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Year 12 Chemistry Module 7 · Organic Chemistry ⏱ ~45 min 5 MC · 3 Short Answer Lesson 1 of 23

Introduction to Organic Chemistry & IUPAC Nomenclature I

Unlock the language of organic chemistry — learn how carbon's tetravalency and IUPAC naming rules let you encode and decode the structure of any organic molecule up to C8.

Today's hook: Every time you read "hexyl acetate" or "2-methylpropan-1-ol" on a label, you're reading a code that tells you exactly how many carbons the molecule has and what kind of compound it is. By the end of this lesson you'll be able to decode — and write — those names yourself.
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Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Before You Read

Before you read on, write down what you think each part of the name "hex-1-ene" is telling you. How many carbons do you think it has? What do you think the "1" means? What does the "-ene" ending suggest?

Hold your answer — you will return to test and revise it at the end of the lesson.

Learning Intentions
goals

Know

  • The general formulas for alkanes (CnH2n+2), alkenes (CnH2n), alkynes (CnH2n-2)
  • The molecular shape and bond angle around carbon for single, double, and triple bonds (tetrahedral/109.5°, trigonal planar/120°, linear/180°)
  • The eight chain-length IUPAC prefixes: meth– to oct–
  • The seven major functional group classes and their suffixes
  • What a homologous series is and how it differs from isomers

Understand

  • Why carbon's tetravalency creates the structural diversity of organic chemistry
  • Why the number of bonds between two carbons determines the shape around them, and how to predict bond angles from bond type alone
  • Why members of a homologous series share chemical behaviour but differ in physical properties
  • When each type of structural formula (full, condensed, skeletal) is appropriate

Can Do

  • Name any straight-chain or branched alkane up to C8
  • Name any alkene or alkyne with correct double/triple bond locant
  • Draw full structural, condensed, and skeletal formulas and convert between them
  • Verify a molecular formula using the relevant general formula
  • Identify the functional group class from a structural formula
Scan these before reading
vocab
HydrocarbonAn organic compound containing only carbon and hydrogen atoms.
Functional groupA specific atom arrangement that determines characteristic chemical reactions.
Homologous seriesA family of compounds with the same functional group, differing by CH₂.
IsomerCompounds with the same molecular formula but different structural arrangements.
Addition reactionA reaction where atoms add across a carbon-carbon multiple bond.
Substitution reactionA reaction where one atom or group replaces another in a molecule.
1
Why Carbon is the Backbone of Organic Chemistry

Tetravalency · carbon's structural diversity

Misconception to fix: Organic chemistry only studies molecules produced by living organisms. Correct: Organic chemistry is the study of carbon-containing compounds, regardless of origin. Many organic compounds are synthesised industrially. The "organic" label refers to carbon-based structure, not biological source.

Carbon's ability to form four stable covalent bonds — with itself and with H, O, N, and halogens — creates a structural diversity no other element comes close to matching.

Organic chemistry is the study of carbon compounds (with conventional exceptions such as carbonates, oxides and carbides). Carbon sits in Group 14 and has four valence electrons, so it forms four covalent bonds to complete its octet. This tetravalency means carbon can bond to itself in straight chains, branched chains, and rings of essentially unlimited length while simultaneously bonding to other elements. The result is millions of known organic compounds and new ones synthesised every year.

Quick check: A carbon atom involved in a triple bond has what shape and bond angle?
2
Types of Structural Formulae

Full structural · Condensed · Skeletal · Converting between them

Organic structures can be drawn at three levels of detail, and knowing which one a question is asking for — and how to read each — is a practical exam skill that affects marks in almost every Module 7 question.

A full structural formula (displayed formula) shows every atom and every bond explicitly as individual lines. For propane: each C–H and C–C bond is drawn separately. This is required when a question says "draw the structural formula" without further specification, especially for short molecules (≤ C4).

