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?


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

  • 10
    $\begingroup$ Principally solubility. Sodium borohydride is essentially insoluble in aprotic solvents. $\endgroup$
    – Waylander
    Aug 10, 2018 at 17:51
  • $\begingroup$ @Waylander What about the need for the ions? These ions help to increase polarity of solvent, increasing the solubility of NaBH4? $\endgroup$ Aug 11, 2018 at 1:28
  • 4
    $\begingroup$ Li and Mg cations act as Lewis acids coordinating to carbonyl oxygen, Na is much weaker in this respect. $\endgroup$
    – Mithoron
    Aug 11, 2018 at 17:54
  • $\begingroup$ Apparently, THF (a non-protic solvent) can also be used to conduct reductions with $\ce {NaBH_4}$, according to this: commonorganicchemistry.com/Common_Reagents/Sodium_Borohydride/…. $\endgroup$ Dec 24, 2023 at 13:33
  • $\begingroup$ As Mithoron states a labile Lewis acid is needed in the reaction similar to the role of Mg++ in a Grignard; Na+ is not quite up to the job in NaBH4. $\endgroup$
    – jimchmst
    Dec 24, 2023 at 19:55

2 Answers 2


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.


  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

  • $\begingroup$ "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 is interesting: do you have a source that goes more into detail about this? [I'm sorry if it's the reference above, I don't have access any more.] $\endgroup$ Dec 24, 2023 at 13:39
  • $\begingroup$ @orthocresol note my new reference. $\endgroup$ Dec 24, 2023 at 14:14



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!


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.

enter image description here

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.


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

  • $\begingroup$ I'm not convinced by the first part of your answer. You say in the second part of the answer that $\ce{LiAlH4}$ should not be used with a protic solvent. Instead, the intermediate alkoxide is quenched with water. You could do the same in the reaction with $\ce{NaBH4}$, so you would not need a protic solvent if that were the only issue. $\endgroup$
    – Karsten
    Dec 23, 2023 at 21:59

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