FIGURE 22.3 MECHANISM - 22.5 Acidity of Alpha Hydrogen Atoms: Enolate Ion Formation
| Functional group | Example | $\mathrm{p} K_{\mathrm{a}}$ |
| :---: | :---: | :---: |
| Thioester | <smiles>C[AsH2]=O</smiles> | 21 |
| Ester | <smiles>CC(C)=O</smiles> | 25 |
| Nitrile | $\mathrm{CH}_{3} \mathrm{C}=\mathrm{N}$ | 25 |
| $N, N$-Dialkylamide | <smiles>CCCCCC(C)=O</smiles> | 30 |
| Dialkylamine | $\mathrm{HN}\left(\mathrm{i}-\mathrm{C}_{3} \mathrm{H}_{7}\right)_{2}$ | 36 |

When a hydrogen atom is flanked by two carbonyl groups, its acidity is enhanced further. TABLE 22.1 thus shows that 1,3 -diketones ( $\beta$-diketones), 3 -oxo esters ( $\beta$-keto esters), and 1,3 -diesters (malonic esters) are more acidic than water. The enolate ions derived from these $\beta$-dicarbonyl compounds are stabilized by sharing the negative charge of the two neighboring carbonyl oxygens. The enolate ion of 2,4 -pentanedione, for instance, has three resonance forms. Similar resonance forms can be drawn for other doubly stabilized enolate ions.


Identifying the Acidic Hydrogens in a Compound -
Identify the most acidic hydrogens in each of the following compounds, and rank the compounds in order of increasing acidity:
(a)

(b)

(c)


Strategy -
Hydrogens on carbon next to a carbonyl group are acidic. In general, a $\beta$-dicarbonyl compound is most acidic, a ketone or aldehyde is next most acidic, and a carboxylic acid derivative is least acidic. Remember that alcohols, phenols, and carboxylic acids are also acidic because of their -OH hydrogens.


Solution -
The acidity order is $(\mathbf{a})>(\mathbf{c})>(\mathbf{b})$. Acidic hydrogens are shown in red.
(a)

(b)

(c)

PROBLEM Identify the most acidic hydrogens in each of the following molecules:
22-7 (a)
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$
(b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{COCH}_{3}$
(c) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$
(d) Benzamide
(e) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CN}$
(f) $\mathrm{CH}_{3} \mathrm{CON}\left(\mathrm{CH}_{3}\right)_{2}$

PROBLEM Draw a resonance structure of the acetonitrile anion, ${ }^{-}: \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}$, and account for the acidity of 22-8 nitriles.


Solution - 22.6 Reactivity of Enolate Ions
Enolate ions are more useful than enols for two reasons. First, pure enols can't normally be isolated but are instead generated only as short-lived intermediates in low concentration. By contrast, stable solutions of pure enolate ions are easily prepared from most carbonyl compounds by reaction with a strong base. Second, enolate ions are more reactive than enols and undergo many reactions that enols don't. Whereas enols are neutral, enolate ions are negatively charged, making them much better nucleophiles.

Because they are resonance hybrids of two nonequivalent forms, enolate ions can be looked at either as vinylic alkoxides ( $\mathrm{C}=\mathrm{C}-\mathrm{O}^{-}$) or as $\alpha$-keto carbanions ( ${ }^{-} \mathrm{C}-\mathrm{C}=\mathrm{O}$ ). Thus, enolate ions can react with electrophiles either on oxygen or on carbon. Reaction on oxygen yields an enol derivative, while reaction on carbon yields an $\alpha$-substituted carbonyl compound (FIGURE 22.6). Both kinds of reactivity are known, but reaction on carbon is more common.

An enol derivative

An $\alpha$-substituted
carbonyl compound

FIGURE 22.6 The electrostatic potential map of acetone enolate ion shows how the negative charge is delocalized over both the oxygen and the $\boldsymbol{\alpha}$ carbon. As a result, two modes of reaction of an enolate ion with an electrophile $\mathrm{E}^{+}$are possible. Reaction on carbon to yield an $\alpha$-substituted carbonyl product is more common.

As an example of enolate ion reactivity, aldehydes and ketones undergo base-promoted $\alpha$ halogenation. Even relatively weak bases such as hydroxide ion are effective for halogenation because it's not necessary to convert the ketone completely into its enolate ion. As soon as a small amount of enolate is generated, halogen reacts with it immediately, removing it from the reaction and driving the equilibrium toward further enolate ion
formation.


Solution - 22.6 Reactivity of Enolate Ions
Base-promoted halogenation of aldehydes and ketones is seldom used in practice because it's difficult to stop the reaction at the monosubstituted product. An $\alpha$-halogenated ketone is generally more acidic than the starting, unsubstituted ketone because of the electron-withdrawing inductive effect of the halogen atom. Thus, the monohalogenated products are themselves rapidly turned into enolate ions and further halogenated.

If excess base and halogen are used, a methyl ketone is triply halogenated and then cleaved by base in the haloform reaction. The products are a carboxylic acid plus a so-called haloform (chloroform, $\mathrm{CHCl}_{3}$; bromoform, $\mathrm{CHBr}_{3}$; or iodoform, $\mathrm{CHI}_{3}$ ). Note that the second step of the reaction is a nucleophilic acyl substitution of ${ }^{-} \mathrm{CX}_{3}$ by ${ }^{-} \mathrm{OH}$. That is, a halogen-stabilized carbanion acts as a leaving group.

PROBLEM Why do you suppose ketone halogenations in acidic media are referred to as being acid-catalyzed, 22-9 whereas halogenations in basic media are base-promoted? In other words, why is a full equivalent of base required for halogenation?


Solution - 22.7 Alkylation of Enolate lons
Perhaps the most useful reaction of enolate ions is their alkylation by treatment with an alkyl halide, thereby forming a new $\mathrm{C}-\mathrm{C}$ bond and joining two smaller pieces into one larger molecule. Alkylation occurs when the nucleophilic enolate ion reacts with the electrophilic alkyl halide in an $\mathrm{S}_{\mathrm{N}} 2$ reaction and displaces the leaving group by backside attack.

Alkylation reactions are subject to the same constraints that affect all $S_{N} 2$ reactions (Section 11.3). Thus, the leaving group X in the alkylating agent $\mathrm{R}-\mathrm{X}$ can be chloride, bromide, or iodide. The alkyl group R should be primary or methyl, and preferably should be allylic or benzylic. Secondary halides react poorly, and tertiary halides don't react at all because a competing E2 elimination of HX occurs instead. Vinylic and aryl halides are also unreactive because a backside approach is sterically prevented.

$$
\mathrm{R}-\mathrm{X}\left\{\begin{array}{l}
-\mathrm{X}: \text { Tosylate }>-\mathrm{I}>-\mathrm{Br}>-\mathrm{Cl} \\
\mathrm{R}-: \text { Allylic } \approx \text { Benzylic }>\mathrm{H}_{3} \mathrm{C}->\mathrm{RCH}_{2}-
\end{array}\right.
$$


The Malonic Ester Synthesis -
One of the oldest and best known carbonyl alkylation reactions is the malonic ester synthesis, a method for preparing a carboxylic acid from an alkyl halide while lengthening the carbon chain by two atoms.

Diethyl propanedioate, commonly called diethyl malonate, or malonic ester, is relatively acidic ( $\mathrm{p} K_{\mathrm{a}}=13$ ) because its $\alpha$ hydrogens are flanked by two carbonyl groups. Thus, malonic ester is easily converted into its enolate ion by reaction with sodium ethoxide in ethanol. The enolate ion, in turn, is a good nucleophile that reacts rapidly with an alkyl halide to give an $\alpha$-substituted malonic ester. Note in the following examples that the abbreviation "Et" is used for an ethyl group, $-\mathrm{CH}_{2} \mathrm{CH}_{3}$.

The product of a malonic ester alkylation has one acidic $\alpha$ hydrogen remaining, so the alkylation process can be repeated to yield a dialkylated malonic ester.

On heating with aqueous hydrochloric acid, the alkylated (or dialkylated) malonic ester undergoes hydrolysis of its two ester groups followed by decarboxylation (loss of $\mathrm{CO}_{2}$ ) to yield a substituted monocarboxylic acid.

$$
\begin{array}{lc}
\text { An alkylated } & \text { A carboxylic } \\
\text { malonic ester } & \text { acid }
\end{array}
$$

Decarboxylation is not a general reaction of carboxylic acids. Rather, it is unique to compounds that have a second carbonyl group two atoms away from the $-\mathrm{CO}_{2} \mathrm{H}$. That is, only substituted malonic acids and $\beta$-keto acids undergo loss of $\mathrm{CO}_{2}$ on heating. The decarboxylation reaction occurs by a cyclic mechanism and involves initial formation of an enol, thereby accounting for the need to have a second carbonyl group appropriately positioned.


The Malonic Ester Synthesis -
As noted previously, the overall effect of malonic ester synthesis is to convert an alkyl halide into a carboxylic acid while lengthening the carbon chain by two atoms ( $\mathrm{RX} \rightarrow \mathrm{RCH}_{2} \mathrm{CO}_{2} \mathrm{H}$ ).

A malonic ester synthesis can also be used to prepare cycloalkanecarboxylic acids. For example, when 1,4 -dibromobutane is treated with diethyl malonate in the presence of two equivalents of sodium ethoxide base, the second alkylation step occurs intramolecularly to yield a cyclic product. Hydrolysis and decarboxylation then give cyclopentanecarboxylic acid. Three-, four-, five-, and six-membered rings can be prepared in this way, but yields decrease for larger ring sizes.

1,4-Dibromobutane


Using Malonic Ester Synthesis to Prepare a Carboxylic Acid -
How would you prepare heptanoic acid using a malonic ester synthesis?


Strategy -
A malonic ester synthesis converts an alkyl halide into a carboxylic acid having two more carbons. Thus, a seven-carbon acid chain must be derived from the five-carbon alkyl halide 1-bromopentane.

Solution

$$
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{CH}_{2}\left(\mathrm{CO}_{2} \mathrm{Et}\right)_{2} \xrightarrow[\text { 2. } \mathrm{H}_{3} \mathrm{O}^{+} \text {, heat }]{\text { 1. }{ }^{+}{ }^{-} \text {OEt }} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COH}
$$

PROBLEM How could you use a malonic ester synthesis to prepare the following compounds? Show all steps.


22-10 (a) -
(b)

(c)

PROBLEM Monoalkylated and dialkylated acetic acids can be prepared by malonic ester synthesis, but 22-11 trialkylated acetic acids $\left(\mathrm{R}_{3} \mathrm{C}-\mathrm{CO}_{2} \mathrm{H}\right)$ can't be prepared. Explain.

PROBLEM How could you use a malonic ester synthesis to prepare the following compound?
22-12


The Acetoacetic Ester Synthesis -
Just as the malonic ester synthesis converts an alkyl halide into a carboxylic acid, the acetoacetic ester synthesis converts an alkyl halide into a methyl ketone having three more carbons.

Ethyl 3-oxobutanoate, commonly called ethyl acetoacetate, or acetoacetic ester, is much like malonic ester in that its $\alpha$ hydrogens are flanked by two carbonyl groups. It is therefore readily converted into its enolate ion, which can be alkylated by reaction with an alkyl halide. A second alkylation can also be carried out if desired, since acetoacetic ester has two acidic $\alpha$ hydrogens.

On heating with aqueous HCl , the alkylated (or dialkylated) acetoacetic ester is hydrolyzed to a $\beta$-keto acid, which then undergoes decarboxylation to yield a ketone product. The decarboxylation occurs in the same way as in the malonic ester synthesis and involves a ketone enol as the initial product.

The three-step sequence of (1) enolate ion formation, (2) alkylation, and (3) hydrolysis/decarboxylation is applicable to all $\beta$-keto esters with acidic $\alpha$ hydrogens, not just to acetoacetic ester itself. For example, cyclic $\beta$-keto esters, such as ethyl 2-oxocyclohexanecarboxylate, can be alkylated and decarboxylated to give 2 -substituted cyclohexanones.


Using Acetoacetic Ester Synthesis to Prepare a Ketone -
How would you prepare 2-pentanone by an acetoacetic ester synthesis?


Strategy -
An acetoacetic ester synthesis yields a methyl ketone by adding three carbons to an alkyl halide.

