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

Use these notes in the last weeks of revision for Structure 3.2, Functional groups: Classification of organic compounds, understandings 3.2.1 to 3.2.6. Their reference point is the Diploma Programme Chemistry guide the IB released in February 2023, first examined in 2025. Every point here applies to both SL and HL.

Each idea is explained at length, with full worked examples, in the Structure 3.2 study guide. Afterwards, try the practice set. For the wider course, open the hub page or print the topic checklist; to find your weaker areas, try a free diagnostic.

Formula types (3.2.1)

Type	Shows	Example: hexane-2,5-dione
Empirical	Simplest ratio of atoms	C₃H₅O
Molecular	Actual atom count	C₆H₁₀O₂
Full structural	Every bond drawn	all C–H, C–C and C=O lines
Condensed structural	Groups listed along the chain	CH₃COCH₂CH₂COCH₃
Skeletal	Zigzag of C–C bonds; C and C–H not written	six-carbon zigzag with =O on carbons 2 and 5
Stereochemical	3D shape with wedges and dashes	recognise; draw only when asked

Skeletal check, in steps

1. Count line ends and corners: each one is a carbon (unless a symbol such as Br or OH is written there).
2. Give each carbon four bonds in total by adding hidden H atoms.
3. Keep every H that is bonded to O or N; these are always drawn.
4. Write the molecular formula, then divide by any common factor for the empirical formula.

3D models (real or virtual) show the tetrahedral carbon that flat drawings hide.

Functional groups (3.2.2)

Group	Structure	Class
halogeno	–F, –Cl, –Br, –I	halogenoalkane
hydroxyl	–OH	alcohol
carbonyl	C=O	aldehyde (–CHO, end of chain) or ketone (inside chain)
carboxyl	–COOH	carboxylic acid
alkoxy	–OR	ether
amino	–NH₂	amine
amido	–CONH₂ / –CONH–	amide
ester	–COOR	ester
phenyl	–C₆H₅	aromatic compound

• Saturated: only C–C single bonds.
• Unsaturated: at least one C=C or C≡C.

Homologous series (3.2.3)

Same functional group, same general formula, successive members differ by CH₂, similar chemistry, gradual trend in physical properties.

Series	General formula	Smallest member
alkanes	CₙH₂ₙ₊₂	CH₄
alkenes	CₙH₂ₙ	C₂H₄
alkynes	CₙH₂ₙ₋₂	C₂H₂
halogenoalkanes	CₙH₂ₙ₊₁X	CH₃X
alcohols	CₙH₂ₙ₊₁OH	CH₃OH
ethers	CₙH₂ₙ₊₂O	CH₃OCH₃
aldehydes	CₙH₂ₙO	HCHO
ketones	CₙH₂ₙO	CH₃COCH₃
carboxylic acids	CₙH₂ₙO₂	HCOOH
esters	CₙH₂ₙO₂	HCOOCH₃
amines	CₙH₂ₙ₊₁NH₂	CH₃NH₂
amides	CₙH₂ₙ₊₁CONH₂	HCONH₂

Pairs that share a general formula: alcohol/ether, aldehyde/ketone, acid/ester. These pairs are functional group isomers of each other.

Quick use. The alkyne with eight carbons is C₈H₁₄; the alkane with 20 carbons is C₂₀H₄₂.

Melting and boiling point trends (3.2.4)

Explaining the trend, in steps

1. Each successive member has one more CH₂ unit.
2. More electrons and a larger surface area per molecule.
3. Stronger London (dispersion) forces between molecules.
4. More energy needed to separate the molecules, so a higher boiling point.

• Boiling points rise smoothly. Melting points rise less regularly, because packing in the solid also matters.
• Branching lowers the boiling point: less surface contact, weaker London forces.
• Functional group raises it: for similar numbers of electrons, hydrogen bonding (–OH, –COOH) > dipole–dipole (C=O, C–X) > London forces only (alkanes).

Worked reminder. Rank butan-1-ol, butanal and 2,2-dimethylpropane by boiling point. Their molecules have similar numbers of electrons (42, 40 and 42), so the type of force decides. Butan-1-ol hydrogen bonds, so it is highest. Butanal has a polar C=O and dipole–dipole forces, so it comes next. 2,2-dimethylpropane has London forces only, and its compact branched shape weakens even those, so it is lowest. Order: butan-1-ol > butanal > 2,2-dimethylpropane.

IUPAC naming (3.2.5)

Scope: parent chain of up to six carbons; saturated or one C=C or C≡C; one type of group from halogeno, hydroxyl, carbonyl and carboxyl; straight or branched chains.

Stem	meth-	eth-	prop-	but-	pent-	hex-
Carbons	1	2	3	4	5	6

Ending or prefix	Meaning
-ane / -ene / -yne	all single / one C=C / one C≡C
-ol	hydroxyl
-al	aldehyde (carbon 1, no locant)
-one	ketone (give its locant)
-oic acid	carboxyl (carbon 1, no locant)
fluoro-, chloro-, bromo-, iodo-	halogeno, always a prefix

Method, in steps

1. Longest chain that contains the functional group and any C=C or C≡C.
2. Number it so the suffix group (-ol, -al, -one, -oic acid) has the lowest locant.
3. Then the lowest locant for the C=C or C≡C.
4. Then the lowest set of locants for prefixes; if the sets tie, the prefix first in the alphabet gets the lower number.
5. Prefixes in alphabetical order (ignore di-, tri-); commas between numbers, hyphens between numbers and words.

