Acylation of amines: practice and mechanism
The amine attacks the acyl chloride with its lone pair: an amide R–NH–C(=O)CH₃ forms.
Example
The amine's lone pair attacks the acyl chloride and chloride leaves: an amide forms. This is how aniline is "protected" as acetanilide so that it does not over-react in further substitutions.
Mechanism step by step
- 1
The lone pair of the amine nitrogen attacks the carbonyl carbon of acetyl chloride.
- 2
Cl⁻ is expelled from the tetrahedral intermediate.
- 3
Pyridine removes the proton from nitrogen: amide (e.g. acetanilide).
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- The acetyl group CH₃–C=O is transferred, not just CH₃. The product is an amide, not an N-methyl amine.
- That is the tetrahedral intermediate. It loses chloride and the C=O re-forms: the amide.
- A second acetylation hardly happens: the amide nitrogen is barely nucleophilic because of the C=O group.
- Without a Lewis acid the ring is not acylated, and the NH₂ reacts much faster: the acetyl group goes to the nitrogen.
Often confused with: SN2 with ammonia
Frequently asked questions
Why protect aniline as acetanilide?
The NH₂ group activates the ring so strongly that multiple substitution and oxidation threaten. As an amide it activates less but still directs o/p.
How do you remove the acetyl group again?
By acidic or basic hydrolysis of the amide.