Secondary alkylamine -
TABLE 24.1 Basicity of Some Common Amines

| Name | Structure | $\mathrm{p} K_{\mathrm{a}}$ of ammonium ion |
| :---: | :---: | :---: |
| Diethylamine | $\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{NH}$ | 10.98 |
| Pyrrolidine | <smiles>C1CCNC1</smiles> | 11.27 |
| Tertiary alkylamine |  |  |
| Triethylamine | $\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}$ | 10.76 |
| Arylamine |  |  |
| Aniline | <smiles>Nc1ccccc1</smiles> | 4.63 |
| Heterocyclic amine |  |  |
| Pyridine | <smiles>c1ccncc1</smiles> | 5.25 |
| Pyrimidine | <smiles>c1cncnc1</smiles> | 1.3 |
| Pyrrole | <smiles>c1cc[nH]c1</smiles> | 0.4 |
| Imidazole | <smiles>c1c[nH]cn1</smiles> | 6.95 |

In contrast with amines, amides $\left(\mathrm{RCONH}_{2}\right)$ are nonbasic. Amides aren't protonated by aqueous acids, and they are poor nucleophiles. The main reason for this difference in basicity between amines and amides is that an amide is stabilized by delocalization of the nitrogen lone-pair electrons through orbital overlap with the carbonyl group. In resonance terms, amides are more stable and less reactive than amines because they are hybrids of two resonance forms. This amide resonance stabilization is lost when the nitrogen atom is protonated, so protonation is disfavored. The following electrostatic potential maps clearly show a reduced electron density on the amide nitrogen.


Secondary alkylamine -
To purify amines, it's often possible to take advantage of their basicity. For example, if a mixture of a basic amine and a neutral compound such as a ketone or alcohol is dissolved in an organic solvent and aqueous acid is added, the basic amine dissolves in the water layer as its protonated salt, while the neutral compound remains in the organic solvent layer. Separation of the water layer and neutralization of the ammonium ion by addition of NaOH then provides the pure amine (FIGURE 24.4).

FIGURE 24.4 Separation and purification of an amine component from a mixture by extraction of its ammonium salt into water.
In addition to their behavior as bases, primary and secondary amines can also act as very weak acids because an $\mathrm{N}-\mathrm{H}$ proton can be removed by a sufficiently strong base. We've seen, for example, how diisopropylamine ( $\mathrm{p} K_{\mathrm{a}} \approx 36$ ) reacts with butyllithium to yield lithium diisopropylamide (LDA; Section 22.5). Dialkylamine anions like LDA are very strong bases that are often used in laboratory organic chemistry for the generation of enolate ions from carbonyl compounds (Section 22.7). They are not, however, encountered in biological chemistry.

PROBLEM Which compound in each of the following pairs is more basic?
24-4 (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}$ or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CONH}_{2}$
(b) NaOH or $\mathrm{CH}_{3} \mathrm{NH}_{2}$
(c) $\mathrm{CH}_{3} \mathrm{NHCH}_{3}$ or pyridine


Secondary alkylamine -
PROBLEM The benzylammonium ion $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{3}{ }^{+}\right)$has $\mathrm{p} K_{\mathrm{a}}=9.33$, and the propylammonium ion has $\mathrm{p} K_{\mathrm{a}}=$
24-5 10.71. Which is the stronger base, benzylamine or propylamine? What are the $\mathrm{p} K_{\mathrm{b}}$ 's of benzylamine and propylamine?


Secondary alkylamine - 24.4 Basicity of Arylamines
As noted previously, arylamines are generally less basic than alkylamines. Anilinium ion has $\mathrm{p} K_{\mathrm{a}}=4.63$, for instance, whereas methylammonium ion has $\mathrm{p} K_{\mathrm{a}}=10.64$. Arylamines are less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring's $\pi$ electron system and are less available for bonding to $\mathrm{H}^{+}$. In resonance terms, arylamines are stabilized relative to alkylamines
because of their five resonance forms.

Much of the resonance stabilization is lost on protonation, however, so the energy difference between protonated and nonprotonated forms is higher for arylamines than it is for alkylamines. As a result, arylamines are less basic. FIGURE 24.5 illustrates this difference.

Aniline (delocalized electrons)

Anilinium ion (localized charge)

FIGURE 24.5 Arylamines have a larger positive $\Delta G^{\circ}$ for protonation and are therefore less basic than alkylamines, primarily because of resonance stabilization of the ground state. Electrostatic potential maps show that lone-pair electron density is delocalized in the amine but the charge is localized in the corresponding ammonium ion.

Substituted arylamines can be either more basic or less basic than aniline, depending on the substituent. Electron-donating substituents, such as $-\mathrm{CH}_{3},-\mathrm{NH}_{2}$, and $-\mathrm{OCH}_{3}$, which increase the reactivity of an aromatic ring toward electrophilic substitution (Section 16.4), also increase the basicity of the corresponding arylamine. Electron-withdrawing substituents, such as $-\mathrm{Cl},-\mathrm{NO}_{2}$, and -CN , which decrease ring reactivity toward electrophilic substitution, also decrease arylamine basicity. TABLE 24.2 considers only $p$-substituted anilines, but similar trends are observed for ortho and meta derivatives.

TABLE 24.2 Base Strength of Some p-Substituted Anilines


Secondary alkylamine - 24.4 Basicity of Arylamines
TABLE 24.2 Base Strength of Some p-Substituted Anilines

|  | Substit | $\mathrm{p} K_{\mathrm{a}}$ |  |
| :---: | :---: | :---: | :---: |
| Stronger base | $-\mathrm{NH}_{2}$ | 6.15 | Activating groups |
|  | $-\mathrm{OCH}_{3}$ | 5.34 |  |
|  | $-\mathrm{CH}_{3}$ | 5.08 |  |
|  | -H | 4.63 |  |
| Weaker base | $-\mathrm{Cl}$ | 3.98 | Deactivating groups |
|  | $-\mathrm{Br}$ | 3.86 |  |
|  | -CN | 1.74 |  |
|  | $-\mathrm{NO}_{2}$ | 1.00 |  |

PROBLEM Without looking at Table 24.2, rank the following compounds in order of ascending basicity.
24-6 (a) $p$-Nitroaniline, $p$-aminobenzaldehyde, $p$-bromoaniline
(b) $p$-Chloroaniline, $p$-aminoacetophenone, $p$-methylaniline
(c) $p$-(Trifluoromethyl)aniline, $p$-methylaniline, $p$-(fluoromethyl)aniline


Secondary alkylamine - 24.5 Biological Amines and the Henderson-Hasselbalch Equation
We saw in Section 20.3 that the extent of dissociation of a carboxylic acid HA in an aqueous solution buffered to a given pH can be calculated with the Henderson-Hasselbalch equation. Furthermore, we concluded that at the physiological pH of 7.3 inside living cells, carboxylic acids are almost entirely dissociated into their carboxylate anions, $\mathrm{RCO}_{2}{ }^{-}$.


Henderson-Hasselbalch equation: -
$$
\mathrm{pH}=\mathrm{p} K \mathrm{a}+\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \text { so } \log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\mathrm{pH}-\mathrm{p} K \mathrm{a}
$$

What about amine bases? In what form do they exist at physiological pH ? As the amine $\left(\mathrm{A}^{-}=\mathrm{RNH}_{2}\right)$, or as the ammonium ion $\left(\mathrm{HA}=\mathrm{RNH}_{3}{ }^{+}\right)$? Let's take a 0.0010 M solution of methylamine at $\mathrm{pH}=7.3$, for example. According to TABLE 24.1, the $\mathrm{p} K_{\mathrm{a}}$ of methylammonium ion is 10.64 , so from the Henderson-Hasselbalch equation, we have

$$
\begin{aligned}
& \log \frac{\left[\mathrm{RNH}_{2}\right]}{\left[\mathrm{RNH}_{3}+\right]}=\mathrm{pH}-\mathrm{p} K_{\mathrm{a}}=7.3-10.64=-3.34 \\
& \frac{\left[\mathrm{RNH}_{2}\right]}{\left[\mathrm{RNH}_{3}{ }^{+}\right]}=\operatorname{antilog}(-3.34)=4.6 \times 10^{-4} \text { so }\left[\mathrm{RNH}_{2}\right]=\left(4.6 \times 10^{-4}\right)\left[\mathrm{RNH}_{3}^{+}\right]
\end{aligned}
$$

In addition, we know that

$$
\left[\mathrm{RNH}_{2}\right]+\left[\mathrm{RNH}_{3}{ }^{+}\right]=0.0010 \mathrm{M}
$$


Henderson-Hasselbalch equation: -
In addition, we know that

$$
\left[\mathrm{RNH}_{2}\right]+\left[\mathrm{RNH}_{3}{ }^{+}\right]=0.0010 \mathrm{M}
$$

Solving the two simultaneous equations gives $\left[\mathrm{RNH}_{3}{ }^{+}\right]=0.0010 \mathrm{M}$ and $\left[\mathrm{RNH}_{2}\right]=5 \times 10^{-7} \mathrm{M}$. In other words, at a physiological pH of 7.3 , essentially $100 \%$ of the methylamine in a 0.0010 M solution exists in its protonated form as methylammonium ion. The same is true of other amine bases, so we always write cellular amines in their protonated form and amino acids in their ammonium carboxylate form to reflect their structures at
physiological pH.

PROBLEM Calculate the percentages of neutral and protonated forms present in a solution of 0.0010 Molar 24-7 pyrimidine at $\mathrm{pH}=7.3$. The $\mathrm{p} K_{\mathrm{a}}$ of pyrimidinium ion is 1.3.


Reduction of Nitriles, Amides, and Nitro Compounds -
We've already seen in Section 20.7 and Section 21.7 how amines can be prepared by reduction of nitriles and amides with $\mathrm{LiAlH}_{4}$. The two-step sequence of $\mathrm{S}_{\mathrm{N}} 2$ displacement with $\mathrm{CN}^{-}$followed by reduction thus converts an alkyl halide into a primary alkylamine having an additional carbon atom. Amide reduction converts carboxylic acids and their derivatives into amines with the same number of carbon atoms.

Arylamines are usually prepared by nitration of an aromatic starting material, followed by reduction of the nitro group (Section 16.2). The reduction step can be carried out in many different ways, depending on the circumstances. Catalytic hydrogenation over platinum works well but is often incompatible with the presence elsewhere in the molecule of other reducible groups, such as $\mathrm{C}=\mathrm{C}$ bonds or carbonyl groups. Iron, zinc, tin, and tin(II) chloride $\left(\mathrm{SnCl}_{2}\right)$ are also effective when used in acidic aqueous solution. Tin(II) chloride is particularly mild and is often used when other reducible functional groups are present.

p-tert-Butylnitrobenzene

m-Nitrobenzaldehyde
p-tert-Butylaniline (100\%)

m-Aminobenzaldehyde (90\%)

PROBLEM Propose structures for either a nitrile or an amide that might be a precursor of each of the following 24-8 amines
(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$
(b) $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}\right)_{2} \mathrm{NH}$
(c) Benzylamine, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}$
(d) N -Ethylaniline


$\mathrm{S}_{\mathrm{N}} 2$ Reactions of Alkyl Halides -
Ammonia and other amines are good nucleophiles in $S_{N} 2$ reactions. As a result, the simplest method of alkylamine synthesis is by $\mathrm{S}_{\mathrm{N}} 2$ alkylation of ammonia or an alkylamine with an alkyl halide. If ammonia is used, a primary amine results; if a primary amine is used, a secondary amine results; and so on. Even tertiary amines react rapidly with alkyl halides to yield quaternary ammonium salts, $\mathrm{R}_{4} \mathrm{~N}^{+} \mathrm{X}^{-}$.

Unfortunately, these reactions don't stop cleanly after a single alkylation has occurred. Because ammonia and primary amines have similar reactivity, the initially formed monoalkylated substance often undergoes further reaction to yield a mixture of products. Even secondary and tertiary amines undergo further alkylation, although to a lesser extent. For example, treatment of 1-bromooctane with a twofold excess of ammonia leads to a mixture containing only $45 \%$ octylamine. A nearly equal amount of dioctylamine is produced by double alkylation, along with smaller amounts of trioctylamine and tetraoctylammonium bromide.

