JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Haloalkanes and Haloarenes

2-Bromopentane has molar mass $12\times 5+11+80=151{\text{g mol}}^{-1}$, so 151 g is exactly one mole. Alcoholic KOH brings about dehydrohalogenation to give pent-2-ene as the major (Saytzeff) product P in 80% yield, that is $0.8$ mol. Pent-2-ene then adds bromine across its double bond to give 2,3-dibromopentane (Q) quantitatively, so $0.8$ mol of Q is formed. The molar mass of 2,3-dibromopentane $\mathrm{C}{}_5\mathrm{H}{}_{10}\mathrm{Br}{}_2$ is $12\times 5+10+80\times 2=230{\text{g mol}}^{-1}$. The mass of Q obtained is $0.8\times 230=184\text{g}$.