Q6(a). Curtius rearrangement; the isocyanate closes to a cyclic urea
Complete the following reaction and identify the products A and B: 2-aminobenzoyl azide, Δ → A → B. [structure of the acyl azide given in the paper]
Answer in brief
A = 2-aminophenyl isocyanate (Curtius rearrangement, N₂ lost). B = benzimidazol-2-one, the cyclic urea formed when the ortho NH₂ group adds to N=C=O.
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CoversCurtius rearrangementacyl azideisocyanatebenzimidazolonecyclic urea
All parts of Question 6
- 6(a) Curtius rearrangement; the isocyanate closes to a cyclic urea4 marks
- 6(b) HBr and a conjugated diene: 1,2- and 1,4-addition4 marks
- 6(c) Phenylglycine and mandelic acid from benzaldehyde4 marks
- 6(d) Coumarin and 3-acetylcoumarin from salicylaldehyde4 marks
- 6(e) Intramolecular oxymercuration–demercuration: a cyclic ether4 marks
- 6(f) trans- and cis-Cyclohexane-1,2-diol from cyclohexanone4 marks
- 6(g) [1,5]-Carbon shift and Cope rearrangement4 marks
- 6(h) Four C₁₀H₁₄ isomers: the one marked by a McLafferty peak4 marks
- 6(i) C₃H₇NO with two N–methyl singlets: N,N-dimethylformamide4 marks
- 6(j) ν(C=C) of methylidenecycloalkanes and ring size4 marks