SN1 Mechanism for Tertiary Alkyl Halide

Steps:

• Identify substrate: 2-bromo-2-methylpropane is (CH3)3CBr, a tertiary alkyl halide.
• Determine reaction type: Aqueous NaOH favors SN1 for tertiary halides due to carbocation stability and polar protic solvent.
• Recall SN1 steps: Rate-determining first step is unimolecular loss of leaving group (Br-) to form planar carbocation.
• Match to diagram: First step shows heterolytic C-Br cleavage, producing (CH3)3C+ and Br-.

Why B is correct:

• B illustrates carbocation formation via ionization, aligning with SN1 rate law (rate = k[RX]) where leaving group departs without nucleophile involvement.

Why the others are wrong:

• A: Depicts nucleophilic attack on carbon, characteristic of SN2 bimolecular mechanism unsuitable for sterically hindered tertiary carbon.
• C: Shows simultaneous bond breaking and forming, resembling concerted SN2 or E2, not stepwise SN1.
• D: Illustrates Br- abstraction by base, suggesting E2 elimination rather than substitution.

Final answer: B