# How to determine the order of nucleophilicity for given chemical species?

How to determine the order of nucleophilicity for given chemical species?

Like I came across this question, to rearrange $\ce{RCOO-, OR-, OH-, H2O}$ (alkyl acetate, alkoxide, hydroxide, and water) in decreasing order of nucleophilicity and it ranked $\ce{OR-}$ first.

Shouldn't it be $\ce{OH-}$ since $\ce{OR-}$ would be more sterically crowded?

• It might be but if it's simple alkyl group then inductive effect is usually more important. – Mithoron Aug 12 '16 at 20:32
• I guess if this question attempts to become a canonical question on how to determine the order of nucleophilicity for a given chemical species, then a better answer with more elaboration on steric effects/electronic effects/etc. would be necessary, at least more than the currently accepted answer. – Gaurang Tandon May 18 '18 at 4:27

Within a group of nucleophiles with the same atom, the nucleophilicity decrease with decreasing basicity of nucleophile.

Decreasing basicity means the decreasing affinity of electron pair for a proton. The decreasing order of nucleophilicity is shown in the figure.

But the relationship between nucleophilicity and basicity can be reversed by steric effect. Less basic but steric unhindered nucleophile therefore have higher nucleophilicity than strong basic but strically hindered nucleophile.

In this case you can look at the pKa values of their conjugate acids. Higher pKa of conjugate acid means a weaker acid hence a stronger base. The pKa of water is 15.7, while most alcohols have pKa of 16-18.

I guess this question is asking for a general trend. Although alkoxide could be sterically hindered, for which an extreme case would be a tertiary alkoxide, but if the electrophile is not hindered at all (e.g. primary alkyl halide), then OR⁻ is still stronger than OH⁻.

In addition to looking at pKa's, you must also consider the stabilizing effects of the surrounding groups. RO- has an alkyl group attached, allowing a greater amount of polarizability. This means Oxygen's loan pairs will be more readily available to react (ie they will not be held in as close and as tight) in RO- than in HO-. Consider Fluoride vs Iodide, the same concept applies. Fluorine has a smaller atomic radius and thus less polarizability (less area to distribute the charge over) than Iodide. Fluroide holds onto it's loan pairs much tighter than Iodide, and therefore the loan pairs are much less available to react.

Another great example of this, consider sulfides. They are generally better nucleophiles than similar compounds that contain oxygen instead of sulfur. This is due to the larger atomic radius of sulfur and therefore greater polarizability/charge distribution.

Also, negatively charged species are almost always more nucleophilic than neutral species (when considering the charged form vs the neutral form of the same molecule, such as water vs hydroxide or ammonia vs amide).

OR¯ is NOT RO¯ !!! Clearly, R¯ is an unknown species and nothing else can be said about it. OR¯ would be an unusual structure, given oxygen's electronegativity compared to carbon's. In the case of RO¯, the R "group" would have to be very large and bulky to dramatically sterically affect (hide) the O¯ from other molecules. So for simple organic compounds, the effect will be quite small. As already said pKa of the conjugate acid would be an excellent guide to nucleophilicity (in the relevant solvent).

• sigh OH- = HO- doesn't it? So where did you get this idea? – Mithoron Aug 13 '16 at 0:00