Sodium Borohydride (NaBH4 ) | Chemistry Optional Notes for UPSC PDF Download

Introduction

• Sodium borohydride (NaBH₄) is a convenient source of hydride ion (H^-) for the reduction of aldehydes and ketones.
• Aldehydes are reduced to primary alcohols and ketones are reduced to secondary alcohols.
• Esters (including lactones) and amides are not reduced.
• As a source of hydride ion, NaBH₄ will also act as a strong base, deprotonating water, alcohols, and carboxylic acids.
• NaBH₄ also sees use in the reduction of organomercury bonds after oxymercuration reactions.

Sodium Borohydride (NaBH₄)

• Sodium borohydride (NaBH₄) can be made through the addition of sodium hydride (NaH) to our old friend borane in an appropriately chosen solvent.   We generally don’t think of the hydride ion (NaH) as being a very good nucleophile, but the empty p-orbital of BH₃ makes this addition much easier.
  
• In contrast to BH₃, which is a highly air-sensitive liquid requiring special inert-atmosphere (Schlenk line) techniques, sodium borohydride NaBH₄ is a white crystalline solid generally dispensed in the form of pellets, very easily handled and weighed on the benchtop.
• It’s worth a reminder about the properties of the B-H bond because this can be a common source of confusion.
• NaBH₄ has a tetrahedral arrangement of hydrogen atoms about the central boron atom, and a formal charge of -1 on the boron.
• That negative charge on boron does not represent a lone pair on boron, however!
• Because hydrogen is more electronegative (2.20) than boron (2.04) the electrons in the B–H bond are polarized towards the hydrogen.
• So where are the electrons, if they’re not on the boron?
• They’re on the hydrogens!
• True to its name,  sodium borohydride acts as a source of hydride ion, H(-).
• You may recall that hydride is the conjugate base of hydrogen (H2) (pKa about 36), making it a very strong base. NaBH₄ reacts with water and other weak acids (such as methanol) to generate hydrogen gas (H₂).

NaBH₄ For The Reduction of Aldehydes and Ketones

• The most important reaction of NaBH4 is its use in the reduction of aldehydes and ketones to give alcohols.
• The reduction of  aldehydes with sodium borohydride gives primary alcohols.   Note the bonds that form and break here – a new C-H bond is formed, and a C-O (pi) is broken. An additional O-H bond forms during during a workup step with mild acid.
  
• The reduction of ketones follows a similar pattern, and results in the formation of secondary alcohols.  A C-H bond is formed and a C–O (pi) bond is broken.
• Note that if the two R groups flanking the C=O bond are different, a new chiral center will be created. In the case of a simple ketone such as acetophenone (phenyl methyl ketone) this will result in a racemic mixture.
  
• If the molecule already contains one or more stereogenic centers, a mixture of diastereomers will form. One notable example is the reduction of the bicyclic ketone [2.2.1]bicycloheptanone (above). Addition of the hydride ion occurs preferentially from the least hindered face (where there is only one bridging carbon) to give an 86:14 ratio of diastereomers.
• Chiral reducing agents similar in reactivity to NaBH₄ have been developed that are capable of performing enantioselective reductions of ketones. One prominent example is the CBS (Corey-Bakshi-Shibata) family of reagents.

Mechanism for the Reduction of Aldehydes and Ketones With NaBH₄

• The mechanism for these reductions follows the very common two-step addition-protonation pattern often found in reactions of aldehydes and ketones.
• The first step in this reaction is nucleophilic addition to the carbonyl carbon. Forming a C-H bond and breaking a C–O (pi) bond.
  
• The addition is followed by protonation of the oxygen with a mild acid (leading to the formation of O–H).
• In practice, this reaction is usually performed in an alcoholic solvent like CH₃OH and the reaction is quenched with a mild acid such as a saturated solution of ammonium chloride (NH₄Cl).

NaBH₄ Will Not Reduce Esters or Amides

• NaBH4 will not generally reduce esters or amides.
• This reaction can be done by LiAlH4, however.
  

Why are esters and amides so unreactive? After all, shouldn’t these functional groups be more reactive than aldehydes and ketones since the carbonyl is attached to the electronegative oxygen and nitrogen atoms?

It’s actually the opposite! The lone pairs from oxygen and nitrogen are capable of donating electron density to the carbonyl carbon through forming a pi bond. This makes the carbonyl carbon less electrophilic and less reactive with nucleophiles.

What about anhydrides and acid halides?

NaBH₄ will reduce anhydrides and acid halides, but in practice, these functional groups will react with the solvent (CH₃OH) before they have a chance to react with NaBH₄.

Reduction of Hemiacetals

• Aldehydes and ketones, check. Esters and amides, no go. So what else can be reduced by sodium borohydride.
• The molecule below might look familiar. It’s glucose!
• When treated with NaBH₄, glucose is reduced to the alcohol sorbitol. 

Hold on for a second. What happened here? There’s no aldehyde or ketone.  Or is there? 

• There actually is an aldehyde present here, but it is in equilibrium with a cyclic hemiacetal. (In cyclic molecules such as sugars, this equilibrium process is known as ring-chain tautomerism.)
• Although the open-chain aldehyde form only comprises 0.02% of an aqueous mixture of glucose at equilibrium, NaBH₄ will quickly reduce any aldehyde that is present to give sorbitol.  Via Le Chatelier’s principle, equilibrium between the cyclic hemiacetal and the aldehyde will eventually result in all of the cyclic hemiacetal being reduced to the alcohol.

Reduction of Organomercury Compounds

• There’s one more use of NaBH₄ worth noting.
• You may recall that alkenes can undergo oxymercuration when treated with water (or alcohols) in the presence of mercuric acetate Hg(OAc)₂ or similar which results in net Markovnikov addition of water to an alkene.
  
• The resulting organomercury compound can then be treated with NaBH4 to give an alcohol.
• The precise details of this “demercuration” step are often skipped over, but for completeness we’ll briefly go through it here.
• The first step is addition of hydride to mercury, giving NaOAc and a new Hg-H bond. Carbon-mercury bonds are extremely weak (this is part of the reason why organomercury compounds are extremely toxic) and upon homolytic cleavage of Hg-C, the resulting carbon radical is then reduced with Hg-H to give C-H and metallic mercury (Hg⁰). On large enough scale, this results in a little puddle of mercury forming at the bottom of the reaction flask.

Summary

• Sodium borohydride will reduce aldehydes to primary alcohols and ketones to secondary alcohols.
• This proceeds via a two-step mechanism consisting of 1) nucleophilic addition, followed by 2) protonation.
• Esters and amides are not reduced by NaBH₄ under normal conditions. (They can be reduced by lithium aluminum hydride (LiAlH₄) however).
• NaBH₄ is also used in the demercuration step of oxymercuration-demercuration.