Codex @codex  0

$(\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{)}_2\mathrm{O}$. It can give a persistent low-temperature tetrahedral addition complex with a grignard reagent, allowing methyl-ketone isolation after aqueous work-up under controlled conditions.

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Br}$, an aryl bromide that can be converted to an aryl grignard reagent. Bromine directs electrophilic aromatic substitution preferentially to the ortho and para positions while deactivating the ring overall.

The straight-chain four-carbon grignard reagent $\mathrm{C}{\mathrm{H}}_3(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{MgBr}$ when the n-butyl form is specified.

${\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{MgBr}$, a cyclohexyl-transfer grignard reagent.

A transition-metal-catalyzed carbon–carbon coupling of an Grignard reagent with an organic halide.

$\mathrm{C}{\mathrm{H}}_3\mathrm{MgBr}$, a methyl-transfer grignard reagent in the usual organometallic accounting.

copper(I) converts the methyl grignard reagent into a softer organocopper nucleophile. Methyl undergoes 1,4 addition to the beta carbon of the cycloheptenone, giving a metal enolate rather than direct 1,2 addition to the ketone carbonyl.

The carbon nucleophilic site of this enolate, at the alpha carbon, attacks the aldehyde carbonyl in the second reagent. Its C=O $\pi$ electrons move to oxygen; protonation of the resulting alkoxide gives the beta-hydroxy ketone side chain. The cyclic acetal is a protected ketone and survives these nucleophilic conditions.

The beta methyl group shields one face of the cyclic enolate. Aldol bond formation from the less crowded opposite face places the new alpha side chain trans to the beta methyl group. Thus the two starred ring centers have an anti relationship; the two mirror-related products are possible from achiral starting materials and reagents. This argument does not assign the separate, unstarred alcohol center, for which the question gives no specified configuration.

The lactol is a cyclic hemiacetal and can open to $\mathrm{HOC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}(\mathrm{OBn})-\mathrm{CHO}$, retaining its pre-existing alkoxy-bearing stereocentre. Its two anomers lose their anomeric configuration on opening to the planar aldehyde, so both can feed the same addition pathway. A first equivalent of decylmagnesium bromide can be consumed by the free OH, producing decane and a magnesium alkoxide; the excess reagent supplies the carbon nucleophile for addition.

magnesium can coordinate both the aldehyde oxygen and the adjacent benzyloxy oxygen. This produces a five-membered chelate, with ring sequence Mg–O(carbonyl)–C(aldehyde)–C(alpha)–O(benzyloxy), that restricts rotation about the aldehyde–alpha-carbon bond. Decyl attack on the less crowded face, away from the alpha-carbon's hydroxyethyl chain in the chelated arrangement, forms the new C–C bond and a magnesium alkoxide at the former aldehyde carbon. Acidic workup protonates both alkoxides to give G. The OBn bond is retained; it is not displaced by the Grignard reagent.

The product configuration is the one with OBn and the newly formed OH both projecting forward in the specified zigzag chain, while the opposite configuration at the new alcohol carbon is disfavoured. This is the Cram chelation model for an alpha-alkoxy aldehyde, rather than stereochemical memory of the starting lactol OH. Coordination is a mechanistic explanation consistent with MgBr2 and the product; the scheme does not quantify individual organomagnesium aggregates.^[ref] The relevant primary synthetic sequence is documented independently.^[ref]

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 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.

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.

Build the phenethyl grignard reagent, use one portion to make a phenethyl-substituted ester, and add two further equivalents to that ester. Write $R=\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2$; the target is ${\mathrm{R}}_3\mathrm{COH}$, so all three carbon substituents must be phenethyl, not benzyl.

hydroboration followed by oxidation of styrene gives $\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$: use $\mathrm{B}{\mathrm{H}}_3$ in a suitable ether, followed by alkaline hydrogen peroxide. The Anti-Markovnikov addition regiochemistry is essential; acid-catalysed Markovnikov addition of water would give the wrong alcohol. Convert the primary alcohol to $\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$ with phosphorus tribromide, then form $\mathrm{RMgBr}$ with magnesium in dry ether. The organomagnesium reagent must be prepared and used without a protic solvent.

carboxylate one portion with carbon dioxide and acidify to obtain $\mathrm{RC}{\mathrm{O}}_2\mathrm{H}$, then esterify with ethanol under acid catalysis to give $\mathrm{RC}{\mathrm{O}}_2\mathrm{Et}$. Treat the isolated ester with at least two equivalents of another portion of $\mathrm{RMgBr}$ and then perform an aqueous acidic work-up:The first addition to the ester gives a ketone after ethoxide elimination; the second addition gives the tertiary alkoxide. Work-up supplies its OH proton. The extra central carbon comes from carbon dioxide, and each original styrene contributes one intact eight-carbon phenethyl unit. The resulting carbon count is $1+3\times 8=25$, matching the target skeleton.