Thus, the acetoacetic ester synthesis of 2-pentanone must involve reaction of bromoethane.
Solution

PROBLEM What alkyl halides would you use to prepare the following ketones by an acetoacetic ester
22-13 synthesis?
(a)

(b)

PROBLEM Which of the following compounds can't be prepared by an acetoacetic ester synthesis? Explain.
22-14
(a) Phenylacetone
(b) Acetophenone
(c) 3,3-Dimethyl-2-butanone

PROBLEM How would you prepare the following compound using an acetoacetic ester synthesis?
22-15

Direct Alkylation of Ketones, Esters, and Nitriles
Both the malonic ester synthesis and the acetoacetic ester synthesis are easy to carry out because they involve relatively acidic dicarbonyl compounds. As a result, sodium ethoxide in ethanol can be used to prepare the
necessary enolate ions. Alternatively, however, it's also possible in many cases to directly alkylate the $\alpha$ position of monocarbonyl compounds. A strong, sterically hindered base such as LDA is needed so that complete conversion to the enolate ion takes place rather than a nucleophilic addition, and a nonprotic solvent must be used.

Ketones, esters, and nitriles can all be alkylated using LDA or related dialkylamide bases in tetrahydrofuran (THF). Aldehydes, however, rarely give high yields of pure products because their enolate ions undergo carbonyl condensation reactions instead of alkylation. (We'll study this condensation reaction in the next chapter.) Some specific examples of alkylation reactions are shown.

Lactone

> Ethyl 2-methylpropanoate

Ethyl 2,2-dimethylpropanoate (87\%)
Ketone

2,2-Dimethylcyclohexanone (6\%)


Nitrile -
Phenylacetonitrile
2-Phenylpropanenitrile (71\%)


Nitrile -
Phenylacetonitrile

2-Phenylpropanenitrile (71\%)

Note in the ketone example that alkylation of 2-methylcyclohexanone leads to a mixture of products because both possible enolate ions are formed. In general, the major product in such cases occurs by alkylation at the less hindered, more accessible position. Thus, alkylation of 2-methylcyclohexanone occurs primarily at C6 (secondary) rather than C 2 (tertiary).


Using an Alkylation Reaction to Prepare a Substituted Ester -
How might you use an alkylation reaction to prepare ethyl 1-methylcyclohexanecarboxylate?

Ethyl 1-methylcyclohexanecarboxylate


Strategy -
An alkylation reaction is used to introduce a methyl or primary alkyl group onto the $\alpha$ position of a ketone, ester, or nitrile by $\mathrm{S}_{\mathrm{N}} 2$ reaction of an enolate ion with an alkyl halide. Thus, we need to look at the target molecule and identify any methyl or primary alkyl groups attached to an $\alpha$ carbon. In the present instance, the target has an $\alpha$ methyl group, which might be introduced by alkylation of an ester enolate ion with iodomethane.


Solution -
Ethyl cyclohexanecarboxylate

Ethyl 1-methylcyclo-
hexanecarboxylate

PROBLEM Show how you might prepare the following compounds using an alkylation reaction as the key step:
22-16 (a)

(b)

(c)

(d)

(e)

(f)


Biological Alkylations -
Alkylations are rare but not unknown in biological chemistry. One example occurs during biosynthesis of the antibiotic indolmycin from indolylpyruvate when a base abstracts an acidic hydrogen from an $\alpha$ position and the resultant enolate ion carries out an $\mathrm{S}_{\mathrm{N}} 2$ alkylation reaction on the methyl group of $S$-adenosylmethionine (SAM; Section 11.6). Although it's convenient to speak of "enolate ion" intermediates in biological pathways, it's unlikely that they exist for long in an aqueous cellular environment. Rather, proton removal and alkylation
probably occur at essentially the same time (FIGURE 22.7).

Indolylpyruvate

Indolmycin (an antibiotic)
FIGURE 22.7 The biosynthesis of indolmycin from indolylpyruvate occurs through a pathway that includes an alkylation reaction of a short-lived enolate ion intermediate.


Barbiturates -
The use of herbal remedies to treat illness and disease goes back thousands of years, but the medical use of chemicals prepared in the laboratory has a much shorter history. Barbiturates, a large class of drugs with a wide variety of uses, constitute one of the earliest successes of medicinal chemistry. The synthesis and medical use of barbiturates goes back to 1904 when Bayer, a German chemical company, first marketed a compound called barbital, trade named Veronal, as a treatment for insomnia. Since that time, more than 2500 different barbiturate analogs have been synthesized by drug companies, more than 50 have been used medicinally, and about a dozen are still in use as anesthetics, anticonvulsants, sedatives, and anxiolytics.

FIGURE 22.8 Different barbiturates come in a multitude of colors, giving rise to similarly colorful street names when the drugs are abused. (credit: "Medications" by freestocks.org/Flickr, Public Domain)

The synthesis of barbiturates is relatively simple and relies on reactions that are now familiar: enolate alkylations and nucleophilic acyl substitutions. Starting with diethyl malonate, alkylation of the corresponding enolate ion with simple alkyl halides provides a wealth of different disubstituted malonic esters. Reaction with urea, $\left(\mathrm{H}_{2} \mathrm{~N}\right)_{2} \mathrm{C}=\mathrm{O}$, then gives the product barbiturates by a twofold nucleophilic acyl substitution reaction of the ester groups with the $-\mathrm{NH}_{2}$ groups of urea (FIGURE 22.9). Amobarbital (Amytal), pentobarbital (Nembutal), and secobarbital (Seconal) are typical examples.


Diethyl malonate -
1. $\mathrm{Na}^{+-}{ }^{-} \mathrm{OEt}$
2. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}$
3. $\mathrm{Na}^{+}-\mathrm{OEt}$
4. $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{Br}$

Amobarbital

Secobarbital

FIGURE 22.9 The synthesis of barbiturates relies on malonic ester alkylations and nucleophilic acyl substitution reactions. More than 2500 different barbiturates have been synthesized over the past 100 years.

Barbital (Veronal), the first barbiturate

In addition to their prescribed medical uses, many barbiturates have also found widespread illegal use as street drugs. Each barbiturate often comes as a tablet of regulated size, shape, and color. Although still used today, most barbiturates have been replaced by safer, more potent alternatives with markedly different structures.


Key Terms -
```
- acetoacetic ester synthesis
- enolate ion
- \alpha-substitution reaction
- malonic ester synthesis
- decarboxylation
- tautomer
- enol
```


Summary -
The $\alpha$-substitution reaction of a carbonyl compound through either an enol or enolate ion intermediate is one of the four fundamental reaction types in carbonyl-group chemistry.

An enol
Carbonyl compounds are in an equilibrium with their enols, a process called keto-enol tautomerism. Although enol tautomers are normally present to only a small extent at equilibrium and can't usually be isolated in pure form, they nevertheless contain a highly nucleophilic double bond and react with electrophiles in an $\boldsymbol{\sigma}$-substitution reaction. An example is the $\alpha$ halogenation of ketones on treatment with $\mathrm{Cl}_{2}, \mathrm{Br}_{2}$, or $\mathrm{I}_{2}$ in acid solution. Alpha bromination of carboxylic acids can be similarly accomplished by the Hell-Volhard-Zelinskii (HVZ) reaction, in which an acid is treated with $\mathrm{Br}_{2}$ and $\mathrm{PBr}_{3}$. The $\alpha$-halogenated products can then undergo base-induced E2 elimination to yield $\alpha, \beta$-unsaturated carbonyl compounds.

Alpha hydrogen atoms of carbonyl compounds are weakly acidic and can be removed by strong bases, such as lithium diisopropylamide (LDA), to yield nucleophilic enolate ions. The most useful reaction of enolate ions is their $\mathrm{S}_{\mathrm{N}} 2$ alkylation with alkyl halides. The Malonic ester synthesis converts an alkyl halide into a carboxylic acid with the addition of two carbon atoms ( $\mathrm{RX} \rightarrow \mathrm{RCH}_{2} \mathrm{CO}_{2} \mathrm{H}$ ). Similarly, the acetoacetic ester synthesis converts an alkyl halide into a methyl ketone with the addition of three carbon atoms (RX $\rightarrow$ $\mathrm{RCH}_{2} \mathrm{COCH}_{3}$ ). In addition, many carbonyl compounds, including ketones, esters, and nitriles, can be directly alkylated by treatment with LDA and an alkyl halide.


Summary of Reactions -
1. Aldehyde/ketone halogenation (Section 22.3)

2. Hell-Volhard-Zelinskii bromination of acids (Section 22.4)

3. Dehydrobromination of $\alpha$-bromo ketones (Section 22.3)

4. Haloform reaction (Section 22.6)

5. Alkylation of enolate ions (Section 22.7)
a. Malonic ester synthesis

b. Acetoacetic ester synthesis

c. Direct alkylation of ketones

d. Direct alkylation of esters

e. Direct alkylation of nitriles


Visualizing Chemistry -
PROBLEM Show the steps in preparing each of the following substances using either a malonic ester synthesis 22-17 or an acetoacetic ester synthesis:
(a)

(b)

PROBLEM Unlike most $\beta$-diketones, the following $\beta$-diketone has no detectable enol content and is about as 22-18 acidic as acetone. Explain.

PROBLEM For a given $\alpha$ hydrogen atom to be acidic, the $\mathrm{C}-\mathrm{H}$ bond must be parallel to the $p$ orbitals of the
22-19 $\mathrm{C}=\mathrm{O}$ double bond (that is, perpendicular to the plane of the adjacent carbonyl group). Identify the most acidic hydrogen atom in the conformation shown for the following structure. Is it axial or equatorial?


Mechanism Problems -
PROBLEM Draw the corresponding keto or enol tautomer for each of the following molecules.

22-20 (a)

(b)

PROBLEM Predict the product(s) and show the mechanism for each of the following reactions:

22-21 (a)

(b)

PROBLEM Predict the product(s) and show the mechanism for each of the following reactions:

22-22 (a)

(b)

PROBLEM The following two optically active $\beta$-keto acids were decarboxylated under the conditions typically
22-23 used for the acetoacetic ester synthesis. Will the ketone products be optically active? Explain.
(a)

(b)

PROBLEM In the Hell-Volhard-Zelinskii reaction, only a catalytic amount of $\mathrm{PBr}_{3}$ is necessary because of the
22-24 following equilibrium. Propose a mechanism for formation of the equilibrium.

PROBLEM When a ketone is treated with a halogen under acidic conditions, the $\alpha$-monohalogenated product
22-25 can be obtained in high yield. Under basic conditions however it is difficult to isolate the monohalogenated product. Explain.

PROBLEM Nonconjugated $\beta, \gamma$-unsaturated ketones, such as 3 -cyclohexenone, are in an acid-catalyzed
22-26 equilibrium with their conjugated $\alpha, \beta$-unsaturated isomers. Propose a mechanism for the isomerization.

PROBLEM 2-substituted 2-cyclopentenones can be interconverted with 5-substituted 2-cyclopentenones 22-27 under basic conditions. Propose a mechanism for this isomerization.

PROBLEM Using curved arrows, propose a mechanism for the following reaction, one of the steps in the
22-28 metabolism of the amino acid alanine.

PROBLEM Using curved arrows, propose a mechanism for the following reaction, one of the steps in the
22-29 biosynthesis of the amino acid tyrosine.

PROBLEM One of the later steps in glucose biosynthesis is the isomerization of fructose 6-phosphate to 22-30 glucose 6-phosphate. Propose a mechanism, using acid or base catalysis as needed.


Mechanism Problems -
PROBLEM One of the later steps in glucose biosynthesis is the isomerization of fructose 6-phosphate to 22-30 glucose 6-phosphate. Propose a mechanism, using acid or base catalysis as needed.

Fructose 6-phosphate

Glucose
6-phosphate

PROBLEM The Favorskii reaction involves treatment of an $\alpha$-bromo ketone with base to yield a ring-contracted
22-31 product. For example, reaction of 2 -bromocyclohexanone with aqueous NaOH yields cyclopentanecarboxylic acid. Propose a mechanism.

PROBLEM Treatment of a cyclic ketone with diazomethane is a method for accomplishing a ring-expansion
22-32 reaction. For example, treatment of cyclohexanone with diazomethane yields cycloheptanone. Propose a mechanism.

