Question

The reactions taking place with 2 - Phenyl - 2 - bromopropane as the starting material is shown below. Identify [A] and [B] formed in the reaction.
C\textsubscript{6}H\textsubscript{5} - C(Br)C\textsubscript{2}H\textsubscript{5} \xrightarrow{\text{KOH Alcoholic}} [A]
[A] + HBr $\to$ [B]

• A. [A] = 2-Phenylpropene [B] = 2-Phenyl-1-bromopropane
• B. [A] = 2-Phenylpropene-2-ol [B] = 2-Phenyl-2-bromopropene
• C. [A] = 2-Bromopropene [B] = 1-Bromopropane
• D. [A] = 4-Hydroxyphenyl-2-bromopropane [B] = 4-Hydroxyphenylpropene

Hint:

In elimination reactions (E2), alcoholic KOH induces the removal of a hydrogen and a halide from adjacent carbons, leading to the formation of an alkene. In addition reactions (like with HBr), the double bond is broken, and a halide adds to the most substituted carbon.

Answer 1

The Correct Option is A

This reaction involves the elimination of a bromine atom from the 2-phenyl-2-bromopropane in the presence of alcoholic KOH. The alcoholic KOH causes a dehydrohalogenation reaction, which leads to the formation of an alkene. - The product formed after this elimination reaction is 2-Phenylpropene, which is an alkene with the phenyl group attached to the second carbon of the propene chain. - Upon treatment with HBr, the alkene undergoes an electrophilic addition reaction, leading to the addition of HBr across the double bon(D) The resulting product is 2-Phenyl-1-bromopropane, where the bromine atom is added to the first carbon of the propene chain. Thus, the correct identification is: - [A] = 2-Phenylpropene - [B] = 2-Phenyl-1-bromopropane