Q7(a). Hofmann rearrangement; proof that it is intramolecular
Complete and outline the mechanism of the following reaction: the amide, with NaOBr, gives A, which on heating (Δ) gives B. How will you establish that Hofmann rearrangement is intermolecular in nature? [Structure of the starting cis amide-acid given in the paper]
Answer in brief
A = cis-3-amino-2,2-dimethylcyclopentane-1-carboxylic acid (–CONH₂ → –NH₂, Hofmann rearrangement). B = the bicyclic lactam 7,7-dimethyl-2-azabicyclo[2.2.1]heptan-3-one. The shift is intramolecular: no cross-over products, and retention of configuration.
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CoversHofmann rearrangementisocyanatecrossover experimentintramolecular rearrangementretention of configurationlactam
All parts of Question 7
- 7(a) Hofmann rearrangement; proof that it is intramolecular10 marks
- 7(b) Pentan-1-ol and pentan-2-ol by mass spectrometry5 marks
- 7(c) Photochemical ring opening of 8-methyl-1-decalone: Norrish Type II5 marks
- 7(d) Intramolecular aldol condensation of a dialdehyde10 marks