The chiral alcohol 3-methylbutan-2-ol is heated with concentrated aqueous HBr and the major organic product is isolated. Considering this reaction…

Protonation of the alcohol and loss of water gives the secondary carbocation $\mathrm{(CH_3)_2CH{-}C^{+}H{-}CH_3}$. A 1,2-hydride shift from the adjacent tertiary carbon converts it into the far more stable tertiary cation $\mathrm{(CH_3)_2C^{+}{-}CH_2CH_3}$, which is captured by bromide to give 2-bromo-2-methylbutane. So (B) is correct and (A) is wrong, since the unrearranged bromide is only a minor product. The tertiary bromide has no stereocentre (the carbon carrying Br bears two identical methyl groups), and in any case the planar cation destroys the original configuration, so the product is achiral and optically inactive: (C) is wrong. A neopentyl-type primary alcohol cannot undergo direct $\mathrm{S_N2}$ because of steric hindrance; the oxonium ion instead ionises with a concerted 1,2-methyl shift, giving the tertiary cation and then 2-bromo-2-methylbutane, not the unrearranged bromide: (D) is wrong. Answer: B only.