SCH 2303: SYNTHETIC ORGANIC CHEMISTRY. Course Outline Methods of forming cyclic and alicyclic carbon...

SCH 2303: SYNTHETIC ORGANIC CHEMISTRY

Course Outline• Methods of forming cyclic and alicyclic carbon

bonds.

• Use of Organometallic reagents,

• Aldol and acyloin condensation, The Dieckmann Cyclisation.

• Annulation methods, alkylation methods and use of protecting and activating groups.

• The Wittig reaction.

• Reduction: dissolving metal reductions, metal hydride reductions, di-imide reductions, catalytic reduction.

Reference books

• Fundamentals of organic Chemistry by John McMurry

• Organic Chemistry by Morisson and Boyd

• Organic Chemistry by Graham Solomons

• Organic Chemistry by L.G. Wade

Introduction

• Aliphatic hydrocarbons- They consist of chains of carbon atoms that do not involve cyclic structures and are often referred to as open chain hydrocarbons or acyclic hydrocarbons.

Introduction

• Alicyclic compound: is an organic compound that is both aliphatic and cyclic.

• They contain one or more all-carbon rings which may be either saturated or unsaturated, but do not have aromatic character.

• Alicyclic compounds may have one or more aliphatic side chains attached.

• Cyclic compound: is a compound in which a series of atoms is connected to form a loop or ring.

• While the vast majority of cyclic compounds are organic, a few inorganic substances form cyclic compounds as well e.g. triboric acid.

O

B

O

B

O

B

OH

HOOH

• Cyclic compounds may or may not be aromatic. Benzene is aromatic while cyclohexane is not.

Valence bond theory • According to valence bond theory, a

covalent bond forms when two atoms approach each other closely and a singly occupied orbital on one atom overlaps a singly occupied orbital on the other atom.

• The electrons are now paired in the overlapping orbitals and are attracted to the nuclei of both atoms, thus bonding the atoms together.

• E.g. H2 molecule, the H-H bond results from the overlap of two singly occupied hydrogen 1s orbitals.

Valence bond theory

H2 molecule• During the bond-forming reaction 2 H· H2, 436

kJ/mol (104 kcal/mol) of energy is released.• Because the product H2 molecule has 436 kJ/mol less

energy than the starting 2 H· atoms, thus the product is more stable than the reactant and that the new H-H bond has a bond strength of 436 kJ/mol.

• If they are too close- they will repel each other because both are positively charged, yet if they are too far apart, they won’t be able to share the bonding electrons.

• Thus, there is an optimum distance between nuclei that leads to maximum stability. Called the bond length, this distance is 74 pm in the H2 molecule. Every covalent bond has both a characteristic bond strength and bond length.

Optimum distance between nuclei that leads to maximum stability

Sp3 hybrid orbitals• The bonding in the H2 molecule is fairly

straightforward, but the situation is more complicated in organic molecules with tetravalent carbon atoms, e.g. methane, CH4.

• Carbon has four valence electrons (2s2 2p2) and forms four bonds.

• All four C-H bonds in methane are identical and are spatially oriented toward the corners of a regular tetrahedron.

Sp3

SP3

• This was explained by Linus Pauling in 1931, who proposed that an s orbital and three p orbitals can combine, or hybridize, to form four equivalent atomic orbitals with tetrahedral orientation, shown in previous slide.

• These tetrahedrally oriented orbitals are called sp3 hybrids.

Note: • The superscript 3 in the name sp3 gives how many

of each type of atomic orbital combine to form the hybrid, not how many electrons occupy it.

• When an s orbital hybridizes with three p orbitals, the resultant sp3 hybrid orbitals are unsymmetrical about the nucleus.

• One of the two lobes is much larger than the other and can therefore overlap better with another orbital when it forms a bond.

• As a result, sp3 hybrid orbitals form stronger bonds than do unhybridized s or p orbitals.

• The asymmetry of sp3 orbitals arises because the two lobes of a p orbital have different algebraic signs, + and -.

• Thus, when a p orbital hybridizes with an s orbital, the positive p lobe adds to the s orbital but the negative p lobe subtracts from the s orbital.

• The resultant hybrid orbital is therefore unsymmetrical about the nucleus and is strongly oriented in one direction.

• When each of the four identical sp3 hybrid orbitals of a carbon atom overlaps with the 1s orbital of a hydrogen atom, four identical C-H bonds are formed and methane results.

• Each C-H bond in methane has a strength of 439 kJ/mol (105 kcal/mol) and a length of 109 pm. Because the four bonds have a specific geometry, and a property called the bond angle can also be defined.

• The angle formed by each H-C-H is 109.5° - tetrahedral angle.

Tetrahedral angle

Structure of ethane• The ethane molecule has the two carbon atoms

bonded to each other by overlap of an sp3 hybrid orbital from each of the carbon atoms.

• The remaining three sp3 hybrid orbitals of each carbon overlap with the 1s orbitals of three hydrogens to form the six C-H bonds.

• The C-H bonds in ethane are similar to those in methane, although a bit weaker-421 kJ/mol (101 kcal/mol) for ethane versus 439 kJ/mol for methane.

• The C-C bond is 154 pm long and has a strength of 377 kJ/mol (90 kcal/mol).

