Structure and properties of hydrocarbons

Q: What is the definition of a hydrocarbon?

A hydrocarbon is a compound containing carbon and hydrogen atoms only .

Q: What is the key difference between a saturated hydrocarbon and an unsaturated hydrocarbon?

A saturated hydrocarbon contains only single (\text{C}-\text{C}\) bonds, whereas an unsaturated hydrocarbon contains at least one multiple carbon-carbon bond, such as a \(\text{C}=\text{C}\) double bond.

Q: What are the bonding geometry and bond angle around each carbon atom in an alkane?

Each carbon atom has a tetrahedral geometry with a bond angle of approximately \(109.5^\circ\) .

Q: How is a carbon-carbon double bond constructed in an alkene, and what is the shape around each double-bonded carbon?

The \(\text{C}=\text{C}\) double bond consists of one \(\sigma\)-bond and one \(\pi\)-bond , creating a trigonal planar geometry with bond angles of approximately \(120^\circ\) .

Q: What is the difference between a displayed formula and a skeletal formula?

A displayed formula shows every atom and every covalent bond as an individual line, whereas a skeletal formula shows only the carbon skeleton as lines/vertices with hydrogen atoms and carbon symbols omitted.

Q: Why does the boiling point of hydrocarbons increase as the carbon chain length increases?

Longer chains contain more electrons and have a greater surface area of contact , resulting in stronger instantaneous dipole–induced dipole forces that require more energy to break.

Q: Why do branched alkanes have lower boiling points than their straight-chain structural isomers?

Branching makes molecules more compact, reducing the surface area of contact and preventing close molecular packing , which weakens the instantaneous dipole–induced dipole forces.

Q: Why are hydrocarbons insoluble in water but soluble in non-polar organic solvents?

Hydrocarbons cannot form hydrogen bonds with water to disrupt water's strong hydrogen-bonded network, but they form favorable instantaneous dipole–induced dipole forces with non-polar solvents.

Q: How do volatility and viscosity change as hydrocarbon chain length increases?

As chain length increases, volatility decreases and viscosity increases due to stronger intermolecular forces and molecular tangling.

Chemistry B (Salters) - H433 · Cambridge OCR A Level · 8 min read · Updated: 23 Oct 2026

Structure and Properties of Hydrocarbons

Welcome to this guide on the structure and properties of hydrocarbons! Hydrocarbons are the fundamental building blocks of organic chemistry. Whether we are discussing petroleum, natural gas, fuels for transport, or feedstock for modern materials, hydrocarbons are at the centre of it all. In this chapter, we will unpack how carbon and hydrogen atoms link together, how to name and draw them, and how their three-dimensional structures govern their physical properties.

Note: Detailed discussions of intermolecular bonding, molecular shapes, and isomerism are developed in companion chapters in this section. We will cross-reference them where helpful!

1. What is a Hydrocarbon?

Let's start with the absolute foundation. A hydrocarbon is a compound that contains carbon and hydrogen atoms only.

If a molecule contains oxygen, nitrogen, a halogen, or any other element, it is not a hydrocarbon, even if it is mostly made of carbon and hydrogen.

Saturated vs Unsaturated Hydrocarbons

Hydrocarbons fall into two primary families based on the types of carbon-carbon bonds they contain:

1. Saturated hydrocarbons: Molecules in which all carbon-carbon bonds are single covalent bonds. Carbon is "saturated" with hydrogen atoms because it holds the maximum possible number of hydrogens per carbon atom. The most common saturated hydrocarbons are the alkanes (general formula \(\text{C}_n\text{H}_{2n+2}\) for open-chain alkanes) and cycloalkanes (\(\text{C}_n\text{H}_{2n}\)).

2. Unsaturated hydrocarbons: Molecules that contain at least one carbon-carbon multiple bond, such as a double bond (\(\text{C}=\text{C}\)) or a triple bond. The most common open-chain unsaturated hydrocarbons met in this unit are the alkenes (general formula \(\text{C}_n\text{H}_{2n}\) with one double bond). Because of the double bond, they possess fewer hydrogen atoms than the corresponding alkane.

Quick Review: Saturated = only single \(\text{C}-\text{C}\) bonds. Unsaturated = contains one or more \(\text{C}=\text{C}\) double bonds.

