Identify the reaction which is not given by aniline .

The correct answer is Option (3) → Friedel Crafts reaction

The –\( \ce{NH2} \) group of aniline is a Lewis base and a powerful activating, ortho/para-directing group. Aniline therefore reacts readily at nitrogen (acetylation, diazotisation) and on the ring (bromination). The Friedel-Crafts reaction needs anhydrous \( \ce{AlCl3} \), a Lewis acid, which forms a salt with aniline instead:

    \( \displaystyle \ce{C6H5NH2 + AlCl3 -> C6H5NH2-AlCl3}\ (\text{salt}) \)

• Option (1) Bromination — Given by aniline. The ring is so activated that bromine water at room temperature gives a white precipitate of 2,4,6-tribromoaniline:

      \( \displaystyle \ce{C6H5NH2 + 3Br2} \xrightarrow{\ce{H2O}} 2,4,6\text{-}\ce{Br3C6H2NH2 v + 3HBr} \)

• Option (2) Acetylation — Given by aniline (N-acetylation). Aniline reacts with acid chlorides or anhydrides by nucleophilic substitution at nitrogen; a base such as pyridine removes the HCl formed:

      \( \displaystyle \ce{C6H5NH2 + CH3COCl} \xrightarrow{\text{pyridine}} \ce{C6H5NHCOCH3 + HCl} \)

• Option (3) Friedel Crafts reaction — Not given by aniline. In the salt the nitrogen carries a positive charge and becomes a strongly deactivating group, so the ring does not undergo Friedel-Crafts alkylation or acylation.

• Option (4) Diazotisation reaction — Given by aniline. Aniline reacts with nitrous acid (from sodium nitrite and hydrochloric acid) at 273–278 K to give benzenediazonium chloride:

      \( \displaystyle \ce{C6H5NH2 + NaNO2 + 2HCl} \xrightarrow{\text{273-278 K}} \ce{C6H5N2^+Cl^- + NaCl + 2H2O} \)

NCERT (2026-27 edition):
Electrophilic Substitution – …

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