Unsaturated hydrocarbons

Alkenes (olefins) are unsaturated hydrocarbons whose molecules contain one C=C double bond. The π-bond within the double bond is weaker than a σ-bond and its electrons are exposed, so alkenes are very reactive: they undergo addition, are easily oxidised and polymerise. Ethene is the organic substance produced in the largest amounts worldwide: it is turned into poly(ethene), ethanol, ethane-1,2-diol and chloroethene. This topic also covers alkadienes (hydrocarbons with two double bonds), because rubbers are made from them.

§1. Structure: σ- and π-bonds

A double bond consists of two different bonds: a σ-bond (hybrid orbitals overlapping along the axis) and a π-bond (unhybridised p-orbitals overlapping sideways). The C=C bond length is 0.134 nm, shorter than a single bond (0.154 nm). Because of the π-bond there is no free rotation about the double bond, which is the cause of geometric isomerism.

Homologous series: ethene (ethylene) C₂H₄, propene (propylene) C₃H₆, butenes C₄H₈, pentenes C₅H₁₀. To name them, replace the -ane ending of the alkane with -ene and number the chain from the end that gives the double bond the lowest number: CH₂=CH–CH₂–CH₃ is but-1-ene, CH₃–CH=CH–CH₃ is but-2-ene.

Physical properties resemble those of alkanes: C₂–C₄ are gases, C₅–C₁₇ liquids and higher members solids; they dissolve poorly in water.

§2. Isomerism

Alkenes show four kinds of isomerism:

• chain isomerism: but-1-ene and 2-methylpropene;
• position isomerism of the double bond: but-1-ene and but-2-ene;
• geometric (cis–trans) isomerism: cis-but-2-ene and trans-but-2-ene;
• functional-group isomerism with cycloalkanes: butene and cyclobutane or methylcyclopropane.

For geometric isomerism each carbon of the double bond must carry two different groups. In propene CH₂=CH–CH₃ the first carbon has two identical H atoms, so there are no cis–trans isomers; in but-2-ene each carbon carries H and CH₃, so the isomers exist. In the cis isomer identical groups are on the same side of the double bond, in the trans isomer on opposite sides.

§3. Addition reactions and Markovnikov’s rule

Alkenes typically undergo electrophilic addition: an electrophile (H⁺, Br⁺) first attacks the π-electrons, a carbocation forms, and then an anion adds.

• Hydrogenation: C₂H₄ + H₂ → C₂H₆ (Ni, Pt, Pd).
• Halogenation: C₂H₄ + Br₂ → CH₂Br–CH₂Br — bromine water is decolourised, the test for a double bond.
• Addition of hydrogen halides: C₂H₄ + HCl → C₂H₅Cl.
• Hydration (H⁺, heat): C₂H₄ + H₂O → C₂H₅OH, the industrial route to ethanol.

For unsymmetrical alkenes Markovnikov’s rule applies: the hydrogen atom adds to the carbon of the double bond that already has more hydrogens, and the halogen or OH adds to the carbon with fewer. The reason is that the more stable (secondary or tertiary) carbocation forms.

In the presence of peroxides HBr adds against the rule (the Kharasch effect): the reaction goes by a radical mechanism and 1-bromopropane forms.

Worked example

7 g of an alkene reacted completely with 16 g of bromine. Find the formula of the alkene.

1. n(Br₂) = 16/160 = 0.1 mol; alkene and bromine react 1 : 1, so n(alkene) = 0.1 mol.
2. M = 7/0.1 = 70 g/mol.
3. 14n = 70 → n = 5: C₅H₁₀ (pentene).