2. Bonding and Shapes in Hydrocarbon Molecules

Why do hydrocarbons adopt their particular shapes? It all comes down to the arrangement of electron pairs around each carbon atom.

Bonding in Alkanes

In an alkane such as methane (\(\text{CH}_4\)) or ethane (\(\text{C}_2\text{H}_6\)):

- Each carbon atom forms four single covalent bonds (shared pairs of electrons).
- These single bonds are known as \(\sigma\)-bonds (sigma bonds), formed by the end-on overlap of atomic orbitals directly along the internuclear axis.
- Because electron pairs repel each other to get as far apart as possible, four bonding electron pairs arrange themselves in a tetrahedral geometry.
- The bond angle around each carbon atom in an alkane is approximately \(109.5^\circ\).
- Single \(\sigma\)-bonds allow free rotation around the \(\text{C}-\text{C}\) axis, meaning alkane chains are flexible and can twist into various conformations.

Bonding in Alkenes

In an alkene such as ethene (\(\text{C}_2\text{H}_4\)):

- The double bond consists of two distinct components: one \(\sigma\)-bond and one \(\pi\)-bond (pi bond).
- The \(\sigma\)-bond is formed by the end-on overlap of orbitals between the two carbon atoms.
- The \(\pi\)-bond is formed by the sideways overlap of adjacent p-orbitals above and below the plane of the carbon and hydrogen nuclei.
- Around each double-bonded carbon atom, there are three regions of negative charge (two single bonds and one double bond). These repel each other into a trigonal planar arrangement.
- The bond angles around each carbon atom in ethene are approximately \(120^\circ\).
- Crucial consequence: Unlike a \(\sigma\)-bond, a \(\pi\)-bond prevents rotation around the double bond at normal temperatures because rotating the bond would break the sideways overlap. This rigidity leads directly to E/Z isomerism (covered in our chapter on Structural and E/Z isomerism).

Key Takeaway: Carbon with four single bonds = tetrahedral (\(109.5^\circ\)) with free rotation. Carbon with a double bond = trigonal planar (\(120^\circ\)) around that carbon, with restricted rotation.

3. Representing Hydrocarbons: Formulas and Conventions

In chemistry examinations, you must be confident interpreting and drawing organic molecules using several standard conventions:

1. Molecular formula: Shows the actual number of atoms of each element in a molecule, but gives no structural information. For example, butane is \(\text{C}_4\text{H}_{10}\), and propene is \(\text{C}_3\text{H}_6\).

2. Structural formula: Shows the arrangement of atoms carbon by carbon, with the attached hydrogens and functional groups written clearly without necessarily drawing all the individual bonds. For example, butane is \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\) or \(\text{CH}_3(\text{CH}_2)_2\text{CH}_3\); 2-methylpropane is \(\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_3\).

3. Displayed formula: Shows all the atoms and all the covalent bonds connecting them as individual lines. Every \(\text{C}-\text{H}\) bond must be explicitly shown.

4. Skeletal formula: A simplified visual representation where the carbon skeleton is drawn as a zig-zag line. Hydrogen atoms attached to carbon are omitted; vertices and line ends represent carbon atoms together with enough hydrogen atoms to satisfy carbon's valency of four. A double line represents a \(\text{C}=\text{C}\) double bond.

Common Mistake to Avoid: When asked for a displayed formula, students often write a \(\text{CH}_3\) group with the bonds to hydrogen omitted. Remember: on a displayed formula, every single bond must be drawn out fully!

4. Systematic Nomenclature: Naming Alkanes and Alkenes

The International Union of Pure and Applied Chemistry (IUPAC) naming rules provide an unambiguous name for every hydrocarbon.

Step-by-Step Naming Method

Step 1: Find the longest continuous carbon chain. This gives the root name:
- \(1\text{ C}\) = meth-
- \(2\text{ C}\) = eth-
- \(3\text{ C}\) = prop-
- \(4\text{ C}\) = but-
- \(5\text{ C}\) = pent-
- \(6\text{ C}\) = hex-
- \(7\text{ C}\) = hept-
- \(8\text{ C}\) = oct-
For alkenes, the longest continuous chain must contain the double bond.

