Jones M., Fleming S.A. Organic Chemistry

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23.2 Amino Acids 1179

As you can see from Table 23.1, amino acids are high-melting solids. Does this

not seem curious? Common amino acids have molecular weights between 75 and

204, and this molecular weight does not seem sufficient to produce molecules that

are solids. This conundrum leads directly to Section 23.2c.

23.2c Acid–Base Properties of Amino Acids Why are amino acids solids?

To answer this question, let’s ask another. What reaction do you expect between the

generic amine and the generic carboxylic acid of Figure 23.4?

A simple organic acid, such as acetic acid, has a pK

of about 4.5, and an ammo-

nium ion has a pK

of about 8–10. An amine and a carboxylic acid must react to

give the ammonium salt of Figure 23.5,because the carboxylic acid is a much stronger

acid than an ammonium ion. Exactly the same reaction must take place within a

single amino acid to form the “inner salt”or zwitterion, a molecule containing both

positively and negatively charged atoms.

FIGURE 23.4 What reaction must

occur between these two molecules?

~ 4.5 pK

~ 9

HNH

–

A zwitterion

FIGURE 23.5 An amine and a carboxylic acid must undergo acid–base chemistry to give an ammonium

salt of the acid. In amino acids, an intramolecular version of the same reaction gives a zwitterion.

Amino acids are ionic compounds under normal con-

ditions,and therefore have high melting points.Other prop-

erties are in accord with this notion.They are miscible with

water, but quite insoluble in nonpolar organic solvents.They

have high dipole moments. Of course, the charge state of

an amino acid depends on the pH of the medium in which

it finds itself. Under neutral or nearly neutral conditions,

amino acids exist as the zwitterions of Figure 23.5. If we

lower the pH, eventually the carboxylate anion will proto-

nate, and the amino acid will exist primarily in the ammo-

nium ion form. Conversely, if we raise the pH, eventually

the ammonium ion site will be deprotonated, and the mol-

ecule will exist as the carboxylate anion (Fig. 23.6).

–

Zwitterion

–

Carboxylate formAmmonium ion form

Neutral At high pHAt low pH

pH = 15pH = 1.0

FIGURE 23.6 At low pH (strongly acidic solution), the

zwitterion is protonated to give a net positively charged

molecule. At high pH (strongly basic solution), the zwitterion is

deprotonated to give a net negatively charged molecule.

PROBLEM 23.3 The pK

for a simple organic acid is 4–5. Note the pK

values of

the amino acids (column 5 in Table 23.1). Why are amino acids much more

acidic?

Amino acids can be described by a series of pK

values. The first refers to the

deprotonation of the carboxylic acid group, and it is in the range 1.7–2.4. The sec-

ond pK

value is for deprotonation of the ammonium ion, and is in the range

8.8–10.8 (Table 23.1).

The pH at which the concentration of the ammonium ion equals that of the car-

boxylate ion is called the isoelectric point (pI).At this pH the molecule will have a net

1180 CHAPTER 23 Amino Acids and Polyamino Acids (Peptides and Proteins)

zero charge.The zwitterion is the predominant form at this pH.This pH is defined

as the pI, which is a symbol for isoelectric point. For the amino acids with nonacidic

or basic side chains, the pI can be calculated by taking the midpoint between the pK

of the carboxylic acid and the pK

of the ammonium group.Thus,pI  (pK

 pK

)/2.

PROBLEM 23.4 Use the pK

data in Table 23.1 to calculate the pI (the isoelectric

point) for alanine.

As you can see from Table 23.1, many amino acids bear acidic or basic side chains,

which complicates the picture somewhat and can make the determination of the iso-

electric point a bit more difficult. These side groups are also involved in acid–base

equilibria and there are pK

values associated with them. For example, arginine will

be doubly protonated at very low pH (Fig. 23.7).

“H

”

Arginine (Arg)

at pH = 7

lower pH

= 9.1

= 12.5

= 2.2

–

.. ..

FIGURE 23.7 Arginine is twice

protonated at low pH. Because the

side chain has a basic group, this

amino acid has three pK

values.

PROBLEM 23.5 Notice which nitrogen on the arginine side chain is protonated

under acidic conditions. Explain why it is the nitrogen shown that is protonated.

The most acidic group is the carboxylic acid, pK

 2.2. Second is the ammo-

nium ion, pK

 9.1. The side chain group has a pK

of 12.5 and will be the last

position deprotonated as the solution becomes less acidic.