A condensed structural formula groups hydrogen atoms with their carbon: propane = CH₃CH₂CH₃; but-1-ene = CH₂=CHCH₂CH₃. Branches are shown in brackets — e.g. 2-methylpropane = CH₃CH(CH₃)CH₃.

A skeletal (line) formula represents the carbon chain as a zigzag, where each vertex and each endpoint is an implied carbon atom, and all H atoms on carbon are implied. Heteroatoms (O, N, Cl, Br) and their attached H atoms must be shown explicitly. Skeletal formulas are fully accepted in NSW HSC responses.

What is shown
  • Every atom and bond explicitly
  • Atoms grouped by carbon; branches in brackets
  • Zigzag; C and H implied; heteroatoms shown
When to use
  • Short molecules ≤ C4; "draw the structural formula"
  • Medium chains; reaction equations
  • Longer molecules (C5+); complex organic reactions
HSC Tip: When an HSC question says "draw the structural formula," draw the FULL structural formula unless the question specifies "condensed" or "skeletal."
Common Error: In skeletal formulas, students miscount implied hydrogens. A terminal C has ONE chain bond → needs THREE implied H atoms (–CH₃). A vertex C has TWO chain bonds → needs TWO implied H (–CH₂–). A vertex with a branch has THREE chain bonds → needs ONE implied H (–CH–). Miscounting changes the molecular formula and will be penalised.
In a skeletal formula, a vertex carbon with two chain bonds has _____ implied hydrogen atoms. A terminal carbon (end of zigzag) has _____ implied hydrogen atoms.
3
Homologous Series and Functional Groups

Families of compounds · –CH₂– increment · predictable properties

A functional group is the part of a molecule that reacts; a homologous series is a family of molecules that share a functional group and differ only in chain length — once you know the family rules, you know how any member will behave.

A functional group is a specific atom or group of atoms responsible for the characteristic reactions of a molecule. Two molecules with the same functional group will undergo the same types of reactions regardless of chain length. The seven major classes:

Class Functional group Suffix Simple example
AlkaneC–C only (no functional group)–aneEthane
AlkeneC=C double bond–eneEthene
AlkyneC≡C triple bond–yneEthyne
Alcohol–OH (hydroxyl)–olEthanol
Aldehyde–CHO (at chain end)–alEthanal
KetoneC=O (in chain, not terminal)–onePropanone
Carboxylic acid–COOH (at chain end)–oic acidEthanoic acid

A homologous series is a sequence of compounds sharing the same functional group and general formula, differing by one –CH₂– unit between consecutive members. Physical properties shift gradually and predictably: boiling point, viscosity, and melting point all increase because dispersion forces strengthen with molecular size. Chemical properties remain similar because the functional group is unchanged.

Method: Learn to identify a functional group from a structural formula before applying any naming rule. Functional group identification is the first move in every naming, property, and reaction question in Module 7.
Common Error — Critical: Students confuse "homologous series" with "isomers." Members of a homologous series have different molecular formulae (C₂H₆, C₃H₈, C₄H₁₀…). Isomers have the same molecular formula arranged differently. These are opposite concepts.
True or False: Members of a homologous series have the same molecular formula but different structural arrangements.
4
IUPAC Naming Rules for Alkanes

Longest chain · Lowest locants · Alphabetical substituents

IUPAC names are built from a small set of rules applied in a fixed order — learn the rules, and you can name or decode any straight-chain or branched alkane you encounter.

An IUPAC name for an alkane is built in three steps:

  1. Find the LONGEST continuous carbon chain — this is the parent chain and gives the base name.
  2. Number the chain from the end CLOSEST to the first branch — giving substituents the lowest possible locants.
  3. Name each substituent as a prefix with its locant, listed alphabetically (ignoring di–, tri– multiplying prefixes when alphabetising).

Chain-length prefixes to memorise: meth– (1C), eth– (2C), prop– (3C), but– (4C), pent– (5C), hex– (6C), hept– (7C), oct– (8C). All alkane names end in –ane.

