Proton transfer to the OH group to form an oxconium ion (add a proton) followed ly loss of \(\mathrm{H}_{2} \mathrm{O}\) (break a bond to give stable molecules or ions) gives a \(2^{\text {" carbocation }}\) intermediate. Miggration of a methyl group with its bonding pair of electrons from the adjacent carbon to the positively charged carbon gives a more stable \(3^{\text {" }}\) carbocation (1,2shift). Proton transfer from this intermediate to a base, here shown as \(\mathrm{H}_{2} \mathrm{O}\), gives the ohserved product (take a proton away). CC1=CC(C)=C(C)CC1 A \(2^{\mathrm{a}}\) carbocation internediate A \(9^{\circ}\) carbocation intermediate

Question: Explain the reaction mechanism involving a proton transfer to an OH group, a methyl group migration from an adjacent carbon, and a proton transfer to form the observed product in a given reaction with the starting molecule SMILES notation: CC1=CC(C)=C(C)CC1. Answer: The reaction mechanism proceeds in four steps: (1) proton transfer to the OH group, forming an oxonium ion with a 2° carbocation intermediate; (2) loss of H2O, creating a stable 2° carbocation intermediate; (3) methyl group migration from the adjacent carbon to form a more stable 3° carbocation intermediate; and (4) proton transfer to a base (H2O) to yield the final neutral product. This reaction sequence involves the migration of a methyl group and the transfer of protons to ultimately generate the observed product.