Welcome to Hydrocarbons: Alkanes
Welcome to your study notes for Unit A2 2: Organic Chemistry! In this chapter, we explore alkanes, the simplest yet most fundamental family of organic compounds. Alkanes are everywhere in modern life—they fuel our central heating, power our cars, and serve as the raw starting materials for countless everyday plastics and pharmaceuticals.
Don't worry if organic chemistry has felt overwhelming in the past! We will break down every concept into small, easy-to-digest steps with clear explanations, memory aids, and examiner tips so you can tackle exam questions with total confidence.
---1. Definitions and Classification
Before looking at how alkanes behave, let us define the key vocabulary you need to master for your CCEA examination.
• Hydrocarbon: A compound composed exclusively of carbon and hydrogen atoms.
• Saturated Hydrocarbon: A hydrocarbon containing only single covalent carbon–carbon bonds (\(\text{C–C}\)). There are no double or triple bonds present.
• Homologous Series: A "family" of organic compounds that shares:
1. The same general formula.
2. The same functional group.
3. Similar chemical properties.
4. A gradual trend (graduation) in physical properties (such as boiling point).
5. Successive members that differ from each other by a \(-\text{CH}_2-\) unit.
The General Formula and the First Six Alkanes
Acyclic (straight-chain and branched) alkanes follow the general formula \(\text{C}_n\text{H}_{2n+2}\), where \(n\) is the number of carbon atoms.
• \(n = 1\): Methane — \(\text{CH}_4\)
• \(n = 2\): Ethane — \(\text{C}_2\text{H}_6\)
• \(n = 3\): Propane — \(\text{C}_3\text{H}_8\)
• \(n = 4\): Butane — \(\text{C}_4\text{H}_{10}\)
• \(n = 5\): Pentane — \(\text{C}_5\text{H}_{12}\)
• \(n = 6\): Hexane — \(\text{C}_6\text{H}_{14}\)
Memory Trick for the first 4 prefixes: Monkeys Eat Peanut Butter (Methane, Ethane, Propane, Butane).
Key Takeaway
Alkanes are saturated hydrocarbons with the general formula \(\text{C}_n\text{H}_{2n+2}\). Every carbon atom forms four single bonds.
---2. Structure, Shape, and Bonding
Sigma (\(\sigma\)) Bonds
Every single covalent bond in an alkane is a sigma (\(\sigma\)) bond. A sigma bond is formed by the direct, head-on overlap of atomic orbitals between bonding atoms. These bonds have high electron density concentrated directly along the axis between the two nuclei, making them strong and stable.
Molecular Geometry
Each carbon atom in an alkane is surrounded by 4 bonding pairs of electrons and 0 lone pairs. According to electron-pair repulsion theory, these four pairs push as far apart as possible:
• Shape around each Carbon: Tetrahedral
• Bond Angle: Approximately \(109.5^\circ\)
Why Are Alkanes Generally Unreactive?
Compared to alkenes and other functional groups, alkanes are relatively unreactive. There are two primary reasons for this:
1. High Bond Enthalpy: The \(\text{C–C}\) and \(\text{C–H}\) \(\sigma\) bonds are very strong and require a large amount of energy to break.
2. Lack of Polarity: Carbon and hydrogen have very similar electronegativities, making the bonds non-polar. Furthermore, alkanes lack regions of high electron density (such as a weaker \(\pi\) bond) that would attract electrophiles or nucleophiles.
Key Takeaway
Alkanes contain strong \(\sigma\) bonds, adopt a tetrahedral shape (\(109.5^\circ\) bond angle) around each carbon, and are unreactive due to strong, non-polar bonds.
---3. Sources of Alkanes: Fractional Distillation and Cracking
Fractional Distillation of Crude Oil
Crude oil is a complex mixture of mostly unbranched and branched alkanes. It is separated into simpler, useful mixtures called fractions using fractional distillation. This physical separation process works because hydrocarbons of different chain lengths have different boiling points.
