Leaving Group Tendency of Iodide

  • $\ce{I-}$ size is big.

  • $\ce{I-}$ has stable negative charge.

$\implies$ $\ce{I-}$ is good leaving group.

Nucleophilicity of Iodide

  • $\ce{I-}$ size is big.

  • $\ce{I-}$ has high polarizability because it has many electrons that are distant from the nucleus.

$\implies$ $\ce{I-}$ is good nucleophile.

$\implies$ $\ce{I-}$ 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?

  • 2
    $\begingroup$ Well, $\ce{F2}$ is perfectly stable. And highly reactive. $\endgroup$
    – Poutnik
    Commented Jun 5, 2023 at 9:47

2 Answers 2



Because smaller anions are very well solvated in polar protic solvents, such as water and methanol, iodide is a very good nucleophile:

$$ \textbf{Polar Protic Solvents}\\ \begin{align} \text{Solvation: }& \ce{I- < Cl- < Br- < F-}\\ \text{Nucleophilicity: }& \ce{I- > Cl- > Br- > F-} \end{align} $$

For polar aprotic solvents such as DMSO, DMF, and acetonitrile that do not solvate anions nearly as well and the cations, still, to an appreciable extent, the order of nucleophilicity reverses:

$$ \textbf{Polar Aprotic Solvents}\\ \begin{align} \text{Solvation: }& \ce{I- > Cl- > Br- > F-}\\ \text{Nucleophilicity: }& \ce{I- < Cl- < Br- < F-} \end{align} $$

Leaving Group Tendency

Leaving group tendency is largely decided on the basicity of the group. Less basic groups make better leaving groups. Thus, order of leaving group tendency follows the same trend in any solvent.

$$ \begin{align} \text{Basicity: }& \ce{I- < Cl- < Br- < F-}\\ \text{Leaving Group Tendency: }& \ce{I- > Cl- > Br- > F-} \end{align} $$


  1. Solomons T. W. G., Fryhle C. B. (2011). Organic Chemistry, 10th ed. John Wiley & Sons.
  2. LibreTexts Chemistry: Nucleophilicity and Solvent Effects

Iodide can be both a good leaving group and also a good nucleophile. By heterolytic cleavage, the iodide ion would leave off the molecule with a net –1 charge.

Take the example of $\ce{H-I}$ which may explain both these effects. Since iodine has a larger size, the longer bond length of $\ce{H-I}$ causes an easy dissociation of it and iodide can easily leave. After leaving, it acquires an extra negative charge. Iodide has larger size to carry this extra charge that provides some stability. Moreover, the extra electron resides in 5p orbital which is outermost shell for iodine atom. This causes a significantly lesser $\ce Z_{eff}$ on it and readily available for an electrophile to attack. You may say that the combination of larger size and low $\ce Z_{eff}$ (consequently lesser negative $\ce \Delta_{eg} =-296$ kJ/mol for iodine) in iodide makes it a better nucleophile.

'stable' and 'can be highly polarized'?

It can be stable and highly polarised if the conditions are just right. Aqueous medium is way to stabilize the polar molecules such as hydronium ion that goes too far to stabilize by making very large associations.

  • 2
    $\begingroup$ Re. your first sentence, that's not correct — iodide is a good leaving group. Where did you get this from? $\endgroup$ Commented Jun 5, 2023 at 12:29
  • $\begingroup$ I was referring to molecules where iodine has either no charge or partial charge such as HI. That is also my example. Maybe I would edit that to include iodide too. $\endgroup$ Commented Jun 5, 2023 at 12:33
  • 1
    $\begingroup$ Yes, but it leaves with a negative charge, so iodide is the leaving group. This is standard usage in every textbook and paper I've seen. $\endgroup$ Commented Jun 5, 2023 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.