When 2-methylbutan-2-ol undergoes an E1 elimination reaction with concentrated sulfuric acid, the major product formed is 2-methylbut-2-ene.Here's the step-by-step process of the reaction:1. Protonation of the alcohol: The oxygen atom in the hydroxyl group -OH of 2-methylbutan-2-ol gets protonated by the sulfuric acid H2SO4 , forming a good leaving group water molecule .2. Formation of carbocation: The water molecule leaves, resulting in the formation of a carbocation at the tertiary carbon C3 of the 2-methylbutan-2-ol molecule. This carbocation is relatively stable due to the hyperconjugation effect.3. Elimination of a proton: A base HSO4- from the reaction mixture abstracts a proton from the adjacent carbon C2 to the carbocation, forming a double bond between C2 and C3.The major product of this E1 elimination reaction is 2-methylbut-2-ene.