IUPAC Nomenclature and Functional Groups

Organic chemistry catalogues millions of molecules, and a universal naming system is the only way to keep that catalogue tractable. The International Union of Pure and Applied Chemistry (IUPAC) developed the systematic nomenclature you will use throughout A-Level: a strict but logical procedure that converts any structure into a single unambiguous name and any name back into a single unambiguous structure. This opening lesson lays down the toolkit you will rely on for every later topic — homologous series and their general formulae; the principal functional groups and the priority order that decides which one supplies the suffix; the five IUPAC naming rules (longest chain, lowest locant set, substituent prefixes, alphabetical ordering with multiplying prefixes, and stereochemistry markers); and the four common structural representations — skeletal, displayed, condensed, and molecular formulae — together with the conversions between them. Master this lesson and the rest of organic chemistry becomes a chain of small, well-defined steps.

Spec mapping (AQA 7405): This lesson anchors §3.3.1 (introduction to organic chemistry — nomenclature, homologous series, functional groups, structural representation) and is the foundation on which every subsequent organic-chemistry lesson on this course is built. The advanced nomenclature for aromatic compounds (§3.3.8) and amines/amino acids/amides (§3.3.9) reuses the IUPAC framework introduced here. Refer to the official AQA specification document for the exact wording of each section.

Assessment objectives: Recall of homologous-series general formulae, suffix conventions, and functional-group priority order is an AO1 task that recurs as opening-question marks throughout Papers 2 and 3. Naming a given compound and drawing a structure from a given IUPAC name is AO2 and appears at least once on every organic paper. The hardest application — AO3 — asks students to deduce a structure from spectroscopic data (mass spectrum, infrared, NMR) and then name it correctly, or to identify every functional group in a complex multi-group molecule such as a drug, a natural product, or an industrial intermediate.

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Homologous Series — the Twelve You Must Know

A homologous series is a family of organic compounds that:

• share a single general formula (CₙH₂ₙ₊₂ for alkanes, CₙH₂ₙ for alkenes and so on);
• contain the same principal functional group;
• differ from one member to the next by a –CH₂– unit (14 g mol⁻¹);
• display a smooth gradation of physical properties (boiling point rises with chain length as London dispersion forces increase);
• undergo broadly the same characteristic reactions (because the chemistry is dictated by the functional group, not the alkyl backbone).

For AQA A-Level chemistry, twelve homologous series cover almost every compound you will encounter.

Series	General formula	Functional group	Suffix	Example	Example name
Alkanes	CₙH₂ₙ₊₂	C–C, C–H only (saturated)	-ane	CH₃CH₃	ethane
Alkenes	CₙH₂ₙ	C=C	-ene	CH₂=CHCH₃	propene
Halogenoalkanes	CₙH₂ₙ₊₁X	C–X (X = F, Cl, Br, I)	(prefix only)	CH₃CH₂Br	bromoethane
Alcohols	CₙH₂ₙ₊₁OH	–OH	-ol	CH₃CH₂OH	ethanol
Aldehydes	CₙH₂ₙO	–CHO (terminal)	-al	CH₃CHO	ethanal
Ketones	CₙH₂ₙO	C=O (internal)	-one	CH₃COCH₃	propan-2-one
Carboxylic acids	CₙH₂ₙO₂	–COOH	-oic acid	CH₃COOH	ethanoic acid
Esters	CₙH₂ₙO₂	–COO–	alkyl …-oate	CH₃COOCH₃	methyl ethanoate
Amines	CₙH₂ₙ₊₃N	–NH₂ (primary)	-amine	CH₃CH₂NH₂	ethylamine
Amides	CₙH₂ₙ₊₁NO	–CONH₂	-amide	CH₃CONH₂	ethanamide
Nitriles	CₙH₂ₙ₋₁N	–C≡N	-nitrile	CH₃CN	ethanenitrile
Arenes	CₙH₂ₙ₋₆ (benzene homologues)	aromatic ring	-benzene / phenyl-	C₆H₅CH₃	methylbenzene

A few subtleties matter at A-Level:

• Aldehydes and ketones share the same general formula (CₙH₂ₙO) and are therefore functional-group isomers of each other. The same is true of carboxylic acids and esters (CₙH₂ₙO₂). Spotting this is a routine AO2 task.
• Nitriles count the C≡N carbon as part of the longest chain. Ethanenitrile (CH₃CN) is a two-carbon compound: the CH₃ carbon is C-2, the CN carbon is C-1.
• Aldehydes are always terminal — the –CHO carbon is C-1 by convention and never receives a locant in the name.
• Amines come in primary (RNH₂), secondary (R₂NH), and tertiary (R₃N) varieties — only primary amines follow the simple general formula above.

