Identify the correct set of reagents or reaction conditions 'X' and 'Y' in the following set of transformation.

Step 1: Identify the transformation

Starting material: 1-Bromopropane ($\ce{CH3-CH2-CH2Br}$)

Final product: 2-Bromopropane ($\ce{CH3-CHBr-CH3}$)

This is a positional isomerization of the bromine from C-1 to C-2.

Step 2: Determine intermediate

Direct 1° → 2° halide conversion doesn't occur. We need an elimination-addition sequence:

1. Elimination: 1-bromopropane → propene
2. Addition: propene → 2-bromopropane (Markovnikov)

Step 3: Identify reagent X (elimination)

For elimination of 1-bromopropane:

Option: Concentrated alcoholic NaOH at 80°C

• Strong base in alcoholic medium
• High temperature favors elimination (E2)
• Product: Propene ($\ce{CH3-CH=CH2}$)

$\ce{CH3-CH2-CH2Br + NaOH(alc) ->[80°C] CH3-CH=CH2 + NaBr + H2O}$

Option: Dilute aqueous NaOH at 20°C

• Weak base in aqueous medium
• Low temperature favors substitution ($S_N2$)
• Product: 1-Propanol (not useful here)

X = Concentrated alcoholic NaOH, 80°C ✓

Step 4: Identify reagent Y (addition)

For addition to propene to give 2-bromopropane:

Option: HBr/acetic acid

• Markovnikov addition
• H goes to C-1 (more H), Br goes to C-2 (more substituted)
• Product: 2-Bromopropane ✓

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

Option: Br₂/CHCl₃

• Addition of Br₂ gives dibromide
• Product: 1,2-dibromopropane (not desired)

Y = HBr/acetic acid ✓

Answer: (c) X = conc.alc. NaOH, 80°C; Y = HBr/acetic acid

Key Concepts:

• Elimination conditions: Strong base (NaOH), alcoholic solvent, high temperature
• Substitution conditions: Weak base/nucleophile, aqueous solvent, low temperature
• Markovnikov's rule: In HX addition, H goes to C with more H, X goes to more substituted C
• Positional isomerization: Often requires elimination → addition sequence