PROBLEM 23.6 Which ring nitrogen in histidine (Table 23.1) will be protonated

first as the solution becomes more acidic?

These acid–base properties give rise to an especially powerful separation method

widely used in analysis of amino acid mixtures, called electrophoresis. It works this

way. A mixture of amino acids is placed on moistened paper. At a given pH, each

amino acid exists as either a positive ion, a neutral

zwitterion, or a negative ion. When a current is

applied to the paper, the positively charged ions will

migrate to the cathode (negative pole), the neutral

zwitterions remain stationary, and the negatively

charged ions move to the anode (positive pole) at

rates depending on the pH and the strength of the

applied current. The net result is a separation, as the

amino acids in the mixture migrate at different rates.

For example, at pH 5.5, the amino acids Lys, Phe,

and Glu exist in the forms shown in Figure 23.8.

Lys

(cation = 1+)

COO

–

Phe

(neutral)

Glu

(anion = 1–)

COO

–

COO

–

COO

–

FIGURE 23.8 The charge states of Lys, Phe, and Glu at pH 5.5.

23.2 Amino Acids 1181

When a current is applied, Phe will not move, Lys will migrate to the cathode, and

Glu to the anode (Fig. 23.9).

Start (current off) Turn current on

Cathode

–

Anode

Mix of Lys, Phe, Glu

Cathode

–

Anode

Lys (moves

to cathode)

Phe (no

movement)

Glu (moves

to anode)

FIGURE 23.9 Separation of three amino acids by electrophoresis.

23.2d Syntheses of Amino Acids One very simple method for making

an amino acid is to alkylate ammonia using the S

2 reaction of ammonia with an

α-halo acid (Fig. 23.10).

THE GENERAL CASE

SPECIFIC EXAMPLES

O1.

2. neutralization

48 h, 25 °C

excess

Glycine

(65%)

2-Aminohexanoic acid

(norleucine)

(65%)

COOH

COO

–

COO

–+

25 h, 55 °C

COO

–

COOH

FIGURE 23.10 Simple alkylation of

an α-halo acid will give an α-amino

acid.

In Chapter 7 (p. 312), a closely related alkylation method appeared as a means

of producing alkylamines. A problem for this synthetic procedure is the ease of

overalkylation.The simple alkylation reaction is not generally a practical method for

1182 CHAPTER 23 Amino Acids and Polyamino Acids (Peptides and Proteins)

PROBLEM 23.7 Devise simple syntheses of the following α-bromo acids starting

from alcohols containing no more than four carbon atoms and inorganic reagents

of your choice (p. 950).

(a)

(b)

PROBLEM 23.8 Why are α-bromo acids (or their carboxylate salts) more easily

alkylated than simple alkyl bromides in the S

2 reaction (p. 542)?

A successful variation of the simple alkylation synthesis involves a protect-

ed, disguised amine, and is called the Gabriel synthesis (Siegmund Gabriel,

1851–1924). In this procedure, bromomalonic ester is first treated with

potassium phthalimide in an S

2 displacement of bromide ion (Fig. 23.11).

COOH

25–110 °C, 1 h

EtOOC

(97%)

Methionine

(85%)

(59%)

1. NaOH/H

2. HCl, Δ

–

SCH

base

(alkylation step)

hydrolysis and

decarboxylation

SCH

EtOOC

COOH

Bromomalonic

ester

Potassium

phthalimide

Phthalic acid

WEB 3D

FIGURE 23.11 The Gabriel synthesis uses bromomalonic ester in the first step.

producing alkylamines,but it is useful in making α-amino acids.The α-bromo acids

are easily available,and a vast excess of ammonia can be used to minimize overalkyl-

ation.Moreover, the first formed α-amino acids are not especially effective competi-

tors in the S

2 reaction, because they are sterically congested and less basic than

simple amines. This technique is an effective way of making both “natural” and

“unnatural” amino acids.

PROBLEM 23.9 Provide detailed mechanisms for the reactions of Figure 23.11.

WORKED PROBLEM 23.10 In a simpler version of the Gabriel synthesis,

acetamidomalonic ester is used rather than bromomalonic ester. Sketch out the

steps in this related amino acid synthesis.

Acetamidomalonic ester Phenylalanine

OOC

CH CH

HOOC

PhCH

1. NaOEt/EtOH

(87%)

2. PhCH

EtOOC

Doubly α

ANSWER It is essentially the same process as that of Figure 23.11 and Problem

23.9. Advantage is first taken of the acidity of the doubly α hydrogen to introduce

an R group.

