For 3-methylcyclohexanone (methyl on the ring carbon two positions from C=O), which option gives both the total number of alpha-hydrogens and the number of constitutionally distinct enolates that can form by removing an alpha-hydrogen?

A3 alpha-hydrogens; 2 enolates
B4 alpha-hydrogens; 1 enolate
C4 alpha-hydrogens; 2 enolates
D5 alpha-hydrogens; 3 enolates

Answer: C

The alpha carbons are C2 and C6. Both are

CH X_{2}

groups, giving

2 + 2 = 4

alpha-hydrogens. The methyl is on C3, a beta carbon, so it contributes none. C2 is adjacent to the methyl-bearing C3 while C6 is not, so enolisation toward C2 or toward C6 gives 2 distinct enolates.

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For 3-methylcyclohexanone (methyl on the ring carbon two positions from C=O), which option gives both the total number of alpha-hydrogens and the number of constitutionally distinct enolates that can form by removing an alpha-hydrogen?

Answer: C

More questions on Aldol, Cannizzaro, Perkin, Claisen and related condensations