Hydrocarbons

0. What Examiners Want

Where this topic appears. Recent AS papers have asked for: the product of bromine reacting with a diene, the products when 2-methylbut-2-ene is heated with hot concentrated acidified KMnO₄, a comparison of the products of but-2-ene with cold and hot KMnO₄, the monomer for a given addition-polymer repeat unit, the repeat unit that forms from a given monomer. Percentage-yield calculations from Topic 2 are often set on these reactions.

1. Alkanes: Unreactivity and Combustion

Alkanes are saturated hydrocarbons: they contain only carbon and hydrogen and only single bonds, so they have the maximum number of hydrogen atoms. The general formula is C_nH_2n+2 (cycloalkanes, C_nH_2n, are also saturated). Each carbon is sp³ with bond angles of about 109.5°. Crude oil is the main source, separated into fractions by fractional distillation.

Why alkanes are unreactive. The C–C and C–H bonds are strong, and the electronegativity difference between C and H is very small, so the bonds are non-polar. There are no δ+ carbon atoms to attract nucleophiles and no region of high electron density to attract electrophiles. Alkanes do not react with acids, alkalis, water, oxidising agents or reducing agents (octane added to NaOH(aq) simply forms a separate layer). They do react with oxygen (combustion) and, in UV light, with halogens.

Combustion

• Complete combustion (plenty of oxygen): carbon dioxide and water. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O (or C₈H₁₈ + 12½O₂ → 8CO₂ + 9H₂O).
• Incomplete combustion (limited oxygen, as in a car engine): carbon monoxide (and sometimes carbon) is formed. 2C₈H₁₈ + 17O₂ → 16CO + 18H₂O.
• Other examples: C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O; 2CH₄ + 3O₂ → 2CO + 4H₂O; 2C₉H₂₀ + 19O₂ → 18CO + 20H₂O.

Carbon monoxide is toxic: it bonds to haemoglobin so the blood cannot carry oxygen. It is odourless, which makes faulty gas heaters dangerous.

2. Pollutants from Engines and Catalytic Converters

Burning hydrocarbon fuels in a car engine releases several pollutants:

• CO from incomplete combustion (toxic).
• Unburnt hydrocarbons (volatile organic compounds, VOCs): some are carcinogenic and they form PAN and photochemical smog.
• Oxides of nitrogen, NO and NO₂: nitrogen and oxygen in the air react at the high temperature of the engine (N₂ + O₂ → 2NO; 2NO + O₂ → 2NO₂). They cause acid rain and smog (Topic 12).
• CO₂: a greenhouse gas, which contributes to climate change.

A catalytic converter (a honeycomb coated with platinum, palladium or rhodium) removes three pollutants once it is hot: CO is oxidised to CO₂; NO is reduced to N₂; unburnt hydrocarbons are oxidised to CO₂ and H₂O. The key equation is 2CO + 2NO → 2CO₂ + N₂.

A converter cannot reduce CO₂; in fact removing CO makes more CO₂.

3. Free-Radical Substitution

Alkanes react with chlorine or bromine in UV light (or sunlight) by free-radical substitution: a hydrogen atom is replaced by a halogen atom. There is no reaction in the dark, because the first step needs energy to break the halogen–halogen bond.

Overall (methane and bromine): CH₄ + Br₂ → CH₃Br + HBr. The same three stages apply with Br (Br₂ → 2Br•), or with ethane and chlorine: C₂H₆ + Cl₂ → C₂H₅Cl + HCl.

• Initiation: UV light breaks Cl–Cl homolytically into two radicals. One molecule in, two radicals out.
• Propagation: a radical attacks a molecule to give a different molecule and a different radical (two steps). The radical Cl• is regenerated, so a chain reaction runs.
• Termination: two radicals combine to give one molecule, with no radical left.

A mixture of products. A chlorine radical can also attack a chloroalkane that has already formed (for example CH₃CH₂Cl + Cl• → •CH₂CH₂Cl + HCl). More chlorine gives dichloro-, trichloro- and tetrachloro- products. So free-radical substitution is not a good way to make a pure halogenoalkane.

4. Cracking and Making Alkenes

Fractional distillation gives more large alkane molecules than the market needs. Cracking converts them into smaller, more useful alkanes and alkenes. The large molecules are heated to a high temperature and passed over an aluminium oxide (Al₂O₃) catalyst with no oxygen (so there is no combustion).

• C₁₀H₂₂ → C₆H₁₄ + C₄H₈
• C₉H₂₀ → C₇H₁₆ + C₂H₄ (nonane to heptane and ethene)

The shorter alkanes make good fuels (petrol). The alkenes are more reactive than alkanes because of the π bond, which makes them valuable feedstocks for plastics and other chemicals. Alkenes can also be made by:

• elimination of hydrogen halide from a halogenoalkane, heating with ethanolic NaOH;
• dehydration of an alcohol, using a heated Al₂O₃ catalyst or concentrated acid (for example C₂H₅OH → C₂H₄ + H₂O).

5. Addition Reactions of Alkenes

Alkenes are unsaturated: they contain a C=C double bond (one σ and one π bond), general formula C_nH_2n. The π bond is a region of high electron density, and it is weaker than a σ bond, so alkenes undergo addition: the π bond breaks and a new single bond forms to each carbon.

Equations: C₂H₄ + H₂ → C₂H₆; C₂H₄ + Br₂ → C₂H₄Br₂; C₂H₄ + HBr → C₂H₅Br; C₂H₄ + H₂O → C₂H₅OH.