Structure of ethane

Sp2 and sp hybrid orbitals• The bonds in methane and ethane are called

single bonds because they result from the sharing of one electron pair between bonded atoms.

• Carbon atoms can also form a double bond by sharing two electron pairs between atoms or a triple bond by sharing three electron pairs.

• E.g. ethylene has the structure H2C=CH2 and contains a carbon–carbon double bond, while acetylene has the structure HC CH and contains a carbon–carbon triple bond

• Like sp3 hybrids, sp2 and sp hybrid orbitals are unsymmetrical about the nucleus and are strongly oriented in a specific direction so they can form strong bonds.

• E.g. in an sp2-hybridized carbon atom, the three sp2 orbitals lie in a plane at angles of 120° to one another, with the remaining p orbital perpendicular to the sp2 plane (Figure (a) next slide).

• In an sp-hybridized carbon atom, the two sp orbitals are oriented 180° apart, with the remaining two p orbitals perpendicular both to the sp hybrids and to each other (Figure (b) next slide).

Sp2 and sp hydridization

• When two sp2-hybridized carbon atoms approach each other, they form a strong bond by sp2–sp2 head-on overlap.

• The unhybridized p orbitals interact by sideways overlap to form a second bond.

• Head-on overlap gives what is called a sigma (δ) bond, while sideways overlap gives a pi () bond.

• The combination of sp2–sp2 δ overlap and 2p–2p overlap results in the net sharing of two electron pairs and the formation of a carbon–carbon double bond as shown in the figure next slide.

• The electrons in a δ bond occupy the region centered between nuclei, while the electrons in a bond occupy regions on either side of a line drawn between nuclei.

Sigma and pi bonds

Organic synthesis and its application in science

• Organic synthesis: is the preparation of organic compounds from other organic compounds.

Why organic synthesis is possible

• Ability of carbon atoms to form bonds with itself and other resulting in cyclic and alicyclic with single or multiple bonds.

• These compounds have a variety of physical and chemical properties and react leading to preparation of compounds of many uses.

Uses of synthesized compounds• Pesticides

• Perfumes

• Food colours

• Flavours

• Detergents

• Disinfectants

• Medicine

• Compounds are also synthesized in nature and are referred to as natural products

• Natural products are obtained in small quantities hence most of them are synthesized artificially

Scientists synthesize natural products:

• Whose structures are uncertain for complete structure elucidation

• To be used as possible intermediates in chemical and biological processes

• For modification of the structure in the study of biological activities

Organometallic reagents

• Organometallic reagents contain a carbon atom bonded to a metal

• Lithium, magnesium, and copper are the most commonly used metals in organometallic reagents, but others (such as Sn, Si, Tl, Al, Ti, and Hg) are known.

• General structures of the three common organometallic reagents are shown below.

• R can be alkyl, aryl, allyl, benzyl and with M = Li or Mg.

• Because metals are more electropositive (less electronegative) than carbon, they donate electron density towards carbon, so that carbon bears a partial negative charge.

• The more polar the carbon–metal bond, the more reactive the organometallic reagent.

Organometallic reagents• Note: Electronegativity values for carbon and the

common metals in R - M reagents are C (2.5), Li (1.0), Mg (1.3), and Cu (1.8).

• Because both Li and Mg are very electropositive metals, organolithium (RLi) and organomagnesium reagents (RMgX) contain very polar carbon–metal bonds and are therefore very reactive reagents.

• Organomagnesium reagents are called Grignard reagents, after Victor Grignard, who received the Nobel Prize in Chemistry in 1912 for his work with these reagents.

• Organocopper reagents (R2CuLi), also called organocuprates, have a less polar carbon–metal bond and are therefore less reactive. Although organocuprates contain two alkyl groups bonded to copper, only one R group is utilized in a reaction.

• Regardless of the metal, organometallic reagents are useful synthetically because they react as if they were free carbanions; i.e., carbon bears a partial negative charge, so the reagents react as bases and nucleophiles.

Preparation of Organometallic Reagents

• Organolithium and Grignard reagents are typically prepared by reaction of an alkyl halide with the corresponding metal, as shown in the equations below.

• With lithium, the halogen and metal exchange to form the organolithium reagent.

• With magnesium, the metal inserts in the carbon–halogen bond, forming the Grignard reagent. Grignard reagents are usually prepared in diethyl ether (CH3CH2OCH2CH3) as solvent.

• It is thought that two ether oxygen atoms complex with the magnesium atom, stabilizing the reagent.

Organocuprates are prepared from organolithium reagents by reaction with a Cu+ salt, often CuI.

Organometallic reagents: Reaction as a base

• Organometallic reagents are strong bases that readily abstract a proton from water to form hydrocarbons.

• The electron pair in the carbon–metal bond is used to form a new bond to the proton.

• Equilibrium favors the products of this acid–base reaction because H2O is a much stronger acid than the alkane product.

• Because organolithium and Grignard reagents are themselves prepared from alkyl halides, a two step method converts an alkyl halide into an alkane (or another hydrocarbon).

Organometallic reagents: Reaction as a nucleophile

• Organometallic reagents are also strong nucleophiles that react with electrophilic carbon atoms to form new carbon-carbon bonds.

