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Antiperiplanar geometry places two specified bonds at a torsion near $180^{\circ }$. It often enables favorable orbital alignment in E2 elimination and supports large vicinal proton couplings in an appropriate calibrated Karplus relation.

Both reactions are E2 eliminations requiring an axial chloride and an axial beta hydrogen in an antiperiplanar conformation relative to the C–Cl bond. The reacting C–H bond supplies the alkene $\pi$ bond as the C–Cl bond breaks; hydroxide accepts the proton.

For the trans-1,2 substrate, the low-energy chair has methyl and chlorine both equatorial. ring inversion is needed to place chlorine axial, but this also places methyl axial. The population of the reactive chair is consequently small, explaining the slow reaction. At the methyl-bearing neighboring carbon the axial position is occupied by methyl, not hydrogen. Only the other beta carbon has the needed anti axial H, so elimination gives 3-methylcyclohexene rather than the more substituted alkene.

For the cis substrate, a favorable chair has methyl equatorial and chlorine axial. It can react without paying the axial-methyl penalty, explaining the faster rate. Both adjacent carbons now supply an anti axial hydrogen. Removing the H on the methyl-bearing carbon gives 1-methylcyclohexene, the major product; removing the other gives 3-methylcyclohexene, the minor product. The former has the more substituted double bond.

Thus alkene stability selects between accessible pathways in the cis case, but it cannot create the missing antiperiplanar conformation H in the trans case.

Label the carbon next to the carbonyl $C_{\alpha }$ and the OH-bearing carbon $C_{\beta }$. The three limiting routes are E1, E1cb and E2 elimination. They differ in whether C${}_{\beta }$–O cleavage, C${}_{\alpha }$–H cleavage, or both occur in the first elementary event. In the following original transition-state structures, dotted lines denote changing bonds, $\delta$ marks developing partial charge, and the bracketed structures carry $‡$. Unmarked skeletal bonds remain present. The dotted line parallel to C${}_{\alpha }$–C${}_{\beta }$ denotes the developing pi bond, in addition to its existing sigma bond.

For E1, C${}_{\beta }$–OH heterolysis occurs first. The electron pair in that bond goes to O, producing $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2{\mathrm{C}}^{+}(\mathrm{H})\mathrm{C}{\mathrm{H}}_3+\mathrm{O}{\mathrm{H}}^{-}$. Its first transition state has a partly broken C${}_{\beta }$–O bond, developing positive charge on C${}_{\beta }$ and negative charge on the leaving O. Next, water uses a lone pair to remove the alpha proton; the C${}_{\alpha }$–H electron pair forms the C${}_{\alpha }$=C${}_{\beta }$ pi bond. This second transition state has partly broken C${}_{\alpha }$–H, partly formed water-O–H, and a partly formed C${}_{\alpha }$–C${}_{\beta }$ pi bond. Positive charge is transferred from the organic cation toward the incoming water/hydronium unit. hydroxide from the first step is a spectator in the second.

For E1cb, water first removes the alpha proton to form a resonance-stabilized enolate plus hydronium:The first transition state has partly broken C${}_{\alpha }$–H and partly formed O${}_{\mathrm{water}}$–H, with developing negative charge delocalized over the alpha-carbon/carbonyl-oxygen system and positive charge on the proton-accepting water unit. The enolate then expels hydroxide: use the carbon-centred contributor to send its alpha-carbon lone pair into C${}_{\alpha }$–C${}_{\beta }$, while the C${}_{\beta }$–O pair goes to oxygen. This second transition state has a partly formed alkene pi bond and partly broken C${}_{\beta }$–O, with negative charge moving from the enolate framework onto the departing hydroxide. hydronium is a spectator in this second step.

For E2, water removes the alpha proton as hydroxide leaves in the same elementary step. Three electron-pair movements are simultaneous: water lone pair to H${}_{\alpha }$, C${}_{\alpha }$–H to the C${}_{\alpha }$–C${}_{\beta }$ pi bond, and C${}_{\beta }$–O to O. Its single transition state therefore contains all three changing bonds: C${}_{\alpha }$–H, water-O–H and C${}_{\beta }$–O, together with the developing pi bond. The departing O becomes negative, the proton-accepting water unit becomes positive, and partial charge at carbon depends on the synchrony of proton loss and leaving-group departure. The reacting conformation should align C${}_{\alpha }$–H and C${}_{\beta }$–O approximately in an antiperiplanar conformation for efficient elimination.

These are formal limiting pathways for the displayed net process, not three claims of equal reaction rates in neutral water. In particular, unassisted loss of OH${}^{-}$ to a secondary carbocation is strongly disfavoured, and water is a weak base for making an enolate. The enolate pathway benefits from resonance stabilization and the final product is a conjugated enone, but actual rate and dominant pathway depend on chemical conditions. The displayed hydroxide and hydronium also rapidly neutralize each other; retaining them separately here tracks which bond cleavage supplied each species. No biological catalysis is needed to describe these chemical mechanisms.

Number the cyclopentanone ring $p_1$ at its carbonyl, then $p_2$ at the methyl-bearing carbon, $p_3$ at the mesylate-bearing carbon, $p_4$ at the carbon joined to cyclohexane $h_2$, and $p_5$ at the remaining methylene. B has a $p_2=p_3$ enone and the $h_1=h_6$ enamine. triethylamine removes the hydrogen at $p_2$ while the $p_3$–O bond breaks to release mesylate. An E2 elimination, with approximately antiperiplanar geometry in the puckered ring, gives the conjugated cyclopentenone. Elimination towards $p_4$ instead would give a nonconjugated alkene.

