Organic chemistry is the study of carbon-containing compounds. Carbon's ability to form four covalent bonds and long chains makes millions of different molecules possible. In this chapter you will learn to name and draw the major functional group families, understand how intermolecular forces determine physical properties, and describe the key reaction types that transform one organic compound into another.
3.1 Introducing Organic Molecules
Definition
Organic molecule
An organic molecule (or organic compound) is a molecule that contains carbon, typically with hydrogen, and frequently also oxygen, nitrogen, sulfur, phosphorus, and halogens. Carbon's ability to form four covalent bonds and to bond to other carbon atoms allows the enormous variety of organic molecules (over 10 million known compounds).
Definition
Functional group
A functional group is a specific atom or group of atoms within a molecule that is responsible for the characteristic chemical properties and reactions of that molecule. All molecules in the same homologous series share the same functional group. Examples: –OH (hydroxyl/alcohol), –COOH (carboxyl/acid), C=C (alkene).
Definition
Homologous series
A homologous series is a series of organic compounds that: (1) have the same general formula and the same functional group, (2) differ from adjacent members by one CH₂ unit (14 g·mol⁻¹), (3) show similar chemical properties, and (4) show a gradual, predictable change in physical properties (e.g. boiling point increases with chain length). Examples: alkanes (CₙH₂ₙ₊₂), alkenes (CₙH₂ₙ), alkanols.
Definition
Isomer
Molecules with the same molecular formula but different structural formula.
Definition
Saturated compounds
A saturated compound has no double or triple bonds (single bonds only).
Definition
Unsaturated compounds
An unsaturated compound contains double or triple bonds.
Definition
Hydrocarbon
An organic molecule which contains only carbon and hydrogen atoms.
Saturated vs Unsaturated Compounds
| Property | Saturated | Unsaturated |
|---|---|---|
| Bond types between carbons | Single bonds only (C–C) | At least one double (C=C) or triple (C≡C) bond |
| General formula (open chain) | Alkanes: CₙH₂ₙ₊₂ | Alkenes: CₙH₂ₙ; Alkynes: CₙH₂ₙ₋₂ |
| Bromine water test | No decolourisation (does not react) | Decolourises bromine water (addition reaction) |
| Example | Ethane (C₂H₆) | Ethene (C₂H₄), Ethyne (C₂H₂) |
A homologous series has a constant difference of CH₂ between successive members. All members share the same functional group, have similar chemical properties, and show a gradual trend in physical properties (e.g. boiling point increases with chain length because dispersion forces increase). Isomers have the same molecular formula but different arrangements of atoms, leading to different physical and sometimes chemical properties.
Exam Tip
EXAM TIP: Learn the prefixes for carbon chain length: meth- (1C), eth- (2C), prop- (3C), but- (4C), pent- (5C), hex- (6C). These are used in IUPAC naming for every functional group family.
Worked Example
Write the structural formula for butane and state whether it is saturated or unsaturated. Then write a structural formula for one isomer of butane (C₄H₁₀).
Given
- Molecular formula of butane: C₄H₁₀
Find
Structural formula of butane; one isomer
Solution
- 1Butane: CH₃–CH₂–CH₂–CH₃ (straight chain, all single bonds)
- 2Saturated — all C–C single bonds only
- 3Isomer: 2-methylpropane — CH₃–CH(CH₃)–CH₃ (branched chain; same formula C₄H₁₀)
Practice Question
State whether each of the following is saturated or unsaturated and give a reason: (a) propane C₃H₈, (b) propene C₃H₆.
(4 marks)
Practice Question
Two compounds have molecular formula C₃H₈O. Explain the term 'isomers' and state how these two compounds could differ despite having the same formula.
(4 marks)
3.2 Functional Group Families and IUPAC Naming
Functional Group Families — Summary
- Alkanes: –C–C– (single bonds only; suffix -ane)
- Alkenes: C=C (double bond; suffix -ene; position number required)
- Alkynes: C≡C (triple bond; suffix -yne)
- Haloalkanes: C–X where X = F, Cl, Br, I (named as halo- prefix, e.g. chloro-)
- Alcohols: –OH (suffix -ol; position number required)
- Carboxylic acids: –COOH (suffix -oic acid; always C1)
- Esters: –COO– (suffix -anoate; named alkyl alkanoate)
- Aldehydes: –CHO (suffix -al; always C1)
- Ketones: C=O in middle of chain (suffix -one; position number required)
- Amines: –NH₂ (suffix -amine or prefix amino-)
- Amides: –CONH₂ (suffix -amide)
Key Functional Groups at a Glance
| Property | Family | Functional Group / Suffix |
|---|---|---|
| Alcohol | –OH | -ol |
| Carboxylic acid | –COOH | -oic acid |
| Ester | –COO– | alkyl -anoate |
| Aldehyde | –CHO | -al |
| Ketone | >C=O (mid-chain) | -one |
| Amine | –NH₂ | -amine |
| Amide | –CONH₂ | -amide |
| Haloalkane | –X (halogen) | halo- prefix |
Exam Tip
EXAM TIP: IUPAC naming steps — (1) Find the longest carbon chain containing the principal functional group. (2) Number from the end giving the functional group the lowest number. (3) Name substituents as prefixes (e.g. methyl-, chloro-). (4) Combine: substituents + parent chain + suffix.
