A Primer On Organic Reactions

By James Ashenhurst

Leaving Groups Are Nucleophiles Acting In Reverse

Last updated: March 27th, 2021 |

What’s A Leaving Group? A Nucleophile Acting In Reverse

We’ve covered “what makes a good leaving group” before – the weaker the base, the better the leaving group. In this short post, I just want to make an observation. Note how similar nucleophiles are to leaving groups:

Both contain a pair or electrons, and therefore both can act as Lewis bases. What determines whether one group is a nucleophile or a leaving group in a given reaction is often a question of basicity.

Just like in acid base reactions, where we go from stronger base to weaker base, in nucleophilic substitution reactions, we generally go from stronger base (nucleophile) to weaker base (leaving group) as well.


So you can think of a leaving group as a nucleophile acting in reverse.

In fact, another name for a leaving group is a nucleofuge; (“nucleus-fleeing” in Latin, as opposed to the “nucleus-loving” nucleophile.

Next: now that we’ve talked about the components of substitution reactions, let’s look into what experiments tell us about how they might work.

Next Post: Two Types of Substitution Reactions


P.S. There are a few exceptions to substitution reactions being based solely on “base strength” considerations, of course. One interesting one is based on different solubilities. The Finkelstein reaction is a way of making alkyl iodides from alkyl chlorides. There is not a tremendous difference between the leaving group ability of chloride ion or iodide ion. However, there is a significant difference between the solubility of NaCl (or NaBr) and NaI in acetone. NaI is soluble in acetone; NaCl is not. Therefore as NaCl is formed, it precipitates out of solution, driving the reaction to the right. Another example of Le Chatelier’s principle in action!



Comment section

7 thoughts on “Leaving Groups Are Nucleophiles Acting In Reverse

  1. Imagine two large entrances to a building through which people are going through one after the other continuously, i.e. in and out of the building. Now visualize that police are on the outside and are capturing people and keeping them outside. In that case, there will be fewer people going into the building than are coming out. That’s what’s happening here. The insolubility of NaCl in acetone removes the Cl- species from solution and it can no longer participate in the reaction. I- however can remain in solution and Cl- is substituted on the alkyl by I- because there are less Cl- ions in solution than I- which disturbs the equilibrium.

  2. Hi. I am going to do an alkylation on the lower rim of a calix[4]arene derivative and I cannot use a strong base due to the presence of an unstable functional group on my starting material. As a result, I am going to try a finkelstein method using Potassium carbonate and Sodium iodide, but unfortunately my compound is not soluble in acetone which is the main solvent for that reaction. I just wanted to know if it is possible to do it this way? What other choice do I have?

  3. Would the Finkelstein reaction be feasible in a polar protic solvent, since the iodide ion is a stronger nucleophile than the chloride ion (owing to solvation effect)?

    1. Maybe, but why would you want to do that? In acetone, the reaction goes to completion because NaCl crashes out of solution. If both NaCl and NaI are present in solution, my guess is that you’ll likely get a mixture. The difference in nucleophilicity isn’t that big. That said, I should look this up in March before shooting off my mouth.

  4. Why is leaving tendency related to basicity and not nucleophilicity?It is seen that, though I- is a good nucleophile it is a very good leaving group.How can this behaviour be explained in terms of nucleophilicity?

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