Properties
Alkanes are saturated hydrocarbons consisting entirely of carbon–carbon single bonds (σ-bonds) and carbon–hydrogen bonds. Because all bonding electrons are localised in strong σ-bonds and alkanes possess no π-electrons or lone pairs, they are chemically unreactive toward most polar reagents such as acids, bases, and oxidising agents under ordinary conditions. This relative inertness earned them the historical name 'paraffins' (from Latin parum affinis, meaning 'little affinity'). Despite this, alkanes undergo three commercially and industrially significant classes of reaction: free-radical halogenation, combustion, and thermal cracking (pyrolysis). Free-radical halogenation replaces one or more hydrogen atoms with halogen atoms (typically Cl or Br) via a chain mechanism initiated by ultraviolet light or heat. The reaction is governed by bond dissociation energies and exhibits a selectivity that depends on both the halogen used and the type of C–H bond (primary, secondary, or tertiary). Combustion—the complete or incomplete oxidation of alkanes—releases large quantities of energy and is the basis of fossil-fuel technology, though it produces CO₂ and H₂O (and CO or soot in incomplete combustion). Pyrolysis involves high-temperature bond cleavage in the absence of oxygen, producing smaller alkanes, alkenes, and hydrogen—reactions central to petroleum refining and the production of petrochemical feedstocks. Understanding these three reaction types requires knowledge of thermodynamics (bond enthalpies, activation energies), radical stability, and reaction kinetics. Together they illustrate the principle that even 'unreactive' molecules participate in energetically favourable processes once the appropriate activation barrier is overcome, whether by photons, high temperatures, or the presence of catalysts.
- Free-Radical Halogenation: Initiation
- Free-Radical Halogenation: Propagation
- Free-Radical Halogenation: Termination and Selectivity
- Combustion of Alkanes
- Pyrolysis (Thermal Cracking)
Free-Radical Halogenation: Initiation
Halogenation begins when ultraviolet radiation (or heat) provides enough energy to homolytically cleave the X–X bond of a halogen molecule (Cl₂ or Br₂). Homolytic cleavage means each atom retains one electron, producing two highly reactive halogen radicals: Cl₂ → 2 Cl•. The bond dissociation energy of Cl–Cl is 243 kJ mol⁻¹ and of Br–Br is 193 kJ mol⁻¹. Because Br₂ has a lower BDE, bromine radicals form more easily, but this alone does not determine overall reactivity.
Free-Radical Halogenation: Propagation
In the first propagation step, a halogen radical abstracts a hydrogen from the alkane: Cl• + R–H → R• + HCl. This step's activation energy depends on the C–H bond strength: tertiary C–H bonds (∼400 kJ mol⁻¹) are weaker than secondary (∼410 kJ mol⁻¹) or primary (∼423 kJ mol⁻¹), so tertiary positions are attacked preferentially. In the second propagation step, the carbon radical reacts with a halogen molecule: R• + Cl₂ → R–Cl + Cl•, regenerating the halogen radical and continuing the chain. Each complete propagation cycle produces one molecule of alkyl halide.
Free-Radical Halogenation: Termination and Selectivity
Termination occurs when two radicals combine: Cl• + Cl• → Cl₂, R• + Cl• → RCl, or R• + R• → R–R. Selectivity is the tendency to react preferentially at a specific C–H position. Chlorination is relatively unselective: at 25 °C the relative rates of attack on tertiary : secondary : primary C–H bonds are approximately 5 : 3.8 : 1. Bromination is far more selective (approximately 1600 : 82 : 1 for tertiary : secondary : primary) because the H-abstraction step by Br• is endothermic and therefore has a later, more product-like transition state (Hammond's Postulate), making it more sensitive to radical stability. Tertiary radicals are most stable due to hyperconjugation and inductive donation from three alkyl groups.