Now acid (or base) is used to convert the esters into carboxylic acids and to

hydrolyze the amide. Heating decarboxylates the malonic acid and liberates the

amino acid.

EtOOC

HOOC

(80%)

1. NaOH/H

2. H

O/H

(neutralize)

HOOC

23.2 Amino Acids 1183

The nitrogen atom destined to become the amine nitrogen of the amino acid is

introduced at this point, and rendered nonnucleophilic by the flanking carbonyl

groups.In the next step,the malonic ester product is treated with base and an active

alkyl halide to introduce the R group of the amino acid.Finally, treatment with acid

or base, followed by neutralization, uncovers the amine. The phthalimide group is

hydrolyzed to phthalic acid, and the malonic ester is hydrolyzed to a malonic acid.

Gentle heating induces decarboxylation to give the amino acid.There are many vari-

ations on this general theme.

1184 CHAPTER 23 Amino Acids and Polyamino Acids (Peptides and Proteins)

In the Strecker synthesis (Adolph Strecker, 1822–1871), a cyanide is used as

the source of the carboxylic acid portion of an amino acid (Fig. 23.12). An aldehyde

is treated with ammonia, the ultimate source of the amino group, and cyanide ion.

In the first step of the reaction, a carbinolamine is formed (p. 790) and water is lost

to give an imine.The imine is then attacked by cyanide to give an α-amino cyanide.

Acid hydrolysis of the nitrile gives the acid (p. 904).

THE GENERAL CASE

A SPECIFIC EXAMPLE

KCN

Carbinolamine

Overall yield

(36%)

COO

–

NaCN

KCN

1. H

O/CH

45 °C, 2 h

HCl

2 h

–

Imine

COOH

FIGURE 23.12 The Strecker amino

acid synthesis.

The first step in this reaction, imine formation, is analogous to the first step in

the Mannich condensation in which the role of the amine is also to produce a reac-

tive imine (p. 1003).This imine can next be attacked by the nucleophilic cyanide to

give the tetrahedral α-amino nitrile.

As we have seen, amino acids in Nature are almost all optically active (S) enan-

tiomers, or

L forms (save glycine, which is achiral). None of the syntheses outlined

in this section automatically produces the natural (S) enantiomer specifically. In order

to obtain the single enantiomer, we either need a method for resolution (separation)

of the racemic mixture of enantiomers formed in these syntheses or a method of

selectively producing the (S) enantiomers.

23.2e Resolution of Amino Acids The standard method is the one we first

discussed in our chapter on chirality (p.169) and involves transformation of the pair

of enantiomers, which have identical physical properties, into a pair of diastere-

omers, which do not, and can be separated by physical means (most often crystal-

lization or chromatography).

PROBLEM 23.11 There actually is one physical property not shared by enantiomers.

What is it?

Optically active amines are used to transform the enantiomeric amino acids in

a racemic mixture into diastereomeric ammonium salts that can be separated by

23.2 Amino Acids 1185

COO

–

(S) =

COOH

HAcNH

Brucine

COO

–

(R) =

COOH

NHAcH

NaOH/H

O NaOH/H

COO

–

COO

–

COO

–

COO

–

AcNH

NHAc

These ammonium salts of

the (R) and (S) carboxylate

anions are diastereomeric

and can be separated by

fractional crystallization

FIGURE 23.13 Resolution

of a pair of enantiomeric

amino acids.

enzyme

“glutamate dehydrogenase”

␣-Ketoglutaric acid

COO

–

L-Glutamic acid

Sugar

NAD

NADH

FIGURE 23.14 Synthesis of an optically active L-amino acid through reductive amination.

However, it seems that there must be a way to avoid the tedium of fractional

crystallization and make the optically active compounds directly. After all, Nature

does this somehow, and we should be clever enough to find out how this happens

and imitate the process. One method Nature uses to make optically active amino

acids involves a reductive amination (Fig. 23.14). Ammonia and an α-keto carboxylic

crystallization.The chiral amines of choice are optically pure alkaloids (p.255), which

are readily available in Nature. The process of separation is outlined in Figure 23.13.

1186 CHAPTER 23 Amino Acids and Polyamino Acids (Peptides and Proteins)

acid react in the presence of an enzyme and a source of hydride, usually NADH

(p. 814), to produce an optically active chiral amino acid.

There are two procedures used by chemists that imitate the natural pathway.

