---
title: "Wittig and Other C=C Forming Reactions"
book: "University Chemistry — Year 2"
subject: chemistry
language: en
chapter: 27
exercises: 12
source: https://one-course.com/books/chemistry/3/en/chapter/27-wittig-and-other-c-c-forming-reactions
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 27 — Wittig and Other C=C Forming Reactions

A female housefly attracts males with a hydrocarbon of 23 carbons carrying a single double bond, between carbons 9 and 10, with the (*Z*) geometry. To make it in quantity, a chemist must put that one double bond exactly there, and with that geometry: an elimination would scatter it over several positions and give both geometries. This chapter builds double bonds the other way, by joining two pieces at a carbonyl group, so that the double bond appears where the $\ce{C=O}$ was.

**You already know.**

[Chapter 25](https://one-course.com/books/chemistry/3/en/chapter/25-enols-enolates-and-the-aldol-reaction#ch-b2-enolates-aldol): the [aldol condensation](https://one-course.com/books/chemistry/3/en/chapter/25-enols-enolates-and-the-aldol-reaction#def-b2-enolates-aldol-aldol). [Chapter 21](https://one-course.com/books/chemistry/3/en/chapter/21-oxidation-and-reduction-of-alkenes#ch-b2-alkene-redox): the hydrogenation of an alkyne to a (*Z*)-alkene over a poisoned catalyst. The Year 1 volume: E1, E2 and Zaitsev’s rule, bimolecular substitution, Grignard reagents and polarity inversion, the (*Z*)/(*E*) descriptors. The school volume: atom economy.

## 27.1 Phosphonium salts and ylides

**Definition 27.1 (Phosphonium ylides).**

A *phosphonium salt* $\mathrm{R_3P^+{-}CH_2R'\ X^-}$ is made by bimolecular substitution of a halogenoalkane by a phosphine, usually triphenylphosphine $\ce{Ph3P}$. Removing a hydrogen from the carbon bound to phosphorus gives a *phosphonium ylide*, a neutral molecule with adjacent opposite charges, $\mathrm{Ph_3P^+{-}CH^-{-}R'}$, also written $\mathrm{Ph_3P{=}CHR'}$. A *stabilised ylide* carries on that carbon an electron-withdrawing group (an ester, a ketone, a [nitrile](https://one-course.com/books/chemistry/3/en/chapter/23-amines-and-nitrogen-compounds#def-b2-amines-nitrile)) that delocalises the negative charge.

![Methylidenetriphenylphosphorane, the simplest ylide. Triphenylphosphine displaces iodide from iodomethane; butyllithium removes a proton from the methyl group. The two structures on the right are two ways of writing the same molecule: the charge-separated one is the closer to reality.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-7e442b2557ef.svg)

*Methylidenetriphenylphosphorane, the simplest ylide. Triphenylphosphine displaces iodide from iodomethane; butyllithium removes a proton from the methyl group. The two structures on the right are two ways of writing the same molecule: the charge-separated one is the closer to reality.*

**Proposition 27.2 (Making the ylide).**

A [phosphonium salt](#def-b2-wittig-ylide) is far more acidic at the carbon next to phosphorus than an alkane. A simple alkyl [phosphonium salt](#def-b2-wittig-ylide) still needs a very strong base (butyllithium, sodium hydride, sodium amide) under dry, inert conditions; a salt whose $\ce{CH2}$ also carries a carbonyl group is deprotonated by a weak base (sodium carbonate, sodium hydroxide), and the [stabilised ylide](#def-b2-wittig-ylide) formed can be isolated and stored.

