I have learnt that sodium borohydride can reduce aldehydes, ketones and acyl chlorides into alcohols. However, it fails to react with esters or acid anhydrides.

On the other hand, alcohols — supposedly weaker nucleophiles — can successfully react with both acyl chlorides and acid anhydrides to give esters. Moreover, my book (Solomons and Fryhle) mentions that acidic conditions are not required for the mentioned reactions.

Why are alcohols able to do what sodium borohydride cannot, even in the absence of acidic conditions?

I have come up with a possible explanation, but I am not sure if it is correct: When an alcohol reacts with an acid anhydrides to produce an ester, a carboxylic acid is formed as a side product. Maybe this provides the required acidic conditions for further reaction?

In general, can a reaction be catalyzed by the formation of a side product initially not added in the reaction mixture?

  • $\begingroup$ Under appropriate conditions sodium borohydride can also reduce esters to alcohols. $\endgroup$
    – aventurin
    Jun 19, 2016 at 9:23

3 Answers 3


General considerations

It’s not the basicity that is the important factor here, it is the nucleophilicity. The hydride ion, whether present as tetrahydridoborate or as a lone hydride is a very lousy nucleophile while an alcohol is an okay-ish nucleophile. The general mechanism of the addition is given in scheme 1.

general nucleophilic addition mechanism for carbonyls
Scheme 1: General mechanism of a nucleophilic attack on a carbonyl.

Now in the case of a hydride — a very weak nucleophile — this reaction is generally slow. It is faster, the more reactive — electrophilic — the carbonyl species is. If $\ce{X}$ is a carbon or hydrogen atom, then all we have is a localised double bond without any significant reduction in reactivity.[1] Reactions with those should thus be fast. If $\ce{X}$ is an element that has a lone pair, such as oxygen or chlorine, then there is a notable mesomeric stabilisation as shown in scheme 2.

mesomeric stabilisation of esters etc.
Scheme 2: Mesomeric stabilisation by lone pairs next to a $\ce{C=O}$ double bond.

This stabilisation is not strong if a chlorine atom is $\ce{X}$ since chlorine is much larger and its lone pairs are more disperse. The stabilisation is stronger in the case of acid anhydrides; however, since a single oxygen has to stabilise two carbonyls it is still not too strong. In esters, carboxylic acids (and their anions) and most so in amides this stabilisation is very strong ad nucleophilic attacks difficult. This explains the general electrophilicity trend:

$$\text{aldehydes} > \text{ketones} > \text{acid chlorides} > \text {acid anhydrides} > \text{esters} > \text{amides}$$

Now let’s look at the reactions in detail. Throughout this post I will assume neutral conditions only; no acid or base catalysis.

The hydride attack

As I already noted, hydride is a bad nucleophile. It makes sense that it can attack the first members of the series — aldehydes, ketones and acid chlorides — but fails to do so for more stabilised members. For completeness, the mechanism of a hydride attack onto an acid chloride is given in scheme 3.

hydride attack mechanism on acid chlorides
Scheme 3: Mechanism of a hydride attack onto an acid chloride.

We see another key point here: Once the carbonyl has been attacked, the oxide anion has a tendency to ‘rebound’ a lone pair and displace one of the bonded atoms. Generally, the displaced anion must be reasonably favourable for the rebound to occur. Chloride ions are good leaving groups and thus a displacement of chloride is likely to happen. Hydride ions are absolutely absymal leaving groups, so the hydride attack is not reversible.

After the inital attack and displacement of the chloride, we are left with an aldehyde which of course will react quicker and lead to full reduction to the alcohol.

Acid anhydrides are not attacked this way because the initial step does hot happen. If you accidentally got a hydride to attack, the $\ce{^{-}OOC-R}$ leaving group is great and thus a reaction would occur.

The alcohol attack

Alcohol hydroxy groups are still not the best nucleophile but they are better nucleophiles than hydride is. Thus, they even manage to attack acid anhydrides and — to a limited extent — esters and acids. The general mechanism of attack is shown in scheme 4.

mechanism of the attack of an alcohol onto an acid chloride
Scheme 4: Reaction mechanism of an alcohol attacking an acid chloride including a potential second attack.

The key difference here is that the alcohol is a neutral nucleophile, not an anion like hydride is. Thus, when the rebound occurs, it is susceptible to redisplacement. This is shown in the final equilibrium of the mechanism above. This is also the reason why — under neutral conditions! — an alcohol may attack an aldehyde or a ketone but the attack is not productive. The alcohol will simply be rebounded back out since the hydride is a terrible leaving group.