A better method for preparing primary amines is to use azide ion, $\mathrm{N}_{3}{ }^{-}$, rather than ammonia, as the nucleophile for $\mathrm{S}_{\mathrm{N}} 2$ reaction with a primary or secondary alkyl halide. The product is an alkyl azide, which is not nucleophilic, so overalkylation can't occur. Subsequent reduction of the alkyl azide with $\mathrm{LiAlH}_{4}$ then leads to the desired primary amine. Although this method works well, low-molecular-weight alkyl azides are explosive and must be handled carefully.


$\mathrm{S}_{\mathrm{N}} 2$ Reactions of Alkyl Halides -
Another alternative for preparing a primary amine from an alkyl halide is the Gabriel amine synthesis, which uses a phthalimide alkylation. An imide $(\mathbf{O}=\mathbf{C}-\mathbf{N}-\mathbf{C}=\mathbf{O})$ is similar to a $\beta$-keto ester in that the acidic $\mathrm{N}-\mathrm{H}$ hydrogen is flanked by two carbonyl groups. Thus, imides are deprotonated by such bases as KOH, and the resultant anions are readily alkylated in a reaction similar to acetoacetic ester synthesis (Section 22.7). Basic hydrolysis of the $N$-alkylated imide then yields a primary amine product. The imide hydrolysis step is analogous to the hydrolysis of an amide (Section 21.7).

PROBLEM Write the mechanism of the last step in Gabriel amine synthesis, the base-promoted hydrolysis of a 24-9 phthalimide to yield an amine plus phthalate ion.

PROBLEM Show two methods for the synthesis of dopamine, a neurotransmitter involved in regulation of the
24-10 central nervous system. Use any alkyl halide needed.


Reductive Amination of Aldehydes and Ketones -
Amines can be synthesized in a single step by treatment of an aldehyde or ketone with ammonia or an amine in the presence of a reducing agent, a process called reductive amination. For example, amphetamine, a central nervous system stimulant, is prepared commercially by reductive amination of phenyl-2-propanone with ammonia using hydrogen gas over a nickel catalyst as the reducing agent. In the laboratory, either $\mathrm{NaBH}_{4}$ or the related $\mathrm{NaBH}(\mathrm{OAc})_{3}$ is commonly used ( $\mathrm{OAc}=$ acetate ).

Reductive amination takes place by the pathway shown in FIGURE 24.6. An imine intermediate is first formed by a nucleophilic addition reaction (Section 19.8), and the $\mathrm{C}=\mathrm{N}$ bond of the imine is then reduced to the amine, much as the $\mathrm{C}=\mathrm{O}$ bond of a ketone can be reduced to an alcohol.


FIGURE 24.6 MECHANISM -
Mechanism for reductive amination of a ketone to yield an amine. Details of the imine-forming step are shown in FIGURE 19.7.
(1) Ammonia adds to the ketone carbonyl group in a nucleophilic addition reaction to yield an intermediate carbinolamine.

2 The carbinolamine loses water to give an imine.
(3) The imine is reduced by $\mathrm{NaBH}_{4}$ or $\mathrm{H}_{2} /$ Ni to yield the amine product.

(1)

(2.

(3) $\mathrm{NaBH}_{4}$ or $\mathrm{H}_{2} / \mathrm{Ni}$

Ammonia, primary amines, and secondary amines can all be used in the reductive amination reaction, yielding primary, secondary, and tertiary amines, respectively.

Reductive aminations also occur in various biological pathways. In the biosynthesis of the amino acid proline, for instance, glutamate 5 -semialdehyde undergoes internal imine formation to give 1-pyrrolinium 5 -carboxylate, which is then reduced by nucleophilic addition of hydride ion to the $\mathrm{C}=\mathrm{N}$ bond. Reduced nicotinamide adenine dinucleotide, NADH, acts as the biological reducing agent.


Using a Reductive Amination Reaction -
How might you prepare $N$-methyl-2-phenylethylamine using a reductive amination reaction?


Strategy -
Look at the target molecule, and identify the groups attached to nitrogen. One of the groups must be derived from the aldehyde or ketone component, and the other must be derived from the amine component. In the case of N -methyl-2-phenylethylamine, two combinations can lead to the product: phenylacetaldehyde plus methylamine or formaldehyde plus 2-phenylethylamine. It's usually better to choose the combination with the simpler amine component-methylamine in this case-and to use an excess of that amine as reactant.

Solution

PROBLEM How might the following amines be prepared using reductive amination reactions? Show all 24-11 precursors if more than one is possible.
(a)

(b)

(c)

PROBLEM How could you prepare the following amine using a reductive amination reaction?
24-12


Hofmann and Curtius Rearrangements -
Carboxylic acid derivatives can be converted into primary amines with loss of one carbon atom by both the Hofmann rearrangement and the Curtius rearrangement. Although the Hofmann rearrangement involves a primary amide and the Curtius rearrangement involves an acyl azide, both proceed through similar mechanisms.
Hofmann
rearrangement

An amide


An acyl azide -
Hofmann rearrangement occurs when a primary amide, $\mathrm{RCONH}_{2}$, is treated with $\mathrm{Br}_{2}$ and base (FIGURE 24.7). The overall mechanism is lengthy, but most of the individual steps have been encountered before. Thus, the bromination of an amide in steps 1 and 2 is analogous to the base-promoted bromination of a ketone enolate ion (Section 22.6), and the rearrangement of the bromoamide anion in step 4 is analogous to a carbocation rearrangement (Section 7.11). Nucleophilic addition of water to the isocyanate carbonyl group in step 5 is a typical carbonyl-group process (Section 19.4), as is the final decarboxylation step 6 (Section 22.7).


FIGURE 24.7 MECHANISM -
Mechanism of the Hofmann rearrangement of an amide to an amine. Each step is analogous to a reaction studied previously.

Amide

© || proton, yielding an amide anion.
(2) The anion reacts with bromine in an $\alpha$-substitution reaction to give an N -bromoamide.

Bromoamide

(3) $\downarrow$

(4)

(5)

Carbamic acid
(6) The carbamic acid
spontaneously loses $\mathrm{CO}_{2}$ to give an amine.


(6) -
alkylamines. For example, the appetite-suppressant drug phentermine is prepared commercially by Hofmann rearrangement of a primary amide. Commonly known by the name Fen-Phen, the combination of phentermine with another appetite-suppressant, fenfluramine, is suspected of causing heart damage.

2,2-Dimethyl-3-phenyl-
Phentermine propanamide
The Curtius rearrangement, like the Hofmann rearrangement, involves migration of an -R group from the $\mathrm{C}=\mathrm{O}$ carbon atom to the neighboring nitrogen with simultaneous loss of a leaving group. The reaction takes place on heating an acyl azide that is itself prepared by nucleophilic acyl substitution of an acid chloride.

Also like the Hofmann rearrangement, the Curtius rearrangement is often used commercially. The antidepressant drug tranylcypromine, for instance, is made by Curtius rearrangement of 2-phenylcyclopropanecarbonyl chloride.

trans-2-Phenylcyclo-
Tranylcypromine propanecarbonyl chloride


Using the Hofmann and Curtius Rearrangements -
How would you prepare o-methylbenzylamine from a carboxylic acid, using both Hofmann and Curtius rearrangements?


Strategy -
Both Hofmann and Curtius rearrangements convert a carboxylic acid derivative-either an amide (Hofmann) or an acid chloride (Curtius)-into a primary amine with loss of one carbon, RCOY $\rightarrow \mathrm{RNH}_{2}$. Both reactions begin with the same carboxylic acid, which can be identified by replacing the $-\mathrm{NH}_{2}$ group of the amine product by a $-\mathrm{CO}_{2} \mathrm{H}$ group. In the present instance, $o$-methylphenylacetic acid is needed.


Solution -
PROBLEM How would you prepare the following amines, using both Hofmann and Curtius rearrangements on
24-13 a carboxylic acid derivative?
(a)

(b)


Alkylation and Acylation -
We've already studied the two most general reactions of amines-alkylation and acylation. As we saw earlier in this chapter, primary, secondary, and tertiary amines can be alkylated by reaction with a primary alkyl halide. Alkylations of primary and secondary amines are difficult to control and often give mixtures of products, but tertiary amines are cleanly alkylated to give quaternary ammonium salts. Primary and secondary (but not tertiary) amines can also be acylated by nucleophilic acyl substitution reaction with an acid chloride or an acid anhydride to yield an amide (Section 21.4 and Section 21.5). Note that overacylation of the nitrogen does not occur because the amide product is much less nucleophilic and less reactive than the starting amine.


Hofmann Elimination Reaction -
Like alcohols, amines can be converted into alkenes by an elimination reaction. But because an amide ion, $\mathrm{NH}_{2}{ }^{-}$, is such a poor leaving group, it must first be converted into a better leaving group. In the Hofmann elimination reaction, an amine is completely methylated by reaction with an excess amount of iodomethane to produce the corresponding quaternary ammonium salt. This salt then undergoes elimination to give an alkene on heating with a base, typically silver oxide, $\mathrm{Ag}_{2} \mathrm{O}$. For example, 1-methylpentylamine is converted into 1-hexene.

1-Methylpentylamine
(1-Methylpentyl)trimethyl-
1-Hexene (60\%) ammonium iodide
Silver oxide acts by exchanging iodide ion for hydroxide ion in the quaternary salt, thus providing the base necessary for elimination. The actual elimination step is an E2 reaction (Section 11.8) in which hydroxide ion removes a proton while the positively charged nitrogen atom leaves.

Unlike what happens in other E2 reactions, the major product of Hofmann elimination is the less highly substituted alkene rather than the more highly substituted one, as shown by the reaction of (1-methylbutyl)trimethylammonium hydroxide to give 1-pentene rather than the alternative 2 -pentene. The reason for this non-Zaitsev result is probably steric. Because of the large size of the trialkylamine leaving group, the base must abstract a hydrogen from the more accessible, least hindered position.

The Hofmann elimination reaction is not often currently used in the laboratory, but analogous biological eliminations occur frequently, although usually with protonated ammonium ions rather than quaternary ammonium salts. In the biosynthesis of nucleic acids, for instance, a substance called adenylosuccinate undergoes an elimination of a positively charged nitrogen to give fumarate plus adenosine monophosphate.


Predicting the Product of a Hofmann Elimination -
What product would you expect from Hofmann elimination of the following amine?


Strategy -
The Hofmann elimination is an E2 reaction that converts an amine into an alkene and occurs with non-Zaitsev regiochemistry to form the less highly substituted double bond. To predict the product, look at the reactant and identify the positions from which elimination might occur (the positions two carbons away from nitrogen). Then, carry out an elimination using the most accessible hydrogen. In the present instance, there are three possible positions from which elimination might occur-one primary, one secondary, and one tertiary. The primary position is the most accessible and leads to the least highly substituted alkene, ethylene.

Solution

PROBLEM What products would you expect from Hofmann elimination of the following amines? If more than 24-14 one product is formed, indicate which is major.
(a)

(b)

(c)

(d)

PROBLEM What product would you expect from Hofmann elimination of a heterocyclic amine such as 24-15 piperidine? Write all the steps.


Electrophilic Aromatic Substitutions -
An amino group is strongly activating and ortho- and para-directing in electrophilic aromatic substitution reactions (Section 16.4). This high reactivity of amino-substituted benzenes can be a drawback at times because it's often difficult to prevent polysubstitution. Reaction of aniline with $\mathrm{Br}_{2}$, for instance, takes place rapidly and yields the 2,4,6-tribrominated product. The amino group is so strongly activating that it's not possible to stop at the monobromo stage.


Aniline $\quad$ 2,4,6-Tribromoaniline <br> (100\%) -
Another drawback to the use of amino-substituted benzenes in electrophilic aromatic substitution reactions is that Friedel-Crafts reactions are not successful (Section 16.3). The amino group forms an acid-base complex with the $\mathrm{AlCl}_{3}$ catalyst, which prevents further reaction. Both drawbacks can be overcome, however, by carrying out electrophilic aromatic substitution reactions on the corresponding amide rather than on the free amine.

As we saw in Section 21.5, treatment of an amine with acetic anhydride yields the corresponding acetyl amide, or acetamide. Although still activating and ortho-, para-directing, amido substituents ( $\mathrm{R}-\mathrm{N}-\mathrm{C}=\mathrm{O}$ ) are less strongly activating and less basic than amino groups because their nitrogen lone-pair electrons are delocalized by the neighboring carbonyl group. As a result, bromination of an $N$-arylamide occurs cleanly to give a monobromo product, and hydrolysis of the amide with aqueous base then gives the free amine. For example, p-toluidine (4-methylaniline) can be acetylated, brominated, and hydrolyzed to yield 2-bromo-4-methylaniline. None of the 2,6-dibrominated product is obtained.

