Final answer:
In the reaction of 2-Methyl-2-butanol with a strong acid, the alcohol is first protonated to form a reactive species, then undergoes elimination to form a tertiary carbocation, and finally, elimination of a proton generates the alkene, likely 2-methyl-2-butene.
Step-by-step explanation:
The reaction of 2-Methyl-2-butanol with a strong acid, typically involves a protonation of the alcohol to form a more reactive species, which then undergoes a loss of water to generate a carbocation intermediate. This carbocation may then undergo various reactions, including nucleophilic substitution (SN1) or elimination (E1). Due to the tertiary nature of 2-Methyl-2-butanol, the predominant reaction is likely to be an E1 elimination leading to the formation of an alkene.
To explain the reaction mechanism step-by-step:
- The alcohol oxygen is protonated by the strong acid, increasing the ability of the alcohol to leave and forming water as a good leaving group.
- Water leaves, forming a tertiary carbocation.
- The carbocation intermediate may then undergo an elimination reaction, where a nearby hydrogen is removed by a base present in the solution (often the conjugate base of the strong acid used to protonate the alcohol), resulting in the formation of an alkene.
In this case, the elimination product would likely be 2-methyl-2-butene due to the more substituted and therefore more stable alkene forming. Remember, reactions can also be influenced by the specific conditions, such as temperature and solvent, which might shift the reaction towards substitution (SN1) or elimination (E1).