Test for a C=C double bond: shake with bromine water (or bubble the gas through it). Orange to colourless if unsaturated; no change for an alkane.

Percentage yield (worked example). 4.20 g of propene reacts with HBr. 9.83 g of 2-bromopropane is obtained. (A_r: H 1.0, C 12.0, Br 79.9.)

• Amount of propene = 4.20 ÷ 42.0 = 0.100 mol (C₃H₆ + HBr → C₃H₇Br, 1 : 1)
• M_r of C₃H₇Br = 36.0 + 7.0 + 79.9 = 122.9; theoretical mass = 0.100 × 122.9 = 12.29 g
• Percentage yield = 9.83 ÷ 12.29 × 100 = 80.0%

6. Electrophilic Addition and Markovnikov's Rule

An electrophile accepts a pair of electrons. The electron-rich C=C attacks electrophiles. HBr is polar (H^δ+–Br^δ−), so H acts as the electrophile. Br₂ is non-polar, but as it approaches the C=C the π electrons repel the Br–Br pair, giving an induced dipole: the nearer Br becomes δ+ and acts as the electrophile.

Mechanism for ethene and bromine (write it in this order)

1. A curly arrow from the C=C (the π bond) to the δ+ Br; a second curly arrow from the Br–Br bond to the δ− Br. Show the dipole on Br–Br.
2. Intermediate: a carbocation (BrCH₂CH₂⁺) and a Br⁻ ion (heterolytic fission of Br–Br).
3. A curly arrow from a lone pair on Br⁻ to C⁺. Product: 1,2-dibromoethane.

Mechanism for HBr: the same pattern, with the curly arrow from C=C to the H of H–Br and a second arrow from the H–Br bond to Br.

Markovnikov's rule. With an unsymmetrical alkene such as propene there are two possible carbocations.

H⁺ adding to the CH₂ carbon gives the secondary carbocation CH₃C⁺HCH₃, stabilised by two electron-donating alkyl groups (positive inductive effect). H⁺ adding to the other carbon gives the primary carbocation, which is less stable. The more stable carbocation forms faster, so 2-bromopropane is the major product and 1-bromopropane the minor product. The halogen ends up on the carbon with more alkyl groups attached.

7. Oxidation of Alkenes with KMnO₄

• Cold, dilute, acidified KMnO₄: the alkene is oxidised to a diol, and the purple solution turns colourless. C₂H₄ + [O] + H₂O → HOCH₂CH₂OH.
• Hot, concentrated, acidified KMnO₄: the C=C is broken and each carbon is oxidised further: a carbon with two H atoms gives CO₂; with one H and one alkyl group it gives a carboxylic acid; with two alkyl groups it gives a ketone.
  • C₅H₁₀ (pent-2-ene) + 4[O] → CH₃COOH + CH₃CH₂COOH
  • C₅H₁₀ (2-methylbut-2-ene) + 3[O] → CH₃COCH₃ + CH₃COOH
  • C₃H₆ (propene) + 5[O] → CH₃COOH + CO₂ + H₂O

Finding the alkene. Identify the products, remove the oxygen atoms and join the carbon atoms with a double bond. If CO₂ forms, the double bond was at the end of the chain. For example, an alkene known to be but-1-ene or but-2-ene gives CO₂ with hot KMnO₄ only if it is but-1-ene (but-2-ene gives only ethanoic acid).

8. Addition Polymers and Waste

Addition polymerisation: many unsaturated monomers (containing C=C) join to form a long-chain polymer, and the polymer is the only product. The repeat unit is the monomer with C=C changed to C–C and an open bond at each end, shown inside brackets with n.

• nCH₂=CH₂ → [−CH₂−CH₂−]_n, poly(ethene)
• nCH₂=CHCH₃ → [−CH₂−CH(CH₃)−]_n, poly(propene)
• nCH₂=CHCl → [−CH₂−CHCl−]_n, poly(chloroethene) (PVC)
• nCF₂=CF₂ → [−CF₂−CF₂−]_n, poly(tetrafluoroethene) (PTFE, non-stick coatings)

To find the monomer from a section of polymer: divide the chain into repeat units (two carbon atoms each in the main chain) and put the C=C back. A polymer made from one monomer has the same atoms in the monomer and repeat unit.

Disposal. Poly(alkene)s are effectively very large alkanes, so they are unreactive and non-biodegradable. They remain for hundreds of years in landfill and pollute the environment.

• Burning poly(alkene)s releases energy for electricity and saves fossil fuels, but gives CO₂ (a greenhouse gas) and, if oxygen is limited, toxic CO. Burning PVC also gives acidic HCl and toxic dioxins, so the gases must be neutralised and the incinerator run at a very high temperature.
• Recycling is difficult because different plastics must be sorted first.

9. Quick Sheet and Checklist

Before you leave the question, check:

• Every alkene reaction has its reagent and its condition.
• Radicals are drawn with a dot; initiation, propagation and termination have the right numbers of radicals.
• Curly arrows start at a bond or lone pair; the intermediate is a carbocation; the dipole on Br–Br or H–Br is shown.
• The major product is explained using carbocation stability and the inductive effect.
• Oxidation products follow the rule (CO₂, carboxylic acid, ketone) and equations balance with [O].
• Repeat units show the open bonds and n; the polymer is not described as unsaturated.