[asascl1992]

The question is to draw a detailed mechanism for the following reaction. I drew out the mechanism I came up with (excuse the bad arrows...it was my first time using chem draw!)

My first question is if the mechanism itself is correct? The part i'm unsure of the 3rd step: will the stereo chemistry of the methyl group switch once the elimination reaction occurs or after the double bond is broken in the last step assuming my mechanism is correct. And for reference, in the 3rd step, couldn't elimination result in the same sterochemistry as the reactant?

[Dan]

The part i'm unsure of the 3rd step: will the stereo chemistry of the methyl group switch once the elimination reaction occurs or after the double bond is broken in the last step assuming my mechanism is correct.

In the enol intermediate that methyl froup has no stereochemistry - the sp² centre is planar. It is only on tautomerisation back to the ketone (last step) where that centre becomes asymmetric again.

And for reference, in the 3rd step, couldn't elimination result in the same sterochemistry as the reactant?

The enol intermediate can and will protonate on either face of the double bond - but all the steps here are reversible, a dynamic equilibrium, which means the most thermodynamically stable product will form. The product with both methyl groups on the same side of the ring is the most thermodynamically stable as in this isomer both methyl groups occupy equatorial positions of a chair conformation - or at least pseudoequatorial given that the sp² centre of the ketone will distort the chair to some extent. For more info on this read about 1,3-diaxial strain in cyclohexanes.