• These reactions are very valuable in forming the carbon skeletons of complex organic molecules.

Reaction of organometallic reagents with aldehydes and ketones

• Treatment of an aldehyde or ketone with either an organolithium or Grignard reagent followed by water forms an alcohol with a new carbon–carbon bond.

• This reaction is an addition reaction because the elements of R'' and H are added across the π bond.

This reaction above follows the general mechanism for nucleophilic addition i.e.,nucleophilic attack by a carbanion followed by protonation.

Nucleophilic Addition of R''MgX to RCHO and R2C=O

Mechanism• Step 1: The nucleophile (R")- attacks the

carbonyl carbon and the π bond cleaves, forming an alkoxide. This step forms a new carbon–carbon bond.

• Step 2: Protonation of the alkoxide by H2O forms the alcohol addition product. This acid-base reaction forms a new O-H bond.

• The overall result is addition of (R'')- (from R''MgX) and H+ (from H2O) to the carbonyl group.

• This reaction is used to prepare 1°, 2°, and 3° alcohols, depending on the number of alkyl groups bonded to the carbonyl carbon of the aldehyde or ketone.

1. Addition of R"MgX to formaldehyde (CH2=O) forms a 1° alcohol.2. Addition of R"MgX to all other aldehydes forms a 2° alcohol.3. Addition of R"MgX to ketones forms a 3° alcohol.

•Each reaction results in addition of one new alkyl group to the carbonyl carbon, and forms one new carbon–carbon bond. •The reaction is general for all organolithium and Grignard reagents.

• Organometallic reagents are strong bases that rapidly react with H2O, therefore the addition of the new alkyl group must be carried out under anhydrous conditions to prevent traces of water from reacting with the reagent, thus reducing the yield of the desired alcohol.

• Water is added after the addition to protonate the alkoxide.

Applications in Synthesis• Many syntheses of useful compounds

utilize the nucleophilic addition of a Grignard or organolithium reagent to form carbon–carbon bonds.

• E.g., a key step in the synthesis of ethynylestradiol an oral contraceptive component, is the addition of lithium acetylide to a ketone.

Retrosynthetic Analysis of Grignard Products

• To use the Grignard addition in synthesis, you must be able to determine what carbonyl and Grignard components are needed to prepare a given compound

• You must work backwards, in the retrosynthetic direction.

• This involves a two-step process:

Two step process

1) Find the carbon bonded to the OH group in the product.

2) Break the molecule into two components: One alkyl group bonded to the carbon with the OH group comes from the organometallic reagent. The rest of the molecule comes from the carbonyl component.

Example• To synthesize pentan-3-ol

[(CH3CH2)2CHOH] by a Grignard reaction: locate the carbon bonded to the OH group, and then break the molecule into two components at this carbon.

• Thus, retrosynthetic analysis shows that one of the ethyl groups on this carbon comes from a Grignard reagent (CH3CH2MgX), and the rest of the molecule comes from the carbonyl component, a three-carbon aldehyde.

Reaction in the synthetic direction

• A three- carbon aldehyde reacts with CH3CH2MgBr to form an alkoxide, which can then be protonated by H2O to form 3-pentanol, the desired alcohol.

Assgn

• Show the method of synthesizing butan-2-ol using a Grignard reaction

Protecting groups

• Although the addition of organometallic reagents to carbonyls is a very versatile reaction, it cannot be used with molecules that contain both a carbonyl group and N – H or O – H bonds.

Carbonyl compounds that also contain N – H or O – H bonds undergo an acid–base reaction with organometallic reagents, not nucleophilic addition.

• For instance, addition of methylmagnesium chloride (CH3MgCl) to the carbonyl group of 5-hydroxy-2-pentanone to form a diol.

• Nucleophilic addition will not occur with this substrate.

• Instead, because Grignard reagents are strong bases and proton transfer reactions are fast, CH3MgCl removes the O-H proton before nucleophilic addition takes place.

• The stronger acid and base react to form the weaker conjugate acid and conjugate base,

• To solve the problem in the previous slide, a three step strategy is required;

1) Convert the OH group into another functional group that does not interfere with the desired reaction. This new blocking group is called a protecting group, and the reaction that creates it is called protection.

2) Carry out the desired reaction

3) Remove the protecting group. This reaction is called deprotection

• A common OH protecting group is a silyl ether. A silyl ether has a new O – Si bond in place of the O – H bond of the alcohol. The most widely used silyl ether protecting group is the tertbutyldimethylsilyl ether.

•tert-Butyldimethylsilyl ethers are prepared from alcohols by reaction with tert-butyldimethylsilylchloride and an amine base, usually imidazole.

The silyl ether is typically removed with a fluoride salt, usually tetrabutylammonium fluoride(CH3CH2CH2CH2)4N+ F–.

Strategy for using a protecting group

PG-Protecting group

Three steps involved

• Step 1, the OH proton in 5-hydroxy-2-pentanone is replaced with a protecting group, written as PG. Because the product of Step 1 no longer has an OH proton, it can now undergo nucleophilic addition.

• Step 2, CH3MgCl adds to the carbonyl group to yield a 3° alcohol after protonation with water.

• Removal of the protecting group in Step 3 forms the desired product, 4-methyl-1,4-pentanediol.

