Two hydrocarbons, $\mathrm{R{-}CH_3}$ and $\mathrm{R'{-}CH_3}$, each react on their own with bromine when ultraviolet light is present. Among these hydrocarbons, one is unsaturated. In both cases, free radical substitution occurs. $\mathrm{R}$ is $\mathrm{CH_3CHC(CH_3)}$. $\mathrm{R'}$ is $\mathrm{CH_3CH(CH_3)CH_2}$. Which row gives the correct answer?
- APropagation stage for the saturated hydrocarbon: $\mathrm{R{-}CH_2^{\bullet} + Br^{\bullet} \rightarrow R{-}CH_2Br}$ ; Termination stage for the unsaturated hydrocarbon: $\mathrm{R'{-}CH_2^{\bullet} + Br^{\bullet} \rightarrow R'{-}CH_2Br}$
- BPropagation stage for the saturated hydrocarbon: $\mathrm{R{-}CH_2^{\bullet} + Br_2 \rightarrow R{-}CH_2Br + Br^{\bullet}}$ ; Termination stage for the unsaturated hydrocarbon: $\mathrm{R'{-}CH_2^{\bullet} + Br_2 \rightarrow R'{-}CH_2Br + Br^{\bullet}}$
- CPropagation stage for the saturated hydrocarbon: $\mathrm{R{-}CH_3 + Br^{\bullet} \rightarrow R{-}CH_2^{\bullet} + HBr}$ ; Termination stage for the unsaturated hydrocarbon: $\mathrm{2R'{-}CH_2^{\bullet} \rightarrow R'{-}CH_2CH_2{-}R'}$
- DPropagation stage for the saturated hydrocarbon: $\mathrm{2R'{-}CH_2^{\bullet} \rightarrow R'{-}CH_2CH_2{-}R'}$ ; Termination stage for the unsaturated hydrocarbon: $\mathrm{R{-}CH_3 + Br^{\bullet} \rightarrow R{-}CH_2^{\bullet} + HBr}$