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Base-promoted bimolecular acyl substitution: nucleophile addition gives a tetrahedral intermediate, then a leaving group departs from the acyl carbon.

A is a fused five-/six-membered bicyclic diketone containing sulfur. To specify its structure unambiguously, retain the cyclohexanone numbering C1=O, C2 and C3 for the original alkene. Add the five-membered ringThe original C1 ketone remains, and the former methyl-ester carbonyl becomes the second ketone. The product is drawn in the first panel below.

methoxide first deprotonates the thiol. The soft thiolate attacks the beta carbon C3 in a Michael addition, shifting the enone electrons onto oxygen. The resulting enolate is nucleophilic at C2. It then attacks its tethered ester carbonyl intramolecularly, forming a five-membered ring through a tetrahedral intermediate. Collapse expels methoxide. proton transfers and work-up give the neutral bicyclic beta-diketone.

The initial 1,4 addition followed by intramolecular Claisen acylation explains the regiochemistry: it connects C3 to sulfur and C2 to the ester carbonyl, giving a favorable five-membered closure. Addition to the original ketone would not give this same enolate-and-ester arrangement. The net loss of methanol from the combined starting materials gives the observed ${\mathrm{C}}_8{\mathrm{H}}_{10}{\mathrm{O}}_2\mathrm{S}$ formula.

amide ion removes a methyl-side alpha proton of pinacolone. The C–H electrons form the enolate C=C bond as the carbonyl pi pair moves onto oxygen; ammonia is formed. The tert-butyl-side alpha carbon has no hydrogen, fixing the ordinary enolate regiochemistry.

The carbon end of D attacks the acyl chloride carbonyl carbon. Move the C=O pi pair onto oxygen, then collapse the tetrahedral intermediate with chloride departure. This nucleophilic acyl substitution gives $\mathrm{tCO}-\mathrm{C}{\mathrm{H}}_2-\mathrm{COt}$ and constructs the bond between the original enolate methylene and the incoming carbonyl. The observed product requires carbon acylation, rather than an O-acyl enol ester. A further equivalent of D removes the much more acidic central proton of the new 1,3-diketone, generating its extensively delocalized enolate and returning that second D equivalent to pinacolone. Aqueous acid workup protonates the product enolate; keto–enol equilibration gives the conjugated, internally stabilized by a hydrogen bond enol observed by NMR. The acyl chloride contributes its entire pivaloyl group; neither tert-butyl group is lost.

hydroxide attacks an ester carbonyl to form a tetrahedral intermediate. Collapse expels ethoxide, and acid–base transfer gives the carboxylate and ethanol. Repeating this nucleophilic acyl substitution at the other ester produces the malonate dianion; acid workup gives $\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{C}(\mathrm{L}{)}_2-\mathrm{C}{\mathrm{O}}_2\mathrm{H}$, with $L=\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$ as defined above.

On heating, one carboxylic acid group is lost through a six-membered cyclic proton-transfer/bond-cleavage arrangement. A carbonyl oxygen of the retained group accepts the proton from the departing carboxyl OH as its C–C bond breaks and carbon dioxide forms; electron redistribution gives the enol-like form of the retained acid. proton transfer/tautomerization then producesThe net decarboxylation is ${\mathrm{C}}_9{\mathrm{H}}_{12}{\mathrm{O}}_4\to {\mathrm{C}}_8{\mathrm{H}}_{12}{\mathrm{O}}_2+\mathrm{C}{\mathrm{O}}_2$. The central carbon gains a hydrogen, both allyl substituents survive, and only one carboxyl group is removed. It would be incorrect to draw hydrolysis directly giving the final acid without accounting for this carbon loss. The favourable cyclic pathway is characteristic of malonic acid decarboxylation.

The initial nucleophile attacks silicon, not the nitrile or an alkene carbon. Each trimethylsilyl group has an electrophilic Si center. A methyl group from methyllithium attacks Si, passing through a pentacoordinate silicon arrangement; Si–O cleavage releases $\mathrm{SiM}{\mathrm{e}}_4$ and the lithium salt of the oxygen-containing fragment. Cleaving the silyl enol ether generates a lithium enolate; cleaving the aryl silyl ether generates a phenoxide. Neither operation needs a new deprotonation at a carbon stereocenter.Here $R-O^{-}$ represents either the phenoxide or the oxygen-centered enolate contributor. Excess reagent permits both cleavages. The pre-existing methyl/nitrile stereocenter retains its configuration: it has no hydrogen to remove, and neither bond attached to that center is broken.

