S3.2 Part A Functional groups: classification of organic compounds — SL core

Nine teaching hours at both levels, outcomes 3.2.1 to 3.2.6. The additional higher level outcomes, 3.2.7 to 3.2.12, are in Part B.

Guiding question: How does the classification of organic molecules help us to predict their properties?

Structure 3.2 · Representing and classifying organic compounds

1Six ways of writing one molecule 3.2.1 SL + HL Repeated

The syllabus statement: Organic compounds can be represented by different types of formulas. These include empirical, molecular, structural (full and condensed), stereochemical and skeletal. The skill: identify different formulas and interconvert molecular, skeletal and structural formulas.

Carbon forms more compounds than all the other elements combined. Three features of the carbon atom make this possible: it forms four strong covalent bonds; it bonds strongly to other carbon atoms, producing chains and rings of almost any length (catenation); and it forms stable single, double and triple bonds, as well as strong bonds to hydrogen, oxygen, nitrogen and the halogens (Structure 2.2). Because so many compounds are possible, and because many share the same molecular formula, organic chemistry needs several kinds of formula, each showing a different amount of information.

**Table 3.8** The types of formula, illustrated for butan-1-ol.
Formula	What it shows	Butan-1-ol
Empirical	Simplest whole-number ratio of atoms of each element	C₄H₁₀O
Molecular	Actual number of atoms of each element in one molecule	C₄H₁₀O
Structural (condensed)	The arrangement of atoms, written in sequence without showing every bond	CH₃CH₂CH₂CH₂OH
Structural (full, or displayed)	Every atom and every bond, drawn in two dimensions	all 14 bonds drawn, including each C–H and the O–H
Skeletal	A zig-zag of carbon–carbon bonds; carbon atoms at corners and ends; H atoms on carbon omitted; other atoms (and H on them) shown	a four-carbon zig-zag with OH at one end
Stereochemical	The three-dimensional arrangement, using wedges (towards the viewer) and dashes (away)	not expected to be drawn except where specified (Part B)

The empirical and molecular formulas can be the same (butan-1-ol, C₄H₁₀O) or different (ethane, molecular C₂H₆, empirical CH₃; glucose, C₆H₁₂O₆, empirical CH₂O). Neither shows the structure, so neither can distinguish between isomers. In a condensed structural formula, groups can be combined: (CH₃)₂CHOH is propan-2-ol, and brackets around a repeated group, CH₃(CH₂)₄CH₃, mean the unit is repeated.

Figure 3.5 Propan-1-ol written five ways. The syllabus counts full and condensed structural formulas as two different kinds, and a published mark was lost for offering one where the other was asked for.

How to think · converting between formulas

Skeletal → molecular. Count a carbon at every corner and every line end. Give each carbon enough hydrogens to make four bonds in total. Then add the atoms that are drawn.

Structural → skeletal. Draw the longest chain as a zig-zag, attach branches at the correct corners, then draw every heteroatom (O, N, halogen) with any hydrogens attached to it. Double bonds are drawn as double lines.

AnimationTypes of formula

Click a compound to see its empirical, molecular, structural, displayed and skeletal formulas.

AnimationDisplayed and structural formulas

Match each displayed formula to its condensed structural formula.

Worked example 3.8 · from a skeletal formula to three other formulas

Structure	A skeletal formula shows a five-carbon zig-zag with a C=O at the second carbon (pentan-2-one).
Molecular formula	5 C. H: C1 has 3, C3 and C4 have 2 each, C5 has 3, C2 (carbonyl) has none: 10 H. One O. C₅H₁₀O.
Empirical formula	5 : 10 : 1 has no common factor: C₅H₁₀O.
Condensed structural	CH₃COCH₂CH₂CH₃

Exam focus · what the published papers show

Common trap · giving the wrong kind of formula

The structure can be right and the mark still lost. Feedback: 60% of the candidates drew the full structural formula of the carboxylic acid isomer of methyl methanoate. Some candidates lost the mark for giving a condensed structural formula. The examining team accepted OH without showing the bond.

Read the adjective. Full means every bond. Condensed means groups. Skeletal means no hydrogens on carbon. Skeletal formulas also make questions harder to read: on a naming item, feedback records that a key factor in not getting the second name correct was the use of skeletal formula, and a published difficulty list carries Naming organic compounds from skeletal formulae.

Worked example · two formulas read off one structure

The structures of three acid–base indicators are printed. Determine the molecular and empirical formulas of phenolphthalein. [2]

Count the atoms from the drawing — that is the molecular formula, C₂₀H₁₄O₄. Then divide through by the highest common factor of the subscripts. The HCF of 20, 14 and 4 is 2:

C₂₀H₁₄O₄ ÷ 2 → C₁₀H₇O₂ the published answer

Notice that the empirical formula has seven hydrogens — an odd number. No molecule can have that composition with normal valences. That is not an error; it is the whole point of an empirical formula. It is a ratio, not a substance.

Exam alert · reading a structure is a reliable mark

Where candidates are asked to read rather than convert, the reports are positive: Almost 90% of students could determine the geometry around specified atoms, and their hybridization, from the skeletal formula of an organic molecule, and About 75% of the candidates correctly deduced the numbers of sigma and pi bonds present from the structural formula of a compound. Reading structures is a strength. Converting between kinds of formula is not.

2Functional groups 3.2.2 SL + HL Strong

The syllabus statement: Functional groups give characteristic physical and chemical properties to a compound. Organic compounds are divided into classes according to the functional groups present in their molecules. The skill names all nine explicitly: identify the following functional groups by name and structure: halogeno, hydroxyl, carbonyl, carboxyl, alkoxy, amino, amido, ester, phenyl. It adds that the terms “saturated” and “unsaturated” should be included.