PROBLEM The final step in an attempted synthesis of laurene, a hydrocarbon isolated from the marine alga
22-33 Laurencia glandulifera, involved the Wittig reaction shown. The product obtained, however, was not laurene but an isomer. Propose a mechanism to account for the unexpected results.

Laurene
(Not formed)
PROBLEM Amino acids can be prepared by reaction of alkyl halides with diethyl acetamidomalonate, followed
22-34 by heating the initial alkylation product with aqueous HCl . Show how you would prepare alanine, $\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{NH}_{2}\right) \mathrm{CO}_{2} \mathrm{H}$, one of the twenty amino acids found in proteins, and propose a mechanism for acid-catalyzed conversion of the initial alkylation product to the amino acid.

Diethyl acetamidomalonate


Mechanism Problems -
Diethyl acetamidomalonate

PROBLEM Amino acids can also be prepared by a two-step sequence that involves Hell-Volhard-Zelinskii
22-35 reaction of a carboxylic acid followed by treatment with ammonia. Show how you would prepare leucine, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}\left(\mathrm{NH}_{2}\right) \mathrm{CO}_{2} \mathrm{H}$, and identify the mechanism of the second step.

PROBLEM Heating carvone with aqueous sulfuric acid converts it into carvacrol. Propose a mechanism for the 22-36 isomerization.


Acidity of Carbonyl Compounds -
PROBLEM Identify all the acidic hydrogens ( $\mathrm{p} K_{\mathrm{a}}<25$ ) in the following molecules:
22-37 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM Rank the following compounds in order of increasing acidity:
22-38
(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$
(c) $\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{NH}$
(d) $\mathrm{CH}_{3} \mathrm{COCH}_{3}$

(f) $\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}$

PROBLEM Write resonance structures for the following anions:
22-39 (a)

(b)

(c)

(d)

(e)

PROBLEM Base treatment of the following $\alpha, \beta$-unsaturated carbonyl compound yields an anion by removal of
$\mathbf{2 2 - 4 0} \mathrm{H}^{+}$from the $\gamma$ carbon. Why are hydrogens on the $\gamma$ carbon atom acidic?

PROBLEM Treatment of 1-phenyl-2-propenone with a strong base such as LDA does not yield an anion, even
22-41 though it contains a hydrogen on the carbon atom next to the carbonyl group. Explain.

1-Phenyl-2-propenone


$\boldsymbol{\sigma}$-Substitution Reactions -
PROBLEM Predict the product(s) of the following reactions:

22-42 (a)

(c)

(b)

(d)

PROBLEM Which, if any, of the following compounds can be prepared by a malonic ester synthesis? Show the 22-43 alkyl halide you would use in each case.
(a) Ethyl pentanoate
(b) Ethyl 3-methylbutanoate
(c) Ethyl 2-methylbutanoate
(d) Ethyl 2,2-dimethylpropanoate

PROBLEM Which, if any, of the following compounds can be prepared by an acetoacetic ester synthesis? 22-44 Explain.
(a) ${ }_{B}$

(b)

(c)

PROBLEM How would you prepare the following ketones using an acetoacetic ester synthesis?
22-45 (a)

(b)

PROBLEM How would you prepare the following compounds using either an acetoacetic ester synthesis or a
22-46 malonic ester synthesis?
(a)

(b)

(c)

(d)

PROBLEM Which of the following substances would undergo the haloform reaction?
22-47
(a) $\mathrm{CH}_{3} \mathrm{COCH}_{3}$
(b) Acetophenone
(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$
(d) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$
(e) $\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{N}$

PROBLEM How might you convert geraniol into either ethyl geranylacetate or geranylacetone?
22-48

PROBLEM Aprobarbital, a barbiturate once used in treating insomnia, is synthesized in three steps from 22-49 diethyl malonate. Show how you would synthesize the necessary dialkylated intermediate, and then propose a mechanism for the reaction of this intermediate with urea to give aprobarbital.

Aprobarbital


General Problems -
PROBLEM One way to determine the number of acidic hydrogens in a molecule is to treat the compound with
22-50 NaOD in $\mathrm{D}_{2} \mathrm{O}$, isolate the product, and determine its molecular weight by mass spectrometry. For example, if cyclohexanone is treated with NaOD in $\mathrm{D}_{2} \mathrm{O}$, the product has $\mathrm{MW}=102$. Explain how this method works.

PROBLEM When optically active ( $R$ )-2-methylcyclohexanone is treated with either aqueous base or acid, 22-51 racemization occurs. Explain.

PROBLEM Would you expect optically active ( $S$ )-3-methylcyclohexanone to be racemized on acid or base 22-52 treatment in the same way as 2-methylcyclohexanone (Problem 22-51)? Explain.

PROBLEM When an optically active carboxylic acid such as ( $R$ )-2-phenylpropanoic acid is brominated under 22-53 Hell-Volhard-Zelinskii conditions, is the product optically active or racemic? Explain.

PROBLEM Fill in the reagents a-c that are missing from the following scheme:
22-54

PROBLEM Although 2-substituted 2-cyclopentenones are in a base-catalyzed equilibrium with their
22-55 5-substituted 2-cyclopentenone isomers, the analogous isomerization is not observed for 2-substituted 2-cyclohexenones. Explain.

PROBLEM How would you synthesize the following compounds from cyclohexanone? More than one step may 22-56 be required.
(a)

(b)

(c)

(d)

(e)

(f)

PROBLEM The two isomers cis- and trans-4-tert-butyl-2-methylcyclohexanone are interconverted by base 22-57 treatment. Which isomer do you think is more stable, and why?

PROBLEM The following synthetic routes are incorrect. What is wrong with each?
22-58 (a)

(b)

(c)


General Problems -
PROBLEM The following synthetic routes are incorrect. What is wrong with each?
22-58 (a)

(b)

(c)

PROBLEM Attempted Grignard reaction of cyclohexanone with tert-butylmagnesium bromide yields only
22-59 about $1 \%$ of the expected addition product along with $99 \%$ unreacted cyclohexanone. If $\mathrm{D}_{3} \mathrm{O}^{+}$is added to the reaction mixture after a suitable period, however, the "unreacted" cyclohexanone is found to have one deuterium atom incorporated into it. Explain.

PROBLEM Ketones react slowly with benzeneselenenyl chloride in the presence of HCl to yield $\alpha$-phenylseleno
22-60 ketones. Propose a mechanism for this acid-catalyzed $\alpha$-substitution reaction.

PROBLEM South American Incas chewed the leaves of the coca bush, Erythroxylon coca, to combat fatigue.
22-61 Chemical studies of Erythroxylon coca by Friedrich Wöhler in 1862 resulted in the discovery of cocaine, $\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}$, as the active component. Basic hydrolysis of cocaine leads to methanol, benzoic acid, and another compound called ecgonine, $\mathrm{C}_{9} \mathrm{H}_{15} \mathrm{NO}_{3}$. Oxidation of ecgonine with $\mathrm{CrO}_{3}$ yields a keto acid that readily loses $\mathrm{CO}_{2}$ on heating, giving tropinone.


Tropinone -
(a) What is a likely structure for the keto acid?
(b) What is a likely structure for ecgonine, neglecting stereochemistry?
(c) What is a likely structure for cocaine, neglecting stereochemistry?

PROBLEM The key step in a reported laboratory synthesis of sativene, a hydrocarbon isolated from the mold
22-62 Helminthosporium sativum, involves the following base treatment of a keto tosylate. What kind of reaction is occurring? How would you complete the synthesis?

PROBLEM Sodium pentothal is a short-acting barbiturate derivative used as a general anesthetic and known 22-63 in popular culture as a truth serum. It is synthesized like other barbiturates (see the Chemistry Matters at the end of this chapter), using thiourea, $\left(\mathrm{H}_{2} \mathrm{~N}\right)_{2} \mathrm{C}=\mathrm{S}$, in place of urea. How would you synthesize sodium pentothal?


CHAPTER 23 <br> Carbonyl Condensation Reactions -
FIGURE 23.1 Many of the molecules needed by all growing organisms are biosynthesized using carbonyl condensation reactions. (credit: modification of work "Newborn Sea Lion" by Steven Bedard/Flickr, CC BY 2.0)


CHAPTER CONTENTS -
23.1 Carbonyl Condensations: The Aldol Reaction
23.2 Carbonyl Condensations versus Alpha Substitutions
23.3 Dehydration of Aldol Products: Synthesis of Enones
23.4 Using Aldol Reactions in Synthesis
23.5 Mixed Aldol Reactions
23.6 Intramolecular Aldol Reactions
23.7 The Claisen Condensation Reaction
23.8 Mixed Claisen Condensations
23.9 Intramolecular Claisen Condensations: The Dieckmann Cyclization
23.10 Conjugate Carbonyl Additions: The Michael Reaction
23.11 Carbonyl Condensations with Enamines: The Stork Enamine Reaction
23.12 The Robinson Annulation Reaction
23.13 Some Biological Carbonyl Condensation Reactions

WHY THIS CHAPTER? We'll see later in this chapter and again in Chapter 29 that carbonyl condensation reactions occur in a large number of metabolic pathways. In fact, almost all classes of biomolecules-carbohydrates, lipids, proteins, nucleic acids, and many others-are biosynthesized through pathways that involve carbonyl condensation reactions. As with the $\alpha$-substitution reaction discussed in the previous chapter, the great value of carbonyl condensations is that they are one of the few general methods for forming carbon-carbon bonds, thereby making it possible to build larger molecules from smaller precursors. In this chapter, we'll see how and why these reactions occur.

We've now studied three of the four general kinds of carbonyl-group reactions and have seen two general
kinds of behavior. In nucleophilic addition and nucleophilic acyl substitution reactions, a carbonyl compound behaves as an electrophile when an electron-rich reagent adds to it. In $\alpha$-substitution reactions, however, a carbonyl compound behaves as a nucleophile when it is converted into its enol or enolate ion. In the carbonyl condensation reactions that we'll study in this chapter, the carbonyl compound behaves both as an electrophile and as a nucleophile.

Electrophilic carbonyl group reacts with nucleophiles.

Nucleophilic enolate ion reacts with electrophiles.


CHAPTER CONTENTS - 23.1 Carbonyl Condensations: The Aldol Reaction
Carbonyl condensation reactions take place between two carbonyl partners and involve a combination of nucleophilic addition and $\alpha$-substitution steps. One partner is converted into an enolate-ion nucleophile and adds to the electrophilic carbonyl group of the second partner. In so doing, the nucleophilic partner undergoes an $\alpha$-substitution reaction and the electrophilic partner undergoes a nucleophilic addition. The general mechanism of the process is shown in FIGURE 23.2.


FIGURE 23.2 MECHANISM -
The general mechanism of a carbonyl condensation reaction. One partner becomes a nucleophilic donor and adds to the second partner as an electrophilic acceptor. After protonation, the final product is a $\beta$-hydroxy carbonyl compound.

(1) A carbonyl compound with an $\alpha$ hydrogen atom is converted by base into its enolate ion.

(2) The enolate ion acts as a nucleophilic donor and adds to the electrophilic carbonyl group of a second carbonyl (2 \| compound.
(3) Protonation of the tetrahedral alkoxide ion intermediate gives the neutral condensation product and regenerates the base catalyst.

(3) $\downarrow$ New C-C bond

A $\boldsymbol{\beta}$-hydroxy
carbonyl compound

Aldehydes and ketones with an $\alpha$ hydrogen atom undergo a base-catalyzed carbonyl condensation reaction called the aldol reaction. For example, treatment of acetaldehyde with a base such as sodium ethoxide or sodium hydroxide in a protic solvent leads to rapid and reversible formation of 3-hydroxybutanal, known commonly as aldol (aldehyde + alcohol), hence the general name of the reaction.

3-Hydroxybutanal
(aldol-a $\beta$-hydroxy carbonyl compound)
The exact position of the aldol equilibrium depends both on reaction conditions and on substrate structure. The equilibrium generally favors the condensation product in the case of aldehydes with no $\alpha$ substituent
$\left(\mathrm{RCH}_{2} \mathrm{CHO}\right)$ but favors the reactant for disubstituted aldehydes $\left(\mathrm{R}_{2} \mathrm{CHCHO}\right)$ and for most ketones. Steric factors are probably responsible for these trends, since increased substitution near the reaction site increases steric congestion in the aldol product.

