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The major product of the following reaction is :

Nopinone (6,6-dimethylbicyclo[3.1.1]heptan-2-one) and methyl bromoacetate treated with Zn in toluene and Et2O, then aqueous work up

  • A.Polycyclic skeleton with a short stub on the upper bridge, HO on the lower left carbon and a CO2Me group on a two-carbon branch at the lower right
  • B.Polycyclic skeleton with a short stub on the upper bridge, HO on the left carbon and a CO2Me group on a two-carbon branch at the lower right
  • C.Polycyclic skeleton with a two-pronged stub on the upper bridge, HO on a vertical left edge and a CO2Me group on a branch below it
  • D.Polycyclic skeleton with a short stub on the upper bridge pointing down, HO on a vertical left edge and a CO2Me group on a branch below it

Answer: C

The major product is (C), the β-hydroxy ester with OH and CH2CO2Me on the former carbonyl carbon.

Zinc inserts into methyl bromoacetate to give the Reformatsky reagent, BrZnCH2CO2Me. This zinc enolate adds to the ketone carbonyl of nopinone (6,6-dimethylbicyclo[3.1.1]heptan-2-one), and work-up gives a tertiary alcohol carrying the CH2CO2Me group on the same carbon, C-2, next to a bridgehead.

(C) keeps the bicyclo[3.1.1] skeleton with the gem-dimethyl bridge and places both new groups on C-2. (A) puts them on C-3, the middle of the three-carbon bridge, (B) puts OH and the ester side chain on different carbons, and (D) is drawn on a different (bicyclo[2.2.1]) skeleton.

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