= Solution
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> $\mathrm{o\!-!MeO_2C-C_6H_4-C(=O)Me}$. A second methyl addition produces $\mathrm{o\!-!MeO_2C-C_6H_4-C(O^-)Me_2}$, 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 \b[<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.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-phthalide-mechanism.png]
{title=Original sequence of two methyl additions followed by five-membered lactone closure, with oxygen provenance retained}
Back to article page
