How to determine order of nucleophilicity among commonly used nucleophiles? For example, in Finklestein reaction, $\ce{NaI}$ (reagent) - taken in acetone (solvent) - displaces $\ce{SH-}$ ion from alkyl group via SN2 mechanism to form an alkyl iodide. But in a separate question we may be asked to write the product formed when $\ce{NaSH}$ reacts with an alkyl iodide in the presence of acetone. Since I do not know which of the two is a stronger nucleophile, I will displace $\ce{I-}$ and form $\ce{alkyl-SH}$. But definitely only one of the reactions shall occur.

So, my question is that is there an order of strength of nucleophilicity (like a metal reactivity series)? I am preparing for JEE Advanced 2015 (college entrance exam in India), so I am expecting answers within its scope.

  • $\begingroup$ "(...) in Finklestein reaction, NaI (reagent) taken in acetone(solvent) displaces SH− ion from alkyl group via SN2 mechanism to form alkyl iodide" this certainly won't happen. en.wikipedia.org/wiki/Finkelstein_reaction $\endgroup$
    – Mithoron
    Aug 14, 2015 at 15:45
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    $\begingroup$ IMO your problem is not that much order of nucleophilicity as difference betweennucleophile and leaving group $\endgroup$
    – Mithoron
    Aug 14, 2015 at 16:33

3 Answers 3


The relative strengths of nucleophiles can be correlated with three structural features:

  • A negatively charged nucleophile is always a more reactive nucleophile than its conjugate acid. Thus $\ce{HO^-}$ is a better nucleophile than $\ce{H_2O}$ and $\ce{RO^-}$ is better than $\ce{ROH}$.
  • In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicities parallel basicities. Oxygen compounds, for example, show the following order of reactivity: $$\ce{RO^- > HO^- >> RCO_2^- > ROH > H2O}$$ This is also their order of basicity. An alkoxide ion ($\ce{RO-}$) is a slightly stronger base than a hydroxide ion ($\ce{HO^-}$), a hydroxide ion is a much stronger base than a carboxylate ion ($\ce{RCO2^-}$), and so on.
  • When the nucleophilic atoms are different, nucleophilicities may not parallel basicities. For example, in protic solvents $\ce{HS, CN}$, and $\ce{I}$ are all weaker bases than $\ce{HO^-}$, yet they are stronger nucleophiles than $\ce{HO^-}$. $$\ce{HS^- > CN^- > I^- > HO^-}$$

Nucleophilicity versus Basicity: While nucleophilicity and basicity are related, they are not measured in the same way. Basicity, as expressed by $\mathrm{p}K_\mathrm{a}$, is measured by the position of an equilibrium involving an electron pair donor (base), a proton, the conjugate acid, and the conjugate base. Nucleophilicity is measured by relative rates of reaction, by how rapidly an electron pair donor reacts at an atom (usually carbon) bearing a leaving group.

For example, the hydroxide ion $\ce{OH^-}$ is a stronger base than a cyanide ion $\ce{CN^-}$; at equilibrium it has the greater affinity for a proton ($\mathrm{p}K_\mathrm{a}(\ce{H_2O})= 16$, while $\mathrm{p}K_\mathrm{a}(\ce{HCN})= 10$). Nevertheless, cyanide ion is a stronger nucleophile; it reacts more rapidly with a carbon bearing a leaving group than does hydroxide ion.

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    $\begingroup$ The answer is from Organic Chemistry by Solomons, Fryhle, Snyder. $\endgroup$
    – Vishnu
    Oct 13, 2019 at 13:50
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    $\begingroup$ Did you use an OCR or typed the whole thing from the book? $\endgroup$
    – Vishnu
    Oct 13, 2019 at 13:51

This article most probably clear all your doubts on the topic: Types Of Reactions with Alkyl Halide

As a supplement :

  1. This would clear your concept of strength of nucleophile- What Makes a good nucleophile
  2. And this will teach you how to compare nucleophilicity Comparison of nucleophilicity

I would recommend you go through these links and clarify your doubts.


Hint: nucleophilicity is different from leaving group strength.

In fact, both are possible it depends what drives the reaction. Though it usually is not possible to displace $\ce{SH^-}$ ion, but precipitation drives it.

UCDavis article on nucleophilicity and solvent effects

  • $\begingroup$ Yes, I do know that precipitation of Na salt is the dr $\endgroup$
    – Kinjal
    Aug 4, 2014 at 16:44

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