The enamine carbon $h_6$ now attacks enone beta carbon $p_3$ intramolecularly. nitrogen donates into $h_1$ to give the iminium ion, while the enone electrons move onto the ketone oxygen. protonation of the ketone enolate gives C. The second new bond is $h_6-p_3$; the first was $h_2-p_4$. These two connections account for the bridged skeleton without a concerted cycloaddition or an aldol step. The formula of B is ${\mathrm{C}}_{16}{\mathrm{H}}_{23}\mathrm{NO}$: relative to A, elimination removes $\mathrm{C}{\mathrm{H}}_4{\mathrm{O}}_3\mathrm{S}$ overall, namely methanesulfonic acid after proton transfers.

The alcohol oxygen attacks sulfur of methanesulfonic anhydride. Cleavage of the anhydride S–O bond releases methanesulfonate, and the amine base removes the O–H proton, giving the corresponding mesylate. This activation preserves the substrate C–O bond while replacing the poor hydroxide leaving group by $\mathrm{OS}{\mathrm{O}}_2\mathrm{Me}$.

The base then removes a hydrogen from the adjacent methylene on the side toward the vinyl chloride. In a favorable antiperiplanar arrangement, C–H electrons form the new C=C bond while the C–OMs bond breaks. An E2 elimination creates the alkene between the former alcohol carbon and that methylene. It links the two neighboring unsaturated regions into the extended conjugated triene segment, while the pre-existing carbon skeleton and vinyl chloride are retained.The left-hand polyenoate portion is unchanged. The base is diisopropylethylamine; it accepts acid during mesylation and promotes elimination. The product's E/Z mixture concerns the newly formed alkene, and no single geometry needs to be assigned.

Disconnect the new bond between the two six-carbon units. One portion of bromocyclohexane supplies a grignard reagent; another supplies cyclohexene oxide:A bulky strong base can favour E2 elimination in the alkene-forming step; a peracid such as meta-chloroperoxybenzoic acid provides epoxidation. None of its carbons enters the target.

Treat cyclohexene oxide with cyclohexylmagnesium bromide, then perform acidic aqueous work-up. Backside attack on one epoxide carbon breaks its C–O bond and installs cyclohexyl opposite the retained oxygen. Protonation of the oxygen gives trans-2-cyclohexylcyclohexan-1-ol, the required relative configuration. The two six-carbon reactant portions supply all 12 target carbons. An achiral synthesis produces a racemic mixture trans pair; obtaining only the particular enantiomer represented by the drawing would require resolution or an enantioselective step. The useful stable chair can place the two large trans-1,2 substituents approximately diequatorial.

The first transformation is two successive dehydrohalogenations; the second opens the lactone by two methyl additions and then closes a six-membered cyclic ether. Both schemes must be accounted for within this part.

For the dichloride, write $R=\mathrm{cyclohexyl}$. In the product-forming path, tert-butoxide removes a hydrogen from the methyl group of $\mathrm{R}-\mathrm{CC}{\mathrm{l}}_2-\mathrm{C}{\mathrm{H}}_3$. Draw arrows from the base lone pair to H, from C–H into the C–C bond, and from C–Cl to Cl. This E2 elimination gives $\mathrm{R}-\mathrm{C}(\mathrm{Cl})=\mathrm{C}{\mathrm{H}}_2$. A second base-induced elimination removes a vinyl hydrogen and the remaining chloride, forming $\mathrm{R}-\mathrm{C}\equiv \mathrm{CH}$. The second elimination is more demanding because a vinylic bond is involved, consistent with the strong-base, heated conditions. This route explains the displayed product without asserting that no other elimination regioisomer can ever form.

For the lactone, a methyl equivalent from methylmagnesium iodide attacks the ester carbonyl carbon, with its pi pair moving to oxygen. Collapse of the tetrahedral intermediate reforms C=O and cleaves the acyl–oxygen bond, opening the ring to the magnesium alkoxide corresponding to $\mathrm{HO}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}(=\mathrm{O})\mathrm{C}{\mathrm{H}}_3$. The newly formed ketone is more susceptible to a second methyl addition; its pi pair again moves to oxygen. After water workup, the product is the diol $\mathrm{HO}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}(\mathrm{OH})(\mathrm{C}{\mathrm{H}}_3{)}_2$.

Under acid and heat, protonate its tertiary OH, then move the C–O pair onto oxygen as water leaves. The resulting $\mathrm{HO}(\mathrm{C}{\mathrm{H}}_2{)}_4{\mathrm{C}}^{+}(\mathrm{C}{\mathrm{H}}_3{)}_2$ is trapped intramolecularly by the remaining primary-alcohol oxygen. Its lone pair forms the closing O–C bond, producing a six-membered cyclic oxonium ion. Deprotonation yields 2,2-dimethyloxane, also called 2,2-dimethyltetrahydropyran. Ring closure retains four chain methylenes; it does not form a five-membered ring. The two methyl groups both derive from the two equivalents of Grignard reagent.

Elimination from A givesRemoving a beta hydrogen from either equivalent methyl group while losing chloride creates the same conjugated enyne. The strongly basic ethoxide, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2{\mathrm{O}}^{-}$, promotes this E2 elimination from the tertiary substrate more strongly than neutral ethanol does. Ethanol instead favors the intended solvolytic capture, although some elimination can still compete.

The formula drawn and used in the rate table is ethoxide. The name “ethanoate” would mean acetate, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{O}}_2^{-}$, a different and resonance-stabilized anion; it should not be silently treated as the same base.

In Reaction 2 the primary benzylic carbon favors rapid substitution. The adjacent aromatic ipso carbon has no beta hydrogen for ordinary side-chain E2 elimination, so the analogous alkene-forming path is unavailable. This does not assert that all benzyl compounds are immune to every side reaction.