SN1 Mechanism Initiation for Tertiary Alkyl Halide

Steps:

• Recognize 2-bromo-2-methylpropane as (CH₃)₃CBr, a tertiary alkyl halide.
• Note aqueous NaOH promotes nucleophilic substitution to form (CH₃)₃COH.
• Determine mechanism: tertiary halide in polar protic solvent favors SN1 over SN2.
• Identify first step: rate-determining heterolytic cleavage of C-Br bond, forming carbocation intermediate.

Why B is correct:

• B depicts carbocation formation via Br⁻ departure, matching SN1's unimolecular rate law (rate = k[RX]).

Why the others are wrong:

• A shows backside nucleophilic attack, characteristic of SN2 bimolecular mechanism unsuitable for tertiary halides.
• C illustrates proton abstraction, typical of E2 elimination, not substitution.
• D represents concerted OH⁻ attack with Br⁻ departure, resembling SN2, which is sterically hindered here.

Final answer: B