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This lesson covers the nomenclature, properties, and reactions of alkanes, alkenes, and halogenoalkanes, including key reaction mechanisms, stereoisomerism, addition polymerisation, and the environmental impact of halogenated hydrocarbons. Organic chemistry at A-Level requires a thorough understanding of functional groups, reaction types, and how to draw curly-arrow mechanisms. This material aligns with the AQA and OCR A specifications for AS and A-Level Chemistry.
Organic molecules are named systematically using IUPAC rules:
Key Definition: A homologous series is a family of organic compounds with the same general formula, the same functional group, and similar chemical properties. Each successive member differs by –CH₂–.
Alkanes are saturated hydrocarbons with the general formula CₙH₂ₙ₊₂. They contain only C–C single bonds and C–H bonds. All carbon atoms are sp³ hybridised, giving a tetrahedral geometry with bond angles of approximately 109.5°.
Structural isomers have the same molecular formula but different structural formulae. For example, C₄H₁₀ has two structural isomers: butane and 2-methylpropane (methylpropane). As chain length increases, the number of possible structural isomers increases dramatically.
Alkanes burn in excess oxygen to produce carbon dioxide and water (complete combustion):
CₙH₂ₙ₊₂ + (3n+1)/2 O₂ → nCO₂ + (n+1)H₂O
For example: CH₄ + 2O₂ → CO₂ + 2H₂O
In limited oxygen, incomplete combustion produces carbon monoxide (CO), carbon (soot/particulates), and/or water. Carbon monoxide is toxic because it binds irreversibly to haemoglobin, reducing its capacity to carry oxygen. Particulates contribute to respiratory problems and smog. Oxides of nitrogen (NOₓ) form at high temperatures in vehicle engines and contribute to photochemical smog and acid rain.
Exam Tip: In combustion equation questions, balance carbon first, then hydrogen, then oxygen last. If you get a fractional coefficient for O₂, multiply the whole equation through by 2 to give whole numbers.
Alkanes react with halogens (e.g. Cl₂, Br₂) in the presence of UV light via a free radical substitution mechanism. This has three stages:
graph TD
A["INITIATION<br/>UV light breaks Cl₂<br/>Cl₂ → 2Cl•"] --> B["PROPAGATION<br/>Self-sustaining chain"]
B --> C["Step 1: CH₄ + Cl• → •CH₃ + HCl"]
C --> D["Step 2: •CH₃ + Cl₂ → CH₃Cl + Cl•"]
D -->|"Cl• feeds back<br/>into Step 1"| C
B --> E["TERMINATION<br/>Two radicals combine"]
E --> F["Cl• + Cl• → Cl₂"]
E --> G["•CH₃ + Cl• → CH₃Cl"]
E --> H["•CH₃ + •CH₃ → C₂H₆"]
1. Initiation: The halogen molecule undergoes homolytic fission to form two free radicals. Homolytic fission means the bond breaks evenly, with one electron going to each atom.
Cl₂ → 2Cl•
Curly arrow description: A single-headed (fish-hook) curly arrow starts from the Cl–Cl bond and points to one Cl atom; a second fish-hook arrow points from the bond to the other Cl atom. Each Cl gains one electron and becomes a radical.
2. Propagation: Two steps that form a self-sustaining chain reaction.
Step 1: CH₄ + Cl• → •CH₃ + HCl Step 2: •CH₃ + Cl₂ → CH₃Cl + Cl•
Curly arrow description for Step 1: A fish-hook arrow goes from the C–H bond to the carbon atom (forming the methyl radical). A second fish-hook arrow goes from the C–H bond to the hydrogen atom. The hydrogen radical then pairs with the chlorine radical to form HCl. In Step 2, a fish-hook arrow goes from the Cl–Cl bond to the carbon radical.
The chlorine radical produced in Step 2 feeds back into Step 1, sustaining the chain.