There is a productive reaction mechanism for alcohols with aldehydes or ketones — the formation of acetals and ketals — however, that generally requires acid catalysis.

Finally, there is also the possibility of base catalysis when alcohols react. It would deprotonate them to give the alcoholate which is a much better nucleophile and even able to attack esters. However, base catalysis is outside of the scope of this question.

  • $\begingroup$ Why is hydride a bad nucleophile? I thought it was one of the best.. $\endgroup$
    – Newton
    Jun 19, 2016 at 12:32
  • 2
    $\begingroup$ @Kalyan Recheck your sources ;) $\endgroup$
    – Jan
    Jun 19, 2016 at 12:32
  • $\begingroup$ It has high charge density, and least steric hindrance possible. Doesn't that mean it is a good nucleophile ? $\endgroup$
    – Newton
    Jun 19, 2016 at 12:33
  • $\begingroup$ @Kalyan Its charge density is not that high. And it is really not that good of a nucleophile. If it were, it would easily displace halides. However, SciFinder is full of examples of hydrides that do not displace the alkyl bromide. It gets even worse if we talk about tetrahydridoborate, which is technically a 2+2 cycloaddition mechanism. $\endgroup$
    – Jan
    Jun 19, 2016 at 12:40
  • $\begingroup$ Hydride is an excellent nucleophile, dear. It so happens, that it rarely occur in solutions, though. However, hydride solutions stabilized by TMEDA are strong reducing agents, capable of reducing carboxilic acid esters pubs.acs.org/doi/abs/10.1021/jo00228a027 $\endgroup$
    – permeakra
    Jun 19, 2016 at 14:52

There are three parts to this answer:

  1. Sodium borohydride is more basic than alcohols
  2. Sodium borohydride reacts with anhydrides under some conditions
  3. Reactions can be autocatalytic

Sodium borohydride is more basic than alcohols

Sodium borohydride will react with water and alcohols to form small amounts of hydrogen gas. If you doubt this, add some to water or ethanol or try to find a video of this phenomenon.

$$\ce{NaBH4 + ROH -> NaOR + BH3 + H2 ^}$$

If sodium borohydride can deprotonate an alcohol, it is more basic than an alcohol.

Sodium borohydride reacts with anhyrides under some conditions

Here is an exmaple In this article in the Journal of the American Chemical Society, sodium borohydride in THF is used to reduce cyclic anhydrides to lactones (esters), not your expected product! However, this reaction might not be general to all anhydrides. Introductory textbooks tend to try to make things simpler to help you learn the material. Think about how maddening it would be to try to learn all of the conditions in which sodium borohydride reacts with anhydrides and all of the ones that do not work.

Reactions can be autocatalytic

If a reaction produces a species as a product that can catalyze the reaction, the rate may increase over time instead of decrease. These reactions are autocatalytic.

Consider the reaction of an alcohol with an anhydride.

$$\ce{ROH + (R'CO)2O -> ROCO2R' + R'CO2H}$$

As you mention, acid is not necessary. However, the substitution reactions of carboxylic acid derivatives are all accelerated by acid.

Rate law without acid $$\mathrm{rate}=k[\ce{ROH}][\ce{(R'CO)2O}]$$

Rate law with acid $$\mathrm{rate}=k[\ce{ROH}][\ce{(R'CO)2O}][\ce{HA}]$$

As the reaction proceeds, a carboxylic acid is produced. While it is not a particularly strong acid, the reaction mixture does become more acidic over time, and the reaction rate increases as a result. This is one reason why alcohols (seemingly poor nucleophiles) react with anhydrides on their own.

  • $\begingroup$ By the way, are hydrides considered good nucleophiles or not? $\endgroup$
    – Newton
    Jun 19, 2016 at 15:06
  • $\begingroup$ The hydride anion $\ce{H-}$ is not a good nucleophile. The hydride equivalent of borohydride is a different matter. $\endgroup$
    – Ben Norris
    Jun 19, 2016 at 15:35

In acidic conditions sodium borohydride is neutralized. Some of the reactions indicated are acidic environments.

In general, sodium borohydride cannot attack/reduce esters in neutral conditions. Neither do alcohols. However, in acid, esters become much more nucleophilic such that alcohols can react/transesterify. Sodium borohydride does not react because it gets consumed by the acidic reaction conditions.

  • $\begingroup$ Please view the edited question. $\endgroup$
    – Newton
    Jun 19, 2016 at 7:55

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