Free-radical monochlorination of 2-methylbutane gives how many different structural isomers of $\mathrm{C}{}_5\mathrm{H}{}_{11}\mathrm{Cl}$ (stereoisomers not counted)?

Environments of H in $\mathrm{CH}{}_3-\mathrm{CH}(\mathrm{CH}{}_3)-\mathrm{CH}{}_2-\mathrm{CH}{}_3$: the two methyls on C2 (equivalent), C2-H, C3-H2 and C4-H3. That gives 4: 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane, 1-chloro-3-methylbutane. Confirmed by substituting every H and de-duplicating.