3. Termination: Two free radicals combine, ending the chain.
Cl• + Cl• → Cl₂ •CH₃ + Cl• → CH₃Cl •CH₃ + •CH₃ → C₂H₆
A major limitation of this mechanism is that it produces a mixture of products (mono-, di-, tri-, and tetra-substituted products, as well as longer-chain by-products such as ethane). This makes it impractical for selective synthesis. Further substitution also occurs because CH₃Cl can undergo the same process to form CH₂Cl₂, CHCl₃, and CCl₄.
Key Definition: Homolytic fission is the breaking of a covalent bond so that each atom receives one electron from the bonding pair, forming two free radicals.
Alkenes are unsaturated hydrocarbons with the general formula CₙH₂ₙ. They contain at least one C=C double bond, which consists of a sigma (σ) bond and a pi (π) bond. The carbon atoms of the double bond are sp² hybridised, giving a trigonal planar geometry with bond angles of approximately 120°.
The σ bond is formed by end-on overlap of sp² hybrid orbitals. The π bond is formed by sideways overlap of unhybridised p orbitals above and below the plane of the molecule. The π bond is weaker than the σ bond and is the region of high electron density that makes alkenes reactive towards electrophiles. The C=C bond cannot rotate (unlike a C–C single bond) because rotation would break the π bond. This restricted rotation gives rise to E/Z isomerism.
Key Definition: E/Z isomers (a type of stereoisomer) arise when there is restricted rotation about a C=C double bond and each carbon of the double bond has two different groups attached.
To assign E or Z labels, use the Cahn-Ingold-Prelog (CIP) priority rules:
Worked Example: Assigning E/Z Configuration
Consider but-2-ene, CH₃CH=CHCH₃.
On C-2: the attached groups are –CH₃ (C, atomic number 6) and –H (atomic number 1). The –CH₃ group has higher priority. On C-3: the attached groups are –CH₃ (C, atomic number 6) and –H (atomic number 1). Again, –CH₃ has higher priority.
If both –CH₃ groups are on the same side of the double bond → Z-but-2-ene (the cis form). If they are on opposite sides → E-but-2-ene (the trans form).
Exam Tip: The terms cis and trans can only be used when there are two identical groups. The E/Z system is more general and works for all cases of geometric isomerism. Always use E/Z in A-Level exams unless specifically asked otherwise.
The C=C double bond is an area of high electron density, making alkenes susceptible to attack by electrophiles (electron-pair acceptors).
Key Definition: An electrophile is an electron-pair acceptor — a species that is attracted to regions of high electron density and accepts a pair of electrons to form a new covalent bond.
1. Addition of hydrogen halides (e.g. HBr) — Electrophilic Addition
Overall: CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (major) or CH₃CH₂CH₂Br (minor)
Step-by-step mechanism (curly arrow description):
Step 1: The π electrons of the C=C bond are attracted to the δ+ hydrogen of HBr. A curly arrow is drawn from the C=C π bond to the H atom of HBr, forming a new C–H bond. Simultaneously, a second curly arrow goes from the H–Br bond to the Br atom, showing heterolytic fission. The Br⁻ ion is released.
Step 2: A carbocation intermediate is formed. For propene, the H can add to either carbon. If H adds to C-1 (the terminal carbon), a secondary carbocation forms on C-2, which is more stable. If H adds to C-2, a primary carbocation forms on C-1, which is less stable. The more stable secondary carbocation forms preferentially.
Step 3: A curly arrow goes from a lone pair on the Br⁻ ion to the positively charged carbon of the carbocation, forming the C–Br bond.
Markownikoff's rule: When HBr adds to an unsymmetrical alkene, the major product has the hydrogen adding to the carbon with more hydrogen atoms already attached. The underlying reason is that the more substituted carbocation intermediate is more stable due to the inductive (electron-donating) effect of alkyl groups, which stabilise the positive charge.
