The trend of halide nucleophilicity in polar protic solvents is $$\ce{I- > Br- > Cl- > F-}$$

The reasons given by Solomons and Fryhle[1], and by Wade[2] are basically as follows.

  1. Smaller anions are more solvated than bigger ones because of their 'charge to size ratio'. Thus, smaller ions are strongly held by the hydrogen bonds of the solvent.

  2. Larger atoms/ions are more polarizable. Thus, larger nucleophilic atoms/ions can donate a greater degree of electron density to the substrate than a smaller nucleophile whose electrons are more tightly held. This helps to lower the energy of the transition state.

Clayden et al.[3] give the reason that in SN2 reactions, the HOMO–LUMO interaction is more important than the electrostatic attraction. The lone pairs of a bigger anion are higher in energy compared to a smaller anion. So, the lone pairs of a bigger anion interact more effectively with the LUMO σ*. Thus, soft nucleophiles react well with saturated carbon.

What I don't understand is the trend of halide nucleophilicity in polar aprotic solvents.

$$\ce{F- > Cl- > Br- > I-}$$

In aprotic solvents, only the solvation factor is absent. Bigger anions are still more polarizable and can interact in a better way with the LUMO. Then, what is the reason for the trend reversal in aprotic solvents? Solomons and Fryhle[1] say

In these[aprotic] solvents anions are unencumbered by a layer of solvent molecules and they are therefore poorly stabilized by solvation. These “naked” anions are highly reactive both as bases and nucleophiles. In DMSO, for example, the relative order of reactivity of halide ions is opposite to that in protic solvents, and it follows the same trend as their relative basicity.

Shouldn't the orbital interactions matter more than the electrostatic attractions even in polar aprotic solvents? Why does nucleophilicity follow the basicity trend in polar aprotic solvents?


  1. Solomons, T. W. Graham; Fryhle, C. B. Organic Chemistry, 10th ed.; Wiley: Hoboken, NJ, 2011; pp. 258–260.

  2. Wade, L. G. Organic Chemistry, 8th ed.; Pearson Education: Glenview, IL, 2013; pp. 237–240.

  3. Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012; pp. 355–357.

  • $\begingroup$ I have resigned to the fact that in aprotic solvent the trend is explained by orbital sizes. $\endgroup$ – Avyansh Katiyar May 11 '18 at 7:52
  • $\begingroup$ What do you think of my answer @ApoorvPotnis ? $\endgroup$ – Tan Yong Boon Jan 5 '19 at 2:23
  • $\begingroup$ @TanYongBoon I really like it. I'll award you a bounty once I gain some reputation. So, can one say that the unequal solvation in polar protic solvents makes the trend reversal? $\endgroup$ – Apoorv Potnis Jan 6 '19 at 13:03
  • $\begingroup$ @TanYongBoon Also, sorry for the late reply. $\endgroup$ – Apoorv Potnis Jan 6 '19 at 13:14
  • 1
    $\begingroup$ Yes that is what I'm saying. The cause of the trend reversal is the different solvation effects in the 2 different environments. $\endgroup$ – Tan Yong Boon Jan 6 '19 at 22:56

This is a rather intellectually-stimulating question and one that is also very difficult to answer. You have constructed a very good case for why the nucleophilicity order would not be expected to reverse in polar aprotic solvents, i.e. the order of intrinsic nucleophilicities of the halide ions should be $\ce {I^- > Br^- > Cl^- > F^-}$, based on the concepts of frontier molecular orbital interactions, as well as electronegativity and polarisability of the nucleophilic atom.

Affirming your viewpoint

Admittedly, the saturated carbon atom is a soft centre and based on electronegativity considerations, the energy of the $\ce {HOMO}$ of the halide ions should increase from $\ce {F^-}$ to $\ce {I^-}$, i.e. the hardness of the the halide ions as nucleophiles decreases from $\ce {F^-}$ to $\ce {I^-}$. Based on the hard-soft acid-base (HSAB) principle, we would expect the the strength of the interaction between the halide ion and the carbon centre to increase from $\ce {F^-}$ to $\ce {I^-}$.

