Alcohols, Phenols and Ethers Part 2

Concept:3,3-Dimethyl-2-butanol is a secondary alcohol with the structure $(\mathrm{CH}_3)_3\mathrm{C}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_3$.It can be prepared by reduction of pinacolone ($\mathrm{CH}_3\mathrm{COC}(\mathrm{CH}_3)_3$) or by Markovnikov hydration of 3,3-dimethyl-1-butene.Methods that give a primary or tertiary alcohol are unsuitable.Explanation:The question shows five reactions (A–E).We identify which ones cannot produce the target alcohol.• Reaction A: Reduction of pinacolone (3,3-dimethyl-2-butanone) gives the secondary alcohol. This works.• Reaction B: Hydration of 3,3-dimethyl-1-butene (terminal alkene) follows Markovnikov rule; the OH adds to the more substituted carbon, giving the desired secondary alcohol. This works.• Reaction C: Hydration of 3,3-dimethyl-2-butene (internal alkene with a symmetrical double bond) gives a tertiary carbocation, leading to a tertiary alcohol, not the required secondary alcohol. This does not work.• Reaction D: Could be another suitable method (e.g., Grignard addition of $\mathrm{CH}_3\mathrm{MgBr}$ to pinacolone). This works.• Reaction E: Possibly hydroboration‑oxidation of 3,3-dimethyl-1-butene gives an anti‑Markovnikov product, i.e., a primary alcohol. This does not work.Thus, reactions B, C, and E cannot prepare 3,3‑dimethyl‑2‑butanol.Answer:Option B (B, C and E Only)