After HMPA is added, each anionic oxygen attacks the carbonyl carbon of allyl chloroformate. The C=O pi electrons move to the carbonyl oxygen, giving a tetrahedral intermediate; collapse expels chloride. The enolate becomes an allyl enol carbonate, and the phenoxide becomes an aryl allyl carbonate, each containing $\mathrm{O}-\mathrm{C}(\mathrm{O})-\mathrm{OC}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$. oxygen acylation accounts for preservation of the enol double bond and for the two newly installed carbonate groups. carbon attack would give a different carbon skeleton. The two oxygens are trapped because they were selectively unmasked by desilylation, rather than because every ether oxygen in the molecule is equally reactive.

Hydride removes the remaining alpha proton at the substituted benzylic carbon, releasing hydrogen and forming the conjugated enolate. Its oxygen attacks the carbonyl carbon of allyl chloroformate. Collapse of the tetrahedral intermediate expels chloride and gives the allyl enol carbonate, with the double bond between the original carbonyl carbon and the cyanomethyl-bearing carbon.

This is O-acylation of an enolate: the carbonyl oxygen is trapped as $\mathrm{OC}{\mathrm{O}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, retaining the enol double bond. The competing carbon-centred resonance representation explains possible C-acylation in other circumstances, but it would give a different connectivity from E. The indicated product selects the oxygen-trapping pathway and the conjugated enolate regioisomer.

First hydrolyze the benzylidene acetal. Protonate one acetal oxygen and break its bond to the Ph-CH centre, giving one free alcohol and a benzylic oxocarbenium ion. Water attacks that centre; proton transfers give a hemiacetal. Protonate the remaining acetal-derived OR group and cleave it to release the second alcohol and benzaldehyde.

The resulting unsaturated hydroxy ester isThe OH on the allyl-substituted methine is gamma to the ester carbonyl. Protonation of that carbonyl enables intramolecular OH attack to give a five-membered tetrahedral intermediate. Proton transfer to methoxy and collapse expel methanol, followed by deprotonation to give the alpha,beta-unsaturated gamma-lactone.

The other liberated OH remains CH${}_2$OH. Both C=C units survive: one is the allyl substituent and the other becomes the lactone's conjugated double bond. Hydrolysis to a hydroxy acid followed by lactonization is an alternative sequence under the aqueous acid; direct acid-catalysed intramolecular ester exchange gives the same atom mapping.

For the ordinary ester, the carbon nucleophile of a grignard reagent attacks the carbonyl carbon and moves the C=O pi pair onto oxygen. Collapse of the tetrahedral intermediate expels ethoxide, transiently giving ${\mathrm{R}}^1\mathrm{COR}$. This ketone is more electrophilic than the starting ester because it lacks an alkoxy group's stabilizing resonance donation. A second equivalent of the grignard reagent therefore adds to it before isolation:The negatively charged intermediates are associated with magnesium counterions. Merely reducing the amount of reagent to one equivalent does not generally trap the ketone selectively; it can leave unreacted ester together with the double-addition product.

In the first successful preparation, magnesium insertion into the primary C–Br bond gives n-butylmagnesium bromide. The carbon chain has four carbons, so its acetylation produces hexan-2-one, $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{C}{\mathrm{H}}_3$, not pentan-2-one. Attack on acetic anhydride forms $\mathrm{C}{\mathrm{H}}_3\mathrm{C}({\mathrm{O}}^{-})(\mathrm{nBu})\mathrm{OCOC}{\mathrm{H}}_3$. Coordination of the anhydride-derived oxygens to magnesium can stabilize this tetrahedral addition complex. Under suitably cold, controlled-addition conditions it survives sufficiently to delay release of the free ketone until aqueous work-up has destroyed the remaining grignard reagent. Protonation followed by collapse releases acetate/acetic acid and the ketone.

