Nucleophilicity with respect to the solvent

[Conservation]

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.

The link below 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.http://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/

However, during my lectures, my professor said that there is no such change in trend: the larger atoms such as I- will always be more nucleophilic than 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 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 H2O and H2S in polar aprotic solvent; similarly, the correct answer provided was that H2S would be a stronger nucleophile as it is larger and more polarizable.

I would appreciate it if someone could clarify this matter.

Thanks.

 

[TeethWhitener]

The overall point is one that trips up a lot of chemistry students: thermodynamics (how deep is the well?) versus kinetics (how high is the wall?). Trends in nucleophilicity might reflect thermodynamic trends, but nucleophilicity itself is kinetic, and one must keep that in mind when thinking about these problems.

For example, typically nucleophilicity tracks basicity: the more basic a species is, the better it is as a nucleophile. This makes sense from the point of view of Lewis basicity, where a base is simply an electron pair donor. (I believe this was probably your professor's point.) But remember that basicity is thermodynamic, whereas nucleophilicity is kinetic. This means that even though the product with the more basic species may ultimately be the most favored species thermodynamically (lowest overall energy), there might be a large kinetic effect (activation barrier) to prevent the reaction from happening. This is, for example, what you usually observe in the case of fluoride as a nucleophile. Fluoride ion is quite a strong base compared with the other halides (think about the strengths of their conjugate acids). In addition, the formation of an extremely strong C-F bond is usually highly energetically favorable. But fluoride can also form very strong intermolecular bonds (and equilibrium proton-sharing) with protic solvents. This means that even though, from a thermodynamic viewpoint, nucleophilic substitution should be favorable, in reality, fluoride doesn't react readily because it's locked up in its solvent cage, which has to be broken before the fluoride can react (a kinetic barrier to reaction). On the other hand, in an aprotic solvent, there's a much lower kinetic barrier because the solvent molecules are not as tightly bound to the fluoride, so fluoride tends to react faster.

(Brief aside: the unsolvated fluoride ion is predicted to be an incredibly strong base/nucleophile and quite interesting from a theoretical standpoint. Trying to realize an essentially "naked" fluoride ion for use in reactions is an active area of research right now.)

 

[DrDu]

I found a database with nucleophilicity indices:
http://www.cup.lmu.de/oc/mayr/reaktionsdatenbank/

There are also data for bromide and chloride in water, ethanol and MeCN.
While Br- is more nucleophilic in water or ethanol than Cl-, this is only marginally so in MeCN.

 

[Half Infinity]

The overall point is one that trips up a lot of chemistry students: thermodynamics (how deep is the well?) versus kinetics (how high is the wall?). Trends in nucleophilicity might reflect thermodynamic trends, but nucleophilicity itself is kinetic, and one must keep that in mind when thinking about these problems.

For example, typically nucleophilicity tracks basicity: the more basic a species is, the better it is as a nucleophile. This makes sense from the point of view of Lewis basicity, where a base is simply an electron pair donor. (I believe this was probably your professor's point.) But remember that basicity is thermodynamic, whereas nucleophilicity is kinetic. This means that even though the product with the more basic species may ultimately be the most favored species thermodynamically (lowest overall energy), there might be a large kinetic effect (activation barrier) to prevent the reaction from happening. This is, for example, what you usually observe in the case of fluoride as a nucleophile. Fluoride ion is quite a strong base compared with the other halides (think about the strengths of their conjugate acids). In addition, the formation of an extremely strong C-F bond is usually highly energetically favorable. But fluoride can also form very strong intermolecular bonds (and equilibrium proton-sharing) with protic solvents. This means that even though, from a thermodynamic viewpoint, nucleophilic substitution should be favorable, in reality, fluoride doesn't react readily because it's locked up in its solvent cage, which has to be broken before the fluoride can react (a kinetic barrier to reaction). On the other hand, in an aprotic solvent, there's a much lower kinetic barrier because the solvent molecules are not as tightly bound to the fluoride, so fluoride tends to react faster.

(Brief aside: the unsolvated fluoride ion is predicted to be an incredibly strong base/nucleophile and quite interesting from a theoretical standpoint. Trying to realize an essentially "naked" fluoride ion for use in reactions is an active area of research right now.)

Hi. Could you please resolve the case of H2O vs H2S nucleophilicity in aprotic medium? Some sources state that H2S is stronger as its electron cloud can be distorted more easily. But some state that it is H2O because O forms stronger bond. Please provide one ultimate answer.

 

[TeethWhitener]

1) Please provide the sources you allude to.

2) The words “stronger bond” imply thermodynamic considerations. As indicated in my post above, nucleophilicity is kinetic, not thermodynamic.

