Which reaction sequence adds water to an alkene in Markovnikov fashion without carbocation rearrangement?

• A. Halohydrin formation
• B. Hydration under acid
• C. Hydrogenation
• D. Oxymercuration-demercuration

Answer: (D)

This question hinges on adding water to an alkene in a way that gives the Markovnikov alcohol without forming a free carbocation that could rearrange. Oxymercuration-demercuration does just that. First, Hg(OAc)2 adds to the double bond to form a bridged mercurinium ion. Water then attacks the more substituted carbon of this intermediate, opening the ring and delivering an OH to the more substituted carbon while mercury ends up on the other carbon. Because a discrete carbocation never forms, there’s no rearrangement pathway (no hydride or alkyl shifts) to scramble the product. The final step with NaBH4 removes the mercury, replacing it with hydrogen and yielding the Markovnikov alcohol. Other approaches either go through a carbocation and can rearrange, or give different products entirely (halohydrin formation or simple hydrogenation), so they don’t achieve the same selective hydration without rearrangement.