Match List-I with List-II.
| List-I (Reaction of aniline) | List-II (Observation) |
|---|---|
| (A) Treatment with bromine water at room temperature | (I) The nitrogen acquires a positive charge and acts as a strong deactivating group, so no substitution takes place |
| (B) Direct nitration in a strongly acidic medium | (II) Anilinium hydrogensulphate is formed first, which on heating gives the para substituted product as the major product |
| (C) Friedel-Crafts alkylation with aluminium chloride as catalyst | (III) A white precipitate is formed, with substitution at both the ortho positions and the para position |
| (D) Heating with concentrated sulphuric acid at 453–473 K | (IV) A significant amount of the meta derivative is formed besides the ortho and para derivatives |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-IV, C-I, D-II
The –NH2 group makes the ortho and para positions of aniline centres of high electron density, so it is a powerful activating, ortho and para directing group. In acid, or with a Lewis acid, the nitrogen takes on a positive charge and this behaviour changes.
(A) Treatment with bromine water at room temperature — This matches (III). The ring is so reactive that all three ortho and para positions are substituted and a white precipitate of 2,4,6-tribromoaniline is formed:
\( \displaystyle \ce{C6H5NH2 + 3Br2} \xrightarrow{\ce{H2O}} \ce{C6H2Br3NH2 v + 3HBr} \)
(B) Direct nitration in a strongly acidic medium — This matches (IV). Aniline is protonated to the anilinium ion, which is meta directing, so a significant amount of m-nitroaniline forms along with the o- and p-nitro products (and tarry oxidation products).
\( \displaystyle \ce{C6H5NH2 + H^+ -> C6H5NH3^+} \)
(C) Friedel-Crafts alkylation with aluminium chloride — This matches (I). Aniline, a base, forms a salt with AlCl3, the Lewis acid. The nitrogen becomes positively charged and strongly deactivates the ring, so the reaction does not take place.
\( \displaystyle \ce{C6H5NH2 + AlCl3 -> [C6H5NH2AlCl3]} \text{ (salt)} \)
(D) Heating with concentrated sulphuric acid at 453–473 K — This matches (II). Anilinium hydrogensulphate forms first; heating it gives p-aminobenzenesulphonic acid (sulphanilic acid) as the major product:
\( \displaystyle \ce{C6H5NH2 + H2SO4 -> C6H5NH3^+HSO4^-} \)
\( \displaystyle \ce{C6H5NH3^+HSO4^-} \xrightarrow{\text{453–473 K}} \ce{H2N-C6H4-SO3H + H2O} \text{ (para)} \)
NCERT (2026-27 edition):
9.6 Chemical Reactions – “(a) Bromination: Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline.”
“(b) Nitration: Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. …in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.”
“(c) Sulphonation: Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid at 453-473K produces p-aminobenzene sulphonic acid, commonly known as sulphanilic acid, as the major product.”
“Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst. Due to this, nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group for further reaction.”