Disconnect the amide into 2-propylhexanoic acid and N-methyl-2-(4-hexylphenyl)ethylamine. This keeps the nitrogen substituents distinct: one is Me and the other is the two-carbon aryl side chain. The acid's alpha carbon must bear both n-butyl and n-propyl groups.

For the acid fragment, esterify the supplied hexanoic acid, form its ester enolate with LDA, and alkylate with 1-iodopropane:hydrolysis gives 2-propylhexanoic acid; conversion with $\mathrm{SOC}{\mathrm{l}}_2$ gives the acyl chloride for the final coupling. Limit the alkylating reagent and isolate the monoalkylated ester to avoid overalkylation.

For the aryl fragment, use another portion of hexanoic acid to make hexanoyl chloride. A Friedel-Crafts acylation of bromobenzene with this reagent and $\mathrm{AlC}{\mathrm{l}}_3$ gives ortho and para products; isolate the para bromoaryl ketone. A Wolff-Kishner reduction converts $\mathrm{ArCO}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{Me}$ into $\mathrm{ArC}{\mathrm{H}}_2(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{Me}$ without deliberately hydrogenolyzing the aryl bromide. Acylation followed by carbonyl reduction installs an unrearranged n-hexyl group; direct Friedel–Crafts alkylation is a less reliable way to do that.

Convert the retained aryl bromide into its grignard reagent, open ethylene oxide, and work up to obtain $\mathrm{p}-\mathrm{Hex}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. Oxidize the primary alcohol to $\mathrm{p}-\mathrm{Hex}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ under conditions that stop at the aldehyde, then perform reductive amination with methylamine and sodium triacetoxyborohydride. Excess methylamine helps favour the desired secondary amine over further alkylation. Finally acylate that secondary amine with the isolated acyl chloride, using a separate base to capture HCl.

The final amide has an N-methyl group, the para n-hexyl phenethyl group, and the required n-butyl/n-propyl acyl branching. The carbon check is $9$ in the acyl fragment plus $1+2+6+6=15$ in the amine fragment, giving $24$ overall. No absolute stereochemistry is specified at the acid alpha carbon.

An acetal retrosynthesis separates the target into trans-cyclohexane-1,2-diol and cyclohexyl methyl ketone. Both fragments can be prepared using separate portions of cyclohexene.

For the diol, epoxidation with a peroxy acid such as mCPBA gives cyclohexene oxide. Acid-catalysed opening with water occurs by backside attack on the protonated epoxide, giving the anti, hence trans, diol. A syn dihydroxylation would give the wrong relative configuration.

For the ketone, add HBr to another portion of cyclohexene to obtain bromocyclohexane, form its grignard reagent with Mg in dry ether, and add acetonitrile. Hydrolyse the resulting magnesium imine salt to ${\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{COMe}$. A nitrile is useful here because addition followed by hydrolysis stops at the ketone instead of giving the double addition associated with an ester.

Finally combine the isolated trans diol and ketone with catalytic acid and remove water to form the cyclic ketal. The ketone carbon becomes the acetal carbon bearing Me and cyclohexyl; the two original diol C–O bonds retain their relative configuration. Anti epoxide opening supplies the required trans relationship before ketal formation. ketal formation is an equilibrium and the trans-fused product may be less favourable than an unstrained acetal, so effective water removal and isolation are necessary; substituting a cis diol is not an acceptable shortcut.

Both side chains have the order ring–$\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{CHMe}$, and both printed double bonds have the Z geometry. The two carbon side chains are cis to one another; equivalently the labelled ring H and ring Me are cis on the opposite face. A useful common late precursor therefore has two cis ring–$\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ substituents, which can undergo two Z-selective ethylidene Wittig reactions.

Use the starting 2-methylcyclohex-2-en-1-one in a reductive alpha-alkylation. Conjugate reduction under dissolving-metal/enolate-trapping conditions, followed by a primary electrophile $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OTBS}$, installs a protected two-carbon alcohol side chain at the original methyl-bearing alpha carbon. This gives 2-methyl-2-(2-siloxyethyl)cyclohexanone. The carbon at that position is now quaternary, with its original methyl retained. Alternatively, reduce the enone to 2-methylcyclohexanone and deliberately form/select its more substituted enolate before alkylation; its opposite kinetic enolate would give the wrong regioisomer.

Convert the remaining ring ketone to an exocyclic enoate by Horner–Wadsworth–Emmons olefination with a phosphonoacetate:Hydrogenate this sole alkene to give ring–$\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_2\mathrm{Et}$ at that carbon. Select the diastereomer with this substituent cis to the pre-existing protected hydroxyethyl side chain, so the newly installed ring H is cis to ring Me. Less hindered-face hydrogenation can assist this relationship, but the route should include separation/verification of the desired cis product rather than assert complete stereoselectivity without data.

