JAM Aliphatic Hydrocarbons — Preparation and Reactions
Addition reactions to alkenes are covered thoroughly in the companion article on core reaction mechanisms, linked below — this one deliberately starts somewhere else: how alkanes, alkenes and alkynes are actually made, the mechanism and selectivity of free-radical halogenation, and the reactions unique to the carbon-carbon triple bond, which get far less attention than they deserve on the exam.
Preparing alkanes
| Method | Reagents / conditions | Note |
|---|---|---|
| Wurtz reaction | 2 R–X + 2Na → R–R + 2NaX (dry ether) | Best for identical alkyl halides; a "crossed" Wurtz between two different halides gives a statistical mixture of three coupling products and is a poor synthetic choice |
| Decarboxylation | Sodium salt of a carboxylic acid + soda lime (NaOH/CaO), heat | RCOONa → RH + Na₂CO₃; removes one carbon as CO₂ equivalent |
| Kolbe electrolysis | Electrolysis of a concentrated aqueous carboxylate salt | 2 RCOO⁻ → R–R (at the anode, via radical coupling) + 2CO₂ + H₂ (at the cathode) |
| Catalytic hydrogenation | Alkene or alkyne + H₂ / Ni, Pd or Pt | Simplest route when the corresponding unsaturated hydrocarbon is available |
Worked example 1 — Wurtz reaction stoichiometry and yield. 21.8 g of ethyl bromide (C₂H₅Br, M = 108.97 g/mol) is treated with sodium in dry ether. The isolated butane (C₄H₁₀, M = 58.12 g/mol) weighs 4.65 g. Find the percentage yield.
Moles of ethyl bromide = 21.8 ÷ 108.97 = 0.200 mol.
The equation is 2 RBr : 1 R–R, so theoretical moles of butane = 0.200 ÷ 2 =
0.100 mol.
Theoretical mass of butane = 0.100 × 58.12 = 5.81 g.
% yield = (4.65 ÷ 5.81) × 100 = 80.0%.
Free-radical halogenation of alkanes — mechanism and selectivity
Propagation: X• + R–H → R• + HX then R• + X₂ → R–X + X•
Termination: any two radicals combine (X• + X•, R• + X•, or R• + R•)
The rate-determining, selectivity-deciding step is hydrogen abstraction — how easily a given C–H bond breaks homolytically to give a stabilised radical. Because radical stability follows 3° > 2° > 1°, a tertiary hydrogen is abstracted preferentially over a primary one. Bromination is far more selective than chlorination for tertiary hydrogens: hydrogen abstraction by Br• is a distinctly endothermic, later, more product-like transition state (Hammond postulate), so it responds much more strongly to how stable the resulting radical is. Abstraction by Cl• is far less endothermic, its transition state resembles the starting materials more than the radical, and chlorination is consequently much less selective — it substitutes primary, secondary and tertiary positions at closer to their statistical ratio.
Preparing alkenes and alkynes
Alkenes come mainly from acid-catalysed dehydration of alcohols (Zaitsev product favoured) or dehydrohalogenation of a haloalkane with alcoholic KOH (elimination — see the companion article on haloalkane mechanisms for the full E1/E2 treatment). Alkynes have two standard routes:
- From calcium carbide — CaC₂ + 2H₂O → C₂H₂ (ethyne) + Ca(OH)₂, the classic industrial/laboratory source of the parent alkyne.
- Double dehydrohalogenation of a vicinal or geminal dihalide — two successive eliminations with excess alcoholic then molten/strong KOH (or NaNH₂) convert a dihaloalkane into an alkyne via the intermediate vinyl halide.
Reactions specific to the carbon-carbon triple bond
Terminal alkyne acidity (the C–H bond on sp-hybridised carbon) is covered in the hybridisation article; here is what that acidity, and the extra π bond, actually let you do:
| Reagent | Outcome on an alkyne |
|---|---|
| NaNH₂ (strong base) | Deprotonates a terminal alkyne's C–H to give a nucleophilic acetylide ion, useful for alkylation to build a longer chain |
| H₂O / Hg²⁺ / H₂SO₄ | Markovnikov hydration — water adds to the more substituted carbon, giving an enol that tautomerises instantly to a ketone (acetylene itself is the one exception, giving acetaldehyde, not a ketone) |
| BH₃ then H₂O₂/OH⁻ | Anti-Markovnikov hydration — gives the enol at the less substituted carbon, tautomerising to an aldehyde for a terminal alkyne |
| H₂ / Lindlar's catalyst (Pd–CaCO₃, poisoned with a trace of quinoline) | Partial reduction, stopping at the alkene — delivers syn addition, giving the cis alkene |
| Na / liquid NH₃ | Dissolving-metal reduction by a radical-anion mechanism, giving the trans alkene — the opposite stereochemical outcome to Lindlar's catalyst |
Worked example 2 — Markovnikov hydration of propyne, with tautomerisation. Propyne, CH₃–C≡CH, is treated with H₂O/Hg²⁺/H₂SO₄. Predict the product.
Water adds Markovnikov-fashion: the –OH lands on the more substituted carbon of the triple bond. This first gives the enol CH₃–C(OH)=CH₂, which tautomerises immediately (keto-enol equilibrium lies almost entirely towards the keto form for a simple enol) to propan-2-one (acetone), CH₃COCH₃.
Common mistakes
- Planning a Wurtz synthesis between two different alkyl halides to get one specific unsymmetrical alkane — the statistical mixture of products makes this a poor choice; it only works cleanly for a symmetrical product from identical starting halides.
- Forgetting the tautomerisation step after alkyne hydration and writing an enol as the final answer instead of the ketone or aldehyde it converts to.
- Mixing up which catalyst gives which alkene geometry — Lindlar gives cis (syn addition of H₂ to one face), Na/liquid NH₃ gives trans.
- Assuming Markovnikov and anti-Markovnikov hydration of an alkyne give the same functional group. Markovnikov hydration of a terminal alkyne gives a ketone; hydroboration-oxidation of the same alkyne gives an aldehyde instead.
- Applying the peroxide (anti-Markovnikov) effect to HCl or HI addition. It is specific to HBr; the chain mechanism is not thermodynamically favourable for the other two hydrogen halides.
Exam relevance
| Question style | What to check first |
|---|---|
| Propose a synthesis of a symmetrical alkane | Wurtz reaction, with identical alkyl halides |
| Predict the major product of radical halogenation | Whether the reagent is Cl₂ (low selectivity) or Br₂ (high selectivity for the most stable radical) |
| Predict the product of alkyne hydration | Markovnikov vs anti-Markovnikov reagent, then remember the tautomerisation step |
| Predict alkene geometry from alkyne reduction | Lindlar (cis) vs Na/liquid NH₃ (trans) |
| Numerical (NAT) | Stoichiometry and percentage yield for a stated preparation route |
Confirm molar masses before working a preparation or yield problem. The molar mass tool accepts any formula, which is the fastest check on a stoichiometry calculation involving alkyl halides, alkenes or alkynes.
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