We're reacting 1-bromo-1-methylcyclohexane with (CH3)3COK in the solvent (CH3)3COH.

Alright, so we're dealing with a fairly strong base, and a polar protic solvent.

The answer states that one of the hydrogens on the methyl is being removed to form the elimination product methylene cyclohexane.

My question is, why isn't a hydrogen at carbon #2 on the ring removed forming the more stable product with the more substitued alkene?

We're reacting 1-bromo-1-methylcyclohexane with (CH3)3COK in the solvent (CH3)3COH.

Alright, so we're dealing with a fairly strong base, and a polar protic solvent.

The answer states that one of the hydrogens on the methyl is being removed to form the elimination product methylene cyclohexane.

My question is, why isn't a hydrogen at carbon #2 on the ring removed forming the more stable product with the more substitued alkene?

Bulky base in E2 conditions will always give the less substituted alkene.

i agree with Shunwei
you are using a bulky base, so its E2 Hoffmann elimination..
you'll get 6-methyl,1-cyclohexene

Can't believe I didn't catch the structure of the base.

Thanks guys.