We can provide more quantitative justification for this using the Klopman-Salem equation$\ce {^1}$:

$$\Delta E=\frac{Q_{\text{Nu}}Q_{\text{El}}}{\varepsilon R}+\frac{2(c_{\text{Nu}}c_{\text{El}}\beta)^2}{E_{\text{HOMO}(\text{Nu})}\pm E_{\text{LUMO}(\text{El})}}$$

The first term is indicative of the strength of the Coulombic interaction while the second term is indicative of the strength of the orbital interactions. In the reaction with the soft saturated carbon, it is actually expected for the reactivity to increase from $\ce {F^-}$ to $\ce {I^-}$ (ref 1, p. 117).

At the moment, all the evidence presented points us towards the conclusion that you have posited. However, we have overlooked the importance of a thermodynamic factor - the strength of the $\ce {C-X}$ bond.

A not-particularly-relevant add-on

Actually, if you think about it the considerations made in the Klopman-Salem equation are essentially kinetic considerations. We cannot conclude anything about reaction energetics from such considerations. The strength of the $\ce {HOMO-LUMO}$ interaction or that of the Coulombic attraction tells us nothing about how the electrons would organise themselves subsequently, what would be the strength of the resultant bonds etc.

A comment made on this by Olmstead & Brauman (1977)

On this topic of explaining the relative order of intrinsic nucleophilicity of the halide ions, Olmstead & Brauman (1977) had this to say$\ce {^2}$:

The intrinsic nucleophilicities follow the reverse order of the polarizabilities (e.g., $\ce {CH3O^-}$ > $\ce {CH3S^-}$ and $\ce {F^-}$ > $\ce {Cl^-}$ > $\ce {Br^-}$). This could be due to a stronger interaction between the more concentrated molecular orbitals of the anion with the carbon center. It could also be simply a reflection of the greater thermodynamic methyl cation affinities of the smaller anions.

It is important to note that Olmstead & Brauman are merely postulating the reasons. However, I do feel that the point on thermodynamics leads us on the right track to answering this question. On the other point of "more concentrated MOs", I am unclear of how it can be interpreted.

Let us focus now on this point.

How does thermodynamics even influence kinetics?

Well... It certainly does here. What are we concerned with when dealing with nucleophilicity? As it is a kinetic phenomenon, we are essentially concerned with the rate of the substitution reaction. This makes the consideration of the activation energy ($E_\mathrm{a}$) a particularly important one because the higher the $E_\mathrm{a}$, the lower the rate of the substitution reaction.

Here, the $E_\mathrm{a}$ can be taken to be the difference in energy between the reactants and the transition state. Thus, we can decrease the $E_\mathrm{a}$ by either raising the energy level of reactants or decreasing the energy level of the transition state. The reaction energy profile for an $\ce {S_N2}$ reaction is shown below, taken from ref. 3 (p. 394).

enter image description here

There are two ways to see how the strength of the $\ce {C-X}$ bond can affect the $E_\mathrm{a}$. I am not sure if they may be considered equivalent perspectives. Nonetheless, I will present both.

The first perspective is well-explained by Francis & Sundberg (2007) with regard to the $\ce {S_N2}$ reaction$\ce {^3}$:

Because the $\ce {S_N2}$ process is concerted, the strength of the partially formed new bond is reflected in the TS. A stronger bond between the nucleophilic atom and carbon atom results in a more stable TS and a reduced activation energy.

Another perspective (which may be somewhat similar to the first, or even possibly the same) is that the transition state structure bears resemblance to the product structure, based on the Hammond's postulate. However, this would be more applicable for an endothermic $\ce {S_N2}$ reaction where the transition state structure does indeed bear more resemblance to the product. When the product is more stable, i.e. has bonds that form more exothermically, the energy of the transition state would thus be lowered, and the $E_\mathrm{a}$ would also decrease, although not necessarily proportionally.


The argument you have presented failed to consider the important thermodynamics factor. Thus, it resulted in the predicted reactivity of the halide ions in the reversed order. The takeaway here is that the application of the $\ce {HSAB}$ concept alone does not always produce the correct prediction.