Worked Example
Name the following compound using IUPAC rules: CH₃–CH(OH)–CH₂–CH₃
Given
- Structure: CH₃–CH(OH)–CH₂–CH₃
Find
IUPAC name
Solution
- 1Longest chain containing –OH: 4 carbons → parent = butane → suffix -ol → butanol
- 2Number from end closest to –OH: –OH is on C2 (numbering from left: C1=CH₃, C2=CH(OH), C3=CH₂, C4=CH₃)
- 3Name: butan-2-ol
Worked Example
Draw the structural formula for ethyl propanoate and name the functional group present.
Given
- Name: ethyl propanoate
Find
Structural formula; functional group
Solution
- 1Propanoate = 3C acid part: CH₃CH₂COO–
- 2Ethyl = 2C alcohol part: –CH₂CH₃
- 3Combined: CH₃CH₂–COO–CH₂CH₃
Watch Out
COMMON MISTAKE: Students confuse aldehydes and ketones. An aldehyde (–CHO) ALWAYS has its carbonyl group at the end of the chain (C1). A ketone (>C=O) ALWAYS has its carbonyl group in the middle of the chain. They are never found at the same position.
Practice Question
Give the IUPAC name for CH₃–CH₂–CH₂–COOH.
(2 marks)
Practice Question
Draw the structural formula for 2-chloropropane.
(3 marks)
Practice Question
Distinguish between an aldehyde and a ketone with reference to the position of the carbonyl group.
(3 marks)
3.3 Physical Properties and Intermolecular Forces
Intermolecular forces (IMF) are the forces of attraction between molecules. The stronger the IMF, the more energy (higher temperature) is needed to separate molecules, so the higher the boiling point. The three main types relevant to organic chemistry are: dispersion forces (all molecules), dipole–dipole forces (polar molecules), and hydrogen bonds (–OH, –NH₂, –COOH groups).
IMF Types Ranked by Strength (weakest → strongest)
- Dispersion forces (London forces) — present in ALL molecules; increase with molecular size/mass
- Dipole–dipole forces — polar molecules (e.g. haloalkanes, aldehydes, ketones)
- Hydrogen bonds — molecules with O–H or N–H groups (alcohols, carboxylic acids, amines, amides)
Effect of IMF on Boiling Point
| Property | Compound type | Dominant IMF and boiling point trend |
|---|---|---|
| Alkanes (non-polar) | Dispersion forces only | Low BP; increases with chain length as dispersion forces increase |
| Haloalkanes / Ketones (polar) | Dipole–dipole forces + dispersion | Higher BP than similar-mass alkanes |
| Alcohols / Carboxylic acids | Hydrogen bonding + dispersion | Highest BP for given molecular mass; carboxylic acids form dimers → exceptionally high BP |
Note
SOLUBILITY RULE — 'like dissolves like': Polar organic compounds (alcohols, carboxylic acids) dissolve well in water because they form hydrogen bonds with water molecules. Non-polar hydrocarbons (alkanes) do not dissolve in water.
Worked Example
Arrange the following in order of increasing boiling point and justify your answer: hexane (C₆H₁₄), hexan-1-ol (C₆H₁₃OH), hexanal (C₆H₁₂CHO).
Given
- Hexane — non-polar, dispersion forces only
- Hexanal — polar carbonyl group, dipole–dipole forces + dispersion
- Hexan-1-ol — –OH group, hydrogen bonds + dispersion
Find
Order of increasing boiling point
Solution
- 1Hexane: weakest IMF (dispersion only) → lowest BP
- 2Hexanal: dipole–dipole > dispersion → intermediate BP
- 3Hexan-1-ol: hydrogen bonds → highest BP among these three
Watch Out
COMMON MISTAKE: Students assume the compound with the highest molecular mass always has the highest boiling point. While mass does affect dispersion forces, hydrogen bonding can give a lighter molecule a much higher boiling point than a heavier non-polar molecule. Always identify the dominant IMF first.
Practice Question
Explain why ethanol (C₂H₅OH, M = 46 g·mol⁻¹) has a much higher boiling point (78 °C) than dimethyl ether (CH₃OCH₃, M = 46 g·mol⁻¹, bp = −24 °C) even though both have the same molar mass.
