Pick the incorrect statement about aniline.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Does undergo alkylation or acylation
Note: The given answer is as per the official NTA Answer Key. However, the statement in Option (2) is true for alkylation and acylation at the nitrogen of aniline and false only for Friedel-Crafts alkylation and acylation of the ring, as explained below.
Aniline is a primary aromatic amine. The lone pair on nitrogen makes the –\( \ce{NH2} \) group a nucleophile, so the nitrogen is alkylated by alkyl halides and acylated by acid chlorides or anhydrides:
\( \ce{C6H5NH2 + CH3COCl} \xrightarrow{\ce{pyridine}} \ce{C6H5NHCOCH3 + HCl} \)
The same lone pair makes aniline a Lewis base. In a Friedel-Crafts reaction it gives its electron pair to the catalyst \( \ce{AlCl3} \) and forms a salt. The nitrogen then carries a positive charge, which deactivates the ring, so aniline does not undergo Friedel-Crafts alkylation or acylation.
Option (1) Forms Zwitter ion when heated with conc. H2SO4 — A true statement about aniline. Aniline first forms anilinium hydrogensulphate; heating at 453-473 K gives sulphanilic acid, which exists as a zwitter ion.
\( \ce{C6H5NH2 + H2SO4 -> C6H5NH3^+HSO4^-} \xrightarrow{\text{453-473 K}} \ce{H3N^+-C6H4-SO3^- + H2O} \)
Option (2) Does undergo alkylation or acylation — Not a true statement about aniline (for Friedel-Crafts reactions). Aniline forms a salt with \( \ce{AlCl3} \), and the positively charged nitrogen deactivates the ring, so no Friedel-Crafts alkylation or acylation takes place.
Option (3) Undergoes diazotisation — A true statement about aniline. With \( \ce{NaNO2} \) and HCl at 273-278 K, aniline gives benzenediazonium chloride.
\( \ce{C6H5NH2 + NaNO2 + 2HCl} \xrightarrow{\text{273-278 K}} \ce{C6H5N2^+Cl^- + NaCl + 2H2O} \)
Option (4) Produces foul smell when heated with chloroform and ethanolic KOH — A true statement about aniline. This is the carbylamine reaction of primary amines; phenyl isocyanide has a foul smell.
\( \ce{C6H5NH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{C6H5NC + 3KCl + 3H2O} \)
NCERT (2026-27 edition):
Electrophilic Substitution – …
