The Wittig Reaction

Introduction

The Wittig reaction replaces the oxygen of an aldehyde or ketone carbonyl with a carbon fragment, producing an alkene. Its value in synthesis is precise carbon-skeleton construction: the carbonyl carbon becomes one alkene carbon, and the carbon of a phosphorus ylide becomes the other. Predicting the product is easier when those two atoms are marked before any bonds are rearranged.

Core explanation

A common ylide representation is Ph3P+–C−R2. The carbon next to phosphorus carries nucleophilic character, while phosphorus carries positive character in that Lewis form. Another resonance contributor places a P=C double bond. Neither drawing requires a free carbon anion floating separately from phosphorus; the ylide is one connected reagent. It can be prepared from a phosphonium salt by deprotonation at the carbon attached to phosphorus, provided that carbon has a removable proton in the precursor.

The carbonyl carbon is electrophilic, so the ylide carbon and carbonyl carbon form a new C–C bond. The reaction proceeds through intermediates often represented with a four-membered ring containing O, P and the two carbons, called an oxaphosphetane. Fragmentation of this ring gives an alkene between the former carbonyl carbon and ylide carbon, plus triphenylphosphine oxide, Ph3P=O. Formation of the strong P–O bond helps drive the overall transformation. Detailed pathways can vary with ylide type and conditions, so a course-level mechanism should be presented as a useful mechanistic model rather than a claim that every system has exactly the same elementary steps.

For product mapping, write the carbonyl as R1R2C=O and the ylide as Ph3P=CR3R4. The alkene is R1R2C=CR3R4. The carbonyl substituents R1 and R2 stay attached to their carbon; ylide substituents R3 and R4 stay attached to theirs. Carbonyl oxygen does not become an alcohol oxygen in this reaction; it is incorporated in phosphine oxide. This is fundamentally different from Grignard addition, where oxygen remains on the organic product as OH after work-up.

An especially simple case is methylenation with Ph3P=CH2. Benzaldehyde, PhCHO, becomes styrene, PhCH=CH2. The ylide contributes the terminal CH2 carbon, and the aldehyde contributes PhCH. Acetone plus the same ylide becomes 2-methylpropene because acetone's two methyl groups remain attached to one alkene carbon. Checking the four substituent positions prevents accidental chain-length errors.

Alkene stereochemistry requires care. If each new double-bond carbon bears two distinct substituents, E and Z products are possible. Stabilised and unstabilised ylides often show different stereochemical tendencies under conventional conditions, but these are not universal guarantees; solvent, counterion, substituents and reaction protocol matter. For an exam lacking specific selectivity data, draw the constitutional alkene correctly and state possible E/Z isomers rather than claiming a single geometry from the name “Wittig” alone. If one alkene carbon has two identical groups, no E/Z distinction exists.

The Wittig reaction is particularly useful when the synthetic target specifies the exact position of a C=C bond. Retrosynthetically, disconnect that bond into a carbonyl carbon and a ylide carbon. Two different disconnections may be possible, but their reagent availability and steric compatibility can differ. The reaction is not a general way to change every carbonyl to every alkene without planning; crowded substrates and ylide preparation can limit the route.

Step-by-step reasoning

Circle carbonyl carbon and ylide carbon. Copy every substituent attached to each into two separate product alkene carbons. Join those carbons with a double bond. Put the carbonyl oxygen with phosphorus in Ph3P=O, not on the organic alkene. Check whether E/Z is meaningful and whether conditions justify predicting one geometry. In reverse planning, partition the target alkene into a plausible carbonyl plus ylide.

Visual explanation

Draw R1R2C=O on the left and Ph3P=CR3R4 on the right. Use matching colours for the two carbon atoms that will form the double bond. Between reactants and products, sketch a small four-membered square labelled C–C–P–O for the oxaphosphetane. On the product side, show R1R2C=CR3R4 plus Ph3P=O, with carbonyl oxygen highlighted in the oxide.

Real-world analogy

Imagine combining two half-frames along a new hinge: one frame comes from the carbonyl and the other from the ylide. The phosphorus component carries away oxygen as a separate piece. This mental image helps assign which substituents stay with which carbon, but the actual reaction is driven by electron distribution and phosphorus–oxygen bond formation.

