Which of the following reaction conditions results in Markovnikov addition to an unsymmetric alkene?

• A. Addition of $H_{2}O$ in the presence of peroxides
• B. Addition of $H_{2}O$ in the presence of ${\mathrm{Hg}}_{\mathrm{2+}}$ and ${\mathrm{OAc}}_2$ (oxymercuration-demercuration)
• C. Addition of $H_{2}O$ in the presence of $B_{2}H{6}_{}$ followed by $H_{2}O{2}_{}$ (hydroboration-oxidation)
• D. Addition of $H_{2}O$ in the presence of dilute $H_2\mathrm{SO}{4}_{}$

Answer: (D)

Verified step by step guidance

1. Understand that Markovnikov addition refers to the regioselectivity where the proton (H\+) adds to the carbon of the double bond with more hydrogens, and the other part (e.g., OH) adds to the more substituted carbon.
2. Analyze the reaction conditions given: Addition of H2O in the presence of peroxides typically leads to anti-Markovnikov addition due to a radical mechanism.
3. Oxymercuration-demercuration (Hg2+ and OAc2) is known to give Markovnikov addition without carbocation rearrangement, so this is a strong candidate.
4. Hydroboration-oxidation (B2H6 followed by H2O2) results in anti-Markovnikov addition because the boron adds to the less substituted carbon.
5. Addition of H2O in the presence of dilute H2SO4 proceeds via a carbocation intermediate, which leads to Markovnikov addition due to the stability of the carbocation formed on the more substituted carbon.