Aldehydes


Predicting the Product of an Aldol Reaction -
What is the structure of the aldol product from propanal?


Strategy -
An aldol reaction combines two molecules of reactant by forming a bond between the $\alpha$ carbon of one partner and the carbonyl carbon of the second partner. The product is a $\beta$-hydroxy aldehyde or ketone, meaning that the two oxygen atoms in the product have a 1,3 relationship.


Solution -
PROBLEM Predict the aldol reaction product of the following compounds:
23-1 (a)

(c)

PROBLEM Using curved arrows to indicate the electron flow in each step, show how the base-catalyzed
23-2 reverse-aldol reaction of 4-hydroxy-4-methyl-2-pentanone takes place to yield 2 equivalents of acetone.


Solution - 23.2 Carbonyl Condensations versus Alpha Substitutions
Two of the four general carbonyl-group reactions-carbonyl condensations and $\alpha$ substitutions-take place under basic conditions and involve enolate-ion intermediates. Because the experimental conditions for the two reactions are similar, how can we predict which will occur in a given case? When we generate an enolate ion
with the intention of carrying out an $\alpha$ alkylation, how can we be sure that a carbonyl condensation reaction won't occur instead?

There is no simple answer to this question, but the exact experimental conditions usually have much to do with the result. Alpha-substitution reactions require a full equivalent of strong base and are normally carried out so that the carbonyl compound is rapidly and completely converted into its enolate ion at a low temperature. An electrophile is then added rapidly to ensure that the reactive enolate ion is quenched quickly. In a ketone alkylation reaction, for instance, we might use 1 equivalent of lithium diisopropylamide (LDA) in tetrahydrofuran solution at $-78^{\circ} \mathrm{C}$. Rapid and complete generation of the ketone enolate ion would occur, and no unreacted ketone would remain, meaning that no condensation reaction could take place. We would then immediately add an alkyl halide to complete the alkylation reaction.

On the other hand, carbonyl condensation reactions require only a catalytic amount of a relatively weak base, rather than a full equivalent, so that a small amount of enolate ion is generated in the presence of unreacted carbonyl compound. Once a condensation occurs, the basic catalyst is regenerated. To carry out an aldol reaction on propanal, for instance, we might dissolve the aldehyde in methanol, add 0.05 equivalent of sodium methoxide, and then warm the mixture to give the aldol product.


Solution - 23.3 Dehydration of Aldol Products: Synthesis of Enones
When heated under either acidic or basic conditions, the $\beta$-hydroxy aldehydes or ketones formed in aldol reactions can be dehydrated to yield $\alpha, \beta$-unsaturated products, or conjugated enones (ene + ketone). In fact, it's this loss of water that gives the carbonyl condensation reaction its name, because water condenses out when an enone product forms.

A $\boldsymbol{\beta}$-hydroxy ketone or aldehyde

A conjugated enone

Most alcohols are resistant to dehydration by base (Section 17.6) because hydroxide ion is a poor leaving group, but aldol products dehydrate easily because of their carbonyl group. Under basic conditions, an acidic $\alpha$ hydrogen is removed, yielding an enolate ion that expels the nearby ${ }^{-} \mathrm{OH}$ leaving group in an E1cB reaction (Section 11.10). Under acidic conditions, an enol is formed, the -OH group is protonated, and water is expelled in an E1 or E2 reaction.

The reaction conditions needed for aldol dehydration are often only a bit more vigorous (slightly higher temperature, for instance) than the conditions needed for the aldol formation itself. As a result, conjugated enones are usually obtained directly from aldol reactions without isolating the intermediate $\beta$-hydroxy carbonyl compounds.

Conjugated enones are more stable than nonconjugated enones for the same reason that conjugated dienes are more stable than nonconjugated dienes (Section 14.1). Interaction between the $\pi$ electrons of the $\mathrm{C}=\mathrm{C}$ bond and the $\pi$ electrons of the $\mathrm{C}=\mathrm{O}$ group leads to a molecular-orbital description for a conjugated enone with an interaction of the $\pi$ electrons over all four atomic centers (FIGURE 23.3).

Propenal

1,3-Butadiene

FIGURE 23.3 The $\boldsymbol{\pi}$ bonding molecular orbitals of a conjugated enone (propenal) and a conjugated diene (1,3-butadiene) are similar in shape and are spread over the entire $\pi$ system.


Solution - 23.3 Dehydration of Aldol Products: Synthesis of Enones
The real value of aldol dehydration is that removal of water from the reaction mixture can be used to drive the aldol equilibrium toward the product. Even though the initial aldol step itself may be unfavorable, as it usually is for ketones, the subsequent dehydration step nevertheless allows many aldol condensations to be carried out in good yield. Cyclohexanone, for example, gives cyclohexylidenecyclohexanone in $92 \%$ yield, even though the initial equilibrium is unfavorable.

Cyclohexanone
Cyclohexylidenecyclohexanone
(92\%)


Predicting the Product of an Aldol Reaction -
What is the structure of the enone obtained from aldol condensation of acetaldehyde?


Strategy -
In the aldol reaction, $\mathrm{H}_{2} \mathrm{O}$ is eliminated and a double bond is formed by removing two hydrogens from the acidic $\alpha$ position of one partner and the carbonyl oxygen from the second partner. The product is thus an $\alpha, \beta$-unsaturated aldehyde or ketone.


Solution -
PROBLEM What enone product would you expect from aldol condensation of each of the following 23-3 compounds?
(a)

(b)

(c)

PROBLEM Aldol condensation of 3-methylcyclohexanone leads to a mixture of two enone products, not 23-4 counting double-bond isomers. Draw them.


Solution - 23.4 Using Aldol Reactions in Synthesis
The aldol reaction yields either a $\beta$-hydroxy aldehyde/ketone or an $\alpha, \beta$-unsaturated aldehyde/ketone, depending on the experimental conditions. By learning how to work backward, it's possible to predict when the aldol reaction might be useful in synthesis. Whenever the target molecule contains either a $\beta$-hydroxy aldehyde/ ketone or a conjugated enone functional group, it might come from an aldol reaction.

We can extend this kind of reasoning even further by imagining that subsequent transformations might be carried out on the aldol products. For example, a saturated ketone might be prepared by catalytic hydrogenation of the enone product. A good example can be found in the industrial preparation of 2-ethyl-1-hexanol, an alcohol used in the synthesis of plasticizers for polymers. Although 2-ethyl-1-hexanol bears little resemblance to an aldol product at first glance, it is in fact prepared commercially from butanal by an aldol reaction. Working backward, we can reason that 2 -ethyl-1-hexanol might come from 2 -ethylhexanal by a reduction. 2-Ethylhexanal, in turn, might be prepared by catalytic reduction of 2 -ethyl-2-hexenal, which is the aldol condensation product of butanal. The reactions that follow show the sequence in reverse order.

Butanal

2-Ethyl-2-hexenal

PROBLEM Which of the following compounds are aldol condensation products? What is the aldehyde or ketone 23-5 precursor of each?
(a) 2-Hydroxy-2-methylpentanal
(b) 5-Ethyl-4-methyl-4-hepten-3-one

PROBLEM 1-Butanol is prepared commercially by a route that begins with an aldol reaction. Show the steps 23-6 that are likely to be involved.

PROBLEM Show how you would synthesize the following compound using an aldol reaction:
23-7


Solution - 23.5 Mixed Aldol Reactions
Until now, we've considered only symmetrical aldol reactions, in which the two carbonyl components have been the same. What would happen, though, if an aldol reaction were carried out between two different carbonyl partners?

In general, a mixed aldol reaction between two similar aldehyde or ketone partners leads to a mixture of four possible products. For example, base treatment of a mixture of acetaldehyde and propanal gives a complex product mixture containing two "symmetrical" aldol products and two "mixed" aldol products. Clearly, such a reaction is of no practical value.

On the other hand, mixed aldol reactions can lead cleanly to a single product if either of two conditions is met:

- If one of the carbonyl partners contains no $\alpha$ hydrogens (and thus can't form an enolate ion to become a donor), but does contain an unhindered carbonyl group (and so is a good acceptor of nucleophiles), then a mixed aldol reaction is likely to be successful. This is the case, for instance, when either benzaldehyde or
formaldehyde is used as one of the carbonyl partners.
Neither benzaldehyde nor formaldehyde can form an enolate ion to add to another partner, yet both compounds have an unhindered carbonyl group. The ketone 2-methylcyclohexanone, for instance, gives the mixed aldol product on reaction with benzaldehyde.

- If one of the carbonyl partners is much more acidic than the other, and is thus transformed into its enolate ion in preference to the other, then a mixed aldol reaction is likely to be successful. Ethyl acetoacetate, for instance, is completely converted into its enolate ion in preference to enolate ion formation from monocarbonyl partners. Thus, aldol condensations of monoketones with ethyl acetoacetate occur preferentially to give the mixed product.

The situation can be summarized by saying that a mixed aldol reaction leads to a mixture of products unless one of the partners either has no $\alpha$ hydrogens but is a good electrophilic acceptor (such as benzaldehyde) or is an unusually acidic nucleophilic donor (such as ethyl acetoacetate).


Solution - 23.5 Mixed Aldol Reactions
PROBLEM Which of the following compounds can probably be prepared by a mixed aldol reaction? Show the 23-8 reactants you would use in each case.
(a)

(b)

(c)


Solution - 23.6 Intramolecular Aldol Reactions
The aldol reactions we've seen thus far have all been intermolecular, meaning that they have taken place between two different molecules. When certain dicarbonyl compounds are treated with base, however, an intramolecular aldol reaction can occur, leading to the formation of a cyclic product. For example, base treatment of a 1,4-diketone such as 2,5-hexanedione yields a cyclopentenone product, and base treatment of a 1,5-diketone such as 2,6-heptanedione yields a cyclohexenone.

$$
\begin{aligned}
& \text { 2,5-Hexanedione } \\
& \text { (a 1,4-diketone) }
\end{aligned}
$$


2,6-Heptanedione (a 1,5-diketone) -
3-Methyl-2-cyclohexenone

The mechanism of intramolecular aldol reactions is similar to that of intermolecular reactions. The only difference is that both the nucleophilic carbonyl anion donor and the electrophilic carbonyl acceptor are now in the same molecule. One complication, however, is that intramolecular aldol reactions might lead to a mixture of products, depending on which enolate ion is formed. For example, 2,5-hexanedione might yield either the five-membered-ring product 3-methyl-2-cyclopentenone or the three-membered-ring product (2-methylcyclopropenyl) ethanone (FIGURE 23.4). In practice, though, only the cyclopentenone is formed.

FIGURE 23.4 Intramolecular aldol reaction of 2,5-hexanedione yields 3-methyl-2-cyclopentenone rather than the alternative cyclopropene.

The selectivity observed in the intramolecular aldol reaction of 2,5-hexanedione is due to the fact that all steps in the mechanism are reversible, so an equilibrium is reached. Thus, the relatively strain-free cyclopentenone product is considerably more stable than the highly strained cyclopropene alternative. For similar reasons, intramolecular aldol reactions of 1,5-diketones lead only to cyclohexenone products rather than to the more strained acylcyclobutenes.

PROBLEM Treatment of a 1,3-diketone such as 2,4-pentanedione with base does not give an aldol
23-9 condensation product. Explain.
PROBLEM What product would you expect to obtain from base treatment of 1,6-cyclodecanedione? 23-10

1,6-Cyclodecanedione


2,6-Heptanedione (a 1,5-diketone) - 23.7 The Claisen Condensation Reaction
Esters, like aldehydes and ketones, are weakly acidic. When an ester with an $\alpha$ hydrogen is treated with 1 equivalent of a base such as sodium ethoxide, a reversible carbonyl condensation reaction occurs to yield a $\beta$-keto ester. For instance, ethyl acetate yields ethyl acetoacetate on base treatment. This reaction between two ester molecules is known as the Claisen condensation reaction. (We'll use ethyl esters, abbreviated "Et," for consistency, but other esters will also work.)

