The product (C) in the below mentioned reaction is: \(\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{Br} \xrightarrow{\text{KOH}_{(alc)} \Delta} A \xrightarrow{\text{HBr}} B \xrightarrow{\text{KOH}_{(alc)}\Delta} C\)
To determine the product (C) in the given reaction sequence, let's analyze each step:
The first reaction involves converting \(\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{Br}\) (1-bromopropane) to an alkene using alcoholic KOH and heat. This is a dehydrohalogenation reaction, where the bromine (Br) and a hydrogen atom from an adjacent carbon are eliminated, leading to the formation of propene:
The second step is the addition of HBr to propene. In this step, Markovnikov's rule applies, where the hydrogen atom from HBr adds to the carbon with more hydrogen atoms (the terminal end), and the bromine attaches to the other carbon in the double bond. This results in the formation of 2-bromopropane:
Finally, 2-bromopropane undergoes another dehydrohalogenation with alcoholic KOH. Since it is a secondary alkyl halide, it will form an alcohol through an elimination reaction. Here, 2-bromopropane is converted to propene again:
The product of this reaction sequence is yet again propene. However, due to the options provided, propene being the most likely target, it seems there's some misunderstanding in the provided correct answer. A direct dehydrohalogenation might not produce propan-2-ol, expect an incorrect choice might have been made trying to correlate reaction proficiency.
Upon reviewing based on the given choices and reactions:
The route is logically expected to lead back to propene but might inadvertently suggest propan-2-ol based on incorrect answers by examiners, possibly seeking knowledge out of provided set options.
