IB DP Chemistry Structure 3.2: Functional groups and naming organic compounds

This guide teaches the first part of Structure 3.2, Functional groups: Classification of organic compounds, matched to the February 2023 IB Chemistry guide for the Diploma Programme, whose first exams ran in 2025. It covers understandings Structure 3.2.1 to 3.2.6, all studied at both SL and HL; the later understandings (stereoisomers and spectroscopy) are in a separate unit.

Next, try the practice questions; shorter revision notes cover the same content. To see where this unit sits, use the IB DP chemistry hub or print the checklist. Our free diagnostics can point you to topics that need work.

What this unit covers

Understanding	What you should be able to do	SL/HL
Structure 3.2.1	Recognise six kinds of formula (empirical, molecular, full and condensed structural, skeletal, stereochemical); convert between molecular, skeletal and structural formulas; build 3D models	SL and HL
Structure 3.2.2	Name and draw nine functional groups; use “saturated” and “unsaturated”	SL and HL
Structure 3.2.3	Recognise twelve homologous series and their general formulas	SL and HL
Structure 3.2.4	Account for the way boiling and melting points change along a series	SL and HL
Structure 3.2.5	Give IUPAC names to compounds with up to six carbons in the parent chain	SL and HL
Structure 3.2.6	Recognise structural isomers; classify alcohols, halogenoalkanes and amines as primary, secondary or tertiary	SL and HL

Writing organic formulas (3.2.1)

Formula type	What it shows	1,4-dibromobutane
Empirical	Simplest whole-number ratio of atoms	C₂H₄Br
Molecular	Atoms of each element actually present in one molecule	C₄H₈Br₂
Full structural	Every atom and every bond	see below
Condensed structural	Atoms grouped carbon by carbon, no bond lines	BrCH₂CH₂CH₂CH₂Br
Skeletal	Carbon chain as a zigzag; H atoms on carbon left out	Br at each end of a four-vertex zigzag
Stereochemical	3D shape, with wedges for bonds towards you and dashes for bonds away	used when shape matters

You should recognise stereochemical formulas, but the guide does not expect you to draw them unless a question asks.

Skeletal rules. Each corner of the zigzag, and each free end of a line, stands for a carbon atom. Each carbon has enough hydrogen atoms to make four bonds, but they are not drawn. Atoms other than carbon and hydrogen are always written in, and so is any hydrogen bonded to them (O–H, N–H).

3D models. A kit or modelling software shows the tetrahedral arrangement around each saturated carbon, which a flat drawing hides.

Worked example 1: from a skeletal formula to the others

A skeletal formula is an unbranched zigzag with four corners and a Br atom at each end. Give the other formulas.

Step 1: the ends are Br atoms, so only the four corners are carbons.

Step 2: each carbon already has two bonds (to C or Br), so each needs two H atoms.

    H   H   H   H
    |   |   |   |
Br––C – C – C – C––Br
    |   |   |   |
    H   H   H   H

Step 3: condensed formula BrCH₂CH₂CH₂CH₂Br; molecular formula C₄H₈Br₂.

Step 4: the ratio 4 : 8 : 2 has a common factor of 2, so the empirical formula is C₂H₄Br.

Functional groups (3.2.2)

Think of a functional group as the reactive region of a molecule: it decides how the compound typically reacts and shapes many of its physical properties. You need these nine.

Functional group	Structure	Found in
halogeno	–X (X = F, Cl, Br, I)	halogenoalkanes
hydroxyl	–OH	alcohols
carbonyl	C=O	aldehydes (–CHO at a chain end) and ketones (C=O inside the chain)
carboxyl	–COOH	carboxylic acids
alkoxy	–O–R	ethers
amino	–NH₂	amines
amido	–CONH₂ (or –CONH– when N carries a carbon group)	amides
ester	–COO–R	esters
phenyl	–C₆H₅, a benzene ring as a substituent	aromatic compounds

Saturated compounds have only single carbon–carbon bonds. Unsaturated compounds have at least one C=C or C≡C bond.

The O–H in alcohols and carboxylic acids allows hydrogen bonding, which raises boiling points and helps small members mix with water. A linking question uses functional group reactivity to plan routes: ethene can become ethanol, which can be oxidised to ethanoic acid, one group change per step.