Common substituents: methyl (–CH₃), ethyl (–C₂H₅). If the same substituent appears more than once, add di–, tri–, or tetra– and list all locants separated by commas — e.g. 2,3-dimethylbutane.

Action
  • Find the longest continuous carbon chain
  • Number from the end nearest the first branch
  • Name substituents alphabetically (not by chain length)
Common failure
  • Counting from a branch rather than tracing the longest path
  • Numbering from the wrong end, giving unnecessarily high locants
  • Listing ethyl after methyl because methyl appears shorter
Method — Step 1 is Critical: The FIRST step before writing any name is to find the longest continuous chain. Redraw the molecule as a straight chain with branches if the original drawing is confusing.
Common Error: "3-methylbutane" — students number from the wrong end and give the methyl branch locant 3. The correct name is 2-methylbutane because numbering from the nearer end gives the branch at C2. Always use the lowest possible locants.
Insight: IUPAC names are fully reversible — from a name you reconstruct a unique structure; from a structure you derive a unique name. Practise drawing a structure FROM a name (not just naming drawn structures).
Quick check: What is the correct IUPAC name for CH₃CH(CH₃)CH₂CH₃?
5
IUPAC Naming Rules for Alkenes and Alkynes

Double/triple bond locant · Priority numbering · Suffix –ene / –yne

Alkenes and alkynes follow the same chain-length and branch-naming rules as alkanes, with two additions: the suffix changes and a locant is required to show where the double or triple bond begins.

For alkenes:

  • Find the longest chain that includes the C=C double bond — this is the parent chain.
  • Change the suffix from –ane to –ene.
  • Add a locant immediately before –ene showing the lower-numbered carbon of the C=C pair.
  • Number the chain from the end closer to the double bond (the double bond takes priority over branches in numbering).

For alkynes: Same rules; suffix = –yne.

When a molecule has both a branch and a double bond: the double bond takes priority — give the double bond the lowest possible locant, even if this means a branch receives a higher locant.

IUPAC name
  • Ethene
  • Prop-1-ene
  • But-2-ene
  • Ethyne
  • Pent-1-yne
  • Pent-2-yne
Reasoning
  • 2C, one C=C — no locant needed
  • 3C, C=C at C1 from nearer end
  • 4C, C=C at C2 (same from either end)
  • 2C, triple bond — no locant needed
  • 5C, C≡C starts at C1
  • 5C, C≡C starts at C2 from nearer end
IUPAC Format: The locant in a modern IUPAC alkene name must appear immediately before the suffix, hyphenated: "but-1-ene" is correct; "1-butene" is the older format still seen in some textbooks and accepted in NSW HSC.
Common Error: Students forget to include the double bond carbon in the parent chain. If the C=C does not sit in the longest chain, you have chosen the wrong parent chain.
Odd one out: Which of these IUPAC names is NOT correct?
6
Molecular Shape and Bond Angles

Shape from bond type · Predict angles · Observable pattern

The number of other atoms a carbon is bonded to determines the shape around it — and you can predict bond angles directly from the type of bond present.

  • Single bonds only (as in alkanes): the carbon is bonded to four atoms, which spread out as far apart as possible — pointing to the corners of a tetrahedron, with bond angles of 109.5°. Methane (CH₄) is the classic example.
  • A double bond (as in alkenes): the two double-bonded carbons and the atoms attached to them all lie in a flat plane, with bond angles of about 120°. Ethene (CH₂=CH₂) is flat.
  • A triple bond (as in alkynes): the triple-bonded carbons and their attached atoms lie in a straight line, with a bond angle of 180°. Ethyne (HC≡CH) is linear.

The pattern: more bonds between two carbons pulls the molecule into a flatter, then straighter shape. You can state the shape and bond angle for any carbon just by spotting whether it sits at a single, double, or triple bond.

