Mission 1: Why does bromine in hexane stay orange in the dark but lose its colour in sunlight?

Radicals

• A radical is a species with an unpaired electron.
• A dot marks the unpaired electron: Cl•, •CH₃.
• The dot is not a charge. These radicals are neutral.
• Odd number of outer electrons → one is unpaired. •CH₃: 4 + 3 = 7.
• Radicals are highly reactive: the unpaired electron pairs up by forming a new bond.

Homolytic fission and fish-hook arrows

• Homolytic fission: a bond breaks with one electron to each atom.
• Two radicals form, and no ions.
• A fish-hook arrow (one barb) moves one electron.
• A curly arrow with two barbs moves an electron pair.
• For Cl₂: two fish-hooks, each from the Cl–Cl bond to a different Cl atom.

Initiation

• Initiation: UV light or heat splits the halogen: Br₂ → 2Br•.
• The halogen–halogen bond breaks because it is the weakest bond present.
• Data booklet: Br–Br 193 kJ mol⁻¹, C–C 346 kJ mol⁻¹. C–H is stronger still.
• So initiation makes no H• and no alkyl radical.

Propagation: the chain

• Propagation: radical + molecule → new molecule + new radical.
• Br• removes an H atom from cyclohexane: C₆H₁₂ + Br• → •C₆H₁₁ + HBr.
• Fish-hooks: one electron from Br• and one from the C–H bond form the H–Br bond.
• The other C–H electron stays on carbon, giving the cyclohexyl radical, •C₆H₁₁.
• Then •C₆H₁₁ + Br₂ → C₆H₁₁Br + Br•. Br• is regenerated, so the chain repeats.
• Br• never swaps directly for H, and no H• forms.

Termination and the product mixture

• Termination: two radicals combine. Radicals are removed; none are made.
• Br• + Br• → Br₂ · •C₆H₁₁ + Br• → C₆H₁₁Br · •C₆H₁₁ + •C₆H₁₁ → C₁₂H₂₂.
• Radicals are scarce, so termination is rare: most product forms in propagation.
• Br• also attacks C₆H₁₁Br, so further substitution gives C₆H₁₀Br₂ and beyond.
• The result is a mixture of products.

Why alkanes are so unreactive

• C–C and C–H bonds are strong: breaking them needs a large energy input.
• C and H have similar electronegativities: the bonds are essentially non-polar.
• There are no significant δ+ or δ− sites, so polar reagents are not attracted.
• Only very reactive species, such as radicals, attack alkanes.

Mission 2: How does warm sodium hydroxide turn a bromoalkane into an alcohol?

Nucleophiles

• A nucleophile donates both electrons of a new bond to its partner.
• It needs an available electron pair, usually a lone pair: OH⁻, CN⁻, Cl⁻, Br⁻.
• No charge needed: H₂O and NH₃ are neutral nucleophiles, with lone pairs on O and N.
• Its partner, the electrophile, accepts the pair.

Heterolytic fission and curly arrows

• Heterolytic fission: both bonding electrons go to one fragment, so ions form.
• H–Br → H⁺ + Br⁻. The pair goes to Br, the more electronegative atom.
• The fragment that takes the pair becomes negative.
• A curly arrow (two barbs) moves an electron pair.
• Start it at the pair's source, a lone pair or a bond. End it where the pair goes.

Nucleophilic substitution

• In a halogenoalkane the C–X bond is polar: Cδ+–Xδ−.
• The nucleophile donates a lone pair to the δ+ carbon.
• The C–X pair moves onto X, which leaves as X⁻: the leaving group.
• CH₃Cl + OH⁻ → CH₃OH + Cl⁻.
• Two curly arrows: lone pair on O → C; C–Cl bond → Cl.

Mission 3: Why does an alkene take bromine's colour away in the dark, when an alkane needs light?

Electrophiles

• An electrophile accepts both electrons of a new bond from a nucleophile.
• It is electron-deficient: H⁺, NO₂⁺, or the δ+ atom of a polar molecule.
• No charge needed: HBr (through Hδ+) and Br₂ act as electrophiles.
• Pair donor = nucleophile. Pair acceptor = electrophile.

Why alkenes attract electrophiles

• The C=C double bond is a region of high electron density.
• It attracts electrophiles and donates an electron pair to them.
• C=C is stronger than C–C. Alkenes react because of electron density, not weakness.
• The reagent adds across C=C. The double bond becomes single; nothing is lost.
• This is electrophilic addition.

Three addition reactions

• Halogen: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br. Fast in the dark, at room temperature.
• Bromine turns orange-brown → colourless: the test for C=C.
• Hydrogen halide: C₆H₁₀ + HBr → C₆H₁₁Br (cyclohexene → bromocyclohexane).
• Water, with an acid catalyst (e.g. H₃PO₄): C₆H₁₀ + H₂O → C₆H₁₁OH (cyclohexanol).
• With HX or H₂O, H goes to one carbon; X or OH to the other.

The big ideas

• Radicals have an unpaired electron. UV light or heat splits a halogen homolytically: fish-hook arrows.
• Radical substitution: initiation, propagation (radical regenerated), termination. The product is a mixture.
• Nucleophiles donate an electron pair; electrophiles accept one. A curly arrow runs from the pair to where it goes.
• Halogenoalkanes undergo nucleophilic substitution. Alkenes add halogens, hydrogen halides and water.