CHAPTER CONTENTS - 18.2 Preparing Ethers
Diethyl ether and other simple symmetrical ethers are prepared industrially by the sulfuric-acid-catalyzed reaction of alcohols. The reaction occurs by $\mathrm{S}_{\mathrm{N}} 2$ displacement of water from a protonated ethanol molecule by the oxygen atom of a second ethanol. Unfortunately, this method is limited to use with primary alcohols because secondary and tertiary alcohols dehydrate by an E1 mechanism to yield alkenes (Section 17.6).


The Williamson Ether Synthesis -
The most generally useful method of preparing ethers is the Williamson ether synthesis, in which an alkoxide ion reacts with a primary alkyl halide or tosylate in an $\mathrm{S}_{\mathrm{N}} 2$ reaction. As we saw in Section 17.2, the alkoxide ion is normally prepared by reaction of an alcohol with a strong base such as sodium hydride, NaH .

Cyclopentanol

Alkoxide ion

Cyclopentyl methyl ether (74\%)

A useful variation of the Williamson synthesis involves silver oxide, $\mathrm{Ag}_{2} \mathrm{O}$, as a mild base rather than NaH . Under these conditions, the free alcohol reacts directly with the alkyl halide, so there is no need to preform the metal
alkoxide intermediate. Sugars react particularly well; glucose, for example, reacts with excess iodomethane in the presence of $\mathrm{Ag}_{2} \mathrm{O}$ to generate a pentaether in $85 \%$ yield.

Because the Williamson synthesis is an $\mathrm{S}_{\mathrm{N}} 2$ reaction, it is subject to all the usual constraints, as discussed in Section 11.3. Primary halides and tosylates work best because competitive E2 elimination can occur with more hindered substrates. Unsymmetrical ethers should therefore be synthesized by reaction between the more hindered alkoxide partner and less hindered halide partner rather than vice versa. For example, tert-butyl methyl ether, a substance used in the 1990s as an octane booster in gasoline, is best prepared by reaction of tert-butoxide ion with iodomethane rather than by reaction of methoxide ion with 2-chloro-2-methylpropane.

tert-Butoxide Iodomethane tert-Butyl methyl ether

PROBLEM Why do you suppose only symmetrical ethers are prepared by the sulfuric-acid-catalyzed 18-2 dehydration procedure? What product(s) would you expect if ethanol and 1-propanol were allowed to react together? In what ratio would the products be formed if the two alcohols were of equal reactivity?


The Williamson Ether Synthesis -
PROBLEM How would you prepare the following ethers using a Williamson synthesis?
18-3 (a) Methyl propyl ether (b) Anisole (methyl phenyl ether) (c) Benzyl isopropyl ether
(d) Ethyl 2,2-dimethylpropyl ether


Alkoxymercuration of Alkenes -
We saw in Section 8.4 that alkenes react with water in the presence of mercuric acetate to yield a hydroxymercuration product. Subsequent treatment with $\mathrm{NaBH}_{4}$ breaks the $\mathrm{C}-\mathrm{Hg}$ bond and yields the alcohol. A similar alkoxymercuration reaction occurs when an alkene is treated with an alcohol in the presence of mercuric acetate or, even better, mercuric trifluoroacetate, $\left(\mathrm{CF}_{3} \mathrm{CO}_{2}\right)_{2} \mathrm{Hg}$. Demercuration by reaction with $\mathrm{NaBH}_{4}$ then yields an ether. The net result is Markovnikov addition of the alcohol to the alkene.

The mechanism of the alkoxymercuration reaction is similar to that described in Section 8.4 for hydroxymercuration. The reaction is initiated by electrophilic addition of $\mathrm{Hg}^{2+}$ to the alkene, followed by reaction of the intermediate cation with alcohol and reduction of the $\mathrm{C}-\mathrm{Hg}$ bond by $\mathrm{NaBH}_{4}$. A variety of alcohols and alkenes can be used in alkoxymercuration. Primary, secondary, and even tertiary alcohols react well, but ditertiary ethers can't be prepared because of steric hindrance.


Synthesizing an Ether -
How would you prepare ethyl phenyl ether? Use whichever method you think is more appropriate, Williamson synthesis or the alkoxymercuration reaction.


Strategy -
Draw the target ether, identify the two groups attached to oxygen, and recall the limitations of the two methods for preparing ethers. Williamson synthesis uses an $S_{N} 2$ reaction and requires that one of the two groups attached to oxygen be either secondary or (preferably) primary. The alkoxymercuration reaction requires that one of the two groups come from an alkene precursor. Ethyl phenyl ether could be made by either method.

Ethyl phenyl ether

Solution

Phenol

Ethyl phenyl ether

PROBLEM Review the mechanism of oxymercuration shown in Figure 8.4, and then write the mechanism of 18-4 the alkoxymercuration reaction of 1-methylcyclopentene with ethanol. Use curved arrows to show the electron flow in each step.