The mechanism of the Claisen condensation is similar to that of the aldol condensation and involves the nucleophilic addition of an ester enolate ion to the carbonyl group of a second ester molecule (FIGURE 23.5). The only difference between the aldol condensation of an aldehyde or ketone and the Claisen condensation of an ester involves the fate of the initially formed tetrahedral intermediate. The tetrahedral intermediate in the aldol reaction is protonated to give an alcohol product-exactly the behavior previously seen for aldehydes and ketones (Section 19.4). The tetrahedral intermediate in the Claisen reaction, however, expels an alkoxide leaving group to yield an acyl substitution product-as previously seen for esters (Section 21.6).


FIGURE 23.5 MECHANISM -
Mechanism of the Claisen condensation reaction.

(1) Base abstracts an acidic alpha hydrogen atom from an ester molecule, yielding an ester enolate ion.
(2) The enolate ion is added in a nucleophilic addition reaction to a second ester molecule, giving a tetrahedral alkoxide intermediate.

(2)

(3) The tetrahedral intermediate expels ethoxide ion to yield a new carbonyl compound, ethyl acetoacetate.
4) But ethoxide ion is a strong enough base to deprotonate ethyl acetoacetate, shifting the equilibrium and driving the overall
reaction to completion.

(5) Protonation of the enolate ion by addition of aqueous acid in a separate step yields the final $\beta$-keto ester product.

If the starting ester has more than one acidic $\alpha$ hydrogen, the product $\beta$-keto ester has a highly acidic, doubly activated hydrogen atom that can be abstracted by base. This deprotonation of the product requires the use of a full equivalent of base rather than a catalytic amount. Furthermore, the deprotonation serves to drive the equilibrium completely to the product side so that high yields are usually obtained in Claisen condensations.


Predicting the Product of a Claisen Condensation Reaction -
What product would you obtain from Claisen condensation of ethyl propanoate?


Strategy -
The Claisen condensation of an ester results in loss of one molecule of alcohol and formation of a product in which an acyl group of one reactant bonds to the $\alpha$ carbon of the second reactant. The product is a $\beta$-keto ester.

Solution

PROBLEM Show the products you would expect to obtain by Claisen condensation of the following esters:
23-11
(a) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CO}_{2} \mathrm{Et}$
(b) Ethyl phenylacetate
(c) Ethyl cyclohexylacetate

PROBLEM As shown in Figure 23.5, the Claisen reaction is reversible. That is, a $\beta$-keto ester can be cleaved 23-12 by base into two fragments. Using curved arrows to indicate electron flow, show the mechanism by which this cleavage occurs.


Strategy - 23.8 Mixed Claisen Condensations
The mixed Claisen condensation of two different esters is similar to the mixed aldol condensation of two different aldehydes or ketones (Section 23.5). Mixed Claisen reactions are successful only when one of the two ester components has no $\alpha$ hydrogens and thus can't form an enolate ion. For example, ethyl benzoate and ethyl formate can't form enolate ions and thus can't serve as donors. They can, however, act as the electrophilic acceptor components in reactions with other ester anions to give mixed $\beta$-keto ester products.

Mixed Claisen-like reactions can also be carried out between an ester and a ketone, resulting in the synthesis of a $\beta$-diketone. The reaction works best when the ester component has no $\alpha$ hydrogens and thus can't act as the nucleophilic donor. For example, ethyl formate gives high yields in mixed Claisen condensations with ketones.


2,2-Dimethylcyclohexanone (donor) -
$+$

Ethyl formate (acceptor)

A $\boldsymbol{b}$-keto aldehyde (91\%)


Predicting the Product of a Mixed Claisen Condensation Reaction -
Diethyl oxalate, $\left(\mathrm{CO}_{2} \mathrm{Et}\right)_{2}$, often gives high yields in mixed Claisen reactions. What product would you expect to obtain from a mixed Claisen reaction of ethyl acetate with diethyl oxalate?


Strategy -
A mixed Claisen reaction is effective when only one of the two partners has an acidic $\alpha$ hydrogen atom. In the present case, ethyl acetate can be converted into its enolate ion, but diethyl oxalate cannot. Thus, ethyl acetate acts as the donor and diethyl oxalate as the acceptor.

Solution

PROBLEM What product would you expect from the following mixed Claisen-like reaction?
23-13


Strategy - 23.9 Intramolecular Claisen Condensations: The Dieckmann Cyclization
Intramolecular Claisen condensations can be carried out with diesters, just as intramolecular aldol condensations can be carried out with diketones (Section 23.6). Called the Dieckmann cyclization, this reaction works best on 1,6-diesters and 1,7-diesters. Intramolecular Claisen cyclization of a 1,6-diester gives a five-membered cyclic $\beta$-keto ester, and cyclization of a 1,7-diester gives a six-membered cyclic $\beta$-keto ester.

The mechanism of the Dieckmann cyclization, shown in FIGURE 23.6, is the same as that of the Claisen condensation. One of the two ester groups is converted into an enolate ion, which carries out a nucleophilic acyl substitution on the second ester group at the other end of the molecule. A cyclic $\beta$-keto ester product results.

FIGURE 23.6 MECHANISM
Mechanism of the Dieckmann cyclization of a 1,7-diester to yield a cyclic $\beta$-keto ester product.

(1) Base abstracts an acidic $\alpha$ proton from
the carbon atom next to one of the ester groups, yielding an enolate ion.
(1) $\downarrow \mathrm{Na}^{+- \text {OEt }}$

O |

(3) $\downarrow$

(4) Deprotonation of the acidic $\beta$-keto ester gives an enolate ion...
-

(5) $\downarrow \mathrm{H}_{3} \mathrm{O}^{+}$


Strategy - 23.9 Intramolecular Claisen Condensations: The Dieckmann Cyclization
(4) Deprotonation of the acidic $\beta$-keto ester gives an enolate ion...
-

(5) $\downarrow \mathrm{H}_{3} \mathrm{O}^{+}$

The cyclic $\beta$-keto ester produced in a Dieckmann cyclization can be further alkylated and decarboxylated by a series of reactions analogous to those used in the acetoacetic ester synthesis (Section 22.7). Alkylation and
subsequent decarboxylation of ethyl 2-oxocyclohexanecarboxylate, for instance, yields a 2-alkylcyclohexanone. The overall sequence of (1) Dieckmann cyclization, (2) $\beta$-keto ester alkylation, and (3) decarboxylation is a powerful method for preparing 2 -substituted cyclopentanones and cyclohexanones.

PROBLEM What product would you expect from the following reaction?

PROBLEM Dieckmann cyclization of diethyl 3-methylheptanedioate gives a mixture of two $\beta$-keto ester
23-15 products. What are their structures, and why is a mixture formed?


Strategy - 23.10 Conjugate Carbonyl Additions: The Michael Reaction
We saw in Section 19.13 that certain nucleophiles, such as amines, react with $\alpha, \beta$-unsaturated aldehydes and ketones to give a conjugate addition product, rather than a direct addition product.

Exactly the same kind of conjugate addition can occur when a nucleophilic enolate ion reacts with an $\alpha, \beta$-unsaturated carbonyl compound-a process known as the Michael reaction after Arthur Michael at Tufts College and Harvard University.

The best Michael reactions are those that take place when a particularly stable enolate ion, such as that derived from a $\beta$-keto ester or other 1,3-dicarbonyl compound, adds to an unhindered $\alpha, \beta$-unsaturated ketone. For example, ethyl acetoacetate reacts with 3-buten-2-one in the presence of sodium ethoxide to yield the conjugate addition product.

Ethyl
3-Buten-2-one


acetoacetate -
Michael reactions take place by addition of a nucleophilic enolate ion donor to the $\beta$ carbon of an $\alpha, \beta$-unsaturated carbonyl acceptor, according to the mechanism shown in FIGURE 23.7.


FIGURE 23.7 MECHANISM -
Mechanism of the Michael reaction between a $\boldsymbol{\beta}$-keto ester and an
$\boldsymbol{\alpha}, \boldsymbol{\beta}$-unsaturated ketone. The reaction is a conjugate addition of an enolate ion to
the unsaturated carbonyl compound.

The base catalyst removes an acidic alpha proton from the starting $\beta$-keto ester to generate a stabilized enolate ion nucleophile.

2 The nucleophile adds to the $\alpha, \beta$-unsaturated ketone electrophile in a Michael reaction to generate a new enolate as product.
(3) The enolate product abstracts an acidic proton, either from solvent or from starting keto ester, to yield the final addition product.

(1) $\| \mathrm{Na}^{+}{ }^{+}$-OEt

(3) $\downarrow \mathrm{EtOH}$

The Michael reaction occurs with a variety of $\alpha, \beta$-unsaturated carbonyl compounds, not just conjugated ketones. Unsaturated aldehydes, esters, thioesters, nitriles, amides, and nitro compounds can all act as the electrophilic acceptor component in Michael reactions (TABLE 23.1). Similarly, a variety of different donors can be used, including $\beta$-diketones, $\beta$-keto esters, malonic esters, $\beta$-keto nitriles, and nitro compounds.

TABLE 23.1 Some Michael Acceptors and Michael Donors
Michael acceptors Michael donors

| <smiles>C=[C]C=O</smiles> | Propenal | <smiles>O=[14C]C=[18O]</smiles> | $\beta$-Diketone |
| :---: | :---: | :---: | :---: |
| <smiles>C=CC(C)=O</smiles> | 3-Buten-2-one | <smiles>CCOC(=O)CC=O</smiles> | $\beta$-Keto ester |
| <smiles>C=C[C-]O[O-]</smiles> | Ethyl propenoate | <smiles></smiles> | Diethyl malonate |


FIGURE 23.7 MECHANISM -
TABLE 23.1 Some Michael Acceptors and Michael Donors

| Michael accep |  | Michael donors |  |
| :---: | :---: | :---: | :---: |
| <smiles>C=C(N)O</smiles> | Propenamide | $\stackrel{\mathrm{O}}{\mathrm{RCCH}}{ }_{2} \mathrm{C} \equiv \mathrm{~N}$ | $\beta$-Keto nitrile |
| $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHC} \equiv \mathrm{N}$ | Propenenitrile | $\mathrm{RCH}_{2} \mathrm{NO}_{2}$ | Nitro compound |
| <smiles>C=CN+[O-]</smiles> | Nitroethylene |  |  |


Using the Michael Reaction -
How might you obtain the following compound using a Michael reaction?


Strategy -
A Michael reaction involves the conjugate addition of a stable enolate ion donor to an $\alpha, \beta$-unsaturated carbonyl acceptor, yielding a 1,5-dicarbonyl product. Usually, the stable enolate ion is derived from a $\beta$-diketone, $\beta$-keto ester, malonic ester, or similar compound. The $C-C$ bond formed in the conjugate addition step is the one between the $\alpha$ carbon of the acidic donor and the $\beta$ carbon of the unsaturated acceptor.


Solution -
PROBLEM What product would you obtain from a base-catalyzed Michael reaction of 2,4-pentanedione with 23-16 each of the following $\alpha, \beta$-unsaturated acceptors?
(a) 2-Cyclohexenone
(b) Propenenitrile
(c) Ethyl 2-butenoate

PROBLEM What product would you obtain from a base-catalyzed Michael reaction of 3-buten-2-one with each 23-17 of the following nucleophilic donors?
(a)

(b)

PROBLEM How would you prepare the following compound using a Michael reaction (gray = H, black = C, red 23-18 = O, blue = N)?


Solution - 23.11 Carbonyl Condensations with Enamines: The Stork Enamine Reaction
In addition to enolate ions, other kinds of carbon nucleophiles also add to $\alpha, \beta$-unsaturated acceptors in Michaellike reactions. Among the most important and useful of such nucleophiles, particularly in biological chemistry, are enamines, which are readily prepared by reaction between a ketone and a secondary amine (Section 19.8). For example:

As the following resonance structures indicate, enamines are electronically similar to enolate ions. Overlap of the nitrogen lone-pair orbital with the double-bond $p$ orbitals leads to an increase in electron density on the $\alpha$ carbon atom, making that carbon nucleophilic. An electrostatic potential map of $N, N$-dimethylaminoethylene shows this shift of electron density (red) toward the $\alpha$ position.

