In the free radical monohalogenation of 2-methylpropane at one hydrogen of each kind, the relative reactivity per hydrogen (tertiary : primary) is 5 : 1 for chlorination and 1600 : 1 for bromination. The percentages of $(CH X_{3}) X_{3} C - X$ in the monohalogenated mixture for $Cl X_{2}$ and $Br X_{2}$ respectively are about

Answer: C

2-Methylpropane has 1 tertiary H and 9 primary H. Chlorination:

5/ (5 + 9) = 35.7%

. Bromination:

1600/ (1600 + 9) = 99.4%

. The bromine radical is less reactive and more selective because its hydrogen abstraction is endothermic with a late, product-like transition state.