A functional group is the atom or group of atoms in a molecule that is responsible for its characteristic chemical reactions. The rest of the molecule — the hydrocarbon chain — is comparatively unreactive, so compounds with the same functional group react in the same way whatever the length of their chain. Butan-1-ol and ethanol are both oxidized by acidified potassium dichromate(VI) because both contain a hydroxyl group on a carbon with two hydrogens. The functional group therefore defines the class of compound and allows its reactions to be predicted.

The functional group also shapes physical properties. Polar groups such as hydroxyl and carboxyl allow hydrogen bonding, raising boiling points and giving solubility in water; a halogeno group introduces a permanent dipole; a hydrocarbon chain contributes only London dispersion forces.

Definition · saturated and unsaturated

A saturated compound contains only single carbon–carbon bonds (alkanes, and the chains of most alcohols and halogenoalkanes). An unsaturated compound contains at least one carbon–carbon double or triple bond (alkenes, alkynes). Unsaturated compounds decolourize bromine water because the halogen adds across the multiple bond.

Definition · the nine groups, and the classes they create

Group (the name asked for)	Structure	Class of compound it makes
halogeno	–F, –Cl, –Br, –I	halogenoalkane
hydroxyl	–OH	alcohol
carbonyl	C=O	aldehyde (at the chain end) or ketone (within the chain)
carboxyl	–COOH	carboxylic acid
alkoxy	–OR	ether
amino	–NH₂	amine
amido	–CONH–	amide
ester	–COO–	ester
phenyl	–C₆H₅	an arene substituent

One group can give two classes — carbonyl gives both aldehydes and ketones — which is exactly why the group name is what a scheme asks for. And saturated means only single carbon–carbon bonds; unsaturated means at least one double or triple bond.

Figure 3.6 The nine names the syllabus asks for, against the names candidates offer instead. Every word in the right-hand panel was listed by one published report as an incorrect answer to a single two-mark question.

The phenyl group, C₆H₅–, is a benzene ring with one hydrogen replaced. Its six carbon atoms are joined by delocalized bonding (Structure 2.2), so it is not an alkene and does not undergo addition reactions readily. The amido group, –CONH–, links a carbonyl carbon to a nitrogen; it forms the peptide link between amino acids in proteins.

Worked example 3.9 · finding every group in a larger molecule

Lactic acid, CH₃CH(OH)COOH	–OH on a chain carbon: hydroxyl. –COOH: carboxyl (not “carbonyl + hydroxyl”, which are pieces of it).
Glycinamide, H₂NCH₂CONH₂	–NH₂ on a chain carbon: amino. –CONH₂: amido — the NH₂ attached to the C=O is part of the amido group, not a second amine.
Aspirin	–COOH: carboxyl; –COO– linking the ring to CH₃CO: ester; the ring: phenyl.

AnimationFunctional groups and classes

Match each general structure to the class of compound it defines.

Exam focus · what the published papers show

Common trap · naming the class when the group was asked for

This is the most sharply documented error in the topic, because a question, its mark scheme and its examiner report all say the same thing. The question: State the name of the functional group present in X. [1]. The scheme: carbonyl, with the note Do not accept aldehyde. The report: a majority gave “aldehyde” as the functional group, instead of carbonyl.

A majority gave the refused answer. And on a different item, asked for two functional groups in aspirin, a report lists the wrong answers in full: Many candidates listed alcohol, aldehyde, carboxylic acid, benzene, ether, ketone, and alkene, which are incorrect. Please refer to the syllabus for the correct names of functional groups. Seven words, not one of them among the nine.

Worked example · class and group, in one question

Ethanol and propan-1-ol are members of a homologous series. State the names of the class of compound and the functional group of this series. [2]

Two marks, one for each — and the question has separated them deliberately. The class is alcohol; the group is hydroxyl. Feedback on a comparable item: the identification of organic class and functional group-alcohol/hydroxyl was quite well answered but some had hydroxide as a response and others inverted the answers. Inverting them scores nothing, and “hydroxide” is an ion, not a group. Elsewhere: Generally, candidates were well-versed in the class and functional group names; however, some did not understand the difference or answered with formulas.

Common trap · reading a fragment out of its group

A functional group has to be taken in whole. Feedback on a harder item: only just over 40% of candidates being able to deduce the functional groups present from the given structure of aspartame. The majority thought that the -NH- present was a secondary amine, not noticing it was part of the -CO-NH- amide linkage.

A carboxyl contains a hydroxyl and a carbonyl inside it; an amide contains an N–H. Look at what a fragment is attached to before you name it. The same pattern shows in the wrong answers elsewhere: 46% of the candidates identified the functional group as carboxyl. Common incorrect answers were ketone, aldehyde and hydroxyl. Every one of those three is a piece of a carboxyl group — which is exactly why they are tempting, and exactly why they are wrong.

3Homologous series 3.2.3 SL + HL Repeated

The syllabus statement: A homologous series is a family of compounds in which successive members differ by a common structural unit, typically CH2. Each homologous series can be described by a general formula. Twelve series are named: alkanes, alkenes, alkynes, halogenoalkanes, alcohols, aldehydes, ketones, carboxylic acids, ethers, amines, amides and esters.

A homologous series is a family of compounds with the same functional group in which successive members differ by a common structural unit, usually –CH₂–. Members of a series share a general formula, show similar chemical properties (because they have the same functional group) and show a gradual trend in physical properties as the chain lengthens. Each series can be written with R standing for the alkyl chain: R–OH (alcohols), R–CHO (aldehydes), R–COOH (carboxylic acids).