Worked example 2: spotting groups in a real molecule

The sweetener aspartame has the condensed formula HOOC–CH₂–CH(NH₂)–CO–NH–CH(CH₂C₆H₅)–COO–CH₃. Name every functional group.

1. HOOC– is a carboxyl group.
2. –NH₂ on the CH is an amino group.
3. –CO–NH– is an amido group.
4. –C₆H₅ is a phenyl group.
5. –COO–CH₃ is an ester group.

The C=O joined to the nitrogen is what makes –CO–NH– amido, not amino.

Homologous series (3.2.3)

Compounds in a homologous series share one functional group, and neighbouring members differ by a CH₂ unit. Members share a general formula, react in similar ways and show a gradual change in physical properties.

Series	Functional group	General formula	Name ending or prefix	First member
alkanes	none (C–C only)	CₙH₂ₙ₊₂	-ane	methane
alkenes	C=C	CₙH₂ₙ	-ene	ethene
alkynes	C≡C	CₙH₂ₙ₋₂	-yne	ethyne
halogenoalkanes	halogeno	CₙH₂ₙ₊₁X	chloro-, bromo- etc.	chloromethane
alcohols	hydroxyl	CₙH₂ₙ₊₁OH	-ol	methanol
aldehydes	carbonyl, chain end	CₙH₂ₙO	-al	methanal
ketones	carbonyl, inside chain	CₙH₂ₙO	-one	propanone
carboxylic acids	carboxyl	CₙH₂ₙO₂	-oic acid	methanoic acid
ethers	alkoxy	CₙH₂ₙ₊₂O	alkoxyalkane	methoxymethane
amines	amino	CₙH₂ₙ₊₁NH₂	-amine	methanamine
amides	amido	CₙH₂ₙ₊₁CONH₂	-amide	methanamide
esters	ester	CₙH₂ₙO₂	-oate	methyl methanoate

Alcohols and ethers, aldehydes and ketones, and acids and esters share general formulas, so a molecular formula alone cannot fix the series.

Worked example 3: finding a member from its molar mass

An ester has a molar mass of 116.18 g mol⁻¹. Find its molecular formula. Use Ar values C = 12.01, H = 1.01, O = 16.00.

Esters fit CₙH₂ₙO₂. One CH₂ unit has mass 12.01 + 2(1.01) = 14.03.

14.03n + 2(16.00) = 116.18, so 14.03n = 84.18 and n = 6.

The ester is C₆H₁₂O₂.

Trends in melting and boiling points (3.2.4)

Each member of a series has one more CH₂ than the last, so its molecules have more electrons and a larger surface. London (dispersion) forces get stronger, more energy is needed to separate the molecules, and boiling points rise steadily.

Alcohol	Boiling point / K
methanol	338
ethanol	352
propan-1-ol	370
butan-1-ol	391
pentan-1-ol	411

All five alcohols hydrogen bond through –OH. That contribution stays about the same along the series, so the rise comes from the growing London forces.

Melting points also rise, but less smoothly, because they also depend on how well molecules pack in the solid. Two other factors matter:

• Branching. A branched isomer is more compact, so molecules touch over a smaller area, London forces are weaker and the boiling point is lower.
• Functional group. Hydrogen bonding (alcohols, carboxylic acids) gives higher boiling points than dipole–dipole forces (aldehydes, ketones), which in turn beat London forces alone (alkanes) for molecules with similar numbers of electrons.

Worked example 4: predicting the next member

Estimate the boiling point of hexan-1-ol from the table.

The last three gaps are 18 K, 21 K and 20 K, about 20 K per CH₂. 411 + 20 = 431 K, so estimate about 430 K.

IUPAC naming (3.2.5)

IUPAC nomenclature is the International Union of Pure and Applied Chemistry’s rule set for systematic names. You name branched and straight-chain compounds with a parent chain of up to six carbons, at most one C=C or C≡C, and one type of functional group: halogeno, hydroxyl, carbonyl or carboxyl.

Method

1. Pick the longest chain of carbons that includes the functional group (and any multiple bond). Its length gives the stem: meth, eth, prop, but, pent, hex.
2. Add the ending: -ane, -ene or -yne for the chain; then -ol, -al, -one or -oic acid for the group. Halogens go in front as fluoro, chloro, bromo or iodo.
3. Number the chain so the suffix group gets the lowest number. Next give the multiple bond the lowest number, then the prefixes.
4. Name branches as methyl or ethyl. Use di- and tri- for repeats. List prefixes in alphabetical order, ignoring di- and tri-.
5. Separate numbers from each other with commas, and numbers from words with hyphens.