§4. Oxidation and polymerisation

• Mild oxidation (Wagner reaction, cold neutral aqueous KMnO₄): 3C₂H₄ + 2KMnO₄ + 4H₂O → 3CH₂OH–CH₂OH + 2MnO₂ + 2KOH — the purple colour disappears, a brown precipitate of MnO₂ forms and a diol (ethane-1,2-diol) is produced. This is another test for a double bond.
• Vigorous oxidation (KMnO₄ or K₂Cr₂O₇ in acid, heat) breaks the double bond completely: a =CH₂ fragment becomes CO₂, a =CH–R fragment becomes RCOOH and a =CR₂ fragment becomes a ketone. The products reveal where the double bond was.
• Combustion: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O; alkenes give n(CO₂) = n(H₂O).
• Catalytic oxidation (Ag, 250 °C): 2C₂H₄ + O₂ → 2C₂H₄O (epoxyethane).
• Polymerisation: nCH₂=CH₂ → (–CH₂–CH₂–)ₙ (poly(ethene)), nCH₂=CH–CH₃ → (–CH₂–CH(CH₃)–)ₙ (poly(propene)).

In practice, ethene is converted into poly(ethene), ethanol, ethane-1,2-diol (antifreeze), epoxyethane and chloroethene; propene into poly(propene) and propan-2-ol. Ethene is also a plant hormone that speeds up fruit ripening, so unripe fruit in warehouses is treated with it.

Worked example

Which alkene gives only ethanoic acid on vigorous oxidation?

1. Ethanoic acid CH₃COOH comes from a =CH–CH₃ fragment.
2. Such a fragment must be on both sides of the double bond: CH₃–CH=CH–CH₃.
3. Answer: but-2-ene.

§5. Preparation

Zaitsev’s rule: in elimination, hydrogen leaves from the neighbouring carbon that has fewer hydrogens, so the more substituted alkene forms: 2-bromobutane gives mainly but-2-ene. Below 140 °C, dehydration of an alcohol gives an ether, not an alkene.

§6. Alkadienes

By the position of the double bonds, dienes are cumulated (CH₂=C=CH₂, allene), conjugated (buta-1,3-diene CH₂=CH–CH=CH₂, isoprene — 2-methylbuta-1,3-diene) or isolated. In conjugated dienes the π-electrons are spread over four carbons, so they give both 1,2- and 1,4-addition; in 1,4-addition the double bond moves to the middle of the chain. Conjugated dienes polymerise to rubbers: natural rubber is poly(isoprene), a common synthetic one is poly(butadiene). Lebedev’s process:

Worked example

An alkadiene whose numbers of H and C atoms differ by 2 was passed through 150 g of 3.2 % bromine solution; dibromo and tetrabromo products formed in equal molar amounts. Find the mass of the alkadiene.

1. CₙH₂ₙ₋₂: (2n − 2) − n = 2 → n = 4, C₄H₆ (M = 54).
2. n(Br₂) = 150·0.032/160 = 0.03 mol.
3. Dibromide x mol (x mol Br₂), tetrabromide x mol (2x mol Br₂): 3x = 0.03 → x = 0.01.
4. Diene 2x = 0.02 mol → m = 0.02·54 = 1.08 g.

Check yourself

1. What is the main product when HCl adds to propene?

2. What mass of bromine adds to 5.6 g of ethene?

3. Which alkene shows cis–trans isomerism: propene, but-1-ene, but-2-ene?

4. What products form when 2-methylbut-2-ene is vigorously oxidised?

5. What volume of ethene (STP, 100 % yield) is obtained from 23 g of ethanol?

6. What forms when 1 mol of H₂ adds 1,4 to buta-1,3-diene?

7. What volume of H₂ (STP) is needed to hydrogenate 0.3 mol of propene completely?

8. Which reagent distinguishes ethane from ethene?

Common mistakes

• Applying Markovnikov’s rule backwards: H goes to the carbon that already has more hydrogens.
• Drawing cis–trans isomers for propene or but-1-ene.
• Mixing up dehydration temperatures: below 140 °C an ether forms.
• Giving alkadienes the alkene formula CₙH₂ₙ: theirs is CₙH₂ₙ₋₂.
• Confusing mild and vigorous oxidation products: a diol in mild conditions; an acid, ketone or CO₂ in vigorous ones.