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The major products M and N in the given reaction sequence are

4-Ethylcyclohexanone treated with (a) NaH, (MeO)2CO, benzene; (b) Br2 (2 equiv), Et2O; (c) NaOMe, MeOH to give M, then excess DIBAL-H to give N

  • A.M: 4-ethylcyclopentane-1,2-dicarboxylic acid dimethyl ester; N: the corresponding diol with two CH2OH groups
  • B.M: dimethyl 4-ethylcyclopent-1-ene-1,2-dicarboxylate; N: the corresponding diol with a ring double bond
  • C.M: dimethyl 4-ethylcyclopent-2-ene-1,1-dicarboxylate; N: the corresponding geminal bis(hydroxymethyl) alkene
  • D.M: cyclopentane with one ethyl group and two methyl ester groups on three separate ring carbons; N: the same ring with two CH2OH groups in place of the esters

Answer: B

The answer is (B): M is dimethyl 4-ethylcyclopent-1-ene-1,2-dicarboxylate and N is the corresponding diol, with the ring double bond kept.

  1. NaH and dimethyl carbonate put a CO2Me group on C2 of 4-ethylcyclohexanone, giving the beta-keto ester.
  2. Two equivalents of Br2 brominate both alpha carbons: C2, between the two carbonyls, and C6.
  3. NaOMe starts a Favorskii rearrangement: the enolate at C6 displaces the C2 bromide to form a cyclopropanone, methoxide adds to the carbonyl, and the ring opens towards the ester-stabilised carbanion. The ring contracts to a cyclopentane carrying two adjacent ester groups, one carbon still bearing Br. Elimination of HBr then puts the double bond between the two ester-bearing carbons, giving the conjugated cyclopentene-1,2-diester M.
  4. Excess DIBAL-H reduces both esters to CH2OH groups and leaves the C=C bond alone, giving N.

(A) is the saturated diester and diol, which would need the C-Br bond to be reduced rather than eliminated. (C) puts both esters on one carbon and (D) places the esters 1,3 with the ethyl group moved, neither of which the Favorskii ring contraction produces.

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