JEE Main 2026 — 4 April, Morning Shift — Question 65

$\mathrm{n}_{\mathrm{Br}}=\frac{3.36}{(108+80)}$ $\mathrm{r}=$ no. of Br atoms in organic compound $\mathrm{n}_{\text {organic compound }}=\frac{3.36}{(108+80) \times \mathrm{x}}=\frac{2}{\mathrm{M}}$ $\mathrm{M}=\frac{2 \times 188 \times \mathrm{x}}{3.36}$ $M=\frac{1}{2} \times x$ ( x is integer) $\mathrm{W}_{\mathrm{C}}=\frac{112 \times \mathrm{x} \times 26.7}{100}=30 \times \mathrm{x}$ $\mathrm{n}_{\mathrm{C}}=\frac{30 \times \mathrm{x}}{12}=2.5 \times \mathrm{x}$ take $\mathrm{x}=2, \mathrm{n}_{\mathrm{C}}=5$ (Empirical formula is $\mathrm{C}_{5} \mathrm{H}_{4} \mathrm{Br}$ )