The product (C) in the below mentioned reaction is :
 $C{H}_3-C{H}_2-C{H}_2-Br\underset{\Delta }{\overset{KO{H}_{\left(alc\right)}}{\to }}A\stackrel{HBr}{\to }B\underset{KO{H}_{\left(aq\right)}}{\overset{\Delta }{\to }}C$

Reaction Product Sequence: Propan-2-ol

Product (C) in the Reaction Sequence

CH₃–CH₂–CH₂–Br

alc. KOH, Δ → A

HBr → B

aq. KOH → C

Product C is propan-2-ol (CH₃–CH(OH)–CH₃)

Stepwise mechanism:

1. Alcoholic KOH eliminates HBr to yield propene (CH₃–CH=CH₂).

2. HBr adds to propene (Markovnikov rule) to give 2-bromopropane (CH₃–CHBr–CH₃).

3. Aqueous KOH replaces Br with OH to yield propan-2-ol (CH₃–CH(OH)–CH₃).