Reaction of organometallic reagents with carboxylic acid derivatives

• Organometallic reagents react with carboxylic acid derivatives (RCOZ) to form two different products, depending on the identity of both the leaving group Z and the reagent R - M.

• The most useful reactions are carried out with esters and acid chlorides, forming either ketones or 3° alcohols.

Reaction of organometallic reagents with carboxylic acid derivatives

Reaction of RLi and RMgX with Esters and Acid Chlorides

• Both esters and acid chlorides form 3° alcohols when treated with two equivalents of either Grignard or organolithium reagents.

• Two new carbon–carbon bonds are formed in the product.

Reaction of R"MgX or R"Li with RCOCl and RCOOR'

Mechanism

• Nucleophilic attack of (R")– (from R''MgX) in Step 1 forms a tetrahedral intermediate with a leaving group Z.

• In Step 2, the π bond is re-formed and Z– comes off. The overall result of the addition of (R")– and elimination of Z– is the substitution of R" for Z.

• Because the product of Part 1 is a ketone, it can react with a second equivalent of R"MgX to form an alcohol by nucleophilic addition in Part 2.

Nucleophilic attack of (R")– (from R"MgX) in Step 3 forms an alkoxide.• Protonation of the alkoxide by H2O in Step 4 forms a 3° alcohol.

• Organolithium and Grignard reagents always form 3° alcohols when they react with esters and acid chlorides.

• As soon as the ketone forms by addition of one equivalent of reagent to RCOZ (Part 1 of the mechanism), it reacts with a second equivalent of reagent to form the 3° alcohol.

• This reaction is more limited than the Grignard addition to aldehydes and ketones, because only 3° alcohols having two identical alkyl groups can be prepared.

• Nonetheless, it is still a valuable reaction because it forms two new carbon–carbon bonds.

Reaction of R2CuLi with Acid Chlorides

• To form a ketone from a carboxylic acid derivative, a less reactive organometallic reagent-an organocuprate - is used.

• Acid chlorides, which have the best leaving group (Cl–) of the carboxylic acid derivatives, react with R'2CuLi, to give a ketone as product.

• Esters, which contain a poorer leaving group (–OR), do not react with R'2CuLi.

Reaction of Grignard Reagents with Carbon Dioxide

• Grignard reagents react with CO2 to give carboxylic acids after protonation with aqueous acid.

• This reaction, called carboxylation, forms a carboxylic acid with one more carbon atom than the Grignard reagent from which it is prepared.

•Because Grignard reagents are made from alkyl halides, an alkyl halide can be converted to a carboxylic acid having one more carbon atom by a two-step reaction sequence: formation of a Grignard reagent, followed by reaction with CO2.

•In Step 1, the nucleophilic Grignard reagent attacks the electrophilic carbon atom of CO2, cleaving a π bond and forming a new carbon–carbon bond.• The carboxylate anion is protonated with aqueous acid in Step 2 to form the carboxylic acid.

α,β-Unsaturated Carbonyl Compounds

• α,β-Unsaturated carbonyl compounds are conjugated molecules containing a carbonyl group and a carbon-carbon double bond, separated by a single σ bond.

• Both functional groups of α,β-unsaturated carbonyl compounds have π bonds, but individually, they react with very different kinds of reagents.

• Carbon-carbon double bonds react with electrophiles and carbonyl groups react with nucleophiles.

• Because the two π bonds are conjugated, the electron density in an α,β-unsaturated carbonyl compound is delocalized over four atoms.

• Three resonance structures show that the carbonyl carbon and the β carbon bear a partial positive charge.

• This means that α,β-unsaturated carbonyl compounds can react with nucleophiles at two different sites.

• Addition of a nucleophile to the carbonyl carbon, called 1,2-addition, adds the elements of H and Nu across the C=O, forming an allylic alcohol.

• Addition of a nucleophile to the β carbon, called 1,4-addition or conjugate addition, forms a carbonyl compound.

The Mechanisms for 1,2-Addition

The Mechanisms for 1,4-Addition

Part 1: Nucleophillic attack at the β carbon

In Part 1, nucleophilic attack at the β carbon forms a resonance-stabilized anion called an enolate. Either resonance structure can be used to continue the mechanism in Part 2.

Part 2: Protonation and tautomerization

• Protonation on the carbon end of the enolate forms the 1,4-addition product directly.

• Protonation of the oxygen end of the enolate forms an enol.

• Enols are unstable and tautomerize (by a two-step process) to carbonyl compounds.

• Tautomerization forms the same 1,4-addition product that results from protonation on carbon.

Reaction of α,β -Unsaturated Carbonyl Compounds with Organometallic Reagents• The identity of the metal in an

organometallic reagent determines whether it reacts with an α,β-unsaturated aldehyde or ketone by 1,2-addition or 1,4-addition.

• Organolithium and Grignard reagents form 1,2-addition products.

Organocuprate reagents form 1,4-addition products.

Carbonyl condensation

• Carbonyl condensations - Reactions between two carbonyl compounds

• One carbonyl component serves as the nucleophile and the other one serves as the electrophile, and a new carbon–carbon bond is formed.

The presence or absence of a leaving group on the electrophilic carbonyl carbon determines the structure of the product.

