(A) The chlorination of isobutane $(CH_3)_3CH$ yields two isomeric monochloroalkanes: $2$-chloro-$2$-methylpropane $(A)$ and $1$-chloro-$2$-methylpropane $(B)$.
The reaction proceeds via a free radical mechanism. The stability of the intermediate free radical determines the rate of formation of the product.
The $3^{\circ}$ free radical (formed at the $3^{\circ}$ carbon) is more stable than the $1^{\circ}$ free radical (formed at the $1^{\circ}$ carbon).
Although there are nine $1^{\circ}$ hydrogen atoms and only one $3^{\circ}$ hydrogen atom,the higher reactivity of the $3^{\circ}$ hydrogen towards chlorination results in a significant yield of the $3^{\circ}$ product $(A)$.
Therefore,the composition of the products is determined by both the number of equivalent hydrogen atoms and the relative reactivity of the $C-H$ bonds $(3^{\circ} > 2^{\circ} > 1^{\circ})$.