The Wittig reaction — Organic Chemistry Notes

Cyclohexanone and the reagent Ph₃P=CH₂ give methylenecyclohexane: the ring’s C=O becomes C=CH₂. The oxygen leaves on the phosphorus, as triphenylphosphine oxide, Ph₃P=O. This is the Wittig reaction: a carbonyl compound and a phosphorus reagent join to make an alkene.

Ph is short for a phenyl group, C₆H₅: a benzene ring attached through one of its carbons. So Ph₃P is a phosphorus atom carrying three benzene rings.

The new double bond forms exactly where the C=O was. The carbonyl carbon keeps its two groups and gains the CH₂.

So far you have two other ways to make an alkene. The first is to reduce an alkyne: Lindlar’s catalyst gives the cis alkene and Na in NH₃ gives the trans one (Alkynes). That controls the geometry well, but the alkyne, with the right carbon skeleton, has to exist first.

The second, and the general one, is elimination: E1, E2, dehydration of an alcohol, dehydrohalogenation. Every elimination forms the C=C between two carbons that were already bonded to each other. Zaitsev’s rule then usually decides which two, and the most substituted alkene wins whether or not you wanted it. A bulky base can tilt an E2 toward the less substituted alkene, but rarely cleanly. The Wittig builds the C=C from two separate pieces instead, so there is no choice to lose.

Making the ylide

Start from methyl iodide, CH₃I. Two reactions you already know turn it into the Wittig reagent.

Step 1: an S_N2. Triphenylphosphine, Ph₃P (also written PPh₃), has a lone pair on phosphorus. It is an excellent nucleophile because it is large and polarizable: its lone pair sits far from the nucleus and is loosely held, so it reaches a carbon easily. Chemists call a nucleophile like this soft. It attacks the carbon of CH₃I from the back, and iodide leaves, as in any S_N2.

The product, Ph₃P⁺–CH₃ I⁻, is a phosphonium salt: a positive phosphorus carrying four carbon groups, paired with the halide ion.

Step 2: a deprotonation. The hydrogens on the carbon next to P⁺ are now acidic. When one leaves as H⁺, its bonding pair stays on carbon, right beside the positive phosphorus, and that neighbor stabilizes the negative charge. A strong base removes one. Sodium hydride, NaH, works. So does butyllithium, n-BuLi, which acts as a butyl carbanion, Bu⁻, strong enough to take almost any C–H proton. Organolithiums get their own treatment in Organolithiums, Cuprates & Cross-Coupling, later.

Panel 1 is an ordinary S_N2: one arrow from the phosphorus lone pair to carbon, one from the C–I bond to iodine. In panel 2, follow the second arrow: the C–H bonding pair stays behind on carbon as its lone pair.

The product, Ph₃P⁺–CH₂⁻, is an ylide: a neutral molecule with a negative carbon bonded directly to a positive phosphorus. Both charges are real and sit side by side. The lone pair on carbon makes that carbon a strong nucleophile.

You will usually see the ylide written Ph₃P=CH₂ instead. Move the carbon lone pair into the P–C bond and one drawing becomes the other, so the two are resonance forms of one molecule.

Left, the charges drawn out. Right, the shorthand you will see in most reaction schemes.

Treat the P=C form as shorthand. The charged form is the better picture of where the electrons are: most of the negative charge stays on carbon. When you push arrows, draw the charges, because the first arrow starts at that carbon lone pair.

Step 1 is an S_N2, so the halide must be methyl or primary; a secondary halide reacts only slowly. A tertiary halide fails twice: the S_N2 does not happen, and even if it did, the carbon on phosphorus would have no hydrogen left to remove. An aryl halide, such as bromobenzene, does no S_N2 at all. This one constraint decides which alkenes a Wittig can reach, so check the halide first whenever a route proposes one.

The reaction itself

Take acetone, (CH₃)₂C=O, and the ylide Ph₃P⁺–CH₂⁻. The ylide carbon is a nucleophile and the carbonyl carbon is an electrophile, as in every nucleophilic addition. Here, though, the reaction does not stop at an addition.

Two bonds form in one step. The carbon lone pair bonds to the carbonyl carbon. At the same time, the C=O π electrons swing over to phosphorus and make an O–P bond. The four atoms P, C, C and O are now joined in a ring.

That ring is an oxaphosphetane: a four-membered ring of phosphorus, two carbons and oxygen (“oxa” for the oxygen, “phosph” for the phosphorus, “-etane” for a four-membered ring). It carries no charges. Its ring angles are near 90°, far from the usual ones, so it is strained: it holds extra energy and breaks apart easily.