Step 2: Identify the main functional group / saturation.
- If only single bonds: suffix is -ane.
- If a \(\text{C}=\text{C}\) double bond is present: suffix is -ene, with a locant number showing the lowest-numbered carbon of the double bond (e.g., \(\text{but-1-ene}\), \(\text{but-2-ene}\)).

Step 3: Number the carbon chain. Number from the end that gives substituents (branches) or the double bond the lowest possible numbers. The double bond takes priority over alkyl branches when numbering.

Step 4: Identify and name alkyl side chains.
- One-carbon branch: methyl- (\(-\text{CH}_3\))
- Two-carbon branch: ethyl- (\(-\text{CH}_2\text{CH}_3\))

Step 5: Assemble the complete name.
- Use prefixes di-, tri-, or tetra- if more than one identical alkyl branch is present (e.g., 2,2-dimethylpropane).
- List different branches in alphabetical order (e.g., ethyl before methyl; multiplying prefixes like di- do not affect alphabetical order).
- Separate numbers from numbers with commas (e.g., \(2,3\)), and numbers from letters with hyphens (e.g., \(2\text{-methylbutane}\)).

5. Physical Properties of Hydrocarbons

Hydrocarbons are non-polar molecules because carbon and hydrogen have very similar electronegativities. Consequently, the only intermolecular forces operating between hydrocarbon molecules are instantaneous dipole–induced dipole forces (often referred to as London dispersion forces).

Trend 1: Boiling Point and Carbon Chain Length

As the chain length (and thus relative molecular mass, \(M_r\)) of an alkane or alkene increases, its boiling point increases systematically.

Why?
1. As the number of carbon and hydrogen atoms increases, each molecule has more electrons.
2. More electrons allow larger fluctuations in electron density, resulting in stronger instantaneous dipole–induced dipole interactions.
3. In addition, longer chains offer a greater surface area of contact between adjacent molecules.
4. More thermal energy is required to overcome these stronger attractive forces, so the boiling point rises.

Example: Methane (\(\text{CH}_4\)), ethane (\(\text{C}_2\text{H}_6\)), propane (\(\text{C}_3\text{H}_8\)), and butane (\(\text{C}_4\text{H}_{10}\)) are gases at room temperature and pressure, whereas pentane (\(\text{C}_5\text{H}_{12}\)) to heptadecane (\(\text{C}_{17}\text{H}_{36}\)) are liquids, and longer chains are waxy solids.

Trend 2: The Effect of Chain Branching on Boiling Point

If you compare structural isomers of an alkane (molecules with the same molecular formula but different structural formulas), the branched isomer always has a lower boiling point than the straight-chain isomer.

Example: Compare pentane and 2,2-dimethylpropane (both \(\text{C}_5\text{H}_{12}\)):
- Pentane: straight chain, boiling point \(\approx 36^\circ\text{C}\) (\(309\text{ K}\))
- 2,2-dimethylpropane: spherical, branched shape, boiling point \(\approx 9.5^\circ\text{C}\) (\(282.5\text{ K}\))

Why does branching lower the boiling point?
1. Molecular shape and surface area: Branched molecules are more compact and approach a spherical shape. This reduces the exposed surface area of the molecule.
2. Steric hindrance and packing: The branches prevent adjacent molecules from getting close to one another.
3. Weaker intermolecular attraction: Because the molecules cannot pack as closely together and have a smaller surface area of contact, the instantaneous dipole–induced dipole forces are weaker.
4. Less energy is needed to separate the molecules, leading to a lower boiling point.

Trend 3: Solubility

- In water (insoluble): Hydrocarbons cannot form hydrogen bonds with water molecules. The weak attractions formed between hydrocarbon molecules and water molecules are not strong enough to disrupt the strong hydrogen bonds between water molecules. As a result, hydrocarbons are immiscible with water and float on top (owing to their lower density).
- In non-polar organic solvents (soluble): Hydrocarbons readily dissolve in non-polar solvents because the intermolecular forces between solvent and solute are of comparable type and magnitude (instantaneous dipole–induced dipole forces).

Trend 4: Volatility and Viscosity

- Volatility (the ease with which a liquid evaporates) decreases as chain length increases because the intermolecular forces become stronger.
- Viscosity (resistance to flow) increases with chain length. Longer molecules become tangled with one another and experience stronger intermolecular attractions, making long-chain hydrocarbons thick and slow-flowing (like lubricating oil or tar).

