ChemistryCore24 min read

Organic Chemistry

Crude oil, hydrocarbons, and the molecules that make plastics

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01

Crude oil and fractional distillation

Almost all organic chemistry starts with crude oil — a mixture of hydrocarbons (compounds of hydrogen and carbon only), formed from ancient sea life. It is useless as a mixture, so it is separated by fractional distillation: heated in a column, the hydrocarbons condense at different heights according to their boiling points.

Short-chain fractions (petrol, LPG) have low boiling points, are runny, ignite easily, and come off high in the column; long-chain fractions (bitumen) are the opposite. That single trend — chain length sets the properties — runs through the whole topic.

02

Alkanes: the saturated ones

Alkanes are hydrocarbons with only single C–C bonds — methane, ethane, propane, butane — with the general formula CₙH₂ₙ₊₂. They are saturated (every carbon holds as many hydrogens as possible) and fairly unreactive, apart from burning.

Their main use is as fuels. Complete combustion (plenty of oxygen) gives carbon dioxide and water. Incomplete combustion (limited oxygen) gives carbon monoxide — a toxic, colourless gas — and soot, which is why faulty heaters are dangerous.

03

Alkenes: the reactive ones

Alkenes contain a carbon–carbon double bond (C=C) — ethene, propene — with the general formula CₙH₂ₙ. They are unsaturated, and that double bond makes them much more reactive than alkanes.

The classic test uses bromine water: an alkene decolourises it (orange to colourless) as bromine adds across the double bond, while an alkane leaves it orange. Alkenes are made by cracking long alkane chains — breaking them into smaller, more useful molecules.

The bromine-water test in one line

Add bromine water: it stays orange with an alkane (saturated), turns colourless with an alkene (the C=C adds bromine). It is the standard way to tell saturated from unsaturated.

04

Polymers

Alkenes are the raw material for polymers. In addition polymerisation, many small alkene molecules (monomers) join into one very long chain: thousands of ethene molecules become poly(ethene), the plastic in bags and bottles. The double bond opens up and links to the next monomer.

Their strength and cheapness are why plastics are everywhere — and their downside is the same chemistry: the strong, unreactive C–C backbone does not break down, so most are non-biodegradable, which is the heart of the plastic-waste problem.

05

A worked example

Worked example

You have two colourless gases, one ethane and one ethene. How do you tell them apart, and what is happening?

  1. Bubble each through bromine water (orange).The test distinguishes saturated from unsaturated hydrocarbons.
  2. Ethene (alkene, C=C) decolourises the bromine water — orange to colourless.Bromine adds across the double bond, so the coloured bromine is used up.
  3. Ethane (alkane, all single bonds) leaves it orange.A saturated alkane has no double bond to react with the bromine.

The gas that decolourises bromine water is ethene (unsaturated); the one that leaves it orange is ethane (saturated).

06

Homologous series and functional groups

Organic chemistry is tamed by families. A homologous series is a set of compounds with the same general formula and similar chemistry, where each member differs from the next by CH₂ — the alkanes (CₙH₂ₙ₊₂), the alkenes (CₙH₂ₙ), the alcohols and so on. As the chain lengthens, physical properties like boiling point change gradually and predictably.

What gives each family its character is its functional group — the reactive part, such as the C=C of alkenes or the –OH of alcohols. The name tells you both: a stem for the chain length (meth-, eth-, prop-, but-) plus an ending for the functional group (-ane, -ene, -ol).

07

The reactions of alkenes

The C=C double bond makes alkenes far more reactive than alkanes, and it reacts by addition — the double bond opens and atoms add across it. With hydrogen and a nickel catalyst an alkene is hydrogenated to an alkane (this hardens vegetable oils into margarine). With steam and a catalyst it is hydrated to an alcohol. With bromine it adds instantly, decolourising bromine water — the standard test.

Alkene molecules can also add to each other, linking into long chains — that is addition polymerisation, and it is how poly(ethene) and most everyday plastics are made.

Addition vs substitution

Alkenes ADD across their reactive C=C double bond; alkanes, having only single bonds, can only undergo substitution (e.g. with chlorine in UV light). The double bond is the key to alkene chemistry.

08

Plastics and their disposal

Addition polymers such as poly(ethene) are chemically unreactive and non-biodegradable — microbes cannot break their strong carbon backbone — so they persist for centuries in landfill and oceans. Condensation polymers like nylon and polyester join monomers while losing a small molecule (usually water), and some can be broken back down.

Because burning plastics releases carbon dioxide and toxic gases, and landfill wastes both the material and space, recycling and designing biodegradable alternatives are increasingly important.

09

Another worked example

Worked example

Ethene is bubbled through bromine water, and separately reacted with steam over a catalyst. State the product and reaction type in each case.