The Aldol Reaction

Many aldol products contain an aldehyde and an alcohol; hence the name aldol.

A proton on the α carbon atom is abstracted to form a resonance-stabilized enolate ion with the negative charge spread over a carbon atom and an oxygen atom.

General features of aldol reactions

• In the aldol reaction, two molecules of an aldehyde or ketone react with each other in the presence of base to form a β -hydroxy carbonyl compound.

• E.g., treatment of acetaldehyde with aqueous -OH forms 3-hydroxybutanal, a β-hydroxy aldehyde.

Example

The mechanism of the aldol reaction has three steps

• In Step 1, the base removes a proton from the α carbon to form a resonance-stabilized enolate.

• In Step 2, the nucleophilic enolate attacks the electrophilic carbonyl carbon of another molecule of aldehyde, thus forming a new carbon–carbon bond. This joins the α carbon of one aldehyde to the carbonyl carbon of a second aldehyde.

• Protonation of the alkoxide in Step 3 forms the β-hydroxy aldehyde.

The aldol reaction is a reversible equilibrium, so the position of the equilibrium depends on the base and the carbonyl compound.

• Aldol reactions can be carried out with either aldehydes or ketones.

• With aldehydes, the equilibrium usually favors the products, but with ketones the equilibrium favors the starting materials.

• The characteristic reaction of aldehydes and ketones is nucleophilic addition. An aldol reaction is a nucleophilic addition in which an enolate is the nucleophile.

Examples of aldol reactions

• An aldol reaction with propanal as starting material

• The two molecules of the aldehyde that participate in the aldol reaction react in opposite ways.

• One molecule of propanal becomes an enolate - an electron-rich nucleophile.

• Another molecule of propanal serves as the electrophile because its carbonyl carbon is electron deficient.

The α carbon of one carbonyl component becomes bonded to the carbonyl carbon of the other component.

Dehydration of the Aldol Product

• The β-hydroxy carbonyl compounds formed in the aldol reaction dehydrate more readily than other alcohols.

• In fact, under the basic reaction conditions, the initial aldol product is often not isolated.

• Instead, it loses the elements of H2O from the α and β carbons to form an α,β-unsaturated carbonyl compound.

α,β -Unsaturated carbonyl compounds are conjugated molecules containing a carbonyl group and a carbon–carbon double bond, separated by a single σ bond

• It may or may not be possible to isolate the β-hydroxy carbonyl compound under the conditions of the aldol reaction.

• When the α,β-unsaturated carbonyl compound is further conjugated with a carbon–carbon double bond or a benzene ring, as in the case of Reaction [2], (previous slide) elimination of H2O is spontaneous and the β-hydroxy carbonyl compound cannot be isolated.

• All alcohols-including β-hydroxy carbonyl compounds-dehydrate in the presence of acid.

• Only β-hydroxy carbonyl compounds dehydrate in the presence of base.

• An aldol reaction is often called an aldol condensation, because the β-hydroxy carbonyl compound that is initially formed loses H2O by dehydration.

• A condensation reaction is one in which a small molecule, in this case H2O, is eliminated during a reaction.

• The mechanism of dehydration consists of two steps: deprotonation followed by loss of –

OH.

• In Step 1, base removes a proton from the α carbon, thus forming a resonance-stabilized enolate.

• In Step 2, the electron pair of the enolate forms the π bond as –OH is eliminated.

Crossed Aldol Reactions

• An aldol reaction between two different carbonyl compounds is called a crossed aldol or mixed aldol reaction.

• When two different aldehydes, both having α H atoms, are combined in an aldol reaction, four different β-hydroxy carbonyl compounds are formed.

• Four products form, not one, because both aldehydes can lose an acidic α hydrogen atom and form an enolate in the presence of base.

• Both enolates can then react with both carbonyl compounds, as shown for acetaldehyde and propanal in the reaction scheme (next slide).

When two different aldehydes have α hydrogens, a crossed aldol reaction is not synthetically useful.

Synthetically Useful Crossed Aldol Reactions

• Crossed aldols are synthetically useful in two different situations.

1. A crossed aldol when only one carbonyl component has α H atoms. When one carbonyl compound has no α hydrogens, a crossed aldol reaction often leads to one product.

Two common carbonyl compounds with no α hydrogens used for this purpose are formaldehyde (CH2=O) and benzaldehyde (C6H5CHO).

Example• Reaction of C6H5CHO (as the electrophile)

with either acetaldehyde (CH3CHO) or acetone [(CH3)2C=O] in the presence of base forms a single α, β-unsaturated carbonyl compound after dehydration.

2. A crossed aldol when one carbonyl component has acidic α H atoms.

Another useful crossed aldol reaction takes place between an aldehyde or ketone and a β-dicarbonyl compound.

• α hydrogens between two carbonyl groups are acidic, and so they are more readily removed than other α H atoms.

• As a result, the β-dicarbonyl compound always becomes the enolate component of the aldol reaction.

• β-Dicarbonyl compounds are sometimes called active methylene compounds because they are more reactive towards base than other carbonyl compounds.

• 1,3-Dinitriles are also active methylene compounds.