3) In my experience, thiols tend to be much better nucleophiles than alcohols, and hydrogen sulfide is generally a better nucleophile than water. But alkoxides (and OH-) can be incredibly powerful nucleophiles, often more so than thiolates (and SH-), and especially in aprotic solvents. But of course most of this is reaction dependent.

 

[Half Infinity]

1) Please provide the sources you allude to.

2) The words “stronger bond” imply thermodynamic considerations. As indicated in my post above, nucleophilicity is kinetic, not thermodynamic.

3) In my experience, thiols tend to be much better nucleophiles than alcohols, and hydrogen sulfide is generally a better nucleophile than water. But alkoxides (and OH-) can be incredibly powerful nucleophiles, often more so than thiolates (and SH-), and especially in aprotic solvents. But of course most of this is reaction dependent.

Thanks a lot. I got it.
One has to just check the activation energy needed to form the transition state with rhe electrophile. Not the latter considerations of how much energy WOULD be given off if reaction COULD jump the kinetic barrier.

I am writing whatever I have understood finally. Please correct me whereever I am wrong.
The following cases are in aprotic solvent:

In case of OH- and SH- , OH- has a negative charge on a MUCH smaller surface. So, there are MUCH more repulsions in the electron cloud ( these repulsions dominate over the greater attractions to the nucleus of O)(one can say that OH- is less stable than SH-). Hence it doesn't need much energy to be distorted and proceed to the transition state of bonding with an electrophile. So, the "wall" (activation energy, i.e. kinetic consideration) is lower.

In case of H2O and H2S, no negative charge initially. Now, electron cloud of S can more easily be distorted (less electronegativity i.e. lesser attractions to its nucleus). So, here H2S needs lesser activation energy for proceeding to transition state. (Although after binding with the electrophile, both can lose an H+ and become neutral. Then, bond of O would be more stable. But that is the thermodynamic part. Molecules like to be at deeper point , but only a few are capable of climbing up the higher wall. Mostly will just jump of the lower one).

In cases like R2O and R2S, R2O isn't even thermodynamically stable as it can't lose any proton after bonding. And positive on O would be thermodynamically less stable than on S ( Although this has nothing to do with nucleophilicity).

In case of OH- and NH2- , the size difference is not that large as size changes lesser along the period than down the group. So, here the greater attractions to the nucleus of O dominate over the greater repulsions among the electron cloud of O (one can say that OH- is more stable than NH2-). Hence, here NH2-needs lessee activation energy or a lower "wall". So, NH2- is stronger nucleophile.

Same in case of H2O vs NH3. N clearly needs lesser activation energy. So, better nucleophile.

If polar solvent was present then H2O would be more solvated than H2S or NH3. And OH- would be more solvated than SH- or NH2- . So, its kinetic barrier increase.

 

[TeethWhitener]

That looks fine, except I would say that it’s very unlikely you’ll get NH2- in any appreciable concentration in protic solvents.

 

[snorkack]

That looks fine, except I would say that it’s very unlikely you’ll get NH2- in any appreciable concentration in protic solvents.

That is ambiguous.
Ammonia itself is sometimes, with good reason, classified as "polar aprotic" solvent, and yet it is commonly included among the "protic" solvents. If you call it "protic" then you can have nucleophiles of NH₂^-, RNH^-, R₂N^- in solvents of ammonia and primary and secondary amines, which are arguably protic.

 

[TeethWhitener]

That is ambiguous.
Ammonia itself is sometimes, with good reason, classified as "polar aprotic" solvent, and yet it is commonly included among the "protic" solvents. If you call it "protic" then you can have nucleophiles of NH₂^-, RNH^-, R₂N^- in solvents of ammonia and primary and secondary amines, which are arguably protic.

Terms that are not quantitatively defined are bound to be at least somewhat ambiguous. For reference, DMSO is more acidic than neutral ammonia (depending on solvent, DMSO pKa~33, NH3 pKa~40), and acetone is many orders of magnitude more acidic than ammonia (pKa~27). Most people definitely consider DMSO and acetone aprotic solvents.

Intro textbooks will generally tell you that protic solvents are ones with labile protons, but the important determiner of “protic” vs “aprotic” from a kinetic standpoint is usually how likely it is to engage in hydrogen bonding. This is what determines the kinetic stability of the solvent shells. It’s also why you sometimes see solvents such as formamide described as protic, even though formamide doesn’t have labile protons (aside: formamide is even more acidic than acetone).

(Also, from a practical standpoint, I don’t think it’s a particularly good idea to run a reaction with a NH2- nucleophile in a solvent with potentially competing NR2- species.)