# How do you rank nucleophiles?

Oct 12, 2015

It depends on context, but to keep it simple, let's use consistent contexts. Normally you would have considered solvent conditions (polar protic, polar aprotic, nonpolar), so let's use one particular solvent.

So, let's just assume we're comparing things in water. The types or "classes" of nucleophiles you can have are the following:

• neutral

• SOME examples: ${H}_{2} O$, $N {H}_{3}$, $R N {H}_{2}$, $R O H$, etc.
• $\left[R N {H}_{2} \approx N {H}_{3}\right] \text{ > } \left[R O H \approx {H}_{2} O\right]$, depending on $R$
• anionic

• SOME examples: $O {H}^{-}$, $N {a}^{+} N {H}_{2}^{-}$, ${R}^{-} L {i}^{+}$, $R {O}^{-}$, etc.
• ${R}^{-} > N {H}_{2}^{-} > \left[O {H}^{-} \approx R {O}^{-}\right]$, depending on $R$

From experimental data, chemists have established that anionic nucleophiles are always better nucleophiles than their neutral counterparts, due to simply the recorded reaction rate.

In terms of distinctive "classes":

• Clearly, $O {H}^{-}$ is a better nucleophile than ${H}_{2} O$ due to a comparison of neutral vs. anionic

Within the same "class":

...you might want to consider the basicity of the head atom (using pKas can help).

• Perhaps less clearly, ${R}^{-} L {i}^{+}$ is a way better nucleophile than $R {O}^{-}$ due to a comparison of basicity in particular, between the electronegativity of $C$ vs. $O$, because its higher tendency to act as a Lewis base promotes its nucleophilicity
(note that basicity does not correlate with nucleophilicity
necessarily
)

...you might want to consider steric hindrance of the nucleophile.

• $E t {O}^{-}$ is better nucleophile than $t \text{-} B u {O}^{-}$ due to the high steric hindrance of $t \text{-} B u {O}^{-}$ with regards to backside-attack; this often is the difference between reactions being substitution or elimination, besides temperature

...you might want to consider the size of the nucleophile (not the same as steric hindrance):

• ${I}^{-}$ is a better nucleophile than ${F}^{-}$, but is also less basic than ${F}^{-}$, because ${I}^{-}$ is much bigger and thus moves more slowly