Note for students: you are not required to explain why these shapes occur using orbital theory — only to identify the shape and bond angle from the bond type present.
Single bond (alkane)
Shape: Tetrahedral
Bond angle: 109.5°
Example: Methane (CH₄)
Double bond (alkene)
Shape: Trigonal planar
Bond angle: ~120°
Example: Ethene (CH₂=CH₂)
Triple bond (alkyne)
Shape: Linear
Bond angle: 180°
Example: Ethyne (HC≡CH)
Common Error: Students state the bond angle without the shape name, or vice versa. Always give both — "trigonal planar with bond angles of approximately 120°." Giving only one of the two will cost a mark.
Quick check: The two carbons in ethene (CH₂=CH₂) and their attached atoms all lie in the same flat plane. What bond angle would you measure between adjacent bonds at one of those carbons?
2-methylbutane — full structural formula C C C C C H H H H H H H H H H H H C1 C2 C3 C4 2-methylbutane (C5H12) — full structural formula, all bonds shown
but-1-ene — full structural formula showing C=C double bond; flat (trigonal planar) geometry at C1 and C2 C C C C H H H H H H H H C1 (C=C, ~120°) C2 (C=C, ~120°) C3 (single, 109.5°) C4 (single, 109.5°) but-1-ene (C4H8) — double bond shown by two parallel lines; C1 and C2 have trigonal planar geometry, ~120°
hex-1-ene — skeletal formula and condensed formula comparison SKELETAL FORMULA CONDENSED FORMULA C1 CH2=CHCH2CH2CH2CH3 double bond shown explicitly; H atoms grouped with their carbon hex-1-ene (C6H12) — in the skeletal formula, every vertex and endpoint is an implied carbon
Worked Example 1 — Naming a branched alkane (straightforward)

Name the compound CH₃CH₂CH(CH₃)CH₂CH₃ and draw its full structural formula.

1
Find the longest continuous carbon chain. Trace: CH₃–CH₂–CH–CH₂–CH₃ = 5 carbons in the main chain. The –CH₃ branch hangs off the middle carbon. Parent chain = pentane (5C).
2
Number from the end closer to the branch. From left: C1–C2–C3(branch)–C4–C5 → branch at C3. From right: also C3. Both directions give C3, so either direction is valid.
3
Name the substituent. –CH₃ = methyl, locant = 3. Full IUPAC name: 3-methylpentane.
4
Verify molecular formula. Parent chain = 5C; methyl branch = 1C; total = 6C. Using CnH2n+2: n = 6 → H = 2(6)+2 = 14. Molecular formula = C₆H₁₄ ✓.
Worked Example 2 — Naming an alkene with a branch (intermediate)

Write the IUPAC name for: CH₃CH₂CH(CH₃)CH=CH₂

1
Identify the functional group. C=C double bond → alkene. Find the longest chain that includes the C=C: CH₂=CH–CH(CH₃)–CH₂–CH₃ = 5 carbons including both C=C carbons. Parent chain = pentene.
2
Number from the end closer to the double bond. From the CH₂= end: C1=C2–C3–C4–C5. Double bond at C1. From the other end: double bond at C3. Choose C1 (lower locant).
3
Locate the branch. With C1 at CH₂=, the CH₃ branch is at C3. Substituent = 3-methyl.
4
Assemble the name. 5C chain + alkene + double bond at C1 + methyl at C3 = 3-methylpent-1-ene. Verify: 6C total, one C=C → C₆H₁₂ ✓.
Worked Example 3 — Evaluating a student's structure and molecular formula (hard)

A student draws CH₃–C(CH₃)₂–CH₂–CH₂–CH₃ for "2,2-dimethylpentane" and writes molecular formula C₆H₁₄. Assess whether both are correct.

