To determine the major product C in the given reaction, let's analyze each step in the sequence.
The reaction starts with \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\), which is 1-bromopropane.
The first step involves heating with alcoholic KOH. This is a dehydrohalogenation reaction, leading to the elimination of HBr and forming an alkene. The product A is propene: \(\text{CH}_3\text{CH}=\text{CH}_2\).
In the next step, propene reacts with HBr. According to Markovnikov's rule, hydrogen attaches to the carbon with more hydrogen atoms, leading to the formation of isopropyl bromide (2-bromopropane): \(\text{CH}_3\text{CHBr}\text{CH}_3\).
Finally, the second treatment with aqueous KOH results in a substitution reaction, where the bromine atom is replaced by a hydroxyl group (-OH). This converts 2-bromopropane into propan-2-ol.