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).
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: