Question

Leaving Group Tendency of Iodide

I− size is big.
I− has stable negative charge.
⟹ I− is good leaving group.
Nucleophilicity of Iodide
I− size is big.
I− has high polarizability because it has many electrons that are distant from the nucleus.
⟹ I− is good nucleophile.
⟹ I− is good leaving group and also good nucleophile because of big size. How can both be correct and valid; 'stable' and 'can be highly polarized'? Am I thinking wrong?

Answer 1

Iodide (I-) is a good leaving group due to its low basicity and stability with a negative charge, and it's also a good nucleophile as its large size and high polarizability enable it to freely form new bonds with electrophiles.

Step-by-step explanation:

The observation that iodide (I-) is both a good leaving group and a good nucleophile might seem counterintuitive, but it's correct due to its unique properties. As a leaving group, stability of the iodide ion is paramount. Iodide is less basic compared to other halides, rendering it more stable with a negative charge.

Let's take the reaction of methyl iodide (CH3-I) with a nucleophile in an S2 reaction as an example. Here, iodide is the best-leaving group due to its low basicity and high stability when it gains a negative charge. On the other hand, the large size and high polarizability of iodide also make it highly reactive towards positively charged or electron-deficient centers, qualifying it as a good nucleophile.