As carbon chain length increases:
• The molecules become larger with more surface contact area.
• Intermolecular forces (van der Waals / London dispersion forces) become stronger.
• More thermal energy is required to separate the molecules, resulting in a higher boiling point.
Order of Fractions (from lowest boiling point at the top to highest at the bottom):
1. Refinery / Liquefied Petroleum Gases (e.g., bottled camping gas)
2. Petrol / Gasoline (fuel for motor cars)
3. Kerosene / Paraffin (jet fuel and heating)
4. Diesel (fuel for diesel engines and lorries)
5. Lubricating Oil (waxes and engine lubricants)
6. Fuel Oil (fuel for ships and power stations)
7. Bitumen / Residue (surfacing roads and roofing)
Cracking
Fractional distillation produces more long-chain alkanes than the market demands, and fewer short-chain alkanes (like petrol) than needed. Industry solves this through cracking.
Definition: Cracking is the thermal decomposition of long-chain saturated hydrocarbons into shorter, more economically useful alkanes and unsaturated alkenes.
There are two main cracking methods:
• Thermal Cracking: Involves high temperatures and high pressures. It produces a high proportion of alkenes (used to make polymers).
• Catalytic Cracking: Involves moderate to high temperatures and a zeolite catalyst. It produces branched alkanes, cycloalkanes, and aromatic hydrocarbons, which burn more efficiently in motor fuels.
Example Cracking Equation:
\(\text{C}_{10}\text{H}_{22} \rightarrow \text{C}_8\text{H}_{18} + \text{C}_2\text{H}_4\)
Notice that the equation must balance for both carbon (\(10 = 8 + 2\)) and hydrogen (\(22 = 18 + 4\)). At least one product is always unsaturated (an alkene).
Key Takeaway
Fractional distillation separates hydrocarbons by boiling point. Cracking breaks long, less useful alkanes into shorter, in-demand alkanes and alkenes.
---4. Chemical Reactions of Alkanes
Although alkanes are generally unreactive, they participate in two key reaction types: combustion and free radical substitution.
A. Combustion Reactions
1. Complete Combustion
Occurs when alkanes burn in a plentiful supply of oxygen. The only products are carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)).
• Methane example: \(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)
• Propane example: \(\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}\)
2. Incomplete Combustion
Occurs when the oxygen supply is limited. Because there is not enough oxygen to fully oxidise all the carbon, the products include carbon monoxide (\(\text{CO}\)), particulate carbon / soot (\(\text{C}\)), alongside water (\(\text{H}_2\text{O}\)).
• Forming Carbon Monoxide: \(\text{CH}_4 + \frac{3}{2}\text{O}_2 \rightarrow \text{CO} + 2\text{H}_2\text{O}\) (or \(2\text{CH}_4 + 3\text{O}_2 \rightarrow 2\text{CO} + 4\text{H}_2\text{O}\))
• Forming Soot / Carbon: \(\text{CH}_4 + \text{O}_2 \rightarrow \text{C} + 2\text{H}_2\text{O}\)
Toxicological and Environmental Dangers
• Carbon Monoxide (\(\text{CO}\)): A colourless, odourless, toxic gas. It binds irreversibly to hemoglobin in red blood cells to form carboxyhemoglobin, drastically reducing the blood's capacity to transport oxygen around the body.
• Carbon particulates (Soot / \(\text{C}\)): Cause respiratory problems, exacerbate asthma, and contribute to global dimming.
---B. Free Radical Substitution (Halogenation)
Alkanes react with halogens (such as chlorine, \(\text{Cl}_2\), or bromine, \(\text{Br}_2\)) in the presence of ultraviolet (UV) light or sunlight. This reaction is a substitution reaction because a hydrogen atom is replaced by a halogen atom.
A free radical is a chemical species with an unpaired electron, represented by a single dot (\(\bullet\)). Radicals are extremely reactive.