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Functional-Group Priority for IUPAC Naming

When a molecule contains more than one functional group, the highest-priority group is the principal functional group: it supplies the suffix and is given the lowest possible locant. All other functional groups are demoted to prefixes. The A-Level priority order (highest to lowest) is:

Rank	Functional group	Suffix (when principal)	Prefix (when not principal)
1	Carboxylic acid (–COOH)	-oic acid	carboxy-
2	Ester (–COO–)	alkyl …-oate	alkoxycarbonyl-
3	Acid chloride / acyl chloride (–COCl)	-oyl chloride	chlorocarbonyl-
4	Amide (–CONH₂)	-amide	carbamoyl-
5	Nitrile (–C≡N)	-nitrile	cyano-
6	Aldehyde (–CHO)	-al	oxo-
7	Ketone (C=O)	-one	oxo-
8	Alcohol (–OH)	-ol	hydroxy-
9	Amine (–NH₂)	-amine	amino-
10	Alkene (C=C)	-ene	(rare as prefix)
11	Halogen (F, Cl, Br, I)	(never the suffix)	fluoro-, chloro-, bromo-, iodo-
12	Alkane backbone	-ane	(alkyl: methyl-, ethyl-, propyl-, …)

Halogens never act as the principal group — they are always named as prefixes regardless of which other groups are present. Alkenes and alkanes appear in the suffix slot only when nothing higher-ranking is present.

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The Five IUPAC Rules

Every organic name follows the same five-step recipe. Memorise this sequence; once internalised it removes ambiguity from every naming question.

Rule 1 — Find the longest carbon chain that contains the principal functional group

The longest continuous carbon chain that includes the principal functional group fixes the root (meth-, eth-, prop-, but-, pent-, hex-, hept-, oct-, non-, dec-). The chain need not be drawn as a straight line — count carbons in the skeletal structure, allowing the chain to zig-zag and turn corners.

If two chains of equal maximum length exist, choose the one with the greater number of substituents (this is "second-tier" rule that occasionally matters).

Rule 2 — Number the chain to give the principal group the lowest locant

Number the carbons from whichever end gives the principal functional group the lowest locant. If the principal group is symmetric or terminal (e.g. carboxylic acid, aldehyde, nitrile), it is automatically C-1 and the question of numbering is decided.

If the principal-group locant is the same from either direction (true of ketones, alcohols, alkenes in symmetric chains), break the tie by giving the lowest locant set to substituents at the first point of difference.

Rule 3 — Identify all substituents and name them as prefixes

Each substituent (alkyl branch, halogen, demoted functional group) is named and given the locant of the carbon to which it is attached. Common substituent names:

• Alkyl groups: methyl- (CH₃–), ethyl- (CH₃CH₂–), propyl- (CH₃CH₂CH₂–), 1-methylethyl- or isopropyl- ((CH₃)₂CH–), butyl-, etc.
• Halogens: fluoro-, chloro-, bromo-, iodo-.
• Demoted oxygen groups: hydroxy- (–OH), oxo- (C=O), carboxy- (–COOH).
• Demoted nitrogen groups: amino- (–NH₂), cyano- (–C≡N), nitro- (–NO₂).

Rule 4 — List substituents alphabetically with multiplying prefixes

Substituent prefixes are listed in alphabetical order in the final name, regardless of the order of their locants on the chain. When a substituent appears more than once, prefix it with di-, tri-, tetra-, penta- as appropriate. Multiplying prefixes (di-, tri-, tetra-) are ignored when alphabetising — "dimethyl" is filed under "m", not "d". So 3-ethyl-2,2-dimethylpentane lists ethyl before dimethyl (e before m, ignoring the "di-").

Punctuation matters: commas separate locants, hyphens separate locants from words, and there are no spaces within the name except where they are explicitly required (e.g. between the two words of "methyl ethanoate").