**Argument.** The positive phosphorus next to the carbanion stabilises it by its charge and by the polarisability of the large phosphorus atom: a simple ylide is a stabilised carbanion, but less so than an [enolate](https://one-course.com/books/chemistry/3/en/chapter/25-enols-enolates-and-the-aldol-reaction#def-b2-enolates-aldol-enolate), hence the strong base. An adjacent carbonyl spreads the negative charge onto oxygen, as in an [enolate](https://one-course.com/books/chemistry/3/en/chapter/25-enols-enolates-and-the-aldol-reaction#def-b2-enolates-aldol-enolate) ([Proposition 25.4](https://one-course.com/books/chemistry/3/en/chapter/25-enols-enolates-and-the-aldol-reaction#prop-b2-enolates-aldol-alpha-acidity)), on top of the phosphorus: the conjugate acid becomes acidic enough for a weak base, and the ylide, much less basic, no longer reacts with water or air. ∎

## 27.2 The Wittig reaction

**Definition 27.3 (Wittig reaction).**

The *Wittig reaction* of a [phosphonium ylide](#def-b2-wittig-ylide) with an aldehyde or a ketone gives an alkene and triphenylphosphine oxide:

$$
\mathrm{Ph_3P{=}CHR + R'CHO \longrightarrow R'CH{=}CHR + Ph_3P{=}O}.
$$

It goes through an *oxaphosphetane*, a four-membered ring of phosphorus, two carbons and oxygen, formed by the ylide carbon bonding to the carbonyl carbon while the carbonyl oxygen bonds to phosphorus.

![The Wittig reaction. The ylide carbon and the carbonyl carbon bond together while oxygen bonds to phosphorus, in one step: the oxaphosphetane. Its ring then opens the other way, breaking the P-C and C-O bonds: the C=C bond forms between the two former reacting carbons, and the P=O bond of triphenylphosphine oxide forms at the same time.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-7eacd0b06315.svg)

![The Wittig reaction. The ylide carbon and the carbonyl carbon bond together while oxygen bonds to phosphorus, in one step: the oxaphosphetane. Its ring then opens the other way, breaking the P-C and C-O bonds: the C=C bond forms between the two former reacting carbons, and the P=O bond of triphenylphosphine oxide forms at the same time.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-f9a599943c9e.svg)

*The [Wittig reaction](#def-b2-wittig-wittig). The ylide carbon and the carbonyl carbon bond together while oxygen bonds to phosphorus, in one step: the [oxaphosphetane](#def-b2-wittig-wittig). Its ring then opens the other way, breaking the $\ce{P-C}$ and $\ce{C-O}$ bonds: the $\ce{C=C}$ bond forms between the two former reacting carbons, and the $\ce{P=O}$ bond of triphenylphosphine oxide forms at the same time.*

**Proposition 27.4 (A double bond exactly in place).**

In a [Wittig reaction](#def-b2-wittig-wittig) the new $\ce{C=C}$ bond joins the former carbonyl carbon to the former ylide carbon, and nowhere else.

**Argument.** The only bonds made and broken are those of the four-membered ring: the ylide carbon to the carbonyl carbon, then the $\ce{C-P}$ and $\ce{C-O}$ bonds. No hydrogen moves and no carbocation or carbanion forms along the chain, so the double bond cannot shift. An elimination, by contrast, chooses between the hydrogens on several neighbouring carbons (Zaitsev’s rule), and an E1 reaction may rearrange. ∎

**Proposition 27.5 (Driving force).**

The formation of the $\ce{P=O}$ bond of triphenylphosphine oxide makes the [Wittig reaction](#def-b2-wittig-wittig) strongly favourable and irreversible.

**Argument.** The reaction trades a $\ce{C=O}$ and a $\ce{P-C}$ bond (in the ylide, the $\ce{P=C}$ is mostly a $\ce{P+-C-}$ single bond) for a $\ce{C=C}$ and a $\ce{P=O}$ bond. The $\ce{P=O}$ bond of a phosphine oxide is very strong, much stronger than the $\ce{P-C}$ bond lost, and the $\ce{C=O}$ lost is compensated in part by the $\ce{C=C}$ formed: the balance is clearly downhill. Phosphorus is the oxygen sink of the reaction. ∎

## 27.3 Stereoselectivity

**Proposition 27.6 (Geometry of the alkene).**

With an aldehyde, a non-stabilised ylide ($\ce{Ph3P=CHR}$, R an alkyl group) gives mainly the (*Z*)-alkene, under salt-free conditions. A [stabilised ylide](#def-b2-wittig-ylide) ($\ce{Ph3P=CH-COOR}$, $\ce{Ph3P=CH-COR}$) gives mainly the (*E*)-alkene. An ylide stabilised only by an aryl group gives mixtures.