In one, actual microorganisms are used in fermentation reactions that can direct-

ly produce some

L-amino acids. In the other, enzymes are used to react with one

enantiomer of a racemic pair. As you might guess, it is usually the

L isomer that

is “eaten” by the enzyme, because enzymes have evolved in an environment of

isomers. So it is usually the D isomers that are ignored by enzymes. In an extraor-

dinarily clever procedure, this preference for natural,

L, enantiomers can be used

to achieve a kinetic resolution (Fig. 23.15). The mixture of enantiomers is first

acetylated to create amide linkages. An enzyme, hog-kidney acylase, then

hydrolyzes the amide linkages of only

L-amino acids. So, treatment of the pair of

acetylated amino acids leads to formation of the free

L-amino acid. The acylat-

D enantiomer is unaffected by the enzyme, which has evolved to react only

with natural

L-amino acids. The free L-amino acid can then easily be separated

from the residual

D acetylated material.

Free L enantiomer Acetylated

D enantiomer

HOAc

COO

–

(S) =

COO

–

(R) =

COO

–

(S) =

COOH

(R) =

hog-kidney

acylase

hog-kidney acylase

(no reaction with

D enantiomer)

COOH

(S) =

NHAc

AcNH

COOH

(R) =

A racemic mixture

of amino acids

A racemic mixture of

acetylated amino acids

Easily separable mixture

FIGURE 23.15 A kinetic resolution of a pair of enantiomeric amino acids.

23.3 Reactions of Amino Acids

The simple expectation that the chemical reactions of the amino acids are a

combination of carboxylic acid (Chapter 17) and amine (Chapter 6) chemistry

is right. Indeed, we have already seen some of these reactions in the acid–base

23.3 Reactions of Amino Acids 1187

1. Ac

HOOC

HOOCHOOC

HOOC

(91%)

(>72%)

2. H

20 min

FIGURE 23.16 The amino group of an amino acid can be acylated.

78 °C

HCl

OCH

(~ 100%)

COO

–

FIGURE 23.17 Fischer esterification of the acid group of an

amino acid.

chemistry described in Section 23.2c. A few more important and useful exam-

ples follow.

23.3a Acylation Reactions The amino group of amino acids can be

acylated using all manner of acylating agents. Typical examples of effective

acylating agents are acetic anhydride, benzoyl chloride, and acetyl chloride

(Fig. 23.16).

23.3b Esterification Reactions The carboxylic acid group of amino acids

can be esterified using typical esterification reactions. Fischer esterification is the

simplest process (Fig. 23.17).

PROBLEM 23.12 Write a mechanism for the acylation reactions of Figure 23.16.

PROBLEM 23.13 If amino acids are typically in their zwitterionic form, how are

acylation and esterification reactions,which depend on the presence of free amine

or acid groups, possible?

23.3c Reaction with Ninhydrin The hydrate of indan-1,2,3-trione is

called ninhydrin. Amino acids react with ninhydrin to form a deep purple mol-

ecule. It is this purple color that is used to detect amino acids. Although it may

1188 CHAPTER 23 Amino Acids and Polyamino Acids (Peptides and Proteins)

seem curious at first, all amino acids except proline, regardless of the identity of

the R group, form the same purple molecule on reaction with ninhydrin. How

can this be?

Like all hydrates, ninhydrin is in equilibrium with the related carbonyl com-

pound (Fig. 23.18).The amino acid first reacts with the trione to form an imine

through the small amount of the free amine form of the amino acid present

at equilibrium. We have seen this kind of reaction many times (p. 793). Next,

decarboxylation of the imine produces a second imine that is hydrolyzed under

the aqueous conditions to give an amine. It is in this hydrolysis step that the

R group of the original amino acid is lost. Still another imine, the third in this

sequence, is then formed through reaction of the amine with another molecule

of the ninhydrin trione. The extensive conjugation in the product leads to a

very low energy absorption in the visible region (p. 529), and the observed

purple color.

Ninhydrin Ninhydrin trione

Second imine

R Group of original

amino acid lost here

COO

–

(–)

(–) (–)

(–)

–

hydrolysis

decarboxylation

First imine

Third imine—purple!

WEB 3D

FIGURE 23.18 Formation of a purple molecule through reaction of an amino acid with the ninhydrin trione.

Each amino acid in Table 23.1 except proline gives the same purple product.

PROBLEM 23.14 Why is the trione of Figure 23.18 hydrated at the middle car-

bonyl rather than either of the side carbonyl groups?