1.Step 1: \({CH3CH2CH2Br} \xrightarrow[\Delta]{\text{alc. KOH}} {A}\) Alcoholic KOH and heat promote elimination reactions. 1-Bromopropane reacts with alcoholic KOH to form propene (A) via a dehydrohalogenation reaction. \({CH3CH2CH2Br} \xrightarrow[\Delta]{\text{alc. KOH}} {CH3CH=CH2(A)}\)
2. Step 2:\({A} \xrightarrow{\text{HBr}} {B}\) Propene (A) reacts with HBr to form 2-bromopropane (B) as the major product according to Markovnikov's rule (the hydrogen atom adds to the carbon with more hydrogen atoms already attached). \({CH3CH=CH2(A)} \xrightarrow{\text{HBr}} {CH3CHBrCH3(B)}\)
3. Step 3:\({B} \xrightarrow[\Delta]{\text{aq. KOH}} {C}\) 2-Bromopropane (B) reacts with aqueous KOH and heat to form propan-2-ol (C) via a nucleophilic substitution reaction (specifically, an S$_N$1 reaction is favored due to secondary alkyl halide and aqueous KOH). \({CH3CHBrCH3(B)} \xrightarrow[\Delta]{\text{aq. KOH}} {CH3CH(OH)CH3(C)}\)
