Alkanes, alkenes and their structures

Key idea: Carbon forms four bonds and hydrogen one. Alkanes have only single carbon-carbon bonds; alkenes have a carbon-carbon double bond and are unsaturated.

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Methane, ethane and propane are alkanes; ethene and propene are alkenes. Every structure in this lesson follows from two bonding rules: in a covalent molecule, carbon forms four bonds and hydrogen forms one. A double bond is two shared pairs, so it counts as two of carbon’s four bonds.

This lesson uses open-chain hydrocarbons: the carbon atoms form a chain, not a ring. When a question gives a structure, check the bonds as well as counting atoms.

Build a structure from bonding rules

A displayed structure (displayed formula) shows every atom and every bond as a line. You can build one without memorising it:

  1. Join the carbon atoms in a chain.
  2. Count the bonds each carbon already has.
  3. Give each carbon hydrogen atoms until it has four bonds.
  4. Count the atoms to get the molecular formula.

Worked example 1

Build the structure of propane

Problem

Propane is a hydrocarbon with a chain of three carbon atoms joined by single bonds. Build its displayed structure and find its molecular formula.

Show solution step by step
  1. Count bonds already used

    Method

    Draw C–C–C and count each carbon’s bonds.

    Reason

    Each carbon needs four bonds in total.

    Working

    Answer

    End carbons: 1 bond each. Middle carbon: 2 bonds.

  2. Add hydrogen atoms

    Method

    Fill each carbon up to four bonds with hydrogen.

    Reason

    Each hydrogen forms one bond.

    Working

    Answer

    End carbons: 3 H each. Middle carbon: 2 H. Total: 3 + 2 + 3 = 8 H.

  3. Write the formulae

    Working

    Answer

    Molecular formula C₃H₈; condensed formula CH₃CH₂CH₃. The displayed structure is the third one in the figure below.

Alkanes: only single bonds

An alkane is a hydrocarbon in which every carbon–carbon bond is a single bond. Each carbon holds as many hydrogen atoms as it can, so alkanes are saturated. The alkanes follow the general formula CₙH₂ₙ₊₂, where n is the number of carbon atoms. Their names end in -ane; the start of the name counts the carbons: meth- (1), eth- (2), prop- (3).

Methane, ethane and propane

Methane is one carbon with four hydrogens. Ethane is two carbons joined by a single bond with three hydrogens on each carbon. Propane is a chain of three carbons joined by single bonds, with three hydrogens on each end carbon and two on the middle carbon.

Scroll across the figure to read all labels.

Displayed structures of methane, ethane and propane, with every hydrogen atom and every single bond drawnDisplayed structures of methane, ethane and propane, with every hydrogen atom and every single bond drawn
Every carbon has four bonds and every hydrogen has one. All the bonds are single bonds.
View figure data
Atoms and bonds in each displayed structure
AlkaneMolecular formulaCarbon–carbon bondsHydrogen atoms
methaneCH₄none4
ethaneC₂H₆1 single6
propaneC₃H₈2 single8
Carbon atomsNameMolecular formulaCondensed formula
1methaneCH₄CH₄
2ethaneC₂H₆CH₃CH₃
3propaneC₃H₈CH₃CH₂CH₃

Alkenes: one carbon–carbon double bond

The alkenes studied here are open-chain hydrocarbons containing one carbon–carbon double bond, C = C. The double bond uses two of each carbon’s four bonds, so those carbons hold fewer hydrogen atoms. Alkenes are unsaturated. They follow the general formula CₙH₂ₙ and their names end in -ene.

There is no one-carbon alkene: a double bond between carbon atoms needs at least two carbons. The first two alkenes are ethene and propene.

Ethene and propene

Ethene is two carbons joined by a double bond, with two hydrogens on each carbon. Propene is a chain of three carbons. A double bond joins an end carbon carrying two hydrogens to the middle carbon, which carries one hydrogen; a single bond joins the middle carbon to the other end carbon, which carries three hydrogens.

Scroll across the figure to read all labels.

Displayed structures of ethene and propene, each with one carbon-carbon double bond and every hydrogen atom drawnDisplayed structures of ethene and propene, each with one carbon-carbon double bond and every hydrogen atom drawn
In each alkene, the two carbons of the double bond use two of their four bonds on each other.
View figure data
Atoms and bonds in each displayed structure
AlkeneMolecular formulaCarbon–carbon bondsHydrogen atoms
etheneC₂H₄1 double4
propeneC₃H₆1 double, 1 single6
Carbon atomsNameMolecular formulaCondensed formula
1none——
2etheneC₂H₄CH₂ = CH₂
3propeneC₃H₆CH₂ = CHCH₃

Check your understanding 1 Fill each carbon to four bonds

A propene chain is C=C–C. Count the hydrogen atoms needed on the first, middle and last carbon. A double bond counts as two bonds.

Show answer

The hydrogen counts are 2, 1 and 3. Together they give C₃H₆. A count alone is not a displayed structure: draw every atom and bond in the next activity.

Check your understanding 2 Draw the complete propene structure

On paper, draw every atom and bond in propene from the C=C–C chain. Then give its molecular formula and explain why it is unsaturated. You can keep the drawing on paper and use optional notes for comparison.

