Q3(a). Cycloocta-1,4-diene to a tricyclic isomer: heat or light
Would you expect the above conversion to require heat or light? Explain using molecular orbital diagram. [conversion of cycloocta-1,4-diene to a tricyclic isomer given in the paper]
Answer in brief
Light (hν). The two C=C bonds add to each other in an intramolecular [π2s + π2s] cycloaddition: a 4-electron (4n) process, thermally forbidden and photochemically allowed.
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Covers[2+2] cycloadditionphotochemical cycloadditionWoodward–Hoffmann rulesfrontier molecular orbitalscycloocta-1,4-diene
All parts of Question 3
- 3(a) Cycloocta-1,4-diene to a tricyclic isomer: heat or light15 marks
- 3(b) 3-Phenyl-4-pentenal from 3-(cinnamyloxy)acrylic acid15 marks
- 3(c) Regioselectivity of Hofmann elimination by Newman projections10 marks
- 3(d) Methoxybutadienes with acrolein10 marks