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Multiple Choice

Predict the product of dehydration of 2-methyl-2-butanol with concentrated sulfuric acid.

Dehydration of a tertiary alcohol with concentrated sulfuric acid follows an E1 mechanism: the alcohol is protonated, water leaves to give a tertiary carbocation, and then a beta-hydrogen is lost to form a carbon–carbon double bond. Among the possible beta-hydrogens, the reaction favors forming the most substituted, most stable alkene (Zaitsev’s rule). In 2-methyl-2-butanol, removing a beta-hydrogen can happen from three positions, leading to three possible alkenes. Elimination from the ethyl side (the right-hand side) gives the double bond between the central carbon and that carbon, producing 2-methyl-2-butene, the most substituted of the options. The other eliminations would yield less substituted alkenes such as 2-methyl-1-butene or 3-methyl-1-butene, which are less favored. So the major product is the alkene with the double bond between the central carbon and the right-hand carbon: (CH3)2C=CH-CH3, i.e., 2-methyl-2-butene.

Dehydration of a tertiary alcohol with concentrated sulfuric acid follows an E1 mechanism: the alcohol is protonated, water leaves to give a tertiary carbocation, and then a beta-hydrogen is lost to form a carbon–carbon double bond. Among the possible beta-hydrogens, the reaction favors forming the most substituted, most stable alkene (Zaitsev’s rule).

In 2-methyl-2-butanol, removing a beta-hydrogen can happen from three positions, leading to three possible alkenes. Elimination from the ethyl side (the right-hand side) gives the double bond between the central carbon and that carbon, producing 2-methyl-2-butene, the most substituted of the options. The other eliminations would yield less substituted alkenes such as 2-methyl-1-butene or 3-methyl-1-butene, which are less favored.

So the major product is the alkene with the double bond between the central carbon and the right-hand carbon: (CH3)2C=CH-CH3, i.e., 2-methyl-2-butene.