Hydrocarbons

📚The chemistry

14.1 Alkanes

Making alkanes

• Hydrogenation of an alkene: H₂(g) with a platinum or nickel catalyst and heat. CH₂=CH₂ + H₂ → CH₃CH₃. (This is how vegetable oils are hardened into margarine.)
• Cracking a longer-chain alkane: heat with an Al₂O₃ catalyst. The long molecule breaks into a shorter alkane and at least one alkene, e.g. C₁₀H₂₂ → C₈H₁₈ + C₂H₄.

Why crack? Fractional distillation of crude oil gives more heavy, long-chain fractions than are wanted, and too little petrol. Cracking converts the surplus heavy fractions into shorter alkanes for fuel, which are in greater demand, and alkenes, which are the feedstock for polymers and other chemicals.

Why alkanes are unreactive

Alkanes react with very little, and not at all with polar reagents such as acids, alkalis or nucleophiles. The C–C and C–H bonds are strong (C–H about 410 kJ mol⁻¹) and almost non-polar (C 2.5, H 2.1), so there is no δ+ or δ− site to attract a nucleophile or an electrophile. They burn, and they react with halogens in UV light, and that is almost all.

Combustion

With limited oxygen, carbon monoxide (and even soot, C) forms instead of CO₂. In an internal combustion engine three pollutants matter:

• carbon monoxide — toxic: it binds to haemoglobin and stops it carrying oxygen;
• oxides of nitrogen — formed from N₂ and O₂ at the high engine temperature; they cause acid rain and photochemical smog (topic 12);
• unburnt hydrocarbons — react with NO_x in sunlight to form photochemical smog; some are carcinogenic.

A catalytic converter removes all three: 2CO + 2NO → 2CO₂ + N₂, and hydrocarbons are oxidised to CO₂ and H₂O.

Free-radical substitution

In ultraviolet light, alkanes react with chlorine or bromine: a hydrogen atom is replaced by a halogen atom. For ethane and chlorine, overall C₂H₆ + Cl₂ → C₂H₅Cl + HCl. The mechanism has three stages:

1. Initiation. UV light supplies the energy for homolytic fission of the halogen, the weakest bond present: \[ \mathrm{Cl_2 \rightarrow 2Cl{\bullet}} \]
2. Propagation. A radical reacts with a molecule, making a product and a new radical, so the chain continues: \[ \mathrm{Cl{\bullet} + C_2H_6 \rightarrow C_2H_5{\bullet} + HCl} \] \[ \mathrm{C_2H_5{\bullet} + Cl_2 \rightarrow C_2H_5Cl + Cl{\bullet}} \]
3. Termination. Two radicals combine, removing radicals from the chain: \[ \mathrm{Cl{\bullet} + Cl{\bullet} \rightarrow Cl_2} \] \[ \mathrm{C_2H_5{\bullet} + Cl{\bullet} \rightarrow C_2H_5Cl} \] \[ \mathrm{C_2H_5{\bullet} + C_2H_5{\bullet} \rightarrow C_4H_{10}} \]

Free-radical substitution gives a mixture. Chlorine radicals also attack the chloroethane already formed, giving dichloroethanes and beyond; and the termination step can make butane. So it is a poor way to make one pure product, and the products must be separated by fractional distillation. Using excess ethane makes monosubstitution more likely.

14.2 Alkenes

Making alkenes

• Elimination of HX from a halogenoalkane: NaOH in ethanol, heat. CH₃CH₂Br + NaOH → CH₂=CH₂ + NaBr + H₂O. (In water, NaOH substitutes instead; see topic 15.)
• Dehydration of an alcohol: pass the vapour over a heated catalyst such as Al₂O₃, or heat with a concentrated acid (H₂SO₄ or H₃PO₄). C₂H₅OH → CH₂=CH₂ + H₂O.
• Cracking a longer alkane, as above.

Electrophilic addition

The C=C double bond is a region of high electron density, held in the π bond above and below the plane. The π bond is weaker than the σ bond and its electrons are exposed, so it attracts electrophiles and the π bond opens up: two new single bonds form, and the alkene becomes saturated.

• Hydrogen (hydrogenation): H₂(g), Pt or Ni catalyst, heat → alkane.
• Steam (hydration): H₂O(g), H₃PO₄ catalyst, high temperature and pressure → alcohol. CH₂=CH₂ + H₂O → CH₃CH₂OH — the industrial route to ethanol.
• Hydrogen halide: HX(g) at room temperature → halogenoalkane.
• Halogen: X₂ at room temperature → dihalogenoalkane. CH₂=CH₂ + Br₂ → CH₂BrCH₂Br.

Mechanism: bromine and ethene

1. As Br₂ approaches the π electrons, it is polarised: the nearer Br becomes δ+ and the further one δ− (an induced dipole).
2. Curly arrow from the C=C π bond to the Br^δ+; a second curly arrow from the Br–Br bond to the Br^δ−. The Br–Br bond breaks heterolytically.
3. This gives a carbocation, CH₂Br–CH₂⁺, and a bromide ion, Br⁻.
4. Curly arrow from a lone pair on Br⁻ to the positive carbon. Product: 1,2-dibromoethane.

Mechanism: hydrogen bromide and propene

1. H–Br is already polar: H^δ+–Br^δ−.
2. Curly arrow from the C=C π bond to the H^δ+; curly arrow from the H–Br bond to the Br.
3. The H can join either carbon, so two carbocations are possible: the secondary CH₃C⁺HCH₃ (H added to the end carbon) or the primary CH₃CH₂CH₂⁺.
4. Br⁻ attacks the positive carbon. The major product is 2-bromopropane, from the secondary carbocation; 1-bromopropane is the minor product.

Why: carbocation stability and Markovnikov addition

Alkyl groups are electron-donating: they push electron density towards the carbon they are attached to — a positive inductive effect. The more alkyl groups attached to the positive carbon, the more its charge is spread out and the more stable the carbocation:

A more stable carbocation forms more readily, so the product from it predominates. This is Markovnikov addition: when HX adds to an unsymmetrical alkene, the H goes to the carbon that already has more hydrogens, and X to the carbon with fewer, because that route passes through the more substituted, more stable carbocation.

Oxidation with potassium manganate(VII)

• Cold, dilute, acidified KMnO₄ adds two OH groups across the double bond, forming a diol. Purple → colourless. CH₂=CH₂ + [O] + H₂O → HOCH₂CH₂OH (ethane-1,2-diol).
• Hot, concentrated, acidified KMnO₄ breaks the C=C bond completely. Each end becomes a C=O, and the products depend on what was attached:
  • =CH₂ (end of chain) → CO₂ + H₂O;
  • =CHR (one H, one alkyl) → a carboxylic acid, RCOOH (an aldehyde would form, but is oxidised further);
  • =CRR′ (two alkyl groups) → a ketone, RCOR′.
  Identifying the products therefore tells you where the double bond was.

Test for C=C

Shake with aqueous bromine (bromine water): an alkene turns it from orange to colourless, as the bromine adds across the double bond. Alkanes do not react in the dark, so the colour stays.

Addition polymerisation

Under suitable conditions, many alkene molecules (monomers) join by opening their double bonds, forming a long saturated chain (the polymer) with no other product:

Poly(ethene) and poly(propene). The repeat unit is the monomer with its double bond opened: two carbons in the main chain, with any groups on the monomer hanging off it as side groups. The polymerisation of these monomers is dealt with further in topic 20.