Consider the following reaction sequence
The percentage of nitrogen in product ' T ' formed is ____ %. (Nearest integer)
(Given molar mass in gmol⁻¹H:1,C:12, N:14,O:16 )

Solution

Chemistry ma’am · 4 steps

Protect the amine, nitrate at para, then take the acetyl off: T is p-nitroaniline.

1. Benzene to aniline. Nitration at 333 K gives nitrobenzene, C₆H₅NO₂ (P). Sn/HCl reduces -NO₂ to -NH₂, and neutralising frees aniline, C₆H₅NH₂ (Q).
2. Protect, then nitrate. Acetic anhydride turns aniline into acetanilide, C₆H₅NHCOCH₃ (R). The -NHCOCH₃ group is still an ortho/para director but milder, and it is bulky. So nitration gives mainly p-nitroacetanilide (S).
3. Remove the acetyl. HCl in ethanol with heat hydrolyses the amide back to -NH₂. After neutralising, T is p-nitroaniline, H₂N-C₆H₄-NO₂, which is C₆H₆N₂O₂.

   % N = (2 × 14)/(6(12) + 6(1) + 2(14) + 2(16)) × 100 = 28/138 × 100 = 20.29

4. So. The percentage of nitrogen in T is 20.29%, which rounds to 20.

The common slip: 16. Stopping at S and using p-nitroacetanilide, C₈H₈N₂O₃ with M = 180, gives 28/180 × 100 = 15.6, which rounds to 16. The last step removes the acetyl group.

The idea to keep. Acetylating aniline before nitration is a protect-direct-deprotect move: it stops oxidation and anilinium formation, gives the para product, and the acid hydrolysis at the end gives back the free amine.

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