Key Takeaway: Longer straight chains = more electrons and greater surface contact \(\implies\) stronger instantaneous dipole–induced dipole forces \(\implies\) higher boiling point, higher viscosity, lower volatility. Branching weakens packing and contact area \(\implies\) lowers boiling point.

Summary and Quick Revision Checklist

Before moving on to chemical reactions and mechanisms, ensure you can:

- Define a hydrocarbon as a compound of carbon and hydrogen only.
- Distinguish between saturated (\(\text{C}-\text{C}\) single bonds only) and unsaturated (\(\text{C}=\text{C}\) double bonds present) hydrocarbons.
- State and explain the tetrahedral shape (\(109.5^\circ\)) around \(\text{C}\) in alkanes and the trigonal planar shape (\(120^\circ\)) around \(\text{C}\) in alkenes.
- Describe the nature of a single bond (\(\sigma\)-bond) versus a double bond (one \(\sigma\)-bond and one \(\pi\)-bond).
- Draw and convert between molecular, structural, displayed, and skeletal formulas.
- Name alkanes and alkenes systematically following IUPAC nomenclature.
- Explain boiling point trends in terms of chain length (electron count) and branching (surface contact area and packing) using instantaneous dipole–induced dipole forces.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the definition of a hydrocarbon?

A hydrocarbon is a compound containing carbon and hydrogen atoms only.

What is the key difference between a saturated hydrocarbon and an unsaturated hydrocarbon?

A saturated hydrocarbon contains only single (\text{C}-\text{C}\) bonds, whereas an unsaturated hydrocarbon contains at least one multiple carbon-carbon bond, such as a \(\text{C}=\text{C}\) double bond.

What are the bonding geometry and bond angle around each carbon atom in an alkane?

Each carbon atom has a tetrahedral geometry with a bond angle of approximately \(109.5^\circ\).

How is a carbon-carbon double bond constructed in an alkene, and what is the shape around each double-bonded carbon?

The \(\text{C}=\text{C}\) double bond consists of one \(\sigma\)-bond and one \(\pi\)-bond, creating a trigonal planar geometry with bond angles of approximately \(120^\circ\).

What is the difference between a displayed formula and a skeletal formula?

A displayed formula shows every atom and every covalent bond as an individual line, whereas a skeletal formula shows only the carbon skeleton as lines/vertices with hydrogen atoms and carbon symbols omitted.

How are branched alkanes and alkenes named according to IUPAC nomenclature?

Identify the longest carbon chain, number it to give substituents or double bonds the lowest locants, and list alkyl branches in alphabetical order with appropriate numerical prefixes and locants.

Why does the boiling point of hydrocarbons increase as the carbon chain length increases?

Longer chains contain more electrons and have a greater surface area of contact, resulting in stronger instantaneous dipole–induced dipole forces that require more energy to break.

Why do branched alkanes have lower boiling points than their straight-chain structural isomers?

Branching makes molecules more compact, reducing the surface area of contact and preventing close molecular packing, which weakens the instantaneous dipole–induced dipole forces.

Why are hydrocarbons insoluble in water but soluble in non-polar organic solvents?

Hydrocarbons cannot form hydrogen bonds with water to disrupt water's strong hydrogen-bonded network, but they form favorable instantaneous dipole–induced dipole forces with non-polar solvents.

How do volatility and viscosity change as hydrocarbon chain length increases?

As chain length increases, volatility decreases and viscosity increases due to stronger intermolecular forces and molecular tangling.

Ready to test yourself?

Turn these notes into exam-style practice. Get unlimited AI questions on this topic with instant marking and explanations.

Practise This Topic

More Chemistry B (Salters) - H433 chapters

* The content provided by thinka is generated by AI and may not always be accurate or up-to-date. Please use it as a supplementary resource and verify with official materials.

Put These Notes into Practice

Reading the notes is a great start. Now practise with unlimited AI-generated questions and get instant feedback. 100,000+ students are already improving their grades.

Start Practising Now View Pricing

Done reading? Test yourself with AI practice questions

Practise This Topic Now