Real-world example

Styrene can be represented as a Wittig product of benzaldehyde and a methylene ylide. The aromatic ring stays on the benzaldehyde carbon, which also keeps its original H, while the ylide supplies terminal CH2. The product is PhCH=CH2, a useful monomeric alkene. The example shows a clear one-carbon addition to the unsaturated skeleton without retaining carbonyl oxygen in the organic molecule.

Why?

Ylide carbon is electron-rich enough to bond to electrophilic carbonyl carbon. The ensuing arrangement places oxygen near phosphorus, allowing formation of phosphine oxide. The strong P–O interaction and formation of an alkene provide a favourable product combination. Mechanistic orbital details explain how these bonds reorganise, but carbon and oxygen tracking already predicts the main constitutional outcome reliably.

Common misconception

The Wittig reaction does not give an alcohol by protonating an alkoxide in its standard net transformation. Carbonyl oxygen leaves the organic framework with phosphorus. Another common error is putting a substituent from the ylide on the carbonyl-derived alkene carbon. Keep each reactant carbon's original attached groups together, then make the C=C bond between those two carbons.

Worked example

Question: Predict the organic product of cyclohexanone with the ylide Ph3P=CHCH3. Is E/Z stereochemistry assigned to the new exocyclic double bond?

Reasoning: Cyclohexanone carbonyl carbon remains in the ring and retains its two bonds to neighbouring ring carbons. The ylide carbon carries H and CH3. Joining those carbons by a double bond gives an exocyclic ethylidene group, cyclohexane–C(=CHCH3) at the former ketone position. On the ring alkene carbon, the two paths around an unsubstituted cyclohexane are constitutionally equivalent, so the usual E/Z distinction is not assigned in this simple symmetric case.

Answer: Ethylidenecyclohexane, with the former carbonyl carbon double-bonded to CHCH3; no distinct E/Z pair arises from the two equivalent ring paths.

Quick check

1. Which two atoms form the alkene C=C bond in a Wittig product? Answer: The starting carbonyl carbon and the carbon atom of the phosphorus ylide.

Exam focus

Mark atom origins before drawing the product. Write the alkene and phosphine oxide, and do not leave oxygen in the organic alkene. Check E/Z eligibility using substituent identity on each alkene carbon. If a specific geometric isomer is requested, use the given ylide class and conditions rather than guessing from a broad rule. Distinguish Wittig alkene formation from ordinary organometallic alcohol formation.

Advanced insight

Ylide stabilisation changes both its preparation and reaction behaviour. Electron-withdrawing groups attached to the ylide carbon can delocalise negative charge and influence transition-state energy and alkene selectivity. The familiar phosphonium ylide model is therefore a family of reagents, not one fixed nucleophile. Computational and mechanistic studies examine several possible early intermediates; the oxaphosphetane-to-products mapping remains especially useful for product prediction.

Summary

The Wittig reaction joins carbonyl carbon to ylide carbon and converts their connection into C=C. The original carbonyl oxygen ends in triphenylphosphine oxide, while substituents on each carbon remain attached to that carbon. A four-membered oxaphosphetane provides a useful mechanistic model. Correct prediction requires atom tracking and, when appropriate, consideration of possible E/Z alkene geometries.

Practice questions

1. What alkene forms from formaldehyde and Ph3P=CHCH3? Answer: Propene, CH2=CHCH3, because formaldehyde supplies terminal CH2 and the ylide supplies CHCH3. 2. What alkene forms from acetone and Ph3P=CH2? Answer: 2-Methylpropene, (CH3)2C=CH2. 3. Why is Ph3P=O a characteristic co-product? Answer: Carbonyl oxygen becomes bonded to phosphorus during collapse of the reaction intermediate. 4. Does CH2=CHPh have E/Z isomers? Answer: No. The terminal CH2 alkene carbon has two identical H substituents. 5. Where does the carbonyl oxygen go in the overall reaction? Answer: Into a phosphorus oxide by-product, commonly triphenylphosphine oxide.