Show answer

Your drawing should show two H on the first carbon, one H on the middle carbon and three H on the last carbon, each joined by a single bond. Retain the double bond between the first two carbons and the single bond between the last two. Every C has four bonds and every H one. There are three C and six H: C₃H₆. The C=C bond makes it unsaturated. CH₂ = CHCH₃ summarises the connections but is a condensed formula; a displayed structure draws every hydrogen and bond. Compare your paper drawing with the propene figure above.

Homologous series

The alkanes form one homologous series and the alkenes form another. A homologous series is a family of compounds that:

  • share one general formula, so each member differs from the next by CH₂;
  • have similar chemical properties, because they have the same kind of bonding (for alkenes, the C = C bond);
  • show a gradual change in physical properties as the molecules get larger and heavier.
AlkaneBoiling point / °CAlkeneBoiling point / °C
methane−162——
ethane−89ethene−104
propane−42propene−48

Boiling point rises down each series. Larger molecules attract one another more strongly, so more energy is needed to separate them. Viscosity also generally increases for comparable liquid hydrocarbons at the same temperature: the larger-molecule fractions of crude oil flow less readily. Melting points do not increase smoothly for every successive member; how molecules pack in a solid also matters.

Worked example 2

Classify a molecular formula

Problem

An open-chain hydrocarbon has the molecular formula C₃H₆. Is it an alkane or an alkene? Name it.
Show full solution
  1. Test both general formulae

    Method

    Substitute n = 3 into each general formula.

    Reason

    For the open-chain molecules considered here, the hydrogen count tells you which series fits.

    Working

    Answer

    Alkane: C₃H₈. Alkene: C₃H₆.
  2. Classify and name

    Working

    Answer

    C₃H₆ matches CₙH₂ₙ, so it is an alkene with a C = C bond: propene.

Guided practice 3

Read a structure before naming it

About 5 min

Problem

A molecule has the condensed structure CH₃-CH₃. Decide its series, whether it is saturated, its name and its molecular formula.

Use the bond and the carbon count

Homologous series
Saturated or unsaturated
Name

Hints

Hint 1

bond

Look for a double bond first.
Hint 2

formula

Choose the general formula for the series you identified.
Show solution step by step
  1. Identify the series

    Method

    Inspect the carbon–carbon bond.

    Reason

    It is single, and every carbon has its full complement of hydrogens.

    Working

    Answer

    Alkane; saturated.
  2. Name it and check the formula

    Method

    Count carbons, then hydrogens.

    Reason

    Two carbons give eth-; for n = 2, CₙH₂ₙ₊₂ gives six hydrogens.

    Working

    Answer

    Ethane, C₂H₆.

Spot the mistake 4

A propene drawing with too many hydrogens

About 4 min

Learner working

A learner draws propene as a chain C=C–C and puts two hydrogens on each of the three carbons, then writes the formula C₃H₆. The formula is right but the drawing is wrong. First count each carbon’s bonds. Then correct the complete displayed structure in the next activity.

Count the bonds on each carbon

Show solution step by step
  1. Count bonds carbon by carbon

    Method

    Add the double bond, the single bond and the hydrogens on each carbon.

    Reason

    Every carbon must have exactly four bonds.

    Working

    Answer

    First carbon: 2 (double bond) + 2 H = 4. Middle carbon: 2 (double bond)

    • 1 (single bond) + 2 H = 5. Third carbon: 1 + 2 H = 3.
  2. Correct the drawing

    Working

    Answer

    The middle carbon has five bonds and the end carbon only three. Give the middle carbon one hydrogen and the end carbon three: (CH₂ = CH - CH₃).

Check your understanding 3 Correct the drawing and its bonds

A C=C–C chain was drawn with two hydrogens on every carbon. Correct the displayed drawing on paper and explain why the right total formula did not check each carbon’s bonds. Use optional notes if helpful.

Show answer

The middle carbon has five bonds while the last has only three. Move one hydrogen from the middle to the last carbon: the counts become 2, 1 and 3. Draw each C–H bond so every carbon has four bonds. The condensed structure is CH₂ = CHCH₃ and the molecular formula remains C₃H₆. Counting six hydrogens in total did not check which carbon they were joined to; both the formula and the local bonding must be right.

Check your understanding 4 Can this molecule have a double bond?

A learner draws a hydrocarbon with formula C₃H₈ and includes one (C = C) bond. Use the bonding rules to decide whether this is possible.

Work on paper or in the optional notes. Compare the carbon-by-carbon bond limits, hydrogen total and saturated/unsaturated conclusion.

Show answer

It is not possible. With a C = C bond in a three-carbon chain, the carbons have room for only 6 hydrogens in total, giving C₃H₆ (propene). Eight hydrogens need every carbon–carbon bond to be single, so C₃H₈ must be the saturated alkane propane.

Check your understanding 5 Build a second structure independently

Draw the displayed structure of ethene on paper, showing every atom and bond. Give its molecular formula and explain why there is no one-carbon alkene. Use optional notes for your comparison.

Show answer

Ethene has two carbons joined by a double bond. Each carbon carries two hydrogens, with every C–H bond drawn: C₂H₄. Every carbon has four bonds and every hydrogen one. There is no one-carbon alkene because the defining carbon–carbon double bond requires two carbon atoms. The ethene figure above provides the complete displayed model for comparison.

Common mistakes

  • Drawing a carbon atom with three or five bonds.
  • Writing “methene”. There is no one-carbon alkene.
  • Using the -ene ending for a molecule with only single bonds.
  • Using “saturated” to mean “dissolved”. Here it means having only single carbon–carbon bonds.

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Syllabus and review details

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