Additional information

More examples are cited for which the HSAB concept fails. These include:

  • The favourable combination of $\ce {H^+}$, a hard acid, and $\ce {H^-}$, a soft base (discussed here as well) (ref. 4).
  • The fact that the rate of reaction between $\ce {Ag^+}$, soft acid, and an alkene, soft base, is slower than that between $\ce {Ag^+}$ and $\ce {OH^-}$, a hard base (ref. 1, p. 110)


  1. Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Student Edition). John Wiley & Sons, Ltd. United Kingdom, 2009.
  2. Olmstead, W. N.; Brauman, J. I. Gas-phase nucleophilic displacement reactions. J. Am. Chem. Soc. 1977, 99(13), 4219–4228.
  3. Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.
  4. Pearson, R. G.; Songstad, J. Application of the Principle of Hard and Soft Acids and Bases to Organic Chemistry. J. Am. Chem. Soc. 1967, 89(8), 1827-1836.
| improve this answer | |
  • $\begingroup$ Upvoted for invoking Salem-Klopman. $\endgroup$ – Zhe Jan 5 '19 at 2:56

In polar aprotic solvent, there is no possibility of formation of a hydrogen bond, to reduce the nucleophilicity of a nucleophile.

The order of basicity of $\ce{F-,Cl-,Br-,I-}$ are, $\ce{F- > Cl- > Br- > I-}$ (since the order of acidity of the hydrogen halides are $\ce{HI > HBr > HCl > HF}$).

So, among halide ions, $\ce{F-}$ has the highest ability to donate its electron pair, and thus, act as a nucleophile.

Also, $\ce{F-}$ being the smallest faces less steric hindrance as a nucleophile.

So, the order of nucleophilicity in polar aprotic solvent is, $$\ce{F- > Cl- > Br- > I-}$$

| improve this answer | |
  • $\begingroup$ Why does $\ce{F-}$ have the maximum capacity to donate its electron pair? Larger ions should be more polarizable and polarizability favours $\mathrm{S_N 2}$ reactions. $\endgroup$ – Apoorv Potnis Feb 24 '18 at 13:22
  • $\begingroup$ If you consider the Bronsted-Lowry definition of bases, the stronger bases have higher ability to donate electron pair $\endgroup$ – Archisman Panigrahi Feb 24 '18 at 13:24
  • $\begingroup$ But basicity doesn't always parallel nucleophilicity; e.g. in polar protic solvents. Does it in polar aprotic solvent? If yes, why? $\endgroup$ – Apoorv Potnis Feb 24 '18 at 13:27
  • $\begingroup$ First of all, the basicity judgement by the ability of donating electron lone pairs is Lewis definition. Secondly, being most electronegative shouldn't $\ce{F^-}$ try to keep electrons on itself rather than donating (according to the concept of electronegativity). How would you account for that ? $\endgroup$ – Soumik Das Feb 24 '18 at 13:30
  • $\begingroup$ @ApoorvPotnis Generally, in polar aprotic solvents, nucleophilicity follows basicity. $\endgroup$ – Archisman Panigrahi Feb 24 '18 at 13:46

In polar protic solvents since cations and anions are well solvated , attractive forces among the ions rupture due to which anions with higher solvation , i.e. high charge density have lower nucleophilicity than the anions with lower solvation, i.e. of lower charge density.

e.g., in $\ce{H2O}$ order of nucleophilicity of halides is $\ce{F− < Cl− < Br− < I−}$.

In polar aprotic solvents, only cations get solvated but not anions hence anion which forms stronger bond with carbon has higher nucleophilicty.

e.g., in $\ce{DMF}$ and $\ce{DMSO}$ order of nucleophilicity of halides is $\ce{F− >Cl− >Br− >I−}$.

| improve this answer | |
  • $\begingroup$ Acetone is also an aprotic polar solvent, then why is the order $\ce{Cl^-< Br^-<I- }$ in it? $\endgroup$ – Archer Aug 5 '18 at 16:49
  • $\begingroup$ I don't think your answer is answering the question. You are essentially answering a different question, which Apoorv Potnis has already answered in his question write-up. Your point on solvation is the bullet point 1 in the write-up. $\endgroup$ – Tan Yong Boon Jan 6 '19 at 0:43

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.