Rule 5 — Add stereochemistry markers

If the structure contains a stereocentre, indicate stereochemistry with the appropriate italicised prefix:

• (E)- or (Z)- for an alkene with different priority groups on each end of the C=C (geometric isomerism). Z = zusammen = highest-priority groups on the same side; E = entgegen = highest-priority groups on opposite sides.
• (R)- or (S)- for a chiral carbon with four different substituents (optical isomerism). R = rectus = priority falls clockwise when the lowest-priority group points away; S = sinister = anticlockwise.

The "priority" of substituents at a double bond or a chiral centre is set by the Cahn–Ingold–Prelog (CIP) priority rules: rank the four atoms bonded to the stereocentre by atomic number, with higher atomic number outranking lower. If two attached atoms are the same element, look at the next shell of atoms outward, and continue until a difference is found. Isotopes break ties at equal atomic number — higher mass number outranks lower (so ²H outranks ¹H).

The older cis- / trans- convention still appears in textbooks and exam mark schemes for simple alkenes such as cis-but-2-ene and trans-but-2-ene. AQA accepts cis/trans where it is unambiguous, but expects E/Z notation when the four substituents on the C=C are non-trivial. Always use E/Z for tri- or tetra-substituted alkenes.

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Structural Representations and Conversions

Four representations of an organic structure are routinely interchanged in A-Level questions, and you must be fluent in moving between them.

Molecular formula

Total count of each element in the molecule, with no positional information. Examples: C₄H₁₀, C₃H₈O, C₂H₄O₂. Useful for combustion calculations and degree-of-unsaturation analysis (degree of unsaturation = (2C + 2 + N − H − X) / 2, counting one degree per ring or π-bond) but tells you nothing about connectivity.

Empirical formula

Simplest whole-number ratio of atoms. C₆H₁₂O₆ (glucose) has empirical formula CH₂O. Routinely tested in combustion-analysis problems.

Condensed structural formula

Written linearly with parentheses for branches. Carbon and hydrogen counts are explicit but bonds are not drawn. Examples: CH₃CH(CH₃)CH₂OH (2-methylpropan-1-ol), (CH₃)₃CCl (2-chloro-2-methylpropane), CH₃COCH₂CH₃ (butan-2-one). Best for compact text.

Displayed (full structural) formula

Every atom, every bond is drawn. A "displayed" ethanol shows two carbons, six hydrogens, one oxygen, and all eight C–H/C–O/O–H bonds explicitly. Used when the question wants you to be unambiguous about all bonds — particularly common in reaction-mechanism questions where every bond change must be tracked.

Skeletal formula

The standard professional representation for medium and large molecules. Rules:

• Each line end and each line–line vertex represents a carbon atom.
• Hydrogens on carbon are implied — every carbon is assumed to carry as many hydrogens as needed to make four bonds total.
• Heteroatoms (O, N, S, halogen, P) and their attached hydrogens are drawn explicitly.
• Multiple bonds (C=C, C=O, C≡N) are shown as double or triple lines at the relevant position.

Conversions: when asked for a skeletal formula from a name, draw the longest chain as a zig-zag, mark the principal functional group at the correct numbered carbon, and add substituents at their numbered positions. When asked to name a skeletal structure, identify the longest chain that contains the principal group, number from the end closest to that group, and read off prefixes and locants.

Exam tip: When counting carbons in a skeletal formula, mark each vertex with a small dot or tick to avoid miscounting. Skeletal-structure misidentification is the single most common error in low-grade AS papers.

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Worked Examples — From Simple to Complex

Worked Example 1 — Pure alkane with one branch

Structure: CH₃CH(CH₃)CH₂CH₂CH₃

Working:

1. Longest chain: 5 carbons (pentane backbone). Numbering from the right: C-1 = right-hand CH₃, C-2 = CH₂, C-3 = CH₂, C-4 = CH(CH₃), C-5 = left-hand CH₃. The methyl substituent is on C-4. Numbering from the left would put the methyl on C-2 — lower, so use that direction.
2. Renumber: C-1 = left-hand CH₃, C-2 = CH(CH₃), … Methyl substituent locant = 2.
3. Name: 2-methylpentane.

