The major product C in the below mentioned reaction is:

Question

Given below are two statements:
Statement I: Phenol on treatment with \( \mathrm{CHCl_3/aq.\ KOH} \) under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydroxy benzaldehyde as the minor product.
Statement II: The mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be easily separated through steam distillation.
In the light of the above statements, choose the correct answer from the options given below

Answer 1

The reaction proceeds through the following stages:

1. Step 1: \({CH3CH2CH2Br} \xrightarrow[\Delta]{\text{alc. KOH}} {A}\)
Alcoholic KOH under heating conditions facilitates an elimination reaction. Specifically, 1-bromopropane undergoes dehydrohalogenation with alcoholic KOH to yield propene, designated as A.
\({CH3CH2CH2Br} \xrightarrow[\Delta]{\text{alc. KOH}} {CH3CH=CH2(A)}\)

2. Step 2: \({A} \xrightarrow{\text{HBr}} {B}\)
Propene (A) reacts with hydrogen bromide (HBr). Adherence to Markovnikov's rule dictates the addition of the hydrogen atom to the carbon with a greater number of existing hydrogen atoms, resulting in 2-bromopropane as the predominant product, labeled B.
\({CH3CH=CH2(A)} \xrightarrow{\text{HBr}} {CH3CHBrCH3(B)}\)

3. Step 3: \({B} \xrightarrow[\Delta]{\text{aq. KOH}} {C}\)
2-Bromopropane (B) reacts with aqueous potassium hydroxide (KOH) in the presence of heat. This process drives a nucleophilic substitution reaction, favoring an S_N1 mechanism due to the secondary alkyl halide nature and the use of aqueous KOH, ultimately forming propan-2-ol, denoted as C.
\({CH3CHBrCH3(B)} \xrightarrow[\Delta]{\text{aq. KOH}} {CH3CH(OH)CH3(C)}\)

Consequently, propan-2-ol is identified as the principal product C.

The major product C in the below mentioned reaction is:

The Correct Option is B

Solution and Explanation

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