[hs764]

I know that for E2 elimination, the cis-trans isomerism of a cyclic molecule matters for the final product because the leaving group and the hydrogen have to be antiperiplanar.  But does that matter for E1 elimination?  Would cis and trans 2-methylcyclohexanol both result in the same mixture of products assuming E1 elimination?

[Dan]

Draw the carbocation intermediates of E1 elimination from cis- and trans-2-methylcyclohexanol. Are these carbocations the same or different? Do you expect them to lead to the same or different products?

[hs764]

The carbocations are the same, right?  So they would lead to identical products?

[Dan]

The carbocations are the same, right?  So they would lead to identical products?

Yes

[orgopete]

The carbocations are the same, right?  So they would lead to identical products?

Identical? I'm inclined to say similar, but maybe not identical. If the question were, "Do cis and trans-2-methylcyclohexanol have the same kinetics for the rate determining step?", then I think we should expect them to differ. I did a Google search (the poor man’s CAS search), and the nearest I could find was, http://www.markedbyteachers.com/as-and-a-level/maths/dehydration-and-gas-chromatography-of-methylcyclohexanols.html. That isn't a real reference, but it suggests they will differ.

I've hinted at why they may differ by considering the kinetics of the reaction (or activation energy), but I think the formation of the carbocation will be influenced by the groups and stereochemistry nearby. If the carbocation was long-lived, then I'd expect the same products (probably from the tertiary carbocation), but otherwise maybe not. I also searched for the dehydration of the isomers of menthol which could give a single racemic mixture, but I couldn't find anything. Perhaps someone could correct my speculation by posting some real data.