Q6(c)(i). OsO₄, then HIO₄, on an alkenyl cyclopropane lactone
Write the structures of the products X, Y and Z in the following reactions and indicate the mechanism for the formation of X: the alkenyl lactone with (1) OsO₄, pyridine, (2) H⁺, H₂O gives X; X with HIO₄ gives Y + Z. [structure of the alkenyl cyclopropane lactone given in the paper]
Answer in brief
X = the syn (erythro) vicinal diol from the chain C=C; Y = the cyclopropanecarbaldehyde of the lactone; Z = 4-phenylbutanal; lactone and cyclopropane rings are unchanged.
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CoversOsO₄ syn-dihydroxylationperiodic acid cleavageosmate esterMalaprade reactionLemieux–Johnson
All parts of Question 6
- 6(a)(i) Photolysis of a mixture of valerophenone and 5,5-dideuteriohexan-2-one5 marks
- 6(a)(ii) Photolysis of 4,5-dimethylheptan-2-one10 marks
- 6(b)(i) Polypropylene by Ziegler–Natta catalysis10 marks
- 6(b)(ii) Perlon (nylon-6) from ε-caprolactam5 marks
- 6(c)(i) OsO₄, then HIO₄, on an alkenyl cyclopropane lactone10 marks
- 6(c)(ii) (S)-3-Phenylbutan-2-one with m-CPBA10 marks