**Table 3.9** The twelve homologous series named in the syllabus.
Series	Functional group	General formula	First member (name, formula)	Suffix / prefix
Alkanes	— (C–C, C–H only)	C_nH_2n+2	methane, CH₄	-ane
Alkenes	C=C	C_nH_2n	ethene, C₂H₄	-ene
Alkynes	C≡C	C_nH_2n−2	ethyne, C₂H₂	-yne
Halogenoalkanes	halogeno, –X	C_nH_2n+1X	chloromethane, CH₃Cl	chloro-, bromo-
Alcohols	hydroxyl, –OH	C_nH_2n+1OH	methanol, CH₃OH	-ol
Ethers	alkoxy, –O–R	C_nH_2n+2O	methoxymethane, CH₃OCH₃	alkoxy-
Aldehydes	carbonyl at chain end, –CHO	C_nH_2nO	methanal, HCHO	-al
Ketones	carbonyl within the chain, C=O	C_nH_2nO (n ≥ 3)	propanone, CH₃COCH₃	-one
Carboxylic acids	carboxyl, –COOH	C_nH_2nO₂	methanoic acid, HCOOH	-oic acid
Esters	ester, –COO–	C_nH_2nO₂	methyl methanoate, HCOOCH₃	-oate
Amines	amino, –NH₂	C_nH_2n+1NH₂	methanamine, CH₃NH₂	-amine
Amides	amido, –CONH₂	C_nH_2n+1NO	methanamide, HCONH₂	-amide

Three pairs of series share a general formula: alcohols and ethers; aldehydes and ketones; carboxylic acids and esters. A general formula alone therefore never identifies a series, and members of the paired series with the same n are functional group isomers of each other (section 6).

AnimationFunctional groups and homologous series

Select each series to see its functional group, general formula and example structures.

Worked example 3.10 · using a general formula

Question	Deduce the molecular formula of the alkene with six carbon atoms and of the next member of the alcohol series after butan-1-ol.
Alkene, n = 6	C_nH_2n → C₆H₁₂ (hexene)
Next alcohol	Butan-1-ol is C₄H₉OH; add CH₂: C₅H₁₁OH (pentan-1-ol)

Exam focus · what the published papers show

Definition · three characteristics, and you need at least two

A published report sets out exactly what a definition may draw on, listing the details (same family/functional group, same general formula and successive members differ by CH2 unit).

Series	General formula	Series
alkanes	C_nH_2n+2
alkenes	C_nH_2n	one degree of unsaturation
alkynes	C_nH_2n−2	two
alcohols · ethers	C_nH_2n+2O	shared
aldehydes · ketones	C_nH_2nO	shared
carboxylic acids · esters	C_nH_2nO₂	shared

Three pairs of series share a general formula. That is not a coincidence — it is the basis of functional group isomerism, and it means a general formula alone can never identify a series.

Common trap · one characteristic is not a definition

Feedback on a definition question: About 40% of the candidates scored this mark. Some candidates were too brief in their answers only stating that homologous series are made up of similar compounds without giving two specific details required for the mark. And on another: About half the candidates gained this mark, with most of the remainder failing to because they only mentioned one characteristic of a homologous series.

Two of the three, every time. Published difficulty lists carry Outlining what is meant by the term homologous series and Defining a homologous series. One report also records the unit itself being misremembered — candidates wrote that subsequent member differ by a CH or CH3 group. It is CH₂.

Exam alert · what gets asked

Short and direct. A multiple-choice item asks Which compounds belong to the same homologous series?, and feedback on a similar one records 61% of the candidates answered the multiple completion question about propanone and butanone that belong to the same homologous series correctly. Another report is simply Good performance on this question identifying a homologous series. A structured task asks State the general formula for the homologous series of alkenes. [1] — one mark for C_nH_2n. And one instruction combines this outcome with the next but one: Draw an isomer of X which belongs to a different homologous series. [1]

Past-paper practice · Practice set S · Formulas, functional groups and homologous series

Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.

S1IB · May 2022 · SL Paper 1 · TZ1 · Q3 · [1]

What is the molecular formula of a compound with an empirical formula of CHO₂ and a relative molecular mass of 90?

A. CHO₂B. C₂H₂O₄C. C₃H₆O₃D. C₄H₁₀O₂

S2IB · May 2022 · SL Paper 1 · TZ1 · Q24 · [1]

A. nitro, carbonyl and carboxylB. amino, hydroxyl and carbonylC. nitro, carboxyl and hydroxylD. amino, carboxyl and hydroxyl

S3IB · May 2017 · SL Paper 1 · TZ2 · Q24 · [1]

A. CarboxylB. AminoC. NitrileD. Hydroxyl

S4IB · November 2016 · SL Paper 1 · Q23 · [1]

A. Hydroxyl and esterB. Hydroxide and etherC. Hydroxyl and etherD. Hydroxide and ester

S5IB · May 2022 · SL Paper 1 · TZ1 · Q25 · [1]

Which compounds are members of the same homologous series?

A. propanal, propanone, propanoic acidB. propane, propene, propyneC. hexan-1-ol, hexan-2-ol, hexan-3-olD. ethanol, propan-1-ol, butan-1-ol

S6IB · November 2022 · SL Paper 1 · Q25 · [1]

Which homologous series has the general formula C_nH_2nO (n > 2)?

A. AlcoholsB. Carboxylic acidsC. EthersD. Ketones

S7IB · May 2022 · SL Paper 1 · TZ2 · Q26 · [1]

Which is a homologous series?

A. C₂H₄, C₃H₅, C₄H₆B. C₂H₂, C₃H₄, C₄H₆C. C₂H₂, C₂H₄, C₂H₆D. C₂H₂, C₄H₄, C₆H₆

S8IB · May 2023 · SL Paper 1 · TZ1 · Q24 · [1]

Which formula represents an ether?

A. C₆H₅OHB. CH₃CHOC. CH₃COCH₃D. CH₃OCH₃

S9IB · May 2022 · SL Paper 2 · TZ2 · Q4(b)(i) · [2]

State two features showing that propane and butane are members of the same homologous series. [2]

S10IB · May 2021 · SL Paper 2 · TZ1 · Q5(a) · [2]

Ethanol is obtained by the hydration of ethene, C₂H₄.

