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For the following reaction, the possible product(s) is/are

(pure enantiomer) 5-ethylcyclohex-2-en-1-one with a wedged ethyl group, treated with (i) MeMgBr, THF, then H3O+; (ii) PCC; (iii) H2, Pd/C.

More than one option may be correct

  • A.Chair cyclohexanone: Et out to the left with H down on its carbon; Me up and H down on the carbon next to C=O
  • B.Chair cyclohexanone drawn the other way: Me up and H down next to C=O on the left; Et out to the right with H down on its carbon
  • C.Chair cyclohexanone: Et out to the left with H down on its carbon; H up and Me down on the carbon next to C=O
  • D.Chair cyclohexanone drawn the other way: H up and Me down next to C=O on the left; Et out to the right with H down on its carbon

Answer: A, C

The possible products are (A) and (C), as the key states.

Step (i): MeMgBr adds 1,2 to the carbonyl of 5-ethylcyclohex-2-enone to give a tertiary allylic alcohol. Step (ii): PCC oxidises this tertiary allylic alcohol with a 1,3-transposition (a Babler-Dauben oxidation): the carbonyl appears at the far end of the allyl system and the methyl-bearing carbon becomes the β-carbon of a new enone, 3-methyl-5-ethylcyclohex-2-enone. Step (iii): H2 over Pd/C saturates the C=C, giving 3-ethyl-5-methylcyclohexanone.

The carbon bearing the ethyl group is never touched, so its configuration is kept. The new methyl-bearing carbon is created by hydrogenation, which can add H2 to either face of the alkene, so both diastereomers (cis and trans) can form. (A) and (C) differ only at that methyl carbon. (B) and (D) are the mirror-image drawings, with the opposite configuration at the ethyl carbon, which the reagents cannot produce from a single enantiomer of the ketone.

I did not assign CIP descriptors to the four chairs independently; the choice of (A)/(C) as the set carrying the retained configuration is the key's.

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