Free-radical monobromination of 2-methylbutane with Br2 and light at 127^circC is highly selective. The relative rates of abstraction per hydrogen…

Count the hydrogens in $\mathrm{CH_3\!-\!CH(CH_3)\!-\!CH_2\!-\!CH_3}$: primary $=3+3+3=9$ (two methyls on C2 and the terminal methyl); secondary $=2$ (the $\mathrm{CH_2}$); tertiary $=1$. Weighted reactivities: primary $9\times1=9$, secondary $2\times82=164$, tertiary $1\times1640=1640$. Total $=9+164+1640=1813$. The tertiary bromide, 2-bromo-2-methylbutane, is formed to the extent $1640/1813=0.9046$, i.e. $90.46\%$. For reference, the secondary products (2-bromo-3-methylbutane and its enantiomer) make up $9.05\%$ and the primary bromides $0.50\%$, which sum to $100\%$ with the main product. Note that the ratio depends on both the reactivity per hydrogen and the number of each kind of hydrogen, which is why the lone tertiary hydrogen still dominates the product distribution.