Active methylene compounds

Example

Useful Transformations of Aldol Products

• The aldol reaction is synthetically useful because it forms new carbon–carbon bonds, generating products with two functional groups.

• β-hydroxy carbonyl compounds formed in aldol reactions are readily transformed into a variety of other compounds.

Directed aldol reactions

• A directed aldol reaction is a variation of the crossed aldol reaction that clearly defines which carbonyl compound becomes the nucleophilic enolate and which reacts at the electrophilic carbonyl carbon.

Directed aldol reaction

• Prepare the enolate of one carbonyl component with Lithium diisopropylamide (LDA), Li+ –N[CH(CH3)2]2,

• Add the second carbonyl compound (the electrophile) to this enolate

• Because the steps are done sequentially and a strong non-nucleophilic base is used to form the enolate of one carbonyl component only, a variety of carbonyl substrates can be used in the reaction.

• Both carbonyl components can have α hydrogens because only one enolate is prepared with LDA.

• Also, when an unsymmetrical ketone is used, LDA selectively forms the less substituted, kinetic enolate.

• E.g product formed by the following directed aldol reaction

2-Methylcyclohexanone forms an enolate on the less substituted carbon, which then reacts with the electrophile, CH3CHO.

Intramolecular Aldol Reactions• Aldol reactions with dicarbonyl compounds

can be used to make five- and six-membered rings.

• The enolate formed from one carbonyl group is the nucleophile, and the carbonyl carbon of the other carbonyl group is the electrophile.

• e.g., treatment of 2,5-hexanedione with base forms a five-membered ring.

Step1: Formation of an enolate

Deprotonation of the CH3 group with base forms a nucleophilic enolate

Step 2: Cyclization and dehydration

•In Step 2, the nucleophilic enolate attacks the electrophilic carbonyl carbon in the same molecule, forming a new carbon–carbon σ bond. •This generates the five membered ring.• Protonation of the alkoxide in Step 3 and loss of H2O by the two steps outlined above form a new C– C π bond.• The overall result is formation of an α,β-unsaturated carbonyl compound in the new five membered ring.

The Claisen Reaction

• The Claisen reaction is the second general reaction of enolates with other carbonyl compounds.

• In the Claisen reaction, two molecules of an ester react with each other in the presence of an alkoxide base to form a β-keto ester.

• E.g., treatment of ethyl acetate with NaOEt forms ethyl acetoacetate after protonation with aqueous acid.

•The mechanism for the Claisen reaction (Next slide) resembles the mechanism of an aldol reaction in that it involves nucleophilic addition of an enolate to an electrophilic carbonyl group.

•Because esters have a leaving group on the carbonyl carbon, however, loss of a leaving group occurs to form the product of substitution, not addition.

Mechanism of Claisen reaction

In Step 1, the base removes a proton from theα carbon to form a resonance-stabilized enolate.

•In Step 2, the nucleophilic enolate attacks the electrophilic carbonyl carbon of another molecule of ester, forming a new carbon–carbon bond. •This joins the α carbon of one ester to the carbonyl carbon of a second ester.•Elimination of the leaving group, EtO–, formsa β-keto ester in Step 3.• Steps 1–3 are reversible equilibria.

•Because the β-keto ester formed in Step 3 has especially acidic protons between its two carbonyl groups, a proton is removed under the basic reaction conditions to form an enolate (Step 4). •The formation of this resonance stabilized enolate drives the equilibrium in the Claisen reaction.

• Deprotonation of the β-keto ester provides a driving force for the Claisen condensation.

• The deprotonation is strongly exothermic, making the overall reaction exothermic and driving the reaction to completion.

• Because the base is consumed in the deprotonation step, a full equivalent of base must be used, and the Claisen condensation is said to be base-promoted rather than base-catalyzed.

• After the reaction is complete, addition of dilute acid converts the enolate back to the β -keto ester.

The Crossed Claisen reactions

• A Claisen reaction between two different carbonyl compounds is called a crossed Claisen reaction.

Useful Crossed Claisen Reactions

1) Between two different esters when only one has α hydrogens.

• When one ester has no α hydrogens, a crossed Claisen reaction often leads to one product.

• Common esters with no α H atoms include ethyl formate (HCO2Et) and ethyl benzoate (C6H5CO2Et).

• E.g, the reaction of ethyl benzoate (as the electrophile) with ethyl acetate (which forms the enolate) in the presence of base forms predominately one β-keto ester.

2) Between a ketone and an ester.

• The reaction of a ketone and an ester in the presence of base also forms the product of a crossed Claisen reaction.

• The enolate is always formed from the ketone component, and the reaction works best when the ester has no α hydrogens.

• The product of this crossed Claisen reaction is a β-dicarbonyl compound, but not a β-keto ester.

Dieckmann reaction• Intramolecular Claisen reactions of diesters

form five and six-membered rings.

• The enolate of one ester is the nucleophile, and the carbonyl carbon of the other is the electrophile.

• An intramolecular Claisen reaction is called a Dieckmann reaction.

• 1,6-Diesters yield five-membered rings by the Dieckmann reaction

• 1,7-Diesters yield six-membered rings by the Dieckmann reaction.

Mechanism of Dieckmann reaction

• In step 1, the base removes a proton to form an enolate, which attacks the carbonyl group of the second ester in step 2, thus forming a new carbon-carbon bond.