PROBLEM How would you prepare the following ethers? Use whichever method you think is more appropriate, 18-5 Williamson synthesis or the alkoxymercuration reaction.
(a) Butyl cyclohexyl ether (b) Benzyl ethyl ether $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}\right)$
(c) sec-Butyl tert-butyl ether (d) Tetrahydrofuran

PROBLEM Rank the following halides in order of their reactivity in the Williamson synthesis:
18-6 (a) Bromoethane, 2-bromopropane, bromobenzene
(b) Chloroethane, bromoethane, 1-Iodopropene


Strategy - 18.3 Reactions of Ethers: Acidic Cleavage
Ethers are unreactive to many reagents used in organic chemistry, a property that accounts for their wide use as reaction solvents. Halogens, dilute acids, bases, and nucleophiles have no effect on most ethers. In fact, ethers undergo only one truly general reaction-they are cleaved by strong acids. Aqueous HBr and HI both work well, but HCl does not cleave ethers.

Acidic ether cleavages are typical nucleophilic substitution reactions and take place by either $\mathrm{S}_{\mathrm{N}} 1$ or $\mathrm{S}_{\mathrm{N}} 2$ mechanisms, depending on the structure of the substrate. Ethers with only primary and secondary alkyl groups react by an $\mathrm{S}_{\mathrm{N}} 2$ mechanism, in which $\mathrm{I}^{-}$or $\mathrm{Br}^{-}$reacts with the protonated ether at the less hindered site. This usually results in a selective cleavage into a single alcohol and a single alkyl halide. For example, ethyl isopropyl ether exclusively yields isopropyl alcohol and iodoethane on cleavage by HI because nucleophilic attack by iodide ion occurs at the less hindered primary site rather than at the more hindered secondary site.

Ethers with a tertiary, benzylic, or allylic group cleave by either an $S_{N} 1$ or E1 mechanism because these substrates can produce stable intermediate carbocations. These reactions are often fast and take place at moderate temperatures. tert-Butyl ethers, for example, react by an E1 mechanism on treatment with trifluoroacetic acid at $0^{\circ} \mathrm{C}$. We'll see in Section 26.7 that this reaction is often used in the laboratory synthesis of peptides.


Predicting the Product of an Ether Cleavage Reaction -
Predict the products of the following reaction:


Strategy -
Identify the substitution pattern of the two groups attached to oxygen-in this case a tertiary alkyl group and a primary alkyl group. Then recall the guidelines for ether cleavages. An ether with only primary and secondary alkyl groups usually undergoes cleavage by $\mathrm{S}_{\mathrm{N}} 2$ attack of a nucleophile on the less hindered alkyl group, but an ether with a tertiary alkyl group usually undergoes cleavage by an $\mathrm{S}_{\mathrm{N}} 1$ mechanism. In this case, an $\mathrm{S}_{\mathrm{N}} 1$ cleavage
of the tertiary $\mathrm{C}-\mathrm{O}$ bond will occur, giving 1-propanol and a tertiary alkyl bromide. In addition, a competitive E1 reaction leading to alkene might occur.


Solution -
PROBLEM Predict the products of the following reactions:

18-7 (a)

(b)

PROBLEM Write the mechanism of the acid-induced cleavage of tert-butyl cyclohexyl ether to yield
18-8 cyclohexanol and 2-methylpropene.
PROBLEM Why are HI and HBr more effective than HCl in cleaving ethers? (See Section 11.3.)
18-9


Solution - 18.4 Cyclic Ethers: Epoxides
For the most part, cyclic ethers behave like acyclic ethers. The chemistry of the ether functional group is the same, whether it's in an open chain or in a ring. Common cyclic ethers such as tetrahydrofuran and dioxane are often used as solvents because of their inertness, yet they can be cleaved by strong acids.

1,4-Dioxane

Tetrahydrofuran

The one group of cyclic ethers that behaves differently from open-chain ethers are the three-membered-ring compounds called epoxides, or oxiranes, which we saw in Section 8.7. The strain of the three-membered ring gives epoxides unique chemical reactivity.

Ethylene oxide, the simplest epoxide, is an intermediate in the manufacture of both ethylene glycol, used for automobile antifreeze, and polyester polymers. Approximately 37 million tons of ethylene oxide is produced worldwide each year, most of it by air oxidation of ethylene over a silver oxide catalyst at $300^{\circ} \mathrm{C}$. This process is not useful for other epoxides, however, and is of little value in the laboratory.

Note that the name ethylene oxide is not a systematic one because the -ene ending implies the presence of a double bond in the molecule. The name is frequently used, however, because ethylene oxide is derived from ethylene by addition of an oxygen atom. Other simple epoxides are named similarly. The systematic name for ethylene oxide is 1,2-epoxyethane.

In the laboratory, as we saw in Section 8.7, epoxides are usually prepared by treatment of an alkene with a
peroxyacid $\left(\mathrm{RCO}_{3} \mathrm{H}\right)$, typically $m$-chloroperoxybenzoic acid.

