Two hydrocarbons, $\text{R-CH}_3$ and $\text{R'-CH}_3$, react separately with bromine in the presence of ultraviolet light. Of these two hydrocarbons, one is unsaturated. In each reaction, free radical substitution takes place. $\text{R}$ is $\text{CH}_3\text{CHC(CH}_3)$ . $\text{R'}$ is $\text{CH}_3\text{CH(CH}_3)\text{CH}_2$ . Which row gives the correct answer?
- ASaturated hydrocarbon propagation: $\text{R-CH}_2^{\bullet} + \text{Br}^{\bullet} \rightarrow \text{R-CH}_2\text{Br}$ ; Unsaturated hydrocarbon termination: $\text{R'-CH}_2^{\bullet} + \text{Br}^{\bullet} \rightarrow \text{R'-CH}_2\text{Br}$
- BSaturated hydrocarbon propagation: $\text{R-CH}_2^{\bullet} + \text{Br}_2 \rightarrow \text{R-CH}_2\text{Br} + \text{Br}^{\bullet}$ ; Unsaturated hydrocarbon termination: $\text{R'-CH}_2^{\bullet} + \text{Br}_2 \rightarrow \text{R'-CH}_2\text{Br} + \text{Br}^{\bullet}$
- CSaturated hydrocarbon propagation: $\text{R-CH}_3 + \text{Br}^{\bullet} \rightarrow \text{R-CH}_2^{\bullet} + \text{HBr}$ ; Unsaturated hydrocarbon termination: $2\text{R'-CH}_2^{\bullet} \rightarrow \text{R'-CH}_2\text{CH}_2\text{-R'}$
- DSaturated hydrocarbon propagation: $2\text{R'-CH}_2^{\bullet} \rightarrow \text{R'-CH}_2\text{CH}_2\text{-R'}$ ; Unsaturated hydrocarbon termination: $\text{R-CH}_3 + \text{Br}^{\bullet} \rightarrow \text{R-CH}_2^{\bullet} + \text{HBr}$