• In Step 3, elimination of -OEt forms the β-keto ester.

• To complete the reaction, the proton between the two carbonyl groups is removed with base, and then protonation of the enolate re-forms the β-keto ester in steps 4 and 5.

The Michael Reaction• Michael reaction involves two carbonyl

components:

• The enolate of one carbonyl compound and an α,β-unsaturated carbonyl compound.

• α,β-unsaturated carbonyl compounds are resonance stabilized and have two electrophilic sites - the carbonyl carbon and the β carbon.

• The Michael reaction involves the conjugate addition (1,4-addition) of a resonance stabilized enolate to the β carbon of an α,β-unsaturated carbonyl system.

• All conjugate additions add the elements of H and Nu across the α and β carbons.

• In the Michael reaction, the nucleophile is an enolate.

• Enolates of active methylene compounds are particularly common.

• The α,β-unsaturated carbonyl component is often called a Michael acceptor.

• Conjugate addition

Michael reaction

• The Michael reaction always forms a new carbon-carbon bond on the β carbon of the Michael acceptor.

• The key step is nucleophilic addition of the enolate to the β carbon of the Michael acceptor in Step 2.

Step 1: Enolate formation

•Base removes the acidic proton between the two carbonyl groups, forming the enolate in Step 1.

Step 2: Nucleophilic attack at the β carbon and protonation

•The nucleophilic enolate adds to the β carbon of the α,β-unsaturated carbonyl compound, forming a new carbon–carbon bond and a resonance-stabilized enolate.• Protonation of the enolate forms the 1,4-addition product in Step 3.

• When the product of a Michael reaction is also a β-keto ester, it can be hydrolyzed and decarboxylated by heating in aqueous acid.

• This forms a 1,5-dicarbonyl compound.

Exercise

1) What product is formed when each pair of compounds is treated with NaOEt in ethanol?

The Robinson Annulation• The Robinson annulation is a ring-forming

reaction that combines a Michael reaction with an intramolecular aldol reaction.

• The word annulation comes from the Greek word annulus for ring.

• It involves enolates and it forms carbon–carbon bonds.

• The two starting materials for a Robinson annulation are an α,β-unsaturated carbonyl compound and an enolate.

•The Robinson annulation forms a six-membered ring and three new carbon – carbon bonds; two σ bonds and one π bond. •The product contains an α,β-unsaturated ketone in a cyclohexane ring- i.e, a 2-cyclohexenone ring. •To generate the enolate component of the Robinson annulation, –OH in H2O and -OEt in EtOH are typically used.

Examples

Mechanism• The mechanism of the Robinson

annulation consists of two parts:

1) A Michael addition to the α,β-unsaturated carbonyl compound to form a 1,5-dicarbonyl compound, followed by

2) An intra molecular aldol reaction to form the six-membered ring.

Step 1: Enolate formation

•Base removes the most acidic Proton – i.e, the proton between the two carbonyl groups - forming the enolate in Step 1.

Part 1: Michael Addition to Form a 1,5-Dicarbonyl Compound

•Steps 2-3: Nucleophilic attack at the β carbon and protonation

•Conjugate addition of the enolate to the β carbon of the α,β-unsaturated carbonyl compound forms a new carbon–carbon bond and a resonance stabilized enolate.• Protonation of the enolate forms the 1,5-dicarbonyl compound.

Part 2: Intramolecular Aldol Reactionto Form a 2-Cyclohexenone

• Steps 4 - 6: Intramolecular aldol reaction to form a β-hydroxy ketone.

• The intramolecular aldol reaction consists of three steps: 4; enolate formation, 5; nucleophilic attack, and 6; protonation.• This forms another carbon–carbon σ bond and a β-hydroxy carbonyl compound.

•Steps 7 - 8: Dehydration to form the α,β-unsaturated ketone

• Dehydration consists of two steps: deprotonation and loss of –OH. This reaction forms the new π bond in the α,β unsaturated ketone.

• To draw the product of Robinson annulation without writing out the mechanism, place the α carbon of the compound that becomes the enolate next to the β carbon of the α,β-unsaturated carbonyl compound.

• Then, join the appropriate carbons together as shown (next slide).

• This method of drawing the starting materials, will have the double bond in the product always in the same position in the six-membered ring.

Product in Robinson annulation method

The Wittig reaction• In the sections above, carbonyl groups undergo

addition by a variety of carbanion-like reagents, including Grignard reagents and organolithium reagents.

• In 1954, Georg Wittig discovered a way of adding a phosphorus-stabilized carbanion to a ketone or aldehyde.

• The product is not an alcohol, because the intermediate undergoes elimination to an alkene.

• In effect, the Wittig reaction converts the carbonyl group of a ketone or an aldehyde into a new C=C double bond where no bond existed before.

• This reaction proved so useful that Wittig received the Nobel Prize in Chemistry in 1979 for this discovery.

A Wittig reaction forms two new carbon–carbon bonds: one new σ bond and one new π bond-as well as a phosphorus by-product, Ph3P-O (triphenylphosphine oxide).