Epoxides can also be prepared from halohydrins, themselves produced by electrophilic addition of $\mathrm{Cl}_{2}$ or Br 2 to an alkene in the presence of water (Section 8.3). When a halohydrin is treated with base, HX is eliminated and an epoxide is produced by an intramolecular Williamson ether synthesis. That is, the nucleophilic alkoxide ion and the electrophilic alkyl halide are in the same molecule.


Solution - 18.4 Cyclic Ethers: Epoxides
PROBLEM What product would you expect from reaction of trans-2-butene with $m$-chloroperoxybenzoic acid?
18-10 Show the stereochemistry.
PROBLEM Reaction of cis-2-butene with m-chloroperoxybenzoic acid yields an epoxide different from that
18-11 obtained by reaction of the trans isomer. Explain.


Acid-Catalyzed Epoxide Opening -
Epoxides are cleaved by treatment with acid just as other ethers are, but under much milder conditions because of ring strain. As we saw in Section 8.7, dilute aqueous acid at room temperature is sufficient for facilitating the hydrolysis of epoxides to give 1,2-diols, also called vicinal glycols. (The word vicinal means "adjacent," and a glycol is a diol.) The epoxide cleavage takes place by $\mathrm{S}_{\mathrm{N}} 2$-like backside attack of a nucleophile on the protonated epoxide, giving a trans-1,2-diol as product.

1,2-Epoxycyclohexane

trans-1,2-Cyclohexanediol
(86\%)

Epoxides can also be opened by reaction with acids other than $\mathrm{H}_{3} \mathrm{O}^{+}$. If anhydrous HX is used, for instance, an epoxide is converted into a trans halohydrin.

A trans 2-halocyclohexanol
where $X=\mathrm{F}, \mathrm{Br}, \mathrm{Cl}$, or I
The regiochemistry of acid-catalyzed ring-opening depends on the epoxide's structure, and a mixture of products is often formed. When both epoxide carbon atoms are either primary or secondary, attack of the nucleophile occurs primarily at the less highly substituted site: an $\mathrm{S}_{\mathrm{N}} 2$-like result. When one of the epoxide carbon atoms is tertiary, however, nucleophilic attack occurs primarily at the more highly substituted site-an $\mathrm{S}_{\mathrm{N}} 1$-like result. Thus, 1,2-epoxypropane reacts with HCl to give primarily 1-chloro-2-propanol, but 2-methyl-1,2-epoxypropane gives 2-chloro-2-methyl-1- propanol as the major product.

1,2-Epoxypropane

2-Methyl-1,2-epoxypropane

$+$


Acid-Catalyzed Epoxide Opening -
1,2-Epoxypropane

2-Methyl-1,2-epoxypropane

$+$

1-Chloro-2-propanol 2-Chloro-1-propanol (90\%)
2-Chloro-2-methyl- 1-Chloro-2-methyl-1-propanol (60\%) 2-propanol (40\%)

The mechanisms of these acid-catalyzed epoxide openings are more complex than they initially appear. They seem to be neither purely $\mathrm{S}_{\mathrm{N}} 1$ nor $\mathrm{S}_{\mathrm{N}} 2$ but instead to be midway between the two extremes and to have characteristics of both. For instance, take the reaction of 1,2-epoxy-1-methylcyclohexane with HBr, shown in FIGURE 18.2. The reaction yields only a single stereoisomer of 2-bromo-2-methylcyclohexanol in which the -Br and -OH groups are trans, an $\mathrm{S}_{\mathrm{N}} 2$-like result caused by backside displacement of the epoxide oxygen. But the fact that $\mathrm{Br}^{-}$attacks the more hindered tertiary side of the epoxide rather than the less hindered secondary side is an $\mathrm{S}_{\mathrm{N}} 1$-like result in which the more stable, tertiary carbocation is involved.

Evidently, the transition state for acid-catalyzed epoxide opening has an $\mathrm{S}_{\mathrm{N}}$ 2-like geometry but also has a high degree of $\mathrm{S}_{\mathrm{N}} 1$-like carbocationic character. Because the positive charge in the protonated epoxide is shared by the more highly substituted carbon atom, backside attack of $\mathrm{Br}^{-}$occurs at the more highly substituted site.


Acid-Catalyzed Epoxide Opening -
FIGURE 18.2 Ring-opening of 1,2-epoxy-1-methylcyclohexane with $\mathbf{H B r}$. There is a high degree of $\mathrm{S}_{\mathrm{N}} 1$-like carbocation character in the transition state, which leads to backside attack of the nucleophile at the tertiary center and to formation of a product isomer that has -Br and -OH groups trans.


Predicting the Product of Epoxide Ring-Opening -
Predict the major product of the following reaction:


Strategy -
Identify the substitution pattern of the two epoxide carbon atoms. In this case, one carbon is secondary and one is primary. Then recall the guidelines for epoxide cleavages. An epoxide with only primary and secondary carbons usually undergoes cleavage by $\mathrm{S}_{\mathrm{N}} 2$-like attack of a nucleophile on the less hindered carbon, but an epoxide with a tertiary carbon atom usually undergoes cleavage by backside attack on the more hindered carbon. In this case, an $\mathrm{S}_{\mathrm{N}} 2$ cleavage of the primary $\mathrm{C}-\mathrm{O}$ epoxide bond will occur.