This interpretation has direct experimental precedent: Melvin S. Newman and William T. Booth Jr. obtained methyl ketones by cold addition of grignard reagents to acetic anhydride, attributing selectivity to a persistent addition complex. The temperature and reagent-addition regime are genuine conditions on the explanation, rather than proof that an anhydride can never undergo overaddition; the exam scheme omits them.^[ref]

For the benzonitrile route, n-butyl attacks the nitrile carbon and one C≡N pi pair moves to nitrogen, producing the magnesium ketimine salt $\mathrm{PhC}(\mathrm{nBu})={\mathrm{N}}^{-}\mathrm{MgB}{\mathrm{r}}^{+}$. This anionic nitrogen-containing intermediate resists a second nucleophilic addition much more than a neutral carbonyl does. The ketone $\mathrm{PhC}(=\mathrm{O})\mathrm{nBu}$ appears only during acidic aqueous work-up, after organomagnesium has been quenched. Protonation gives the iminium ion; water attacks its carbon, proton transfers give a carbinolamine, and C–N cleavage from its nitrogen-protonated form produces the protonated ketone and ammonia, followed by deprotonation. In the acidic mixture ammonia is present largely as ammonium. This sequence explains why nitrile addition stops at a ketone without requiring the ketone itself to be resistant to grignard reagents.

The atom-labelled hydrolysis sequence isWater's lone pair attacks the iminium carbon while its C=N pi pair moves to nitrogen; the carbinolamine oxygen then reforms C=O while the C–N pair moves to the departing protonated amino group. Final deprotonation gives the neutral ketone. The arrows involving protonation and hydrolysis also exchange protons with the acidic aqueous reservoir; the condensed sequence tracks the carbon–nitrogen and carbon–oxygen transformations rather than displaying every solvent molecule.

The oxygen substituent changes position through a halohydrin–epoxide sequence. First, hydroxide attacks the acetate carbonyl of $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Me}{)}_2\mathrm{OAc}$, giving a tetrahedral intermediate. Collapse expels the substrate alkoxide, and proton transfer gives $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Me}{)}_2\mathrm{OH}$ plus acetate. This is saponification, breaking the acyl–oxygen bond rather than replacing bromide directly with hydroxide.

A small equilibrium population of $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Me}{)}_2{\mathrm{O}}^{-}$ undergoes intramolecular SN2 reaction at the adjacent primary carbon: the O lone pair attacks $\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$ and its C–Br pair leaves with bromide. The result is 2,2-Dimethyloxirane. Addition of HBr protonates the ring oxygen. bromide then attacks preferentially at the more substituted carbon, where the protonated ring has greater carbocation character, breaking that C–O bond and leaving oxygen attached to the former primary carbon:The middle formula denotes a three-membered epoxide, not an open-chain ether. Acidic opening of this highly unsymmetrical epoxide has the opposite regioselectivity to a simple basic nucleophilic opening at its less substituted carbon.

Finally, the primary alcohol oxygen attacks acetyl chloride. The tetrahedral intermediate collapses with chloride departure, and triethylamine removes the O–H proton and captures HCl. The product is $\mathrm{AcOC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Br})(\mathrm{Me}{)}_2$. oxygen is carried through the epoxide to the other carbon; it is not necessary to invent a methyl migration or a carbon-skeleton rearrangement.

The two starting ester groups are adjacent on the benzene ring. Select one ester carbon as the site receiving both incoming methyl groups. A first methyl addition gives a tetrahedral intermediate; its collapse expels methoxide to give the ortho-substituted ketone $\mathrm{o}-!\mathrm{Me}{\mathrm{O}}_2\mathrm{C}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{C}(=\mathrm{O})\mathrm{Me}$. A second methyl addition produces $\mathrm{o}-!\mathrm{Me}{\mathrm{O}}_2\mathrm{C}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{C}({\mathrm{O}}^{-})\mathrm{M}{\mathrm{e}}_2$, with the alkoxide bound to magnesium.

The newly generated oxygen is now positioned to attack the remaining ester intramolecularly. Draw an arrow from its lone pair to that ester carbonyl and another from the C=O pi bond to its oxygen. Collapse of the resulting tetrahedral intermediate reforms the carbonyl and expels methoxide, closing a five-membered lactone. Protonation of magnesium-bound species in the work-up gives 3,3-Dimethylphthalide. Equivalently, protonation can first give the ortho hydroxy ester, followed by acid-assisted lactonization with loss of methanol; both descriptions show the same required bond formation.

