Hydrocarbons • Topic 1 of 3

Alkanes

Hydrocarbons are compounds made only of carbon and hydrogen. They are classified into two broad families: aliphatic (open-chain or non-benzenoid rings) and aromatic (containing a benzene-type ring). Aliphatic hydrocarbons split further into saturated (alkanes, only C–C single bonds) and unsaturated (alkenes with C=C, alkynes with C≡C).

Alkanes: general formula and nomenclature

Alkanes have the general formula CnH2n+2 and are the least reactive hydrocarbons, earning the old name paraffins (‘little affinity’). The first four members are methane (CH4), ethane (C2H6), propane (C3H8) and butane (C4H10). In IUPAC naming, the longest continuous chain is the parent; substituents (alkyl groups) are named with locants chosen to give the lowest set.

Isomerism

From butane onward, alkanes show chain isomerism — the same molecular formula with different carbon skeletons. C4H10 has two isomers (n-butane and isobutane); C5H12 has three. Branching lowers the boiling point because the more spherical molecule has less surface contact and weaker van der Waals forces.

Methods of preparation

  • From unsaturated hydrocarbons (hydrogenation): CH2=CH2 + H2 ⟶[Ni] CH3–CH3 (Sabatier–Senderens).
  • From alkyl halides — Wurtz reaction: 2 R–X + 2 Na → R–R + 2 NaX, giving symmetrical alkanes with an even number of carbons.
  • Reduction of alkyl halides: R–X + Zn/H+ (or red P/HI) → R–H.
  • Decarboxylation (soda-lime): CH3COONa + NaOH ⟶[CaO,Δ] CH4 + Na2CO3 — loses one carbon.
  • Kolbe’s electrolysis: electrolysis of aqueous sodium carboxylate gives R–R at the anode plus CO2.

Conformations of ethane

Rotation about the C–C single bond gives an infinite set of spatial arrangements called conformations, shown clearly by Newman projections. The staggered form (H atoms 60° apart) is the most stable; the eclipsed form (H atoms aligned) is least stable, lying about 12.5 kJ mol-1 higher because of torsional strain. The barrier is small, so at room temperature ethane rotates freely.

Chemical reactions

  • Free-radical halogenation: CH4 + Cl2 ⟶[hν] CH3Cl + HCl, proceeding by initiation (Cl· formed), propagation and termination steps.
  • Combustion: CnH2n+2 + O2 → CO2 + H2O + heat (basis of fuels).
  • Controlled oxidation gives alcohols, aldehydes or acids depending on conditions.
  • Isomerisation: n-alkanes rearrange to branched ones over anhydrous AlCl3/HCl.
  • Aromatisation (reforming): n-hexane → benzene over Cr2O3/V2O5 at high T.
  • Pyrolysis (cracking): larger alkanes break into smaller alkanes and alkenes at high temperature.
Newman projections of ethane: staggered (stable) vs eclipsed (less stable)StaggeredEclipsedHHHHHHH atoms 60° apartHHHH atoms aligned (strain)
Solved Examples
1
Worked Example
How many structural (chain) isomers are possible for C5H12? Name each.
Solution
  1. Draw the straight chain of five carbons: this is n-pentane.
  2. Shorten to a four-carbon chain and attach one CH3 branch on C-2: 2-methylbutane (isopentane).
  3. Use a three-carbon chain with two CH3 branches on the central carbon: 2,2-dimethylpropane (neopentane).
  4. No further distinct skeletons are possible.

Answer: 3 isomers — n-pentane, 2-methylbutane and 2,2-dimethylpropane.

2
Worked Example
Predict the product of the Wurtz reaction between two molecules of CH3CH2Br with sodium in dry ether.
Solution
  1. Wurtz couples two alkyl halides: 2 R–X + 2 Na → R–R + 2 NaX.
  2. Here R = CH3CH2– (ethyl), so the two ethyl groups join.
  3. The product is CH3CH2–CH2CH3, i.e. n-butane.

Answer: n-Butane, C4H10 (plus 2 NaBr).

3
Worked Example
Write the steps of the free-radical chlorination of methane and identify the chain-initiation step.
Solution
  1. Initiation: Cl2 ⟶[hν] 2 Cl· — homolysis of the Cl–Cl bond gives chlorine radicals.
  2. Propagation: Cl· + CH4 → CH3· + HCl; then CH3· + Cl2 → CH3Cl + Cl·.
  3. Termination: two radicals combine, e.g. Cl· + Cl· → Cl2 or CH3· + CH3· → C2H6.

Answer: The initiation step is the photolysis Cl2 → 2 Cl·; the overall mechanism is a free-radical chain.

4
Worked Example
What product forms when sodium propanoate (CH3CH2COONa) is heated with soda-lime?
Solution
  1. Soda-lime (NaOH + CaO) decarboxylates the sodium salt of a carboxylic acid.
  2. One carbon is lost as carbonate: R–COONa + NaOH → R–H + Na2CO3.
  3. Here R = CH3CH2–, so R–H = CH3CH3.

Answer: Ethane (CH3CH3) is obtained, with Na2CO3.

5
Worked Example
Explain why the staggered conformation of ethane is more stable than the eclipsed conformation.
Solution
  1. In the eclipsed form the front and back C–H bonds line up, so the bonding electron pairs are closest.
  2. This raises the repulsion between the bonds, called torsional strain.
  3. In the staggered form the H atoms are 60° apart, giving maximum separation and minimum repulsion.
  4. The energy difference is about 12.5 kJ mol-1 in favour of staggered.

Answer: Staggered ethane is more stable because it has the least torsional strain (bonds maximally apart).

6
Worked Example
Arrange n-pentane, 2-methylbutane and 2,2-dimethylpropane in order of increasing boiling point and explain.
Solution
  1. All three have the same molecular formula C5H12 and the same molar mass.
  2. Boiling point depends on van der Waals contact, which falls as branching makes the molecule more spherical.
  3. n-Pentane is linear (most contact, highest b.p.); neopentane is nearly spherical (least contact, lowest b.p.).

Answer: 2,2-dimethylpropane < 2-methylbutane < n-pentane (increasing boiling point).

Key Points

  • Alkanes are saturated hydrocarbons CnH2n+2 (paraffins); they show chain isomerism from butane onward, and branching lowers the boiling point.
  • Preparations: hydrogenation of alkenes/alkynes, Wurtz coupling (2 R-X + 2 Na), reduction of alkyl halides, soda-lime decarboxylation (loses one C) and Kolbe electrolysis.
  • Ethane prefers the staggered conformation (H atoms 60° apart); the eclipsed form is ~12.5 kJ mol-1 higher due to torsional strain.
  • Halogenation of alkanes is a free-radical chain: initiation (X2 → 2X·), propagation and termination.
  • Other reactions: combustion (fuels), controlled oxidation, isomerisation (AlCl3), aromatisation (n-hexane → benzene) and pyrolysis/cracking.
Quick Check — 4 MCQs
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Q1.The general formula of an alkane is:
Explanation: Saturated open-chain hydrocarbons have the formula CnH2n+2.
Q2.The Wurtz reaction of CH3Br with sodium gives mainly:
Explanation: Two methyl groups couple: 2 CH3Br + 2 Na → CH3CH3 (ethane).
Q3.Which conformation of ethane is the most stable?
Explanation: The staggered form has the H atoms 60° apart, giving minimum torsional strain.
Q4.Heating CH3COONa with soda-lime produces:
Explanation: Decarboxylation removes the COONa as carbonate, leaving CH4 (one carbon is lost).
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