Solution -
PROBLEM Predict the major product of each of the following reactions:
18-12 (a)

(b)

PROBLEM How would you prepare the following diols?
18-13


Base-Catalyzed Epoxide Opening -
Unlike other ethers, epoxide rings can be cleaved by bases and nucleophiles as well as by acid. Although an ether oxygen is normally a poor leaving group in an $\mathrm{S}_{\mathrm{N}} 2$ reaction (Section 11.3), the strain of the threemembered ring causes epoxides to react with hydroxide ion at elevated temperatures.


Methylenecyclohexane oxide -
1-Hydroxymethylcyclohexanol (70\%)
Base-catalyzed epoxide opening is a typical $S_{N} 2$ reaction in which attack of the nucleophile takes place at the less hindered epoxide carbon. For example, 1,2-epoxypropane reacts with ethoxide ion exclusively at the less highly substituted, primary carbon to give 1-ethoxy-2-propanol.

1-Ethoxy-2-propanol (83\%)
here ( $2^{\circ}$ )
Many different nucleophiles can be used for epoxide opening, including amines ( $\mathrm{RNH}_{2}$ or $\mathrm{R}_{2} \mathrm{NH}$ ) and Grignard reagents ( RMgX ). An example of an amine reacting with an epoxide occurs in the commercial synthesis of metoprolol, a so-called $\beta$-blocker that is used for treatment of cardiac arrhythmias, hypertension, and heart attacks. $\beta$-Blockers are among the most widely prescribed drugs in the world.

Metoprolol
A similar nucleophilic ring-opening occurs when epoxides are treated with Grignard reagents. Ethylene oxide is frequently used, thereby allowing the conversion of a Grignard reagent into a primary alcohol having two more carbons than the starting alkyl halide. 1-Bromobutane, for example, is converted into 1-hexanol by reaction of its Grignard reagent with ethylene oxide.

PROBLEM Predict the major product of the following reactions:


Methylenecyclohexane oxide - 18.6 Crown Ethers
Crown ethers, discovered in the early 1960s by Charles Pedersen at the DuPont Company, are a relatively recent addition to the ether family. They are named according to the general format $x$-crown- $y$, where $x$ is the total number of atoms in the ring and $y$ is the number of oxygen atoms. Thus, 18-crown-6 ether is an 18-membered ring containing 6 ether oxygen atoms. Note the size and negative (red) character of the crown ether cavity in the following electrostatic potential map.

18-Crown-6 ether
The importance of crown ethers stems from their ability to sequester specific metal cations in the center of the polyether cavity. 18-Crown-6, for example, binds strongly with potassium ion. As a result, a solution of 18 -crown-6 in a nonpolar organic solvent will dissolve many potassium salts. Potassium permanganate, $\mathrm{KMnO}_{4}$, dissolves in toluene in the presence of 18 -crown- 6 , for instance, and the resulting solution is a valuable reagent for oxidizing alkenes.

The effect of using a crown ether to dissolve an inorganic salt in a hydrocarbon or ether solvent is similar to the effect of dissolving the salt in a polar aprotic solvent such as DMSO, DMF, or HMPA (Section 11.3). In both cases, the metal cation is strongly solvated, leaving the anion bare. Thus, the $\mathrm{S}_{\mathrm{N}} 2$ reactivity of an anion is tremendously enhanced in the presence of a crown ether.

Although crown ethers do not occur naturally, a group of compounds called ionophores have similar ionbinding properties. Produced by various microorganisms, ionophores are fat-soluble molecules that bind to specific ions and facilitate transport of the ions through biological membranes. The antibiotic valinomycin, for instance, binds specifically to $\mathrm{K}^{+}$ions with a ten-thousandfold selectivity over $\mathrm{Na}^{+}$.


Methylenecyclohexane oxide - 18.6 Crown Ethers
Valinomycin
PROBLEM 15-Crown-5 and 12-crown-4 ethers complex $\mathrm{Na}^{+}$and $\mathrm{Li}^{+}$, respectively. Make models of these crown 18-15 ethers, and compare the sizes of the cavities.


Thiols -
Thiols, also called mercaptans, are sulfur analogs of alcohols. They are named by the same system used for alcohols, with the suffix -thiol used in place of -ol. The -SH group itself is referred to as a mercapto group. Like alcohols, thiols are weakly acidic; the $\mathrm{p}_{\mathrm{a}}$ of $\mathrm{CH}_{3} \mathrm{SH}$, for instance, is 10.3 . Unlike alcohols, however, thiols don't typically form hydrogen bonds because the sulfur atom is not sufficiently electronegative.

