Concept: Oxymercuration-Demercuration Reaction

This reaction is used to add water (H₂O) across an alkene double bond, following Markovnikov's rule, without rearrangement. The process occurs in two steps: oxymercuration (addition of Hg²⁺ and OH⁻) and demercuration (replacement of Hg with H).

Step-by-Step Solution:

Step 1: Identify the Reactant
The reactant is an alkene: $CH 3 - CH = C (CH 3) 2$ . This is 3-methylbut-1-ene.

Step 2: Apply Markovnikov's Rule
Markovnikov's rule states that the hydrogen atom (from H₂O) adds to the carbon with more hydrogens. In this alkene, the double bond is between C1 and C2. C1 has 2 hydrogens, C2 has 0 hydrogens and is tertiary. Therefore, the H⁺ (from the acid) will add to C1, and the OH⁻ will add to the more substituted carbon, C2.

Step 3: Predict the Product
The major product will be the Markovnikov alcohol: 3-methylbutan-2-ol. $CH 3 - CH (OH) - CH (CH 3) 2$

Step 4: Match with the Options
The correct structure is the one showing a secondary alcohol with the -OH group on the carbon adjacent to the two methyl groups.

Final Answer: The second option is the major product.
Correct Product

Related Topics:

Hydroboration-Oxidation (Anti-Markovnikov addition of water)
Acid-Catalyzed Hydration of Alkenes
Markovnikov's Rule
Alkene Addition Reactions

Key Formulae & Theory:

General Reaction:
$Alkene + 1) {Hg}_{(O A c)} 2, H 2 O; 2) {NaBH}_{4} \to Alcohol$

Markovnikov's Rule: "The rich get richer." In the addition of HX to an unsymmetrical alkene, the hydrogen atom adds to the carbon that already has the greater number of hydrogen atoms.

Mechanism: The reaction proceeds through a mercurinium ion intermediate, which is attacked by water at the more substituted carbon, leading to anti-addition. The demercuration step with NaBH₄ replaces the HgOAc group with a hydrogen atom, resulting in overall syn addition of H and OH.