The Wittig reaction uses a carbon nucleophile, the Wittig reagent, to form alkenes. When a carbonyl compound is treated with a Wittig reagent, the carbonyl oxygen atom is replaced by the negatively charged alkyl group bonded to the phosphorus i.e, the C=O is converted to a C=C.

Examples

The Wittig Reagent

• A Wittig reagent is an organophosphorus reagent, i.e. a reagent that contains a carbon - phosphorus bond.

• A typical Wittig reagent has a phosphorus atom bonded to three phenyl groups, plus another alkyl group that bears a negative charge.

A Wittig reagent is an ylide, a species that contains two oppositely charged atoms bonded to each other, and both atoms have octets. In a Wittig reagent, a negatively charged carbon atom is bonded to a positively charged phosphorus atom.

Note: Phosphorus ylides are also called phosphoranes.

• Phosphorus is a third-row element, therefore, it can be surrounded by more than eight electrons.

• As a result, a second resonance structure can be drawn that places a double bond between carbon and phosphorus. Regardless of which resonance structure is drawn, a Wittig reagent has no net charge.

• In one resonance structure, though, the carbon atom bears a net negative charge, so it is nucleophilic.

Resonance structures for the Wittig reagent

Synthesis of Wittig reagentsStep 1: SN2 reaction of triphenylphosphine with an alkyl halide forms a phosphonium salt.

Triphenylphosphine (Ph3P:), which contains a lone pair of electrons on P, is the nucleophile. Because the reaction follows an SN2 mechanism, it works best with unhindered CH3X and 1° alkyl halides (RCH2X). Secondary alkyl halides (R2CHX) can also be used, although yields are often lower.

Step 2: Deprotonation of the phosphonium salt with a strong base (:B) forms the ylide.

• Because removal of a proton from a carbon bonded to phosphorus generates a resonance stabilized carbanion (the ylide), this proton is somewhat more acidic than other protons on an alkyl group in the phosphonium salt.

• Very strong bases are still needed, to favor the products of this acid–base reaction.

• Common bases used for this reaction are the organolithium reagents such as butyllithium,CH3CH2CH2CH2Li, abbreviated as BuLi.

Synthesis of the Wittig reagent, Ph3P=CH2,

Mechanism of the Wittig Reaction

• The mechanism of the Wittig reaction involves two steps.

• Step 1:Nucleophilic addition forms a four-membered ring.

• Step 1: forms two bonds and generates a four-membered ring.

• The negatively charged carbon atom of the ylide attacks the carbonyl carbon to form a new carbon–carbon σ bond, while the carbonyl O atom attacks the positively charged P atom.

• This process generates an oxaphosphetane, a four-membered ring containing a strong P-O bond.

Step 2: Elimination of Ph3P=O forms the alkene.

In Step 2:, Ph3P=O (triphenylphosphine oxide) iseliminated, forming two new π bonds. The formation of thevery strong P – O double bond provides the driving force forthe Wittig reaction.

• One limitation of the Wittig reaction is that a mixture of alkene stereoisomers sometimes forms.

• E.g., reaction of propanal (CH3CH2CHO) with a Wittig reagent forms the mixture of E and Z isomers.

The Wittig reaction has been used to synthesize many natural products, including β-carotene.

Reductions:Dissolving metal reductions

• All reducing agents provide the equivalent of two hydrogen atoms, but there are three types of reductions, differing in how H2 is added.

• The simplest reducing agent is molecular H2. Reductions of this sort are carried out in the presence of a metal catalyst that acts as a surface on which reaction occurs.

• The second way to deliver H2 in a reduction is to add two protons and two electrons to a substrate, i.e H2 = 2H+ + 2e-.

• Reducing agents of this sort use alkali metals as a source of electrons and liquid ammonia (NH3) as a source of protons.

• Reductions with Na in NH3 are called dissolving metal reductions.

• The third way to deliver the equivalent of two hydrogen atoms is to add hydride (H–) and a proton (H+).

• The most common hydride reducing agents contain a hydrogen atom bonded to boron or aluminum.

• Examples include sodium borohydride (NaBH4) and lithium aluminum hydride (LiAlH4).

• These reagents deliver H– to a substrate, and then a proton is added from H2O or an alcohol.

Metal hydride reagents act as a source of H– because they contain polar metal–hydrogen bonds that place a partial negative charge on hydrogen.

• Reduction of an alkene forms an alkane by addition of H2. Two bonds are broken-the weak π bond of the alkene and the H2 σ bond-and two new C -H σ bonds are formed.

Catalytic reduction

• The addition of H2 occurs only in the presence of a metal catalyst, and thus, the reaction is called catalytic hydrogenation.

• The catalyst consists of a metal—usually Pd, Pt, or Ni - adsorbed onto a finely divided inert solid, such as charcoal.

• E.g, the catalyst 10% Pd on carbon is composed of 10% Pd and 90% carbon, by weight.

• H2 adds in a syn stereochemistry.

Examples

Di-imide reductionsDi-imide, HN=NH, an unstable hydrogen donor has

specialized application in the reduction of carbon-carbon

double bonds.

Simple alkenes are reduced efficiently by diimide, The

mechanism of the reaction is pictured as a transfer of

hydrogen via a nonpolar cyclic transition state.

The addition ooccurs with syn stereochemistry.