Alkanes, cracking and combustion

Alkanes are saturated hydrocarbons (only single bonds), C_nH_2n+2. C–H and C–C bonds are strong and almost non-polar, so alkanes are unreactive towards most reagents — but they burn.

Fractional distillation separates crude oil by boiling point: bigger molecules have more electrons, stronger van der Waals forces and a higher boiling point, so they condense lower down the column.

Cracking converts the surplus long-chain fractions into shorter, more valuable ones:

• Thermal cracking — high temperature (up to 1000 K) and high pressure (up to 7000 kPa); produces a high proportion of alkenes (for polymers) by homolytic fission.
• Catalytic cracking — a zeolite catalyst, slight pressure, about 720 K; produces branched alkanes, cycloalkanes and aromatics for motor fuels. The catalyst cuts the cost by allowing a lower temperature.

Combustion: complete combustion gives CO₂ and H₂O. Incomplete combustion gives toxic carbon monoxide (which binds far more strongly than O₂ to the iron in haemoglobin, and is not readily released) and carbon particulates. High engine temperatures make nitrogen oxides (NO_x), and sulfur impurities burn to SO₂ — both cause acid rain. SO₂ is removed from flue gases with CaO or CaCO₃.

Your turn

Free-radical substitution

In UV light, chlorine substitutes for hydrogen in an alkane. The mechanism has three stages — and you must be able to label them.

• Initiation. UV light causes homolytic fission of the Cl–Cl bond (each atom keeps one electron): Cl₂ → 2Cl•. The Cl–Cl bond is weaker than C–H, so it is the one that breaks.
• Propagation. Two steps that form a chain — each uses up one radical and makes another:
  • CH₄ + Cl• → •CH₃ + HCl
  • •CH₃ + Cl₂ → CH₃Cl + Cl•
• Termination. Any two radicals combine, removing them from the chain: Cl• + Cl• → Cl₂, •CH₃ + Cl• → CH₃Cl, •CH₃ + •CH₃ → C₂H₆.

Why this is a poor way to make a pure product: further substitution gives CH₂Cl₂, CHCl₃ and CCl₄, and termination produces alkane by-products such as ethane. You get a mixture, so the yield of any one product is low.

The ozone connection: CFCs are so stable that they reach the stratosphere, where UV breaks a C–Cl bond (the weakest bond present) to give Cl•. These radicals catalyse ozone destruction: Cl• + O₃ → ClO• + O₂, then ClO• + O₃ → 2O₂ + Cl•. Overall 2O₃ → 3O₂, and the chlorine radical is regenerated — so one radical destroys thousands of ozone molecules.

Sort each step of the chlorination of methane

Tap a step, then tap which stage of the free-radical mechanism it belongs to.

Your turn

Alkenes and electrophilic addition

An alkene has a C=C double bond: one σ bond and one π bond formed by the sideways overlap of p orbitals above and below the plane. The π bond is a region of high electron density, so alkenes are attacked by electrophiles — electron-pair acceptors.

The mechanism (electrophilic addition), in three moves:

• The π electrons attack the electrophile (an arrow from the C=C to the δ+ atom).
• The bond in the electrophile breaks heterolytically (an arrow from that bond to the atom that leaves), forming a carbocation intermediate.
• The negative ion attacks the carbocation (an arrow from the lone pair to the C⁺).

With a non-polar electrophile like Br₂, the alkene’s π electrons induce a dipole in the Br–Br bond as it approaches — which is why bromine water is decolourised instantly and is the test for unsaturation.

Markownikoff’s rule, properly explained. When HBr adds to propene, the major product is 2-bromopropane. The reason is carbocation stability: alkyl groups are electron-releasing, so they spread out the positive charge. Stability runs tertiary > secondary > primary, so the H⁺ adds to give the more stable (here secondary) carbocation, and the bromide attacks that.

Quick check

Your turn

Your turn

Nucleophilic substitution and elimination

The C–X bond is polar (X is more electronegative), so the carbon is δ+ and is attacked by nucleophiles — electron-pair donors with a lone pair (OH⁻, CN⁻, NH₃).

The same halogenoalkane with the same reagent but a different solvent gives a different reaction:

Two mechanisms for substitution:

• S_N2 — primary halogenoalkanes. The nucleophile attacks from the opposite side to the leaving group in a single step, through a transition state. Little steric hindrance.
• S_N1 — tertiary halogenoalkanes. The C–X bond breaks first, forming a relatively stable tertiary carbocation, which is then attacked. The bulky alkyl groups also block a backside attack.

Rate of hydrolysis is governed by BOND ENTHALPY, not polarity. C–I is the weakest bond, so iodoalkanes hydrolyse fastest; C–F is the strongest, so fluoroalkanes are the slowest — even though the C–F bond is by far the most polar. Students who argue from polarity get this exactly backwards.

Quick check

Quick check

Match the reagent and conditions to the reaction type

Tap a reagent on the left, then the type of reaction on the right.

Quick check

Drawing the arrows and choosing the solvent

• Curly arrows start at an electron source — a lone pair or the middle of a bond — and point to where the pair goes. Starting an arrow at a positive charge, or at an atom rather than its electrons, loses the mark.
• Aqueous NaOH substitutes; ethanolic KOH eliminates. Same reagent, different solvent, completely different product.
• Hydrolysis rate follows BOND ENTHALPY, not polarity. C–I is weakest, so iodoalkanes react fastest. The most polar bond (C–F) is the slowest.
• Show the lone pair on the nucleophile (OH⁻, CN⁻, :NH₃) and the δ+ and δ− on the C–X bond. Both are usually explicitly worth marks.

Radical mechanisms use HALF arrows. A fish-hook (single-headed) arrow moves one electron; a normal curly arrow moves a pair. Using a full arrow in a free-radical mechanism is a guaranteed mark lost.

Quick check

Your turn

The big ideas to know

That is AQA 3.3.2, 3.3.3 and 3.3.4. Press Finish.