Final answer:
The reaction of 2,2-dimethyl-1-propanol with HBr is slow due to steric hindrance and involves the formation of a more stable tertiary carbocation followed by a methyl shift, resulting in 2-bromo-2-methylbutane.
Step-by-step explanation:
The reaction of 2,2-dimethyl-1-propanol with HBr to yield 2-bromo-2-methylbutane involves several considerations:
Steric hindrance prevents nucleophilic attack at the more substituted carbon due to the bulky methyl groups present.
The formation of a carbocation is necessary, which is an unstable positively charged species. 2,2-dimethyl-1-propanol forms a tertiary carbocation during the reaction, which is more stable than a primary or secondary carbocation and can therefore form despite the energy barrier.
A carbocation rearrangement occurs with a methyl shift, which explains the formation of 2-bromo-2-methylbutane as the major product. This rearrangement follows an SN1 mechanism, as opposed to an SN2 mechanism, which is not sterically favored in this case.