**Proof.** *Admitted at this level.* ∎

These are observations. Their usual explanation reads the [Wittig reaction](#def-b2-wittig-wittig) as a competition between rates and stabilities. With a reactive, non-stabilised ylide, the [oxaphosphetane](#def-b2-wittig-wittig) forms fast and irreversibly, through the less crowded transition state, in which the two substituents end up on the same side of the ring; it then collapses with retention of that arrangement: kinetic control gives the (*Z*)-alkene. With a [stabilised ylide](#def-b2-wittig-ylide), the first step is slower and reversible; the [oxaphosphetanes](#def-b2-wittig-wittig) interconvert through the starting materials, and the one with the substituents on opposite sides, less strained, wins: the (*E*)-alkene, under thermodynamic control.

![Two ylides, one aldehyde. Top: the non-stabilised propylidene ylide gives mainly (Z)-1-phenylbut-1-ene. Bottom: the stabilised ylide gives mainly ethyl (E)-3-phenylpropenoate (ethyl cinnamate). Triphenylphosphine oxide, formed in both, is not shown.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-31229012a71f.svg)

![Two ylides, one aldehyde. Top: the non-stabilised propylidene ylide gives mainly (Z)-1-phenylbut-1-ene. Bottom: the stabilised ylide gives mainly ethyl (E)-3-phenylpropenoate (ethyl cinnamate). Triphenylphosphine oxide, formed in both, is not shown.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-9f88cf703654.svg)

*Two ylides, one aldehyde. Top: the non-stabilised propylidene ylide gives mainly (*Z*)-1-phenylbut-1-ene. Bottom: the [stabilised ylide](#def-b2-wittig-ylide) gives mainly ethyl (*E*)-3-phenylpropenoate (ethyl cinnamate). Triphenylphosphine oxide, formed in both, is not shown.*

## 27.4 The Horner–Wadsworth–Emmons reaction

**Definition 27.7 (Phosphonates and the HWE reaction).**

A *phosphonate* is an ester of a phosphonic acid, $\mathrm{(RO)_2P({=}O){-}R'}$, with a $\ce{P-C}$ bond. It is made by the Michaelis–Arbuzov reaction of a trialkyl phosphite with a halogenoalkane, $\mathrm{P(OEt)_3 + R'Br \to (EtO)_2P(O)R' + EtBr}$. In the *Horner–Wadsworth–Emmons reaction* (HWE), the anion of a phosphonate carrying an electron-withdrawing group on its $\ce{CH2}$ adds to an aldehyde and gives an alkene and a dialkyl phosphate anion.

![The Horner–Wadsworth–Emmons reaction of triethyl phosphonoacetate with benzaldehyde. Sodium hydride removes the hydrogen between the phosphonate and the ester; the anion adds to the aldehyde, and the four-membered intermediate collapses as in the Wittig reaction. The product is ethyl (E)-cinnamate; the by-product is the diethyl phosphate anion, soluble in water.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-e689bd3a8b6a.svg)

![The Horner–Wadsworth–Emmons reaction of triethyl phosphonoacetate with benzaldehyde. Sodium hydride removes the hydrogen between the phosphonate and the ester; the anion adds to the aldehyde, and the four-membered intermediate collapses as in the Wittig reaction. The product is ethyl (E)-cinnamate; the by-product is the diethyl phosphate anion, soluble in water.](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-cff722165922.svg)