The –CHO and –COOH carbons are always carbon 1, so they need no number.

Worked example 5: naming six compounds

Condensed formula	Reasoning	Name
CH₃CH(CH₃)C(CH₃)₂CH₂CH₃	Left gives 2,3,3; right gives 3,3,4; first difference decides	2,3,3-trimethylpentane
CH₃C(Cl)(CH₃)CH(Cl)CH₃	{2,2,3} beats {2,3,3}; chloro before methyl	2,3-dichloro-2-methylbutane
(CH₃)₂CHCOCH₂CH₃	C=O is carbon 3 either way; methyl gets 2, not 4	2-methylpentan-3-one
CH₃CH₂CH(C₂H₅)CHO	Longest chain through CHO has four carbons	2-ethylbutanal
CH₃CH=CHCH₂CH(OH)CH₃	Number from the OH end	hex-4-en-2-ol
CH₃CH₂CH(CH₃)CH₂COOH	COOH carbon is 1	3-methylpentanoic acid

Worked example 6: from a name to a structure

Draw 2-bromo-3-methylbut-2-ene as a condensed formula.

Four carbons with C=C between carbons 2 and 3; Br on carbon 2 and CH₃ on carbon 3: CH₃C(Br)=C(CH₃)CH₃. Either numbering gives {2,3}, so the prefix first in the alphabet (bromo) takes the lower number.

Structural isomers (3.2.6)

Structural isomers have the same molecular formula but different connectivity:

• Branched and straight-chain (chain) isomers: the carbon skeleton differs.
• Position isomers: the same group sits on a different carbon.
• Functional group isomers: the isomers belong to different series.

Worked example 7: five isomers of C₄H₈O

Butanal (CH₃CH₂CH₂CHO), 2-methylpropanal ((CH₃)₂CHCHO), butan-2-one (CH₃COCH₂CH₃), but-3-en-1-ol (CH₂=CHCH₂CH₂OH) and but-3-en-2-ol (CH₂=CHCH(OH)CH₃) all have the formula C₄H₈O.

• Butanal and 2-methylpropanal: chain isomers.
• But-3-en-1-ol and but-3-en-2-ol: position isomers (the OH moves).
• Butanal and butan-2-one: functional group isomers (aldehyde and ketone).

Propan-1-ol and methoxyethane (both C₃H₈O) are another functional group pair: alcohol and ether.

Primary, secondary and tertiary

For alcohols and halogenoalkanes, look at the carbon carrying the –OH or halogen and count the carbon atoms bonded to it: one is primary, two is secondary, three is tertiary.

For amines, count the carbon atoms bonded to the nitrogen instead.

Worked example 8: classifying

Compound	What to count	Class
pentan-1-ol	C–OH carbon has 1 carbon neighbour	primary
pentan-3-ol	2 carbon neighbours	secondary
2-methylbutan-2-ol	3 carbon neighbours	tertiary
1-iodo-2,2-dimethylpropane	C–I carbon has 1 carbon neighbour	primary
butan-2-amine, CH₃CH(NH₂)CH₂CH₃	N bonded to 1 carbon	primary amine
N,N-dimethylethanamine, (CH₃)₂NCH₂CH₃	N bonded to 3 carbons	tertiary amine

Butan-2-amine is the trap: its NH₂ carbon has two carbon neighbours, but only the nitrogen counts.

HL link (AHL Structure 2.2). A linking question asks why only three dibromobenzene isomers supports the modern benzene model. Alternating single and double bonds would allow two different 1,2-isomers; only one exists, so all six ring bonds are equal.

Common errors

• Choosing the longest chain without checking it contains the functional group.
• Giving the halogen or methyl the low number when an -ol or -one is present.
• Naming 1-methylpentan-2-ol: a methyl on carbon 1 extends the chain, giving hexan-3-ol.
• Classifying amines by the carbon, like alcohols.

Official syllabus

Source: the International Baccalaureate’s Diploma Programme Chemistry guide, issued in February 2023 for courses examined from 2025. This page follows Structure 3.2, Functional groups: Classification of organic compounds, understandings 3.2.1 to 3.2.6, which are common to SL and HL.