Final answer:
Iodide (I-) is the best leaving group in a substitution reaction of an alkyl halide due to its ability to disperse the negative charge better than other halides such as fluoride (F-), which is the least favorable.
Step-by-step explanation:
In a substitution reaction of an alkyl halide, iodide (I-) is considered the best leaving group. This trend in leaving group ability is inverse to the basicity of the group. Hence, among the given options, Cl-, Br-, I-, and F-, iodide (I-) is the most preferred leaving group due to its weakest basicity and ability to stabilize the negative charge well compared to the others. Fluoride (F-) is the worst leaving group as it is the strongest base and does not leave easily. This can be understood because as the size of the halide ion increases from fluoride to iodide, their ability to disperse the negative charge increases, making them better leaving groups.
The trend for the halogen leaving groups in an alkyl halide substitution reaction is therefore I- > Br- > Cl- > F-, with I- being the best and F- being the least favorable.