IB Chemistry · Unit 8 · Organic

The chemistry
of carbon.

Carbon makes more compounds than every other element combined. Get the families, get the mechanisms, get the names.

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C₆H₆ BENZENE
Unit 8 · Standard & Higher Level

8 lessons to work through.

The required syllabus content for Unit 8, in order. Each card is one lesson-sized checkpoint.

Lesson 1

Organic Formulae

Lesson 1 of Unit 8.

Lesson 2

Functional Groups

Organic chemistry focuses on the chemistry of compounds containing carbon.

Lesson 3

Homologous Series

A homologous series is a family of compounds that can be grouped together based on similarities in their structure, functional group and chemical reactivity.

Lesson 4

Naming Molecules

Lesson 4 of Unit 8.

Lesson 5

Structural Isomers

When naming amines, the parent name loses the suffix -e, which is replaced by

Lesson 6

Polymers

Crude oil is a primary source of hydrocarbons used to produce organic compounds.

Lesson 7

Addition Polymerisation

Produce a polymer with a regular ABABAB… structure

Lesson 8

Condensation Polymerisation

In condensation polymerisation, each monomer have two reactive functional groups at each end, we will look at the following examples:

Lessons in detail

The unit, lesson by lesson.

Each lesson card below mirrors the original teacher deck — syllabus refs, content, worked examples and practice questions in order.

Lesson 1

Organic formulae

Structure 3.2.1

Lesson outcomes

Organic molecules can be drawn in several ways depending on how much detail is needed:

TypeExample for propan-1-olUse
EmpiricalC₃H₈OSimplest ratio (here, equal to molecular).
MolecularC₃H₈OActual count of each atom.
Full structuralH₃C–CH₂–CH₂–OHShows every bond.
CondensedCH₃CH₂CH₂OHCompact text form.
Skeletal/\/OHEach vertex = C; H's on C are implicit; only heteroatoms are drawn.

Try these

  1. What is the difference between an empirical, molecular, and structural formula?
    Show answer
    Empirical: simplest whole-number ratio of atoms. Molecular: actual count of each atom. Structural: shows how atoms are connected (every bond drawn — full; abbreviated — condensed; vertices implicit — skeletal).
  2. Convert butan-2-ol to (a) full structural, (b) condensed and (c) skeletal form.
    Show answer
    (a) H₃C–CH(OH)–CH₂–CH₃. (b) CH₃CH(OH)CH₂CH₃. (c) skeletal: a 4-carbon zig-zag with –OH on the second carbon.
  3. What molecular formula does the skeletal structure of cyclohexane represent?
    Show answer
    A regular hexagon with no labelled atoms = C₆H₁₂. Each vertex is a CH₂.
  4. Draw the skeletal structure of propan-2-ol.
    Show answer
    Three-carbon chain (\__/ shape) with an –OH on the middle carbon. Implicit H atoms are not shown.
  5. Convert this skeletal formula to a molecular formula: a 6-carbon ring with one double bond and a –COOH attached to one vertex.
    Show answer
    Cyclohex-1-ene with a carboxylic acid group attached = cyclohex-2-ene-1-carboxylic acid. C₇H₁₀O₂.
Lesson 2

Functional groups

Structure 3.2.2

Lesson outcomes

A functional group is an atom or group of atoms within a molecule that determines its chemistry. Compounds with the same group react similarly — and this is what justifies grouping them into homologous series.

GroupStructureSuffix / Prefix
AlkaneC–C single-ane
AlkeneC=C double-ene
AlkyneC≡C triple-yne
HalogenoalkaneC–X (X = F, Cl, Br, I)halo- prefix
AlcoholC–OH-ol
Aldehyde–CHO (end of chain)-al
KetoneC–CO–C-one
Carboxylic acid–COOH-oic acid
EsterC–COO–C-oate
AmineC–NH₂-amine
Amide–CONH₂-amide
Nitrile–C≡N-nitrile

