O-Level Chemistry · Secondary 3 and 4
Organic Chemistry
Organic chemistry is the chemistry of carbon compounds, from the fuels refined from crude oil to plastics, alcohol and vinegar. In O-Level Chemistry you study homologous series (alkanes, alkenes, alcohols and carboxylic acids), their typical reactions, and how small molecules join into polymers. The topic rewards careful learning of names, structures and reaction conditions, and it uses ideas from bonding, redox and energy changes along the way.
What the syllabus expects
- Describe a homologous series and name and draw the first members of the alkanes, alkenes, alcohols and carboxylic acids
- Describe the separation of crude oil by fractional distillation and state uses of the main fractions
- Describe the combustion of alkanes and their substitution reaction with chlorine in ultraviolet light
- Describe cracking, the addition reactions of alkenes and the bromine water test for unsaturation
- Describe the manufacture of ethanol by fermentation and by the catalytic addition of steam to ethene
- Describe the oxidation of ethanol to ethanoic acid and the formation of esters
- Describe addition polymerisation and deduce the repeat unit of a polymer
Homologous series and naming
A homologous series is a family of organic compounds with the same functional group and the same general formula. Members show similar chemical properties and a gradual change in physical properties, such as boiling point rising as the chain gets longer, and each member differs from the next by CH₂. Alkanes have the general formula CₙH₂ₙ₊₂ and only single bonds, so they are saturated. Alkenes are CₙH₂ₙ and contain a C=C double bond, so they are unsaturated. Alcohols contain the -OH group, CₙH₂ₙ₊₁OH. Carboxylic acids contain the -COOH group. The prefix tells you the total number of carbon atoms: meth- is 1, eth- is 2, prop- is 3 and but- is 4. Ethanoic acid, CH₃COOH, therefore has two carbon atoms, counting the one in the -COOH group. Compounds with the same molecular formula but different structural formulae, such as butane and methylpropane, are isomers.
Crude oil and alkanes
Crude oil is a mixture of hydrocarbons, separated by fractional distillation. The oil is heated and the vapour rises up a fractionating column that is hotter at the bottom and cooler at the top. Fractions with small molecules and low boiling points condense near the top, and those with large molecules and high boiling points condense lower down. Petroleum gas is used for cooking and heating, petrol as fuel for cars, naphtha as a raw material for making chemicals, kerosene as jet fuel, diesel for diesel engines, and bitumen for road surfaces. Alkanes are fairly unreactive. They burn in plenty of oxygen to give carbon dioxide and water, CH₄ + 2O₂ gives CO₂ + 2H₂O, but in limited oxygen they also form carbon monoxide and soot. In ultraviolet light they undergo substitution with chlorine, where a hydrogen atom is replaced: CH₄ + Cl₂ gives CH₃Cl + HCl.
Alkenes and cracking
Cracking breaks long-chain alkanes into shorter alkanes and alkenes by heating them with a catalyst. It matches supply to demand, because there is more demand for petrol than crude oil naturally provides, and it produces alkenes for making polymers. For example, C₁₀H₂₂ gives C₈H₁₈ + C₂H₄. Alkenes are more reactive than alkanes because of the C=C bond, and they undergo addition reactions in which the double bond opens. With bromine, ethene forms 1,2-dibromoethane, and this is the test for unsaturation: reddish-brown bromine water is decolourised by an alkene but not by an alkane. With hydrogen and a nickel catalyst on heating, alkenes form alkanes, which is how vegetable oils are hardened to make margarine. With steam at about 300 °C and 60 atm with a phosphoric acid catalyst, ethene forms ethanol.
Alcohols, carboxylic acids and esters
Ethanol is made in two ways. Fermentation uses yeast to convert glucose into ethanol and carbon dioxide, C₆H₁₂O₆ gives 2C₂H₅OH + 2CO₂, in the absence of oxygen at about 37 °C. The catalytic addition of steam to ethene is faster and gives purer ethanol, but uses a non-renewable raw material. Ethanol burns as a fuel and can be oxidised to ethanoic acid, either by warming with acidified potassium manganate(VII), which turns from purple to colourless, or by bacteria in air, which is why wine left open turns sour. Ethanoic acid is a weak acid. It reacts with metals, bases and carbonates like other acids. Carboxylic acids react with alcohols, when warmed with a few drops of concentrated sulfuric acid as a catalyst, to form an ester and water. Ethanoic acid and ethanol give ethyl ethanoate: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. Esters have sweet, fruity smells and are used in flavourings and perfumes.