Enamines behave in much the same way as enolate ions and enter into many of the same kinds of reactions. In the Stork enamine reaction, named for its discoverer, Gilbert Stork of Columbia University, for example, an enamine adds to an $\alpha, \beta$-unsaturated carbonyl acceptor in a Michael-like process. The initial product is then hydrolyzed by aqueous acid to yield a 1,5 -dicarbonyl compound. The overall reaction is thus a threestep sequence of (1) enamine formation from a ketone, (2) Michael addition to an $\alpha, \beta$-unsaturated carbonyl compound, and (3) enamine hydrolysis back to a ketone.

The net effect of the Stork enamine reaction is the Michael addition of a ketone to an $\alpha, \beta$-unsaturated carbonyl compound. For example, cyclohexanone reacts with the cyclic amine pyrrolidine to yield an enamine; further reaction with an enone such as 3-buten-2-one yields a Michael adduct; and aqueous hydrolysis completes the sequence to give a 1,5-diketone (FIGURE 23.8).


Solution - 23.11 Carbonyl Condensations with Enamines: The Stork Enamine Reaction
FIGURE 23.8 The Stork enamine reaction between cyclohexanone and 3-buten-2-one. (1) Cyclohexanone is first converted into an enamine, (2) the enamine adds to the $\alpha, \beta$-unsaturated ketone in a Michael reaction, and (3) the conjugate addition product is hydrolyzed to yield a 1,5-diketone.

The enamine-Michael reaction has two advantages over the enolate-ion-Michael reaction that makes it particularly useful in biological pathways. First, an enamine is neutral, easily prepared, and easily handled, while an enolate ion is charged, sometimes difficult to prepare, and must be handled carefully. Second, an enamine from a monoketone can be used in the Michael addition, whereas only enolate ions from $\beta$-dicarbonyl compounds can be used.


Using the Stork Enamine Reaction -
How might you use an enamine reaction to prepare the following compound?


Strategy -
The overall result of an enamine reaction is the Michael addition of a ketone as donor to an $\alpha, \beta$-unsaturated carbonyl compound as acceptor, yielding a 1,5-dicarbonyl product. The $\mathrm{C}-\mathrm{C}$ bond formed in the Michael addition step is the one between the $\alpha$ carbon of the ketone donor and the $\beta$ carbon of the unsaturated acceptor.

Solution

PROBLEM What products would result after hydrolysis from reaction of the enamine prepared from
23-19 cyclopentanone and pyrrolidine with the following $\alpha, \beta$-unsaturated acceptors?
(a) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCO}_{2} \mathrm{Et}$
(b) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCHO}$
(c) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCOCH}_{3}$

PROBLEM Show how you might use an enamine reaction to prepare each of the following compounds:
23-20
(a)

(b)


Strategy - 23.12 The Robinson Annulation Reaction
Carbonyl condensation reactions are perhaps the most versatile methods available for synthesizing complex molecules. By putting a few fundamental reactions together in the proper sequence, some remarkably useful transformations can be carried out. One such example is the Robinson annulation reaction for the synthesis of polycyclic molecules. The word annulation derives from the Latin annulus, meaning "ring," so an annulation reaction is one that builds a new ring onto a molecule.

Named after Robert Robinson, a British chemist and 1947 Nobel Prize winner, the Robinson annulation is a two-step process that combines a Michael reaction with an intramolecular aldol reaction. It takes place between a nucleophilic donor, such as a $\beta$-keto ester; an enamine, or a $\beta$-diketone; and an $\alpha, \beta$-unsaturated ketone acceptor, such as 3-buten-2-one. The product is a substituted 2 -cyclohexenone.

The first step of the Robinson annulation is simply a Michael reaction. An enamine or an enolate ion from a $\beta$-keto ester or $\beta$-diketone effects a conjugate addition to an $\alpha, \beta$-unsaturated ketone, yielding a 1,5 -diketone. But as we saw in Section 23.6, 1,5-diketones undergo intramolecular aldol condensation to yield cyclohexenones when treated with base. Thus, the final product contains a six-membered ring, and an annulation has been accomplished. One example of this occurs during a synthesis of the steroid hormone estrone (FIGURE 23.9).


Strategy - 23.12 The Robinson Annulation Reaction
FIGURE 23.9 Synthesis of the steroid hormone estrone using a Robinson annulation reaction. The nucleophilic donor is a $\beta$-diketone.
In this example, the $\beta$-diketone 2 -methyl-1,3-cyclopentanedione is used to generate the enolate ion required for Michael reaction and an aryl-substituted $\alpha, \beta$-unsaturated ketone is used as the acceptor. Base-catalyzed Michael reaction between the two partners yields an intermediate triketone, which then cyclizes in an intramolecular aldol condensation to give a Robinson annulation product. Several further transformations are required to complete the synthesis of estrone.

PROBLEM What product would you expect from a Robinson annulation reaction of
23-21 2-methyl-1,3-cyclopentanedione with 3-buten-2-one?


2-Methyl-1,3-cyclo- 3-Buten-2-one pentanedione -
PROBLEM How would you prepare the following compound using a Robinson annulation reaction between
23-22 a $\beta$-diketone and an $\alpha, \beta$-unsaturated ketone? Draw the structures of both reactants and the intermediate Michael addition product.


Biological Aldol Reactions -
Aldol reactions occur in many biological pathways but are particularly common in carbohydrate metabolism, where enzymes called aldolases catalyze the addition of a ketone enolate ion to an aldehyde. Aldolases occur in all organisms and are of two types. Type I aldolases occur primarily in animals and higher plants; type II aldolases occur primarily in fungi and bacteria. Both types catalyze the same kind of reaction, but type I aldolases operate through an enamine while type II aldolases require a metal ion (usually $\mathrm{Zn}^{2+}$ ) as a Lewis acid and operate through an enolate ion.

An example of an aldolase-catalyzed reaction occurs in glucose biosynthesis when dihydroxyacetone phosphate reacts with glyceraldehyde 3 -phosphate to give fructose 1,6-bisphosphate. In animals and higher plants, dihydroxyacetone phosphate is first converted into an enamine by reaction with the $-\mathrm{NH}_{2}$ group on a lysine amino acid in the enzyme. The enamine then adds to glyceraldehyde 3-phosphate, and the iminium ion that results is hydrolyzed. In bacteria and fungi, the aldol reaction occurs directly, with the ketone carbonyl group of glyceraldehyde 3-phosphate complexed to a $\mathrm{Zn}^{2+}$ ion to make it a better acceptor.


Type II aldolase -
Glyceraldehyde 3-phosphate


Fructose -
1,6-bisphosphate

Note that the aldolase-catalyzed reactions are mixed aldol reactions, which take place between two different partners, as opposed to the symmetrical aldol reactions between identical partners usually carried out in the laboratory. Mixed aldol reactions often give mixtures of products in the laboratory but are successful in living systems because of the selectivity of the enzyme catalysts.


Biological Claisen Condensations -
Claisen condensations, like aldol reactions, also occur in a large number of biological pathways. In fatty-acid biosynthesis, for instance, an enolate ion generated by decarboxylation (Section 22.7) of malonyl ACP adds to the carbonyl group of another acyl group bonded through a thioester linkage to a synthase enzyme. The tetrahedral intermediate that results then expels the synthase, giving acetoacetyl ACP. (The abbreviation ACP stands for acyl carrier protein, which forms thioester bonds to acyl groups.)

Mixed Claisen condensations also occur frequently in living organisms, particularly in the pathway for fattyacid biosynthesis that we'll discuss in Section 29.4. Butyryl synthase, for instance, reacts with malonyl ACP in a mixed Claisen condensation to give 3-ketohexanoyl ACP.


A Prologue to Metabolism -
Biochemistry is carbonyl chemistry. Almost all metabolic pathways used by living organisms involve one or more of the four fundamental carbonyl-group reactions we've seen in the chapters on Aldehydes and Ketones: Nucleophilic Addition Reactions through Carbonyl Condensation Reactions. The digestion and metabolic breakdown of all the major classes of food molecules-fats, carbohydrates, and proteins-take place by nucleophilic addition reactions, nucleophilic acyl substitutions, $\alpha$ substitutions, and carbonyl condensations. Similarly, hormones and other crucial biological molecules are built up from smaller precursors by these same carbonyl-group reactions.

FIGURE 23.10 You are what you eat. Food molecules are metabolized by pathways that involve the four major carbonyl-group reactions. (credit: "Spring" by Giuseppe Arcimboldo, Louvre Museum/Wikimedia Commons, Public Domain)

Take glycolysis, for example, the metabolic pathway by which organisms convert glucose to pyruvate as the first step in extracting energy from carbohydrates.

Glycolysis is a ten-step process that begins with isomerization of glucose from its cyclic hemiacetal form to its open-chain aldehyde form-the reverse of a nucleophilic addition reaction. The aldehyde then undergoes tautomerization to yield an enol, which undergoes yet another tautomerization to give the ketone fructose.

Fructose, a $\beta$-hydroxy ketone, is then cleaved by a retro-aldol reaction into two three-carbon molecules-one ketone and one aldehyde. Further carbonyl-group reactions then occur until pyruvate is formed.

These few examples are only an introduction; we'll look at several of the major metabolic pathways in more detail in Chapter 29. The bottom line, however, is that you haven't seen the end of carbonyl-group chemistry if you have any interest in biological or medical fields. A solid grasp of carbonyl-group reactions is crucial to an understanding of biochemistry.


Key Terms -
- aldol reaction
- carbonyl condensation reaction
- Claisen condensation reaction
- Dieckmann cyclization
- Michael reaction
- Robinson annulation reaction
- Stork enamine reaction


Summary -
In this chapter, we've discussed the fourth and last of the common carbonyl-group reactions-the carbonyl condensation. A carbonyl condensation reaction takes place between two carbonyl partners and involves both nucleophilic addition and $\alpha$-substitution processes. One carbonyl partner is converted by base into a nucleophilic enolate ion, which then adds to the electrophilic carbonyl group of the second partner. The first partner thus undergoes an $\alpha$ substitution, while the second undergoes a nucleophilic addition.

$$
\begin{array}{cc}
\text { Nucleophilic } & \text { Electrophilic } \\
\text { donor } & \text { acceptor }
\end{array}
$$

The aldol reaction is a carbonyl condensation that occurs between two aldehyde or ketone molecules. Aldol reactions are reversible, leading first to $\beta$-hydroxy aldehydes/ketones and then to $\alpha, \beta$-unsaturated products after dehydration. Mixed aldol condensations between two different aldehydes or ketones generally give a mixture of all four possible products. A mixed reaction can be successful, however, if one of the two partners is an unusually good donor (ethyl acetoacetate, for instance) or if it can act only as an acceptor (formaldehyde and benzaldehyde, for instance). Intramolecular aldol condensations of 1,4- and 1,5-diketones are also successful and provide a good way to make five- and six-membered rings.

The Claisen condensation reaction is a carbonyl condensation that occurs between two ester components and gives a $\beta$-keto ester product. Mixed Claisen condensations between two different esters are successful only when one of the two partners has no acidic $\alpha$ hydrogens (ethyl benzoate and ethyl formate, for instance) and thus can function only as the acceptor partner. Intramolecular Claisen condensations, called Dieckmann cyclization reactions, yield five- and six-membered cyclic $\beta$-keto esters starting from 1,6-and 1,7-diesters.


Summary -
The conjugate addition of a carbon nucleophile to an $\alpha, \beta$-unsaturated acceptor is known as the Michael reaction. The best Michael reactions take place between relatively acidic donors ( $\beta$-keto esters or $\beta$-diketones) and unhindered $\alpha, \beta$-unsaturated acceptors. Enamines, prepared by reaction of a ketone with a disubstituted amine, are also good Michael donors in the Stork enamine reaction.

Carbonyl condensation reactions are widely used in synthesis. One example of their versatility is the Robinson annulation reaction, which leads to the formation of a substituted cyclohexenone. Treatment of a $\beta$-diketone or $\beta$-keto ester with an $\alpha, \beta$-unsaturated ketone leads first to a Michael addition, which is followed by intramolecular aldol cyclization. Condensation reactions are also used widely in nature for the biosynthesis of such molecules as fats and steroids.


