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.

Mechanism: alkene + halogen

• Approaching C=C repels the Br–Br electrons: an induced dipole, Brδ+–Brδ−.
• Arrow 1: from the C=C bond to Brδ+.
• Arrow 2: from the Br–Br bond to Brδ−, which leaves as Br⁻.
• A carbocation forms: a C⁺ with only three bonds.
• Arrow 3: from a lone pair on Br⁻ to C⁺. Product: CH₂BrCH₂Br.
• Some books show a three-membered ring, the bromonium ion, in place of C⁺.

Mechanism: alkene + hydrogen halide

• H–Br is already polar: Hδ+–Brδ−.
• Arrow 1: from the C=C bond to Hδ+.
• Arrow 2: from the H–Br bond to Br, which leaves as Br⁻.
• Arrow 3: from a lone pair on Br⁻ to the carbocation.
• Cyclohexene + HBr → bromocyclohexane. HCl and HI react the same way.

Mechanism: alkene + water

• Step 1: the C=C bond donates a pair to H⁺ from the acid. A carbocation forms.
• Step 2: H₂O donates a lone pair from O to C⁺, giving R–OH₂⁺.
• Step 3: an O–H bonding pair moves onto O. H⁺ is released; the alcohol forms.
• H⁺ is used in step 1 and returned in step 3: a catalyst.
• The nucleophile is neutral H₂O, not OH⁻.

Carbocation stability

• A carbocation has a C⁺ with three bonds and six outer electrons. It is planar.
• Primary: C⁺ bonded to one alkyl group. Secondary: two. Tertiary: three.
• Alkyl groups release electron density towards C⁺, spreading the charge.
• Stability: tertiary > secondary > primary.

Which product wins

• An unsymmetrical alkene can form two carbocations, so two products.
• H adds to the C=C carbon that already has more H atoms.
• That gives the more stable carbocation, so X or OH lands on the more substituted carbon.
• (CH₃)₂C=CHCH₃ + HBr → mainly (CH₃)₂CBrCH₂CH₃, via a tertiary carbocation.
• This pattern is Markovnikov's rule. Explain it with carbocation stability.

Mission 4: Benzene does react with bromine, given a catalyst. Why does a hydrogen get swapped rather than bromine adding?

Why benzene substitutes

• Benzene's six π electrons are delocalized round the ring: extra stability.
• The ring is electron-rich, so it attracts electrophiles, E⁺.
• Addition would destroy the delocalized ring. Substitution keeps it.
• Overall: C₆H₆ + E⁺ → C₆H₅E + H⁺.

The substitution mechanism

• Arrow 1: from the delocalized ring to E⁺. A C–E bond forms.
• The intermediate is positive: delocalization over five carbons only, a horseshoe with + inside.
• The attacked carbon now carries both H and E.
• Arrow 2: from that C–H bond into the ring. H⁺ leaves; the full ring is restored.
• Example: E⁺ = NO₂⁺ gives nitrobenzene. How E⁺ forms is not assessed.

Mission 5: Two bromoalkanes share the formula C₄H₉Br. Why do they swap Br for OH by different routes?

SN2: one step

• SN2: substitution, nucleophilic, bimolecular: two species in the rate-determining step.
• Typical of primary halogenoalkanes. One concerted step: bonds form and break together.
• The nucleophile attacks the δ+ carbon from the side opposite the halogen.
• Transition state [HO···C···Br]⁻: an energy maximum, not an intermediate.
• Rate = k[RX][OH⁻].

SN2 turns the molecule inside out

• As the nucleophile bonds, the other three groups flip, like an umbrella in a gale.
• This is inversion of configuration.
• One enantiomer of a chiral halogenoalkane gives one enantiomer of product.
• So SN2 is stereospecific.

SN1: two steps

• SN1: unimolecular: only RX in the rate-determining step. Rate = k[RX].
• Typical of tertiary halogenoalkanes.
• Step 1 (slow): C–X breaks heterolytically, giving a tertiary carbocation and X⁻.
• Step 2 (fast): the nucleophile donates a lone pair to the planar C⁺.
• (CH₃)₃CCl + OH⁻ → (CH₃)₃COH + Cl⁻.
• If that carbon is chiral, attack on either face gives a racemic mixture.

Why the routes differ

• Tertiary: three alkyl groups stabilize the carbocation and block attack from the back → SN1.
• Primary: the carbocation is too unstable; the back is open → SN2.
• Secondary: both mechanisms occur.
• The 1 and 2 are not step counts: SN1 has two steps, SN2 one.

Energy profiles

• SN2: one hump, the transition state. There is no intermediate.
• SN1: two humps, with the carbocation intermediate in the dip between them.
• The dip lies above the reactants and below both humps.
• The first hump, the slow step, is the higher one.

The leaving group

• In both mechanisms the C–X bond must break, so its strength matters.
• Data booklet: C–Cl 324, C–Br 285, C–I 228 kJ mol⁻¹.
• Rate, same R and conditions: RCl < RBr < RI.
• Bond strength decides, not polarity: C–Cl is the most polar, yet the slowest.

Mission 6: Aluminium chloride contains no hydrogen at all. How can it be an acid?

Lewis acids and bases

• Lewis acid: electron-pair acceptor. Lewis base: electron-pair donor.
• No H needed: in AlCl₃, Al has only six outer electrons, so it accepts a pair.
• H⁺ is a Lewis acid too: it accepts a lone pair from H₂O, giving H₃O⁺.
• Every nucleophile is a Lewis base. Every electrophile is a Lewis acid.
• A bonding pair can be donated: a C=C bond or benzene's ring can act as the base.

Coordination bonds

• A Lewis base donating a pair to a Lewis acid makes a coordination bond.
• Both shared electrons came from one atom.
• Once formed, it is an ordinary covalent bond: all three O–H bonds in H₃O⁺ are identical.
• In a Lewis formula the donor's lone pair becomes the bond. Show the overall charge.
• Some books draw it as an arrow, donor → acceptor.

Complex ions

• Transition element cations accept lone pairs from ligands, forming coordination bonds.
• Ligands can be neutral (H₂O, NH₃) or negative (Cl⁻, CN⁻).
• Write the complex in square brackets, with its charge outside.
• Charge = metal ion charge + sum of ligand charges.
• [CuCl₄]²⁻: (+2) + 4(−1) = −2. [Ni(NH₃)₆]²⁺: (+2) + 6(0) = +2.

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.
• AHLH adds to give the more stable carbocation. Benzene substitutes to keep its ring.
• AHLSN2: one step, inversion. SN1: via a carbocation. RI reacts fastest.
• AHLLewis acid = pair acceptor. Complex charge = metal charge + ligand charges.