1
Decode the IUPAC name. "2,2-dimethylpentane" = parent chain 5C (pentane); two methyl groups, both at C2.
2
Check the drawn structure. CH₃–C(CH₃)₂–CH₂–CH₂–CH₃: C1(CH₃)–C2(C(CH₃)₂)–C3(CH₂)–C4(CH₂)–C5(CH₃). C2 has bonds to C1, C3, and two CH₃ branches (four bonds total) — this matches 2,2-dimethylpentane. Structure is correct ✓
3
Check the molecular formula. Count all carbons: parent chain 5C + two methyl branches 2C = 7C total. Using CnH2n+2: n=7 → H = 2(7)+2 = 16. Correct formula = C₇H₁₆.
4
Diagnose the error. The student wrote C₆H₁₄. They counted 5 (parent) + 1 (one methyl) = 6, forgetting the second methyl branch. Molecular formula is wrong ✗ — should be C₇H₁₆.
Interactive Tool — IUPAC Nomenclature Builder Open fullscreen ↗
Safe Handling and Disposal of Organic Substances
+5 XP

Why organics need specific safety and disposal procedures

Many organic substances are flammable, volatile and/or toxic, so they must be handled and disposed of using procedures that control those specific hazards.

Handling: many organics (e.g. alcohols, alkanes) are highly flammable, so keep them away from naked flames and heat sources and warm them with a water bath, not a Bunsen flame. Volatile and toxic organics must be used in a fume cupboard, with gloves and safety glasses, because vapours can be inhaled or absorbed through the skin.

Disposal: organic liquids must NOT be poured down the sink — many are immiscible with water, persistent and toxic to aquatic life. They are collected in labelled organic-waste containers for safe disposal (e.g. by licensed processing or incineration). Always consult the Safety Data Sheet (SDS) and a risk assessment before use.

Copy the highlighted handling and disposal points into your book.

Waste organic liquids should not be poured down the sink; instead they are collected in labelled organic-_____ containers for safe disposal.

The IUPAC Nomenclature tool shows the suffix ‘-ane’ indicates a hydrocarbon that is…
🔬 Predict — Then Reveal +8 XP
An unknown compound has the molecular formula C₅H₁₂. Before reading on, predict: (a) how many structural isomers exist, and (b) which IUPAC suffix (ending) this compound's name will have.
Your predictionExpert answerCompare
Key Formulas & Relationships
Alkane general formula: CnH2n+2
Saturated — single C–C bonds only e.g. n=4: C4H10 (butane) | n=1: CH4 (methane)
Alkene general formula: CnH2n
One C=C double bond (one degree of unsaturation) e.g. n=4: C4H8 (but-1-ene or but-2-ene)
Alkyne general formula: CnH2n-2
One C≡C triple bond (two degrees of unsaturation) e.g. n=5: C5H8 (pent-1-yne)
Bond type → Shape → Bond angle
Single bonds only → tetrahedral → 109.5° (alkanes, e.g. methane) Double bond C=C → trigonal planar → ~120° (alkenes, e.g. ethene) Triple bond C≡C → linear → 180° (alkynes, e.g. ethyne)

No numerical calculation formulas this lesson — nomenclature is conceptual. Memorise general formulas and verify any molecular formula by substituting n.

Complete the Learn phase to unlock Practice.

A1
From Structure to IUPAC Name

For each compound below, write the correct IUPAC name. Show your working (identify the parent chain length, numbering direction, and substituents/locants). Verify each molecular formula using the appropriate general formula.

# Condensed structural formula Your IUPAC name Molecular formula check
1 CH₃CH₂CH₂CH₂CH₃
2 CH₃CH(CH₃)CH₂CH₃
3 CH₂=CHCH₂CH₂CH₃
4 CH₃CH₂C≡CCH₂CH₃
5 CH₃CH₂CH(CH₃)CH(CH₃)CH₃
A2
From IUPAC Name to Structural Formula

For each IUPAC name below: (A) Draw the condensed structural formula. (B) State the molecular shape and bond angle around each carbon in the chain. (C) State the molecular formula and verify it against the appropriate general formula.

1. 3-methylhexane

2. 2,3-dimethylbutane

3. 4-methylpent-2-ene

MC
5 Questions

Click an option to check your answer.

1. What is the correct IUPAC name for CH₃CH₂CH(CH₃)CH₂CH₂CH₃?

2. A carbon atom in an organic molecule has bond angles of approximately 120° to its neighbours. What molecular geometry and bond type does this indicate?