  1. With bromine water: the orange colour disappears as bromine adds across the C=C double bond.Alkenes decolourise bromine water — an addition reaction.
  2. Product: 1,2-dibromoethane (C₂H₄Br₂).One Br adds to each carbon of the former double bond.
  3. With steam and a catalyst (~300 °C, ~60 atm, phosphoric acid): water adds across the double bond.Hydration is an addition reaction of alkenes.
  4. Product: ethanol, C₂H₅OH.This is the industrial route to ethanol from ethene.

Bromine water → 1,2-dibromoethane (addition, decolourises); steam → ethanol (hydration, addition).

10

Free-radical substitution of alkanes

Alkanes are unreactive, but in ultraviolet light they react with halogens by a free-radical chain reaction in which a hydrogen is substituted by a halogen. It runs in three stages. Initiation: UV light splits a halogen molecule into two radicals, Cl₂ → 2Cl•. Propagation: a radical attacks the alkane and regenerates another radical, so the chain continues — Cl• + CH₄ → •CH₃ + HCl, then •CH₃ + Cl₂ → CH₃Cl + Cl•. Termination: two radicals meet and combine, ending a chain (e.g. Cl• + •CH₃ → CH₃Cl).

Because the reaction is hard to control, it gives a mixture of products — further substitution can replace more hydrogens (CH₂Cl₂, CHCl₃ …) — which limits its usefulness for making a single compound.

Initiation, propagation, termination

A radical has an unpaired electron (shown as a dot). Initiation makes radicals (UV breaks the halogen homolytically); propagation uses one radical and makes another, so the chain keeps going; termination joins two radicals and stops it.

11

How alkenes add: the electrophilic mechanism

The C=C double bond is a region of high electron density, so it attracts electrophiles. When hydrogen bromide adds, the π electrons attack the δ+ hydrogen, forming a positively charged carbocation intermediate; the bromide ion then bonds to it. This step-by-step attack is electrophilic addition.

With an unsymmetrical alkene, Markovnikov's rule predicts the major product: the hydrogen adds to the carbon that already has more hydrogens, because this gives the more stable carbocation (a tertiary carbocation is more stable than a secondary, which beats a primary). So propene + HBr gives mainly 2-bromopropane.

12

Alcohols and ethanol

The alcohols are a homologous series with the –OH (hydroxyl) functional group; the most important is ethanol, C₂H₅OH. Ethanol is made two ways: by fermentation, where yeast converts sugars to ethanol and carbon dioxide (C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂) at about 30 °C without air — a slow, batch process from a renewable crop; and by the hydration of ethene with steam and a catalyst — a fast, continuous process from crude oil.

Ethanol burns cleanly (C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O), which makes it a useful fuel and biofuel, and it is also a common solvent. Warmed with an oxidising agent it is oxidised to ethanoic acid — which is what happens, slowly in air, when wine turns to vinegar.

13

Carboxylic acids

The carboxylic acids contain the –COOH (carboxyl) functional group; ethanoic acid, CH₃COOH, is the acid in vinegar. They are weak acids — only partly ionised — but they show all the typical acid reactions: with reactive metals they give a salt and hydrogen, with carbonates a salt, water and carbon dioxide (the fizz that confirms an acid), and with alkalis a salt and water. The salts are called ethanoates.

Ethanoic acid is made industrially, and in the lab by oxidising ethanol (with acidified potassium dichromate) or by the fermentation of ethanol in air.

14

Esters

Warm a carboxylic acid with an alcohol and a little concentrated sulfuric acid catalyst, and they combine to form a sweet-smelling ester plus water — a reaction called esterification. Ethanoic acid and ethanol, for example, give ethyl ethanoate.

Esters are prized for their pleasant, fruity smells, so they are used in flavourings and perfumes, as well as being useful solvents. Naturally occurring esters give many fruits their scent, and animal and vegetable fats are esters too.

The organic family tree

At O Level the organic chapter runs: crude oil → alkanes and alkenes (hydrocarbons) → alcohols → carboxylic acids → esters, plus polymers. Oxidising an alcohol gives a carboxylic acid; an acid plus an alcohol gives an ester.

Practice questions

10 questions · 39 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

6 · 15 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Define the term homologous series.
Model answer

A family of organic compounds with the same general formula and similar chemical properties, where each member differs from the next by CH₂.

Examiner tip. e.g. alkanes CₙH₂ₙ₊₂; members show a gradual trend in physical properties.

SQ2[2 marks]
Describe a chemical test to distinguish an alkane from an alkene.
Model answer

Add bromine water: an alkene decolourises it (orange to colourless); an alkane does not (stays orange).

Examiner tip. Alkenes have a C=C double bond and undergo addition with bromine — the decolourising test.

SQ3[3 marks]
Ethene undergoes addition polymerisation to form poly(ethene). Explain what this means and give the repeat unit.
Model answer

Many ethene monomers (with C=C double bonds) join together, with no other product, to form a long-chain polymer. Repeat unit: –(CH₂–CH₂)–.

Examiner tip. Addition polymerisation: monomers with C=C open up and link, forming only the polymer (no small molecule lost).