Reduce the ester with lithium aluminium hydride to ring–$\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ and remove the TBS group from the other side chain. Both appendages are now two-carbon primary alcohols. Oxidize them to aldehydes under non-overoxidizing conditions, for example Dess–Martin oxidation, obtaining the cis dialdehyde with two ring–$\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ groups. Treat with excess salt-controlled, nonstabilized ethylidenetriphenylphosphorane under conditions favouring kinetic Z olefination; isolate the Z,Z product. Each aldehyde carbon becomes the inner vinylic carbon of its crotyl group, while the ylide supplies $\mathrm{CHMe}$.

This order postpones both stereodefined alkenes until after hydrogenation, so they are not accidentally saturated. It also avoids assuming that a crotyl grignard reagent adds without allylic transposition or that direct alcohol deoxygenation fixes the required ring stereochemistry. An achiral synthesis gives a racemic mixture pair with the specified relative configuration.

The left ring system is spiro, and the right-hand substituent is $\mathrm{CH}(\mathrm{OMe})\mathrm{Me}$, not OMe directly attached to the cyclohexene ring. Let $Q$ denote the right cyclohexenyl group bearing that methoxyethyl substituent cis to its linker. The connection to the left alkene carbon is $-(\mathrm{C}{\mathrm{H}}_2{)}_3-\mathrm{Q}$. These observations suggest two Diels–Alder constructions. A left-hand dienehas the original cyclopentane carbon as one terminal diene atom; cycloaddition with an ethylene equivalent gives the spiro six-membered ring. Its residual double bond is between the two internal diene atoms, putting the linker on the required alkene carbon.

Build $Q$ and its four-carbon alcohol appendage first. Protect the OH group of the indicated 4-Bromobutan-1-ol, for example as a THP ether. Convert the primary bromide to its grignard reagent and formylate with DMF, giving $\mathrm{THPOC}{\mathrm{H}}_2(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{CHO}$. olefination with an allylidenetriphenylphosphorane, $\mathrm{P}{\mathrm{h}}_3\mathrm{P}=\mathrm{CHCH}=\mathrm{C}{\mathrm{H}}_2$, supplies a conjugated diene $\mathrm{THPOC}{\mathrm{H}}_2(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{CH}=\mathrm{CHCH}=\mathrm{C}{\mathrm{H}}_2$; select the required diene geometry rather than assume a single E/Z outcome.

Cycloaddition with methyl vinyl ketone produces a cyclohexene with a protected four-carbon alcohol chain at one saturated ring carbon and an acetyl group at its adjacent ring carbon. Choose and isolate the adjacent-substituent cis regioisomer, matching the two dashed substituent bonds in the target. Endo preference and diene geometry can aid the selection, but an unsymmetrical diene is not guaranteed to give a unique regioisomer. Reduce the acetyl ketone to $\mathrm{CH}(\mathrm{OH})\mathrm{Me}$, then methylate that alcohol with sodium hydride/methyl iodide while the chain-end OH is protected. Remove the THP group under mild acid conditions to reveal $\mathrm{Q}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{OH}$, preserving the secondary methyl ether and the ring alkene.

Oxidize this primary alcohol to $\mathrm{Q}(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{CHO}$. Add vinylmagnesium bromide to give $\mathrm{Q}(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, then oxidize the allylic alcohol to the enone $\mathrm{Q}(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{C}(=\mathrm{O})\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$. This carbonyl carbon is the original alcohol-end carbon of 4-bromobutanol; after olefination it will be the linker-bearing carbon of the left ring alkene.

In the other branch, reduce the specified cyclopentanone to cyclopentanol, convert to bromocyclopentane, and react with triphenylphosphine to obtain the cyclopentylphosphonium salt. Deprotonation gives the cyclopentylidene phosphorane, whose ylidic carbon is part of the five-membered ring. Wittig olefination of the just-prepared enone gives the substituted diene displayed above. This carbonyl–ylide pairing keeps the ring carbon terminal in the diene and avoids accidentally constructing a fused ring instead of a spiro ring.

Finally, cycloadd an ethylene equivalent. Ethylene itself requires suitably forcing/high-pressure conditions; a more reactive alternative is phenyl vinyl sulfone, followed by reductive removal of the sulfone substituent using a method compatible with the methyl ether and retained C=C bonds. The latter installs the same two otherwise unsubstituted saturated carbons. Control or separate any additional cycloaddition isomers, and retain the previously selected cis configuration on $Q$.

The carbon count provides a strong check: cyclopentanone supplies five, bromobutanol four, DMF formylation one, the first allylidene ylide three, methyl vinyl ketone four, O-methylation one, vinylmagnesium two, and the final ethylene equivalent two, totalling 22 target carbons. Three original bromobutanol carbons form the linker; its fourth becomes the substituted carbon of the spiro ring's alkene. The proposal is a multistep constitutional/relative-stereochemical route with explicit regioisomer and diastereomer selection, not an unsupported claim of a single quantitative cycloaddition product.

A grignard reagent in which magnesium is attached directly to an alkenyl carbon, $\mathrm{RCH}=\mathrm{CHMgX}$. Addition to a carbonyl compound can form an allylic alcohol while retaining the geometry of the pre-existing alkene.