(a)(i) State the class of compound to which ethene belongs. [1]
(ii) State the molecular formula of the next member of the homologous series to which ethene belongs. [1]

S11IB · November 2016 · SL Paper 2 · Q5(a) · [1]

Propane and propene are members of different homologous series. Draw the full structural formulas of propane and propene. [1]

Solutions and mark-scheme guidance · Set S

S1 B

M(CHO₂) = 12.01 + 1.01 + 32.00 = 45.02; 90/45.02 = 2, so the molecular formula is (CHO₂)₂ = C₂H₂O₄ (ethanedioic acid).

S2 D

–NH₂ is amino, –OH on CH₂ is hydroxyl, –COOH is carboxyl. There is no nitro group, and the C=O belongs to the carboxyl group rather than being a separate carbonyl.

S3 D

The –OH on the ring is a hydroxyl group. The –NH–C(=O)– is an amido group, not an amino group, and there is no carboxyl or nitrile.

S4 C

–OH is hydroxyl; the O bridging two carbons (and CH₃O–) is an ether (alkoxy) linkage. There is no C=O, so no ester; “hydroxide” is an ion, not a functional group.

S5 D

Ethanol, propan-1-ol and butan-1-ol share the hydroxyl group and differ by CH₂. A has three different groups; B three different series; C three position isomers with the same formula.

S6 D

Aldehydes and ketones are C_nH_2nO; alcohols and ethers C_nH_2n+2O; carboxylic acids C_nH_2nO₂. The condition n > 2 points to ketones, whose smallest member has three carbons.

S7 B

C₂H₂, C₃H₄, C₄H₆ fit C_nH_2n−2 (alkynes) and differ by CH₂. C₃H₅ in A is not a stable molecule; C and D do not differ by CH₂.

S8 D

An ether has an oxygen between two carbon atoms, C–O–C: methoxymethane. A is phenol (hydroxyl), B an aldehyde, C a ketone.

S9 [2]

Same general formula / C_nH_2n+2 ✔; differ by CH₂ / a common structural unit ✔. The scheme also accepts “similar chemical properties” and “gradation / gradual change in physical properties”.

S10 [2]

(i) alkene ✔ (ii) C₃H₆ ✔ (a structural formula is also accepted).

S11 [1]

Propane: CH₃–CH₂–CH₃ with all eight C–H bonds drawn; propene: CH₂=CH–CH₃ with the C=C and all six C–H bonds drawn. Both structures are needed for the mark. “Full” means every bond is shown.

4Trends in physical properties 3.2.4 SL + HL Repeated

The syllabus statement: Successive members of a homologous series show a trend in physical properties. The skill: describe and explain the trend in melting and boiling points of members of a homologous series.

Along a homologous series the functional group stays the same while the chain grows by one CH₂ unit at a time. Each added CH₂ adds eight electrons to the molecule and increases its surface area, so the London (dispersion) forces between neighbouring molecules become stronger. More energy is needed to separate the molecules, and melting and boiling points rise. The increase is largest between the first members and becomes gradually smaller, because each additional CH₂ is a smaller fraction of the whole molecule.

**Table 3.10** Boiling points of the first six straight-chain alkanes and primary alcohols (reference data, °C at 1 atm).
n	1	2	3	4	5	6
Alkane, C_nH_2n+2	−162	−89	−42	−1	36	69
Alcohol, C_nH_2n+1OH	65	78	97	117	138	157

The two series show the same trend, but the alcohols boil much higher than alkanes of similar size because their hydroxyl groups form hydrogen bonds between molecules in addition to London forces. Along the alcohol series the hydrogen bonding stays approximately constant — there is one –OH per molecule in every member — so it is the growing London forces that cause the trend. Physical states follow: the first four alkanes are gases at room temperature, pentane to about C₁₇ are liquids, and longer chains are waxy solids.

Branching lowers the boiling point of isomers. A branched molecule is more compact, closer to spherical, and has a smaller surface area in contact with its neighbours, so the London forces between molecules are weaker. Pentane boils at 36 °C, 2-methylbutane at 28 °C and 2,2-dimethylpropane at 10 °C.

Figure 3.7 The distinction this outcome turns on. Within a series only the dispersion forces change, so they explain the trend. Between series the functional group changes, so hydrogen bonding explains the difference.

Interactive model — S3.2.3 · S3.2.4

S3.2

Homologous series — structure and boiling point

Choose a series and a chain length. The model writes the molecular and condensed formulas and the IUPAC name, draws the skeletal formula, and plots the boiling points of the series against the number of carbon atoms, with the alkanes as a baseline.

Series

Carbon atoms 3

Boiling points are standard reference values for the straight-chain compounds (°C at 1 atm), not data-booklet values.

AnimationTrends in boiling points

Plot the boiling points of the alkanes against number of carbon atoms and describe the curve.

AnimationBoiling points of carboxylic acids

Place the carboxylic acids on the graph, then compare with alcohols of similar molar mass.

Worked example 3.11 · explaining a comparison, not just a trend

Question	Explain why propan-1-ol (M = 60) boils at 97 °C but butane (M = 58) boils at −1 °C.
Same force	The molecules have a similar number of electrons, so their London (dispersion) forces are similar.
Different force	Propan-1-ol has an –OH group and forms hydrogen bonds between molecules; butane cannot.
Consequence	Hydrogen bonds are stronger than London forces, so more energy is needed to separate propan-1-ol molecules: a higher boiling point.

Exam focus · what the published papers show

How to think · only a force that changes can explain a change

Go along the alcohols: methanol, ethanol, propan-1-ol. Every one has exactly one –OH, so every one hydrogen bonds to the same extent. Hydrogen bonding is constant along the series — it cannot explain a rising boiling point.

What does change is the number of electrons. Each added CH₂ makes the molecule larger and more polarisable, so the London dispersion forces grow, and more energy is needed to separate the molecules. That is the trend, and dispersion is its cause.

Common trap · reaching for hydrogen bonding

Feedback, in one sentence: 43% of candidates could explain the increase in the boiling point in the homologous series of alcohols as due to increasing London dispersion forces. However, many incorrectly thought it was related to Hydrogen bonding.

