Look at the arrows below, and draw the intermediate that you get after pushing the arrows. In the second step the species that leaves is :

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Follow the arrows as a reaction roadmap: The first arrow indicates a lone pair on the oxygen of CH₃O^- attacking the carbonyl carbon, forming a new C-O bond. The second arrow shows the C=O π bond breaking and moving to the oxygen. This process results in the tetrahedral intermediate CH₃CH₂C(O^-)(Br)(OCH₃).

When the O^- lone pair re-forms the C=O double bond, the best leaving group is expelled. Br^- is the weakest base among Br^-, CH₃O^-, and CH₃CH₂^-. Therefore, Br^- leaves, and methyl propanoate is formed. The correct answer is (B).

Chemistry - GOC Question with Solution | TestHub

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