The ring then splits along its other two sides. The C–P and C–O bonds break. Their electrons become a new C=C π bond and a new P=O π bond. The products are the alkene, here 2-methylpropene, (CH₃)₂C=CH₂, and triphenylphosphine oxide.

Read the top row, then the bottom row. In panel 1 the dashed line is the O–P bond as it forms. Follow the two highlighted carbons: they are the ones that end up in the C=C.

The ring-forming step is called a [2+2] cycloaddition, because two atoms from each molecule join into one ring. The ring-breaking step is a retro-[2+2]: the same kind of step, run backward. In general:

R₂C=O + Ph₃P=CR′₂ → R₂C=CR′₂ + Ph₃P=O

Older textbooks and exam questions draw a different first step. The ylide carbon adds to the C=O on its own, giving a betaine: an open chain with P⁺ at one end and O⁻ at the other, Ph₃P⁺–CR₂–CR₂–O⁻. The O⁻ then closes onto phosphorus to give the same oxaphosphetane. Studies of the reaction without lithium salts point to the direct ring closure instead. Both routes reach the same ring and the same products, so your predictions do not change. If your course draws the betaine, draw it on its exams.

The P=O bond drives the reaction. The P=O bond of triphenylphosphine oxide is one of the strongest bonds in organic chemistry, about 130–140 kcal/mol; a C–C single bond is about 83 kcal/mol. Forming it pulls the whole sequence forward. The phosphorus is there only to be thrown away, and a full equivalent of Ph₃P is spent to do it.

Why it beats an elimination

Suppose you want methylenecyclohexane. Heat 1-methylcyclohexan-1-ol with sulfuric acid, and water leaves. The C=C can then form from that ring carbon toward any of three neighbors: two ring carbons and the methyl carbon. Zaitsev’s rule favors the most substituted alkene, so you get mostly 1-methylcyclohexene, with its C=C inside the ring. Methylenecyclohexane is the minor product.

Cyclohexanone with Ph₃P=CH₂ gives methylenecyclohexane alone. The C=O carbon is the only carbon that can end up double-bonded to the ylide carbon.

On the right, the highlighted bonds mark the C=O and the C=C that takes its place.

The Wittig product here is the less stable alkene. The reaction does not prefer it. No other position is open, so stability never gets a vote.

	Elimination	Wittig
Where the C=C goes	Zaitsev decides	Exactly at the old carbonyl carbon
Mixtures?	Often several alkenes	One connectivity
Rearrangement	Possible under E1	None, because no carbocation forms
Limitation	Needs a leaving group	The halide must undergo S_N2

Geometry, which is the one thing it does not fully control

Benzaldehyde, PhCHO, with Ph₃P=CH–CH₃ gives mainly the Z alkene PhCH=CH–CH₃, with Ph and CH₃ on the same side. Benzaldehyde with Ph₃P=CH–CO₂Et gives mainly the E alkene PhCH=CH–CO₂Et, ethyl (E)-3-phenylprop-2-enoate, with Ph and CO₂Et on opposite sides. (E and Z are set out in Cis/trans and E/Z.) The position of the C=C is certain in both. Its geometry depends on the ylide.

The difference is the group on the ylide carbon. In Ph₃P=CH–CO₂Et, that carbon is bonded to the C=O of an ester. The negative charge can spread onto the ester oxygen by resonance, and a spread-out charge is a more stable one.

The two arrows in the upper drawing lead to the lower one. The double-headed arrow joins two drawings of one molecule.

An unstabilized ylide has only H or alkyl groups on the ylide carbon, as in Ph₃P=CH–CH₃. It is very reactive and gives mainly the Z alkene.
A stabilized ylide has a C=O group, such as an ester or a ketone, on the ylide carbon, as in Ph₃P=CH–CO₂Et. It is less reactive and gives mainly the E alkene.
A semi-stabilized ylide has an aryl group (a benzene ring) on the ylide carbon, as in Ph₃P=CH–Ph. It sits between the two and often gives a poor ratio, near 50:50. Avoid it when the geometry matters.

Stabilized ylides also behave differently at the bench. Their C–H is acidic enough for a milder base, such as NaOH or an alkoxide, and many are stable enough to buy and store. They react well with aldehydes but often poorly with ketones.

The geometry is set in the oxaphosphetane. The ring holds the two carbons rigidly, and it splits without letting them rotate. So groups on the same face of the ring end up on the same side of the C=C. A cis-oxaphosphetane, with the two large groups on the same face, gives the Z alkene. A trans-oxaphosphetane gives the E alkene.

Wedges point toward you and hashes away. Compare each ring with the alkene drawn under it.

