SN1 mechanism for tertiary alkyl halide

Steps:

• Identify substrate: 2-chloro-2-methylpropane is (CH3)3CCl, a tertiary alkyl halide.
• Recognize reaction type: Nucleophilic substitution with CH3O- favors SN1 due to tertiary carbon and stable carbocation.
• SN1 steps: Dissociation forms (CH3)3C+ carbocation and Cl-; then CH3O- attacks carbocation to yield (CH3)3COCH3.
• Match to options: A shows correct carbocation intermediate and substitution product.

Why A is correct:

• Matches SN1: Tertiary halides ionize to form (CH3)3C+ per rate law k[RX], with nucleophile adding to give (CH3)3COCH3 ether.

Why the others are wrong:

• B: Invalid intermediate; resembles SN2 but substrate is tertiary, not primary.
• C: Wrong substrate; shows secondary isopropyl system, not tertiary.
• D: Wrong substrate; depicts primary ethyl system, irrelevant to 2-methylpropane.

Final answer: A