*The [Horner–Wadsworth–Emmons reaction](#def-b2-wittig-hwe) of triethyl phosphonoacetate with benzaldehyde. Sodium hydride removes the hydrogen between the [phosphonate](#def-b2-wittig-hwe) and the ester; the anion adds to the aldehyde, and the four-membered intermediate collapses as in the [Wittig reaction](#def-b2-wittig-wittig). The product is ethyl (*E*)-cinnamate; the by-product is the diethyl phosphate anion, soluble in water.*

**Proposition 27.8 (Why chemists like the HWE reaction).**

The HWE reaction gives mainly the (*E*)-alkene, and its phosphorus by-product, a water-soluble salt, is removed by a simple aqueous wash; the [phosphonate](#def-b2-wittig-hwe) anion is also more nucleophilic than the corresponding [stabilised ylide](#def-b2-wittig-ylide) and reacts with ketones too.

**Argument.** The [phosphonate](#def-b2-wittig-hwe) anion is a stabilised carbanion, so its addition is reversible and the reaction runs under thermodynamic control: (*E*), as with [stabilised ylides](#def-b2-wittig-ylide). Unlike a [phosphonium ylide](#def-b2-wittig-ylide), it carries a full negative charge, not compensated by a neighbouring cation: it is more reactive. The by-product, $\ce{(EtO)2PO2-}$, is an ionic salt, while triphenylphosphine oxide is a neutral organic solid that has to be separated from the product by chromatography or crystallisation. ∎

## 27.5 Choosing a method

**Method 27.9 (Wittig disconnection).**

1. Cut the target $\ce{C=C}$ bond; one carbon becomes a carbonyl carbon, the other the ylide carbon (a [phosphonium salt](#def-b2-wittig-ylide) , from a halide).
2. Of the two ways of doing it, prefer the one whose halide is primary (bimolecular substitution by $\ce{Ph3P}$ ) and whose carbonyl compound is an aldehyde.
3. Choose the reagent from the geometry wanted: a non-stabilised ylide for ( *Z* ), a [stabilised ylide](#def-b2-wittig-ylide) or a [phosphonate](#def-b2-wittig-hwe) (HWE) for ( *E* ).
4. Check that nothing else in the molecule reacts with the base used.

**Method 27.10 (Choosing a C=C forming reaction).**

| Method | Position of the $\ce{C=C}$ | Geometry |
| --- | --- | --- |
| E2 or E1 elimination | Zaitsev’s rule, mixtures | mostly (*E*), mixtures |
| [Aldol condensation](https://one-course.com/books/chemistry/3/en/chapter/25-enols-enolates-and-the-aldol-reaction#def-b2-enolates-aldol-aldol) | conjugated with $\ce{C=O}$ | (*E*) |
| Wittig, non-stabilised ylide | at the former $\ce{C=O}$ | (*Z*) |
| Wittig, [stabilised ylide](#def-b2-wittig-ylide); HWE | at the former $\ce{C=O}$ | (*E*) |
| Alkyne, then poisoned catalyst | at the former triple bond | (*Z*) |

Joining two pieces of similar size at a $\ce{C=C}$, the [Wittig reaction](#def-b2-wittig-wittig) and its relatives are usually the first choice; olefin metathesis, which exchanges the ends of two alkenes, comes in the Year 3 volume.

**History — The ylide that became a reagent.**

Georg Wittig, studying phosphorus ylides in the early 1950s, found that methylidenetriphenylphosphorane converts benzophenone into 1,1-diphenylethene and triphenylphosphine oxide; with his student Ulrich Schöllkopf he turned the observation into a general method in 1954. Industry adopted it quickly, for vitamin A among others, and Wittig shared the 1979 Nobel Prize in Chemistry.