2. Addition of bromine (Br₂) — Electrophilic Addition
Overall: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br (1,2-dibromoethane)
Step-by-step mechanism (curly arrow description):
Step 1: As the Br₂ molecule approaches the electron-rich C=C bond, the electrons in the C=C repel the nearer bromine's electron cloud. This induces a dipole in the Br–Br bond (the nearer Br becomes δ+ and the farther Br becomes δ−). A curly arrow goes from the C=C π bond to the δ+ Br atom, forming a C–Br bond. A second curly arrow goes from the Br–Br bond to the terminal Br atom, showing heterolytic fission. A carbocation intermediate is formed and Br⁻ is released.
Step 2: A curly arrow goes from a lone pair on the Br⁻ to the δ+ carbon of the carbocation, forming the second C–Br bond.
The decolourisation of bromine water from orange/brown to colourless is the standard test for a C=C double bond.
3. Addition of steam (hydration)
Alkenes react with steam in the presence of an H₃PO₄ catalyst at 300 °C and 60–70 atm pressure to form alcohols. This is an industrial method for producing ethanol.
CH₂=CH₂ + H₂O → CH₃CH₂OH
4. Hydrogenation
Alkenes react with hydrogen gas at 150 °C in the presence of a nickel catalyst to form alkanes. This is used industrially to harden vegetable oils into margarine.
CH₂=CH₂ + H₂ → CH₃CH₃
Under conditions of high pressure and with a suitable catalyst, many alkene monomers join together by addition polymerisation. The C=C double bonds open up and the monomers link together to form a long-chain polymer with no by-product.
For example, ethene polymerises to form poly(ethene):
n CH₂=CH₂ → –(CH₂–CH₂)ₙ–
Other important addition polymers include:
| Monomer | Polymer | Uses |
|---|---|---|
| Ethene (CH₂=CH₂) | Poly(ethene) | Plastic bags, bottles |
| Propene (CH₃CH=CH₂) | Poly(propene) | Rope, crates, carpets |
| Chloroethene (CH₂=CHCl) | Poly(chloroethene), PVC | Window frames, pipes, insulation |
| Tetrafluoroethene (CF₂=CF₂) | PTFE (Teflon) | Non-stick coatings |
| Phenylethene (C₆H₅CH=CH₂) | Polystyrene | Packaging, insulation |
Exam Tip: When drawing the repeat unit of an addition polymer, open the double bond, draw the resulting single-bond unit in square brackets with continuation bonds through the brackets, and write 'n' as a subscript outside the closing bracket.
Addition polymers are non-biodegradable because they are saturated (no functional groups for microorganisms to attack). They can be recycled mechanically (re-melting) or used as fuel (incineration), though incineration may release toxic gases if the polymer contains chlorine (e.g. PVC releases HCl).
Halogenoalkanes contain a halogen atom bonded to a saturated carbon chain. They are classified as primary, secondary, or tertiary depending on the number of carbon atoms bonded to the carbon bearing the halogen.
Key Definition: A primary halogenoalkane has the halogen bonded to a carbon that is attached to one (or zero) other carbon atoms. A secondary has two carbons attached, and a tertiary has three carbons attached.
The C–X bond is polar (carbon is δ+, halogen is δ−), making the carbon susceptible to attack by nucleophiles (electron-pair donors). Common nucleophiles include OH⁻, CN⁻, and NH₃.
Key Definition: A nucleophile is an electron-pair donor — a species with a lone pair (or negative charge) that is attracted to a δ+ carbon and donates a pair of electrons to form a new covalent bond.
A one-step (concerted) mechanism:
Curly arrow description: A curly arrow goes from a lone pair on the nucleophile (e.g. OH⁻) to the δ+ carbon atom. Simultaneously, a second curly arrow goes from the C–X bond to the halogen atom (X), showing the departure of the halide ion (X⁻). The nucleophile attacks from the opposite side (180°) to the leaving group — this is called backside attack. A transition state forms in which the carbon has five groups partially bonded to it (the incoming nucleophile and the outgoing halide are both partially bonded). The product has inverted stereochemistry (Walden inversion) compared to the original molecule.