Worked Example 2 — Alcohol with branching

Structure: CH₃CH(CH₃)CH₂CH₂OH

Working:

1. Principal group: –OH (alcohol, suffix -ol). Longest chain containing the –OH: 4 carbons (butan-1-ol backbone). Number from the OH end so that the OH locant is the lowest possible: C-1 = CH₂OH, C-2 = CH₂, C-3 = CH(CH₃), C-4 = CH₃.
2. Methyl substituent on C-3. OH on C-1.
3. Name: 3-methylbutan-1-ol.

Worked Example 3 — Carboxylic acid with gem-dimethyl substitution

Structure: (CH₃)₃CCH₂COOH

Working:

1. Principal group: –COOH (carboxylic acid). Suffix -oic acid. The carboxyl carbon is always C-1.
2. Longest chain containing the COOH: 4 carbons. C-1 = COOH, C-2 = CH₂, C-3 = C(CH₃)₂(CH₃) — three methyls all on C-3? No: the (CH₃)₃C grouping means there is one quaternary carbon with three methyls and one CH₂ linkage to COOH. So the longest chain is COOH–CH₂–C(CH₃)–CH₃ giving 4 carbons total (butanoic acid backbone), with two methyl substituents on C-3.
3. Name: 3,3-dimethylbutanoic acid.

Worked Example 4 — Compound with two functional groups, lower-priority demoted

Structure: HOCH₂CH₂CH₂COOH

Working:

1. Two functional groups present: –OH and –COOH. Carboxylic acid is higher priority (rank 1 vs rank 8). The –COOH supplies the suffix; the –OH becomes the prefix "hydroxy-".
2. Longest chain containing COOH: 4 carbons. COOH = C-1; CH₂ = C-2; CH₂ = C-3; CH₂OH = C-4.
3. Hydroxy substituent on C-4.
4. Name: 4-hydroxybutanoic acid (gamma-hydroxybutyric acid in the older Greek-letter nomenclature, used recreationally as GHB).

Worked Example 5 — Geometric isomer of an alkene (E/Z)

Structure: CH₃CH=CHCl, with the higher-priority groups on opposite ends.

Working:

1. Longest chain: 3 carbons (propene backbone). C-1 = CHCl, C-2 = CH, C-3 = CH₃ (numbering chosen so that the C=C has the lowest locant — both C-1=C-2 and C-2=C-3 are possible; the chloro substituent locks numbering at the Cl end).
2. C=C between C-1 and C-2. Chloro at C-1.
3. CIP priorities on C-1: Cl (atomic number 17) outranks H (1). On C-2: CH₃ (carbon) outranks H. If Cl and CH₃ are on opposite sides of the double bond, it is (E)-1-chloroprop-1-ene. If on the same side, (Z)-1-chloroprop-1-ene.

Worked Example 6 — Chiral compound (R/S)

Structure: 2-bromobutane, CH₃CHBrCH₂CH₃

Working:

1. Carbon-2 carries four different substituents: Br, CH₃, CH₂CH₃, and H. This is a chiral centre.
2. CIP priorities at C-2: Br (35, highest), CH₂CH₃ (carbon attached to another carbon, then H, H), CH₃ (carbon attached to three H), H (lowest).
3. To assign R or S: orient the molecule so that the lowest-priority group (H) points away from you. Trace a path from highest to next-highest to third-highest priority (Br → CH₂CH₃ → CH₃). If clockwise, it is (R)-; if anticlockwise, (S)-.
4. The two enantiomers are (R)-2-bromobutane and (S)-2-bromobutane — non-superimposable mirror images with otherwise identical properties.

Practical skill — identifying functional groups in a complex molecule: When presented with a structure such as ibuprofen (2-(4-isobutylphenyl)propanoic acid), proceed in three steps. First, scan for the highest-priority group: ibuprofen contains –COOH, which fixes the suffix as -propanoic acid. Second, scan for ring systems: ibuprofen contains a benzene ring (arene) bearing two substituents — the propanoic-acid arm at one position and an isobutyl arm at the para (1,4) position. Third, scan for any remaining sp²/sp³ junctions or heteroatoms: ibuprofen contains a chiral carbon (the carbon attached to COOH, CH₃, H, and the phenyl ring) — so it has R and S enantiomers (and the marketed drug is a racemic mixture, although only the S form is pharmacologically active). This three-pass approach (suffix-group first, ring systems second, residual features third) reliably identifies every functional group in any A-Level structure.