Friedel-Crafts alkylations and acylations of $N$-arylamides also proceed normally. For example, benzoylation of acetanilide ( $N$-acetylaniline) under Friedel-Crafts conditions gives 4-aminobenzophenone in $80 \%$ yield after hydrolysis.

Moderating the reactivity of an amino-substituted benzene by forming an amide is a useful trick that allows many kinds of electrophilic aromatic substitutions to be carried out that would otherwise be impossible. One example is the preparation of the sulfa drugs, such as sulfanilamide.


Aniline $\quad$ 2,4,6-Tribromoaniline <br> (100\%) -
Sulfa drugs were among the first pharmaceutical agents to be used clinically against bacterial infection. Although they have largely been replaced today by safer and more powerful antibiotics, sulfa drugs are credited with saving the lives of thousands of wounded during World War II and are still prescribed for urinary tract infections. They are prepared by chlorosulfonation of acetanilide, followed by reaction of
$p$-( $N$-acetylamino)benzenesulfonyl chloride with ammonia or some other amine to give a sulfonamide. Hydrolysis of the amide then yields the sulfa drug. Note that hydrolysis of the amide can be carried out in the presence of the sulfonamide group because sulfonamides hydrolyze very slowly.

PROBLEM Propose a synthesis of the drug sulfathiazole from benzene and any necessary amine.
24-16


Sulfathiazole -
PROBLEM Propose syntheses of the following compounds from benzene:
24-17 (a) $N, N$-Dimethylaniline
(b) p-Chloroaniline
(c) m-Chloroaniline
(d) 2,4-Dimethylaniline


Diazonium Salts: The Sandmeyer Reaction -
Primary arylamines react with nitrous acid, $\mathrm{HNO}_{2}$, to yield stable arenediazonium salts, $\mathrm{Ar}-\stackrel{+}{\mathrm{N}} \equiv \mathrm{N} \mathrm{X}^{-}$, a process called a diazotization reaction. Alkylamines also react with nitrous acid, but the corresponding alkanediazonium products are so reactive they can't be isolated. Instead, they lose nitrogen instantly to yield carbocations. The analogous loss of $\mathrm{N}_{2}$ from an arenediazonium ion to yield an aryl cation is disfavored by the instability of the cation.

Arenediazonium salts are useful because the diazonio group ( $-\mathrm{N} \equiv \mathrm{N}$ ) can be replaced by a nucleophile in a substitution reaction.

Many different nucleophiles-halide, hydride, cyanide, and hydroxide among others-react with arenediazonium salts, yielding many different kinds of substituted benzenes. The overall sequence of (1) nitration, (2) reduction, (3) diazotization, and (4) nucleophilic substitution is perhaps the single most versatile method of aromatic substitution.

Aryl chlorides and bromides are prepared by reaction of an arenediazonium salt with the corresponding copper(I) halide, CuX , a process called the Sandmeyer reaction. Aryl iodides can be prepared by direct reaction
with NaI without using a copper(I) salt. Yields generally fall between $60 \%$ and $80 \%$.

Aniline
Iodobenzene (67\%)
Similar treatment of an arenediazonium salt with CuCN yields the nitrile ArCN, which can then be further converted into other functional groups such as carboxyl. For example, Sandmeyer reaction of $o$-methylbenzenediazonium bisulfate with CuCN yields o-methylbenzonitrile, which can be hydrolyzed to give $o$-methylbenzoic acid. This product can't be prepared from $o$-xylene by the usual side-chain oxidation route because both methyl groups would be oxidized.


Diazonium Salts: The Sandmeyer Reaction -
The diazonio group can also be replaced by -OH to yield a phenol and by -H to yield an arene. A phenol is prepared by reaction of the arenediazonium salt with copper(I) oxide in an aqueous solution of copper(II) nitrate, a reaction that is especially useful because few other general methods exist for introducing an -OH group onto an aromatic ring.

$\underset{\text { ( } \boldsymbol{p} \text {-Toluidine) }}{\boldsymbol{p} \text {-Methylaniline }}$
p-Cresol
(93\%)
Reduction of a diazonium salt to give an arene occurs on treatment with hypophosphorous acid, $\mathrm{H}_{3} \mathrm{PO}_{2}$. This reaction is used primarily when there is a need for temporarily introducing an amino substituent onto a ring to take advantage of its directing effect. Suppose, for instance, that you needed to make 3,5-dibromotoluene. This product can't be made by direct bromination of toluene because reaction would occur at positions 2 and 4. Starting with $p$-methylaniline ( $p$-toluidine), however, dibromination occurs ortho to the strongly directing amino substituent, and diazotization followed by treatment with $\mathrm{H}_{3} \mathrm{PO}_{2}$ to remove the amino group yields the desired product.

Mechanistically, these diazonio replacement reactions occur through radical rather than polar pathways. In the presence of a copper(I) compound, for instance, it's thought that the arenediazonium ion is first converted to an aryl radical plus copper(II), followed by subsequent reaction to give product plus regenerated copper(I) catalyst.


Using Diazonium Replacement Reactions -
How would you prepare $m$-hydroxyacetophenone from benzene, using a diazonium replacement reaction in your scheme?

m-Hydroxyacetophenone


Strategy -
As always, organic syntheses are planned by working retrosynthetically from the final product, one step at a time. First, identify the functional groups in the product and recall how those groups can be synthesized. $m$-Hydroxyacetophenone has an -OH group and $\mathrm{a}-\mathrm{COCH}_{3}$ group in a meta relationship on a benzene ring. A hydroxyl group is generally introduced onto an aromatic ring by a four-step sequence of nitration, reduction, diazotization, and diazonio replacement. An acetyl group is introduced by a Friedel-Crafts acylation reaction.

Next, ask yourself what an immediate precursor of the target might be. Since an acetyl group is a meta director while a hydroxyl group is an ortho and para director, acetophenone might be a precursor of $m$-hydroxyacetophenone. Benzene, in turn, is a precursor of acetophenone.


Solution -
PROBLEM How would you prepare the following compounds from benzene, using a diazonium replacement 24-18 reaction in your scheme?
(a) $p$-Bromobenzoic acid
(b) $m$-Bromobenzoic acid
(c) m-Bromochlorobenzene
(d) $p$-Methylbenzoic acid
(e) 1,2,4-Tribromobenzene


Diazonium Coupling Reactions -
Arenediazonium salts undergo a coupling reaction with activated aromatic rings such as phenols and arylamines to yield brightly colored azo compounds, $\mathbf{A r}-\mathbf{N}=\mathbf{N}-\mathbf{A r}^{\prime}$.

An azo compound

$$
\text { where } \mathrm{Y}=-\mathrm{OH} \text { or }-\mathrm{NR}_{2}
$$

Diazonium coupling reactions are typical electrophilic aromatic substitutions in which the positively charged diazonium ion is the electrophile that reacts with the electron-rich ring of a phenol or arylamine. Reaction usually occurs at the para position.

Azo-coupled products are widely used as dyes for textiles because their extended conjugated $\pi$ electron system causes them to absorb in the visible region of the electromagnetic spectrum (Section 14.9). p-(Dimethylamino)azobenzene, for instance, is a bright yellow compound that was at one time used as a coloring agent in margarine.

PROBLEM Propose a synthesis of $p$-(dimethylamino)azobenzene with benzene as your only organic starting 24-19 material.


Diazonium Coupling Reactions - 24.9 Heterocyclic Amines
As noted in Section 15.5 in connection with a discussion of aromaticity, a cyclic organic compound that contains atoms of two or more elements in its ring is a called a heterocycle. Heterocyclic amines are particularly common, and many have important biological properties. Pyridoxal phosphate, a coenzyme; sildenafil (Viagra), a well-known pharmaceutical; and heme, the oxygen carrier in blood, are a few examples.

Pyridoxal phosphate (a coenzyme)

Sildenafil
(Viagra)

Heme

Most heterocycles have the same chemistry as their open-chain counterparts. Lactones and acyclic esters behave similarly, lactams and acyclic amides behave similarly, and cyclic and acyclic ethers behave similarly. In certain cases, however, particularly when the ring is unsaturated, heterocycles have unique and interesting properties.


Pyrrole and Imidazole -
Pyrrole, the simplest five-membered unsaturated heterocyclic amine, is obtained commercially by treatment of furan with ammonia over an alumina catalyst at $400^{\circ} \mathrm{C}$. Furan, the oxygen-containing analog of pyrrole, is obtained by acid-catalyzed dehydration of the five-carbon sugars found in oat hulls and corncobs.

Although pyrrole appears to be both an amine and a conjugated diene, its chemical properties are not consistent with either of these structural features. Unlike most other amines, pyrrole is not basic-the $\mathrm{p} K_{\mathrm{a}}$ of the pyrrolinium ion is 0.4 ; unlike most other conjugated dienes, pyrrole undergoes electrophilic substitution reactions rather than additions. The reason for both of these properties, as noted in Section 15.5, is that pyrrole has six $\pi$ electrons and is aromatic. Each of the four carbons contributes one $\pi$ electron, and the $s p^{2}$-hybridized nitrogen contributes two more from its lone pair.

Pyrrole


Six $\pi$ electrons -
Because the nitrogen lone pair is a part of the aromatic sextet, protonation on nitrogen would destroy the aromaticity of the ring. The nitrogen atom in pyrrole is therefore less electron-rich, less basic, and less nucleophilic than the nitrogen in an aliphatic amine. By the same token, the carbon atoms of pyrrole are more electron-rich and more nucleophilic than typical double-bond carbons. The pyrrole ring is therefore reactive toward electrophiles in the same way as enamines (Section 23.11). Electrostatic potential maps show how the pyrrole nitrogen is electron-poor (less red) compared with the nitrogen in its saturated counterpart pyrrolidine, while the pyrrole carbon atoms are electron-rich (more red) compared with the carbons in 1,3-cyclopentadiene.

The chemistry of pyrrole is similar to that of activated benzene rings. In general, however, the heterocycles are more reactive toward electrophiles than benzene rings, and low temperatures are often necessary to control the reactions. Halogenation, nitration, sulfonation, and Friedel-Crafts acylation can all be accomplished. For example:

Electrophilic substitutions normally occur at C2, the position next to the nitrogen, because reaction at this position leads to a more stable intermediate cation having three resonance forms, whereas reaction at C3 gives a less stable cation with only two resonance forms (FIGURE 24.8).

3-Nitropyrrole
(Not formed)
FIGURE 24.8 Electrophilic nitration of pyrrole. The intermediate produced by reaction at C 2 is more stable than that produced by reaction at C3.

Other common five-membered heterocyclic amines include imidazole and thiazole. Imidazole, a constituent of the amino acid histidine, has two nitrogens, only one of which is basic. Thiazole, the five-membered ring system on which the structure of thiamin (vitamin $B_{1}$ ) is based, also contains a basic nitrogen that is alkylated in thiamin to form a quaternary ammonium ion.

Imidazole

Thiazole

Histidine

Thiamin (vitamin $\mathrm{B}_{1}$ )


Six $\pi$ electrons -
Imidazole

Thiazole

Histidine

Thiamin (vitamin $\mathrm{B}_{1}$ )

PROBLEM Draw an orbital picture of thiazole. Assume that both the nitrogen and sulfur atoms are 24-20 $s p^{2}$-hybridized, and show the orbitals that the lone pairs occupy.

PROBLEM What is the percent protonation of the imidazole nitrogen atom in histidine at a physiological pH of 24-21 7.3 (Section 24.5)?


Pyridine and Pyrimidine -
Pyridine is the nitrogen-containing heterocyclic analog of benzene. Like benzene, pyridine is a flat, aromatic molecule, with bond angles of $120^{\circ}$ and $\mathrm{C}-\mathrm{C}$ bond lengths of 139 pm , intermediate between typical single and double bonds. The five carbon atoms and the $s p^{2}$-hybridized nitrogen atom each contribute one $\pi$ electron to the aromatic sextet, and the lone-pair electrons occupy an $s p^{2}$ orbital in the plane of the ring (Section 15.5).