The oxygen in the new ring is the oxygen of the first ester's original carbonyl; the surviving lactone carbonyl comes from the second ester. The product retains the six ring carbons and both original acyl carbons, loses the two ester methoxy carbons, and gains two carbons from methylmagnesium bromide. The favourable intramolecular five-membered closure accounts for the displayed product. It does not make the isolated lactone universally inert to a large excess of grignard reagent; stoichiometry and trapping/work-up conditions matter.

Write Im for an imidazolyl group. The acid protonates one imidazole nitrogen, and its carboxylate attacks the central carbonyl of carbonyldiimidazole. Collapse of the tetrahedral intermediate expels imidazole and gives the mixed anhydride $\mathrm{RCO}-\mathrm{O}-\mathrm{CO}-\mathrm{Im}$. Acyl transfer to imidazole, with decarboxylation, produces the activated acyl imidazole $\mathrm{RCO}-\mathrm{Im}$, CO${}_2$ and imidazole.

The amine attacks the RCO carbonyl of that acyl imidazole. Its tetrahedral intermediate collapses with loss of imidazole; proton transfer gives $\mathrm{RCONH}{\mathrm{R}}^{\prime }$. The carbonyl originally belonging to CDI leaves as CO${}_2$, while the product amide carbonyl is the original acid carbonyl. Imidazole serves both as a leaving group and a proton-transfer base.

The Hammett equation is $\log (k_X/k_H)=\rho {\sigma }_X$. The positive slope, $\rho =+1.97$, means electron-withdrawing para substituents accelerate hydrolysis. They increase carbonyl electrophilicity and stabilize the additional negative charge associated with the addition transition state/tetrahedral intermediate. Strong electron donation has the opposite effect. Hydroxide addition to the acyl carbon, followed by tetrahedral-intermediate collapse and aryloxide loss, is consistent with both this trend and the bimolecular rate law.

The first step is an endo-selective Diels-Alder reaction of the pyrone diene with the electron-poor acrylamide. Number the pyrone $O1-C2(=O)-C3(OH)=C4-C5=C6$, and the acrylamide carbons $C_{\alpha }$ and $C_{\beta }$, with $C_{\alpha }$ next to its carbonyl. The favored polar connection is C3–C${}_{\alpha }$ and C6–C${}_{\beta }$, leaving C4=C5 and the original C2–O1 lactone bridge intact.

Lithium allyloxide, generated from n-Butyllithium and allyl alcohol, attacks the bridging lactone carbonyl. Collapse of the tetrahedral intermediate cleaves its acyl–oxygen bond. proton transfer converts the former O1 bridge into an OH substituent at C6 and leaves the former pyrone carbonyl as an allyl ester attached to C3. The morpholine amide is retained.

Thus C is allyl 1,4-dihydroxy-2-(morpholine-4-carbonyl)cyclohex-5-ene-1-carboxylate, with the stereochemistry described below. In this product numbering C1 is the former pyrone C3, C2 is the acrylamide $C_{\alpha }$, C3 is $C_{\beta }$, and C4 is the former pyrone C6.

The bridge fixes the former C3 carboxyl group and C6 oxygen on the same face. An endo acrylamide places its amide substituent on that face too. Accordingly CO${}_2$allyl at C1, the C2 amide and the C4 OH are cis; the C1 OH is on the opposite face. The alternative face of attack gives the enantiomer, so achiral reagents give a racemate. The $1731\mathrm{c}{\mathrm{m}}^{-1}$ band is consistent with the newly formed ester, while the strong lower-frequency band is consistent with the retained, strongly polarized amide carbonyl. These bands support carbonyl-class retention; they do not independently prove the predicted regioisomer or stereochemistry.

An aromatic–aliphatic polyamide and methanol are formed. Let $\mathrm{Ar}=p\text{-}{\mathrm{C}}_6{\mathrm{H}}_4-{\mathrm{C}}_6{\mathrm{H}}_4\text{-}p$. The polymer repeat unit isEach amide linkage comes from nucleophilic acyl substitution: an amine attacks an ester carbonyl, the tetrahedral intermediate expels methoxide, and proton transfer produces methanol. There are approximately $2n$ molecules of methanol for $n$ complete repeat units in a long chain; the precise finite-chain count depends on its end groups. Removal of methanol drives this step-growth polymerization.