Question

What happens when aniline reacts with the following?
i) Bromine water
ii) (CH₃CO)₂O / Pyridine
iii) HNO₂ + HCl (0°-5°C)

Hint:

Bromine water gives 2,4,6-tribromoaniline; acetic anhydride/pyridine acetylates to acetanilide; cold HNO2/HCl diazotises to benzenediazonium chloride.

Answer 1

Background: In aniline the \(-NH_2\) group is strongly activating and directs incoming groups to the ortho and para positions of the benzene ring.

i) With bromine water.
Step 1: The \(-NH_2\) group activates the ring so strongly that bromine substitutes at all three positions (2, 4 and 6) at once, without any catalyst.
Step 2: Product is 2,4,6-tribromoaniline, which comes out as a white precipitate.
\[ C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2\ (\downarrow) + 3HBr \]

ii) With acetic anhydride (CH₃CO)₂O in pyridine.
Step 1: This is acetylation of the \(-NH_2\) group (an amide is formed). Pyridine removes the acid formed and pushes the reaction forward.
Step 2: Product is acetanilide (N-phenylacetamide).
\[ C_6H_5NH_2 + (CH_3CO)_2O \xrightarrow{\text{pyridine}} C_6H_5NHCOCH_3 + CH_3COOH \]

iii) With HNO₂ + HCl at 0°-5°C.
Step 1: \(HNO_2\) (from \(NaNO_2 + HCl\)) reacts with the \(-NH_2\) group. This is called diazotisation and is done in cold conditions (273-278 K) because the product is unstable when warm.
Step 2: Product is benzenediazonium chloride.
\[ C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{273-278\,K} C_6H_5N_2^{+}Cl^{-} + 2H_2O \]