As shown previously in TABLE 24.1, pyridine ( $\mathrm{p} K_{\mathrm{a}}=5.25$ ) is a stronger base than pyrrole but a weaker base than the alkylamines. The diminished basicity of pyridine compared with that of alkylamines is due to the fact that the lone-pair electrons on the pyridine nitrogen are in an $s p^{2}$ orbital, while those on an alkylamine nitrogen are in an $s p^{3}$ orbital. Because $s$ orbitals have their maximum electron density at the nucleus but $p$ orbitals have a node at the nucleus, electrons in an orbital with more $s$ character are held more closely to the positively charged nucleus and are less available for bonding. As a result, the $s p^{2}$-hybridized nitrogen atom ( $33 \% s$ character) in
pyridine is less basic than the $s p^{3}$-hybridized nitrogen in an alkylamine ( $25 \% s$ character).

Pyridine
Unlike benzene, pyridine undergoes electrophilic aromatic substitution reactions with difficulty. Halogenation can be carried out under drastic conditions, but nitration occurs in very low yield, and Friedel-Crafts reactions are not successful. Reactions usually give the 3 -substituted product.


Pyridine and Pyrimidine -
The low reactivity of pyridine toward electrophilic aromatic substitution is caused by a combination of factors. One is that acid-base complexation between the basic ring's nitrogen atom and the incoming electrophile places a positive charge on the ring, thereby deactivating it. Equally important is that the electron density of the ring is decreased by the electron-withdrawing inductive effect of the electronegative nitrogen atom. Thus, pyridine has a substantial dipole moment ( $\mu=2.26 \mathrm{D}$ ), with the ring carbons acting as the positive end of the dipole. Reaction of an electrophile with the positively polarized carbon atoms is therefore difficult.

In addition to pyridine, the six-membered diamine pyrimidine is also found commonly in biological molecules, particularly as a constituent of nucleic acids. With a $\mathrm{p} K_{\mathrm{a}}$ of 1.3 , pyrimidine is substantially less basic than pyridine because of the inductive effect of the second nitrogen.

PROBLEM Electrophilic aromatic substitution reactions of pyridine normally occur at C3. Draw the 24-22 carbocation intermediates resulting from reaction of an electrophile at C2, C3, and C4, and explain the observed result.


Polycyclic Heterocycles -
As we saw in Section 15.6, quinoline, isoquinoline, indole, and purine are common polycyclic heterocycles. The first three contain both a benzene ring and a heterocyclic aromatic ring, while purine contains two heterocyclic rings joined together. All four ring systems occur commonly in nature, and many compounds with these rings have pronounced physiological activity. The quinoline alkaloid quinine, for instance, is widely used as an antimalarial drug; tryptophan is a common amino acid; and the purine adenine is a constituent of nucleic acids.

Quinine (antimalarial)

Tryptophan
(amino acid)

Adenine (DNA constituent)

The chemistry of these polycyclic heterocycles is just what you might expect from a knowledge of the simpler heterocycles pyridine and pyrrole. Quinoline and isoquinoline both have basic, pyridine-like nitrogen atoms, and both undergo electrophilic substitutions. As with pyridine, both quinoline and isoquinoline are less reactive toward electrophilic substitution than benzene because of the electronegative nitrogen atom that withdraws electrons from the ring. Reaction occurs on the benzene ring rather than on the nitrogen-containing pyridine ring, and a mixture of substitution products is obtained.

Indole has a nonbasic, pyrrole-like nitrogen and undergoes electrophilic substitution more easily than benzene. Substitution occurs at C3 of the electron-rich pyrrole ring rather than on the benzene ring.

Purine has three basic, pyridine-like nitrogens with lone-pair electrons in $s p^{2}$ orbitals in the plane of the ring. The remaining purine nitrogen is nonbasic and pyrrole-like, with its lone-pair electrons as part of the aromatic
$\pi$ electron system.

Purine
PROBLEM Which nitrogen atom in the hallucinogenic indole alkaloid $N, N$-dimethyltryptamine is more basic? 24-23 Explain.


N,N-Dimethyltryptamine -
PROBLEM Indole reacts with electrophiles at C3 rather than at C2. Draw resonance forms of the intermediate 24-24 cations resulting from reaction at C2 and C3, and explain the observed results.


Infrared Spectroscopy -
Primary and secondary amines can be identified by a characteristic N-H stretching absorption in the 3300 to $3500 \mathrm{~cm}^{-1}$ range of the IR spectrum. Alcohols also absorb in this range (Section 17.11), but amine absorption bands are generally sharper and less intense than hydroxyl bands. Primary amines show a pair of bands at about 3350 and $3450 \mathrm{~cm}^{-1}$ from the symmetric and asymmetric stretching modes, and secondary amines show a single band at $3350 \mathrm{~cm}^{-1}$. Tertiary amines have no absorption in this region because they have no $\mathrm{N}-\mathrm{H}$ bonds. FIGURE 24.9 recalls the IR spectrum of cyclohexylamine from Section 12.8.

FIGURE 24.9 IR spectrum of cyclohexylamine.


Nuclear Magnetic Resonance Spectroscopy -
Amines are difficult to identify solely by ${ }^{1} \mathrm{H}$ NMR spectroscopy because $\mathrm{N}-\mathrm{H}$ hydrogens tend to appear as broad signals without clear-cut coupling to neighboring $\mathrm{C}-\mathrm{H}$ hydrogens. As with $\mathrm{O}-\mathrm{H}$ absorptions (Section 17.11), amine $\mathrm{N}-\mathrm{H}$ absorptions can appear over a wide range and are best identified by adding a small amount of $\mathrm{D}_{2} \mathrm{O}$ to the sample. Exchange of $\mathrm{N}-\mathrm{D}$ for $\mathrm{N}-\mathrm{H}$ occurs, and the $\mathrm{N}-\mathrm{H}$ signal disappears from the NMR spectrum.

Hydrogens on the carbon next to nitrogen are deshielded because of the electron-withdrawing effect of the nitrogen, and they therefore absorb further downfield than alkane hydrogens. $N$-Methyl groups are particularly distinctive because they absorb as a sharp three-proton singlet at 2.2 to $2.6 \delta$. This $N$-methyl resonance at 2.42 $\delta$ is easily seen in the ${ }^{1} \mathrm{H}$ NMR spectrum of $N$-methylcyclohexylamine (FIGURE 24.10).

FIGURE 24.10 Proton NMR spectrum of $\boldsymbol{N}$-methylcyclohexylamine.
Carbons next to amine nitrogens are slightly deshielded in the ${ }^{13} \mathrm{C}$ NMR spectrum and absorb about 20 ppm downfield from where they would absorb in an alkane of similar structure. In $N$-methylcyclohexylamine, for example, the ring carbon to which nitrogen is attached absorbs at a position 24 ppm downfield from any other ring carbon.


Nuclear Magnetic Resonance Spectroscopy -
PROBLEM Compound $\mathbf{A}, \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}$, has an IR absorption at $1715 \mathrm{~cm}^{-1}$ and gives compound $\mathbf{B}, \mathrm{C}_{6} \mathrm{H}_{15} \mathrm{~N}$, when 24-25 treated with ammonia and $\mathrm{NaBH}_{4}$. The IR and ${ }^{1} \mathrm{H}$ NMR spectra of $\mathbf{B}$ are shown. What are the structures of $\mathbf{A}$ and $\mathbf{B}$ ?


Mass Spectrometry -
The nitrogen rule of mass spectrometry says that a compound with an odd number of nitrogen atoms has an odd-numbered molecular weight. Thus, the presence of nitrogen in a molecule is detected simply by observing its mass spectrum. An odd-numbered molecular ion usually means that the unknown compound has one or three nitrogen atoms, and an even-numbered molecular ion usually means that a compound has either zero or two nitrogen atoms. The logic behind the rule derives from the fact that nitrogen is trivalent, thus requiring an odd number of hydrogen atoms. For example, morphine has the formula $\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3}$ and a molecular weight of 285 amu .

Alkylamines undergo a characteristic $\alpha$ cleavage in the mass spectrometer, similar to the cleavage observed for alcohols (Section 17.11). $\mathrm{C}-\mathrm{C}$ bond nearest the nitrogen atom is broken, yielding an alkyl radical and a resonance-stabilized, nitrogen-containing cation.

As an example, the mass spectrum of $N$-ethylpropylamine shown in FIGURE 24.11 has peaks at $m / z=58$ and $m / z=72$, corresponding to the two possible modes of $\alpha$ cleavage.

FIGURE 24.11 Mass spectrum of $N$-ethylpropylamine. The two possible modes of $\alpha$ cleavage lead to the observed fragment ions at $m / z=$ 58 and $m / z=72$.


Green Chemistry II: Ionic Liquids -
Liquids made of ions? Usually when we think of ionic compounds, we think of high-melting solids: sodium chloride, magnesium sulfate, lithium carbonate, and so forth. But yes, there are also ionic compounds that are liquid at room temperature, and they are gaining importance as reaction solvents, particularly for use in green chemistry processes (see the Chapter 11 Chemistry Matters). More than 1500 ionic liquids are known, and about 500 are available commercially.

Ionic liquids have been studied for nearly a century; the first to be discovered was ethylammonium nitrate, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}{ }^{+} \mathrm{NO}_{3}{ }^{-}$, with a melting point of $12{ }^{\circ} \mathrm{C}$. More generally, however, the ionic liquids in use today are salts in which the cation is unsymmetrical and in which one or both of the ions are bulky so that the charges are dispersed over a large volume. Both factors minimize the crystal lattice energy and disfavor formation of the solid. Typical cations are quaternary ammonium ions from heterocyclic amines, either 1,3-dialkylimidazolium ions, $N$-alkylpyridinium ions, or ring-substituted $N$-alkylpyridinium ions.

$$
\left[\begin{array}{rl}
\mathrm{R}= & -\mathrm{CH}_{3},-\mathrm{CH}_{2} \mathrm{CH}_{3},-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}, \\
& -\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}
\end{array}\right]
$$

1,3-Dialkylimidazolium ions


Green Chemistry II: Ionic Liquids -
1,3-Dialkylimidazolium ions

$\left[\begin{array}{c}\mathrm{R}=- \\ -\mathrm{CH}_{2} \mathrm{CH}_{3},-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}, \\ \end{array}\right]$
$\boldsymbol{N}$-Alkylpyridinium ions

Anions are just as varied as the cations. Hexafluorophosphate, tetrafluoroborate, alkyl sulfates, trifluoromethanesulfonates (triflates), and halides are some anion possibilities.

Hexafluorophosphate

Tetrafluoroborate

Methyl
sulfate

$$
\mathrm{Cl}^{-}, \mathrm{Br}^{-}, \mathrm{I}^{-}
$$

Trifluoromethane- Halide sulfonate

Ionic liquids have several important features that make them attractive for use, both as solvents in green chemistry and as specialty chemicals in such applications as paint additives and refrigerants:

- They dissolve both polar and nonpolar organic compounds, giving high solute concentrations and thereby minimizing the amount of solvent needed.
- They can be optimized for specific reactions by varying cation and anion structures.
- They are nonflammable.
- They are thermally stable.
- They have negligible vapor pressures and do not evaporate.
- They are generally recoverable and can be reused many times.

As an example of their use in organic chemistry, the analgesic drug pravadoline has been synthesized in two steps using 1-butyl-3-methylimidazolium hexafluorophosphate, abbreviated bmim, as the solvent for both steps. The first step is a base-induced $\mathrm{S}_{\mathrm{N}} 2$ reaction of 2-methylindole with a primary alkyl halide, and the second is a Friedel-Crafts acylation. Both steps take place with $95 \%$ yield, and the ionic solvent is recovered simply by washing the reaction mixture, first with toluene and then with water.


Green Chemistry II: Ionic Liquids -
The first commercial process using an ionic liquid catalyst was introduced by PetroChina in 2008, when they opened a plant producing 65,000 tons per year of alkylate gasoline from isobutane. The aluminum-based ionic liquid catalyst replaced the sulfuric acid and hydrofluoric acid catalysts that had previously been used.


