$\mathrm{(CH_3)_3CCl}$ is classified as a
- primary (1°) haloalkane
- secondary (2°) haloalkane
- tertiary (3°) haloalkane
- vinyl halide
Show answer and solution
Correct answer: C. tertiary (3°) haloalkane
1. Your syllabus: 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: syllabus Chemistry Part 2, Ch 6, §6.1 (Classification — 1°/2°/3°), p. 1–2._