JEE Main 2025 Apr 2 Shift 2, Chemistry Q23: Mechanism of Some Important Reactions of Alkenes

Step 1: Moles of starting material. Molar mass of 2-bromopentane $=151{\text{g mol}}^{-1}$, so 151 g corresponds to 1 mol.

Step 2: Elimination with alcoholic KOH. Beta-elimination governed by Saytzeff's rule gives the more substituted alkene, pent-2-ene, as the major product P. With 80% yield:

moles of P $=1\times 0.80=0.80\text{mol}$.

Step 3: Bromine addition. $\mathrm{B}{\mathrm{r}}_2$ adds across the double bond of pent-2-ene to give 2,3-dibromopentane (Q). With 100% yield, moles of Q = moles of P = 0.80 mol.

Step 4: Mass of Q. Molar mass of 2,3-dibromopentane (${\mathrm{C}}_5{\mathrm{H}}_{10}\mathrm{B}{\mathrm{r}}_2$) $=5(12)+10(1)+2(80)=60+10+160=230{\text{g mol}}^{-1}$.

Mass of Q $=0.80\times 230=184\text{g}$.