14. Hydrocarbons

An alkene can be hydrogenated to an alkane with H₂ over a nickel or platinum catalyst on heating. Long-chain alkanes can be cracked over hot aluminium oxide to form shorter hydrocarbons.

Hydrogenation adds H across C=C and increases saturation. Cracking breaks C–C bonds and usually gives a mixture, often including an alkane and an alkene, so balance atoms rather than guessing one product.

Ethene + H₂ → ethane. A long-chain alkane can crack to octane plus ethene if the carbon and hydrogen totals balance.

A catalyst speeds hydrogenation but is not a reactant, and cracking is not simply “breaking every bond” into identical fragments.

Complete combustion of an alkane gives CO₂ and H₂O when oxygen is sufficient. Limited oxygen causes incomplete combustion, producing CO and/or carbon. Under UV light, Cl₂ or Br₂ can substitute for H in a free-radical chain.

Balance combustion by carbon, hydrogen, then oxygen. Radical substitution proceeds through initiation, propagation and termination; it is not electrophilic addition.

Ethane burns completely as 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. Under UV, CH₃CH₃ + Cl₂ → CH₃CH₂Cl + HCl, with further substitution possible.

Incomplete combustion is not a single fixed product, and UV light initiates the radical chain rather than acting as a catalyst.

UV light breaks Cl₂ or Br₂ homolytically to form radicals. A radical then abstracts hydrogen and a halogen radical substitutes it in a propagation chain.

Initiation creates radicals; propagation consumes one radical and regenerates another; termination combines radicals and removes them. Curly arrows show one-electron movement.

Cl₂ → 2Cl· is initiation. Cl· + CH₄ → HCl + CH₃· and CH₃· + Cl₂ → CH₃Cl + Cl· are propagation steps.

UV light starts the chain but is not a catalyst, and propagation is not the same as electrophilic addition.

Cracking breaks C–C bonds in long-chain hydrocarbons to make shorter alkanes and alkenes with lower relative molecular masses.

Demand for petrol and alkene feedstocks can exceed the natural fraction distribution. Choose a cracking equation that balances carbon and hydrogen atoms and includes the required heat/catalyst conditions.

C₁₀H₂₂ can crack to C₈H₁₈ + C₂H₄. The alkane is a fuel-range product and the alkene is a useful petrochemical feedstock.

Cracking is not complete combustion and does not produce one fixed product from every heavy fraction.

Alkanes have strong C–H and C–C sigma bonds and no strongly polar functional group. Polar reagents therefore have little incentive to react under ordinary conditions.

This is a kinetic and electronic explanation, not an absolute ban. Radical halogenation, combustion and cracking need suitable light, heat, oxygen or catalysts.

Bromine water is not rapidly decolourised by an alkane at room temperature, whereas an alkene reacts by electrophilic addition across C=C.

“Unreactive” does not mean chemically inert, and the explanation is not simply that alkanes are non-polar gases.

Incomplete combustion can produce CO; high-temperature combustion forms NOx; unburned hydrocarbons can escape. These pollutants have different health and atmospheric effects.

A catalytic converter promotes redox reactions: CO and hydrocarbons are oxidised, while NOx is reduced to N₂. The catalyst changes rate and the exhaust composition controls performance.

2CO + 2NO → 2CO₂ + N₂ is a simplified three-way-converter reaction. It reduces CO and NO simultaneously but does not remove every environmental impact of fuel use.

CO is not simply “more CO₂”, and catalytic removal is not the same as preventing pollutant formation in the engine.

Elimination removes atoms from neighbouring carbons to create C=C. Halogenoalkanes eliminate HX with hot ethanolic NaOH; alcohols dehydrate with heat and an acid or heated alumina; cracking also forms alkenes.

The reagent and conditions determine the route. Balance the equation and identify the small molecule removed—HX, H₂O or a fragment from a larger alkane.

CH₃CH₂Br + OH⁻(ethanol) → CH₂=CH₂ + H₂O + Br⁻. Ethanol can also dehydrate to ethene with concentrated sulfuric acid and heat.

Aqueous OH⁻ favours substitution more than hot ethanolic OH⁻, so do not ignore the solvent and temperature.

The electron-rich C=C attacks an electrophile, then a nucleophile bonds to the carbocation-like intermediate. Hydrogenation adds H₂ with Pt/Ni, hydration adds steam with phosphoric acid, and HX adds a hydrogen halide at room temperature.

The π bond is consumed and each alkene carbon gains a new atom or group. Write the reagent, catalyst and product rather than treating every addition as hydrogenation.

CH₂=CH₂ + H₂ → CH₃CH₃; CH₂=CH₂ + H₂O ⇌ CH₃CH₂OH with H₃PO₄. Both use the same alkene but different reagents and products.

Electrophilic addition is not substitution, and steam hydration is reversible whereas catalytic hydrogenation is not described by the same equilibrium.

Bromine water is orange-brown. An alkene decolourises it because the electron-rich C=C reacts with Br₂ in an electrophilic addition reaction.

The test is evidence for a reactive C=C under the stated conditions, not a universal test for every unsaturated substance. Keep the reagent and observation together.

Ethene + Br₂ → 1,2-dibromoethane, so the orange colour disappears. An alkane does not rapidly decolourise bromine water at room temperature without radical conditions.

Decolourisation is not proof that a molecule contains only one double bond, and bromine water is not the same as bromide solution.

The π bond donates an electron pair to an electrophile. In Br₂ addition, a bromine electrophile is attacked and Br⁻ completes the addition; in HBr addition, H⁺ adds first and Br⁻ attacks the carbocation-like intermediate.

Curly arrows begin at the π bond or a lone pair and end at the electron-poor atom. The intermediate and its stability help determine which product predominates.

Ethene gives 1,2-dibromoethane. Propene gives mainly 2-bromopropane with HBr because the more stable secondary carbocation pathway is favoured.

Do not draw a radical mechanism for HBr addition, and do not start a curly arrow at a positively charged atom.

Alkyl groups have a positive inductive effect: they push electron density toward a positively charged carbon. More substituted carbocations are therefore generally more stable: tertiary > secondary > primary.

When HX adds to an unsymmetrical alkene, the pathway that forms the more stable carbocation is favoured. The hydrogen adds so that the halide ends up on the more substituted carbon: the Markovnikov product.

Propene + HBr gives mainly 2-bromopropane, because proton addition can form a secondary rather than primary carbocation.

This is a product-prediction rule, not a claim that the final product contains a free carbocation or that every addition gives one product only.