Key Terms -
- alkylamine
- amine
- arylamine
- azo compound ( $\mathrm{R}-\mathrm{N}=\mathrm{N}-\mathrm{R}^{\prime}$ )
- Curtius rearrangement
- Gabriel amine synthesis
- heterocyclic amine
- Hofmann elimination reaction
- Hofmann rearrangement
- imide $(\mathrm{O}=\mathrm{C}-\mathrm{N}-\mathrm{C}=\mathrm{O})$
- primary amine $\left(\mathrm{RNH}_{2}\right)$
- quaternary ammonium salt
- reductive amination
- Sandmeyer reaction
- secondary amine ( $\mathrm{R}_{2} \mathrm{NH}$ )
- tertiary amine ( $\mathrm{R}_{3} \mathrm{~N}$ )


Summary -
We've now seen all the common functional groups that occur in organic and biological chemistry. Of those groups, amines are among the most abundant and have among 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 reactions are amines.
Amines are organic derivatives of ammonia. They are named in the IUPAC system either by adding the suffix -amine to the name of the alkyl substituent or by considering the amino group as a substituent on a more complex parent molecule.
The chemistry of amines is dominated by the lone-pair electrons on nitrogen, which makes amines both basic and nucleophilic. The basicity of arylamines is generally lower than that of alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic $\pi$ system. Electron-withdrawing substituents on the aromatic ring further weaken the basicity of a substituted aniline, while electron-donating substituents increase basicity. Alkylamines are sufficiently basic that they exist almost entirely in their protonated form at the physiological pH of 7.3.

Heterocyclic amines are compounds that contain one or more nitrogen atoms as part of a ring. Saturated
heterocyclic amines usually have the same chemistry as their open-chain analogs, but unsaturated heterocycles such as pyrrole, imidazole, pyridine, and pyrimidine are aromatic. All four are unusually stable, and all undergo aromatic substitution on reaction with electrophiles. Pyrrole is nonbasic because its nitrogen lone-pair electrons are part of the aromatic $\pi$ system. Fused-ring heterocycles such as quinoline, isoquinoline, indole, and purine are also commonly found in biological molecules.


Summary -
Arylamines are prepared by nitration of an aromatic ring followed by reduction. Alkylamines are prepared by $\mathrm{S}_{\mathrm{N}} 2$ reaction of ammonia or an amine with an alkyl halide or by the Gabriel amine synthesis. Amines can also be prepared by a number of reductive methods, including $\mathrm{LiAlH}_{4}$ reduction of amides, nitriles, and azides. Also important is the reductive amination reaction in which a ketone or an aldehyde is treated with an amine in the presence of a reducing agent such as $\mathrm{NaBH}_{4}$. In addition, amines result from Hofmann and Curtius rearrangements of carboxylic acid derivatives. Both methods involve migration of the -R group bonded to the carbonyl carbon and yield a product that has one less carbon atom than the starting material.

Many of the reactions of amines are familiar from past chapters. Thus, amines react with alkyl halides in $\mathrm{S}_{\mathrm{N}} 2$ reactions and with acid chlorides in nucleophilic acyl substitution reactions. Amines also undergo E2 elimination to yield alkenes if they are first quaternized by treatment with iodomethane and then heated with silver oxide, a process called the Hofmann elimination.

Arylamines are converted by diazotization with nitrous acid into arenediazonium salts, $\mathrm{ArN}_{2}{ }^{+} \mathrm{X}^{-}$. The diazonio group can then be replaced by many other substituents by the Sandmeyer reaction to give a wide variety of substituted aromatic compounds. Aryl chlorides, bromides, iodides, and nitriles can be prepared from arenediazonium salts, as can arenes and phenols. In addition to their reactivity toward substitution reactions, diazonium salts undergo coupling with phenols and arylamines to give brightly colored azo compounds.


Summary of Reactions -
1. Synthesis of amines (Section 24.6)
a. Reduction of nitriles

$$
\mathrm{RCH}_{2} \mathrm{X} \xrightarrow{\mathrm{NaCN}} \mathrm{RCH}_{2} \mathrm{C} \equiv \mathrm{~N} \xrightarrow[\text { 2. } \mathrm{H}_{2} \mathrm{O}]{\text { 1. } \mathrm{LiAlH}_{4} \text {, ether }} \underset{\mathrm{RCH}_{2}}{\text { - }}
$$

b. Reduction of amides

c. Reduction of nitrobenzenes

d. $\mathrm{S}_{\mathrm{N}} 2$ Alkylation of alkyl halides

e. Gabriel amine synthesis

f. Reduction of azides

$$
\mathrm{RCH}_{2}-\mathrm{X} \xrightarrow[\text { ethanol }]{\mathrm{Na}^{+}{ }^{-} \mathrm{N}_{3}} \quad \mathrm{RCH}_{2}-\mathrm{N}=\stackrel{+}{\mathrm{N}}=\stackrel{-}{\mathrm{N}} \quad \xrightarrow[2 . \mathrm{H}_{2} \mathrm{O}]{\text { 1. } \mathrm{LiAlH}_{4} \text {, ether }} \quad \mathrm{R}-\mathrm{CH}_{2}-\mathrm{NH}_{2}
$$

g. Reductive amination of aldehydes/ketones

h. Hofmann rearrangement of amides

i. Curtius rearrangement of acyl azides

2. Reactions of amines
a. Alkylation with alkyl halides; see reaction 1(d)
b. Hofmann elimination (Section 24.7)

c. Diazotization (Section 24.8)

3. Reactions of arenediazonium salts (Section 24.8)
a. Nucleophilic substitutions


Summary of Reactions -
c. Diazotization (Section 24.8)

3. Reactions of arenediazonium salts (Section 24.8)
a. Nucleophilic substitutions

| HCl
 CuCl | HBr
 CuBr | NaI | KCN
 CuCN | $\mathrm{Cu}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{O}$
 $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ |
| :--- | :--- | :--- | :--- | :--- | $\mathrm{H}_{3} \mathrm{PO}_{2}$

b. Diazonium coupling


Visualizing Chemistry -
PROBLEM Name the following amines, and identify each as primary, secondary, or tertiary:

24-26 (a)

(b)

(c)

PROBLEM The following compound contains three nitrogen atoms. Rank them in order of increasing basicity.
24-27

PROBLEM Name the following amine, including $R, S$ stereochemistry, and draw the product of its reaction with
24-28 excess iodomethane followed by heating with $\mathrm{Ag}_{2} \mathrm{O}$ (Hofmann elimination). Is the stereochemistry of the alkene product $Z$ or $E$ ? Explain.

PROBLEM Which nitrogen atom in the following compound is most basic? Explain.
24-29


Mechanism Problems -
PROBLEM Predict the product(s) and write the mechanism for each of the following reactions:
24-30 (a)

(b)

PROBLEM Predict the product(s) and write the mechanism for each of the following reactions:
24-31 (a)

(b)

PROBLEM Predict the product(s) and write the mechanism for each of the following reactions:
24-32 (a)

(b)

(c)

(d)

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

24-33 (a)

(b)

PROBLEM The diazotization of aniline first involves the formation of $\mathrm{NO}^{+}$(nitrosonium ion) by the dehydration
24-34 of nitrous acid with sulfuric acid. The aniline nitrogen then acts as a nucleophile and eventually loses water. Propose a mechanism for the formation of the dizaonium salt of aniline using curved arrows to show all electron movement.

PROBLEM Substituted pyrroles are often prepared by treatment of a 1,4-diketone with ammonia. Propose a 24-35 mechanism.

PROBLEM 3,5-Dimethylisoxazole is prepared by reaction of 2,4-pentanedione with hydroxylamine. Propose a
24-36 mechanism.

3,5-Dimethylisoxazole
PROBLEM One problem with reductive amination as a method of amine synthesis is that by-products are
24-37 sometimes obtained. For example, reductive amination of benzaldehyde with methylamine leads to a mixture of $N$-methylbenzylamine and $N$-methyldibenzylamine. How do you suppose the tertiary amine by-product is formed? Propose a mechanism.


Mechanism Problems -
PROBLEM Chlorophyll, heme, vitamin $\mathrm{B}_{12}$, and a host of other substances are biosynthesized from
24-38 porphobilinogen (PBG), which is itself formed from condensation of two molecules of 5-aminolevulinate. The two 5-aminolevulinates are bound to lysine (Lys) amino acids in the enzyme, one in the enamine form and one in the imine form, and their condensation is thought to occur by the following steps. Use curved arrows to show the mechanism of each step.

Enzyme-bound 5-aminolevulinate

PROBLEM Choline, a component of the phospholipids in cell membranes, can be prepared by $\mathrm{S}_{\mathrm{N}} 2$ reaction of
24-39 trimethylamine with ethylene oxide. Show the structure of choline, and propose a mechanism for the reaction.

PROBLEM The antitumor antibiotic mitomycin C functions by forming cross-links in DNA chains. 24-40

Mitomycin C

(a) The first step is loss of methoxide and formation of an iminium ion intermediate that is deprotonated to give an enamine. Show the mechanism.
(b) The second step is reaction of the enamine with DNA to open the three-membered, nitrogencontaining (aziridine) ring. Show the mechanism.
(c) The third step is loss of carbamate $\left(\mathrm{NH}_{2} \mathrm{CO}_{2}{ }^{-}\right)$and formation of an unsaturated iminium ion, followed by a conjugate addition of another part of the DNA chain. Show the mechanism.
PROBLEM $\alpha$-Amino acids can be prepared by the Strecker synthesis, a two-step process in which an aldehyde
24-41 is treated with ammonium cyanide followed by hydrolysis of the amino nitrile intermediate with aqueous acid. Propose a mechanism for the reaction.

An $\alpha$-amino acid
PROBLEM One of the reactions used in determining the sequence of nucleotides in a strand of DNA is reaction
24-42 with hydrazine. Propose a mechanism for the following reaction, which occurs by an initial conjugate addition followed by internal amide formation.


Mechanism Problems -
PROBLEM When an $\alpha$-hydroxy amide is treated with $\mathrm{Br}_{2}$ in aqueous NaOH under Hofmann rearrangement
24-43 conditions, loss of $\mathrm{CO}_{2}$ occurs and a chain-shortened aldehyde is formed. Propose a mechanism.

PROBLEM The following transformation involves a conjugate nucleophilic addition reaction (Section 19.13)
24-44 followed by an intramolecular nucleophilic acyl substitution reaction (Section 21.2). Show the mechanism.

PROBLEM Propose a mechanism for the following reaction:
24-45

PROBLEM One step in the biosynthesis of morphine is the reaction of dopamine with
24-46 $p$-hydroxyphenylacetaldehyde to give ( $S$ )-norcoclaurine. Assuming that the reaction is acidcatalyzed, propose a mechanism.

Naming Amines
PROBLEM Name the following compounds:
24-47 (a)

(b)

(c)

(d)

(e)

(f) $\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CN}$

PROBLEM Draw structures corresponding to the following IUPAC names:
24-48
(a) $N, N$-Dimethylaniline
(b) (Cyclohexylmethyl)amine
(c) $N$-Methylcyclohexylamine
(d) (2-Methylcyclohexyl)amine (e) 3-(N,N-Dimethylamino)propanoic acid

PROBLEM Classify each of the amine nitrogen atoms in the following substances as primary, secondary, or 24-49 tertiary:
(a)

(b)

(c)

Lysergic acid diethylamide


Amine Basicity -
PROBLEM Although pyrrole is a much weaker base than most other amines, it is a much stronger acid ( $\mathrm{p} K_{\mathrm{a}} \approx$
24-50 15 for the pyrrole versus 35 for diethylamine). The $\mathrm{N}-\mathrm{H}$ hydrogen is readily abstracted by base to yield the pyrrole anion, $\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{~N}^{-}$. Explain.

PROBLEM Histamine, whose release in the body triggers nasal secretions and constricted airways, has three
24-51 nitrogen atoms. List them in order of increasing basicity and explain your ordering.

Histamine

PROBLEM Account for the fact that $p$-nitroaniline ( $\mathrm{p} K_{\mathrm{a}}=1.0$ ) is less basic than $m$-nitroaniline ( $\mathrm{p} K_{\mathrm{a}}=2.5$ ) by
24-52 a factor of 30 . Draw resonance structures to support your argument. (The $\mathrm{p} K_{\mathrm{a}}$ values refer to the corresponding ammonium ions.)