(4 marks)
Practice Question
Will hexane dissolve readily in water? Give a reason using the concept of IMF.
(3 marks)
3.4 Organic Reactions and Polymers
Definition
Monomer
A monomer is a small organic molecule of relatively low molecular mass that can react with many identical (or similar) molecules through polymerisation to form a large polymer chain. Monomers must contain at least one reactive site (e.g. a double bond or functional group pair). Example: ethene (CH₂=CH₂) is the monomer for polyethylene.
Definition
Polymer
A polymer is a large molecule (macromolecule) of high molecular mass consisting of many repeating structural units called monomers, joined by covalent bonds through polymerisation reactions. Polymers can be natural (e.g. starch, rubber, DNA) or synthetic (e.g. polyethylene, nylon, PVC). Their properties depend on the monomer used, chain length, and degree of branching or cross-linking.
Key Organic Reaction Types
- Addition reaction: a molecule adds across a double or triple bond; the π bond breaks and two new σ bonds form. Unsaturated → saturated. Example: ethene + Br₂ → 1,2-dibromoethane.
- Substitution reaction: an atom or group in a molecule is replaced by another atom or group. Example: chloromethane + NaOH → methanol + NaCl (hydrolysis of haloalkane).
- Elimination reaction: a small molecule (e.g. H₂O or HBr) is removed from adjacent carbons, forming a double bond. Saturated → unsaturated. Example: butan-2-ol + H₂SO₄(conc.) → but-2-ene + H₂O.
- Esterification: alcohol + carboxylic acid ⇌ ester + water (reversible; acid catalyst; heated under reflux). Example: ethanol + ethanoic acid → ethyl ethanoate + H₂O.
- Saponification: ester + NaOH(aq) → soap (sodium salt of fatty acid) + alcohol. Irreversible; used in soap manufacture.
Addition vs Condensation Polymerisation
| Property | Addition polymerisation | Condensation polymerisation |
|---|---|---|
| Monomer requirement | Must have C=C double bond (alkene) | Must have TWO functional groups (e.g. –OH and –COOH) |
| By-product | None — all atoms in monomers go into polymer | Small molecule released (usually H₂O or HCl) |
| Example polymer | Polyethylene (from ethene), PVC (from chloroethene) | Nylon, polyester, proteins |
| Repeat unit | –[CH₂–CH₂]ₙ– (from ethene) | Contains ester or amide linkages |
Exam Tip
EXAM TIP: To identify an addition reaction, check whether the product is saturated while the reactant was unsaturated. To identify an elimination reaction, check whether a double bond is FORMED. Esterification always produces water as a by-product.
Worked Example
Propan-1-ol reacts with propanoic acid in the presence of concentrated H₂SO₄. (a) Name the type of reaction. (b) Write the structural formula of the organic product. (c) Name the product.
Given
- Reactant 1: propan-1-ol (CH₃CH₂CH₂OH)
- Reactant 2: propanoic acid (CH₃CH₂COOH)
- Catalyst: H₂SO₄(conc.), heat
Find
(a) Reaction type (b) Structural formula of product (c) Product name
Solution
- 1(a) Alcohol + carboxylic acid → ester + water = esterification
- 2(b) The –OH from the acid and the H from the alcohol combine to form H₂O; the remaining fragments join: CH₃CH₂COO–CH₂CH₂CH₃
- 3(c) Propyl propanoate (propan-1-ol contributes the propyl group; propanoic acid contributes the propanoate part)
Worked Example
Draw 3 repeat units of the addition polymer formed from chloroethene (CH₂=CHCl) and name the polymer.
Given
- Monomer: chloroethene (CH₂=CHCl)
Find
3 repeat units of addition polymer; name
Solution
- 1The C=C double bond breaks; each monomer contributes –CH₂–CHCl– as a repeat unit
- 2Three repeat units: –[CH₂–CHCl]–[CH₂–CHCl]–[CH₂–CHCl]–
Watch Out
COMMON MISTAKE: Students write water as a by-product of addition polymerisation. Water is only a by-product in condensation polymerisation (and esterification). In addition polymerisation there is NO by-product — all atoms in the monomers are incorporated into the polymer chain.
Practice Question
Explain the difference between esterification and saponification. Include reactants, products, and whether each reaction is reversible.
(4 marks)
Practice Question
Classify the following reaction and write the structural formula of the organic product: CH₃–CH₂–CH₂Br + KOH(aq) →
(4 marks)
Practice Question
Nylon-6,6 is formed from a diamine and a dicarboxylic acid. State the type of polymerisation and explain why water is produced.
(4 marks)