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The major product of the following reaction is :

Bicyclo[3.3.0]octane-diol: one secondary OH in the plain ring and one secondary OH next to a gem-dimethyl carbon in the other ring

(i) Me3SiCl (1.0 eq.) Et3N; CH2Cl2 (ii) PDC, CH2Cl2 (iii) dil. HCl

  • A.Bicyclo[3.3.0]octane with a C=O in the left ring and a wedge OH on the right ring beside the gem-dimethyl carbon
  • B.Bicyclo[3.3.0]octane with a C=O in the right ring beside the gem-dimethyl carbon and a hashed OH in the left ring
  • C.Bicyclo[3.3.0]octane-dione with a C=O in each ring and the gem-dimethyl carbon in the right ring
  • D.Bicyclo[3.3.0]octene with a C=O in the right ring beside the gem-dimethyl carbon and a C=C in the left ring

Answer: B

The product is (B): the hindered alcohol is oxidised to a ketone and the other OH is kept.

With only 1.0 equivalent of Me3SiCl and Et3N, the less hindered secondary alcohol is silylated selectively. The OH in the plain cyclopentane ring is far more accessible than the OH next to the gem-dimethyl (neopentyl-like) carbon, so it becomes the TMS ether. PDC then oxidises the free, hindered OH to a ketone, and dilute HCl removes the TMS group, restoring the first OH.

(A) is what you get if the hindered OH were protected instead. (C), the diketone, needs both alcohols free during oxidation, and (D) shows an elimination, which none of these mild steps causes.

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