Synthesis of Amines -
PROBLEM How would you prepare the following substances from 1-butanol?
24-53 (a) Butylamine (b) Dibutylamine (c) Propylamine (d) Pentylamine
(e) $N, N$-Dimethylbutylamine (f) Propene

PROBLEM How would you prepare the following substances from pentanoic acid?
24-54 (a) Pentanamide (b) Butylamine (c) Pentylamine (d) 2-Bromopentanoic acid
(e) Hexanenitrile (f) Hexylamine
PROBLEM How would you prepare aniline from the following starting materials?
24-55 (a) Benzene (b) Benzamide (c) Toluene
PROBLEM How would you prepare benzylamine, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}$, from benzene? More than one step is needed.
24-56
PROBLEM How might you prepare pentylamine from the following starting materials?
24-57 (a) Pentanamide
(b) Pentanenitrile
(c) 1-Butene
(d) Hexanamide
(e) 1-Butanol
(f) 5-Decene (g) Pentanoic acid

PROBLEM How might a reductive amination be used to synthesize ephedrine, an amino alcohol that was
24-58 widely used for the treatment of bronchial asthma?

Ephedrine


Reactions of Amines -
PROBLEM How would you convert aniline into each of the following products?
24-59
(a) Benzene
(b) Benzamide
(c) Toluene

PROBLEM Write the structures of the major organic products you would expect from reaction of $m$-toluidine 24-60 ( $m$-methylaniline) with the following reagents:
(a) $\mathrm{Br}_{2}$ (1 equivalent)
(b) $\mathrm{CH}_{3} \mathrm{I}$ (excess)
(c) $\mathrm{CH}_{3} \mathrm{COCl}$ in pyridine
(d) The product of (c), then $\mathrm{HSO}_{3} \mathrm{Cl}$

PROBLEM Show the products from reaction of $p$-bromoaniline with the following reagents:
24-61 (a) $\mathrm{CH}_{3} \mathrm{I}$ (excess)
(b) HCl
(c) $\mathrm{HNO}_{2}, \mathrm{H}_{2} \mathrm{SO}_{4}$
(d) $\mathrm{CH}_{3} \mathrm{COCl}$
(e) $\mathrm{CH}_{3} \mathrm{MgBr}$
(f) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}, \mathrm{AlCl}_{3}$ (g) Product of (c) with $\mathrm{CuCl}, \mathrm{HCl}$ (h) Product of (d) with $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}, \mathrm{AlCl}_{3}$

PROBLEM What are the major products you would expect from Hofmann elimination of the following amines?

24-62 (a)

(b)

(c)

PROBLEM How would you prepare the following compounds from toluene? A diazonio replacement reaction 24-63 is needed in some instances.
(a)

(b)

(c)

PROBLEM Predict the product(s) of the following reactions. If more than one product is formed, tell which is 24-64 major.
(a)

(b)

(c)

(d)


Reactions of Amines -
(b)

(c)

PROBLEM Predict the product(s) of the following reactions. If more than one product is formed, tell which is 24-64 major.
(a)

(b)

(c)

(d)

Spectroscopy
PROBLEM Phenacetin, a substance formerly used in over-the-counter headache remedies, has the formula
24-65 $\mathrm{C}_{10} \mathrm{H}_{13} \mathrm{NO}_{2}$. Phenacetin is neutral and does not dissolve in either acid or base. When warmed with aqueous NaOH , phenacetin yields an amine, $\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{NO}$, whose ${ }^{1} \mathrm{H}$ NMR spectrum is shown. When heated with HI , the amine is cleaved to an aminophenol, $\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{NO}$. What is the structure of
phenacetin, and what are the structures of the amine and the aminophenol?

PROBLEM Propose structures for amines with the following ${ }^{1} \mathrm{H}$ NMR spectra:
24-66 (a) $\mathrm{C}_{3} \mathrm{H}_{9} \mathrm{NO}$

(b) $\mathrm{C}_{4} \mathrm{H}_{11} \mathrm{NO}_{2}$

(c) $\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{~N}$

PROBLEM Draw the structure of the amine that produced the ${ }^{1} \mathrm{H}$ NMR spectrum shown in Problem 24-66(c).
24-67 This compound has a single strong peak in its IR spectrum at $3280 \mathrm{~cm}^{-1}$.


General Problems -
PROBLEM Fill in the missing reagents a-e in the following scheme:
24-68

PROBLEM Oxazole is a five-membered aromatic heterocycle. Would you expect oxazole to be more basic or 24-69 less basic than pyrrole? Explain.

Oxazole

PROBLEM Protonation of an amide using strong acid occurs on oxygen rather than on nitrogen. Suggest a 24-70 reason for this behavior, taking resonance into account.

PROBLEM What is the structure of the compound with formula $\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{~N}$ that produced the following IR 24-71 spectrum?

PROBLEM Fill in the missing reagents a-d in the following synthesis of racemic methamphetamine from 24-72 benzene.

PROBLEM Cyclopentamine is an amphetamine-like central nervous system stimulant. Propose a synthesis of 24-73 cyclopentamine from materials of five carbons or less.


Cyclopentamine -
PROBLEM Tetracaine is a substance used as a spinal anesthetic.
24-74

(a) How would you prepare tetracaine from the corresponding aniline derivative, $\mathrm{ArNH}_{2}$ ?
(b) How would you prepare tetracaine from $p$-nitrobenzoic acid?
(c) How would you prepare tetracaine from benzene?

PROBLEM Atropine, $\mathrm{C}_{17} \mathrm{H}_{23} \mathrm{NO}_{3}$, is a poisonous alkaloid isolated from the leaves and roots of Atropa
24-75 belladonna, the deadly nightshade. In small doses, atropine acts as a muscle relaxant; 0.5 ng (nanogram, $10^{-9} \mathrm{~g}$ ) is sufficient to cause pupil dilation. On basic hydrolysis, atropine yields tropic acid, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{CH}_{2} \mathrm{OH}\right) \mathrm{CO}_{2} \mathrm{H}$, and tropine, $\mathrm{C}_{8} \mathrm{H}_{15} \mathrm{NO}$. Tropine is an optically inactive alcohol that yields tropidene on dehydration with $\mathrm{H}_{2} \mathrm{SO}_{4}$. Propose a structure for atropine.

PROBLEM Tropidene (Problem 24-75) can be converted by a series of steps into tropilidene 24-76 (1,3,5-cycloheptatriene). How would you accomplish this conversion?

PROBLEM Propose a structure for the product with formula $\mathrm{C}_{9} \mathrm{H}_{17} \mathrm{~N}$ that results when 24-77 2-(2-cyanoethyl)cyclohexanone is reduced catalytically.


Cyclopentamine -
PROBLEM Coniine, $\mathrm{C}_{8} \mathrm{H}_{17} \mathrm{~N}$, is the toxic principle of the poison hemlock drunk by Socrates. When subjected
24-78 to Hofmann elimination, coniine yields 5 -( $N, N$-dimethylamino)-1-octene. If coniine is a secondary amine, what is its structure?

PROBLEM How would you synthesize coniine (Problem 24-78) from acrylonitrile ( $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCN}$ ) and ethyl
24-79 3-oxohexanoate $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CO}_{2} \mathrm{Et}\right)$ ? (See Problem 24-77.)
PROBLEM Tyramine is an alkaloid found, among other places, in mistletoe and ripe cheese. How would you
24-80 synthesize tyramine from benzene? From toluene?

PROBLEM Reaction of anthranilic acid (o-aminobenzoic acid) with $\mathrm{HNO}_{2}$ and $\mathrm{H}_{2} \mathrm{SO}_{4}$ yields a diazonium salt
24-81 that can be treated with base to yield a neutral diazonium carboxylate.
(a) What is the structure of the neutral diazonium carboxylate?
(b) Heating the diazonium carboxylate results in the formation of $\mathrm{CO}_{2}, \mathrm{~N}_{2}$, and an intermediate that reacts with 1,3-cyclopentadiene to yield the following product:

What is the structure of the intermediate, and what kind of reaction does it undergo with cyclopentadiene?

PROBLEM Cyclooctatetraene was first synthesized in 1911 by a route that involved the following 24-82 transformation:

How might you use the Hofmann elimination to accomplish this reaction? How would you finish the synthesis by converting cyclooctatriene into cyclooctatetraene?


Cyclopentamine -
How might you use the Hofmann elimination to accomplish this reaction? How would you finish the synthesis by converting cyclooctatriene into cyclooctatetraene?

PROBLEM Propose structures for compounds that show the following ${ }^{1} \mathrm{H}$ NMR spectra.
24-83 (a) $\mathrm{C}_{9} \mathrm{H}_{13} \mathrm{~N}$

(b) $\mathrm{C}_{15} \mathrm{H}_{17} \mathrm{~N}$

PROBLEM 4-Dimethylaminopyridine (DMAP) acts as a catalyst in acyl transfer reactions. DMAP's catalytic
24-84 activity stems from its nucleophilic character at the pyridine nitrogen, not the dimethylamino group. Explain this behavior, taking resonance into account.


CHAPTER 25 <br> Biomolecules: Carbohydrates -
FIGURE 25.1 Produced by honeybees from the nectar of flowers, honey is primarily a mixture of the two simple sugars fructose and glucose. (credit: modification of work "Apis mellifera" by Thomas Bresson/Flickr, CC BY 2.0)


CHAPTER CONTENTS -
25.1 Classification of Carbohydrates
25.2 Representing Carbohydrate Stereochemistry: Fischer Projections
25.3 D,L Sugars
25.4 Configurations of the Aldoses
25.5 Cyclic Structures of Monosaccharides: Anomers
25.6 Reactions of Monosaccharides
25.7 The Eight Essential Monosaccharides
25.8 Disaccharides
25.9 Polysaccharides and Their Synthesis
25.10 Some Other Important Carbohydrates

WHY THIS CHAPTER? We've now seen all the common functional groups and reaction types that occur in organic and biological chemistry. In this and the next four chapters, we'll focus on the major classes of biological molecules, beginning with a look at the structures and primary biological functions of carbohydrates. Then, in Chapter 29, we'll return to the subject to see how carbohydrates are both synthesized and degraded in organisms.

Carbohydrates occur in every living organism. The sugar and starch in food, and the cellulose in wood, paper, and cotton are nearly pure carbohydrates. Modified carbohydrates form part of the coating around living cells, other carbohydrates are part of the nucleic acids that carry our genetic information, and still others are used as medicines.

The word carbohydrate derives historically from the fact that glucose, the first simple carbohydrate to be obtained in pure form, has the molecular formula $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ and was originally thought to be a "hydrate of
carbon, $\mathrm{C}_{6}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}$." This view was soon abandoned, but the name persisted. Today, the term carbohydrate is used to refer loosely to the broad class of polyhydroxylated aldehydes and ketones commonly called sugars. Glucose, also known as dextrose in medical work, is the most familiar example.


CHAPTER CONTENTS -
Glucose (dextrose),
a pentahydroxyhexanal
Carbohydrates are synthesized by green plants during photosynthesis, a complex process in which sunlight provides the energy to convert $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ into glucose plus oxygen. Many molecules of glucose are then chemically linked for storage by the plant in the form of either cellulose or starch. It has been estimated that more than $50 \%$ of the dry weight of the earth's biomass-all plants and animals-consists of glucose polymers. When eaten and metabolized, carbohydrates then provide animals with a source of readily available energy. Thus, carbohydrates act as the chemical intermediaries by which solar energy is stored and used to support life.

$$
6 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} \xrightarrow{\text { Sunlight }} 6 \mathrm{O}_{2}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \longrightarrow \text { Cellulose, starch }
$$

Glucose
Because humans and most other mammals lack the enzymes needed for digestion of cellulose, they require starch as their dietary source of carbohydrates. Grazing animals such as cows, however, have microorganisms in their first stomach that are able to digest cellulose. The energy stored in cellulose is thus moved up the biological food chain when these ruminant animals eat grass and are themselves used for food.


CHAPTER CONTENTS - 25.1 Classification of Carbohydrates
Carbohydrates are generally classified as either simple or complex. Simple sugars, or monosaccharides, are carbohydrates like glucose and fructose that can't be converted into smaller sugars by hydrolysis. Complex carbohydrates are made of two or more simple sugars linked together by acetal bonds (Section 19.10). Sucrose (table sugar), for instance, consists of one glucose linked to one fructose. Similarly, cellulose is made up of several thousand glucose units linked together. Enzyme-catalyzed hydrolysis of a complex carbohydrate breaks it down into its constituent monosaccharides.

Sucrose (a disaccharide)

Monosaccharides are further classified as either aldoses or ketoses. The -ose suffix designates a carbohydrate, and the prefixes aldo- and keto- identify the kind of carbonyl group in the molecule, whether aldehyde or ketone.

