The Wittig reaction — Carbonyl & Enolate Breadth

You have two ways to make an alkene so far. One is to reduce an alkyne — Lindlar for cis, Na/NH₃ for trans — which controls geometry beautifully but needs an alkyne with the right skeleton already in place. The other, and the general one, is elimination: E1, E2, dehydration of an alcohol, dehydrohalogenation. Every elimination shares a problem. The double bond forms between two carbons that were already joined, and Zaitsev decides where it ends up, so you get the most substituted alkene whether or not that is the one you wanted.

The Wittig reaction is different in the way that matters: it builds the C=C from two separate pieces, and the double bond lands exactly where the old carbonyl carbon was. No rearrangement, no Zaitsev, no ambiguity.

Making the ylide

Two steps, and both are reactions you already know.

S_N2. Triphenylphosphine, PPh₃, is an excellent nucleophile — large, polarizable, and soft. It displaces the halide from a methyl or primary alkyl halide to give a phosphonium salt.
Deprotonation. The hydrogens on the carbon next to P⁺ are now acidic, because the resulting carbanion sits beside a positive charge. A strong base — n-BuLi or NaH — removes one.

What you have made is an ylide: a molecule with a negative carbon bonded directly to a positive phosphorus, neutral overall. Both charges are real and adjacent, which is exactly why the thing is reactive.

The alkyl halide must be methyl, primary, or at a push secondary, because step one is an S_N2. That single constraint decides which alkenes a Wittig can reach, and it is the first thing to check when a route proposes one.

The reaction itself

The ylide carbon is nucleophilic and attacks the carbonyl carbon. The alkoxide that forms closes onto phosphorus to give a four-membered ring — an oxaphosphetane — which then falls apart in the one way it can:

R₂C=O + Ph₃P=CR′₂ → R₂C=CR′₂ + Ph₃P=O

The driving force is the last product. The P=O bond of triphenylphosphine oxide is one of the strongest bonds in organic chemistry, around 130–140 kcal/mol, and forming it is what pulls the whole sequence forward. Phosphorus is in this reaction to be thrown away, expensively.

Why it beats an elimination

	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
Limitation	Needs a leaving group	The halide must survive an S_N2

Retrosynthetically this gives you a clean disconnection. Cut any C=C in the middle, put a carbonyl on one half and an ylide on the other, and ask which assignment leaves the easier alkyl halide. Both directions are usually possible, and the one with the less hindered halide wins.

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

The reason a Wittig is worth the phosphine. An elimination forms the C=C between two carbons that were already bonded, so Zaitsev decides which, and on a ring that means the endocyclic alkene. A Wittig puts the double bond where the carbonyl carbon was, which here is pointing out of the ring.Note that the Wittig product is the less stable of the two. That is the point: it is not that the reaction prefers the exocyclic alkene, it is that no other position is available to it, so stability never gets a vote. Where regiochemistry is concerned, having only one option is better than having a preference.

The connectivity is certain; the cis/trans outcome depends on the ylide:

An unstabilized ylide — the carbanion carries only alkyl groups — is very reactive and gives mainly the cis (Z) alkene.
A stabilized ylide — the carbanion is conjugated to an ester or ketone — is less reactive, has time to equilibrate, and gives mainly the trans (E) alkene.

Note which way round that is, because it is the opposite of the usual intuition. The more reactive ylide gives the less stable alkene. A fast, irreversible addition locks in whichever geometry the transition state happened to have; a slow, reversible one has time to find the more stable arrangement.

What carries forward

A Wittig turns C=O into C=C with no ambiguity about where the double bond lands, at the cost of a stoichiometric phosphine and an S_N2 constraint on the halide. When a target alkene 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 carbonyl carbon.