Year 12 Chemistry Module 7 · Organic Chemistry ⏱ ~45 min 5 MC · 3 Short Answer Lesson 4 of 23 Core Core

Hydrocarbon Structure & Bonding

Build carbon models you can reason from: see how a C–C single, C=C double or C≡C triple bond fixes the local shape, angle and freedom to rotate.

Today's hook: Ethane, ethene and ethyne each contain two carbon atoms. Why do their atoms point in completely different directions?

0/5TASKS

Connect: three molecules, three shapes

Warm up first

Retrieve three ideas from earlier lessons before building the new model.

Think First

Ethane is C₂H₆, ethene is C₂H₄, and ethyne is C₂H₂. Each has two carbons, yet their models are tetrahedral, planar and linear.

Predict: how can changing only the bond between the carbons change the directions in which every other bond points?

Beyond the syllabus. Linking single, double and triple carbon–carbon bonds to tetrahedral, trigonal-planar and linear geometry is core. Sigma and pi orbital language is a deep dive — helpful for understanding why the shapes arise, never required in the exam.

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Learning intentions

goals

Know

Single, double and triple bonds contain different numbers of sigma and pi bonds
The characteristic geometries and bond angles around carbon

Understand

Electron regions arrange as far apart as possible
Pi bonding restricts rotation around multiple bonds

Can Do

Construct models of ethane, ethene and ethyne
Infer shape and angle from a structural formula

Vocabulary that unlocks the lesson

Sigma (σ) bondEnd-on orbital overlap along the line between two nuclei.

Pi (π) bondSideways overlap above and below the internuclear axis.

Bond angleThe angle between two bonds that meet at the same atom.

Misconception to fix: a double bond is not two identical lines. It is one σ bond plus one π bond; the π overlap is why a C=C cannot freely rotate.

Explain: a single bond gives tetrahedral carbon

Build ethane: C₂H₆

Each carbon in ethane makes four σ bonds: three C–H bonds and one C–C bond. Four bonding regions repel into a tetrahedral arrangement.

Read the model

The angle between bonds is approximately 109.5°. The diagram may look flat on paper, but the molecule is three-dimensional. Rotation can occur around the C–C σ bond without destroying the end-on overlap.

Model instruction: Join two four-hole carbon centres with one connector. Use the remaining three positions on each carbon for hydrogen. Rotate the C–C bond and notice that connectivity does not change.

A carbon has four single bonds. Which local model should you choose?

Represent: a double bond fixes a planar arrangement

Build ethene: C₂H₄

At each carbon in ethene there are three bonding regions: two C–H directions and the C=C direction. They arrange in one plane, approximately 120° apart.

What the two lines mean

The C=C contains one σ bond and one π bond. The sideways π overlap sits above and below the molecular plane. Twisting one carbon would destroy that overlap, so rotation is restricted and the six atoms of ethene remain planar.

Represent: draw H₂C=CH₂ with both carbon centres and all four hydrogens in the same plane. Label one C–C component σ and the other π.

True or false: ethene can freely rotate around C=C without changing its bonding.

Respond: a triple bond makes carbon linear

Build ethyne: C₂H₂

Each carbon in ethyne has only two bonding directions: one towards hydrogen and one towards the other carbon. Maximum separation places them 180° apart.

One axis, three bonds

The C≡C triple bond contains one σ bond and two mutually perpendicular π bonds. The whole H–C≡C–H molecule is linear: all four atoms lie on the same line.

Do not count drawn lines as directions: the three lines in C≡C connect the same two atoms, so the triple bond counts as one region when predicting geometry.

H–C≡C–H is _______ because each carbon has _______ bonding directions.

Feedback: compare bond, shape and angle

The comparison worth memorising

Model	C–C bonding	Regions at each C	Local geometry	Angle	Rotation
Ethane	1σ	4	Tetrahedral	109.5°	Free about C–C
Ethene	1σ + 1π	3	Trigonal planar	~120°	Restricted
Ethyne	1σ + 2π	2	Linear	180°	Restricted

Precision: say “geometry around the double-bonded carbon is trigonal planar”, not “every alkene molecule is flat”. Carbons elsewhere in a longer alkene may still be tetrahedral.

Which complete chain is correct?