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Synoptic Links

• isomerism (next lesson, §3.3.1 continued): structural isomerism (chain, position, functional-group) and stereoisomerism (E/Z, optical) all build on the naming conventions and CIP priority rules introduced here. Naming and identifying isomers is impossible without nomenclature fluency.
• alkanes and alkenes (§3.3.2 / §3.3.4): the foundational hydrocarbon chemistry — combustion, free-radical substitution, electrophilic addition — relies on naming alkanes and alkenes precisely, and on identifying which carbon bears which substituent in a reaction mechanism.
• halogenoalkanes (§3.3.3): nucleophilic substitution and elimination reactions are taught with names like "2-bromobutane" or "2-chloro-2-methylpropane", and the SN1/SN2 distinction depends on whether the halide carbon is primary, secondary, or tertiary — a classification made possible by skeletal-structure analysis.
• organic-analysis-and-spectroscopy (§3.3.6 / §3.3.16, A2): mass spectrometry, infrared, and NMR spectra are converted to structures, which are then named. Every spectroscopy question on Paper 2 culminates in an IUPAC name.
• organic-synthesis-and-multistep (§3.3.16, A2): multistep synthetic routes are written as sequences of named intermediates. A weak grasp of nomenclature limits both AO1 retrieval and AO3 application across the whole A2 organic curriculum.

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Specimen Exam Question — modelled on the AQA A-Level Chemistry paper format

Question 1. [13 marks total]

(a) Give the systematic IUPAC name for each of the following three compounds. [4 marks]

(i) CH₃CH(CH₃)CH₂CH(OH)CH₃ [1]

(ii) CH₃CH₂COOCH₂CH₃ [1]

(iii) (CH₃)₂CHCH₂C(=O)CH₃ [2]

(b) Draw the displayed formula of 3-methylbut-2-enenitrile. [3 marks]

(c) The painkiller paracetamol has the structure HOC₆H₄NHCOCH₃ (where C₆H₄ denotes a 1,4-disubstituted benzene ring). Identify, naming each one, every functional group present in the molecule. [3 marks]

(d) A compound has the structure CH₃CH=CHCH₂CH₃ in which the two larger substituents lie on opposite sides of the C=C double bond. Apply the CIP priority rules to each end of the double bond, state which geometric isomer (E or Z) this corresponds to, and give the full IUPAC name. [3 marks]

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Mark Scheme — Specimen Question 1 (modelled, not a real paper)

(a) Three IUPAC names [4 marks, AO2]

(i) 4-methylpentan-2-ol [1 mark]. Longest chain = 5 carbons containing OH; OH on C-2 (lower than C-4); methyl on C-4. Accept "4-methyl-2-pentanol" (US convention) only if explicitly noted.

(ii) ethyl propanoate [1 mark]. Ester from propanoic acid and ethanol; alkyl group from the alcohol named first, then "alkanoate" from the acid.

(iii) 4-methylpentan-2-one [2 marks]: 1 mark for "pentan-2-one" backbone (longest chain = 5 C, C=O on C-2); 1 mark for the "4-methyl-" substituent prefix with correct locant.

(b) Displayed formula of 3-methylbut-2-enenitrile [3 marks, AO2]

• 1 mark: correct 4-carbon backbone with C≡N at one end (C-1).
• 1 mark: C=C double bond between C-2 and C-3.
• 1 mark: methyl substituent on C-3, plus all hydrogen atoms drawn correctly. Final structure: CH₃–C(CH₃)=CH–C≡N with every C–H bond explicit.

(c) Functional groups in paracetamol [3 marks, AO3]

• 1 mark: hydroxyl / phenol group (–OH) attached to the benzene ring (this is a phenol, not a simple alcohol — distinction not required at A-Level but acceptable for the mark).
• 1 mark: amide group (–NHCO–) linking the ring to the methyl group via N–C(=O).
• 1 mark: arene / aromatic ring (the benzene ring itself counts as a functional group; AQA accepts "benzene ring" or "aromatic ring"). Also accept identification of the methyl substituent as alkyl and the N–H as a secondary amide proton, but the three marks are awarded for phenol/–OH, amide, and arene.

