The Wittig Reaction

Organophosphorus ylides react with aldehydes or ketones to give substituted alkenes in a transformation called the Wittig reaction.  This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979.  A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.

Preparation of Phosphorus Ylides

It has been noted that dipolar phosphorus compounds are stabilized by p-d bonding. This bonding stabilization extends to carbanions adjacent to phosphonium centers, and the zwitterionic conjugate bases derived from such cations are known as ylides. An ylide is defined as a compound with opposite charges on adjacent atoms both of which have complete octets. For the Wittig reaction discussed below an organophosphorus ylide, also called Wittig reagents, will be used. The ability of phosphorus to hold more than eight valence electrons allows for a resonance structure to be drawn forming a double bonded structure.

The stabilization of the carbanion provided by the phosphorus causes an increase in acidity (pKa ~35). Very strong bases, such as butyl lithium, are required for complete formation of ylides.

The ylides shown here are all strong bases. Like other strongly basic organic reagents, they are protonated by water and alcohols, and are sensitive to oxygen. Water decomposes phosphorous ylides to hydrocarbons and phosphine oxides, as shown.

Although many ylides are commercially available it is often necessary to create them synthetically. Ylides can be synthesized from an alkyl halide and a trialkyl phosphine. Typically triphenyl phosphine is used to synthesize ylides. Because a SN2 reaction is used in the ylide synthesis methyl and primary halides perform the best.  Secondary halides can also be used but the yields are generally lower.  This should be considered when planning out a synthesis which involves a synthesized Wittig reagent. 

Mechanism of ylide formation

1)  SN2 reaction

2) Deprotonation

Examples of ylide formation

The Wittig Reaction

The most important use of ylides in synthesis comes from their reactions with aldehydes and ketones, which are initiated in every case by a covalent bonding of the nucleophilic alpha-carbon to the electrophilic carbonyl carbon. Ylides react to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979.  A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.

Going from reactants to products simplified

Examples of the Wittig reaction

Mechanism of the Wittig reaction

Following the initial carbon-carbon bond formation, two intermediates have been identified for the Wittig reaction, a dipolar charge-separated species called a betaine and a four-membered heterocyclic structure referred to as an oxaphosphatane. Cleavage of the oxaphosphatane to alkene and phosphine oxide products is exothermic and irreversible.

1) Nucleophillic attack on the carbonyl

2)  Formation of a 4 membered ring

3)  Formation of the alkene

Limitation of the Wittig reaction

If possible both E and Z isomer of the double bond will be formed. This should be considered when planning a synthesis involving a Wittig Reaction.

Problems

1) Please write the product of the following reactions.

 

2) Please indicate the starting material required to produce the product.

3) Please draw the structure of the oxaphosphetane which is made during the mechanism of the reaction given that produces product C.                           

4) Please draw the structure of the betaine which is made during the mechanism of the reaction given that produces product D. 

5) Please give a detailed mechanism and the final product of this reaction

6)  It has been shown that reacting and epoxide with triphenylphosphine forms an alkene. Please propose a mechanism for this reaction.  Review the section on epoxide reactions if you need help.

Answers

1)

2)

3)

4)

 5)

Nucleophillic attack on the carbonyl

Formation of a 4 membered ring

Formation of the alkene

 6) Nucleophillic attack on the epoxide

Formation of a 4 membered ring

Formation of the alkene

aldehydes & ketones problems

Baeyer-Villiger oxidation

Also known as: Baeyer-Villiger rearrangement

The Baeyer-Villiger oxidation is an organic reaction used to convert a ketone to an ester using a peroxyacid (such as mCPBA). The reaction of the ketone with the acid results in a tetrahedral intermediate, with an alkyl migration following to release a carboxylic acid. The more electron rich R group migrates to the oxygen in this concerted process, allowing for accurate prediction of the stereochemistry of the product.

                         

Mechanism

When the two ligands on the carbonyl carbon in the ketone are different, Baeyer-Villiger oxidation is regioselective. Of the two alpha carbons in the ketone, the one that can stabilize a positive charge more effectively, which is the more highly substituted one, migrates from carbon to oxygen preferentially.

Baeyer-Villiger Migratory Aptitude

 the approximate order of decreasing ease of migration is hydrogen > tertiary alkyl > secondary alkyl > phenyl > primary alkyl > methyl

eg. 1:

eg. 2:

Baeyer-Villiger Migratory Aptitude

 the approximate order of decreasing ease of migration is hydrogen > tertiary alkyl > secondary alkyl > phenyl > primary alkyl > methyl

Oppenauer oxidation

method for selectively oxidizing secondary alcohols to ketones.


                                   

The Oppenauer oxidation is an organic reaction used to convert a primary or secondary alcohol to a ketone using another excess ketone reagent (such as acetone) and an aluminium triisopropoxide catalyst. The mechanism begins with the alcohol replacing one of the isopropoxide groups on the aluminum to generate isopropanol. Acetone then coordinates to the aluminum complex and a rearrangement reaction occurs which includes a hydride transfer from the alcohol to acetone. This process, which oxidizes the alcohol and reduces the acetone, results in the formation of the final ketone product and regeneration of the aluminum triisopropoxide catalyst






The reaction is the opposite of Meerwein-Ponndorf-Verley reduction. The alcohol is oxidized with aluminium isopropoxide in excess acetone. This shifts the equilibrium toward the product side.

examples

Oxidation of Aldehydes and Ketones

This page looks at ways of distinguishing between aldehydes and ketones using oxidizing agents such as acidified potassium dichromate(VI) solution, Tollens' reagent, Fehling's solution and Benedict's solution.