Ethanethiol

Cyclohexanethiol

$m$-Mercaptobenzoic acid

The most striking characteristic of thiols is their appalling odor. Skunk scent, for instance, is caused primarily by the simple thiols 3-methyl-1-butanethiol and 2-butene-1-thiol. Volatile thiols such as ethanethiol are also added to natural gas and liquefied propane to serve as an easily detectable warning in case of leaks.

Thiols are usually prepared from alkyl halides by $\mathrm{S}_{\mathrm{N}} 2$ displacement with a sulfur nucleophile such as hydrosulfide anion, ${ }^{-}$SH.

The reaction often works poorly unless an excess of the nucleophile is used because the product thiol can undergo a second $S_{N} 2$ reaction with alkyl halide to give a sulfide ( $\mathbf{R}-\mathbf{S}-\mathbf{R}$ ') as by-product. To circumvent this problem, thiourea, $\left(\mathrm{NH}_{2}\right)_{2} \mathrm{C}=\mathrm{S}$, is often used as the nucleophile in the preparation of a thiol from an alkyl halide. The reaction occurs by displacement of the halide ion to yield an intermediate alkyl isothiourea salt, which is hydrolyzed by subsequent reaction with aqueous base.


Thiols -
Thiols can be oxidized by $\mathrm{Br}_{2}$ or $\mathrm{I}_{2}$ to yield disulfides ( $\mathbf{R}-\mathbf{S}-\mathbf{S}-\mathbf{R}^{\prime}$ ). The reaction is easily reversed, and a disulfide can be reduced back to a thiol by treatment with zinc and acid.

$$
2 \mathrm{R}-\mathrm{SH} \underset{\mathrm{Zn}, \mathrm{H}^{+}}{\stackrel{\mathrm{I}_{2}}{\rightleftarrows}} \mathrm{R}-\mathrm{S}-\mathrm{S}-\mathrm{R}+2 \mathrm{HI}
$$

A thiol
A disulfide
This thiol-disulfide interconversion is a key part of numerous biological processes. We'll see in Chapter 26, for instance, that disulfide formation is involved in defining the structure and three-dimensional conformations of proteins, where disulfide "bridges" often form cross-links between cysteine amino-acid units in the protein chains. Disulfide formation is also involved in the process by which cells protect themselves from oxidative degradation. A cellular component called glutathione removes potentially harmful oxidants and is itself oxidized to glutathione disulfide in the process. Reduction back to the thiol requires the coenzyme reduced flavin adenine dinucleotide, abbreviated $\mathrm{FADH}_{2}$.


Sulfides -
Sulfides are the sulfur analogs of ethers just as thiols are the sulfur analogs of alcohols. Sulfides are named by following the same rules used for ethers, with sulfide used in place of ether for simple compounds and alkylthio used in place of alkoxy for more complex substances.


Dimethyl sulfide -
Methyl phenyl sulfide

3-(Methylthio)cyclohexene

Treatment of a thiol with a base, such as NaH , gives the corresponding thiolate ion ( $\mathrm{RS}^{-}$), which undergoes reaction with a primary or secondary alkyl halide to give a sulfide. The reaction occurs by an $S_{N} 2$ mechanism, analogous to the Williamson synthesis of ethers (Section 18.2).

Despite their close structural similarity, sulfides and ethers differ substantially in their chemistry. Because the valence electrons on sulfur are farther from the nucleus and are less tightly held than those on oxygen ( $3 p$ electrons versus $2 p$ electrons), sulfur compounds are more nucleophilic than their oxygen analogs. Unlike dialkyl ethers, dialkyl sulfides react rapidly with primary alkyl halides by an $\mathrm{S}_{\mathrm{N}} 2$ mechanism to give sulfonium ions $\left(\mathrm{R}_{3} \mathrm{~S}^{+}\right)$.

Dimethyl sulfide Iodomethane Trimethylsulfonium iodide
The most common example of this process in living organisms is the reaction of the amino acid methionine with adenosine triphosphate (ATP; Section 6.8) to give $S$-adenosylmethionine. The reaction is somewhat unusual in that the biological leaving group in this $\mathrm{S}_{\mathrm{N}} 2$ process is the triphosphate ion rather than the more frequently seen diphosphate ion (Section 11.6).

Adenosine triphosphate (ATP)
$S$-Adenosylmethionine
Sulfonium ions are themselves useful alkylating agents because a nucleophile can attack one of the groups bonded to the positively charged sulfur, displacing a neutral sulfide as leaving group. We saw an example of this in Section 11.6 (FIGURE 11.17), in which $S$-adenosylmethionine transferred a methyl group to norepinephrine to give adrenaline.


Dimethyl sulfide -
Another difference between sulfides and ethers is that sulfides are easily oxidized. Treatment of a sulfide with hydrogen peroxide, $\mathrm{H}_{2} \mathrm{O}_{2}$, at room temperature yields the corresponding sulfoxide ( $\mathbf{R}_{\mathbf{2}} \mathbf{S O}$ ), and further oxidation of the sulfoxide with a peroxyacid yields a sulfone ( $\mathbf{R}_{\mathbf{2}} \mathbf{S O}_{\mathbf{2}} \mathbf{)}$.

