When $3-$methylbutan$-2-$ol is treated with $HBr$,the following reaction takes place:
$CH_3-CH(CH_3)-CH(OH)-CH_3 \xrightarrow{HBr} CH_3-C(Br)(CH_3)-CH_2-CH_3$
Give a mechanism for this reaction.
(Hint : The secondary carbocation formed in step $II$ rearranges to a more stable tertiary carbocation by a hydride ion shift from $3^{rd}$ carbon atom.)

(N/A) The mechanism of the given reaction involves the following steps:
Step $1:$ Protonation
$CH_3-CH(CH_3)-CH(OH)-CH_3 + H^+ \rightarrow CH_3-CH(CH_3)-CH(OH_2^+)-CH_3$
Step $2:$ Formation of $2^{\circ}$ carbocation by the elimination of a water molecule
$CH_3-CH(CH_3)-CH(OH_2^+)-CH_3 \rightarrow CH_3-CH(CH_3)-CH^+-CH_3 + H_2O$
Step $3:$ Rearrangement by the hydride-ion shift
The $2^{\circ}$ carbocation undergoes a $1,2-$hydride shift from the $3^{rd}$ carbon to form a more stable $3^{\circ}$ carbocation:
$CH_3-CH(CH_3)-CH^+-CH_3 \rightarrow CH_3-C^+(CH_3)-CH_2-CH_3$
Step $4:$ Nucleophilic attack
The bromide ion $(Br^-)$ attacks the $3^{\circ}$ carbocation to form the final product:
$CH_3-C^+(CH_3)-CH_2-CH_3 + Br^- \rightarrow CH_3-C(Br)(CH_3)-CH_2-CH_3$ ($2-$bromo$-2-$methylbutane)