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Incoming links: Lactonization
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 CHOH. 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.
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 . A second methyl addition produces , 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.
Open the cyclic ester at its acyl–oxygen bond. The resulting delta-hydroxy acid isOxidation-state reversal at the benzylic alcohol gives . This identifies the key Michael disconnection between the central and groups:The nucleophile is the enolate or enamine of the phenyl ketone. The electrophile's propyl substituent must be alpha to its ester carbonyl; using a beta-propyl acrylate would move the substituent to the wrong ring position.
Disconnect further to the indicated acetophenone. Crossed aldol addition to acetone followed by dehydration gives ; selective conjugate reduction or controlled alkene hydrogenation then gives . This avoids requiring an efficient SN2 alkylation with a secondary isopropyl halide. The alpha-propyl acrylate can be obtained by alpha-hydroxymethylation/dehydration of ethyl pentanoate, , using formaldehyde.
In the forward direction, perform the Michael coupling, reduce the phenyl ketone selectively with sodium borohydride while retaining the ordinary ester, hydrolyse the ester, and cyclize the delta-hydroxy acid by controlled acid-catalysed lactonization or an activated-acid coupling. The six-membered ring then has the order carbonyl–––––O. This atom order checks all three substituent positions. No relative configuration is specified, so the scheme does not imply a single diastereomer from the Michael addition or reduction.
