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The Wittig reaction is an important organic chemistry reaction that converts aldehydes or ketones into alkenes. This cheat sheet helps students track how the phosphorus ylide is made, how it reacts with a carbonyl compound, and what products form. It is especially useful because the reaction combines mechanism, reagent choice, and stereochemistry in one process.

A clear reference makes it easier to predict alkene products from common starting materials.

The reaction begins when a phosphine such as PPh3\mathrm{PPh_3} reacts with an alkyl halide to form a phosphonium salt. A strong base removes an acidic proton to make a phosphorus ylide, often written as Ph3P=CHR\mathrm{Ph_3P{=}CHR}. The ylide then reacts with an aldehyde or ketone, and the overall result is replacement of the carbonyl oxygen with the ylide carbon group.

A common summary is R2C=O+Ph3P=CHRR2C=CHR+Ph3P=O\mathrm{R_2C{=}O + Ph_3P{=}CHR' \rightarrow R_2C{=}CHR' + Ph_3P{=}O}.

Key Facts

  • A phosphonium salt forms when triphenylphosphine attacks an alkyl halide: PPh3+RCH2X[Ph3P+CH2R]X\mathrm{PPh_3 + RCH_2X \rightarrow [Ph_3P^+CH_2R]X^-}.
  • A strong base forms the ylide by removing an acidic alpha hydrogen: [Ph3P+CH2R]X+BasePh3P=CHR+HB+X\mathrm{[Ph_3P^+CH_2R]X^- + Base \rightarrow Ph_3P{=}CHR + HB + X^-}.
  • The overall Wittig reaction converts a carbonyl compound into an alkene: R2C=O+Ph3P=CHRR2C=CHR+Ph3P=O\mathrm{R_2C{=}O + Ph_3P{=}CHR' \rightarrow R_2C{=}CHR' + Ph_3P{=}O}.
  • The carbon atom of the ylide becomes one carbon of the new C=C\mathrm{C{=}C} double bond.
  • The original carbonyl carbon becomes the other carbon of the new C=C\mathrm{C{=}C} double bond.
  • Aldehydes usually react faster than ketones because aldehydes are less sterically hindered and more electrophilic.
  • Unstabilized ylides often favor ZZ alkenes, while stabilized ylides often favor EE alkenes.
  • Triphenylphosphine oxide, Ph3P=O\mathrm{Ph_3P{=}O}, is a strong driving-force byproduct because the P=O\mathrm{P{=}O} bond is very stable.

Vocabulary

Wittig reaction
An organic reaction that forms an alkene by reacting an aldehyde or ketone with a phosphorus ylide.
Phosphorus ylide
A neutral molecule with adjacent positive phosphorus and negative carbon character, commonly written as Ph3P=CHR\mathrm{Ph_3P{=}CHR}.
Phosphonium salt
An ionic intermediate such as [Ph3P+CH2R]X\mathrm{[Ph_3P^+CH_2R]X^-} formed before the ylide is made.
Carbonyl compound
An aldehyde or ketone containing a carbon oxygen double bond, written generally as C=O\mathrm{C{=}O}.
Oxaphosphetane
A four-membered intermediate containing oxygen and phosphorus that can form during the Wittig mechanism.
Triphenylphosphine oxide
The stable byproduct Ph3P=O\mathrm{Ph_3P{=}O} formed when oxygen transfers from the carbonyl compound to phosphorus.

Common Mistakes to Avoid

  • Putting the new double bond in the wrong place is wrong because the alkene must form between the original carbonyl carbon and the ylide carbon.
  • Forgetting the oxygen byproduct is wrong because the carbonyl oxygen does not disappear, it becomes part of Ph3P=O\mathrm{Ph_3P{=}O}.
  • Using an alkyl halide that cannot form the needed ylide is wrong because the phosphonium salt must have a removable hydrogen next to phosphorus.
  • Treating all Wittig reactions as stereospecific is wrong because EE and ZZ selectivity depends on whether the ylide is stabilized or unstabilized.
  • Confusing the ylide carbon with the phenyl groups on phosphorus is wrong because the phenyl groups remain attached to phosphorus and do not enter the alkene product.

Practice Questions

  1. 1 Predict the alkene product of CH3CHO+Ph3P=CH2?\mathrm{CH_3CHO + Ph_3P{=}CH_2 \rightarrow ?}.
  2. 2 What phosphorus ylide is needed to convert PhCHO\mathrm{PhCHO} into PhCH=CHCH3\mathrm{PhCH{=}CHCH_3}?
  3. 3 Write the overall product and byproduct for (CH3)2C=O+Ph3P=CHCO2Et?\mathrm{(CH_3)_2C{=}O + Ph_3P{=}CHCO_2Et \rightarrow ?}.
  4. 4 Explain why formation of Ph3P=O\mathrm{Ph_3P{=}O} helps drive the Wittig reaction forward.

Understanding Wittig Reaction Reference

The ylide is unusual because its carbon has both negative character and a bond to positively charged phosphorus. You can picture it in two useful ways. One picture shows a negative charge on carbon and a positive charge on phosphorus.

Another shows a phosphorus to carbon double bond. The first picture helps explain why the carbon attacks an electron poor carbonyl carbon. The oxygen then becomes negatively charged for a short time.

This creates a crowded intermediate in which phosphorus and oxygen are connected through a small four membered ring. The ring breaks apart to give the alkene and the phosphorus oxide. Drawing the charge movement carefully helps students see that no carbon atoms disappear during this process.

Carbon tracking is the most reliable way to predict a product. First, mark the carbonyl carbon in the starting compound. Next, find the carbon that was next to phosphorus in the ylide.

Those are the two atoms joined by the new double bond. Every group originally attached to either of these carbons stays attached in the product. This method prevents a common mistake of placing the ylide group on the oxygen side of the carbonyl.

It is useful to draw each alkene carbon separately before adding the double bond. Count carbons before and after the reaction. The product should contain the carbon skeletons from both reactants, while oxygen leaves with the phosphorus containing material.

Alkene geometry needs careful attention because it is not controlled by one simple rule. Ylides with no group that can spread out the negative charge often react quickly. Their reactions commonly give more of the Z alkene.

Ylides next to groups such as esters, nitriles, or carbonyl groups are stabilized by resonance. They usually react more slowly and commonly give more of the E alkene. These are trends, not promises.

The structure of the aldehyde or ketone, the solvent, the base, and the reaction temperature can change the result. When assigning E or Z, compare the priority of the two groups on each alkene carbon using the usual atomic number rules. Do not decide geometry merely by looking for the larger group.

In real laboratory work, the reaction can be limited by the starting halide used to make the phosphorus salt. Primary halides usually work well because the phosphine can attack them directly. Tertiary halides often give unwanted elimination instead.

Strong bases require dry glassware and dry solvents because water can destroy the ylide before it reacts. Ketones may need more forcing conditions because groups around their carbonyl carbon block approach. After the reaction, the phosphorus oxide byproduct can be difficult to remove since it is polar and often sticks to silica during chromatography.

Students should learn to separate the reaction goal from the practical task of isolating a clean product. A good reaction scheme predicts the alkene, but a good experiment considers side reactions, moisture, purification, and safe handling of strong bases.