Final answer:
Changing the leaving group in 2-chloro-2-methylpropane to bromide would make the reaction go faster because bromide is a weaker base. Hence, the correct answer is option d).
Step-by-step explanation:
If you changed the leaving group in 2-chloro-2-methylpropane to bromide, the reaction would go faster because bromide is a weaker base compared to chloride. According to the general principle in organic chemistry, good leaving groups are usually weaker bases.
The order of leaving group ability among the halogens is I¯ > Br¯ > Cl¯ > F¯, with iodide being the best and fluoride being the worst. The basicity of halide ions follows the reverse order: F¯ > Cl¯ > Br¯ > I¯. Thus, since bromide is a weaker base than chloride.
It serves as a better leaving group, leading to an increased reaction rate for a substitution or elimination process involving 2-bromo-2-methylpropane compared to 2-chloro-2-methylpropane.