The ratio of number of oxygen atoms to bromine atoms in the product Q is _________ × 10⁻¹.

Solution

Chemistry ma’am · 4 steps

Let the ethoxy group steer both steps. Then count every oxygen, including the one in the ether.

1. Nitration gives P. -OC₂H₅ activates the ring and directs ortho and para. The para position is less crowded, so the major product P is p-nitrophenetole, C₂H₅O-C₆H₄-NO₂.
2. Bromination gives Q. In P, the two positions ortho to -OC₂H₅ are also meta to -NO₂, so both groups favour them. With 2 equivalents of Br₂ and Fe, both of these positions are brominated. Q is 2,6-dibromo-4-nitrophenetole.
3. Count atoms. Oxygen: 1 in -OC₂H₅ and 2 in -NO₂, which makes 3. Bromine: 2.

   n_O/n_(Br) = 3/2 = 1.5 = 15 × 10⁻¹

4. So. The ratio of oxygen atoms to bromine atoms in Q is 15 × 10⁻¹, so the answer is 15.

The common slip: 10. Counting only the two oxygens of -NO₂ and forgetting the ether oxygen gives 2/2 = 1, which is reported as 10, not 15.

The idea to keep. In a two-step aromatic substitution, find each new position from the groups already on the ring. A strong activator like -OR usually decides where the group goes, and you should check that the other groups agree.

Answer from the NTA final answer key. Solution by PrepWiser.