The hydrolysis of optically active 2-bromobutane with aqueous NaOH result in the formation of :

Options

A. (`pm`) butan–1–ol
B. (+) butan–2–ol
C. (pm) butan–2–ol
D. (-) butan–2–ol

Answer

C

Explanation

Upon the removal of halogen from the compound, a carbocation `CH_3(C^(+)HBr)CH_2CH_3` is formed. As the carbocation is 2 and stable, it will follow `S_N^1` reaction. `S_N^1` reactions yield a racemic mixture of (+) and (-) optically active compound hence option (c) is correct.