Cannizzaro Reaction

Introduction

What happens when an aldehyde has no alpha hydrogen and is placed in concentrated base? It cannot form the usual enolate that begins an aldol reaction. Some such aldehydes instead react in pairs: one molecule becomes an alcohol while the other becomes a carboxylate. This is the Cannizzaro reaction. Its distinctive step transfers hydride from one aldehyde-derived intermediate directly to a second aldehyde carbonyl.

Core explanation

An aldehyde normally has a carbonyl carbon bonded to hydrogen. If its neighbouring, alpha carbon bears at least one H, base may remove that H to produce an enolate and enable aldol chemistry. Benzaldehyde, C₆H₅CHO, has the carbonyl carbon directly attached to an aromatic ring and has no alpha H on an ordinary saturated carbon. Formaldehyde, HCHO, likewise has no alpha carbon. These are classic Cannizzaro candidates. An aldehyde that can readily enolise may instead undergo aldol reaction under similar conditions, so the absence of alpha H is a useful selection test, not just a naming detail.

Hydroxide first attacks the carbonyl carbon of one aldehyde. The C=O pi pair moves onto oxygen, forming a tetrahedral alkoxide bearing a newly attached OH group. In the key redox step, that tetrahedral intermediate transfers its original aldehydic hydrogen with its electron pair—a hydride equivalent—to the carbonyl carbon of a second aldehyde molecule. As the donor loses hydride, its oxygen electrons restore a carbonyl; its carbon has acquired an OH and becomes a carboxylic-acid derivative. The second aldehyde gains hydride, becoming an alkoxide. Proton transfers in the basic medium leave the oxidised product as carboxylate and the reduced product as alcohol.

For benzaldehyde, the overall organic products in aqueous base are sodium benzoate, C₆H₅COO⁻ Na⁺, and benzyl alcohol, C₆H₅CH₂OH. Acidifying the carboxylate after the reaction gives benzoic acid. Two benzaldehyde molecules are needed because one supplies the hydride and the other accepts it. Do not mistake the hydroxide added to the first carbonyl for the reducing reagent; the aldehyde-derived tetrahedral intermediate is the hydride donor. Hydroxide enables the reaction and appears in the overall base-promoted stoichiometry.

Oxidation state makes the paired change clear. The aldehyde carbon that becomes carboxylate gains a bond to oxygen in place of its C–H bond and is oxidised. The other carbonyl carbon gains a C–H bond as C=O becomes C–O in the alcohol and is reduced. Both changes occur to molecules that began as the same aldehyde, hence disproportionation . A crossed Cannizzaro reaction can use two different aldehydes; formaldehyde is often preferentially oxidised to formate, allowing the other aldehyde to be reduced to its alcohol. Product prediction must then specify which partner donated hydride.

The textbook mechanism is best represented by concerted hydride transfer between the hydroxide-adduct donor and the acceptor aldehyde, rather than by a freely wandering hydride ion. Free H⁻ would be highly reactive in water. In a curved-arrow drawing, show the donor C–H bond delivering its pair to the acceptor carbonyl carbon while the donor O⁻ reforms C=O and the acceptor C=O electrons move to oxygen. Counting electrons and charges prevents an impossible intermediate.

Step-by-step reasoning

Check whether the aldehyde has an alpha H. If it does not, draw hydroxide addition to one carbonyl. Put a second aldehyde next to the tetrahedral intermediate. Move the donor C–H pair to the acceptor carbonyl carbon while adjusting both C–O bonds. Finally show proton transfers and write carboxylate plus alcohol under basic conditions, or carboxylic acid after acid work-up.

Visual explanation

Draw two benzaldehyde molecules in different colours. On the first, add OH beneath the carbonyl and label its original aldehydic H. Curve an arrow from that C–H bond to the second carbonyl carbon. Mark the first molecule C₆H₅COO⁻ and the second C₆H₅CH₂OH after proton transfers; trace the hydrogen from donor to acceptor.

Real-world analogy

Imagine two identical stores exchanging one valuable package. One store gives away the package and becomes depleted, while the receiving store becomes enriched. The stores started alike, but finish in opposite states. In Cannizzaro, the transferable package is a hydride equivalent; the donor is oxidised, and the acceptor is reduced. The analogy describes the paired redox outcome, not a free hydride moving through solution.

Real-world example

Heating benzaldehyde with concentrated aqueous sodium hydroxide produces benzyl alcohol and benzoate. If a laboratory then acidifies the aqueous layer, benzoic acid can precipitate while benzyl alcohol remains a different organic product. Separating those substances is a practical illustration of how a single aldehyde feedstock splits into an acid-family product and an alcohol-family product.

Why?

Why is benzaldehyde a better textbook Cannizzaro example than acetaldehyde? Benzaldehyde lacks an alpha hydrogen needed for ordinary enolate formation, while acetaldehyde has three. In base, acetaldehyde readily enters aldol chemistry, forming a carbon–carbon bond. Choosing the substrate by its alpha-hydrogen inventory predicts the dominant introductory mechanism much better than merely seeing an aldehyde group.

Common misconception

"Cannizzaro converts every aldehyde to an alcohol." One aldehyde is reduced to alcohol, but another is oxidised to carboxylate. Under the reaction's basic conditions, write carboxylate rather than the free acid. Acid work-up changes the form of the oxidised product without changing which molecule was oxidised.

Worked example

Question: What organic products result when two molecules of benzaldehyde react with concentrated NaOH, followed by acid work-up?

Reasoning: Benzaldehyde lacks an alpha H. One molecule supplies hydride from its hydroxide adduct and becomes benzoate; the second receives hydride and becomes benzyl alkoxide, then benzyl alcohol. Acid work-up protonates benzoate.

Answer: One equivalent each of benzoic acid, C₆H₅COOH, and benzyl alcohol, C₆H₅CH₂OH.

Quick check

1. Which product is oxidised in benzaldehyde Cannizzaro disproportionation under basic conditions? Answer: Benzoate is the oxidised product; the other benzaldehyde molecule becomes benzyl alcohol.

Exam focus

First inspect the alpha carbon and state why enolate formation is unavailable. Draw two aldehyde molecules, hydroxide addition, the hydride-transfer arrows, and the correct basic products. If the question includes acid work-up, protonate the carboxylate only then. Name donor and acceptor explicitly.

Advanced insight

The reaction is a useful contrast with biochemical carbonyl reduction, where a dedicated hydride carrier such as NADH provides the reducing equivalent. Cannizzaro instead sacrifices one aldehyde molecule to reduce another. Crossed reactions with formaldehyde exploit its tendency to be the sacrificial oxidised partner, but substrate choice and conditions must still be checked.

Summary

Cannizzaro reaction disproportionates a non-enolizable aldehyde in strong base. Hydroxide addition creates a hydride-donating tetrahedral intermediate, which reduces a second aldehyde as the first is oxidised. The products are an alcohol and a carboxylate in base; acid work-up gives the corresponding carboxylic acid. Benzaldehyde provides a clear two-molecule example.

Practice questions

1. Why does benzaldehyde fit the Cannizzaro substrate test? Answer: Its carbonyl has no enolizable alpha hydrogen on the adjacent aromatic carbon. 2. Where does the hydride accepted by the reduced aldehyde originate? Answer: It comes from the aldehydic C–H bond of the hydroxide-adduct donor molecule. 3. What are the two organic product classes before acid work-up? Answer: An alcohol and a carboxylate salt. 4. Why may acetaldehyde behave differently in base? Answer: It has alpha hydrogens and readily forms an enolate that enters aldol chemistry.