Hydride Reduction of Carbonyls

Introduction

Aldehydes and ketones can be reduced to alcohols by reagents that deliver hydride-like hydrogen to the electrophilic carbonyl carbon. Sodium borohydride and lithium aluminium hydride are standard examples. The first step forms a new C–H bond and moves the C=O pi pair onto oxygen, creating an alkoxide. A later proton source supplies the alcohol O–H. Tracking which carbonyl substituents remain explains why aldehydes give primary alcohols and ketones give secondary alcohols.

Core explanation

In a mechanism sketch, a B–H bond of BH₄⁻ or an Al–H bond of AlH₄⁻ supplies the hydride-like pair to carbonyl carbon. Draw an arrow from that bond toward the C of C=O, with a simultaneous arrow from the C=O pi bond to O. The carbonyl carbon changes from approximately planar to tetrahedral and gains H. The immediate oxygen-containing product is an alkoxide or metal-bound alkoxide, not yet the isolated neutral alcohol. Water, alcohol solvent or acidic work-up then protonates oxygen.

For an aldehyde R–CHO, the carbonyl carbon initially has R and H. Hydride adds another H, giving R–CH₂O⁻, and protonation gives R–CH₂OH, a primary alcohol when R is a carbon group. Formaldehyde, HCHO, gives methanol. For a ketone R–CO–R′, the carbonyl carbon initially has two carbon groups; hydride adds H, and protonation gives R–CH(OH)–R′, a secondary alcohol . Acetone, CH₃COCH₃, becomes propan-2-ol.

NaBH₄ is relatively mild and often reduces aldehydes and ketones while leaving many esters and carboxylic acids unchanged under ordinary conditions. LiAlH₄ is more reactive and can reduce a broader range of carbonyl derivatives, including esters, but it is incompatible with water during the initial reduction stage. The work-up is added after the hydride reaction. These are general laboratory trends, not guarantees of selectivity in every multifunctional molecule.

The carbon skeleton does not gain a carbon atom during hydride reduction. Hydride provides H, not an alkyl group. This distinguishes it from a Grignard addition, where a carbon nucleophile makes a new C–C bond. A carbonyl already bearing a stereogenic environment may show facial selectivity; otherwise, if an unsymmetrical ketone becomes a new stereocentre in an achiral setting, attack from either face can produce a racemic mixture. Acetone gives no new stereocentre because two methyl groups remain identical.

"Hydride" in the arrow drawing does not require a bottle of free H⁻ ions floating in solution. The reactive H is transferred from a covalent B–H or Al–H bond with its electron pair. Formal charge and stoichiometric details depend on the reagent and solvent, but the two-arrow carbonyl attack pattern remains the central predictive tool.

The reduction is an organic oxidation-state change: carbon gains a C–H bond and loses the pi component of C=O. Oxygen remains attached and is protonated rather than removed. If a product has lost oxygen entirely and become an alkane, a different reduction pathway and reagent set is required. Name the specific alcohol expected from the given aldehyde or ketone.

Step-by-step reasoning

Classify the starting C=O as aldehyde or ketone and record its attached R groups and H. Draw hydride delivery to carbonyl carbon and movement of the C=O pi pair to oxygen. Write the tetrahedral O⁻ intermediate with a new C–H bond. Add a separate protonation or aqueous work-up step to form O–H. Count carbon attachments to the alcohol carbon to decide primary or secondary classification and check for a new stereocentre.

Visual explanation

Draw R–C(=O)–H with a wedge-shaped incoming H from a BH₄⁻-derived B–H bond. A curved arrow points to carbonyl carbon, and a second points to oxygen. The middle drawing is R–CH₂O⁻; the final drawing after H₂O/H⁺ is R–CH₂OH. In a parallel ketone drawing, keep both R groups and show R₂CHOH as the final secondary alcohol.

Real-world analogy

A flat three-point hub receives one small connector at its exposed centre while an upper flexible link shifts toward the oxygen side. A later cap neutralises that shifted link. The connector is hydride-like H, and the cap is a proton on oxygen. The analogy helps separate carbon attack from oxygen protonation, which are different stages.

Real-world example

Reducing benzaldehyde with NaBH₄ gives benzyl alcohol after work-up. The aromatic ring remains attached to the same carbonyl carbon, which gains H and becomes CH₂OH. This alcohol can then be converted into other benzylic products. A chemist choosing NaBH₄ for a molecule containing a ketone and ester may exploit its relative mildness to reduce one group preferentially, with actual selectivity checked experimentally.

Why?

Why does aldehyde reduction give a primary alcohol but ketone reduction give a secondary one? Hydride adds H without changing the carbon groups already attached to carbonyl carbon. An aldehyde has one carbon substituent, so the alcohol carbon ends with one carbon neighbour; a ketone has two, so it ends with two. The classification follows connectivity, not the strength of the reducing reagent.

Common misconception

"Hydride attacks oxygen because oxygen is more electronegative." The carbonyl carbon is partially positive and receives the hydride-like pair. The C=O pi electrons move onto oxygen, which is then protonated. Drawing H attached to oxygen in the attack step would not explain formation of the new alcohol C–H bond.

Worked example

Question: Predict the product of propanal, CH₃CH₂CHO, treated with NaBH₄ then aqueous work-up, and identify the intermediate charge after hydride transfer.

Reasoning: Hydride bonds to the aldehyde carbonyl carbon while C=O electrons move to oxygen, giving CH₃CH₂CH₂O⁻. Work-up protonates O⁻ without altering the three-carbon chain.

Answer: Propan-1-ol forms; the intermediate is an alkoxide carrying negative charge on oxygen.

Quick check

1. What alcohol class results from reduction of an ordinary ketone by NaBH₄ followed by work-up? Answer: A secondary alcohol, because the former carbonyl carbon retains two carbon substituents and gains H.

Exam focus

Show H transfer to carbonyl carbon and pi-electron movement to oxygen, then a separate protonation. Classify the starting carbonyl and final alcohol correctly. Distinguish NaBH₄'s relative mildness from LiAlH₄'s greater reactivity, and do not place aqueous work-up before a moisture-sensitive LiAlH₄ step.

Advanced insight

The rate and face of hydride transfer can depend on the metal counterion, solvent and chelation to nearby heteroatoms. In a chiral substrate, one face of the planar carbonyl may be shielded, giving a diastereomeric excess even with an achiral hydride reagent. This turns the simple two-arrow reduction into a powerful stereochemical transformation in complex synthesis.

Summary

NaBH₄ and LiAlH₄ deliver hydride-like H to the electrophilic carbonyl carbon of aldehydes or ketones. C=O pi electrons move to oxygen, forming a tetrahedral alkoxide that is protonated during work-up. Aldehydes yield primary alcohols and ketones secondary alcohols, with no new carbon atom added. Reagent strength and substrate functionality determine practical selectivity.

Practice questions

1. What does ethanal give after NaBH₄ reduction and protonation? Answer: Ethanol, a primary alcohol. 2. What does acetone give after the same sequence? Answer: Propan-2-ol, a secondary alcohol. 3. Where is the negative charge after the initial hydride-addition arrow pair? Answer: On oxygen in the alkoxide intermediate, before protonation. 4. Does hydride reduction form a new C–C bond? Answer: No. It adds a C–H bond; a carbon nucleophile such as a Grignard reagent is needed for carbon-chain extension.