Question

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

• A. $X = \text{conc. alc. NaOH, 80}^\circ \text{C}, \, Y = \text{Br}_2/\text{CHCl}_3$
• B. $X = \text{dil. aq. NaOH, 20}^\circ \text{C}, \, Y = \text{HBr/acetic acid}$
• C. $X = \text{conc. alc. NaOH, 80}^\circ \text{C}, \, Y = \text{HBr/acetic acid}$
• D. $X = \text{dil. aq. NaOH, 20}^\circ \text{C}, \, Y = \text{Br}_2/\text{CHCl}_3$

Answer 1

Answer: (C)

$X = \text{conc. alc. NaOH, 80}^\circ \text{C}, \, Y = \text{HBr/acetic acid}$

Answer 2

Determine the Reaction with Reagent X:
The starting compound, CH3 − CH2 − CH2 − CH2 − Br, is a 1-bromobutane.
When treated with concentrated alcoholic NaOH at 80°C, an elimination reaction (dehydrohalogenation) occurs, leading to the formation of an alkene.
The product after elimination is CH3 − CH = CH − CH3 (1-butene).

Reaction with Reagent Y:
After the formation of 1-butene, adding HBr in the presence of acetic acid will convert it back into an alkyl halide by electrophilic addition.
The final product is 1-bromo-2-butene.

Conclusion:
The correct set of reagents for X and Y is:
X = conc. alc. NaOH, 80°C
Y = HBr/acetic acid
This corresponds to Option (3).