Hydrocarbons · Lesson 4 of 14
Alkane Reactions and Radical Chains
“Explain radical substitution and use reaction conditions to distinguish alkane transformations.”
• Follow initiation, propagation, and termination in alkane halogenation. • Balance complete combustion and distinguish incomplete combustion. • Compare controlled oxidation, isomerisation, aromatisation, reforming, and cracking. • Connect industrial products to their reaction conditions.
Low everyday reactivity does not mean no reactions
Alkanes were historically called paraffins because they show little affinity for many common reagents. Their largely non-polar, strong sigma bonds account for much of this low reactivity under ordinary conditions. Light, high temperature, catalysts, or oxygen can nevertheless open useful reaction routes. Always read the conditions as part of the reaction.
Halogenation replaces one or more alkane hydrogens with halogen. Methane can be chlorinated successively to chloromethane, dichloromethane, trichloromethane, and tetrachloromethane. Heat, typically in the 573–773 K range, or suitable light initiates this process. Further substitution makes mixtures likely unless conditions and proportions are controlled. Lower alkanes do not readily undergo the nitration and sulphonation reactions sometimes listed for hydrocarbons more generally.
Halogen reactivity decreases broadly as . Fluorination can be dangerously vigorous, while iodination is slow and reversible. Removing hydrogen iodide with an oxidising agent can help iodination proceed. For comparable hydrogen abstraction sites, tertiary hydrogens are more readily replaced than secondary, then primary hydrogens, although product proportions also depend on how many hydrogens of each kind are present.
A radical carries the chain forward
A radical contains an unpaired electron. In initiation, light breaks the chlorine–chlorine bond homolytically, giving one electron to each chlorine. In propagation, a chlorine radical abstracts hydrogen from methane and creates a methyl radical. That methyl radical takes chlorine from another chlorine molecule, making chloromethane and regenerating a chlorine radical. A radical is consumed and another created, so the chain continues.
Propagation is not a claim that every collision succeeds. It identifies steps that can repeat while regenerating a radical carrier. Chloromethane can also lose a hydrogen to a chlorine radical, allowing formation of the more highly chlorinated products. In termination, two radicals combine and no replacement radical is produced, ending those chains.
Problem
Explain an ethane byproduct without proposing that methane adds directly to methane.
- 1.The propagation mechanism produces methyl radicals by hydrogen abstraction.
- 2.Occasionally two methyl radicals meet and form a carbon–carbon sigma bond. Each radical contributes one electron to the new bond.
- 3.Ethane is therefore a termination product. The main substitution reaction and this coupling byproduct arise from the same radical pool.
Combustion and controlled oxidation
With sufficient oxygen, complete combustion converts all carbon to carbon dioxide and all hydrogen to water. It releases substantial energy: methane and butane have standard combustion enthalpies of about −890 and −2876 kilojoules per mole respectively when liquid water is the stated product. A balanced general equation follows by counting carbon, hydrogen, and finally oxygen.
Problem
Write complete combustion of .
- 1.Four carbons require four carbon dioxide molecules. Ten hydrogens require five water molecules.
- 2.These products contain oxygen atoms, requiring oxygen molecules per butane molecule.
- 3.Multiply by two for whole coefficients: . The fractional form is equally balanced.
Limited oxygen can produce carbon monoxide and/or carbon particles instead of complete conversion to carbon dioxide. Carbon black is used in inks, pigments, and as a reinforcing filler. Controlled industrial oxidation is a different idea: regulate oxygen, pressure, temperature, and catalysts to favour useful oxygen-containing products rather than burning the hydrocarbon completely.
| Starting material and conditions | Important product | What changes |
|---|---|---|
| Methane; oxygen, copper, 523 K, high pressure | Methanol | Partial oxidation |
| Methane; oxygen and molybdenum oxide catalyst | Methanal | Partial oxidation |
| Ethane; oxygen and manganese ethanoate catalyst | Ethanoic acid | Further controlled oxidation |
| 2-Methylpropane; suitable oxidation conditions | 2-Methylpropan-2-ol | Oxidation at a tertiary C–H site |
Changing the skeleton or splitting it
Anhydrous aluminium chloride with hydrogen chloride can rearrange normal alkanes into branched isomers. Heating a suitable straight chain containing at least six carbons over chromium, vanadium, or molybdenum oxides on alumina can instead cyclise and dehydrogenate it into an aromatic compound. This aromatisation is also called reforming in this context. Hexane yields benzene; heptane provides the corresponding route to toluene.
Steam reforming uses methane and water vapour at high temperature over nickel to make carbon monoxide and hydrogen. Cracking, or pyrolysis, uses heat to split larger hydrocarbon molecules into smaller ones, often an alkane plus an alkene. A heated feed gives a mixture of possible fragments; a displayed equation commonly represents one pathway rather than the only product.
Problem
Compare isomerisation, aromatisation, and cracking of a six-carbon alkane.
- 1.Isomerisation retains six carbons and the saturated molecular formula while changing branching.
- 2.Aromatisation retains six carbons in the organic product but closes a ring and removes hydrogen to give benzene.
- 3.Cracking breaks the skeleton: possible products include butene plus ethane, or propene plus ethene plus methane. Count both carbon and hydrogen to check any proposed pathway.
A radical propagation step regenerates a radical; a termination step does not. Formation of chloromethane can occur in either kind of step, so identifying the product alone is not enough to classify a mechanism step.
Quiz
Which event initiates methane chlorination under light?
Which is a termination step?
What is essential for complete combustion?
Which transformation produces benzene from hexane?
Why may cracking equations show several products?
Practice Problems
- Write the initiation, both principal propagation steps, and three termination steps for methane chlorination.
- Explain how dichloromethane can form after chloromethane has appeared.
- Balance complete combustion of propane and pentane.
- Distinguish incomplete combustion from controlled partial oxidation.
- Write equations for hexane aromatisation and methane steam reforming, including conditions.
- Propose two atom-balanced cracking pathways for hexane and explain why they do not imply a pure single product.
Key Takeaways
• Light or heat can initiate radical substitution in alkanes. • Propagation regenerates radical carriers; termination combines radicals. • Combustion, partial oxidation, reforming, and cracking have different goals and conditions. • Carbon and hydrogen balances help distinguish skeleton rearrangement, ring formation, and fragmentation.