Hydrogen bonding is not a wrong idea — it is the right answer to a different question. Asked why an alcohol boils higher than an alkane of similar size, hydrogen bonding is the answer. Asked why propan-1-ol boils higher than ethanol, it is not. A published mark scheme shows the shape of a good comparison, crediting both have «similar» London/dispersion forces «due to having same number of electrons/similar Mr» and then ethenone has dipole-dipole forces «and carbon dioxide does not» — say which force is the same, then which one differs.

Exam alert · branching runs the other way

Feedback: Over 45% of the students correctly remembered that the boiling point of alkanes increases with chain length and decreases with chain branching, with many opting for the opposite effect of branching. A branched molecule is more compact and closer to spherical, so neighbouring molecules touch over a smaller area, the dispersion forces between them are weaker, and it boils lower. Chain length up, boiling point up. Branching up, boiling point down. Solubility belongs here too — a published difficulty list carries Outlining why ethanal and methanoic acid are both fully soluble in water, which is answered by the same reasoning applied to hydrogen bonding with water rather than with itself.

5IUPAC nomenclature 3.2.5 SL + HL Strong

The syllabus statement: “IUPAC nomenclature” refers to a set of rules used by the International Union of Pure and Applied Chemistry to apply systematic names to organic and inorganic compounds. The scope is precise: saturated or mono-unsaturated compounds that have up to six carbon atoms in the parent chain and contain one type of the following functional groups: halogeno, hydroxyl, carbonyl, carboxyl, and it must include straight-chain and branched-chain isomers.

The International Union of Pure and Applied Chemistry (IUPAC) system gives every compound one unambiguous name from which its structure can be drawn. A name is built from three parts: a stem that counts the carbon atoms in the longest chain containing the functional group, a suffix that names the principal functional group, and prefixes that name and locate substituents.

**Table 3.11** Stems and the functional-group endings within the syllabus scope.
Carbon atoms	1	2	3	4	5	6
Stem	meth-	eth-	prop-	but-	pent-	hex-

Group	Named as	Example
Alkane / alkyl branch	-ane; branch as methyl-, ethyl-	2-methylbutane
C=C	-ene, with locant	but-2-ene, CH₃CH=CHCH₃
Halogeno	prefix fluoro-, chloro-, bromo-, iodo-	2-bromopropane
Hydroxyl	-ol, with locant	propan-2-ol
Carbonyl (aldehyde)	-al; always C1, no locant	butanal, CH₃CH₂CH₂CHO
Carbonyl (ketone)	-one, with locant when needed	pentan-3-one
Carboxyl	-oic acid; always C1	2-methylpropanoic acid

Definition · the naming procedure

Find the longest continuous carbon chain that contains the functional group. Branches may be drawn in any direction; follow the bonds, not the page.
Number the chain from the end that gives the functional group (or the C=C) the lowest locant. If there is no functional group suffix, number to give substituents the lowest locants.
Name the substituents with their locants, in alphabetical order; use di-, tri-, tetra- for repeats, with a locant for every one (2,2-dimethyl…). Multiplying prefixes are ignored when alphabetizing.
Assemble the name: numbers are separated by commas, numbers and letters by hyphens, for example 2,3-dimethylbutan-2-ol.

AnimationNaming branched-chain alkanes

Play to see the longest chain found, numbered and named step by step.

AnimationNaming the alkanes

Type the IUPAC name of each structure shown.

Worked example 3.12 · four names, four decisions

CH₃CH(CH₃)CH₂OH	Longest chain containing C–OH: 3 C. Number from the OH end: OH on C1, methyl on C2. 2-methylpropan-1-ol.
CH₃CHBrCH(CH₃)CH₂CH₃	Longest chain 5 C (the CH₂CH₃ is part of the main chain). Numbering from the bromo end gives 2, 3; from the other end 3, 4. 2-bromo-3-methylpentane (b before m).
(CH₃)₂C=CHCH₃	4 C chain containing the C=C; numbering gives the double bond locant 2; methyl on C2. 2-methylbut-2-ene.
CH₃CH₂CH(CH₃)COOH	Carboxyl carbon is C1; chain of 4; methyl on C2. 2-methylbutanoic acid.

AnimationStructures of the alkenes

Select each name to see its structure, and notice how the locant fixes the position of the double bond.

AnimationNaming aldehydes and ketones

Type the name of each carbonyl compound.

AnimationNaming carboxylic acids

Type the name of each acid, remembering that the carboxyl carbon is always C1.

Figure 3.8 A name read backwards. Each syllable of ethenone names one structural feature, and a published scheme awards only one of its two marks if the features are given without being attached to their syllables.

Exam focus · what the published papers show

Worked example · nomenclature run backwards

Ethenone, CH₂CO, is used in the synthesis of pharmaceutical compounds. Suggest why the compound is given this IUPAC name. [2]

Take the name apart. The scheme accepts any two of three:

eth	contains two carbon atoms
en	contains a carbon-carbon double bond / C=C
one	contains a carbonyl / C=O

The notes are generous about wording — Accept “alkene” for C=C and “ketone” for C=O — and strict about structure:

Award [1 max] if two structural features given without relating to the relevant part of IUPAC name.

Listing the features is one mark. Attaching each to its syllable is the second. Feedback confirms it exactly: About a third of the candidates scored the full mark relating two structural features to the relevant parts of the IUPAC name. Others simply stated two structural features without identifying the relevant part of the IUPAC name, scoring one of the two marks. And on a similar item: Candidates often identified functional groups but failed to link them clearly to the IUPAC structuring of the name.

Marking language · what is forgiven when you build a name

One scheme's answer is 3,5,5-trimethylhexanal, and its notes show where the tolerance lies: Accept 5,5,3 instead of 3,5,5 and do not penalize missing hyphen/dash.

Punctuation and locant order are not the point; the locants themselves are. Feedback on a related item records the real failures — candidates struggled to name the compound accurately, often misnumbering chains or omitting branches. Number the chain from the end that gives the functional group the lowest number, and count the branches before you write anything.

