The elimination reaction of 2-bromo-2-methylbutane with potassium hydroxide in ethanol at 25C is an E2 reaction. In this reaction, the base potassium hydroxide abstracts a proton from the -carbon, and the bromide ion leaves simultaneously, forming a double bond between the - and -carbons.There are two possible -carbons in 2-bromo-2-methylbutane: one is the 3 carbon C3 adjacent to the bromine, and the other is the 2 carbon C1 at the other end of the molecule. The major product will be formed from the most stable alkene, which is the one with the most substituted double bond.1. Elimination at the 3 carbon C3 :If the base abstracts a proton from the 3 carbon C3 , we will get a double bond between C2 and C3. The resulting alkene is 2-methyl-2-butene, which is a tetrasubstituted alkene.2. Elimination at the 2 carbon C1 :If the base abstracts a proton from the 2 carbon C1 , we will get a double bond between C1 and C2. The resulting alkene is 3-methyl-1-butene, which is a trisubstituted alkene.The major product will be the more stable alkene, which is 2-methyl-2-butene the tetrasubstituted alkene . The minor product will be 3-methyl-1-butene the trisubstituted alkene .So, the major product is 2-methyl-2-butene, and the minor product is 3-methyl-1-butene. This is due to the fact that more substituted alkenes are more stable and favored in E2 reactions.