c) Which functional group is principal in each, and why? (2)
Top-band response (~360 words):
(a) Check each:
(b) Displayed formulae:
Both have molecular formula C6H12O2 — but the question states Y has C6H12O. Recheck: each structure contains one -OH and one C=O, so the formula must be C6H12O2. Therefore Y as stated cannot exist with both groups; the question contains an inconsistency, or Y has only one of the two groups.
(c) In both (i) and (iii), the ketone (-one) is principal because ketones outrank alcohols in IUPAC priority (carboxylic acid > aldehyde > ketone > alcohol > amine). The hydroxyl is therefore named as a prefix "hydroxy-".
Examiner commentary: Full marks (9/9) for the IUPAC reasoning, plus a bonus for spotting the molecular-formula inconsistency in (b). The candidate uses lowest-locant comparison, distinguishes (i) and (iii) as distinct isomers, recognises the rule that aldehydes are always C1, and applies the priority order in (c). The spotted error in the stem is the kind of forensic reading that signals examiner-grade fluency. Total: 9/9 with merit.
The errors that distinguish A from A* on naming questions:
Lowest-locant comparison errors. When two numbering options give locant sets such as 2,4,5 versus 3,3,4, students sum (2+4+5 = 11 vs 3+3+4 = 10) and pick the lower sum. The correct rule is first point of difference: compare 2,… vs 3,… and stop at the first unequal pair — 2,4,5 wins because 2 < 3. Sum-comparison is a non-IUPAC heuristic.
Treating "iso-" and "tert-" prefixes as IUPAC. Isopropyl, tert-butyl, sec-butyl are retained names — they are widely recognised but Edexcel only credits them in informal contexts. The IUPAC names are propan-2-yl, 2-methylpropan-2-yl, butan-2-yl.
Mis-locating the C=C in alkenes. "but-2-ene" means the double bond starts at C2; "but-1-ene" starts at C1. Students sometimes use the locant of the higher-numbered alkene carbon — wrong. The locant is always the lower of the two carbons in the double bond.
Counting branches into the parent chain. When a methyl branch could be incorporated into a longer parent chain by re-tracing, always take the longer parent. Methyl-on-pentane is sometimes longer-as-hexane after re-counting. Re-trace before committing.
Forgetting that aldehydes are always at C1. "pentan-2-al" is not a valid name. Aldehydes terminate the chain at C1 and the locant "1" is implied.
Cyclic numbering. In 1,2-dimethyl- vs 1,3-dimethylcyclohexane the position of the second methyl relative to the first matters and cannot be flipped freely. Students sometimes pick whichever numbering gives the alphabetically first substituent locant 1; the correct rule is lowest set of locants, with alphabetical priority used only as a tiebreak.
Alphabetising on the multiplying prefix. "trimethyl" alphabetises under "m"; "dichloro" under "c". The multiplying prefix is invisible for alphabetisation but visible in the written name.
Three patterns repeatedly cost candidates marks on Paper 2 naming questions. They are about presentation, not technique.
This pattern is endemic to Paper 2 nomenclature questions: candidates know the rules, lose marks on punctuation and presentation.
IUPAC nomenclature underpins several undergraduate trajectories:
Oxbridge interview prompt: "Why does IUPAC nomenclature use atomic number rather than alphabetical order to assign stereodescriptors? Construct an example where the two would disagree."
This sub-topic is foundational for all five organic core practicals (CP5–CP7, CP14, CP15). CP5 (rates of hydrolysis of haloalkanes) requires you to name reactants and products correctly so observations can be matched to substrates. CP6 (distillation of an organic product, typically conversion of an alcohol to an alkene or to a halogenoalkane) demands that the distillate be identified by IUPAC name and verified against expected boiling point. CP7 (tests for organic functional groups: bromine water for alkenes, Fehling's/Tollens' for aldehydes vs ketones) requires you to interpret a positive/negative test in terms of the named functional group present in the compound. CP14 (analysis of an organic acid and an organic alcohol by titration and esterification) and CP15 (preparation of an ester) both record the IUPAC names of starting materials and products in the practical write-up. For Edexcel-style 6-mark practical questions, examiners reward candidates who name the species in every observation: "the colourless distillate, identified as butan-1-ol by its boiling point of 117 °C, was redistilled to remove residual water" earns more credit than "the product was distilled". Naming is the operational vocabulary of every organic practical.
This content is aligned with the Pearson Edexcel GCE A Level Chemistry (9CH0) specification, Paper 2 — Core Organic and Physical Chemistry, Topic 6: Organic Chemistry I, sub-topic 6.1 (nomenclature). For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.
graph TD
A["Organic structure<br/>given"] --> B["Identify principal<br/>functional group<br/>(highest priority)"]
B --> C["Find longest chain<br/>containing the<br/>principal group"]
C --> D["Number to give<br/>lowest locant<br/>to principal group"]
D --> E{"Tie?"}
E -->|"Yes"| F["Number to give<br/>lowest locants<br/>to substituents"]
E -->|"No"| G["Locants set"]
F --> G
G --> H["List substituents<br/>alphabetically<br/>(ignore di/tri-)"]
H --> I["Assemble:<br/>locants + prefixes +<br/>root + suffix"]
I --> J["Verify by drawing<br/>structure from name"]
J --> K["Cross-check<br/>molecular formula"]
style B fill:#27ae60,color:#fff
style I fill:#3498db,color:#fff
style K fill:#e67e22,color:#fff