Nucleophilicity with respect to the solvent

Question

I am currently studying nucleophilicity of molecules and am getting a set of conflicting information, so I wanted to clarify nucleophilicity trends with respect to solvent.

Master Organic Chemistry discusses how nucleophilicity increases down the periodic table with protic solvents and decreases down the periodic table with aprotic solvents; the main argument is that a solvating shell will form with the protic solvent, making the smaller atoms less likely to form a bond. When this effect is removed, the trend is reversed.

However, during my lectures, my professor said that there is no such change in the trend: the larger atoms such as $\ce{I-}$ will always be more nucleophilic than $\ce{F-}$ due to the fact that it is more polarizable and likely to donate an electron pair. The textbook that I am using, Organic Chemistry by Wade, also does not discuss this trend reversal depending on the solvent. The only discussion talks about how $\ce{F-}$ would be a poor nucleophile as it is small and holds its electrons "tightly."

A practice problem provided during class also asked to compare the nucleophilic strength between $\ce{H2O}$ and $\ce{H2S}$ in polar aprotic solvent; similarly, the correct answer provided was that $\ce{H2S}$ would be a stronger nucleophile as it is larger and more polarizable.

I would appreciate it if someone could clarify this matter.

Answer 1

Actually nucleophilicity is inversely proportional to electronegativity when comparing compounds across the period, and inversely proportional to size when comparing in the same group. This condition is meant in aprotic solvent.

In protic solvents the more electronegative ions gets easily solvated to a larger extent (i.e., like a ball with the ion in the centre and the water molecules around it) so the order across the period remains same since electronegativity is not effected but the order about the group changes since the now smaller ions become the larger ones.

For example, the trend for nucleophilicity is $\ce{F- > Cl- > Br- > I-}$ in aprotic solvent but in protic solvent it is $\ce{I- > Br- > Cl- > F-}$ ($\ce{F-}$ is now larger than others in size because it is solvated.)

Answer 2

You seem to be describing the trends correctly.

Perhaps what is missing is the fact that only protic solvents form a meaningful shell. In aprotic solvents the nucleophile is "free" without a shell.

As a result of this the solvent identity is often the important factor b/c a protic solvent will "lock up" what is normally a very strong nucleohphile.

So in protic solvents like MeOH you'll usually see the neutral form be the nucleophile (H20) vs the charged, more nucleophilic form (-OH)

There are several competing factors here, solvent, nucleophilicity, charge density, and others. In an academic learning setting you won't be asked to distinguish between examples that might fall into a gray area. In a practical setting you'd just test the examples that might not be predicted well from trends.

These trends are also useful in the context of predicting whether SN1/SN2/E1/E2 will likely occur given a solvent/nucleophile combination. Sometimes this is straightforward. Sometimes it's a mixture of the reactions.