**Safety.**

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-4b496db0d6c3.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-afc03563fb34.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-b3bf7bb196be.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-3/b2-wittig/fig-748906458fb6.svg)

Butyllithium solutions are flammable and corrosive and react violently with water; sodium hydride releases hydrogen with water. Triphenylphosphine is harmful and corrosive to the eyes, triethyl phosphite flammable and harmful; triphenylphosphine oxide is harmful.

## 27.6 Exercises

**Exercise 27.1 ★.**

Write the preparation of the ylide $\ce{Ph3P=CHCH3}$ from triphenylphosphine and a halogenoalkane.

**Solution of Exercise 27.1.**

$\ce{Ph3P + CH3CH2Br -> [Ph3PCH2CH3]+ + Br-}$ (bimolecular substitution, heating in a solvent), then butyllithium in dry ether or THF under inert gas: $\ce{[Ph3PCH2CH3]+ + BuLi -> Ph3P=CHCH3 + BuH + Li+}$.

**Exercise 27.2 ★.**

Give the products of the [Wittig reaction](#def-b2-wittig-wittig) of benzaldehyde with the ylide made from iodomethane.

**Solution of Exercise 27.2.**

The ylide $\ce{Ph3P=CH2}$ and benzaldehyde give phenylethene (styrene), $\ce{C6H5CH=CH2}$, and triphenylphosphine oxide.

**Exercise 27.3 ★.**

Classify as stabilised, non-stabilised or aryl-stabilised: $\ce{Ph3P=CHCOOEt}$, $\ce{Ph3P=CHCH3}$, $\ce{Ph3P=CHCOCH3}$, $\ce{Ph3P=CHC6H5}$. Which needs butyllithium?

**Solution of Exercise 27.3.**

$\ce{Ph3P=CHCOOEt}$ and $\ce{Ph3P=CHCOCH3}$: stabilised. $\ce{Ph3P=CHCH3}$: non-stabilised, the one that needs butyllithium (or sodium hydride, sodium amide). $\ce{Ph3P=CHC6H5}$: aryl-stabilised, made with an alkoxide or hydroxide in practice, and giving mixtures of geometries.

**Exercise 27.4 ★.**

Give the product of triethyl phosphonoacetate, sodium hydride and propanal, with its geometry.

**Solution of Exercise 27.4.**

Ethyl (*E*)-pent-2-enoate, $\ce{CH3CH2CH=CHCOOEt}$, with sodium diethyl phosphate.

**Exercise 27.5 ★★.**

Methylenecyclohexane is wanted. Compare a Wittig route from cyclohexanone with the acid-catalysed dehydration of 1-methylcyclohexan-1-ol.

**Solution of Exercise 27.5.**

Wittig: the ylide $\ce{Ph3P=CH2}$, made from methyltriphenylphosphonium iodide and butyllithium, with cyclohexanone gives only methylenecyclohexane, the double bond outside the ring. Dehydration of 1-methylcyclohexan-1-ol goes through a tertiary carbocation and loses the hydrogen that gives the more substituted alkene (Zaitsev): mainly 1-methylcyclohexene, the wrong isomer.

**Exercise 27.6 ★★.**

Disconnect (*E*)-1,2-diphenylethene (stilbene) by the [Wittig reaction](#def-b2-wittig-wittig). Why is the Wittig route a poor choice for pure (*E*)-stilbene, and what would you use instead?

**Solution of Exercise 27.6.**

Both cuts are the same: benzaldehyde and the benzylidene ylide $\ce{Ph3P=CHC6H5}$. That ylide is only aryl-stabilised and gives an (*E*)/(*Z*) mixture. For pure (*E*), use the HWE reaction of diethyl benzylphosphonate with benzaldehyde.

**Exercise 27.7 ★★.**

Propanal reacts with $\ce{Ph3P=CHCH3}$ on one side and with $\ce{Ph3P=CHCOOEt}$ on the other. Give the products and their main geometry.

**Solution of Exercise 27.7.**

With $\ce{Ph3P=CHCH3}$: (*Z*)-pent-2-ene, $\ce{CH3CH2CH=CHCH3}$, mainly (*Z*). With $\ce{Ph3P=CHCOOEt}$: ethyl (*E*)-pent-2-enoate, mainly (*E*).