Common trap · esters are named backwards

Feedback: 68% of the candidates selected the correct IUPAC name of the compound (methyl propanoate). The most commonly chosen distractor was propyl methanoate.

The first word comes from the alcohol; the second from the acid. Methyl propanoate is made from propanoic acid and methanol. Propyl methanoate is made from methanoic acid and propan-1-ol — a different compound with the same molecular formula. The two names are not interchangeable, and swapping them is the single most popular wrong answer on that item.

Exam alert · this outcome is examined in every form

Multiple choice — What is the IUPAC name of this compound? — twice in this collection, with 61% of the candidates selected the correct IUPAC name on a third. Structured, forwards and backwards. And it is named directly in published lists on both sides: among areas that appeared difficult, Identifying the IUPAC name of a structure and IUPAC naming of organic compounds.; among areas students appeared well prepared, Structure 3.2.5- apply IUPAC nomenclature and simply IUPAC nomenclature. It moves between the two lists from session to session, which is the clearest sign available that it rewards preparation.

6Structural isomers 3.2.6 SL + HL Strong

The syllabus statement: Structural isomers are molecules that have the same molecular formula but different connectivities. The skill: recognize isomers, including branched, straight-chain, position and functional group isomers, and primary, secondary and tertiary alcohols, halogenoalkanes and amines should be included. This outcome appeared in all six sessions in the collection behind these notes.

Structural isomers have the same molecular formula but a different connectivity — the atoms are joined in a different order. Because the structure differs, their physical properties differ, and if the functional group differs their chemical properties differ too. Three kinds are recognized in the syllabus.

Definition · four kinds of structural isomer

Chain (branched vs straight)	The carbon skeleton differs. Butane and 2-methylpropane.
Position	Same skeleton, same group, different place. Butan-1-ol and butan-2-ol.
Functional group	Same molecular formula, different class. An alcohol and an ether; an aldehyde and a ketone; a carboxylic acid and an ester.

Same molecular formula is the entry requirement. If two structures have different formulas they are not isomers, whatever else is true of them — which is why one reported distractor failed: The most commonly chosen distractor was A that included a pair of compounds that differed in the level of unsaturation.

For alcohols, halogenoalkanes and amines, position isomers are classified further as primary, secondary or tertiary. For alcohols and halogenoalkanes the classification depends on the carbon atom bearing the functional group: it is primary (1°) if that carbon is attached to one other carbon atom (or none, as in methanol), secondary (2°) if attached to two, and tertiary (3°) if attached to three. For amines it depends on the nitrogen: primary amines have one alkyl group on N (R–NH₂), secondary two (R₂NH) and tertiary three (R₃N). The classification matters chemically: primary and secondary alcohols are oxidized by acidified potassium dichromate(VI), tertiary alcohols are not (Reactivity 3.2).

Figure 3.9 The four isomers of C₄H₉OH, enumerated: two straight-chain, two branched, and between them all three of primary, secondary and tertiary.

Interactive model — S3.2.6

S3.2

Every structural isomer of a formula

Choose a molecular formula. The model lists every structural isomer within the stated scope, grouped by skeleton, with its IUPAC name, the kind of isomerism relating it to the first entry, and — for alcohols, halogenoalkanes and amines — its primary, secondary or tertiary classification.

Molecular formula

Scope: acyclic compounds only; for C₄H₈O₂ only acids and esters are listed.

AnimationStructural isomers

Identify which pairs are chain, position or functional group isomers.

Worked example 3.13 · isomers of C₃H₉N, and the amine rule

CH₃CH₂CH₂NH₂	propan-1-amine — one alkyl group on N: primary
CH₃CH(NH₂)CH₃	propan-2-amine — still one alkyl group on N: primary (though the carbon is secondary)
CH₃CH₂NHCH₃	N-methylethanamine — two alkyl groups on N: secondary
(CH₃)₃N	N,N-dimethylmethanamine (trimethylamine) — three: tertiary

Exam focus · what the published papers show

Worked example · counting the isomers of C₄H₉OH

How many structural isomers have the molecular formula C₄H₉OH?

Work in two stages, and do not mix them. First the skeleton, then the group.

Skeleton	Where the –OH can go	Isomer
straight, four carbons	on an end carbon	butan-1-ol
straight, four carbons	on an inner carbon	butan-2-ol
branched, three plus a methyl	on an end carbon	2-methylpropan-1-ol
branched, three plus a methyl	on the branch carbon	2-methylpropan-2-ol

Four. Feedback: Only 37% of the candidates determined the number of structural isomers with the molecular formula C4H9OH. The majority of candidate selected distractors with a lower number of isomers. The failure is undercounting, not miscounting — and an answer that only searches the straight chain undercounts by exactly two. Enumerate the skeletons first, every time.

Common trap · using the alcohol rule on an amine

Feedback: Only 25% of students realised that the classification of amines relates to the number of hydrocarbon chains attached to the nitrogen not, as with alcohols, the carbon to which the functional group is attached. Consider the skeleton of 2-methylpropan-2-ol. Its –OH sits on a carbon with three carbon neighbours, so the alcohol is tertiary. Put an –NH₂ on the same carbon instead and the nitrogen carries just one chain, so the amine is primary. Same skeleton, opposite classification.

Exam alert · what gets asked, and what is accepted

Sometimes any answer will do. One scheme reads simply any structural isomer of CH3CHBrC(CH3)3. — the mark is for understanding what an isomer is, not for finding a particular one. Elsewhere the question is a count, or a match: Which pair of compounds represents the correct type of isomerism? Published lists carry Identifying structural isomers and Determining the number of structural isomers with the molecular formula C 4H9OH. And isomers arrive inside longer questions too — one item required candidates first to draw the isomers of C3H8O, then work out the oxidation product for each one, which is primary-versus-secondary doing real work.

Past-paper practice · Practice set T · Physical trends, nomenclature and isomers

Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.