Try these

  1. What is a 'functional group'? Why is it useful to identify them?
    Show answer
    An atom or group of atoms within a molecule that gives it characteristic chemistry. Compounds with the same FG react similarly. So once you can identify the FG, you can predict reactions, names, and properties — the whole logic of organic chemistry.
  2. Identify the functional group(s) in: (a) CH₃CH₂COOH, (b) CH₃COCH₃, (c) CH₃CH=CH₂, (d) CH₃CH₂NH₂.
    Show answer
    (a) carboxylic acid (propanoic acid). (b) ketone (propanone/acetone). (c) alkene (propene). (d) primary amine (ethylamine).
  3. Identify the functional groups in: (a) CH₃CH₂Cl, (b) HCOOH, (c) CH₃CONH₂, (d) HOCH₂CH₂COOH.
    Show answer
    (a) halogenoalkane (chloroethane). (b) carboxylic acid (methanoic acid). (c) amide (ethanamide). (d) two FGs — alcohol AND carboxylic acid (3-hydroxypropanoic acid).
  4. How would you distinguish an aldehyde from a ketone chemically?
    Show answer
    Use Tollens' reagent or Fehling's solution — both are mild oxidising agents. Aldehydes are oxidised (Tollens' gives silver mirror; Fehling's gives red-brown Cu₂O precipitate). Ketones are not oxidised by either, so no change.
  5. Why is acidified K₂Cr₂O₇ orange-to-green colour change used as a test for oxidisable compounds (alcohols, aldehydes)?
    Show answer
    Orange Cr₂O₇²⁻ (Cr +6) is reduced to green Cr³⁺ during oxidation of the substrate. Colour change is a visible confirmation that oxidation has occurred — and that the substrate is therefore a primary/secondary alcohol or aldehyde.
Lesson 3

Homologous series

Structure 3.2.3

Lesson outcomes

A homologous series is a family of organic compounds with:

Common general formulae

Try these

  1. Define a homologous series. List the five defining features.
    Show answer
    Family of organic compounds with: (1) same general formula, (2) same functional group, (3) successive members differ by –CH₂–, (4) gradually changing physical properties (b.p., m.p., density, viscosity rise with chain length), (5) similar chemical reactions.
  2. Give the general formula for: (a) alkanes, (b) alkenes, (c) alkynes, (d) alcohols, (e) carboxylic acids.
    Show answer
    (a) CnH2n+2. (b) CnH2n (one C=C). (c) CnH2n−2 (one C≡C). (d) CnH2n+1OH or CnH2n+2O. (e) CnH2n+1COOH or CnH2nO₂.
  3. Why do boiling points of alkanes increase up the series (methane gas to hexadecane liquid)?
    Show answer
    More carbons → more electrons → larger and more polarisable electron cloud → stronger London dispersion forces between molecules → more energy needed to overcome them.
  4. Which has the higher b.p.: butane (CH₃CH₂CH₂CH₃) or 2-methylpropane ((CH₃)₃CH)? Why?
    Show answer
    Butane (b.p. ≈ −0.5 °C) > 2-methylpropane (≈ −11.7 °C). Same molecular formula C₄H₁₀, but the straight chain has greater surface contact and so stronger London forces. Branched isomers have lower b.p.
  5. What is the next member of the alcohols series after butan-1-ol? Write its molecular formula.
    Show answer
    Pentan-1-ol, CH₃CH₂CH₂CH₂CH₂OH = C₅H₁₂O. Each next member adds one –CH₂– → +CH₂.
Lesson 4

Naming molecules (IUPAC)

Structure 3.2.4

Lesson outcomes

The five-step IUPAC procedure

  1. Find the longest carbon chain containing the principal functional group. Length = root name (meth- 1, eth- 2, prop- 3, but- 4, pent- 5, hex- 6, hept- 7, oct- 8).
  2. Number the chain from the end nearest the principal functional group (lowest locant rule). If no FG, number from the end nearest the first substituent.
  3. Identify substituents: methyl (–CH₃), ethyl (–C₂H₅), chloro (–Cl), etc.
  4. Alphabetise substituents in the name (ignore di-, tri-, etc. when ordering).
  5. Assemble: locant-substituent-locant-substituent-root-suffix. Hyphens between numbers and letters; commas between numbers.
Worked example

