In LG Wade's textbook, he has written that a good leaving group must be a weak base. Since substitution reactions (not too sure about elimination reactions) tend to be kinetically controlled, i.e irreversible, why has the concept of a weak base (which is defined for thermodynamically controlled reactions) been used here? Why does the leaving group have to be a weak base rather than a weak nucleophile?


It is a common misconception that good leaving groups should be weak nucleophiles. However, there is a simple counterexample: iodide is known to be both a good nucleophile and a good leaving group in $\mathrm{S_N2}$ reactions.

However, it does make sense for the leaving group to be defined in thermodynamic properties. Consider the reaction equation for a nucleophilic substitution following an $\mathrm{S_N2}$ mechanism:

$$\ce{Nu- + R3C-Lg -> [Nu\bond{...}CR3\bond{...}Lg]^\ddagger -> Nu-CR3 + Lg-}\tag{1}$$

Nucleophilicity, i.e. the ability of $\ce{Nu-}$ to attack the reaction centre is, as you noticed, a kinetic property: the nucleophile has to approach the electrophilic carbon atom from a certain direction in a certain angle with appropriately constructed orbitals. The leaving group itself does play a slight role here in determining the energy of the $\unicode{x3c3}^*$ orbital and thus whether it is energetically favourable for the nucleophile to attack.

However, the leaving group’s principal characteristic is to diffuse away after the reaction. If it is diffusing away, its kinetic properties no longer influence how the reaction proceeds as it is not attacking anything anymore. Rather, a favourable overall reaction enthalpy is achieved if $\ce{Lg-}$ is able to stabilise the negative charge well (or lone pair, if e.g. $\ce{Lg} = \ce{NR3+}$). This is indeed a thermodynamic property, not a kinetic one. And stabilising a negative charge directly correlates with being a weak base.

| improve this answer | |
  • 1
    $\begingroup$ What exactly do you mean by "diffuse away?" Also, if the leaving group is a strong nucleophile but weak base like iodide, what stops is from attacking the nucleophilic centre? $\endgroup$ – xasthor Jun 22 '17 at 14:11
  • 1
    $\begingroup$ @xasthor Nothing — which is why nucleophilic attacks that insert an iodide functionality are often re-attacked by iodide. However, in most substitutions that involve iodide, whatever the nucleophile is will be a much worse leaving group and thus the iodide cannot re-attack to induce the back reaction. ‘Diffuse away’ just means ‘stays in solution’, i.e. does not participate in the immediate reaction any more. $\endgroup$ – Jan Jun 22 '17 at 14:13
  • $\begingroup$ Is this correct: The entire reason substitution reactions are irreversible is because having a much weaker base than the attacking nucleophile as a leaving group, drives the reaction entirely to the right & hence good leaving groups are weak bases. This LG is so much worse a base than the nucleophile that the reaction is always driven to completion and only the kinetics of the reagent(i.e its nucleophilicity) matter. Hence we check the nucleophilicity of the reagent, but basicity of the LG (But this statement also implies that bad leaving groups make substitution reactions reversible) $\endgroup$ – xasthor Jun 22 '17 at 14:26
  • $\begingroup$ Side question: are elimination reactions irreversible too? $\endgroup$ – xasthor Jun 22 '17 at 14:27
  • $\begingroup$ @xasthor Let me start by pointing out that $\text{kinetically controlled} \ne \text{irreversible}$. Substitution reactions are typically irreversible either because the leaving group is a bad nucleophile or because the nucleophile is a bad leaving group. But forward and backward reactions need to be examined separately. It may even be that upon change of conditions (e.g. lowering of pH) a substitution can be selectively reversed. $\endgroup$ – Jan Jun 22 '17 at 14:37

LG Wade has put those concepts in another direction, but his explanation is valid and have put it across in the following discussion:

What makes a good leaving group?

When the C-X bond breaks in a nucleophilic substitution, the pair of electrons in the bond goes with the leaving group. In this way, the leaving group is analogous to the conjugate base in a Brønsted-Lowry acid-base reaction. (In order to act as a proton acceptor, a base must have a reactive pair of electrons)

enter image description here

Image illustrating the congujate base-conjugate acid formation

When evaluating the stability of the conjugate base that resulted from the proton transfer, basic concepts used include:

  • resonance - delocalization of electron density has a stabilizing effect, and the greater area over which the delocalization is possible, the greater the stabilization.

  • inductive effects e.g. electron withdrawing groups such as chlorine help to further spread out the electron density of the conjugate base, has a stabilizing effect.

These are the same principles when dealing with leaving groups.

In other words, the trends in basicity are parallel to the trends in leaving group potential - the weaker the base, the better the leaving group. Just as with conjugate bases, the most important question regarding leaving groups is this: when a leaving group leaves and takes a pair of electrons with it, how well is the extra electron density stabilized?

In laboratory synthesis reactions, halides often act as leaving groups. Iodide, which is the least basic of the four main halides, is also the best leaving group – it is the most stable asa negative ion. Fluoride is the least effective leaving group among the halides, because fluoride anion is the most basic.

Anomaly trend of fluorine explained

The more electronegative an atom, the better it is able to bear a negative charge. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion.

But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom outdoes its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodine ion, the negative charge is spread out over a significantly larger volume)

Some examples:

Chlorides, bromides, and tosylate / mesylate groups are excellent leaving groups in nucleophilic substitution reactions, due to resonance delocalization of the developing negative charge on the leaving oxygen.

enter image description here


Organic Chemistry with a Biological Emphasis (T Soderberg)

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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