The reaction of 1-bromo-1-methylcyclohexane with sodium ethoxide in ethanol gives 2 products: - major and minor as shown below:

And,

The 2 products are obtained because of the presence of two types of beta (β) hydrogen.

According to saytzeff rule, the alkene formed in greatest amount (major) is the one that corresponds to removal of the hydrogen from the β-carbon having the fewest hydrogen substituents.

When the 1-bromo-1-methylcyclohexane reacts with sodium ethoxide in ethanol, removal of HBr takes place giving two products one major and the other minor.

(In the major product number of alpha hydrogen=5 and in the minor product no. of alpha product=4)