Electrophilic substitution in aniline

Why -NH₂ activates the ring so strongly

The lone pair on the nitrogen of aniline is delocalised into the benzene ring by resonance. This raises the electron density at the ortho and para positions. So -NH₂ is a powerful activating group and an ortho-para director. Aniline reacts with electrophiles so fast that it is hard to stop after one substitution. Strong acids and oxidising agents also attack the -NH₂ group itself. To control this, chemists protect the -NH₂ group by acetylation, carry out the substitution, and then hydrolyse the protecting group off.

Bromination: uncontrolled and controlled

1. 1★
   C₆H₅NH₂ + 3Br₂ (bromine water) → 2,4,6-Br₃C₆H₂NH₂ + 3HBr

   At room temperature, a white precipitate of 2,4,6-tribromoaniline forms at once; no catalyst is needed
2. 2
   C₆H₅NH₂ + (CH₃CO)₂O → (pyridine) C₆H₅NHCOCH₃ + CH₃COOH

   Protect -NH₂ as acetanilide
3. 3
   C₆H₅NHCOCH₃ + Br₂ → (CH₃COOH) 4-BrC₆H₄NHCOCH₃ + HBr

   4-Bromoacetanilide is the major product; the milder group allows a single substitution
4. 4
   4-BrC₆H₄NHCOCH₃ + H₂O → (H⁺ or OH⁻) 4-BrC₆H₄NH₂ + CH₃COOH

   Hydrolysis gives 4-bromoaniline (p-bromoaniline)

★ key step

-NH₂ vs -NHCOCH₃

After the substitution, hydrolysis with acid or alkali removes the acetyl group and gives back -NH₂.

Nitration

1. 1★
   C₆H₅NH₂ → (conc. HNO₃ + conc. H₂SO₄, 288 K) nitroanilines (51% para, 47% meta, 2% ortho) + tarry products

   HNO₃ also oxidises aniline. In the strongly acidic medium, aniline is protonated to the anilinium ion, C₆H₅NH₃⁺, which is meta-directing. So a large amount of m-nitroaniline forms
2. 2★
   C₆H₅NHCOCH₃ + HNO₃ → (conc. H₂SO₄, 288 K) 4-O₂NC₆H₄NHCOCH₃ + H₂O

   4-Nitroacetanilide is the major product
3. 3
   4-O₂NC₆H₄NHCOCH₃ + H₂O → (H⁺ or OH⁻) 4-O₂NC₆H₄NH₂ + CH₃COOH

   Hydrolysis gives 4-nitroaniline (p-nitroaniline)

★ key step

Sulphonation, and why there is no Friedel-Crafts reaction

1. 1
   C₆H₅NH₂ + H₂SO₄ (conc.) → C₆H₅NH₃⁺HSO₄⁻

   Anilinium hydrogensulphate
2. 2★
   C₆H₅NH₃⁺HSO₄⁻ → (453 to 473 K) 4-H₂NC₆H₄SO₃H + H₂O

   Sulphanilic acid (4-aminobenzenesulphonic acid) is the major product. It exists as a zwitterion, ⁺H₃NC₆H₄SO₃⁻
3. 3
   C₆H₅NH₂ + AlCl₃ → C₆H₅N⁺H₂-Al⁻Cl₃ (a salt); no Friedel-Crafts alkylation or acylation

   The N lone pair bonds to the Lewis acid AlCl₃. The nitrogen becomes positively charged and strongly deactivates the ring

★ key step

Activity: prepare acetanilide and p-nitroacetanilide

1. 1
   Acetanilide: in a 100 mL round-bottom flask, mix 5 mL of aniline with 5 mL of acetic anhydride and 5 mL of glacial acetic acid. Add a few pumice stones, fit an air condenser and reflux gently on a sand bath for 10 to 15 minutes.

   Acetyl chloride with dry pyridine can be used in place of the anhydride and acid
2. 2★
   Cool, then pour the mixture slowly into 150 to 200 mL of ice-cold water with stirring.

   White crystals of acetanilide separate
3. 3
   Filter, wash with cold water until the washings are neutral to litmus, and recrystallise from hot water with a few drops of ethanol. Record the yield and the melting point (pure acetanilide melts at about 114 °C).
4. 4
   p-Nitroacetanilide: dissolve 2 g of acetanilide in 2 mL of glacial acetic acid, add 4 mL of conc. H₂SO₄ and cool to 0 to 5 °C in ice. Add a cold mixture of 1 mL conc. HNO₃ and 1 mL conc. H₂SO₄ drop by drop, so the temperature stays below 10 °C.

   Cooling stops further nitration and oxidation
5. 5★
   Leave at room temperature for about 30 minutes, then pour onto about 100 g of crushed ice. Filter, wash with cold water, dry, and recrystallise from ethanol. Record the yield and the melting point (about 215 °C).

   Colourless crystals of p-nitroacetanilide form; the yellow ortho isomer stays dissolved in the ethanol

★ key step