3-Methylbut-1-ene is treated with $HBr$ (no peroxide) to give A. A is heated with alcoholic $KOH$ (Zaitsev elimination) to give B. B is treated with $BH X_{3} \cdot THF$ and then $H X_{2} O X_{2} / NaOH$ to give C. C is

A3-methylbutan-2-ol
B3-methylbutan-1-ol
C2-methylbutan-2-ol
D2-methylbutan-1-ol

Answer: A

A is 2-bromo-2-methylbutane, formed after a hydride shift. Zaitsev elimination gives B, 2-methylbut-2-ene. Hydroboration puts boron on the less substituted carbon C3, so C is

(CH X_{3}) X_{2} CH - CH (OH) CH X_{3}

, 3-methylbutan-2-ol.

Practice Organic Reaction Mechanisms II: Electrophilic and Radical Reactions More on Hydroboration, oxymercuration, halohydrin formation

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3-Methylbut-1-ene is treated with HBr (no peroxide) to give A. A is heated with alcoholic KOH (Zaitsev elimination) to give B. B is treated with BHX3​⋅THF and then HX2​OX2​/NaOH to give C. C is

Answer: A

More questions on Hydroboration, oxymercuration, halohydrin formation

3-Methylbut-1-ene is treated with $HBr$ (no peroxide) to give A. A is heated with alcoholic $KOH$ (Zaitsev elimination) to give B. B is treated with $BH X_{3} \cdot THF$ and then $H X_{2} O X_{2} / NaOH$ to give C. C is