Identify the set of reagents/reaction conditions ‘X’ and ‘Y’ in the following set of transformation: ${CH}_{3} - {CH}_{2} - {CH}_{2} Br \overset{X}{⟶}$ Product

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X = dilute aqueous NaOH, 20°C; Y = HBr/acetic acid, 20°C

X = concentrated alcoholic NaOH, 80°C; Y = Br₂/CHCl₃, 0°C

X = dilute aqueous NaOH, 20°C; Y = Br₂/CHCl₃, 0°C

X = concentrated alcoholic NaOH, 80°C; Y = HBr/acetic acid 20°C

answer is B.

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Detailed Solution

E₂reaction (strong base) formed by Markovnikov addition reaction.

The dehydrohalogenation of 1-bromo propane with alco. NaOH or KOH gives propene which on again hydrohalogenation with HBr gives 2-bromopropane due to addition of Markovnikov rule for addition.

${CH}_{3} {CH}_{2} {CH}_{2} Br \overset{80 ° C alc . NaOH / KOH}{\to} {CH}_{3} CH = {CH}_{2} \overset{HBr}{\to} {CH}_{3} {CHBrCH}_{3} $

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