**Exercise 27.8 ★★.**

Compute the atom economy of the methylenation of cyclohexanone by $\ce{Ph3P=CH2}$.

**Solution of Exercise 27.8.**

$\ce{Ph3P=CH2 + C6H10O -> C7H12 + Ph3PO}$. Masses: methylenecyclohexane $96.173\,\mathrm{g}/\mathrm{mol}$, triphenylphosphine oxide $278.291\,\mathrm{g}/\mathrm{mol}$; atom economy $96.173/(96.173 + 278.291) =
96.173/374.464 \approx 25.7~\%$. Three quarters of the mass of the reagents leaves as phosphine oxide.

**Exercise 27.9 ★★.**

Write the Michaelis–Arbuzov reaction of triethyl phosphite with ethyl bromoethanoate, in two steps: substitution by phosphorus, then dealkylation by bromide.

**Solution of Exercise 27.9.**

Phosphorus, nucleophilic, displaces bromide:

$$
\ce{P(OEt)3 + BrCH2COOEt -> [(EtO)3PCH2COOEt]+ + Br-}.
$$

Bromide then attacks a carbon of an ethyl group of the phosphonium ion (bimolecular substitution), releasing bromoethane and forming the $\ce{P=O}$ bond:

$$
\ce{[(EtO)3PCH2COOEt]+ + Br- -> (EtO)2P(O)CH2COOEt + EtBr}.
$$

**Exercise 27.10 ★★★.**

Propose a synthesis of 1,6-diphenylhexa-1,3,5-triene from (*E*)-but-2-enedial $\ce{OHC-CH=CH-CHO}$ by two [Wittig reactions](#def-b2-wittig-wittig), and say which ylide and how much of it.

**Solution of Exercise 27.10.**

$\ce{OHC-CH=CH-CHO + 2Ph3P=CHC6H5 -> C6H5CH=CH-CH=CH-CH=CHC6H5 + 2Ph3PO}$: two equivalents of the benzylidene ylide (from benzyltriphenylphosphonium chloride and a base), one for each aldehyde group. The new double bonds are formed as mixtures of geometries; the all-(*E*) isomer, the most stable, is isolated by crystallisation, or the mixture is isomerised.

**Exercise 27.11 ★★★.**

(*E*)-4-Phenylbut-3-en-2-one can be made by an [aldol condensation](https://one-course.com/books/chemistry/3/en/chapter/25-enols-enolates-and-the-aldol-reaction#def-b2-enolates-aldol-aldol) or by a [Wittig reaction](#def-b2-wittig-wittig). Write both routes and compare them.

**Solution of Exercise 27.11.**

[Aldol](https://one-course.com/books/chemistry/3/en/chapter/25-enols-enolates-and-the-aldol-reaction#def-b2-enolates-aldol-aldol): benzaldehyde with propanone in excess and sodium hydroxide ([Chapter 25](https://one-course.com/books/chemistry/3/en/chapter/25-enols-enolates-and-the-aldol-reaction#ch-b2-enolates-aldol)); cheap reagents, water as by-product, (*E*) product, but the condensation can run twice (dibenzalacetone) if propanone is not in excess. Wittig: benzaldehyde with the [stabilised ylide](#def-b2-wittig-ylide) $\ce{Ph3P=CHCOCH3}$; (*E*), one product, no double condensation, but the ylide is costly and triphenylphosphine oxide must be removed. The [aldol](https://one-course.com/books/chemistry/3/en/chapter/25-enols-enolates-and-the-aldol-reaction#def-b2-enolates-aldol-aldol) is the better route here.

**Exercise 27.12 ★★★.**

Propose two routes to (*Z*)-oct-4-ene, one by the [Wittig reaction](#def-b2-wittig-wittig) and one through an alkyne, and say which by-products each leaves.