Why does each ylide make its own ring? Most courses teach an answer based on reversibility. An unstabilized ylide is so reactive that it closes the ring fast and for good, and the cis ring is the one that forms faster. A stabilized ylide adds slowly and reversibly, so a cis ring can reopen and try again, and the more stable trans ring builds up. This is kinetic against thermodynamic control.

Notice the surprise in that account. The more reactive ylide gives the less stable alkene. A fast, one-way step locks in whichever product forms fastest; a slow, reversible one has time to reach the more stable product.

The reversibility explanation is debated. Detailed studies of the reaction without lithium salts suggest that the ring forms irreversibly with both kinds of ylide, and that the shape of the ring-forming transition state picks cis or trans directly. The outcomes are not in doubt: unstabilized gives Z, stabilized gives E. For the reason, give the account your course teaches.

Lithium salts matter for the same reason. When BuLi is the base, the lithium left in the flask lowers the Z preference of an unstabilized ylide. Recipes that need a high Z ratio therefore use sodium or potassium bases instead.

Planning a Wittig backward

Chemists often plan a synthesis backward, starting from the target. This is called retrosynthesis, and each imagined cut of a bond is a disconnection. Synthesis & Retrosynthesis treats the method in full. For a Wittig the recipe is short. Cut the C=C. Put a C=O on one half and the ylide on the other. Then find the halide each ylide needs, and judge it. Both assignments are usually possible; pick the one whose halide is less hindered.

Worked example — make (3-methylbut-1-en-1-yl)benzene, PhCH=CH–CH(CH₃)₂

1. Cut the C=C. The two halves are Ph–CH= and =CH–CH(CH₃)₂.
2. Write both assignments. (a) Benzaldehyde, PhCHO, with the ylide Ph₃P=CH–CH(CH₃)₂. (b) 2-Methylpropanal, (CH₃)₂CH–CHO, with the ylide Ph₃P=CH–Ph.
3. Trace each ylide back to its halide. Put the proton back on the ylide carbon, so it is a CH₂ again, and you have the phosphonium salt. Take off the Ph₃P and put a Br there, and you have the halide. Route (a) needs (CH₃)₂CHCH₂Br, 1-bromo-2-methylpropane, which is primary. The branch one carbon further along slows its S_N2 somewhat but does not stop it. Route (b) needs PhCH₂Br, benzyl bromide. It is benzylic: its C–Br carbon is bonded directly to a benzene ring. Benzylic halides are very good at S_N2.
4. Judge the halides. Both are good S_N2 substrates, so this time the halide does not settle it.
5. Judge the geometry instead. Route (b) goes through a semi-stabilized ylide, the case with the worst E/Z ratio. Route (a) uses an unstabilized ylide and gives mostly the Z alkene. So choose (a) for the Z alkene. If the target is the E alkene, leave the Wittig alone: use the phosphonate variant below, or reduce the alkyne PhC≡C–CH(CH₃)₂ with Na in NH₃.

Each column is one way to split the target. In each, the ylide row sits above the halide it is made from: the halide’s CH₂ becomes the ylide’s CH.

The phosphonate variant

Swap the three phenyls on phosphorus for two ethoxy groups (EtO, CH₃CH₂O–) and a P=O, and you have a phosphonate ester. The usual one is triethyl phosphonoacetate, (EtO)₂P(=O)–CH₂–CO₂Et.

Its CH₂ sits between the phosphorus and the ester C=O, so its hydrogens are acidic enough for NaH or sodium ethoxide, NaOEt. No butyllithium is needed. The anion then reacts with an aldehyde much as a stabilized ylide does. This is the Horner–Wadsworth–Emmons (HWE) reaction.

Left, the highlighted CH₂ is the carbon that loses a proton. Right, the charged by-product is the part that washes out.

There are two practical reasons to choose it when the target is an α,β-unsaturated ester (also called a conjugated ester): an ester whose C=C joins the carbon next to the C=O (the α carbon) to the carbon after it (the β carbon). First, it strongly favors the E alkene. Second, its by-product is a phosphate salt that dissolves in water and washes out in a separatory funnel. Ph₃P=O, by contrast, usually has to be separated from the product by chromatography.

If a question asks for the (E) isomer of an α,β-unsaturated ester, the HWE is the expected answer.

What carries forward

A Wittig turns C=O into C=C with no doubt about where the double bond lands. The cost is a full equivalent of phosphine and an S_N2 limit on the halide. When the alkene you need is the “wrong” one by Zaitsev, this is normally the answer. And when a question gives you an aldehyde and a phosphonium salt, the product is simply the two halves joined by a double bond at the old carbonyl carbon.