$\mathrm{(CH_3)_3CCl}$ is classified as a

1. NCERT §6.1: haloalkanes are classified by the type of carbon bearing the halogen. 2. PRIMARY: the carbon attached to halogen is connected to ONE other carbon. SECONDARY: two. TERTIARY: three. 3. In $\mathrm{(CH_3)_3CCl}$, the C bonded to Cl is connected to THREE other carbons (three methyl groups). 4. So this is a TERTIARY (3°) haloalkane. Common name: tert-butyl chloride; IUPAC: 2-chloro-2-methylpropane. 5. Tertiary haloalkanes favour SN1 mechanisms (stable tertiary carbocations). 6. Options A and B are wrong substitution counts. Option D requires C=C adjacent to the halogen, not present here. _Source: NCERT Class 12 Chemistry Part 2, Ch 6, §6.1 (Classification — 1°/2°/3°), p. 1–2._