**Solution of Exercise 27.12.**

Wittig: butanal with $\ce{Ph3P=CHCH2CH2CH3}$ (from 1-bromobutane, then butyllithium), salt-free: mainly (*Z*); by-product triphenylphosphine oxide, and some (*E*) isomer. Alkyne: oct-4-yne (from ethyne by two alkylations with 1-bromopropane and sodium amide) hydrogenated over a poisoned palladium catalyst ([Proposition 21.2](https://one-course.com/books/chemistry/3/en/chapter/21-oxidation-and-reduction-of-alkenes#prop-b2-alkene-redox-syn-hydrogenation)): (*Z*) only; the by-products are sodium bromide and ammonia from the alkylations.

## 27.7 Problem: A Pheromone by Wittig

**Problem 27.1.**

Weekend problem — the retrosynthesis of a (Z)-alkene, the non-stabilised ylide and its selectivity, the cost in triphenylphosphine oxide, and the alkyne alternative

Target: (*Z*)-tricos-9-ene, $\ce{CH3(CH2)7CH=CH(CH2)12CH3}$, the housefly pheromone of the opening. Molar masses ($\mathrm{g}/\mathrm{mol}$): C 12.011, H 1.008, O 15.999, P 30.974, Br 79.904.

**Part I — Retrosynthesis.**

1. Write the molecular formula of the target and check that it is an alkene.
2. Which bond does a Wittig disconnection cut?
3. Give the two possible pairs (carbonyl compound, [phosphonium salt](#def-b2-wittig-ylide) ).
4. For the pair nonanal + ylide, which halogenoalkane makes the [phosphonium salt](#def-b2-wittig-ylide) ?
5. Why is a primary halogenoalkane suitable for this step?
6. Write the formation of the [phosphonium salt](#def-b2-wittig-ylide) .

**Part II — The ylide and the geometry.**

7. Which base deprotonates this salt? Why would sodium hydroxide not do?
8. Is the ylide stabilised? What geometry does that predict?
9. Write the [oxaphosphetane](#def-b2-wittig-wittig) formed with nonanal.
10. Why must the reaction be run dry and under inert gas?
11. Why does the double bond end up exactly between carbons 9 and 10?
12. What would the acid-catalysed dehydration of tricosan-9-ol give instead?

**Part III — Triphenylphosphine oxide.**

13. Write the overall equation from the ylide and nonanal.
14. Compute the molar masses of the target, of nonanal, of the [phosphonium salt](#def-b2-wittig-ylide) and of triphenylphosphine oxide.
15. Compute the atom economy of the Wittig step (ylide + aldehyde).
16. How is triphenylphosphine oxide separated from a hydrocarbon?
17. Why is the formation of triphenylphosphine oxide the driving force?
18. Why would the HWE reaction not suit this target?

**Part IV — The alkyne route.**

19. Which alkyne, hydrogenated, gives the target, and with which catalyst?
20. Build that alkyne from dec-1-yne and a halogenoalkane: which halide, which base?
21. Why must the hydrogenation stop at the alkene, and how does the catalyst ensure it?
22. Compare the two routes: number of steps, geometry control, by-products.
23. What is the atom economy of the hydrogenation step?
24. State the mass of triphenylphosphine oxide co-produced per gram of pheromone by the Wittig route.

