Step 1: Understanding the Figure:
The reactant is 2-bromopropane, \(CH_3-CH(Br)-CH_3\). It is heated with alcoholic KOH to give compound A. Then A reacts with HBr in the absence of peroxide to give B.
Step 2: Step A: Elimination:
Alcoholic KOH removes HBr from an alkyl halide (dehydrohalogenation). The H comes from a carbon next to the one holding Br. \[ CH_3-CH(Br)-CH_3 \xrightarrow{\text{Alc. KOH}, \Delta} CH_3-CH=CH_2 + KBr + H_2O \] So A is propene.
Step 3: Step B: Addition of HBr:
Without peroxide, HBr adds by Markovnikov's rule. The hydrogen goes to the double-bond carbon that already has more hydrogens (the terminal \(CH_2\)). The bromine goes to the middle carbon, which gives the more stable secondary carbocation. \[ CH_3-CH=CH_2 + HBr \to CH_3-CH(Br)-CH_3 \]
Step 4: Name of B:
B is \(CH_3-CH(Br)-CH_3\), which is 2-bromopropane.
Step 5: Why the other options are wrong:
1-Bromopropane would form only by anti-Markovnikov addition, which needs peroxide. Bromoethane has only two carbons and carbon tetrabromide has no chain, so neither can form from a three-carbon alkene.
Final Answer:
The major product B is 2-bromopropane, option (C).
\[ \boxed{\text{2-Bromopropane}} \]