Dimethyl sulfoxide (DMSO) is a particularly well-known sulfoxide that is often used as a polar aprotic solvent. It must be handled with care, however, because it has a remarkable ability to penetrate the skin, carrying along whatever is dissolved in it.

Dimethyl sulfoxide (a polar aprotic solvent)

PROBLEM Name the following compounds:
18-16 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM 2-Butene-1-thiol is one component of skunk spray. How would you synthesize this substance 18-17 (either cis or trans) from methyl 2-butenoate? From 1,3-butadiene?

$\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{SH}$
Methyl 2-butenoate
2-Butene-1-thiol


Infrared Spectroscopy -
Ethers are difficult to identify by IR spectroscopy. Although they show an absorption due to C-O single-bond stretching in the range 1050 to $1150 \mathrm{~cm}^{-1}$, many other kinds of absorptions occur in the same range. FIGURE 18.3 shows the IR spectrum of diethyl ether and identifies the $\mathrm{C}-\mathrm{O}$ stretch. Phenyl alkyl ethers show two strong absorbances for $\mathrm{C}-\mathrm{O}$ stretching at 1050 and $1250 \mathrm{~cm}^{-1}$. FIGURE 18.4 shows the IR spectrum of anisole.

FIGURE 18.3 The infrared spectrum of diethyl ether, $\mathrm{CH}_{3} \mathbf{C H}_{2} \mathbf{O C H}_{2} \mathbf{C H}_{3}$.

FIGURE 18.4 The infrared spectrum of anisole.


Nuclear Magnetic Resonance Spectroscopy -
Hydrogens on carbon next to an ether oxygen are shifted downfield from the normal alkane resonance and show ${ }^{1}$ H NMR absorptions in the region 3.4 to $4.5 \delta$. This downfield shift is clearly seen in the spectrum of dipropyl ether shown in FIGURE 18.5.

FIGURE 18.5 The ${ }^{\mathbf{1}} \mathbf{H}$ NMR spectrum of dipropyl ether. Protons on carbon next to oxygen are shifted downfield to $3.4 \delta$.
Epoxides absorb at a slightly higher field than other ethers and show characteristic resonances at 2.5 to 3.5 $\delta$ in their ${ }^{1}$ H NMR spectra, as indicated for 1,2-epoxypropane in FIGURE 18.6. The methylene protons of this epoxide are diastereotopic, and display complex splitting (see Section 13.7 and Section 13.8).

Ether carbon atoms also exhibit a downfield shift in the ${ }^{13}$ C NMR spectrum, where they usually absorb in the 50 to $80 \delta$ range. For example, the carbon atoms next to oxygen in methyl propyl ether absorb at 58.5 and $74.8 \delta$. Similarly, the methyl carbon in anisole absorbs at $54.8 \delta$.

PROBLEM The ${ }^{1} \mathrm{H}$ NMR spectrum shown is that of a cyclic ether with the formula $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$. Propose a structure. 18-18


(30) CHEMISTRY MATTERS -
Epoxy Resins and Adhesives
Few nonchemists know exactly what an epoxide is, but practically everyone has used an "epoxy glue" for household repairs or an epoxy resin for a protective coating. Worldwide, approximately 14 billion dollars' worth of epoxy resins and adhesives are used annually for a vast number of applications, including many in the aerospace industry. Much of the Boeing 787 Dreamliner, for instance, is held together with epoxy-based adhesives.

Epoxy resins and adhesives generally consist of two components that are mixed just prior to use. One component is a liquid "prepolymer," and the second is a "curing agent" that reacts with the prepolymer and causes it to solidify.

The most widely used epoxy resins and adhesives are based on a prepolymer made from bisphenol A and epichlorohydrin. On treatment with base, bisphenol A is converted into its anion, which acts as a nucleophile in an $\mathrm{S}_{\mathrm{N}} 2$ reaction with epichlorohydrin. Each epichlorohydrin molecule can react with two molecules of bisphenol $A$, once by $S_{N} 2$ displacement of chloride ion and once by nucleophilic opening of the epoxide ring. At the same time, each bisphenol A molecule can react with two epichlorohydrins, leading to a long polymer chain. Each end of a prepolymer chain has an unreacted epoxy group, and each chain has numerous secondary alcohol groups spaced regularly along its midsection.

FIGURE 18.7 Kayaks are often made of a high-strength polymer coated with epoxy resin. (credit: "Potomac River Festival Virginia 2017 Great Falls Kayak Race" by Watts/Flickr, CC BY 2.0)

Bisphenol A
Epichlorohydrin


(30) CHEMISTRY MATTERS -
Bisphenol A
Epichlorohydrin

When an epoxide is to be used, a basic curing agent such as a tertiary amine, $\mathrm{R}_{3} \mathrm{~N}$, is added to cause the individual prepolymer chains to link together. This cross-linking of chains is simply a base-catalyzed, $\mathrm{S}_{\mathrm{N}} 2$ epoxide ring-opening of an -OH group in the middle of one chain with an epoxide group on the end of another chain. The result of such cross-linking is formation of a vast, three-dimensional tangle that has enormous strength and chemical resistance.