T1IB · May 2023 · SL Paper 1 · TZ1 · Q25 · [1]

Which statement explains the increase in boiling point for the homologous series of the primary alcohols?

A. London dispersion forces become strongerB. Hydrogen bonds become strongerC. Polarity of hydroxyl group increasesD. Standard state of the compound changes to solid

T2IB · May 2019 · SL Paper 1 · TZ2 · Q24 · [1]

Which compound has the lowest boiling point?

A. CH₃CH₂CH₂CH₂CH₂CH₃B. CH₃CH₂CH₂CH₂CH₃C. CH₃CH(CH₃)CH₂CH₃D. CH₃C(CH₃)₂CH₃

T3IB · November 2021 · SL Paper 1 · Q12 · [1]

The following compounds have similar relative molecular masses. What is the order of increasing boiling point?

A. CH₃CH₂CH₂OH < CH₃CH₂CHO < CH₃COOHB. CH₃CH₂CHO < CH₃CH₂CH₂OH < CH₃COOHC. CH₃CH₂CHO < CH₃COOH < CH₃CH₂CH₂OHD. CH₃COOH < CH₃CH₂CHO < CH₃CH₂CH₂OH

T4IB · May 2019 · SL Paper 1 · TZ1 · Q25 · [1]

A. 2-bromo-3-ethylbutaneB. 3-methyl-4-bromopentaneC. 2-ethyl-3-bromobutaneD. 2-bromo-3-methylpentane

T5IB · May 2019 · SL Paper 1 · TZ2 · Q27 · [1]

A. 2,3-diethylbutaneB. 2-ethyl-3-methylpentaneC. 3-methyl-4-ethylpentaneD. 3,4-dimethylhexane

T6IB · November 2017 · SL Paper 1 · Q28 · [1]

How many structural isomers of C₆H₁₄ exist?

A. 4B. 5C. 6D. 7

T7IB · November 2021 · SL Paper 1 · Q26 · [1]

Which pair of compounds are structural isomers?

A. Propane and propeneB. Propanal and propanoneC. Propan-1-ol and propanalD. Propyl propanoate and propanoic acid

T8IB · November 2023 · SL Paper 1 · TZ1 · Q26 · [1]

Question figure: an amine and an alcohol

	Type of amine	Type of alcohol
A.	Primary	Primary
B.	Tertiary	Tertiary
C.	Tertiary	Primary
D.	Primary	Tertiary

T9IB · May 2016 · SL Paper 1 · Q23 · [1]

How many alcohols have the general formula C₄H₁₀O?

A. 3B. 4C. 5D. 6

T10IB · May 2023 · SL Paper 2 · TZ1 · Q3(a)(i)–(ii) · [3]

Alkanes form a homologous series.

(i) Outline the meaning of homologous series. [1]

T11IB · May 2019 · SL Paper 2 · TZ2 · Q1(c)(ii) · [1]

Ethyne reacts with steam: C₂H₂(g) + H₂O(g) → C₂H₄O(g).

State the name of product B, applying IUPAC rules. [1]

Solutions and mark-scheme guidance · Set T

T1 A

Each additional CH₂ adds electrons and surface area, so London forces increase. Every primary alcohol has one –OH, so hydrogen bonding does not change along the series.

T2 D

D (2,2-dimethylpropane) has only five carbons and is the most branched: smallest surface contact, weakest London forces.

T3 B

Propanal has dipole–dipole forces but no hydrogen bonding; propan-1-ol forms hydrogen bonds; ethanoic acid forms more extensive hydrogen bonding (it can form hydrogen-bonded dimers) and boils highest.

T4 D

The longest chain runs through the CH₂CH₃ group: five carbons. Numbering from the bromo end gives Br on C2 and CH₃ on C3; substituents in alphabetical order: 2-bromo-3-methylpentane.

T5 D

Following both ethyl branches into the chain gives six carbons with methyl groups on C3 and C4: 3,4-dimethylhexane. A name ending in “ethylbutane” means the longest chain was missed.

T6 B

Hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane and 2,3-dimethylbutane.

T7 B

Propanal and propanone are both C₃H₆O: functional group isomers. The other pairs have different molecular formulas.

T8 D

The nitrogen carries one alkyl group, so the amine is primary, even though the carbon it is attached to has three carbon neighbours. The –OH is on a carbon bonded to three other carbons: a tertiary alcohol.

T9 B

Butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol. The three ethers with formula C₄H₁₀O are not alcohols.

T10 [3]

(i) Compounds of the same family AND same general formula, OR same family AND differ by a common structural unit/CH₂ ✔ (“contain the same functional group” accepted for “same family”).
(ii) 2-chlorobutane ✔; 1-chloro-2-methylpropane ✔ (1-chloromethylpropane accepted; “2-methyl-1-chloropropane” not accepted).

T11 [1]

Ethanal ✔. B is CH₃CHO: two carbons, carbonyl at the chain end.

Review · Structure 3.2, SL core

7Misconceptions, the examiner’s view, and the question types

Misconceptions to correct

Naming the class when the group is asked for (“aldehyde” for carbonyl, “alcohol” for hydroxyl, “hydroxide” for hydroxyl). Classes are families of compounds; groups are parts of molecules.
Splitting a functional group into pieces: the O–H in –COOH is not a hydroxyl group of an alcohol, and the N–H in –CONH– is not an amine.
“Successive members differ by CH or CH₃.” They differ by CH₂.
“Boiling point rises along the alcohols because hydrogen bonding increases.” Hydrogen bonding is constant (one –OH each); London forces increase.
“Branching raises boiling point.” Branching lowers it: smaller contact area, weaker London forces.
Numbering from the wrong end, or choosing a chain that is not the longest because it was drawn bent.
Naming esters backwards: the alcohol part comes first (methyl propanoate is made from methanol and propanoic acid).
Classifying amines by the carbon instead of the number of alkyl groups on nitrogen.

Examiner’s Overall Observation · classifying organic compounds (SL core)

Evidence base: the IB Diploma chemistry subject reports quoted in the exam-focus sections of this page.

