The product (C) in the below mentioned reaction is:

Step 1: Identify product A (elimination)

1-Bromopropane + heat (Δ) → elimination

With heat alone (no base), this likely means dehydrohalogenation or the question implies a base.

Assuming elimination with heat: $\ce{CH3-CH2-CH2Br ->[\Delta] CH3-CH=CH2 + HBr}$

Product A = Propene ($\ce{CH3-CH=CH2}$)

Step 2: Identify product B (addition of HBr)

Propene + HBr → addition following Markovnikov's rule

Markovnikov: H goes to carbon with more H, Br goes to more substituted carbon

$\ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3}$

Product B = 2-Bromopropane

Step 3: Identify product C (hydrolysis)

2-Bromopropane + aqueous KOH → nucleophilic substitution

$\ce{CH3-CHBr-CH3 + KOH(aq) -> CH3-CH(OH)-CH3 + KBr}$

Product C = Propan-2-ol (isopropanol)

Overall transformation: $\ce{CH3CH2CH2Br -> CH3CH=CH2 -> CH3CHBrCH3 -> CH3CH(OH)CH3}$

1-Bromopropane → Propene → 2-Bromopropane → Propan-2-ol

Answer: (d)

Key Reactions:

1. Elimination (Δ): Alkyl halide → Alkene
2. Markovnikov addition: HX adds to alkene (H to less substituted C)
3. Nucleophilic substitution: Alkyl halide + OH⁻ → Alcohol

Note: This sequence converts a primary alcohol position to a secondary alcohol position via elimination-addition rearrangement.