Rate = k[halogenoalkane][nucleophile] — the rate depends on the concentration of both reactants because they are both involved in the single rate-determining step.
A two-step mechanism:
Step 1 (slow, rate-determining): The C–X bond breaks heterolytically. A curly arrow goes from the C–X bond to the halogen atom. The halide ion (X⁻) departs, leaving a tertiary carbocation. This is the slow step because bond breaking requires energy.
Step 2 (fast): The nucleophile (e.g. OH⁻) rapidly attacks the planar carbocation. A curly arrow goes from a lone pair on the nucleophile to the positively charged carbon, forming the new C–nucleophile bond. Because the carbocation is planar, the nucleophile can attack from either side, producing a racemic mixture if a chiral centre is formed.
Rate = k[halogenoalkane] — the rate depends only on the concentration of the halogenoalkane because only the halogenoalkane is involved in the slow (rate-determining) step.
| Feature | SN1 | SN2 |
|---|---|---|
| Number of steps | 2 (via carbocation) | 1 (concerted) |
| Rate equation | Rate = k[RX] | Rate = k[RX][Nu⁻] |
| Favoured by | Tertiary halogenoalkanes | Primary halogenoalkanes |
| Intermediate | Carbocation | Transition state (no true intermediate) |
| Stereochemical outcome | Racemic mixture (if chiral) | Inversion of configuration |
| Effect of nucleophile concentration | No effect on rate | Rate increases with [Nu⁻] |
| Solvent preference | Polar protic (e.g. water) | Polar aprotic (e.g. DMSO) |
Secondary halogenoalkanes can react by either SN1 or SN2, depending on reaction conditions. With a strong nucleophile (e.g. OH⁻, CN⁻) in polar aprotic solvent, SN2 is favoured. With a weak nucleophile in a polar protic solvent, SN1 is favoured.
flowchart TD
A["Halogenoalkane<br/>undergoing substitution"] --> B{"What type of<br/>halogenoalkane?"}
B -->|"Primary"| C["SN2 mechanism<br/>One step, concerted<br/>Rate = k·RX·Nu"]
B -->|"Tertiary"| D["SN1 mechanism<br/>Two steps, via carbocation<br/>Rate = k·RX"]
B -->|"Secondary"| E{"Conditions?"}
E -->|"Strong nucleophile<br/>Polar aprotic solvent"| C
E -->|"Weak nucleophile<br/>Polar protic solvent"| D
C --> F["Inversion of configuration"]
D --> G["Racemic mixture"]
Worked Example: Identifying Mechanism Type from Structure
Question: Predict the mechanism of the reaction between 2-bromo-2-methylpropane and aqueous sodium hydroxide. Justify your answer.
Answer: 2-Bromo-2-methylpropane is (CH₃)₃CBr, a tertiary halogenoalkane. The carbon bearing the bromine is bonded to three other carbon atoms. The three methyl groups create steric hindrance, preventing the nucleophile from approaching for backside attack. Additionally, the tertiary carbocation formed in step 1 is stabilised by the inductive effect of three alkyl groups donating electron density towards the positive charge. Therefore, the reaction proceeds via an SN1 mechanism. Rate = k[(CH₃)₃CBr].
The product depends on the nucleophile used:
| Nucleophile | Conditions | Product | Functional group formed |
|---|---|---|---|
| OH⁻ (aq) | Heat under reflux with NaOH(aq) | Alcohol (R–OH) | Hydroxyl |
| CN⁻ | Heat under reflux with KCN in ethanol/water | Nitrile (R–CN) | Nitrile (extends C chain by 1) |
| NH₃ | Heat in sealed tube with excess NH₃ in ethanol | Primary amine (R–NH₂) | Amine |
Exam Tip: The reaction with CN⁻ is important in synthesis because it increases the carbon chain length by one. The nitrile can then be hydrolysed to a carboxylic acid or reduced to an amine, making it a versatile synthetic step.