Polymers
Polymers are long-chain molecules made by joining many small molecules called monomers. In addition polymerisation, alkene monomers join when their double bonds open, and no other product is formed. Ethene forms poly(ethene), whose repeat unit is -CH₂-CH₂- with bonds extending out at each end, and the polymer contains no C=C bonds. In condensation polymerisation, monomers join and a small molecule such as water is eliminated each time. Nylon, a polyamide, and polyesters such as PET are examples. Many plastics are not biodegradable, so they persist in landfills and oceans, and burning them can release toxic gases, which is why reducing, reusing and recycling plastics matters.
Worked example
Compound A has the formula C₃H₆ and decolourises bromine water. (a) Name A and its homologous series. (b) Name the product with bromine and write its formula. (c) Calculate the mass of product formed when 4.2 g of A reacts completely with bromine. (Ar: C = 12, H = 1, Br = 80)
- C₃H₆ fits CₙH₂ₙ, and decolourising bromine water shows a C=C bond, so A is propene, an alkene.
- Bromine adds across the double bond: C₃H₆ + Br₂ gives C₃H₆Br₂, which is 1,2-dibromopropane.
- Mr of C₃H₆ = 3 × 12 + 6 × 1 = 42, so moles of A = 4.2 ÷ 42 = 0.10 mol.
- The ratio of A to product is 1 : 1, so 0.10 mol of C₃H₆Br₂ forms.
- Mr of C₃H₆Br₂ = 42 + 2 × 80 = 202, so mass = 0.10 × 202 = 20.2 g.
Answer: A is propene, an alkene. It forms 1,2-dibromopropane, C₃H₆Br₂, and 4.2 g of propene gives 20.2 g of product.
Common mistakes
- Saying alkanes decolourise bromine water. Alkanes are saturated, so bromine water stays reddish-brown. Only compounds with a C=C bond decolourise it quickly.
- Naming esters the wrong way round. The alcohol part comes first: ethanol and methanoic acid give ethyl methanoate, not methyl ethanoate.
- Drawing the repeat unit of poly(ethene) with a C=C double bond. The double bond opens during polymerisation, so the repeat unit has only single bonds.
- Forgetting that fermentation must be in the absence of oxygen. With oxygen present, the ethanol is oxidised to ethanoic acid.
- Forgetting to count the carbon in the -COOH group. Propanoic acid is C₂H₅COOH, with three carbon atoms in total.
- Calling cracking a way to make more fuel from nothing. It converts less useful long-chain fractions into more useful shorter alkanes and alkenes.
Exam tips
- Make a one-page map of the reactions: alkane to chloroalkane, alkene to alkane, alcohol, dibromo compound or polymer, alcohol to acid, acid plus alcohol to ester. Write the conditions beside each arrow on your own notes.
- Always state conditions when a reaction needs them, such as ultraviolet light, a nickel catalyst, concentrated sulfuric acid or yeast. They often carry a mark of their own.
- Check every organic equation by counting carbon, hydrogen and oxygen atoms on both sides.
Practice questions
1. Easy: Give the molecular formulae of butane and butene, and say which one decolourises bromine water.
Show answer
Butane is C₄H₁₀ (CₙH₂ₙ₊₂ with n = 4) and butene is C₄H₈ (CₙH₂ₙ). Butene decolourises bromine water because it has a C=C bond.
2. Medium: Name the ester formed from methanol and ethanoic acid, state the conditions and write the equation.
Show answer
Methyl ethanoate. Warm the alcohol and acid with a few drops of concentrated sulfuric acid as catalyst. CH₃COOH + CH₃OH ⇌ CH₃COOCH₃ + H₂O.
3. Harder: Decane, C₁₀H₂₂, is cracked to give octane and one other product. Write the equation, name the other product, give a test for it and draw the repeat unit of the polymer it forms.
Show answer
C₁₀H₂₂ gives C₈H₁₈ + C₂H₄. Balancing carbon (10 = 8 + 2) and hydrogen (22 = 18 + 4) shows the other product is ethene. It decolourises reddish-brown bromine water. It forms poly(ethene), with repeat unit -CH₂-CH₂- and a bond extending from each end, drawn inside brackets with n.
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