Why do aldehydes and ketones behave differently?

You will remember that the difference between an aldehyde and a ketone is the presence of a hydrogen atom attached to the carbon-oxygen double bond in the aldehyde. Ketones don't have that hydrogen.

The presence of that hydrogen atom makes aldehydes very easy to oxidize. Or, put another way, they are strong reducing agents.

Because ketones do not have that particular hydrogen atom, they are resistant to oxidation, and only very strong oxidizing agents like potassium manganate(VII) solution (potassium permanganate solution) oxidize ketones. However, they do it in a destructive way, breaking carbon-carbon bonds.

Provided you avoid using these powerful oxidizing agents, you can easily tell the difference between an aldehyde and a ketone. Aldehydes are easily oxidized by all sorts of different oxidizing agents: ketones aren't.

What is formed when aldehydes are oxidized?

It depends on whether the reaction is done under acidic or alkaline conditions. Under acidic conditions, the aldehyde is oxidized to a carboxylic acid. Under alkaline conditions, this couldn't form because it would react with the alkali. A salt is formed instead.

Building equations for the oxidation reactions

If you need to work out the equations for these reactions, the only reliable way of building them is to use electron-half-equations. The half-equation for the oxidation of the aldehyde obviously varies depending on whether you are doing the reaction under acidic or alkaline conditions.

Under acidic conditions it is:

                   RCHO+H2O→RCOOH+2H++2e−

. . . and under alkaline conditions:

                    RCHO+3OH−→RCOO−+2H2O+2e−

These half-equations are then combined with the half-equations from whatever oxidizing agent you are using. Examples are given in detail below.

Specific examples

In each of the following examples, we are assuming that you know that you have either an aldehyde or a ketone. There are lots of other things which could also give positive results. Assuming that you know it has to be one or the other, in each case, a ketone does nothing. Only an aldehyde gives a positive result.

Using acidified potassium dichromate(VI) solution

A small amount of potassium dichromate(VI) solution is acidified with dilute sulphuric acid and a few drops of the aldehyde or ketone are added. If nothing happens in the cold, the mixture is warmed gently for a couple of minutes - for example, in a beaker of hot water.

ketone	No change in the orange solution.
aldehyde	Orange solution turns green.

The orange dichromate(VI) ions have been reduced to green chromium(III) ions by the aldehyde. In turn the aldehyde is oxidized to the corresponding carboxylic acid. The electron-half-equation for the reduction of dichromate(VI) ions is:

               Cr2O2−7+14H++6e−→2Cr3++7H2O

Combining that with the half-equation for the oxidation of an aldehyde under acidic conditions:

              RCHO+H2O→RCOOH+2H++2e−

. . . gives the overall equation:

                          2RCHO+Cr2O2−7+8H+→3RCOOH+2Cr3++4H2O

Using Tollens' reagent (the silver mirror test)

Tollens' reagent contains the diamminesilver(I) ion, [Ag(NH3)2]+. This is made from silver(I) nitrate solution. You add a drop of sodium hydroxide solution to give a precipitate of silver(I) oxide, and then add just enough dilute ammonia solution to redissolve the precipitate. To carry out the test, you add a few drops of the aldehyde or ketone to the freshly prepared reagent, and warm gently in a hot water bath for a few minutes.

ketone	No change in the colourless solution.
aldehyde	The colourless solution produces a grey precipitate of silver, or a silver mirror on the test tube.

Aldehydes reduce the diamminesilver(I) ion to metallic silver. Because the solution is alkaline, the aldehyde itself is oxidized to a salt of the corresponding carboxylic acid. The electron-half-equation for the reduction of of the diamminesilver(I) ions to silver is:

              Ag(NH3)+2+e−→Ag+2NH3

Combining that with the half-equation for the oxidation of an aldehyde under alkaline conditions:

             RCHO+3OH−→RCOO−+2H2O+2e−

gives the overall equation:

                             2Ag(NH3)+2+RCHO+3OH−→2Ag+RCOO−+4H2O+2H2O

Using Fehling's solution or Benedict's solution

Fehling's solution and Benedict's solution are variants of essentially the same thing. Both contain complexed copper(II) ions in an alkaline solution.

• Fehling's solution contains copper(II) ions complexed with tartrate ions in sodium hydroxide solution. Complexing the copper(II) ions with tartrate ions prevents precipitation of copper(II) hydroxide.
• Benedict's solution contains copper(II) ions complexed with citrate ions in sodium carbonate solution. Again, complexing the copper(II) ions prevents the formation of a precipitate - this time of copper(II) carbonate.

Both solutions are used in the same way. A few drops of the aldehyde or ketone are added to the reagent, and the mixture is warmed gently in a hot water bath for a few minutes.

ketone	No change in the blue solution.
aldehyde	The blue solution produces a dark red precipitate of copper(I) oxide.

Aldehydes reduce the complexed copper(II) ion to copper(I) oxide. Because the solution is alkaline, the aldehyde itself is oxidized to a salt of the corresponding carboxylic acid. The equations for these reactions are always simplified to avoid having to write in the formulae for the tartrate or citrate ions in the copper complexes. The electron-half-equations for both Fehling's solution and Benedict's solution can be written as:

                   2Cu2+complexed       +     2OH−+2e−→Cu2O+H2O

Combining that with the half-equation for the oxidation of an aldehyde under alkaline conditions:

           RCHO+3OH−→RCOO−+2H2O+2e−

to give the overall equation:

RCHO+2Cu2+complexed      + 5OH−→RCOO−+Cu2O+3H2O