Thanks for this. Sorry for the absurdly late reply, I fixed quiz 2400 so that it shows a ring flip before the elimination reaction. Hopefully it’s more clear now!

On a secondary alkyl halide, especially one adjacent to two methyl groups, elimination will be by far the dominant product.

But in other places, the reason said is because the “transition state for the chair having CH3 and Br in axial position is more unstable, so activation energy is larger”.

Which explanation is correct?

Draw it out. It would still be 4-methyl cyclohexene, so it can’t be a constitutional isomer. Does it look superimposable on the *other* 4-methyl cylohexene? : – )

Glad you find the website helpful Nata!

It’s safe to assume the t-butyl group will prefer to be equatorial. With respect to the 1-bromo group, E2 will only occur when the Br is axial. So you’ll have different rates of reaction for cis- and trans- 1-bromo-4-t-butylcyclohexane, since one will have an axial Br when t-butyl is equatorial, and the other will only have an axial Br when the t-butyl is axial. You can determine for yourself which elimination will be faster. The situation is further complicated with 1-bromo-2-methyl-4-t-butyl cyclohexane depending on the stereochemistry of the methyl group.