3, 3-Dimethyl-2-butanol cannot be prepared by:

Choose the correct answer from the options given below:

A. Grignard Reaction

• Reaction: The nucleophilic methyl group attacks the carbonyl carbon.
• Result: Forms 3,3-Dimethyl-2-butanol. This is a standard method to create this specific secondary alcohol.

B. Acid-Catalyzed Hydration of Alkene

• Reaction: A carbocation is formed at $$C2$$. However, because there is a quaternary carbon next to it, a 1,2-methyl shift occurs to form a more stable tertiary carbocation.
• Result: Does NOT form target. It yields 2,3-Dimethyl-2-butanol due to rearrangement.

C. Ozonolysis followed by Reduction

• Reaction: Ozonolysis breaks the double bond to form Pinacolone (3,3-dimethyl-2-butanone). $$NaBH_4$$ then reduces the ketone to a secondary alcohol.
• Result: Forms 3,3-Dimethyl-2-butanol.

D. Reduction of Ketone

• Reaction: $$LiAlH_4$$ is a strong reducing agent that converts the ketone into a secondary alcohol.
• Result: Forms 3,3-Dimethyl-2-butanol.

E. Hydration of Alkyne

• Reaction: Addition of water using $$Hg^{2+}/H^+$$ follows Markovnikov's rule to form an enol, which tautomerizes to a ketone (Pinacolone).
• Result: Does NOT form target. It stops at the ketone stage; it does not reduce further to the alcohol without an additional reducing agent.