Halogenoalkanes also undergo elimination reactions when heated with a strong base dissolved in ethanol (rather than aqueous conditions, which favour substitution). The base (e.g. KOH in ethanol) removes a hydrogen atom from a carbon adjacent to the halogen (the β-carbon), while the halogen leaves as X⁻, and a C=C double bond forms.
Curly arrow description: The base (OH⁻) abstracts a proton from the β-carbon. A curly arrow goes from the C–H bond to form part of the new C=C double bond. Simultaneously, a curly arrow goes from the C–X bond to the halogen, showing the departure of X⁻.
The preference for substitution vs elimination depends on the structure of the halogenoalkane and the reaction conditions:
| Condition | Favours |
|---|---|
| Aqueous NaOH, warm | Substitution |
| NaOH in ethanol, hot | Elimination |
| Primary halogenoalkane | Substitution |
| Tertiary halogenoalkane | Elimination |
The rate of hydrolysis (substitution with OH⁻) depends on the strength of the C–X bond (not the polarity):
| Bond | Bond enthalpy / kJ mol⁻¹ | Rate of hydrolysis |
|---|---|---|
| C–F | 484 | Slowest (most unreactive) |
| C–Cl | 338 | Intermediate |
| C–Br | 276 | Faster |
| C–I | 238 | Fastest (most reactive) |
Although the C–F bond is the most polar (which would suggest fastest nucleophilic attack), the bond is so strong that it is the hardest to break. Bond strength is the dominant factor, not polarity.
Exam Tip: To experimentally compare the rates of hydrolysis of different halogenoalkanes, add each separately to ethanol (as a co-solvent, since halogenoalkanes are immiscible with water), then add aqueous silver nitrate. The silver ions react with the halide ions produced to form precipitates: AgCl (white), AgBr (cream), AgI (yellow). The iodoalkane produces a precipitate first (fastest reaction).
Chlorofluorocarbons (CFCs) are halogenoalkanes containing chlorine and fluorine but no hydrogen (e.g. CCl₃F, CCl₂F₂). They were formerly used as refrigerants, aerosol propellants, and blowing agents for foams because they are non-toxic, non-flammable, and chemically inert under normal conditions.
However, CFCs persist in the atmosphere because they are so unreactive. In the stratosphere, UV radiation causes homolytic fission of the C–Cl bond:
CCl₃F → •CCl₂F + Cl•
The chlorine radicals catalyse the destruction of ozone (O₃) in a chain reaction:
Step 1: Cl• + O₃ → ClO• + O₂ Step 2: ClO• + O₃ → Cl• + 2O₂
graph TD
A["CFC in stratosphere"] -->|"UV light breaks C–Cl"| B["Cl• radical released"]
B -->|"Cl• + O₃ → ClO• + O₂"| C["ClO• formed"]
C -->|"ClO• + O₃ → Cl• + 2O₂"| B
B -->|"Cl• regenerated:<br/>catalytic cycle"| B
The Cl• radical is regenerated and can destroy thousands of ozone molecules before being removed. This is a catalytic cycle. The depletion of the ozone layer allows more harmful UV radiation to reach the Earth's surface, increasing the risk of skin cancer and cataracts.
CFCs have been replaced by HCFCs (hydrochlorofluorocarbons, which break down more readily in the troposphere) and HFCs (hydrofluorocarbons, which contain no chlorine and do not deplete ozone), as mandated by the Montreal Protocol.
Exam Tip: When asked why CFCs deplete the ozone layer, always mention (1) homolytic fission of C–Cl by UV in the stratosphere, (2) the radical chain mechanism, and (3) the fact that Cl• is regenerated, making it a catalyst. Note that C–F bonds are NOT broken by UV — they are too strong.