Why is the iodide anion a good nucleophile but a poor base?

Question

The $\\mathrm{pK_{aH}}$ of $\\ce I^-$ is very low which indicated that it is not favourable for it to bond with a proton. However, why would it be likely to bond with any other atom (mostly carbon when considering it's nucleophilic properties).

Answer 1

One reason as you have mentioned is the pKa. For haloacids within the same group, HF, HCl, HBr, HI, acid strength increases as you go down the group due to decreasing electronegativity of the halide ion. As a result, HF is a much weaker acid than HI.

For acid-base conjugate pairs, a general rule is:

(a) Strong acid → weak conjugate base

(b) Weak acid → strong conjugate base

As a result, I^- is a much weaker base than F^-.

The nucleophilic strength of halides depends on their ability to donate electrons (lewis base) in a substitution reaction due to size of the anion. The valence electrons (5p subshell) in I^- feel less electrostatic attraction to the protons in the nucleus as compared to F^- (2p subsell). As a result, I^- can easily donate these electrons when forming a bond with a carbon substrate resulting in a greater reactivity.

In addition, the solvent also plays a role in nucleophilicity. In general, all nucleophiles undergo solvation (solvent molecules surrounding the anion) during a reaction. Since I^- is a large anion, it undergoes less solvation compared to F^-, therefore, it is able to bond to the carbon substrate much faster.

Hope the above was helpful!