Answered well: reading structures — geometry and hybridization around atoms, counting sigma and pi bonds; recognising members of a homologous series; in some sessions, IUPAC naming, which appears in both the well-prepared and the difficult lists in different years.

Found difficult: (1) giving the kind of formula asked for (full rather than condensed); (2) naming functional groups with the syllabus names — a majority gave “aldehyde” when “carbonyl” was required, and reports list seven non-syllabus names offered on one item; (3) defining a homologous series with two characteristics rather than one; (4) explaining the boiling-point trend in a series with London forces rather than hydrogen bonding, and the effect of branching; (5) attaching each part of a name to the structural feature it denotes; (6) naming esters in the right order; (7) enumerating all isomers (only just over a third found the four isomers of C₄H₉OH); and (8) classifying amines, which only a quarter of candidates did correctly.

What successful answers did: used the syllabus vocabulary exactly, gave two characteristics in definitions, named the intermolecular force that changes, and worked systematically — skeletons first, then positions — when counting isomers.

Six question types cover this whole document, and each has a fixed opening move.

If the question asks…	…then
Draw / give the formula	Read the adjective first. Full means every bond; condensed means groups; skeletal means no C–H; empirical means divide through.
Name the functional group	Use one of the nine syllabus names. Not the class. And take in the whole group before you name it.
Define a homologous series	Two of: same functional group, same general formula, successive members differ by CH₂.
Explain a boiling point	Within a series → dispersion forces grow. Between series → the functional group differs, so hydrogen bonding. Branching lowers it.
Name the compound / explain the name	Prefix, root, suffix. Running backwards, attach each syllable to the feature it names.
How many isomers?	Skeletons first, then move the group. Check every candidate has the same molecular formula.

8Quick check

Quick check · cover the answers

Give the empirical formula of C₆H₁₂O₆, and say what it tells you about the molecule.
Name the functional group in propanal, and name the class.
Give two characteristics of a homologous series that would earn a definition mark.
Why does butan-1-ol boil higher than propan-1-ol?
Why does butan-1-ol boil higher than butane?
Explain why the name propanone is given to its compound, part by part.
Which is made from ethanoic acid and propan-1-ol: propyl ethanoate or ethyl propanoate?
Draw or name all four structural isomers of C₄H₉OH, and classify each.
An –NH₂ group sits on a carbon that carries three other carbons. Is the amine primary, secondary or tertiary?

Answers

1	CH₂O — dividing through by 6. It gives the ratio of atoms only, and tells you nothing about the structure. Methanal has the same empirical formula.
2	Group: carbonyl. Class: aldehyde. A scheme refuses “aldehyde” when the group is asked for.
3	Any two of: same functional group / family; same general formula; successive members differ by CH₂.
4	It has one more CH₂, so more electrons and stronger London dispersion forces. The hydrogen bonding is the same in both.
5	Butan-1-ol has an –OH and can form hydrogen bonds between its molecules; butane cannot. Their dispersion forces are similar because their sizes are similar.
6	prop — three carbon atoms; an — no carbon–carbon double bond, saturated; one — a carbonyl group, and within the chain rather than at its end. Attach each part to its feature.
7	Propyl ethanoate. The alkyl part comes from the alcohol and is written first; the second word is the acid.
8	Butan-1-ol (primary), butan-2-ol (secondary), 2-methylpropan-1-ol (primary), 2-methylpropan-2-ol (tertiary).
9	Primary. Amines are classified by the chains on the nitrogen, and this nitrogen carries only one. The carbon's three neighbours are irrelevant to the amine.

9Summary and knowledge organiser

Essential knowledge

Formulas: empirical (ratio), molecular (actual numbers), structural (condensed and full), skeletal (zig-zag, no C–H), stereochemical (3D).
Nine functional groups by name and structure: halogeno, hydroxyl, carbonyl, carboxyl, alkoxy, amino, amido, ester, phenyl. Saturated = C–C single bonds only; unsaturated = C=C or C≡C.
Twelve homologous series; members share a functional group and general formula, differ by CH₂, have similar chemical properties and graded physical properties.
Boiling point rises along a series (London forces grow) and falls with branching; hydrogen bonding explains differences between series, not the trend within one.
IUPAC: longest chain containing the functional group; lowest locant for the functional group; substituents alphabetical; up to six carbons and one type of group in scope.
Structural isomers: chain, position, functional group. Alcohols and halogenoalkanes are 1°/2°/3° by the carbon; amines by the nitrogen.

Knowledge organiser

Outcome	Key facts	Must-remember distinctions and common errors
Formulas 3.2.1	C₄H₁₀ → C₂H₅ (empirical); skeletal: carbon at every vertex.	“Full” = every bond; “condensed” = groups. Empirical formula is a ratio.
Functional groups 3.2.2	–OH hydroxyl; C=O carbonyl; –COOH carboxyl; –O– alkoxy; –NH₂ amino; –CONH– amido; –COO– ester; C₆H₅– phenyl; –X halogeno.	Group name ≠ class name. Read the whole group.
Homologous series 3.2.3	C_nH_2n+2, C_nH_2n, C_nH_2n−2, C_nH_2n+1OH, C_nH_2nO, C_nH_2nO₂.	Two characteristics for a definition. Paired series share a general formula.
Physical trends 3.2.4	b.p. ↑ with n; b.p. ↓ with branching.	Trend: London forces. Between series: hydrogen bonding / dipole–dipole.
IUPAC 3.2.5	stem + suffix (-ane, -ene, -ol, -al, -one, -oic acid) + prefixes (methyl-, chloro-).	-al and -oic acid are C1 with no locant. Esters: alkyl (from alcohol) + -oate (from acid).
Isomers 3.2.6	C₄H₁₀ 2; C₅H₁₂ 3; C₆H₁₄ 5; C₄H₉OH 4 alcohols.	Same molecular formula is the entry test. Amines: count alkyl groups on N.