3. What is the correct IUPAC name for HC≡CCH₂CH₂CH₃?

4. A compound has the molecular formula C₅H₁₀. Which homologous series does it most likely belong to?

5. Which condensed structural formula correctly represents 3-methylpent-1-ene?

SA
Extended Questions

6. (a) State the molecular shape and bond angle around each carbon in propene (CH₃CH=CH₂). (b) Identify the bond type responsible for the shape at C1 and C2, and the bond type at C3. (c) Explain, in terms of the bonds present, why C1 and C2 have different geometry to C3. 4 MARKS

7. (a) Write the IUPAC name for the compound with condensed formula CH₃CH₂C(CH₃)₂CH₂CH₃. (b) State its molecular formula and verify it using the appropriate general formula. (c) State the molecular shape and bond angle around all carbon atoms in this compound and explain what structural feature leads to this. 4 MARKS

8. A student claims that propane (C₃H₈), butane (C₄H₁₀), and 2-methylpropane (C₄H₁₀) are all members of the same homologous series. (a) Identify which pair consists of isomers and explain how you know. (b) Identify which pair are consecutive members of a homologous series and explain. (c) Predict, with reasoning, whether propane and 2-methylpropane will have similar chemical reactivity. 5 MARKS

Show All Answers

Activity 1 — Naming Practice

1. CH₃CH₂CH₂CH₂CH₃: Longest chain = 5C → pentane. No branches. Name: pentane. Formula: C₅H₁₂ ✓ (CₙH₂ₙ₊₂, n=5 → 12H).

2. CH₃CH(CH₃)CH₂CH₃: Main chain: CH₃–CH–CH₂–CH₃ = 4C → butane. Branch: –CH₃ at C2. Name: 2-methylbutane. Formula: C₅H₁₂ ✓.

3. CH₂=CHCH₂CH₂CH₃: Functional group = C=C (alkene). Chain including C=C = 5C. Double bond at C1. Name: pent-1-ene. Formula: C₅H₁₀ ✓ (CₙH₂ₙ, n=5 → 10H).

4. CH₃CH₂C≡CCH₂CH₃: Functional group = C≡C (alkyne). Chain = 6C. Triple bond at C3 (from either end). Name: hex-3-yne. Formula: C₆H₁₀ ✓ (CₙH₂ₙ₋₂, n=6 → 10H).

5. CH₃CH₂CH(CH₃)CH(CH₃)CH₃: Longest chain: CH₃–CH₂–CH–CH–CH₃ = 5C. Two methyl branches at C2 and C3 from the CH(CH₃)CH₃ end. Name: 2,3-dimethylpentane. Formula: C₇H₁₆ ✓.

Activity 2 — Structure from Name

1. 3-methylhexane: Condensed: CH₃CH₂CH(CH₃)CH₂CH₂CH₃. All C: tetrahedral geometry, 109.5° (single bonds only). Formula: parent 6C + branch 1C = 7C total → C₇H₁₆ ✓ (n=7 → 16H).

2. 2,3-dimethylbutane: Condensed: CH₃CH(CH₃)CH(CH₃)CH₃. All C: tetrahedral geometry, 109.5° (single bonds only). Formula: parent 4C + 2 methyl branches = 6C → C₆H₁₄ ✓ (n=6 → 14H).

3. 4-methylpent-2-ene: Pent-2-ene = 5C chain, C=C at C2–C3. Methyl at C4: CH₃–CH=CH–CH(CH₃)–CH₃. C2 and C3: trigonal planar, ~120° (C=C present); C1, C4, C5: tetrahedral, 109.5°. Formula: 5C parent + 1C branch = 6C → C₆H₁₂ ✓ (CₙH₂ₙ, n=6 → 12H).

Multiple Choice

1. C — 3-methylhexane. Main chain = 6C (hexane). Methyl branch at C3. Options A and B have 7C in the main chain (heptane) — wrong.