Combustion of Alkanes
Complete combustion of an alkane CₙH₂ₙ₊₂ in excess oxygen produces carbon dioxide and water with release of large quantities of heat. The general equation is: CₙH₂ₙ₊₂ + [(3n+1)/2] O₂ → n CO₂ + (n+1) H₂O. For example, methane: CH₄ + 2 O₂ → CO₂ + 2 H₂O, ΔH° = −890 kJ mol⁻¹; octane: 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O. Incomplete combustion (limited O₂) produces carbon monoxide (CO) and/or soot (C), both hazardous. The energy released per mole increases with chain length because more C–H and C–C bonds are broken and more C=O and O–H bonds (stronger) are formed.
Pyrolysis (Thermal Cracking)
Pyrolysis (from Greek pyr = fire, lysis = breaking) is the thermal decomposition of alkanes at 450–750 °C in the absence of oxygen. C–C bonds (BDE ≈ 347 kJ mol⁻¹), being weaker than C–H bonds, break first, generating two alkyl radicals. These radicals then undergo β-scission: the C–C bond one position away from the radical centre breaks, producing a smaller alkyl radical and an alkene. For example, butane cracking: C₄H₁₀ → C₂H₄ + C₂H₆, or C₄H₁₀ → CH₄ + C₃H₆. The resulting alkenes (ethylene, propylene) are vital feedstocks for the polymer industry. Catalytic cracking uses zeolite catalysts and proceeds via carbocation intermediates at lower temperatures (~500 °C), producing more branched and aromatic products suitable for high-octane petrol.
The Key Equation
- \ce{CH4} — Methane (mol): The simplest alkane; one carbon bonded to four hydrogens.
- \ce{O2} — Dioxygen (mol): Molecular oxygen; the oxidant in combustion reactions.
- \ce{CO2} — Carbon dioxide (mol): Product of complete combustion; a greenhouse gas.
- \ce{H2O} — Water (mol): Product of combustion; formed when H–H and O=O bonds are replaced by O–H bonds.
- \Delta H^\circ — Standard enthalpy change (kJ mol⁻¹): Energy released (negative sign = exothermic) under standard conditions (298 K, 100 kPa).
Worked Example
Problem. Propane (C₃H₈) undergoes free-radical chlorination. (a) Write the two propagation steps. (b) Determine how many distinct monochlorinated products are possible and predict which is the major product, explaining your reasoning using radical stability and selectivity data (relative rates: tertiary 5, secondary 3.8, primary 1 per H atom for Cl•).
Answer. Major product is 2-chloropropane (~56%), minor product is 1-chloropropane (~44%). Two distinct monochlorinated products are possible.
Origins · 1 minMorris Selig Kharasch and how this was discovered
Kharasch was a pioneer in free-radical chemistry who elucidated the mechanism of radical addition reactions and the anti-Markovnikov addition of HBr to alkenes. His systematic investigations of radical chain reactions in the 1930s–1940s laid the theoretical groundwork for understanding why halogenation of alkanes proceeds through radical intermediates initiated by light or heat, and how radical stability governs selectivity. Kharasch discovered the 'peroxide effect'—the reversal of Markovnikov's rule when peroxides (radical initiators) are present—after noticing puzzling inconsistencies in HBr addition results reported in the literature. He suspected the discrepancies were due to trace peroxide contaminants in different reagent samples.
Related Concepts
- Reactions of Alkenes — Pyrolysis of alkanes produces alkenes; understanding electrophilic addition to alkenes contrasts with the radical mechanism of alkane halogenation.
- Alkane Structure and Bonding — The σ-bond framework and tetrahedral geometry of alkanes directly explain why homolytic C–H and C–C bond cleavage is required for reactivity.
- Bond Enthalpy and Hess's Law — Combustion enthalpy calculations and determining whether radical abstraction steps are exo- or endothermic both rely on bond dissociation energies.
- Activation Energy and Transition State Theory — Hammond's Postulate, which explains why Br• is more selective than Cl•, is a direct application of transition-state theory and activation energy concepts.
- IUPAC Nomenclature of Hydrocarbons — Correctly naming the monochlorinated products formed during halogenation requires systematic IUPAC naming of substituted alkanes.