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CIE 9701 Chemistry · AS · Topic 14

Hydrocarbons

CIE 9701 ChemistryASFree revision notes

Contents: 5 sections

Alkanes

General formula CₙH₂ₙ₊₂. Saturated, with only single bonds, and non-polar, so they are unreactive towards most reagents: there is no δ+ carbon for a nucleophile and no electron-rich region for an electrophile.

Boiling point rises with chain length, because larger molecules have more electrons and so stronger van der Waals forces. Branching lowers the boiling point, because branched molecules cannot pack as closely and the contact between them is reduced.

Combustion

Complete combustion, with plenty of oxygen, gives carbon dioxide and water:

CH₄ + 2O₂ → CO₂ + 2H₂O

Incomplete combustion, with a limited supply, gives carbon monoxide and possibly soot. Carbon monoxide is dangerous precisely because it is colourless and odourless: it binds to haemoglobin far more strongly than oxygen does, so the blood cannot carry oxygen.

Burning hydrocarbons also produces carbon dioxide, a greenhouse gas, and where the fuel contains sulfur it produces sulfur dioxide, which causes acid rain. In an engine, the high temperature makes nitrogen and oxygen from the air combine to give oxides of nitrogen.

Free-radical substitution

Methane reacts with chlorine in the presence of ultraviolet light:

Concept explainer · 3 minInitiation, propagation and termination, with the reason for eachGradefruitSays why the ultraviolet light breaks the Cl-Cl bond and not the C-H bonds, which is the bond energy point the equation alone hides. Propagation is then drawn as a cycle that hands the chlorine radical back, so one initiation gives thousands of product molecules, and termination is derived as the only way to remove radicals rather than stated as a third step.

CH₄ + Cl₂ → CH₃Cl + HCl

The mechanism has three stages and the exam expects all three named and shown.

Initiation. UV light breaks the Cl-Cl bond by homolytic fission:

Cl₂ → 2Cl•

Propagation. Two steps, and they form a cycle, which is why one initiation event can lead to thousands of product molecules:

Cl• + CH₄ → •CH₃ + HCl

•CH₃ + Cl₂ → CH₃Cl + Cl•

Termination. Any two radicals combine, which removes them:

Cl• + Cl• → Cl₂

•CH₃ + Cl• → CH₃Cl

•CH₃ + •CH₃ → C₂H₆

The reaction is a poor synthetic method, and the reason is worth understanding rather than just stating. Further substitution gives CH₂Cl₂, CHCl₃ and CCl₄, and termination gives ethane, so the product is a mixture that is difficult to separate.

Alkenes

General formula CₙH₂ₙ. The C=C double bond is a region of high electron density, which makes alkenes far more reactive than alkanes and makes them targets for electrophiles.

The double bond does not rotate, which is why alkenes can show cis-trans isomerism.

Addition reactions

ReagentConditionsProduct
H₂Nickel catalyst, 150 °CAlkane
Br₂Room temperatureDibromoalkane
HBrRoom temperatureBromoalkane
SteamPhosphoric acid catalyst, high temperature and pressureAlcohol
Cold dilute acidified KMnO₄Room temperatureDiol
Hot concentrated acidified KMnO₄HeatCarbon chain broken

The addition of hydrogen is how unsaturated vegetable oils are hardened into margarine. The addition of steam is the industrial route to ethanol.

The test for a C=C bond

Shake with bromine water. An alkene decolourises it from orange to colourless; an alkane does not. This is the standard distinguishing test and it works because the alkene adds bromine across the double bond while the alkane has nothing for it to react with.

The electrophilic addition mechanism

Ethene with HBr:

  1. The H-Br bond is polar, so hydrogen carries δ+ and acts as the electrophile.
  2. A curly arrow runs from the C=C double bond to the hydrogen, and a second from the H-Br bond to the bromine. This gives a carbocation and a bromide ion.
  3. A curly arrow from a lone pair on the bromide ion to the positive carbon forms the product.

With bromine, which is non-polar, the double bond induces a dipole in the Br₂ molecule as it approaches, and the mechanism then runs the same way. That induced dipole step is a common omission.

Markovnikov's rule

When an unsymmetrical alkene adds an unsymmetrical reagent, the hydrogen adds to the carbon that already has more hydrogens. So propene with HBr gives mainly 2-bromopropane, not 1-bromopropane.

Concept explainer · 3 minBoth carbocations from propene and HBr, then picking the winnerChemistorianAdds HBr to propene and forms both carbocations on screen before choosing between them. The primary one has a single alkyl group pushing electron density onto the positive carbon and the secondary has 2, so the secondary is more stable and 2-bromopropane is the major product. The rule arrives as the conclusion rather than the premise.

The reason is carbocation stability. Alkyl groups push electron density towards the positive carbon and stabilise it, so the order of stability is

tertiary > secondary > primary

The route through the more stable carbocation dominates, and that gives the Markovnikov product.

Cracking

Cracking breaks long-chain alkanes from crude oil into shorter, more useful molecules.

C₁₂H₂₆ → C₈H₁₈ + 2C₂H₄

It is done because the fractions from crude oil do not match demand: there is more long-chain material than the market wants and not enough petrol and alkenes.

Common mistakes

Check you have it

Question 1

In polymer G every carbon atom in the polymer chain is bonded to one hydrogen atom and one methyl group.
Which alkene could be polymerised to make polymer G?

Question 2

Aluminium carbide, Al 4C3, reacts readily with aqueous sodium hydroxide. The two products of the reaction are NaAlO2 and a hydrocarbon. Water molecules are also involved as reactants.
What is the formula of the hydrocarbon?

Question 3

Z is a gaseous hydrocarbon which has a density of 3.50 × 10⁻³ g cm⁻³ under room conditions.
Z reacts with an excess of hot concentrated acidified KMnO4. Only one type of carboxylic acid is formed in this reaction.
What is Z?

What the syllabus asks for on this topicSyllabus points

Syllabus points

  • Describe the reactions of alkanes with oxygen and with halogens, including the free-radical substitution mechanism.
  • Explain the environmental consequences of burning hydrocarbons and of cracking.
  • Describe the reactions of alkenes: addition of hydrogen, halogens, hydrogen halides, steam and oxidising agents.
  • Describe the electrophilic addition mechanism and use Markovnikov's rule.
  • Distinguish between alkanes and alkenes chemically.

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