Q73E - Organic Chemistry (Mcmurry)

Question

When 2-methyl cyclohexanone is converted into an enamine, only one product is formed despite the fact that the starting ketone is unsymmetrical. Build molecular models of the two possible products and explain the fact that the sole product is the one with the double bond opposite the methyl-substituted carbon.

Step-by-Step Solution

Answer

Formation of enamine I is not favoured due to the steric crowding between the methyl groups of the six-membered ring and the five-membered ring.  The formation of enamine II is favoured because the steric crowding can be relieved by the flipping of the ring. After the flipping, the methyl of the six-membered rings, is on the axial position, therefore there is no change of steric crowding between the methyl groups of the two rings.

Molecular model of two possible products

Solution of the problem

The steric crowding between the methyl groups of the six-membered ring and the five-membered rings restricts the formation of the enamine.

Explanation of the solution

Among the two enamines, the formation of enamine I is not favoured due to the steric crowding between the methyl groups of the six-membered ring and the five-membered ring. 

 

The formation of enamine II is favored because the steric crowding can be relieved by the flipping of the ring. After the flipping, the methyl of the six-membered ring, is on axial position, therefore there is no change of steric crowding between the methyl groups of the two rings.