Match List-I with List-II.
| List-I | List-II |
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| (B) \( \ce{C6H5NH2} \xrightarrow[\text{(3) }\ce{OH^-}]{\text{(1) }\ce{(CH3CO)2O}\text{, Pyridine; (2) }\ce{HNO3}\text{, }\ce{H2SO4}} \) |  |
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Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Note: NTA dropped this question in the official Answer Key. However, to provide additional question practice to the students, the question has been slightly modified so that the question has one clear answer.
Aniline is very reactive towards electrophiles because the –\( \ce{NH2} \) group pushes electrons into the ring (ortho/para director). Its reactions can be controlled by acetylation, and the diazonium group can be replaced by other groups such as –\( \ce{NO2} \).
(A) Aniline + \( \ce{Br2} \)/\( \ce{H2O} \) — This matches (II) 2,4,6-tribromoaniline. With bromine water all the ortho and para positions are substituted at room temperature and a white precipitate forms.
\( \ce{C6H5NH2 + 3Br2} \xrightarrow{\ce{H2O}} \text{ 2,4,6-}\ce{Br3C6H2NH2 v + 3HBr} \)
(B) (1) \( \ce{(CH3CO)2O} \), pyridine (2) \( \ce{HNO3} \), \( \ce{H2SO4} \) (3) \( \ce{OH^{-}} \) — This matches (III) p-nitroaniline. Acetylation lowers the activating effect of –\( \ce{NH2} \) and protects it from oxidation and protonation; nitration then goes mainly to the para position, and hydrolysis frees the amino group.
\( \ce{C6H5NH2 + (CH3CO)2O} \xrightarrow{\text{pyridine}} \ce{C6H5NHCOCH3 + CH3COOH} \)
\( \ce{C6H5NHCOCH3 + HNO3} \xrightarrow{\ce{H2SO4}} \text{ p-}\ce{O2NC6H4NHCOCH3 + H2O} \)
\( \text{p-}\ce{O2NC6H4NHCOCH3 + OH- ->} \text{ p-}\ce{O2NC6H4NH2 + CH3COO-} \)
(C) Benzenediazonium chloride, (1) \( \ce{HBF4} \) (2) \( \ce{NaNO2} \), Δ — This matches (IV) nitrobenzene. \( \ce{HBF4} \) gives the diazonium fluoroborate, which on heating with aqueous \( \ce{NaNO2} \) (in the presence of copper) replaces –\( \ce{N2^{+}} \) by –\( \ce{NO2} \).
\( \ce{C6H5N2+Cl- + HBF4 -> C6H5N2+BF4- + HCl} \)
\( \ce{C6H5N2+BF4- + NaNO2} \xrightarrow[\Delta]{\ce{Cu}} \ce{C6H5NO2 + N2 + NaBF4} \)
(D) Aniline + \( \ce{H2SO4} \), Δ — This matches (I) sulphanilic acid. Aniline first forms anilinium hydrogensulphate, which on heating at 453–473 K gives p-aminobenzenesulphonic acid; it exists as the zwitterion shown in (I).
\( \ce{C6H5NH2 + H2SO4 -> C6H5NH3+HSO4-} \xrightarrow{\text{453-473 K}} \text{ p-}\ce{H3N+C6H4SO3- + H2O} \)
Shortcut: (A)-(II) leaves Options (1) and (3); (D)-(I) is in both, and (C) must give nitrobenzene (IV), which only Option (3) has.