Name a branched alcohol

Problem. Name CH₃CH(CH₃)CH₂CH(OH)CH₃.
Solution. Longest chain containing –OH = 5 carbons → pentan-...
–OH attached to C2 (lower than C4 from the other end): pentan-2-ol.
Methyl substituent at C4: 4-methylpentan-2-ol. Final: 4-methylpentan-2-ol.
Worked example

Name a branched alkene

Problem. Name CH₂=CH–C(CH₃)₃.
Solution. Longest chain with C=C = 4 carbons → but-...-ene.
Number from the end nearest the double bond: C1=C2–C3(CH₃)₂–CH₃.
But wait — C(CH₃)₃ as a single group means there's a substituent at C3. Three methyls at C3? Let me reread: CH₂=CH–C(CH₃)₃ has C1=C2–C3 with three methyls on C3. So the main chain is 4 carbons (C1, C2, C3, plus one of the methyls = C4): 3,3-dimethylbut-1-ene.
Final: 3,3-dimethylbut-1-ene.

Try these

  1. Name: (a) CH₃CH₂CH₂CH₂CH₃, (b) CH₃CH(CH₃)CH₃, (c) (CH₃)₃CCH₂CH₃.
    Show answer
    (a) pentane. (b) 2-methylpropane. (c) 2,2-dimethylbutane (longest chain = 4 C, then 2 methyls on C2).
  2. Name: (a) CH₂=CHCH₂CH₃, (b) CH₃CH=CHCH₃, (c) CH₃CH(CH₃)CH=CH₂.
    Show answer
    (a) but-1-ene. (b) but-2-ene. (c) 3-methylbut-1-ene.
  3. Draw the structures of: (a) 2-methylpropan-2-ol, (b) 3-chlorobutanoic acid, (c) propanoic acid, (d) ethyl methanoate.
    Show answer
    (a) (CH₃)₃COH — tert-butanol. (b) CH₃CHClCH₂COOH. (c) CH₃CH₂COOH. (d) HCOOCH₂CH₃ — the methanoate ion paired with an ethyl group.
  4. Why is the name 2-ethylbutan-1-ol wrong? Give the correct name.
    Show answer
    The longest continuous chain is 5 carbons (the ethyl substituent and three chain carbons add up to a longer chain). Correct: 2-methylpentan-1-ol.
  5. Name: (a) CH₃CHBrCH(CH₃)CH₂CH₃ — a brominated alkane, (b) ClCH₂CH₂Cl — used in industry.
    Show answer
    (a) 2-bromo-3-methylpentane (number to give Br lowest locant). (b) 1,2-dichloroethane.
Lesson 5

Structural isomers

Structure 3.2.5

Lesson outcomes

Structural isomers have the same molecular formula but different connectivity. Three types:

Number of structural isomers grows rapidly with carbon count. C₅H₁₂ has 3 isomers; C₆H₁₄ has 5; C₁₀H₂₂ has 75.

Worked example

Isomers of C₄H₁₀O

Problem. Draw all structural isomers of C₄H₁₀O. State the FG of each.
Solution. Alcohols (4):
• butan-1-ol — CH₃CH₂CH₂CH₂OH
• butan-2-ol — CH₃CH₂CH(OH)CH₃
• 2-methylpropan-1-ol — (CH₃)₂CHCH₂OH
• 2-methylpropan-2-ol — (CH₃)₃COH
Ethers (3):
• methoxypropane — CH₃O–CH₂CH₂CH₃
• 2-methoxypropane — CH₃O–CH(CH₃)₂
• ethoxyethane — CH₃CH₂–O–CH₂CH₃
Total: 7 isomers (4 alcohols + 3 ethers).