(d) E/Z assignment and full name [3 marks, AO2 + AO3]

• 1 mark: apply CIP priorities to each end of the C=C. At C-2 (the left carbon of the C=C): substituents are CH₃ and H — CH₃ outranks H. At C-3 (the right carbon): substituents are CH₂CH₃ and H — CH₂CH₃ outranks H.
• 1 mark: the two higher-priority groups (CH₃ at C-2, CH₂CH₃ at C-3) lie on opposite sides of the C=C, therefore this is the E-isomer.
• 1 mark: full IUPAC name (E)-pent-2-ene.

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Grade-Band Model Answers (Question 1)

The three responses below cover the meaningful A-Level range: Grade C (the borderline-pass floor), Grade B (solid mark-scheme coverage), and Grade A* (top-band synthesis). No Grade D or E responses are shown — no A-Level student is aiming for those bands, and modelling failure adds nothing pedagogically. The editorial commentary after each response (not a real examiner report) names the marks earned and the specific moves that differentiate from adjacent bands.

Grade C response (~285 words)

(a) (i) The longest chain is 5 carbons with an OH; numbering from the OH end puts OH on C-2 and a methyl on C-4. Name: 4-methylpentan-2-ol. (ii) Ester from propanoic acid and ethanol; the alkyl from the alcohol comes first. Name: ethyl propanoate. (iii) Longest chain is 5 carbons with a ketone C=O on C-2; methyl on C-4. Name: 4-methylpentan-2-one.

(b) 3-methylbut-2-enenitrile: 4-carbon backbone with C≡N on C-1, C=C between C-2 and C-3, methyl on C-3. The displayed formula is H–C(CH₃)=C(CH₃)–C≡N — i.e. (CH₃)₂C=CH–CN with all C–H bonds drawn explicitly.

(c) Paracetamol HOC₆H₄NHCOCH₃ contains: a hydroxyl –OH on the benzene ring; an amide group –NHCO– linking the ring to the methyl carbon; and the benzene (aromatic) ring itself.

(d) For CH₃CH=CHCH₂CH₃ with the two larger groups on opposite sides: at C-2 the substituents are CH₃ (priority) and H; at C-3 the substituents are CH₂CH₃ (priority) and H. The two higher-priority groups lie on opposite sides of the double bond, so this is the E-isomer. Name: (E)-pent-2-ene.

Editorial commentary (Grade C): Correct answers throughout, awarded all 13 marks. The naming reasoning is competent but compressed; the structural drawing in (b) is described in words rather than properly displayed. To progress to B, the response should make the chain-numbering decisions explicit (why number from the OH end, not the methyl end?), and should distinguish a phenol –OH from an alcohol –OH in (c) since this affects acidity and reactivity in later topics.

Grade B response (~325 words)

(a) (i) Number the chain from the end nearest the principal functional group: the –OH on C-2 gives a lower locant set (2,4) than starting from the methyl end (which would give 2,4 the other way around — the OH still wins by being principal). Name: 4-methylpentan-2-ol. (ii) The ester linkage –COO– splits the molecule into an acyl part (from propanoic acid) and an alkyl part (from ethanol). The alkyl group is named first as a separate word; the acyl part takes the -oate suffix. Name: ethyl propanoate. (iii) Carbonyl C=O is internal (a ketone, not aldehyde). Longest chain through the C=O is 5 carbons; numbering from the carbonyl end puts C=O on C-2 and the isopropyl-derived methyl branch on C-4. Name: 4-methylpentan-2-one.

(b) 3-methylbut-2-enenitrile: 4-carbon chain with nitrile (–C≡N) at C-1. Double bond between C-2 and C-3. Methyl substituent on C-3. Displayed structure: CH₃–C(CH₃)=CH–C≡N with every C–H bond drawn out — i.e. (CH₃)₂C=CH–C≡N showing six C–H bonds (three on each methyl) and one alkenyl C–H.

(c) Paracetamol: hydroxyl group attached to the aromatic ring (this is technically a phenol, distinguished from a simple alcohol because phenols are weakly acidic; pKa ≈ 10); secondary amide linkage –NH–C(=O)– connecting the ring to the methyl group; and the benzene/aromatic ring itself.