Summary of Reactions -
1. Aldol reaction (Section 23.1)

2. Mixed aldol reaction (Section 23.5)

3. Intramolecular aldol reaction (Section 23.6)

4. Dehydration of aldol products (Section 23.3)

5. Claisen condensation reaction (Section 23.7)

6. Mixed Claisen condensation reaction (Section 23.8)

7. Intramolecular Claisen condensation (Dieckmann cyclization; Section 23.9)

8. Michael reaction (Section 23.10)

9. Carbonyl condensations with enamines (Stork enamine reaction; Section 23.11)


Additional Problems -
Visualizing Chemistry
PROBLEM What ketones or aldehydes might the following enones have been prepared from by aldol reaction?
23-23 (a)

(b)

PROBLEM The following structure represents an intermediate formed by addition of an ester enolate ion to a
23-24 second ester molecule. Identify the reactant, the leaving group, and the product.

PROBLEM The following molecule was formed by an intramolecular aldol reaction. What dicarbonyl precursor 23-25 was used for its preparation?

PROBLEM The following molecule was formed by a Robinson annulation reaction. What reactants were used? 23-26


Mechanism Problems -
PROBLEM Predict the addition product for each of the following reactions and write the mechanism.

23-27 (a)

(b)

PROBLEM Based on your answers to Problem 23-27, predict the dehydration product for both reactions and 23-28 write the mechanism.

PROBLEM Predict the product(s) and provide the mechanisms for the following reactions.
23-29 (a)

(b)

PROBLEM Predict the product(s) and provide the mechanisms for the following reactions.
23-30 (a)

(b)

PROBLEM Predict the product(s) and provide the mechanisms for the following reactions.
23-31 (a)

(b)

PROBLEM Knoevenagel condensation is a reaction involving an active methylene compound (a $\mathrm{CH}_{2}$ flanked
23-32 by two electron-withdrawing groups) and an aldehyde and ketone. Propose a mechanism for the reaction below.

PROBLEM Azlactones are important starting materials used in the synthesis of dehydro $\alpha$-aminoacids. They
23-33 react with aldehydes to form an intermediate that is hydrolyzed under acidic conditions to give the final amino acid product. Provide the structure of the intermediate and propose a mechanism for its formation.


Azlactone -
Deydro- $\alpha$-amino acid
PROBLEM Leucine, one of the twenty amino acids found in proteins, is metabolized by a pathway that includes 23-34 the following step. Propose a mechanism.

PROBLEM Isoleucine, another of the twenty amino acids found in proteins, is metabolized by a pathway that 23-35 includes the following step. Propose a mechanism.

PROBLEM The first step in the citric acid cycle of food metabolism is reaction of oxaloacetate with acetyl CoA 23-36 to give citrate. Propose a mechanism, using acid or base catalysis as needed.

PROBLEM The amino acid leucine is biosynthesized from $\alpha$-ketoisovalerate by the following sequence of steps.
23-37 Show the mechanism of each.

PROBLEM The Knoevenagel reaction was introduced in Problem 23-2. Show the mechanism for the 23-38 Knoevenagel reaction of diethyl malonate with benzaldehyde.

Benzaldehyde
Cinnamic acid (91\%)
PROBLEM The Darzens reaction involves a two-step, base-catalyzed condensation of ethyl chloroacetate with
23-39 a ketone to yield an epoxy ester. The first step is a carbonyl condensation reaction, and the second step is an $\mathrm{S}_{\mathrm{N}} 2$ reaction. Write both steps, and show their mechanisms.

PROBLEM The following reaction involves a hydrolysis followed by an intramolecular nucleophilic acyl 23-40 substitution reaction. Write both steps, and show their mechanisms.

PROBLEM The following reaction involves an intramolecular Michael reaction followed by an intramolecular 23-41 aldol reaction. Write both steps, and show their mechanisms.

PROBLEM The following reaction involves a conjugate addition reaction followed by an intramolecular Claisen
23-42 condensation. Write both steps, and show their mechanisms.

PROBLEM The following reaction involves an intramolecular aldol reaction followed by a retro aldol-like 23-43 reaction. Write both steps, and show their mechanisms.

PROBLEM Propose a mechanism for the following base-catalyzed isomerization:
23-44


Azlactone -
PROBLEM Propose a mechanism for the following base-catalyzed isomerization:
23-44

PROBLEM The Mannich reaction of a ketone, an amine, and an aldehyde is one of the few three-component
23-45 reactions in organic chemistry. Cyclohexanone, for example, reacts with dimethylamine and acetaldehyde to yield an amino ketone. The reaction takes place in two steps, both of which are typical carbonyl-group reactions.

(a) The first step is reaction between the aldehyde and the amine to yield an intermediate iminium ion $\left(\mathrm{R}_{2} \mathrm{C}=\mathrm{NR}_{2}{ }^{+}\right)$plus water. Propose a mechanism, and show the structure of the intermediate iminium ion.
(b) The second step is reaction between the iminium ion intermediate and the ketone to yield the final product. Propose a mechanism.

PROBLEM Cocaine has been prepared by a sequence beginning with a Mannich reaction (Problem 23-45)
23-46 between dimethyl acetonedicarboxylate, an amine, and a dialdehyde. Show the structures of the amine and dialdehyde.

PROBLEM Propose a mechanism to account for the following reaction of an enamine with an alkyl halide: 23-47


Aldol Reactions -
PROBLEM Which of the following compounds would you expect to undergo aldol self-condensation? Show the 23-48 product of each successful reaction.
(a) Trimethylacetaldehyde
(b) Cyclobutanone
(c) Benzophenone (diphenyl ketone)
(d) 3-Pentanone
(e) Decanal
(f) 3-Phenyl-2-propenal

PROBLEM How might you synthesize each of the following compounds using an aldol reaction? Show the 23-49 structure of the starting aldehyde(s) or ketone(s) you would use in each case.
(a)

(b)

(c)

(d)

PROBLEM What product would you expect to obtain from aldol cyclization of hexanedial, 23-50 $\mathrm{OHCCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$ ?

PROBLEM Intramolecular aldol cyclization of 2,5-heptanedione with aqueous NaOH yields a mixture of two 23-51 enone products in the approximate ratio 9:1. Write their structures, and show how each is formed.

PROBLEM The major product formed by intramolecular aldol cyclization of 2,5-heptanedione (Problem 23-51)
23-52 has two singlet absorptions in the ${ }^{1} \mathrm{H}$ NMR spectrum, at $1.65 \delta$ and $1.90 \delta$, and has no absorptions in the range 3 to $10 \delta$. What is its structure?

PROBLEM Treatment of the minor product formed in the intramolecular aldol cyclization of 2,5-heptanedione 23-53 (Problems 23-51 and 23-52) with aqueous NaOH converts it into the major product. Propose a mechanism to account for this base-catalyzed isomerization.

PROBLEM How can you account for the fact that 2,2,6-trimethylcyclohexanone yields no detectable aldol 23-54 product even though it has an acidic $\alpha$ hydrogen?


Aldol Reactions -
PROBLEM How can you account for the fact that 2,2,6-trimethylcyclohexanone yields no detectable aldol 23-54 product even though it has an acidic $\alpha$ hydrogen?

PROBLEM The aldol reaction is catalyzed by acid as well as base. What is the reactive nucleophile in the acid-23-55 catalyzed aldol reaction? Propose a mechanism.

PROBLEM Cinnamaldehyde, the aromatic constituent of cinnamon oil, can be synthesized by a mixed aldol 23-56 condensation. Show the starting materials you would use, and write the reaction.


Cinnamaldehyde -
PROBLEM The following bicyclic ketone does not undergo aldol self-condensation even though it has two $\alpha$ 23-57 hydrogen atoms. Explain.


Claisen Condensations -
PROBLEM Give the structures of the possible Claisen condensation products from the following reactions. Tell 23-58 which, if any, you would expect to predominate in each case.
(a) $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Et}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{Et}$
(b) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Et}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{Et}$
(c) $\mathrm{EtOCO}_{2} \mathrm{Et}+$ cyclohexanone
(d) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{Et}$

PROBLEM In the mixed Claisen reaction of cyclopentanone with ethyl formate, a much higher yield of the
23-59 desired product is obtained by first mixing the two carbonyl components and then adding base, rather than by first mixing base with cyclopentanone and then adding ethyl formate. Explain.

PROBLEM Ethyl dimethylacetoacetate reacts instantly at room temperature when treated with ethoxide ion
23-60 to yield two products, ethyl acetate and ethyl 2-methylpropanoate. Propose a mechanism for this cleavage reaction.

PROBLEM In contrast to the rapid reaction shown in Problem 23-60, ethyl acetoacetate requires a temperature
23-61 over $150^{\circ} \mathrm{C}$ to undergo the same kind of cleavage reaction. How can you explain the difference in reactivity?


Michael and Enamine Reactions -
PROBLEM How might the following compounds be prepared using Michael reactions? Show the nucleophilic 23-62 donor and the electrophilic acceptor in each case.
(a)

(b)

(c)

(d)

(e)

(f)

PROBLEM The so-called Wieland-Miescher ketone is a valuable starting material used in the synthesis of 23-63 steroid hormones. How might you prepare it from 1,3-cyclohexanedione?

Wieland-Miescher ketone

PROBLEM The Stork enamine reaction and the intramolecular aldol reaction can be carried out in sequence 23-64 to allow the synthesis of cyclohexenones. For example, reaction of the pyrrolidine enamine of cyclohexanone with 3-buten-2-one, followed by enamine hydrolysis and base treatment, yields the product indicated. Write each step, and show the mechanism of each.

PROBLEM How could you prepare the following cyclohexenones by combining a Stork enamine reaction with
23-65 an intramolecular aldol condensation? (See Problem 23-64.)
(a)

(b)

(c)

PROBLEM The following reaction involves two successive intramolecular Michael reactions. Write both steps, 23-66 and show their mechanisms.


General Problems -
PROBLEM What condensation products would you expect to obtain by treatment of the following substances 23-67 with sodium ethoxide in ethanol?
(a) Ethyl butanoate
(b) Cycloheptanone
(c) 3,7-Nonanedione
(d) 3-Phenylpropanal

PROBLEM The following reactions are unlikely to provide the indicated product in high yield. What is wrong 23-68 with each?
(a)

(b)

(c)

PROBLEM Fill in the missing reagents a-h in the following scheme:
23-69

PROBLEM How would you prepare the following compounds from cyclohexanone?

23-70 (a)

(b)

(c)

(d)

PROBLEM The compound known as Hagemann's ester is prepared by treatment of a mixture of formaldehyde 23-71 and ethyl acetoacetate with base, followed by acid-catalyzed decarboxylation.

Hagemann's ester
(a) The first step is an aldol-like condensation between ethyl acetoacetate and formaldehyde to yield an $\alpha, \beta$-unsaturated product. Write the reaction, and show the structure of the product.
(b) The second step is a Michael reaction between ethyl acetoacetate and the unsaturated product of the first step. Show the structure of the product.

PROBLEM The third and fourth steps in the synthesis of Hagemann's ester from ethyl acetoacetate and
23-72 formaldehyde (Problem 23-71) are an intramolecular aldol cyclization to yield a substituted cyclohexenone, and a decarboxylation reaction. Write both reactions, and show the products of each step.

PROBLEM When 2-methylcyclohexanone is converted into an enamine, only one product is formed despite
23-73 the fact that the starting ketone is unsymmetrical. Build molecular models of the two possible products and explain the fact that the sole product is the one with the double bond opposite the methyl-substituted carbon.


Amines and Heterocycles -
FIGURE 24.1 The characteristic and unmistakable odor of fish is due to a mixture of simple alkylamines. (credit: modification of work "Pike Place" by Sharat Ganapati/Flickr, CC BY 2.0)


CHAPTER CONTENTS - 24.1 Naming Amines
24.2 Structure and Properties of Amines
24.3 Basicity of Amines
24.4 Basicity of Arylamines
24.5 Biological Amines and the Henderson-Hasselbalch Equation
24.6 Synthesis of Amines
24.7 Reactions of Amines
24.8 Reactions of Arylamines
24.9 Heterocyclic Amines
24.10 Spectroscopy of Amines

WHY THIS CHAPTER? By the end of this chapter, we will have seen all the common functional groups. Of those groups, carbonyl compounds and amines are the most abundant and have the richest chemistry. In addition to proteins and nucleic acids, the majority of pharmaceutical agents contain amine functional groups, and many of the common coenzymes necessary for biological catalysis are amines.