The number of carbon atoms in the monosaccharide is indicated by the appropriate numerical prefix tri-, tetr-, pent-, hex-, and so forth, in the name. Putting it all together, glucose is an aldohexose, a six-carbon aldehyde sugar; fructose is a ketohexose, a six-carbon keto sugar; ribose is an aldopentose, a five-carbon aldehyde sugar; and sedoheptulose is a ketoheptose, a seven-carbon keto sugar. Most of the common simple sugars are either pentoses or hexoses.

PROBLEM Classify each of the following monosaccharides:
25-1 (a)

Threose
(b)

Ribulose
(c)

Tagatose
(d)

2-Deoxyribose


CHAPTER CONTENTS - 25.2 Representing Carbohydrate Stereochemistry: Fischer Projections
Because most carbohydrates have numerous chirality centers, it was recognized long ago that a quick method for representing their stereochemistry was needed. In 1891, the German chemist Emil Fischer suggested a method based on the projection of a tetrahedral carbon atom onto a flat surface. These Fischer projections were soon adopted and are still a common means of representing stereochemistry at chirality centers in carbohydrate chemistry.

A tetrahedral carbon atom is represented in a Fischer projection by two crossed lines. The horizontal lines represent bonds coming out of the page, and the vertical lines represent bonds going into the page.

For example, (R)-glyceraldehyde, the simplest monosaccharide, can be drawn as in FIGURE 25.2.

FIGURE 25.2 A Fischer projection of (R)-glyceraldehyde.
Because a given chiral molecule can be drawn in many ways, it's sometimes necessary to compare two projections to see if they represent the same or different enantiomers. To test for identity, Fischer projections can be moved around on the paper, but only two kinds of motions are allowed; moving a Fischer projection in any other way inverts its meaning.

- A Fischer projection can be rotated on the page by $180^{\circ}$, but not by $90^{\circ}$ or $270^{\circ}$. Only a $180^{\circ}$ rotation
maintains the Fischer convention by keeping the same substituent groups going into and coming out of the plane. In the following Fischer projection of ( $R$ )-glyceraldehyde, for instance, the -H and -OH groups come out of the plane both before and after a $180^{\circ}$ rotation.

A $90^{\circ}$ rotation breaks the Fischer convention by exchanging the groups that go into the plane with those that come out. In the following Fischer projections of $(R)$-glyceraldehyde, the -H and -OH groups originally come out of the plane but go into the plane after a $90^{\circ}$ rotation. As a result, the rotated projection represents (S)-glyceraldehyde.

(R)-Glyceraldehyde

- A Fischer projection can also have one group held steady while the other three rotate in either a clockwise or a counterclockwise direction. The effect is simply to rotate around a single bond, which does not change the stereochemistry.


CHAPTER CONTENTS - 25.2 Representing Carbohydrate Stereochemistry: Fischer Projections
- A Fischer projection can also have one group held steady while the other three rotate in either a clockwise or a counterclockwise direction. The effect is simply to rotate around a single bond, which does not change the stereochemistry.

(R)-Glyceraldehyde

(R)-Glyceraldehyde

(S)-Glyceraldehyde
(S)-Glyceraidehyde

$R, S$ stereochemical designations (Section 5.5) can be assigned to the chirality center in a Fischer projection by following three steps, as shown in Worked Example 25.1.


STEP 1 -
Rank the four substituents in the usual way (Section 5.5).


STEP 2 -
Place the group of lowest ranking, usually H, at the top of the Fischer projection by using one of the allowed motions. This means that the lowest-ranked group is oriented back, away from the viewer, as required for assigning configuration.


STEP 3 -
Determine the direction of rotation $1 \rightarrow 2 \rightarrow 3$ of the remaining three groups, and assign $R$ or $S$ configuration.
Carbohydrates with more than one chirality center are shown in Fischer projections by stacking the centers on top of one another, with the carbonyl carbon at or near the top. Glucose, for example, has four chirality centers stacked on top of one another in a Fischer projection. Such representations don't, however, give an accurate
picture of the molecule's true three-dimensional conformation, which is curled around on itself like a bracelet.

Glucose
(carbonyl group on top)


Assigning $R$ or $S$ Configuration to a Fischer Projection -
Assign $R$ or $S$ configuration to the following Fischer projection of alanine:

Alanine


Strategy -
Follow the steps in the text. (1) Rank the four substituents on the chiral carbon. (2) Manipulate the Fischer projection to place the group of lowest ranking at the top by carrying out one of the allowed motions. (3) Determine the direction $1 \rightarrow 2 \rightarrow 3$ of the remaining three groups.


Solution -
The rankings of the groups are (1) $-\mathrm{NH}_{2},(\mathbf{2})-\mathrm{CO}_{2} \mathrm{H},(\mathbf{3})-\mathrm{CH}_{3}$, and $(\mathbf{4})-\mathrm{H}$. To bring the group of lowest ranking $(-\mathrm{H})$ to the top, we might want to hold the $-\mathrm{CH}_{3}$ group steady while rotating the other three groups counterclockwise.

Going from first- to second- to third-highest ranking requires a counterclockwise turn, corresponding to $S$ stereochemistry.

PROBLEM Convert each of the following Fischer projections into a tetrahedral representation, and assign $R$ or 25-2 $S$ stereochemistry:
(a)

(b)

(c)

PROBLEM Which of the following Fischer projections of glyceraldehyde represent the same enantiomer?
25-3
(a)

(b)

(c)

(d)

PROBLEM Redraw the following molecule as a Fischer projection, and assign $R$ or $S$ configuration to the
25-4 chirality center (green $=\mathrm{Cl}$ ):

PROBLEM Redraw the following aldotetrose as a Fischer projection, and assign $R$ or $S$ configuration to each
25-5 chirality center:


25.3 d, L Sugars -
Glyceraldehyde, the simplest aldose, has only one chirality center and thus has two enantiomeric (nonidentical mirror-image) forms. Only the dextrorotatory enantiomer occurs naturally, however. That is, a sample of naturally occurring glyceraldehyde placed in a polarimeter rotates plane-polarized light in a clockwise direction, denoted (+). Since (+)-glyceraldehyde has been found to have an $R$ configuration at C2, it can be represented by a Fischer projection as shown previously in FIGURE 25.2. For historical reasons dating back long before the adoption of the $R, S$ system, $(R)-(+)$-glyceraldehyde is also referred to as D-glyceraldehyde (D for dextrorotatory). The other enantiomer, (S)-(-)-glyceraldehyde, is known as L-glyceraldehyde (L for levorotatory).

Because of the way that monosaccharides are biosynthesized in nature, glucose, fructose, and most other naturally occurring monosaccharides all have the same $R$ stereochemical configuration as D-glyceraldehyde at the chirality center farthest from the carbonyl group. In Fischer projections, therefore, most naturally occurring sugars have the hydroxyl group at the bottom chirality center pointing to the right (FIGURE 25.3). Such compounds are referred to as D sugars.

D-Glyceraldehyde [(R)-(+)-glyceraldehyde]

D-Ribose

D-Glucose

D-Fructose

FIGURE 25.3 Some naturally occurring D sugars. The -OH group at the chirality center farthest from the carbonyl group has the same configuration as (R)-(+)-glyceraldehyde and points toward the right in Fischer projections.

In contrast with D sugars, L sugars have an $S$ configuration at the lowest chirality center, with the bottom -OH group pointing to the left in Fischer projections. Thus, an l sugar is the mirror image (enantiomer) of the corresponding D sugar and has the opposite configuration from the $D$ sugar at all chirality centers.


25.3 d, L Sugars -
Note that the $D$ and L notations have no relation to the direction in which a given sugar rotates plane-polarized light. A D sugar can be either dextrorotatory or levorotatory. The prefix D indicates only that the - OH group at the lowest chirality center has $R$ stereochemistry and points to the right when the molecule is drawn in a standard Fischer projection. Note also that the D,L system of carbohydrate nomenclature describes the configuration at only one chirality center and says nothing about the configuration of other chirality centers that may be present.

PROBLEM Assign $R$ or $S$ configuration to each chirality center in the following monosaccharides, and tell
25-6 whether each is a D sugar or an L sugar:
(a)

(b)

(c)

PROBLEM (+)-Arabinose, an aldopentose that is widely distributed in plants, is systematically named
25-7 ( $2 R, 3 S, 4 S$ )-2,3,4,5-tetrahydroxypentanal. Draw a Fischer projection of (+)-arabinose, and identify it as a D sugar or an L sugar.


25.3 d, L Sugars - 25.4 Configurations of the Aldoses
An aldotetrose is a four-carbon sugar with two chirality centers. Thus, there are $2^{2}=4$ possible stereoisomeric aldotetroses, or two D,L pairs of enantiomers named erythrose and threose.

Aldopentoses have three chirality centers and a total of $2^{3}=8$ possible stereoisomers, or four D,L pairs of enantiomers. These four pairs are called ribose, arabinose, xylose, and lyxose. All except lyxose occur widely in nature. D-Ribose is an important constituent of RNA (ribonucleic acid), L-arabinose is found in many plants, and

D-xylose is found in wood.
Aldohexoses have four chirality centers and a total of $2^{4}=16$ possible stereoisomers, or eight D,L pairs of enantiomers. The names of the eight are allose, altrose, glucose, mannose, gulose, idose, galactose, and talose. Only D-glucose, from starch and cellulose, and D-galactose, from gums and fruit pectins, are widely distributed in nature. D-mannose and D-talose also occur naturally but in lesser abundance.

Fischer projections of the four-, five-, and six-carbon D aldoses are shown in FIGURE 25.4. Starting with D-glyceraldehyde, we can imagine constructing the two D aldotetroses by inserting a new chirality center just below the aldehyde carbon. Each of the two D aldotetroses then leads to two D aldopentoses (four total), and each of the four $D$ aldopentoses leads to two $D$ aldohexoses (eight total). In addition, each of the $D$ aldoses in FIGURE 25.4 has an l enantiomer, which is not shown.


25.3 d, L Sugars - 25.4 Configurations of the Aldoses
FIGURE 25.4 Configurations of $\mathbf{D}$ aldoses. The structures are arranged from left to right so that the -OH groups on C2 alternate right/left $(R / L)$ across each series. Similarly, the -OH groups at C 3 alternate two right/two left ( $2 \mathrm{R} / 2 \mathrm{~L}$ ), the -OH groups at C 4 alternate $4 R / 4 \mathrm{~L}$, and the -OH groups at C 5 are to the right in all eight ( 8 R ). Each D aldose has a corresponding L enantiomer that is not shown.

Louis Fieser of Harvard University long ago suggested a simple procedure for remembering the names and structures of the eight $D$ aldohexoses:

STEP 1
Set up eight Fischer projections with the -CHO group on top and the $-\mathrm{CH}_{2} \mathrm{OH}$ group at the bottom.
STEP 2
At C5, place all eight -OH groups to the right (D series).

STEP 3
At C 4 , alternate four -OH groups to the right and four to the left.
STEP 4
At C3, alternate two -OH groups to the right, two to the left.
STEP 5
At C2, alternate - OH groups right, left, right, left.
STEP 6
Name the eight isomers using the mnemonic "All altruists gladly make gum in gallon tanks."
The structures of the four $D$ aldopentoses can be generated in a similar way and named by the memonic suggested by a Cornell University undergraduate: "RIBs ARe eXtra Lean."


Drawing a Fischer Projection -
Draw a Fischer projection of L-fructose.


Strategy -
Because L-fructose is the enantiomer of D-fructose, simply look at the structure of D-fructose and reverse the configuration at each chirality center.


Drawing a Fischer Projection of a Molecular Model -
Draw the following aldotetrose as a Fischer projection, and identify it as a D sugar or an L sugar:


Strategy -
The Fischer projection of a monosaccharide is drawn vertically, with the carbonyl group at or near the top and the $-\mathrm{CH}_{2} \mathrm{OH}$ group at the bottom. The interior -H and -OH are drawn to the sides, pointing out of the page toward the viewer.


Solution -
Fischer projection

PROBLEM Only the D sugars are shown in Figure 25.4. Draw Fischer projections for the following l sugars:
25-8 (a) L-Xylose (b) L-Galactose (c) L-Allose
PROBLEM How many aldoheptoses are there? How many are D sugars, and how many are L sugars?
25-9
PROBLEM The following model is that of an aldopentose. Draw a Fischer projection of the sugar, name it, and
25-10 identify it as a D sugar or an L sugar.