Key Terms -
```
• alkoxymercuration • sulfone
- crown ether
- sulfonium ion
• disulfide (R-S-S-R') . sulfoxide
- ether (R-0-R')
- thiol
- mercapto group
- thiolate ion (RS`)
- sulfide (R-S-R')
- Williamson ether synthesis
```


Summary -
This chapter has finished the coverage of functional groups with $\mathrm{C}-\mathrm{O}$ and $\mathrm{C}-\mathrm{S}$ single bonds, focusing primarily on ethers, epoxides, thiols, and sulfides. Ethers are compounds that have two organic groups bonded to the same oxygen atom, ROR'. The organic groups can be alkyl, vinylic, or aryl, and the oxygen atom can be in a ring or in an open chain. Ethers are prepared by either Williamson ether synthesis, which involves $\mathrm{S}_{\mathrm{N}} 2$ reaction of an alkoxide ion with a primary alkyl halide, or the alkoxymercuration reaction, which involves Markovnikov addition of an alcohol to an alkene.

Ethers are inert to most reagents but react with strong acids to give cleavage products. Both HI and HBr are often used. The cleavage reaction takes place by an $\mathrm{S}_{\mathrm{N}} 2$ mechanism at the less highly substituted site if only primary and secondary alkyl groups are bonded to the ether oxygen, but by an $\mathrm{S}_{\mathrm{N}} 1$ or E1 mechanism if one of the alkyl groups bonded to oxygen is tertiary.

Epoxides are cyclic ethers with a three-membered, oxygen-containing ring. Because of the strain in the ring, epoxides undergo a cleavage reaction with both acids and bases. Acid-catalyzed ring-opening occurs with a regiochemistry that depends on the structure of the epoxide. Cleavage of the $\mathrm{C}-\mathrm{O}$ bond at the less highly substituted site occurs if both epoxide carbons are primary or secondary, but cleavage of the $\mathrm{C}-\mathrm{O}$ bond to the more highly substituted site occurs if one of the epoxide carbons is tertiary. Base-catalyzed epoxide ringopening occurs by $\mathrm{S}_{\mathrm{N}} 2$ reaction of a nucleophile at the less hindered epoxide carbon.


Summary -
Thiols, the sulfur analogs of alcohols, are usually prepared by $S_{N} 2$ reaction of an alkyl halide with thiourea. Mild oxidation of a thiol yields a disulfide, and mild reduction of a disulfide returns the thiol. Sulfides, the sulfur analogs of ethers, are prepared by an $\mathrm{S}_{\mathrm{N}} 2$ reaction between a thiolate anion and a primary or secondary alkyl halide. Sulfides are more nucleophilic than ethers and can be alkylated by reaction with a primary alkyl halide to yield a sulfonium ion. Sulfides can also be oxidized to sulfoxides and to sulfones.


Summary of Reactions -
1. Synthesis of ethers (Section 18.2)
a. Williamson ether synthesis

$$
\mathrm{RO}^{-}+\mathrm{R}^{\prime} \mathrm{CH}_{2} \mathrm{X} \longrightarrow \mathrm{ROCH}_{2} \mathrm{R}^{\prime}+\mathrm{X}^{-}
$$

b. Alkoxymercuration/demercuration

2. Reactions of ethers
a. Cleavage by HBr or HI (Section 18.3)

$$
\mathrm{R}-\mathrm{O}-\mathrm{R}^{\prime} \xrightarrow[\mathrm{H}_{2} \mathrm{O}]{\mathrm{HX}} \mathrm{RX}+\mathrm{R}^{\prime} \mathrm{OH}
$$

b. Acid-catalyzed epoxide opening (Section 18.5)

c. Base-catalyzed epoxide opening (Section 18.5)

3. Synthesis of thiols (Section 18.7)

$$
\mathrm{RCH}_{2} \mathrm{Br} \xrightarrow[\text { 2. } \mathrm{H}_{2} \mathrm{O}, \mathrm{NaOH}]{\text { 1. } \left.\mathrm{H}_{2} \mathrm{~N}\right)_{2} \mathrm{C}=\mathrm{S}} \quad \mathrm{RCH}_{2} \mathrm{SH}
$$

4. Oxidation of thiols to disulfides (Section 18.7)

$$
2 \mathrm{RSH} \xrightarrow{\mathrm{I}_{2}, \mathrm{H}_{2} \mathrm{O}} \quad \mathrm{RS}-\mathrm{SR}
$$

5. Synthesis of sulfides (Section 18.7)

$$
\mathrm{RS}^{-}+\mathrm{R}^{\prime} \mathrm{CH}_{2} \mathrm{Br} \longrightarrow \mathrm{RSCH}_{2} \mathrm{R}^{\prime}+\mathrm{Br}^{-}
$$

