The reaction of 3-methylbutan-2-ol with HBr, will give:

The correct answer is Option (1) → 2-Bromo-2-methylbutane

Secondary and tertiary alcohols react with HX by the S_N1 route, through a carbocation. A carbocation can rearrange by a 1,2-hydride shift to a more stable one before the halide ion attacks.

Protonation of the –OH group and loss of water give a secondary carbocation at C-2:

    \( \ce{CH3-CH(OH)-CH(CH3)-CH3 + H+ -> CH3-CH(OH2+)-CH(CH3)-CH3} \)

    \( \ce{CH3-CH(OH2^{+})-CH(CH3)-CH3} \) → \( \ce{CH3-C^{+}H-CH(CH3)-CH3} \) + \( \ce{H2O} \)  (2° carbocation)

The H atom on the next carbon (C-3) moves with its electron pair to C-2, giving the more stable tertiary carbocation:

    \( \ce{CH3-C^{+}H-CH(CH3)-CH3} \) → \( \ce{CH3-CH2-C^{+}(CH3)-CH3} \)  (3° carbocation)

\( \ce{Br^{-}} \) then attacks the tertiary carbon:

    \( \ce{CH3-CH2-C^{+}(CH3)-CH3} \) + \( \ce{Br^{-}} \) → \( \ce{CH3-CH2-CBr(CH3)-CH3} \)

Numbering from the end nearer the bromine gives 2-bromo-2-methylbutane as the main product.