Q3(d). SOCl₂/dioxane, Li/NH₃ cyclisation, sodium naphthalenide

Introduction. (i) A chlorosulfite decomposes in a nucleophilic solvent with double inversion. (ii) A dissolving-metal reduction of an enone creates a β-carbanion, which is alkylated intramolecularly by the neighbouring CH₂OTs. (iii) Sodium naphthalenide is a single-electron donor; the 5-hexenyl group reveals the radical by cyclising.

(i) (S)-1-Phenylethyl chlorosulfite in dioxane

1. As drawn the substrate is the (S) chlorosulfite (priorities OSOCl > Ph > CH₃ > H, H at the back).
2. A dioxane oxygen attacks C-1 from the back and displaces OSOCl⁻ (which breaks into SO₂ + Cl⁻): first inversion, an alkyloxonium ion.
3. Cl⁻ attacks C-1 from the back of the C–O⁺ bond and dioxane leaves: second inversion.
4. Two inversions give overall retention: (S)-1-chloro-1-phenylethane (the benzylic cation, if formed, would racemise the product; the solvent keeps it covered).

Scheme: Double inversion = retention (SNi with solvent participation).

(ii) Li/NH₃, THF on the 4a-(tosyloxymethyl)octalone

1. Li gives one electron to the enone (radical anion) and then a second: the enolate–β-carbanion (dianion). No alcohol is present, so the carbanion at C-8a (the β-carbon) is not protonated.
2. The C-8a carbanion attacks the CH₂ carrying OTs on the adjacent C-4a: intramolecular SN2 (3-exo-tet) closes a cyclopropane and expels TsO⁻.
3. H₃O⁺ protonates the lithium enolate at carbon: the ketone.
4. Product: tricyclo[4.4.1.0^(1,6)]undecan-3-one, a [4.4.1]propellanone (C₁₁H₁₆O): C-4a and C-8a are shared by three rings.

Scheme: Reductive cyclisation: the β-carbon becomes a cyclopropane carbon.

(iii) Sodium naphthalenide and 6-bromohex-1-ene

1. Na transfers an electron to naphthalene (in THF): green sodium naphthalenide, C₁₀H₈˙⁻ Na⁺.
2. Single electron transfer to the bromide gives RBr˙⁻, which loses Br⁻: the hex-5-enyl radical.
3. Competition: the radical is either reduced at once by a second naphthalenide (carbanion, protonated on work-up) to hex-1-ene, or it first cyclises 5-exo-trig to the cyclopentylmethyl radical, which is then reduced and protonated to methylcyclopentane. Some radicals couple with naphthalenide to give alkyl-dihydronaphthalenes.
4. The cyclised product is the proof that a radical is formed: the 5-hexenyl group acts as a radical clock.

Scheme: Uncyclised radical: reduced to hex-1-ene; cyclised radical: methylcyclopentane.

Conclusion. (i) (S)-1-chloro-1-phenylethane, retention by double inversion through a dioxane oxonium ion. (ii) Reductive cyclisation to tricyclo[4.4.1.0^(1,6)]undecan-3-one. (iii) SET from sodium naphthalenide: hex-1-ene and methylcyclopentane (plus alkylated dihydronaphthalenes); the cyclised product reveals the radical.

Application. SOCl₂ in dioxane (retention) or pyridine (inversion) lets either enantiomer of a chloride be made from one alcohol; the 5-hexenyl radical clock is used to test whether Grignard formation and other reactions pass through radicals.