Amines are organic derivatives of ammonia in the same way that alcohols and ethers are organic derivatives of water. Like ammonia, amines contain a nitrogen atom with a lone pair of electrons, making amines both basic and nucleophilic. We'll soon see, in fact, that most of the chemistry of amines depends on the presence of this lone pair of electrons.

Amines occur widely in all living organisms. Trimethylamine, for instance, occurs in animal tissues and is partially responsible for the distinctive odor of fish; nicotine is found in tobacco; and cocaine is a stimulant found in the leaves of the South American coca bush. In addition, amino acids are the building blocks from
which all proteins are made, and cyclic amine bases are constituents of nucleic acids.

Trimethylamine

Nicotine

Cocaine


CHAPTER CONTENTS - 24.1 Naming Amines
Amines can be either alkyl-substituted (alkylamines) or aryl-substituted (arylamines). Although much of the chemistry of the two classes is similar, there are also substantial differences. Amines are classified as primary ( $\mathbf{R N H}_{\mathbf{2}}$ ), secondary ( $\left.\mathbf{R}_{\mathbf{2}} \mathbf{N H}\right)$, or tertiary $\left(\mathbf{R}_{\mathbf{3}} \mathbf{N}\right)$, depending on the number of organic substituents attached to nitrogen. Thus, methylamine $\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)$ is a primary amine, dimethylamine $\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\right]$ is a secondary amine, and trimethylamine $\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\right]$ is a tertiary amine. Note that this usage of the terms primary, secondary, and tertiary differs from our previous usage. When we speak of a tertiary alcohol or alkyl halide, we refer to the degree of substitution at the alkyl carbon atom, but when we speak of a tertiary amine, we refer to the degree of substitution at the nitrogen atom.

tert-Butyl alcohol (a tertiary alcohol)

Trimethylamine (a tertiary amine)

tert-Butylamine
(a primary amine)

Compounds containing a nitrogen atom with four attached groups also exist, but the nitrogen atom must carry a formal positive charge. Such compounds are called quaternary ammonium salts.


A quaternary ammonium salt -
Primary amines are named in the IUPAC system in several ways. For simple amines, the suffix -amine is added to the name of the alkyl substituent. You might also recall from the chapter on Benzene and Aromaticity that the aromatic phenylamine, $\mathrm{H}_{2} \mathrm{~N}-\mathrm{C}_{6} \mathrm{H}_{5}$, has the common name aniline.

tert-Butylamine

Cyclohexylamine

Alternatively, the suffix -amine can be used in place of the final -e in the name of the parent compound.

4,4-Dimethylcyclohexanamine
$\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$

1,4-Butanediamine

Amines with more than one functional group are named by considering the $-\mathrm{NH}_{2}$ as an amino substituent on the parent molecule.

Symmetrical secondary and tertiary amines are named by adding the prefix di- or tri- to the alkyl group.

Diphenylamine

Triethylamine

Unsymmetrically substituted secondary and tertiary amines are referred to as $N$-substituted primary amines. The largest alkyl group takes the parent name, and the other alkyl groups are considered $N$-substituents on the parent ( $N$ because they're attached to nitrogen).

$\mathbf{N}, \mathbf{N}$-Dimethylpropylamine
$\boldsymbol{N}$-Ethyl- $\boldsymbol{N}$-methylcyclohexylamine
Heterocyclic amines-compounds in which the nitrogen atom occurs as part of a ring-are also common, and each different heterocyclic ring system has its own parent name. The heterocyclic nitrogen atom is always numbered as position 1.

Pyridine

Pyrrole

Quinoline

Imidazole


Indole -
Pyrimidine

Pyrrolidine

Piperidine

PROBLEM Name the following compounds:
24-1 (a) $\mathrm{CH}_{3} \mathrm{NHCH}_{2} \mathrm{CH}_{3}$
(b)

(c)

(d)

PROBLEM Draw structures corresponding to the following IUPAC names:
24-2 (a) Triisopropylamine (b) Triallylamine (c) $N$-Methylaniline
(d) $N$-Ethyl- $N$-methylcyclopentylamine (e) $N$-Isopropylcyclohexylamine (f) $N$-Ethylpyrrole

PROBLEM Draw structures for the following heterocyclic amines:
24-3 (a) 5-Methoxyindole
(b) 1,3-Dimethylpyrrole
(c) 4-( $N, N$-Dimethylamino)pyridine
(d) 5-aminopyrimidine


Indole - 24.2 Structure and Properties of Amines
The bonding in alkylamines is similar to the bonding in ammonia. The nitrogen atom is $s p^{3}$-hybridized, with its three substituents occupying three corners of a regular tetrahedron and the lone pair of electrons occupying the fourth corner. As you might expect, the $\mathrm{C}-\mathrm{N}-\mathrm{C}$ bond angles are close to the $109^{\circ}$ tetrahedral value. For trimethylamine, the $\mathrm{C}-\mathrm{N}-\mathrm{C}$ bond angle is $108^{\circ}$ and the $\mathrm{C}-\mathrm{N}$ bond length is 147 pm .

Trimethylamine
One consequence of tetrahedral geometry is that an amine with three different substituents on nitrogen is chiral, as we saw in Section 5.10. Unlike chiral carbon compounds, however, chiral amines can't usually be resolved because the two enantiomeric forms rapidly interconvert by a pyramidal inversion, much as an alkyl halide inverts in an $\mathrm{S}_{\mathrm{N}} 2$ reaction. Pyramidal inversion occurs by a momentary rehybridization of the nitrogen atom to planar, $s p^{2}$ geometry, followed by rehybridization of the planar intermediate to tetrahedral, $s p^{3}$ geometry (FIGURE 24.2). The barrier to inversion is about $25 \mathrm{~kJ} / \mathrm{mol}$ ( $6 \mathrm{kcal} / \mathrm{mol}$ ), an amount only twice as large as the barrier to rotation about a $\mathrm{C}-\mathrm{C}$ single bond.


Indole - 24.2 Structure and Properties of Amines
FIGURE 24.2 Pyramidal inversion rapidly interconverts the two mirror-image (enantiomeric) forms of an amine.
Alkylamines have a variety of applications in the chemical industry as starting materials for the preparation of insecticides and pharmaceuticals. Labetalol, for instance, a so-called $\beta$-blocker used for the treatment of high blood pressure, is prepared by $\mathrm{S}_{\mathrm{N}} 2$ reaction of an epoxide with a primary amine. The substance marketed for drug use is a mixture of all four possible stereoisomers, but the biological activity results primarily from the $(R, R)$ isomer.

Labetalol
Like alcohols, amines with fewer than five carbon atoms are generally water-soluble. Also like alcohols, primary and secondary amines form hydrogen bonds and are highly associated. As a result, amines have higher boiling points than alkanes of similar molecular weight. Diethylamine ( $\mathrm{MW}=73 \mathrm{amu}$ ) boils at $56.3^{\circ} \mathrm{C}$, for instance, while pentane ( $\mathrm{MW}=72 \mathrm{amu}$ ) boils at $36.1^{\circ} \mathrm{C}$.

FIGURE 24.3 Hydrogen bonding in amines.
One other characteristic of amines is their odor. Low-molecular-weight amines such as trimethylamine have a distinctive fishlike aroma, while diamines such as cadaverine (1,5-pentanediamine) and putrescine (1,4-butanediamine) have the appalling odors you might expect from their common names. Both of these diamines arise from the decomposition of decaying tissue.


Indole - 24.3 Basicity of Amines
The chemistry of amines is dominated by the lone pair of electrons on nitrogen, which makes amines both basic and nucleophilic. They react with acids to form acid-base salts, and they react with electrophiles in many of the polar reactions seen in past chapters. Note in the following electrostatic potential map of trimethylamine how the negative (red) region corresponds to the lone pair of electrons on nitrogen.

Amines are much stronger bases than alcohols and ethers, their oxygen-containing analogs. When an amine is dissolved in water, an equilibrium is established in which water acts as an acid and transfers a proton to the amine. Just as the acid strength of a carboxylic acid can be measured by defining an acidity constant $K_{\mathrm{a}}$ (Section 2.8), the base strength of an amine can be measured by defining an analogous basicity constant $K_{b}$. The larger the value of $K_{\mathrm{b}}$ and the smaller the value of $\mathrm{p} K_{\mathrm{b}}$, the more favorable the proton-transfer equilibrium and the stronger the base.

For the reaction

$$
\begin{aligned}
& \mathrm{RNH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{RNH}_{3}^{+}+\mathrm{OH}^{-} \\
& K_{\mathrm{b}}=\frac{\left[\mathrm{RNH}_{3}{ }^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{RNH}_{2}\right]} \\
& \mathrm{p} K_{\mathrm{b}}=-\log K_{\mathrm{b}}
\end{aligned}
$$

In practice, $K_{\mathrm{b}}$ values are not often used. Instead, the most convenient way to measure the basicity of an amine $\left(\mathrm{RNH}_{2}\right)$ is to look at the acidity of the corresponding ammonium ion $\left(\mathrm{RNH}_{3}{ }^{+}\right)$.

For the reaction


Indole - 24.3 Basicity of Amines
For the reaction

$$
\begin{aligned}
& \mathrm{RNH}_{3}{ }^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{RNH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+} \\
& K_{\mathrm{a}}=\frac{\left[\mathrm{RNH}_{2}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{RNH}_{3}{ }^{+}\right]}
\end{aligned}
$$

so

$$
\begin{aligned}
K_{\mathrm{a}} & \cdot K_{\mathrm{b}}=\left[\frac{\left[\mathrm{RNH}_{2}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{RNH}_{3}{ }^{+}\right]}\right]\left[\frac{\left[\mathrm{RNH}_{3}{ }^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{RNH}_{2}\right]}\right] \\
& =\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=K_{\mathrm{w}}=1.00 \times 10^{-14}
\end{aligned}
$$

Thus

$$
K_{\mathrm{a}}=\frac{K_{\mathrm{w}}}{K_{\mathrm{b}}} \quad \text { and } \quad K_{\mathrm{b}}=\frac{K_{\mathrm{w}}}{K_{\mathrm{a}}}
$$

and

$$
\mathrm{p} K_{\mathrm{a}}+\mathrm{p} K_{\mathrm{b}}=14
$$


Indole - 24.3 Basicity of Amines
and

$$
\mathrm{p} K_{\mathrm{a}}+\mathrm{p} K_{\mathrm{b}}=14
$$

These equations say that the $K_{\mathrm{b}}$ of an amine multiplied by the $K_{\mathrm{a}}$ of the corresponding ammonium ion is equal to $K_{\mathrm{W}}$, the ion-product constant for water ( $1.00 \times 10^{-14}$ ). Thus, if we know $K_{\mathrm{a}}$ for an ammonium ion, we also know $K_{\mathrm{b}}$ for the corresponding amine base because $K_{\mathrm{b}}=K_{\mathrm{w}} / K_{\mathrm{a}}$. The more acidic the ammonium ion, the less tightly the proton is held and the weaker the corresponding base. That is, a weaker base has an ammonium ion with a smaller $\mathrm{p} K_{\mathrm{a}}$ and a stronger base has an ammonium ion with a larger $\mathrm{p} K_{\mathrm{a}}$.


Weaker base Smaller $\mathrm{p} K_{\mathrm{a}}$ for ammonium ion <br> Stronger base Larger $\mathrm{p} K_{\mathrm{a}}$ for ammonium ion -
TABLE 24.1 lists $\mathrm{p} K_{\mathrm{a}}$ values of the ammonium ions from a variety of amines and indicates that there is a substantial range of amine basicities. Most simple alkylamines are similar in their base strength, with $\mathrm{p} K_{\mathrm{a}}$ 's for their ammonium ions in the narrow range 10 to 11. Arylamines, however, are considerably less basic than alkylamines, as are the heterocyclic amines pyridine and pyrrole.

TABLE 24.1 Basicity of Some Common Amines

| Name | Structure | $\mathrm{p} K_{\mathrm{a}}$ of ammonium ion |
| :--- | :--- | :--- |
| Ammonia | $\mathrm{NH}_{3}$ | 9.26 |
| Primary alkylamine |  |  |
| Methylamine |  | $\mathrm{CH}_{3} \mathrm{NH}_{2}$ |