2. B — Trigonal planar, ~120°, double bond present. Bond angles of ~120° → trigonal planar geometry → C=C double bond present. Tetrahedral (109.5°) indicates single bonds only; linear (180°) indicates a triple bond.

3. A — Pent-1-yne. HC≡C–CH₂–CH₂–CH₃: longest chain including C≡C = 5C → pent. Suffix = –yne. Numbered from the HC≡C end gives locant 1. Modern IUPAC: pent-1-yne.

4. B — Alkene (CₙH₂ₙ). C₅H₁₀: test n=5. Alkene: 2(5)=10 → C₅H₁₀ ✓. Alkane: 2(5)+2=12 → C₅H₁₂ ✗.

5. A — CH₂=CHCH(CH₃)CH₂CH₃. 3-methylpent-1-ene: pent-1-ene = 5C with C=C at C1. Methyl at C3. Option B has double bond at C3. Option C has methyl at C4. Option D has methyl at C2.

Short Answer Model Answers

Q6 (4 marks): (a) C1 (CH₂=): trigonal planar, ~120°; C2 (=CH–): trigonal planar, ~120°; C3 (–CH₃): tetrahedral, 109.5° [2 — 1 mark for C1/C2 set, 1 mark for C3]. (b) C1 and C2 are part of the C=C double bond; C3 has only single bonds [1]. (c) C1 and C2 are each bonded to three groups via the C=C arrangement — three groups spread into a flat plane at ~120°. C3 is bonded to four groups via single bonds only — four groups spread to tetrahedral positions at 109.5° [1].

Q7 (4 marks): (a) CH₃–CH₂–C(CH₃)₂–CH₂–CH₃: main chain = 5C (pentane). Two methyl branches at C3. Name: 3,3-dimethylpentane [1]. (b) Total C: 5+2 = 7C. CₙH₂ₙ₊₂, n=7: H=16. Formula = C₇H₁₆ ✓ [1]. (c) All 7 carbons have tetrahedral geometry, 109.5° bond angles [1] because this compound has only single bonds — no C=C or C≡C present, so all carbons bond to four groups and adopt tetrahedral positions [1].

Q8 (5 marks): (a) Isomers = butane and 2-methylpropane (both C₄H₁₀) [1] — same molecular formula but different structural arrangements [1]. (b) Homologous series pair = propane and butane [1] — they are consecutive members of the alkane series, differing by exactly one –CH₂– unit, sharing general formula CₙH₂ₙ₊₂ [1]. (c) Propane and 2-methylpropane will have similar chemical reactivity [1] — they share the alkane functional group class (C–C and C–H single bonds only). Chemical reactivity in a homologous series depends on the functional group, which is identical for both.

How did your thinking change?

Back at the start you predicted what "hex-1-ene" meant. Now you know: "hex-" = 6 carbons, "-ene" = a carbon–carbon double bond, and the "1" tells you the double bond starts at carbon 1. You can now decode any name like this — and that's the same skill that let you read "hexyl acetate" off the shampoo bottle.

Go back to your Think First response. Now that you've studied IUPAC nomenclature:

  • In "hex-1-ene": were you right about the number of carbons? ("hex" = 6 carbons)
  • What does the "1" actually mean — and was your prediction correct? (locant of the double bond — the C=C starts at C1)
  • What does "–ene" tell you about the compound? (contains a C=C double bond; trigonal planar geometry, ~120° bond angles at the C=C carbons; formula C₆H₁₂)
  • Write the condensed structural formula for hex-1-ene from memory and verify it against the alkene general formula.

What is the general formula for alkanes, and what does it mean in terms of structure?

For each bond type (single, double, triple), what shape forms around the carbon and what bond angle results?

What are the three IUPAC naming steps for a branched alkane?

What is the difference between a homologous series and structural isomers?

In IUPAC naming, what takes priority when numbering a chain — a branch or a double bond?

Lesson 2 →
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