Try these

  1. What is structural isomerism? List the three types.
    Show answer
    Compounds with the same molecular formula but different structural arrangements of atoms. Three types: chain (different carbon skeleton), position (different position of FG on same skeleton), functional-group (different FG entirely).
  2. How many chain isomers of C₅H₁₂ exist? Draw and name them.
    Show answer
    Three: pentane (straight chain), 2-methylbutane (one methyl branch), 2,2-dimethylpropane (also called neopentane — two methyls on the central C).
  3. Identify the type of isomerism between: (a) propan-1-ol and propan-2-ol, (b) ethanol and methoxymethane, (c) pentane and 2-methylbutane.
    Show answer
    (a) Position isomerism (same FG, different position). (b) Functional-group isomerism (alcohol vs ether). (c) Chain isomerism (different branching).
  4. Draw all isomers of C₃H₆O₂ that are carboxylic acids or esters.
    Show answer
    Propanoic acid: CH₃CH₂COOH. Methyl methanoate: HCOOCH₃. Ethyl... wait, ethyl methanoate has C₃H₆O₂ too? No — ethyl methanoate is HCOOCH₂CH₃ = C₃H₆O₂ ✓. Actually only methyl methanoate HCOOCH₃ = C₂H₄O₂ — that's smaller. Correctly: just two — propanoic acid and methyl ethanoate (CH₃COOCH₃).
Lesson 6

Polymers · introduction

Reactivity 2.5

Lesson outcomes

A polymer is a long-chain molecule made by joining many small repeating units (monomers). The repeating unit is what's left after polymerisation strips the reactive bits from the monomer.

Two main polymerisation strategies

Common polymers

PolymerMonomerTypeUse
Polythene (LDPE/HDPE)etheneAdditionBags, bottles
Polypropene (PP)propeneAdditionRope, packaging
PVCchloroetheneAdditionPipes, cables
Polystyrenephenylethene (styrene)AdditionInsulation, cups
Nylon 6,6hexanedioic acid + 1,6-diaminohexaneCondensationFibres, ropes
PETbenzene-1,4-dicarboxylic acid + ethane-1,2-diolCondensationBottles, fibres

Try these

  1. Define monomer, polymer, and repeat unit.
    Show answer
    Monomer: small molecule with at least one reactive site (e.g. a C=C or a pair of functional groups) that links to others to form a polymer. Polymer: long molecule made of many repeating monomer units. Repeat unit: the structural unit that appears repeatedly along the polymer chain (typically the monomer with the reactive bonds opened).
  2. Distinguish addition and condensation polymerisation in one line each.
    Show answer
    Addition: monomers with C=C join end-to-end; no atoms lost. Condensation: monomers with two different FGs join with loss of a small molecule (usually H₂O).
  3. Why are most addition polymers non-biodegradable?
    Show answer
    The C–C backbone is a chain of strong σ bonds with no functional groups that microbes or hydrolysis can attack. Without an attackable site, the polymer persists for centuries.
  4. Which is more easily recycled by hydrolysis: polyester (PET) or polythene?
    Show answer
    Polyester. Its ester linkages (–COO–) can be cleaved by water (especially with acid or base catalysis) back to the original di-acid and di-ol monomers. Polythene's C–C backbone has no hydrolysable groups.
Lesson 7

Addition polymerisation

Reactivity 2.5

Lesson outcomes

Addition polymerisation works for any monomer with a C=C double bond. Under heat, pressure and a catalyst, the π bond breaks; each carbon forms one new bond to the next monomer; nothing else is added or removed.

The recipe

n CH₂=CHX → –(CH₂–CHX)n–

The repeat unit shows what's inside the brackets: take the C=C, draw it as a single C–C, and place X (the substituent) on whichever carbon it was on in the monomer.

Worked example

Repeat unit of PVC

Problem. Draw the monomer and repeat unit of poly(chloroethene) (PVC).
Solution. Monomer: CH₂=CHCl (chloroethene, "vinyl chloride").
Repeat unit: –(CH₂–CHCl)n–. Each former double-bond opens up; each unit gains two new single bonds to neighbours.
Worked example

Identifying a monomer from a polymer

Problem. The repeat unit of polystyrene is –(CH₂–CH(C₆H₅))n–. What is the monomer?
Solution. Insert a double bond between the two carbons of the repeat unit: CH₂=CH(C₆H₅) = phenylethene (styrene).