(d) Apply CIP priority at each end of the C=C in CH₃CH=CHCH₂CH₃. At C-2 the substituents are H (priority 1) and CH₃ (priority 6 by atomic number C > H). At C-3 the substituents are H and CH₂CH₃ (CH₂CH₃ wins over H by atomic number). With the two higher-priority groups (CH₃ and CH₂CH₃) on opposite sides of the double bond, this is (E)-pent-2-ene.

Editorial commentary (Grade B): Now A-level-rigorous: chain-numbering choices are justified explicitly, phenol is distinguished from alcohol, CIP priority reasoning is sketched. To progress to A*, the response could expand the CIP logic — explaining what is meant by "compare atomic numbers at the first point of difference" — and link the E/Z assignment to the broader phenomenon of restricted rotation about C=C.

Grade A* response (~370 words)

(a) (i) Principal functional group is the –OH (alcohol, rank 8 in the priority list; no higher-ranking group present). Longest chain through the OH carbon is 5 (pent-); number from the end nearer the OH (puts OH on C-2 vs C-4 the other way), with the methyl substituent landing on C-4. Name: 4-methylpentan-2-ol.

(ii) Ester linkage –COO–. By IUPAC convention the ester is named as "alkyl alkanoate": the alkyl group derives from the alcohol parent (ethanol → ethyl), the alkanoate from the carboxylic acid parent (propanoic acid → propanoate). Name: ethyl propanoate.

(iii) Carbonyl C=O is non-terminal — ketone, suffix -one. Longest chain through C=O = 5 carbons; numbering from the carbonyl end gives C=O on C-2 (lowest locant for the principal group). Methyl branch on C-4. Name: 4-methylpentan-2-one.

(b) 3-methylbut-2-enenitrile: backbone CH₃–C(CH₃)=CH–C≡N. The nitrile carbon is C-1 (always terminal); C-2 carries one H and is doubly bonded to C-3; C-3 carries the methyl substituent and is bonded to C-4 (the second methyl carbon, here actually the substituent methyl since the longest chain runs C-1–C-2–C-3–C-4 where C-4 is the methyl listed at the end of the chain). Full displayed formula shows seven C–H bonds (three on each methyl plus one alkenyl C–H), one C=C, and one C≡N.

(c) Paracetamol contains: (i) a phenolic hydroxyl group (–OH bonded to an sp² aromatic carbon) — weakly acidic (pKₐ ≈ 9.4) due to resonance stabilisation of the phenolate anion across the ring; (ii) a secondary amide group (–NHCO–), planar at the N because of partial C=N double-bond character (resonance with the carbonyl); (iii) the aromatic (benzene) ring itself, 1,4-disubstituted ("para-substituted"). The amide and phenol both interact with hepatic enzymes during paracetamol metabolism (via glucuronidation, sulfation, and CYP-mediated oxidation to NAPQI).

(d) CIP priorities. At C-2: substituents are CH₃ (atomic number 6 at first shell) and H (1) — CH₃ wins. At C-3: substituents are CH₂CH₃ (atomic number 6 at first shell, but on the second shell C > H so the ethyl wins by extension) and H — CH₂CH₃ wins. With the two higher-priority groups on opposite sides of the C=C ("entgegen"), the assignment is (E)-. Restricted rotation about the C=C bond (the π-overlap costs ~250 kJ mol⁻¹ to break) locks the geometry — unlike single C–C bonds, which rotate freely at room temperature. Full IUPAC name: (E)-pent-2-ene.

Editorial commentary (Grade A):* Genuinely A*: explicit application of the priority rules, distinction between phenol and alcohol, acknowledgement of CIP "first point of difference" logic, and a synoptic insight (the energetic basis for E/Z stereochemistry — restricted rotation about C=C). The response also previews real-world relevance (paracetamol metabolism) without losing focus on the technical naming. This is exactly what top-band examiners reward: technically correct, contextually rich, and clearly reasoned.

