THE CONCEPT
Trading a C=O for a C=C, With No Ambiguity About Where
Every other reaction covered so far in this sub-subject either adds a nucleophile across the carbonyl or converts it into a different single-bonded functional group — the Wittig reaction is genuinely different: it directly replaces a C=O double bond with a C=C double bond, converting a carbonyl compound into a specific, predictable alkene. The key reagent is a phosphorus ylide (Ph₃P=CHR) — a species with a carbanion-like carbon directly bonded to a phosphorus atom carrying a positive charge, giving the carbon substantial nucleophilic character.
The mechanism proceeds through a four-membered ring intermediate (an oxaphosphetane) formed when the ylide's carbanion-like carbon attacks the carbonyl carbon, followed by the ring collapsing to eject triphenylphosphine oxide (Ph₃P=O) and form the new C=C double bond in its place. The overall transformation, exactly as the hub's equation states, is Ph₃P=CHR + R'₂C=O → R'₂C=CHR + Ph₃P=O.
💡 Memory Trick
The hub's trick states both the transformation and its key advantage directly: the Wittig reaction converts a carbonyl to an alkene, and it is highly selective — no rearrangements, a specific alkene formed. That selectivity point is worth taking seriously, since it's precisely why the Wittig reaction is chosen so often in synthesis planning: unlike some carbocation-based alkene-forming reactions (like alcohol dehydration) that can suffer from rearrangements or a mixture of regiochemical outcomes, the Wittig places the new double bond in exactly one predictable location, with the carbonyl carbon and the ylide's original carbon becoming the two new alkene carbons and nothing else changing.
PREDICTING E VS. Z STEREOCHEMISTRY FROM THE YLIDE TYPE
Stabilized vs. Non-Stabilized Ylides Give Opposite Geometry
The hub's closing distinction is essential for predicting product stereochemistry correctly: non-stabilized ylides give the Z (cis) alkene, while stabilized ylides give the E (trans) alkene. A 'non-stabilized' ylide has ordinary alkyl or hydrogen substituents on its carbanion carbon, with no additional resonance stabilization available — this type of ylide reacts through the four-membered oxaphosphetane intermediate quickly and with less time for that intermediate to equilibrate toward its most stable arrangement, kinetically favoring the Z alkene.
A 'stabilized' ylide instead has an adjacent group (like a carbonyl or an aromatic ring) that can delocalize the carbanion's negative charge by resonance, making that ylide more stable and less reactive overall. Because a stabilized ylide reacts more slowly and reversibly, the oxaphosphetane intermediate has more opportunity to equilibrate toward its lower-energy, more thermodynamically stable arrangement before collapsing — and that more stable arrangement leads to the E alkene. Recognizing which type of ylide you're given (checking specifically for an adjacent resonance-stabilizing group) is exactly what lets you predict E versus Z stereochemistry correctly, rather than guessing.
🧪 Lab Application
You're running a Wittig reaction between cyclohexanone and a non-stabilized ylide (Ph₃P=CHCH₃) and need to predict both the product and its alkene stereochemistry.
1
Identify the carbonyl carbon and the ylide carbon that will become the new alkene. Cyclohexanone's carbonyl carbon and the ylide's CHCH3 carbon will become the two carbons of the new double bond.
2
Predict the overall product skeleton. The reaction replaces the C=O with a C=C, giving an ethylidenecyclohexane product (a cyclohexane ring with an exocyclic =CHCH3 group) plus triphenylphosphine oxide as the byproduct.
3
Classify the ylide type. Ph₃P=CHCH3 has only an ordinary alkyl substituent on its carbanion carbon, with no adjacent resonance-stabilizing group — classifying it as a non-stabilized ylide.
4
Predict the stereochemical outcome. Since non-stabilized ylides favor the Z alkene, expect the Z-configured ethylidenecyclohexane product to predominate.
📌 Exam Application
Wittig questions frequently test both halves of this lesson together — predicting the correct alkene connectivity from the carbonyl and ylide structures, AND predicting E vs. Z stereochemistry from whether the ylide is stabilized or non-stabilized — always address both parts explicitly rather than stopping at the connectivity.
⚠️ Most Common Wittig Reaction Mistakes
The most common mistake is forgetting to check whether the given ylide is stabilized or non-stabilized before predicting stereochemistry, defaulting to a guess rather than checking for an adjacent resonance-stabilizing group. The other frequent trap is forgetting the byproduct entirely — every Wittig reaction produces triphenylphosphine oxide (Ph3P=O) alongside the alkene product, and a complete mechanism or balanced equation should show it.
✓ Quick Self-Test
1) What is a phosphorus ylide? 2) What is the overall transformation accomplished by the Wittig reaction? 3) What byproduct forms alongside the alkene product? 4) What stereochemistry (E or Z) does a non-stabilized ylide typically give? 5) What stereochemistry does a stabilized ylide typically give, and why does resonance stabilization shift the outcome?
Next Lesson
Keto-Enol Tautomerism
→
← All Aldehydes & Ketones Lessons