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protonation of the alcohol converts its OH group into a water leaving group. Departure of water initially gives the secondary carbocation $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}^{+}-\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3{)}_2$. A 1,2-hydride shift from the neighboring methine group transfers its C–H bonding pair to the electron-deficient carbon:The resulting tertiary carbocation is the tert-amyl electrophile. An aromatic $\pi$ bond of the dimethoxybenzene attacks that carbon, producing a Wheland intermediate. Removal of the proton from the attacked ring carbon restores aromaticity and regenerates acid. Repeating this Friedel-Crafts alkylation at the symmetry-related second site gives the disubstituted product.

The hydride migration explains the new ethyl group and the absence of a benzylic methine group proton. Simply attaching the original alcohol skeleton at its OH-bearing carbon would not explain the twelve-proton NMR singlet.

For the three elementary transformations just proposed, a qualitative free-energy profile has three transition states and two intermediates. The sugar substituents and indole C3 substituent stay unchanged, so the structural shorthand below suppresses them while retaining the reacting atoms.

The first transition state has a partially broken anomeric C1–O(leaving group) bond and developing oxocarbenium character, $[{\mathrm{O}}_{\mathrm{ring}}-{\mathrm{C}}_1^{\delta +}\cdots {\mathrm{O}}_{\mathrm{LG}}{]}^{‡}$. Its following minimum is the glycosyloxocarbenium ion plus counterion. The second transition state has a forming C1–C2 bond, $[{\mathrm{O}}_{\mathrm{ring}}-{\mathrm{C}}_1\cdots {\mathrm{C}}_2(\mathrm{H}{)}_{\mathrm{ind}}{]}^{‡}$, and developing positive charge in the indole. Its following minimum is the covalently joined arenium ion with an intact C2–H bond. The third transition state has simultaneous C2–H cleavage, B–H formation and restoration of the indole pi bond, $[\mathrm{B}\cdots \mathrm{H}\cdots {\mathrm{C}}_2-{\mathrm{C}}_3{]}^{‡}$. It leads to the aromatic C-linked product and protonated base.

The vertical separations are illustrative because no activation free energies or equilibrium constants are supplied. In particular, the highest barrier cannot be identified experimentally from this drawing alone. protonation of the leaving group is treated as an initial rapid equilibrium in the chosen mechanism; explicitly resolving it would add the corresponding proton-transfer state and intermediate.

Treat the reacting carbon skeletons as an activated glycosyl donor and a 3-substituted indole. The new linkage is an anomeric-carbon–indole-C2 carbon–carbon bond, not a bond to indole nitrogen. A plausible purely chemical pathway is a dissociative electrophilic aromatic substitution:

The drawn donor and product have opposite anomeric faces: the anomeric leaving group is on the same face as the hydroxymethyl substituent, while the new carbon substituent is on the opposite face. Thus the target has net inversion at the anomeric carbon. A tight ion pair with the departing group shielding its original face can rationalize opposite-face capture in this proposed pathway. A freely solvated planar oxocarbenium ion would not by itself guarantee exclusive inversion; competing faces, hydrolysis and other products remain possible without an experimentally specified chemical promoter. This is a plausible non-enzymatic mechanism, not a claim that the uncatalysed reaction is rapid or clean in neutral water.

Direct iodide attack at the terminal CH2–OMs would replace the leaving group without disturbing the adjacent stereocentre, so it cannot by itself explain racemization. sulfur participation produces an intermediate with two equivalent methylene paths. Label the original S-adjacent methylene A, the stereogenic carbon C, and its CH2–OMs substituent B. The thioether sulfur attacks B intramolecularly and expels mesylate, forming a four-membered thietanium circuit S–A–C–B–S. The rest of the original six-membered ring remains as the other path between S and C.

In this bridged sulfonium ion, A and B are both CH2 groups joining the same S and C centres. Exchanging them no longer changes the labelled molecular connectivity: the distinction that made C stereogenic has been removed. iodide can open the strained intermediate by backside attack at either A or B. Opening one S–CH2 bond gives one configuration of the six-membered product, while opening its equivalent partner exchanges which methylene is in the ring and which is the iodomethyl substituent, giving the opposite configuration.

The carbon iodinated in each opening is a methylene; racemization appears at the adjacent carbon through exchange of equivalent paths. It is not inversion at a CH2 stereocentre, nor does the mesylate carbon have to form a primary free carbocation. An achiral intermediate in an achiral medium gives equal mirror-related outcomes. nitrogen participation alone in a simple terminal displacement would not provide this specific equivalent-path explanation.

All comparisons assume the same temperature and comparable reagent thermodynamic activities in methanol.

Draw one maximum for the concerted SN2 reaction step, with no intermediate minimum. Its transition structure iswith O-C forming and C-X breaking, both shown as partial bonds. O-C-X is approximately linear, and the three C-H bonds lie in the equatorial plane. The overall charge is $-1$, distributed over the entering and leaving groups.

