Final answer:
The iodide ion (I-) is the best leaving group in a substitution reaction of an alkyl halide due to its low basicity and larger size, which allows for better stabilization of the negative charge upon leaving.
Step-by-step explanation:
The best leaving group in a substitution reaction of an alkyl halide among the choices provided (F-, Cl-, Br-, I-) is iodide (I-). This is due to the effect of the leaving group on the rate of the reaction. Good leaving groups are generally weaker bases, and among halogens, iodide is the weakest base, making it the best leaving group. The propensity for halogens to be good leaving groups increases as you go down the periodic table from fluorine to iodine. The trend in basicity of the halide ions is F- > Cl- > Br- > I-, and thus the ability to be a good leaving group follows the opposite trend: F- < Cl- < Br- < I-.
In addition to its low basicity, iodide is a larger anion compared to fluoride, chloride, and bromide, which leads to better stabilization of the negative charge when it departs from the carbon atom during the substitution reaction.