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

Answer: (B)

Propan-2-ol

Answer 2

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.