Question

The major product C in the below mentioned reaction is:

• A. Propan-1-ol
• B. Propan-2-ol
• C. Propane
• D. Propyne

Answer 1

The Correct Option is B

Let's analyze the reaction step-by-step:

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).
\({CH3CHBrCH3(B)} \xrightarrow[\Delta]{\text{aq. KOH}} {CH3CH(OH)CH3(C)}\)

Therefore, the major product C is propan-2-ol.