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Common Errors

• Numbering the chain from the wrong end. Always start from the end that gives the principal functional group the lowest locant; only after that is fixed do you worry about substituent locants. A surprisingly large number of marks are lost to "2-methylbutan-3-ol" (wrong) instead of "3-methylbutan-2-ol" (right).
• Missing a longer chain that turns a corner. The longest chain need not be drawn horizontally. Always trace every possible path through the carbon skeleton and pick the longest one that contains the principal functional group.
• Forgetting that halogens are always prefixes. A halogen never supplies the suffix, even when it is the only "non-carbon" feature of the molecule. CH₃CH₂Br is "bromoethane", not "ethanobromide".
• Confusing aldehydes and ketones. Aldehyde C=O is terminal (–CHO) with one H on the carbonyl carbon; ketone C=O is internal with two carbons attached. Same general formula CₙH₂ₙO — different chemistry.
• Alphabetising "di-", "tri-", and "tetra-" prefixes incorrectly. Multiplying prefixes are ignored when alphabetising substituents. "Dimethyl" files under "m", "diethyl" under "e", and so on.
• Forgetting the space in ester names. "Methyl ethanoate" has a space between the two words; "methylethanoate" (one word) is wrong.
• Drawing displayed formulae with missing hydrogens. When a question asks for a displayed (full structural) formula, every bond and every atom — including every hydrogen — must be shown explicitly. Skeletal-style shortcuts lose marks.

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A-Level Depth Misconceptions

• "Functional-group priority" is a naming convention, not a measure of chemical reactivity. The priority list dictates which group supplies the suffix; it does not predict which group will react fastest. For example, an amide is higher in the priority list than an alcohol, but an alcohol is far more reactive toward most reagents than an amide.
• CIP priorities are determined by atomic number, not by the size of the substituent. A heavy halogen (Br, atomic number 35) outranks a much larger alkyl group (–C(CH₃)₃, all carbons of atomic number 6). Many students wrongly assume that "bigger" = "higher priority".
• Skeletal formulae assume tetrahedral sp³ carbons unless C=C or C≡C is drawn. The zig-zag pattern is not arbitrary — it reflects the ~109.5° tetrahedral angle. Drawing a skeletal compound as a straight line obscures stereochemistry.
• The "longest chain" rule has tie-breaks beyond chain length. When two equal-length chains exist, the second-tier rules pick the chain with the most substituents (giving a simpler, more economical name). This is rarely tested at A-Level but matters in IUPAC 2013 recommendations.
• E/Z is not synonymous with cis/trans. For 1,2-disubstituted alkenes the two systems agree, but for tri- or tetra-substituted alkenes only CIP-based E/Z is unambiguous. Always default to E/Z at A-Level.

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Going Further

Three undergraduate-adjacent extensions:

• IUPAC 2013 recommendations. The 2013 revision tightened several conventions (locant positioning inside the suffix, stereodescriptor placement, treatment of bridged ring systems) and is the modern reference for chemistry research papers and most graduate-level texts. Reading: IUPAC, Nomenclature of Organic Chemistry: Recommendations and Preferred Names 2013 (Royal Society of Chemistry, 2014).
• SMILES and InChI representations. Computational chemistry replaces drawn structures with linear text strings: SMILES (Simplified Molecular-Input Line-Entry System) writes ibuprofen as CC(C)Cc1ccc(cc1)C(C)C(=O)O; InChI generates a unique algorithmic hash that databases use as a primary key. Every drug listed in DrugBank or PubChem has both a SMILES and an InChI representation in its record.
• Hantzsch–Widman nomenclature for heterocycles. A separate IUPAC system names ring systems containing nitrogen, oxygen, sulfur, or other heteroatoms, using prefixes (oxa-, aza-, thia-, etc.) and stem fragments (-irane, -etane, -olane, -inane) to encode ring size and saturation. Pyridine, furan, and thiophene are everyday heterocycles encountered in pharmaceuticals; the Hantzsch–Widman system is needed for systematic names of less common rings such as oxetane (4-membered ether) and thiirane (3-membered sulfide).

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This lesson establishes the IUPAC framework — homologous series, functional-group priority, the five naming rules, and the four structural representations — on which every subsequent organic-chemistry topic on this course depends. Every reaction, every mechanism, every spectroscopic analysis, and every synthesis question you encounter at AS and A2 will begin with a correctly named compound. Internalise this material now and the rest of the organic curriculum becomes a chain of small, well-defined steps; skimp on it and every later lesson becomes harder than it needs to be.