Try these

  1. Draw the monomer and repeat unit of polythene (poly(ethene)).
    Show answer
    Monomer: CH₂=CH₂ (ethene). Repeat unit: –(CH₂–CH₂)n–. The C=C opens; each former double-bond C makes one new single bond to the next monomer unit.
  2. Draw the monomer and repeat unit of poly(propene).
    Show answer
    Monomer: CH₂=CH(CH₃) (propene). Repeat unit: –(CH₂–CH(CH₃))n–. The methyl group decorates every other carbon along the backbone.
  3. 1.00 kg of polypropene with degree of polymerisation 5000. Calculate the molar mass of the polymer chain.
    Show answer
    M(propene C₃H₆) = 42.08. Polymer M = 5000 × 42.08 = 2.10 × 10⁵ g mol⁻¹. n(polymer chains) = 1000/2.10 × 10⁵ = 4.76 × 10⁻³ mol.
  4. A polymer has the repeat unit –(CH₂–CCl(CH₃))n–. Identify the monomer.
    Show answer
    Insert a C=C in place of the C–C in the backbone: monomer = CH₂=CCl(CH₃) = 2-chloroprop-1-ene.
  5. Why doesn't poly(propene) have head-to-head and tail-to-tail isomers in regularly-prepared samples?
    Show answer
    Industrial Ziegler-Natta or metallocene catalysts give 'isotactic' polypropene where all methyl groups point the same way. Without such catalysts (radical polymerisation), atactic (random) polypropene results — a softer, less ordered polymer. Industry deliberately controls this.
Lesson 8

Condensation polymerisation

Reactivity 2.5

Lesson outcomes

Condensation polymerisation joins two monomers by forming a new bond and expelling a small molecule (typically H₂O). It requires monomers with two reactive functional groups so the chain can extend in both directions.

Two key examples

Polyester (PET):

HOOC–C₆H₄–COOH   +   HO–CH₂CH₂–OH   →   –[OOC–C₆H₄–COO–CH₂CH₂–]n–   +   2n H₂O

Polyamide (Nylon 6,6):

HOOC–(CH₂)₄–COOH   +   H₂N–(CH₂)₆–NH₂   →   –[CO–(CH₂)₄–CO–NH–(CH₂)₆–NH–]n–   +   2n H₂O

Bonds formed

Condensation vs addition — three differences

AdditionCondensation
Monomer FGC=C2× FG per monomer (e.g. di-acid + di-amine)
Small molecule lostNoneH₂O (or HCl)
Atom economy100%<100%
BiodegradabilityPoorBetter (ester/amide bonds can be hydrolysed)

Try these

  1. Why is nylon called a polyamide?
    Show answer
    Each repeat unit contains amide linkages (–CONH–), formed by condensation of a carboxylic acid group with an amine group with loss of water.
  2. What is the small molecule eliminated when (a) a polyester and (b) a polyamide form by condensation?
    Show answer
    Both eliminate H₂O. In a polyester, –COOH + HO– → –COO– + H₂O. In a polyamide, –COOH + H₂N– → –CONH– + H₂O.
  3. Suggest one reason PET can be hydrolysed for recycling, but polythene cannot.
    Show answer
    PET contains ester linkages that hydrolyse with water (especially with acid or base catalysis), regenerating the original di-acid and di-ol monomers. Polythene's C–C backbone has no hydrolysable linkages.
  4. Compare the atom economies of (a) addition polymerisation of ethene, (b) condensation of ethanedioic acid with ethane-1,2-diol.
    Show answer
    (a) 100% — all atoms end up in the polymer. (b) Less than 100% — every link loses one H₂O molecule. For each pair joined: AE = (M_repeat)/(M_monomers) × 100%. The wasted mass is the H₂O eliminated.
Vocabulary

42 terms to own.

If you can't define one of these in a sentence, that's where to revise next. Click any term for its definition.

elementcompoundatomelectronmoleempirical formulamolecular formulaatom economysolutiongraphsurface areagroupcovalent bonddouble bondorganic chemistryhydrocarbonhomologous seriesfunctional groupalkanealkenealkynealcoholhalogenoalkanecarboxylic acidesteraminealdehydeketonestructural isomerchain isomerposition isomerfunctional group isomeradditionpolymerisationmonomerpolymeracidbasephkaoptical isomerenantiomer