Small worked reminders

• ClCH₂CH₂CH₂CH(CH₃)₂: numbering from the Cl end gives {1,4}; the other way gives {2,5}. Name: 1-chloro-4-methylpentane.
• CH₂=CHCOCH₂CH₃: C=O is carbon 3 both ways, so the C=C decides. Name: pent-1-en-3-one.
• (CH₃)₂CHCH₂COOH: COOH carbon is 1, methyl on 3. Name: 3-methylbutanoic acid.

Structural isomers (3.2.6)

Same molecular formula, different connectivity.

Type	What changes	Pair (formula)
Branched vs straight-chain	Carbon skeleton	hexan-1-ol and 3-methylpentan-1-ol (C₆H₁₄O)
Position	Where the group sits	1-bromopentane and 3-bromopentane (C₅H₁₁Br)
Functional group	Which series	propanal and propanone (C₃H₆O)

Primary, secondary, tertiary

Class of compound	Count the carbons bonded to…	1	2	3
alcohols, halogenoalkanes	the carbon carrying –OH or –X	primary	secondary	tertiary
amines	the nitrogen atom	primary	secondary	tertiary

So (CH₃)₃CNH₂, 2-methylpropan-2-amine, is a primary amine, even though the same carbon skeleton with –OH would be a tertiary alcohol. CH₃CH₂NHCH₂CH₃ is a secondary amine.

Linking questions to HL content. Only three dibromobenzene isomers exist (1,2-, 1,3- and 1,4-). With alternating single and double bonds there would be two different 1,2-isomers, so the result supports six equal ring bonds. A second HL link: two amino acids join through an amido link (–CONH–) when they form a dipeptide.

Must-know distinctions

• Empirical vs molecular: divide by the common factor; if there is none, they are the same.
• Full vs condensed structural: full draws every bond; condensed groups atoms carbon by carbon.
• Aldehyde vs ketone: –CHO at the end of the chain vs C=O between two carbons.
• Carboxyl vs ester: –COOH vs –COO– bonded to another carbon group.
• Amino vs amido: –NH₂ on carbon vs N bonded to a C=O carbon.
• Alcohol vs amine classification: count around the carbon vs around the nitrogen.
• Isomer vs homologue: isomers share a molecular formula; homologues differ by CH₂.

Quick self-test

1. Give the empirical formula of butane-1,4-diol, HOCH₂CH₂CH₂CH₂OH.
2. Which series has general formula CₙH₂ₙ₋₂, and what is the formula of its six-carbon member?
3. Name the functional group in CH₃CH₂OCH₂CH₃ and give an alcohol that is its isomer.
4. Name CH₃CH=CHCH(OH)CH₃.
5. Name CH₃CH₂CH(CH₃)CH(Cl)CH₃.
6. Write a condensed formula for 2,2-dimethylbutan-1-ol and classify it.
7. Classify 3-chloro-3-methylhexane.
8. Classify N-ethylethanamine, CH₃CH₂NHCH₂CH₃.
9. State the type of isomerism shown by propanoic acid and methyl ethanoate.
10. Butan-1-ol and ethoxyethane have the same molecular formula. Which has the higher boiling point, and why?
11. Explain why 2,2-dimethylbutane boils at a lower temperature than hexane.

Answers

1. C₄H₁₀O₂ divides by 2: C₂H₅O.
2. Alkynes; C₆H₁₀.
3. Alkoxy (an ether); butan-1-ol (or butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol).
4. Pent-3-en-2-ol: number from the OH end so OH gets 2.
5. 2-chloro-3-methylpentane: {2,3} beats {3,4}.
6. HOCH₂C(CH₃)₂CH₂CH₃; primary, since the C–OH carbon has one carbon neighbour.
7. Tertiary: the C–Cl carbon is bonded to three carbons.
8. Secondary: the nitrogen is bonded to two carbons.
9. Functional group isomerism (carboxylic acid and ester, both C₃H₆O₂).
10. Butan-1-ol: its –OH groups form hydrogen bonds between molecules; the ether cannot hydrogen bond to itself, so its molecules separate more easily.
11. The branched molecule is more compact, so there is less surface contact and the London forces between molecules are weaker.

Where marks are usually lost

• Writing an empirical formula that still has a common factor, or dividing when there is none.
• Missing hidden hydrogens on skeletal carbons, or leaving out the H of an O–H.
• Calling a ketone C=O an aldehyde because it is “near” the end of the chain.
• Writing “amine” for an amide, or “amino” for –CONH₂.
• Picking the longest chain that does not contain the functional group.
• Numbering from the end nearer a methyl group when an -ol or -one is present.
• Adding a locant to -al or -oic acid, or putting prefixes in number order instead of alphabetical order.
• Explaining the boiling point trend with “bonds breaking” rather than weaker or stronger forces between molecules.
• Classifying amines by the carbon atom, as for alcohols.
• Calling two compounds isomers without checking that their molecular formulas match.

Official syllabus

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