Checklist: confirm the Lessons 1 to 4 core

Core Coverage Checklist for Lessons 1 to 4 Core

One page confirming the examinable boundary of the nomenclature and structure block

Nomenclature, formulae, models and structural properties are spread across Lessons 1 to 4, and those lessons carry far more examples than the syllabus requires. Use this list to check the core boundary. If you can do every line, the block is complete; anything beyond it is enrichment.

1 · Naming and formulae, up to C8

Name and draw straight-chain alkanes, alkenes and alkynes from 1 to 8 carbon atoms, placing the multiple bond correctly with a locant.
Name and draw compounds with simple methyl and ethyl branches, numbering the parent chain to give the lowest locants.
Name and draw the required functional-group classes: primary, secondary and tertiary alcohols, aldehydes, ketones, carboxylic acids, amines, amides and halogenated compounds.
Convert freely between molecular, condensed structural and full structural formulae for any of the above.

2 · Isomers — the three kinds you must separate

Chain isomers: same molecular formula, different carbon skeleton (butane and 2-methylpropane).
Position isomers: same skeleton and same functional group, different location (pent-1-ene and pent-2-ene; propan-1-ol and propan-2-ol).
Functional-group isomers: same molecular formula, different functional group entirely (butanal and butan-2-one, both C₄H₈O).
Given a molecular formula, generate a valid isomer of each type and say which type it is.

3 · Structure, bonding and shape

State that carbon is tetravalent and check any structure you draw against it.
Link bond type to geometry: C–C tetrahedral (about 109.5°), C=C trigonal planar (about 120°), C≡C linear (180°).
Explain saturation and unsaturation in terms of multiple bonds, and connect it to reactivity.

4 · Physical properties and the evidence behind them

Explain boiling-point trends within a homologous series using dispersion forces and polarisability, and support it with processed data (Lesson 3 investigation).
Explain why branching lowers boiling point, using reduced surface contact.
Explain hydrocarbon solubility in polar and non-polar solvents.

5 · Safety, which is compulsory and often skipped

State the safe handling and disposal requirements for organic substances: flammability and ignition sources, ventilation, correct waste stream rather than the sink.

Deep dive — Extension: outside the checklist

Benzene and aromatic naming, ester nomenclature, compounds carrying several functional groups, protein and peptide examples, and sigma/pi orbital language all appear in these lessons as enrichment. They are good for transfer but they are not part of the Module 7 nomenclature and structure requirement, and they should not be what you revise first.

Butanal and butan-2-one share the molecular formula C₄H₈O but contain an aldehyde and a ketone group respectively. Which kind of isomerism is this?

Apply: reason from a structural formula

Worked example

Worked Example: analyse propene, CH₃CH=CH₂

State the geometry around each carbon and explain whether the whole molecule is planar.

Mark the bond type at each carbon. Carbon 1 has only single bonds; carbons 2 and 3 form the C=C.

Carbon 1 has four bonding regions, so it is tetrahedral at about 109.5°.

Carbons 2 and 3 each have three regions, so each is trigonal planar at about 120°.

Answer: only the atoms directly arranged around C=C must share a plane. The CH₃ carbon remains tetrahedral, so it is inaccurate to call every atom in propene planar.

✓

Model-building checklist

Write the structural formula and locate each multiple bond.
Count bonding directions around the carbon being analysed.
Assign tetrahedral, trigonal planar or linear geometry.
Add 109.5°, 120° or 180° and check whether rotation is possible.

In CH₃C≡CH, which statement is correct?

Practise: construct, compare and explain

Model and explain

1. Draw displayed structures for ethane, ethene and ethyne. Annotate the geometry and bond angle around each carbon.

2. Explain why rotation is possible around C–C in ethane but restricted around C=C in ethene.

3. Analyse CH₂=CHCH₂CH₃. Identify which carbons are trigonal planar and which are tetrahedral.

Randomised question bank

Complete five questions drawn from the ten aligned structure-and-bonding items.

Exam-style response

Compare the bonding and geometry of ethane, ethene and ethyne. Refer to σ and π bonds, bond angles, molecular shape and rotation. (6 marks)

Review: retrieve the model without notes

Check what actually stuck

Give the shape and angle around carbon in ethane, ethene and ethyne.

How many σ and π bonds form C–C, C=C and C≡C?

Why does a double bond restrict rotation?

Return to the hook

Use the number of bonding regions to explain why ethane, ethene and ethyne have different shapes despite each containing two carbon atoms.