SQ4[2 marks]
Describe the test to distinguish ethane from ethene, giving the result for each.
Model answer

Add bromine water: ethene (an alkene) decolourises it from orange to colourless; ethane (an alkane) leaves it orange (no reaction).

Examiner tip. The C=C of an alkene adds bromine, decolourising the water.

SQ5[3 marks]
Methane reacts with chlorine in ultraviolet light. Name the type of mechanism and write the initiation and the two propagation steps.
Model answer

Free-radical substitution. Initiation: Cl₂ → 2Cl• (UV). Propagation: Cl• + CH₄ → •CH₃ + HCl, then •CH₃ + Cl₂ → CH₃Cl + Cl•.

Examiner tip. Initiation makes radicals; each propagation step uses one radical and makes another.

SQ6[3 marks]
Ethanol can be made by fermentation. Give the equation, the conditions, and one advantage of this method over hydration of ethene.
Model answer

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂, using yeast at about 30 °C in the absence of air. An advantage is that it uses a renewable resource (sugar/plant crops) rather than crude oil.

Examiner tip. Fermentation is renewable but slow and batch; hydration is fast and continuous but uses crude oil.

Long questions

4 · 24 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
Ethanol can be made by fermentation or by the hydration of ethene.
  1. Write the equation for the fermentation of glucose to ethanol. [2]
  2. State the conditions for making ethanol by hydration of ethene, and write the equation. [3]
  3. Give one advantage of the hydration method over fermentation. [1]
Mark scheme
  1. C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂[1]
  2. correct formulae / balanced[1]
  3. Steam / water, ~300 °C, ~60 atm, phosphoric acid catalyst[1]
  4. C₂H₄ + H₂O → C₂H₅OH[1]
  5. balanced equation[1]
  6. It is a continuous process / faster / gives purer ethanol (any one)[1]

Fermentation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂; hydration: C₂H₄ + H₂O → C₂H₅OH (steam, ~300 °C, ~60 atm, H₃PO₄); hydration is continuous/faster/purer.

Examiner tip. Fermentation is a slow batch process from a renewable source; hydration is fast and continuous from crude-oil ethene.

LQ2[6 marks]
Ethene (C₂H₄) is an important product of cracking.
  1. Explain what cracking is and why it is done. [3]
  2. Ethene undergoes addition polymerisation. Describe what this means and name the product. [3]
Mark scheme
  1. Cracking breaks long-chain hydrocarbons into shorter ones[1]
  2. using heat and a catalyst[1]
  3. to make more useful small molecules (fuels) and alkenes for plastics[1]
  4. Many ethene monomers (with C=C double bonds) join together[1]
  5. the double bonds open and link into a long chain, with no other product[1]
  6. The product is poly(ethene)[1]

Cracking breaks long hydrocarbons (heat + catalyst) into useful short molecules and alkenes; addition polymerisation joins many ethene monomers into poly(ethene).

Examiner tip. Cracking supplies alkenes; alkenes polymerise by addition into plastics.

LQ3[6 marks]
But-1-ene reacts with hydrogen bromide.
  1. Name the mechanism and describe the role of the C=C double bond. [2]
  2. Explain, using Markovnikov's rule, which product forms in the greater amount. [4]
Mark scheme
  1. The mechanism is electrophilic addition[1]
  2. the electron-rich C=C double bond attracts the electrophile (the δ+ H of HBr)[1]
  3. The H can add to either carbon, giving either a secondary or a primary carbocation[1]
  4. the secondary carbocation is more stable than the primary[1]
  5. so the H adds to the terminal (CH₂) carbon (Markovnikov's rule)[1]
  6. giving mainly 2-bromobutane[1]

Electrophilic addition; the C=C attracts the electrophile; via the more stable secondary carbocation the major product is 2-bromobutane.

Examiner tip. The more stable carbocation forms preferentially — that is the basis of Markovnikov's rule.

LQ4[6 marks]
Ethanol, ethanoic acid and their reactions are important in the O Level organic chapter.
  1. Describe how ethanoic acid can be made from ethanol. [2]
  2. State what you would observe when ethanoic acid is added to magnesium, and write the equation. [2]
  3. Name the ester and other product formed when ethanoic acid reacts with ethanol, and state the catalyst. [2]
Mark scheme
  1. Oxidise the ethanol with acidified potassium dichromate (warm/reflux)[1]
  2. (or ferment ethanol in air) to give ethanoic acid[1]
  3. The magnesium fizzes / dissolves, giving off hydrogen gas[1]
  4. Mg + 2CH₃COOH → (CH₃COO)₂Mg + H₂[1]
  5. Ethyl ethanoate is formed, plus water[1]
  6. using a concentrated sulfuric acid catalyst[1]

Oxidise ethanol → ethanoic acid; with magnesium it fizzes giving H₂ (Mg + 2CH₃COOH → (CH₃COO)₂Mg + H₂); with ethanol it makes ethyl ethanoate + water (conc. H₂SO₄ catalyst).

Examiner tip. Carboxylic acids show normal acid reactions; acid + alcohol → ester + water.