The sketch illustrates an exergonic case; the elementary-step description alone does not supply a numerical reaction energy or determine the product-reactant energy difference for every possible X.

The ordering is allyl chloride $>$ 1-chloropropane $>$ 2-chloropropane. Iodide is a good nucleophile but a weak base, and polar aprotic solvent acetone favours its SN2 reaction. SN2 reaction and C–Cl cleavage occur in one step; precipitation of sodium chloride can also favour halide exchange without changing the basic steric comparison.

The primary saturated chloride supplies the reference rate. Allyl chloride is also primary, but its neighbouring pi system stabilizes the developing delocalized electronic structure in the substitution transition state. It can therefore react about 64 times faster without requiring a free allylic carbocation. The secondary chloride is sterically crowded at the attacked carbon, making backside approach and the crowded transition state less favourable; its rate is about 0.02 times the primary value. The leaving group and nucleophile are the same in the comparison, so assigning the ordering to different halide leaving groups would be incorrect. If prefactor differences are neglected, $\Delta \Delta G^{‡}=-RT\ln (k/k_{\mathrm{ref}})$ at 333 K gives roughly $-11.5$ and $+10.8\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for the allylic and secondary substrates respectively: modest barrier changes produce large rate ratios.

The first product is $\mathrm{BrCH}=\mathrm{C}(\mathrm{C}{\mathrm{H}}_3)\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OC}{\mathrm{H}}_3$, with the original alkene geometry retained. methoxide attacks the primary saturated carbon bearing bromine from the backside. Its oxygen lone pair forms the new C–O bond as the C–Br bonding pair leaves with bromide. The other bromine is vinylic, directly bonded to an sp${}^2$ carbon: ordinary SN2 reaction is unfavourable there, and ionization would require an unstable vinyl cation. Thus the primary alkyl bromide is replaced selectively, rather than replacing the vinylic halide or moving the double bond. There is no observable inversion label at the attacked carbon because it is a methylene.

The second product is sodium 4-propoxybutane-1-sulfonate, $\mathrm{N}{\mathrm{a}}^{+}{}^{-}{\mathrm{O}}_3\mathrm{S}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{OC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. sodium hydride first removes the alcohol proton, giving propoxide and hydrogen. The propoxide oxygen then attacks the primary ring carbon bonded to the ring oxygen; simultaneous cleavage of that carbon–oxygen bond opens the sultone. The departing ring oxygen remains attached to sulfur as part of the resonance-stabilized sulfonate anion. The carbon attached directly to sulfur is not the substitution site, because cleavage there would not expel the same stable sulfonate leaving group. This is carbon-centred ring opening of a cyclic sulfonate ester, not addition across an S=O bond.

Aqueous base selectively converts the methyl ester to a carboxylate, and brief acidification gives the corresponding monoacid while retaining the tert-Butyl ester. At the methyl-ester end, a hydroxide lone pair attacks the acyl carbon and the carbonyl pi pair moves to oxygen. The tetrahedral intermediate collapses as the oxygen anion reforms C=O and the acyl C–OMe bond breaks to methoxide. proton transfer then gives the carboxylate and methanol. carboxylate formation makes the net saponification effectively irreversible under the basic conditions; protonation protonates that carboxylate.

The tert-Butyl ester reacts more slowly under these conditions because its bulky alkoxy group hinders acyl attack, and tert-butoxide is a stronger base and poorer leaving group than methoxide. The selectivity is kinetic, not an assertion that tert-butyl esters can never undergo base hydrolysis. The work-up should be sufficiently mild and brief to avoid the separate acid-labile pathway.

Anhydrous trifluoroacetic acid instead removes the tert-butyl group, leaving the methyl ester. protonation activates the ester; in a useful reactive protonation form the alkoxy oxygen bears the proton. The O–C(tert-butyl) bond then cleaves with its electron pair going to oxygen, producing the carboxylic acid and a tert-butyl-cation-like fragment. Loss of a methyl proton gives 2-methylpropene and regenerates the acid catalyst. Carbonyl-oxygen protonation and charge delocalization can precede cleavage; drawing alkoxy-oxygen protonation is a transparent way of tracking the bonds broken. The key distinction is alkyl–oxygen cleavage here versus acyl–oxygen cleavage in the base reaction. Methyl-cation formation is strongly unfavourable, and without water ordinary acid-catalysed methyl-ester hydrolysis is not available. No water is required to account for the tert-butyl deprotection or its alkene by-product.

Formation of a carbonyl accompanies migration of an adjacent C–H or C–C bond and departure of a leaving group. An alkoxide-assisted hydride migration can convert a suitably aligned halohydrin into a ketone.

antiperiplanar conformation elimination in a cyclohexane chair with axial leaving group and axial beta hydrogen.