The first transformation is two successive dehydrohalogenations; the second opens the lactone by two methyl additions and then closes a six-membered cyclic ether. Both schemes must be accounted for within this part.
For the dichloride, write . In the product-forming path, tert-butoxide removes a hydrogen from the methyl group of . Draw arrows from the base lone pair to H, from C–H into the C–C bond, and from C–Cl to Cl. This E2 elimination gives . A second base-induced elimination removes a vinyl hydrogen and the remaining chloride, forming . The second elimination is more demanding because a vinylic bond is involved, consistent with the strong-base, heated conditions. This route explains the displayed product without asserting that no other elimination regioisomer can ever form.
For the lactone, a methyl equivalent from methylmagnesium iodide attacks the ester carbonyl carbon, with its pi pair moving to oxygen. Collapse of the tetrahedral intermediate reforms C=O and cleaves the acyl–oxygen bond, opening the ring to the magnesium alkoxide corresponding to . The newly formed ketone is more susceptible to a second methyl addition; its pi pair again moves to oxygen. After water workup, the product is the diol .
Under acid and heat, protonate its tertiary OH, then move the C–O pair onto oxygen as water leaves. The resulting is trapped intramolecularly by the remaining primary-alcohol oxygen. Its lone pair forms the closing O–C bond, producing a six-membered cyclic oxonium ion. Deprotonation yields 2,2-dimethyloxane, also called 2,2-dimethyltetrahydropyran. Ring closure retains four chain methylenes; it does not form a five-membered ring. The two methyl groups both derive from the two equivalents of Grignard reagent.
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