Tertiary Haloalkane SN1 Reaction

Steps:

• Identify 2-chloro-2-methylpropane as (CH₃)₃CCl, with Cl on a carbon attached to three alkyl groups.
• Classify it as tertiary because the carbon bearing Cl has three substituents.
• Recall that tertiary haloalkanes with aqueous NaOH favor SN1 due to carbocation stability and polar protic solvent.
• Select option matching tertiary classification and predominant SN1 mechanism.

Why C is correct:

• Tertiary haloalkanes undergo SN1 via carbocation intermediate, as the tertiary carbocation is stabilized by hyperconjugation and inductive effects.

Why the others are wrong:

• A: Incorrectly labels it secondary; tertiary favors SN1 over SN2.
• B: Incorrectly labels it secondary and claims only SN2, but tertiary sterics block SN2.
• D: Tertiary haloalkanes do react with OH⁻ via SN1 to form alcohols.

Final answer: C