The mechanism proceeds through three distinct stages: Initiation, Propagation, and Termination.
Stage 1: Initiation
UV radiation provides the energy to break the covalent bond in a halogen molecule. The bond breaks evenly, with one electron going to each chlorine atom (a process called homolytic fission):
\(\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}^\bullet\)
Stage 2: Propagation (Chain Reaction)
Propagation steps are a self-sustaining cycle where a radical reacts with a stable molecule to generate a new radical and a new molecule:
1. The chlorine radical removes a hydrogen atom from the alkane:
\(\text{CH}_4 + \text{Cl}^\bullet \rightarrow {}^\bullet\text{CH}_3 + \text{HCl}\)
(Notice the methyl radical \({}^\bullet\text{CH}_3\) is formed along with hydrogen chloride gas.)
2. The newly formed methyl radical reacts with an intact chlorine molecule:
\({}^\bullet\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet\)
(This yields the product chloromethane and regenerates a \(\text{Cl}^\bullet\) radical to continue the chain.)
Stage 3: Termination
Termination removes free radicals from the system when any two radicals collide and combine to form a stable covalent bond:
• \({}^\bullet\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl}\)
• \({}^\bullet\text{CH}_3 + {}^\bullet\text{CH}_3 \rightarrow \text{C}_2\text{H}_6\) (Evidence of methyl radicals combining to form ethane!)
• \(\text{Cl}^\bullet + \text{Cl}^\bullet \rightarrow \text{Cl}_2\)
Limitations of Free Radical Substitution
Free radical substitution is not an efficient way to prepare pure halogenoalkanes in industry because it produces a mixture of products:
1. Further substitution: If excess chlorine is present, chloromethane (\(\text{CH}_3\text{Cl}\)) will undergo further radical attacks to form dichloromethane (\(\text{CH}_2\text{Cl}_2\)), trichloromethane (\(\text{CHCl}_3\)), and tetrachloromethane (\(\text{CCl}_4\)).
2. Chain isomers: With longer alkanes (like propane), substitution can occur on different carbon atoms (e.g., forming both 1-chloropropane and 2-chloropropane).
Key Takeaway
Free radical substitution requires UV light and occurs in three steps: Initiation (\(0 \rightarrow 2\) radicals), Propagation (\(1 \rightarrow 1\) radical), and Termination (\(2 \rightarrow 0\) radicals).
---5. Common Pitfalls and Examiner Tips
• Radical Dot Placement: Always place the radical dot clearly on the atom with the unpaired electron. In the methyl radical, the unpaired electron is on the carbon atom, so write it as \({}^\bullet\text{CH}_3\) or \(\text{H}_3\text{C}^\bullet\), not \(\text{CH}_3^\bullet\) on the hydrogen.
• Addition vs. Substitution: Alkanes cannot undergo addition reactions because they do not have double bonds. They only undergo substitution.
• Incomplete Combustion Products: Students often forget that water (\(\text{H}_2\text{O}\)) is always produced during incomplete combustion alongside \(\text{CO}\) or \(\text{C}\).
• Balancing Cracking Equations: Remember that mass is conserved. Double-check that total carbons and hydrogens on the left equal the total on the right.
---6. Quick Revision Checklist
Can you answer these key revision questions?
1. What is the general formula for alkanes? (\(\text{C}_n\text{H}_{2n+2}\))
2. What is the bond angle and shape around a carbon atom in an alkane? (\(109.5^\circ\), tetrahedral)
3. What essential condition is required for the initiation step of free radical substitution? (UV light / sunlight)
4. Why is carbon monoxide dangerous to human health? (Binds irreversibly to hemoglobin forming carboxyhemoglobin, stopping oxygen transport)
5. Write the two propagation steps for the reaction between methane and chlorine. (\(\text{CH}_4 + \text{Cl}^\bullet \rightarrow {}^\bullet\text{CH}_3 + \text{HCl}\) and \({}^\bullet\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet\))