Why does reduction using sodium borohydride require a protic solvent?

Question

Recently, I was looking at the chapter on chemoselectivity in Clayden et al.'s Organic Chemistry (2nd ed.). On p. 530, it is mentioned that sodium borohydride reduction only occurs in protic solvents or in the presence of electrophilic metal ions, such as $\ce {Li+}$ and $\ce {Mg^2+}$ while the more powerful reducing agent lithium aluminium hydride does not require protic solvents. Why is that the case?

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

Answer 1

The roles of a protic solvent and lithium and magnesium salts can be tied together in their role as electrophiles—and they impact both solubility and reactivity.

Protic Hydrogen's evil twin

Sodium borohydride may require (or at least favor) protic solvents for good solubility, but lithium borohydride is soluble up to several percent in diethyl ether[1]. The presence of lithium ions, as well as protic hydrogen, appears to enhance the solubility of sodium borohydride, and we can expect magnesium with its diagonal-relationship similarity to lithium to do the same.

Among ions that are reasonably stable in most solvents, borohydride is unique in having hydrogen with a significant negative partial charge. With that comes boron-hydrogen bonding orbitals that are relatively strongly concentrated on the hydrogen atoms and relatively high in energy levels. Thus by both electrostatic and molecular-orbital renderings of hydrogen bonding, the hydridic hydrogens can serve as relatively good electron donors for hydrogen bonding. This interaction between hydridic and protic hydrogen in described for a magnesium borohydride-ammonia complex in Yan et al[2]. Hence the preference of borohydride for protic solvents. Lithium and magnesium ions, whose hydride salts have greater covalent character than hydrides of heavier alkali or alkaline earth metals, can also serve as electron acceptors as well. Thereby borohydride ion is drawn into solution by protic hydrogen, lithium ions or magnesium ions.

Both ends against the middle

In this answer the role of magnesium ion is Grignard reactions is described. This ion acts as an electrophile, attaching to the oxygen and polarizing the carbonyl group and thus assisting the nucleophilic attack. Magnesium ion from a salt (e.g., $\ce{MgCl2}$}, lithium from a salt (e.g.,$\ce{LiCl}$) or organolithium compound (e.g., $\ce{R2CuLi}$), or a proton transferred from a protic solvent, can have similar effects in making a carbonyl group more reactive.

References

1. Mikheeva, V.I., Troyanovskaya, E.A. Solubility of lithium aluminum hydride and lithium borohydride in diethyl ether. Russ Chem Bull 20, 2497–2500 (1971). https://doi.org/10.1007/BF00853610

2. Yigang Yan, Jakob B. Grinderslev, Mathias Jo̷rgensen, Lasse N. Skov, Jo̷rgen Skibsted, and Torben R. Jensen. "Ammine Magnesium Borohydride Nanocomposites for All-Solid-State Magnesium Batteries". ACS Appl. Energy Mater. 2020, 3, 9, 9264–9270.https://doi.org/10.1021/acsaem.0c01599

Answer 2

BACKGROUND

This is the mechanism of the sodium borohydride reduction.

The article goes on to say why $\ce{NaBH4}$ is used:

Use of an alcohol as solvent does two things to help accelerate the $\ce{NaBH4}$ reduction.

1. The protic solvent may hydrogen bond with the carbonyl oxygen, further polarizing the pi-bond, making it more receptive to nucleophilic attack by sodium borohydride.
2. The solvent itself may attack the $\ce{BH3}$ formed after step (1) of the reaction, placing 4 groups on boron once again. Since the solvent is constantly surrounding the boron complex, attack of solvent on $\ce{BH3}$ will occur much more quickly than carbonyl attack as had to occur in an aprotic solvent. These two changes aid in increasing the kinetics of hydride addition using sodium borohydride a great deal!

[1]

This was relevant but did not answer the question, so basically 'naan thookiten' (I removed it)

---

Now also it is known that carbonyls love the electrophilic addition reaction.

Any polar compound like $\ce{LiCl, MgCl2}$ you give - it gets added very readily.

This picture here: it tells us how carbonyl groups are affected by Grignard reagents. Metal ions like $\ce{Li+, Mg^2+}$ make the carbonyl undergo this effect. The resulting adduct will make it prone to substitution and on hydrolysis, this adduct will lead to alcohol formation.

I can say that the same holds for protic solvents with $\ce{H+}$.

---

Principally solubility. Sodium borohydride is essentially insoluble in aprotic solvents. - Waylander

As Waylander rightly pointed out: Sodium borohydride is insoluble in aprotic solvents. Now we use its lithium counterpart (i.e., $\ce{LiAlH4}$). It is soluble in ether. The Al-H bond is much weaker, and thus, ionisation much easier [2].

Thus, $\ce{NaBH4}$ needs protic solvents (as it will budge and get ionised only then), while $\ce{LiAlH4}$ does not need them as it readily gets ionised.

---

Sources:

1 - https://healy.create.stedwards.edu/Chemistry/CHEM30/organicCD(Mitzel)/chapter6/pages7and8/page7and8.htm

2 - https://en.wikipedia.org/wiki/Lithium_aluminium_hydride#Use_in_organic_chemistry