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Mahesh Somasundaram

Enamine formation and asymmetric ketones

David Cislo  Follow
Yes, but the equilibrium is dynamic and not static. Remember, that there is a tautomerism between the enamines, as well between their chiral forms. The stronger the base, the faster the tautomerism is.
For your own education and for the same steric hindrance reasons, deprotonation of the more-substituted carbon is more convention via formation of the trimethylsilylenol ether, instead of using the enamine (Please,take a look to the literature, when you have time).

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Allan Ashworth  Follow
I'd say that the second product will be predominant. The double bond is more stable when there are more substituents.
EDIT: I'll think about the stereochemistry a little bit more because now I don't have enough time! I'll write it later  ;D

Ps.
It's right if I say that if I want to deprotonate the less-substituted carbon I have to use LDA and if I want to deprotonate the more-substituted carbon I have to use another base as KH?

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Mbanginu Aurelie Kfe  Follow
You will get a mixture of both.
According your opinion which enamine will be predominant in the mixture and why?
If the methylcyclohexane is chiral, will it be retention or inversion of the chirality or a mixture of both stereoisomers and why?
If you draw the conformers (accoding streochemistry rules), you can access to the answer of the second question.

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