**Solution of Problem 27.1.**

**1.** $\ce{C23H46}$, that is $\ce{C_nH_{2n}}$ with $n = 23$: one double bond (or one ring), here a $\ce{C=C}$. **2.** The $\ce{C9=C10}$ double bond. **3.** Nonanal $\ce{CH3(CH2)7CHO}$ with the ylide from a 1-halotetradecane; or tetradecanal $\ce{CH3(CH2)12CHO}$ with the ylide from a 1-halononane. **4.** 1-Bromotetradecane, $\ce{CH3(CH2)13Br}$. **5.** Bimolecular substitution by the bulky triphenylphosphine is fast at an unhindered primary carbon and does not compete with elimination. **6.** $\ce{Ph3P + C14H29Br -> [Ph3PC14H29]+ + Br-}$, tetradecyltriphenylphosphonium bromide. **7.** Butyllithium (or sodium hydride, sodium amide); hydroxide is too weak a base for a non-stabilised ylide, and the water it brings would protonate the ylide. **8.** No, it carries only an alkyl group: mainly (*Z*) ([Proposition 27.6](#prop-b2-wittig-stereo)). **9.** A four-membered ring $\ce{P-C-C-O}$: phosphorus bonded to the carbon carrying the $\ce{C13H27}$ chain, that carbon to the one carrying the $\ce{C8H17}$ chain, and that one to oxygen, bonded to phosphorus. **10.** Butyllithium and the ylide react with water and oxygen. **11.** The double bond joins the former carbonyl carbon of nonanal (C9 of the product) to the former ylide carbon (C10), and no other bond moves ([Proposition 27.4](#prop-b2-wittig-regiospecific)). **12.** A carbocation at C9, which may lose a hydrogen from C8 or C10 and may shift: a mixture of tricos-8-ene and tricos-9-ene, each as (*E*) and (*Z*), mostly (*E*). **13.** $\ce{Ph3P=CHC13H27 + C8H17CHO -> C8H17CH=CHC13H27 + Ph3PO}$. **14.** Target $\ce{C23H46}$: $322.621\,\mathrm{g}/\mathrm{mol}$; nonanal $\ce{C9H18O}$: $142.242\,\mathrm{g}/\mathrm{mol}$; salt $\ce{C32H44BrP}$: $539.582\,\mathrm{g}/\mathrm{mol}$; $\ce{Ph3PO}$, $\ce{C18H15OP}$: $278.291\,\mathrm{g}/\mathrm{mol}$. **15.** The ylide $\ce{C32H43P}$ weighs $539.582 - 80.912 = 458.670\,\mathrm{g}/\mathrm{mol}$; with nonanal, $600.912\,\mathrm{g}/\mathrm{mol}$ of reagents for $322.621\,\mathrm{g}/\mathrm{mol}$ of product: $322.621/600.912 \approx
53.7~\%$. **16.** By its polarity: the nonpolar hydrocarbon passes through a short silica column with a nonpolar eluent while the oxide stays; or the oxide crystallises out of a cold nonpolar solvent. **17.** The very strong $\ce{P=O}$ bond formed outweighs the $\ce{P-C}$ bond broken ([Proposition 27.5](#prop-b2-wittig-driving)). **18.** It needs a [phosphonate](#def-b2-wittig-hwe) with an electron-withdrawing group on the reacting carbon, absent from a plain alkyl chain, and it gives the (*E*)-alkene. **19.** Tricos-9-yne, $\ce{CH3(CH2)7C#C(CH2)12CH3}$, with hydrogen over a poisoned palladium catalyst (palladium on calcium carbonate with a lead salt and quinoline). **20.** Dec-1-yne, $\ce{CH3(CH2)7C#CH}$, deprotonated by sodium amide, then 1-bromotridecane $\ce{CH3(CH2)12Br}$: $8 + 2 + 13 = 23$ carbons, the triple bond between C9 and C10. **21.** Further hydrogenation would give tricosane, with no double bond; the poisoned catalyst binds the alkyne much more strongly than the alkene, which leaves the surface before a second addition. **22.** Both take two steps from available pieces. The alkyne route gives the (*Z*) isomer cleanly and leaves only salts; the Wittig route gives some (*E*) isomer and a mass of triphenylphosphine oxide. **23.** 100 %: $\ce{C23H44 + H2 -> C23H46}$, every atom ends in the product. **24.** $278.291/322.621 = \boldsymbol{\approx 0.86\,\mathrm{g}}$ of triphenylphosphine oxide per gram of pheromone.