Machine-translated by Google. Questions are authored in English.
Top

Discussion

Consider the reaction sequence shown below:

Reaction sequence: an octahydronaphthalene, then step 1 X and step 2 TsCl in pyridine, gives a decalin bearing a secondary tosylate next to a tertiary OH at the ring fusion; t-BuOK in t-BuOH then gives the product asked for

TsCl = p-toluenesulfonyl chloride

The product is

  • A.Decalin epoxide spanning C1 and the ring-fusion carbon, with the other ring-fusion hydrogen hashed
  • B.Tricyclic alcohol with a cyclopropane bridge across the ring and a tertiary OH at the ring fusion
  • C.Octahydronaphthalenol with a ring double bond next to the ring-fusion carbon bearing OH
  • D.Hydroazulenone, a cyclopentane fused to a cycloheptanone, with a hashed ring-fusion hydrogen

Answer: D

The product is the ring-contracted and ring-expanded ketone, a cyclopentane fused to a cycloheptanone (D).

t-BuOK deprotonates the tertiary OH. The resulting alkoxide pushes a bond on its own carbon across to the neighbouring carbon, which loses tosylate: a semipinacol (pinacol-type) rearrangement. The bond that migrates is the one anti-periplanar to the C-OTs bond. With the tosylate equatorial on the trans-decalin, that is the ring-fusion C-C bond.

When the ring-fusion bond shifts, the ring that held the tosylate loses a member and becomes a cyclopentane, the other ring gains one and becomes a seven-membered ring, and the alkoxide carbon becomes the C=O: a hydroazulenone, structure D.

An epoxide (A) would need the alkoxide to attack from the back of the C-OTs bond, which the cis arrangement prevents. B and C are not products of a base-promoted reaction of this hydroxy tosylate; the anti-periplanar migration outcompetes elimination.

No comment is present. Be the first to comment.
Loading…

Post your comment

Please log in or register to post a comment.