Q10(c). Urea, amine and carbamate from benzoic acid (Curtius route)
Sketch the route for the synthesis of the following from benzoic acid involving common intermediate(s): (i) C₆H₅NHCONHCH₃; (ii) C₆H₅NH₂; (iii) C₆H₅NHCOOC₂H₅
Answer in brief
Benzoic acid → C₆H₅COCl → C₆H₅CON₃ → (heat, Curtius) phenyl isocyanate, C₆H₅–N=C=O, which with CH₃NH₂ gives (i), with water (ii), with ethanol (iii).
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CoversCurtius rearrangementphenyl isocyanatebenzoyl azidecarbamateurea synthesis
All parts of Question 10
- 10(a) Structure from UV, IR, NMR and mass spectra: C₁₂H₁₄O₄10 marks
- 10(b) Dimer of cyclopentadiene; a [1,3] shift of carbon with inversion10 marks
- 10(c) Urea, amine and carbamate from benzoic acid (Curtius route)10 marks