Solution - 25.5 Cyclic Structures of Monosaccharides: Anomers
We said in Section 19.10 that aldehydes and ketones undergo a rapid and reversible nucleophilic addition reaction with alcohols to form hemiacetals.


An aldehyde -
A hemiacetal
If the carbonyl and the hydroxyl group are in the same molecule, an intramolecular nucleophilic addition can take place, leading to the formation of a cyclic hemiacetal. Five- and six-membered cyclic hemiacetals are relatively strain-free and particularly stable, and many carbohydrates therefore exist in an equilibrium between open-chain and cyclic forms. Glucose, for instance, exists in aqueous solution primarily in the six-membered pyranose form resulting from intramolecular nucleophilic addition of the -OH group at C 5 to the C 1 carbonyl group (FIGURE 25.5). The name pyranose is derived from pyran, the name of the unsaturated six-membered cyclic ether.

Like cyclohexane rings (Section 4.6), pyranose rings have a chairlike geometry with axial and equatorial substituents. By convention, the rings are usually drawn by placing the hemiacetal oxygen atom at the right rear,
as shown in FIGURE 25.5. Note that an -OH group on the right in a Fischer projection is on the bottom face of the pyranose ring, and an -OH group on the left in a Fischer projection is on the top face of the ring. For D sugars, the terminal $-\mathrm{CH}_{2} \mathrm{OH}$ group is on the top of the ring, whereas for L sugars, the $-\mathrm{CH}_{2} \mathrm{OH}$ group is on the bottom.


An aldehyde -
When an open-chain monosaccharide cyclizes to a pyranose form, a new chirality center is generated at the former carbonyl carbon and two diastereomers, called anomers, are produced. The hemiacetal carbon atom is referred to as the anomeric center. For example, glucose cyclizes reversibly in aqueous solution to a $37: 63$ mixture of two anomers (FIGURE 25.5). The compound with its newly generated - OH group at C 1 cis to the -OH at the lowest chirality center in a Fischer projection is called the $\boldsymbol{\alpha}$ anomer; its full name is $\alpha$-D-glucopyranose. The compound with its newly generated -OH group trans to the -OH at the lowest chirality center is called the $\boldsymbol{\beta}$ anomer; its full name is $\beta$-D-glucopyranose. Note that in $\beta$-D-glucopyranose, all the substituents on the ring are equatorial. Thus, $\beta$-D-glucopyranose is the least sterically crowded and most stable of the eight D aldohexoses.

$\alpha$-D-Glucopyranose
(37.3\%)

$\beta$-D-Glucopyranose
(62.6\%)

FIGURE 25.5 Glucose in its cyclic pyranose forms. As explained in the text, two anomers are formed by cyclization of glucose. The molecule whose newly formed -OH group at C 1 is cis to the oxygen atom on the lowest chirality center (C5) in a Fischer projection is the $\alpha$ anomer. The molecule whose newly formed -OH group is trans to the oxygen atom on the lowest chirality center in a Fischer projection is the $\beta$ anomer.


An aldehyde -
Some monosaccharides also exist in a five-membered cyclic hemiacetal form called a furanose. D-Fructose, for instance, exists in water solution as $68 \% \beta$-pyranose, $2.7 \% \alpha$-pyranose, $0.5 \%$ open-chain, $22.4 \% \beta$-furanose, and $6.2 \% \alpha$-furanose. The pyranose form results from addition of the - OH at C 6 to the carbonyl group, while the furanose form results from addition of the -OH at C 5 to the carbonyl group (FIGURE 25.6).

FIGURE 25.6 Pyranose and furanose forms of fructose in aqueous solution. The two pyranose anomers result from addition of the $\mathrm{C} 6-\mathrm{OH}$ group to the C2 carbonyl; the two furanose anomers result from addition of the $\mathrm{C} 5-\mathrm{OH}$ group to the C 2 carbonyl.

Both anomers of $D$-glucopyranose can be crystallized and purified. Pure $\alpha$-D-glucopyranose has a melting point of $146{ }^{\circ} \mathrm{C}$ and a specific rotation $[\alpha]_{D}=+112.2$; pure $\beta$-D-glucopyranose has a melting point of 148 to $155{ }^{\circ} \mathrm{C}$ and a specific rotation $[\alpha]_{D}=+18.7$. When a sample of either pure anomer is dissolved in water, however, its optical rotation slowly changes until it reaches a constant value of +52.6 . That is, the specific rotation of the $\alpha$-anomer solution decreases from +112.2 to +52.6 , and the specific rotation of the $\beta$-anomer solution increases from +18.7 to +52.6 . Called mutarotation, this change in optical rotation is due to the slow interconversion of the pure anomers to give the same $37: 63$ equilibrium mixture.


An aldehyde -
Mutarotation occurs by a reversible ring-opening of each anomer to the open-chain aldehyde, followed by reclosure. Although the equilibration is slow at neutral pH , it is catalyzed by both acid and base.


Drawing the Chair Conformation of an Aldohexose -
D-Mannose differs from D-glucose in its stereochemistry at C2. Draw D-mannose in its chairlike pyranose form.


Strategy -
First draw a Fischer projection of D-mannose. Then lay it on its side, and curl it around so that the - CHO group (C1) is on the right front and the $-\mathrm{CH}_{2} \mathrm{OH}$ group (C6) is toward the left rear. Now, connect the -OH at C 5 to the C1 carbonyl group to form the pyranose ring. In drawing the chair form, raise the leftmost carbon (C4) up and drop the rightmost carbon (C1) down.


Drawing the Chair Conformation of a Pyranose -
Draw $\beta$-L-glucopyranose in its more stable chair conformation.


Strategy -
It's probably easiest to begin by drawing the chair conformation of $\beta$-D-glucopyranose. Then draw its mirrorimage L enantiomer by changing the stereochemistry at every position on the ring, and carry out a ring-flip to give the more stable chair conformation. Note that the $-\mathrm{CH}_{2} \mathrm{OH}$ group is on the bottom face of the ring in the L enantiomer, as is the anomeric -OH .


Solution -
$\beta$-D-Glucopyranose
$\beta$-L-Glucopyranose

PROBLEM Ribose exists largely in a furanose form, produced by addition of the $\mathrm{C} 4-\mathrm{OH}$ group to the C 1 25-11 aldehyde. Draw D-ribose in its furanose form.

PROBLEM Figure 25.6 shows only the $\beta$-pyranose and $\beta$-furanose anomers of D -fructose. Draw the $\alpha$-pyranose
$\mathbf{2 5 - 1 2}$ and $\alpha$-furanose anomers.
PROBLEM Draw $\beta$-D-galactopyranose and $\beta$-D-mannopyranose in their more stable chair conformations. Label
25-13 each ring substituent as either axial or equatorial. Which would you expect to be more stable, galactose or mannose?

PROBLEM Draw $\beta$-L-galactopyranose in its more stable chair conformation, and label the substituents as
25-14 either axial or equatorial.
PROBLEM Identify the following monosaccharide, write its full name, and draw its open-chain form as a
25-15 Fischer projection.


Solution - 25.6 Reactions of Monosaccharides
Because monosaccharides have only two kinds of functional groups, hydroxyls and carbonyls, most of the chemistry of monosaccharides is the familiar chemistry of these two groups. As we've seen, alcohols can be converted into esters and ethers and can be oxidized; carbonyl compounds can react with nucleophiles and can be reduced.


Ester and Ether Formation -
Monosaccharides behave as simple alcohols in much of their chemistry. For example, carbohydrate - OH groups can be converted into esters and ethers, which are often easier to work with than the free sugars. Because of their many hydroxyl groups, monosaccharides are usually soluble in water but insoluble in organic solvents such as ether. They are also difficult to purify and have a tendency to form syrups rather than crystals when water is removed. Ester and ether derivatives, however, are soluble in organic solvents and are easily purified and crystallized.

Esterification is normally carried out by treating a carbohydrate with an acid chloride or acid anhydride in the presence of a base (Section 21.4 and Section 21.5). All the -OH groups react, including the anomeric one. For example, $\beta$-D-glucopyranose is converted into its pentaacetate by treatment with acetic anhydride in pyridine solution.

$\beta$-D-Glucopyranose
Penta- $O$-acetyl- $\beta$-D-glucopyranose
(91\%)
Carbohydrates are converted into ethers by treatment with an alkyl halide in the presence of base-the Williamson ether synthesis (Section 18.2). Standard Williamson conditions using a strong base tend to degrade sensitive sugar molecules, but silver oxide works well as a mild base and gives high yields of ethers. For example, $\alpha$-D-glucopyranose is converted into its pentamethyl ether in $85 \%$ yield on reaction with iodomethane and $\mathrm{Ag}_{2} \mathrm{O}$.

PROBLEM Draw the products you would obtain by reaction of $\beta$-d-ribofuranose with: (a) $\mathrm{CH}_{3} \mathrm{I}, \mathrm{Ag}_{2} \mathrm{O}$ (b)
25-16 $\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}$, pyridine


Glycoside Formation -
We saw in Section 19.10 that treatment of a hemiacetal with an alcohol and an acid catalyst yields an acetal.

A hemiacetal
An acetal

In the same way, treatment of a monosaccharide hemiacetal with an alcohol and an acid catalyst yields an acetal called a glycoside, in which the anomeric -OH has been replaced by an -OR group. For example, reaction of $\beta$-D-glucopyranose with methanol gives a mixture of $\alpha$ and $\beta$ methyl D-glucopyranosides. (Note that a glycoside is the functional group name for any sugar, whereas a glucoside is formed specifically from glucose.)

Glycosides are named by first citing the alkyl group and then replacing the -ose ending of the sugar with -oside. Like all acetals, glycosides are stable in neutral water. They aren't in equilibrium with an open-chain form, and they don't show mutarotation. They can, however, be hydrolyzed to give back the free monosaccharide plus alcohol on treatment with aqueous acid (Section 19.10).

Glycosides are abundant in nature, and many biologically important molecules contain glycosidic linkages. For example, digitoxin, the active component of the digitalis preparations used for treatment of heart disease, is a glycoside consisting of a steroid alcohol linked to a trisaccharide. Note also that the three sugars are linked to one another by glycoside bonds.

Digitoxin, a glycoside
The laboratory synthesis of glycosides can be difficult because of the numerous -OH groups on the sugar molecule. One method that is particularly suitable for preparing glucose $\beta$-glycosides involves treatment of glucose pentaacetate with HBr , followed by addition of the appropriate alcohol in the presence of silver oxide. Called the Koenigs-Knorr reaction, the sequence involves formation of a pyranosyl bromide, followed by nucleophilic substitution. For example, methylarbutin, a glycoside found in pears, has been prepared by reaction of tetraacetyl- $\alpha$-D-glucopyranosyl bromide with $p$-methoxyphenol.


Glycoside Formation -
Although the Koenigs-Knorr reaction appears to involve a simple backside $\mathrm{S}_{\mathrm{N}} 2$ displacement of bromide ion by alkoxide ion, the situation is actually more complex. Both $\alpha$ and $\beta$ anomers of tetraacetyl-D-glucopyranosyl bromide give the same $\beta$-glycoside product, implying that they react by a common pathway.

This result can be understood by assuming that tetraacetyl-D-glucopyranosyl bromide (either $\alpha$ or $\beta$ anomer) undergoes a spontaneous $\mathrm{S}_{\mathrm{N}} 1$-like loss of $\mathrm{Br}^{-}$, followed by internal reaction with the ester group at C 2 to form an oxonium ion. Since the acetate at C 2 is on the bottom of the glucose ring, the $\mathrm{C}-\mathrm{O}$ bond also forms from the bottom. Backside $\mathrm{S}_{\mathrm{N}} 2$ displacement of the oxonium ion then occurs with the usual inversion of configuration, yielding a $\beta$-glycoside and regenerating the acetate at C2 (FIGURE 25.7).

FIGURE 25.7 Mechanism of the Koenigs-Knorr reaction, showing the neighboring-group effect of a nearby acetate.
The participation shown by the nearby acetate group in the Koenigs-Knorr reaction is referred to as a neighboring-group effect and is a common occurrence in organic chemistry. Neighboring-group effects are usually noticeable only because they affect the rate or stereochemistry of a reaction; the nearby group itself does not undergo any evident change during the reaction.