6. Oxidation of sulfides to sulfoxides and sulfones (Section 18.7)


Visualizing Chemistry -
PROBLEM Give IUPAC names for the following compounds (red = O; reddish brown = Br; yellow $=\mathrm{S}$ ):

18-19 (a)

(b)

(c)

PROBLEM Show the product, including stereochemistry, that would result from reaction of the following
18-20 epoxide with HBr :

PROBLEM Show the product, including stereochemistry, of the following reaction:
18-21

PROBLEM Treatment of the following alkene with a peroxyacid yields an epoxide different from that obtained
18-22 by reaction with aqueous $\mathrm{Br}_{2}$ followed by base treatment. Propose structures for the two epoxides, and explain the result.


Mechanism Problems -
PROBLEM Predict the product(s) and provide the mechanism for each of the following reactions.

18-23 (a)

(b)

PROBLEM Predict the product(s) and provide the mechanism for each of the following reactions.
18-24
(a)

$\xrightarrow{\mathrm{HBr}}$ ?
(b)

PROBLEM Predict the product(s) and provide the mechanism for each of the following two-step processes.

18-25 (a)

(b)

PROBLEM The alkoxymercuration of alkenes involves the formation of an organomercury intermediate (I),
18-26 which is reduced with $\mathrm{NaBH}_{4}$ to give an ether product. Predict the ether product and provide the mechanism for the following reaction.

PROBLEM Predict the product(s) and provide the mechanism for the following reactions:

18-27 (a)

(b)

PROBLEM Predict the product(s) and provide the mechanism for each of the following reactions.

18-28 (a)

(b)

PROBLEM In the formation of the prepolymer used to make epoxy resins, a bisphenol reacts with
18-29 epichlorohydrin in the presence of a base. Show the product and mechanism when two moles of phenol react with epichlorohydrin.


Epichlorohydrin -
PROBLEM Ethers undergo an acid-catalyzed cleavage reaction when treated with the Lewis acid $\mathrm{BBr}_{3}$ at room 18-30 temperature. Propose a mechanism for the reaction.

PROBLEM Treatment of 1,1-diphenyl-1,2-epoxyethane with aqueous acid yields diphenyl acetaldehyde as the
18-31 major product. Propose a mechanism for the reaction.

PROBLEM Fluoxetine, a heavily prescribed antidepressant marketed under the name Prozac, can be prepared
18-32 by a route that begins with reaction between a phenol and an alkyl chloride.

(a) The rate of the reaction depends on both phenol and alkyl halide. Is this an $\mathrm{S}_{\mathrm{N}} 1$ or an $\mathrm{S}_{\mathrm{N}} 2$ reaction? Show the mechanism.
(b) The physiologically active enantiomer of fluoxetine has $(S)$ stereochemistry. Based on your answer in part (a), draw the structure of the alkyl chloride you would need, showing the correct stereochemistry.

PROBLEM When 2-methyl-2,5-pentanediol is treated with sulfuric acid, dehydration occurs and
18-33 2,2-dimethyltetrahydrofuran is formed. Suggest a mechanism for this reaction. Which of the two oxygen atoms is most likely to be eliminated, and why?


2,2-Dimethyltetrahydrofuran -
PROBLEM Methyl aryl ethers, such as anisole, are cleaved to iodomethane and a phenoxide ion by treatment 18-34 with LiI in hot DMF. Propose a mechanism for this reaction.

PROBLEM The herbicide acifluorfen can be prepared by a route that begins with reaction between a phenol 18-35 and an aryl fluoride. Propose a mechanism.

PROBLEM Aldehydes and ketones undergo acid-catalyzed reaction with alcohols to yield hemiacetals, from 18-36 aldehydes or ketals with ketones compounds that have one alcohol-like oxygen and one ether-like oxygen bonded to the same carbon. Further reaction of a hemiacetal with alcohol then yields an acetal, a compound that has two ether-like oxygens bonded to the same carbon.

(a) Show the structures of the hemiketal and ketal you would obtain by reaction of cyclohexanone with ethanol.
(b) Propose a mechanism for the conversion of a hemiacetal into a ketal.

PROBLEM Propose a mechanism to account for the following transformation. What two kinds of reactions are 18-37 occurring?


Naming Ethers -
PROBLEM Draw structures corresponding to the following IUPAC names:
18-38 (a) Ethyl 1-ethylpropyl ether (b) Di(p-chlorophenyl) ether (c) 3,4-Dimethoxybenzoic acid
(d) Cyclopentyloxycyclohexane (e) 4-Allyl-2-methoxyphenol (eugenol; from oil of cloves)

PROBLEM Give IUPAC names for the following structures:
18-39

Synthesizing Ethers
PROBLEM How would you prepare the following ethers?
18-40 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM How would you prepare the following compounds from 1-phenylethanol?
18-41
(a) Methyl 1-phenylethyl ether
(b) Phenylepoxyethane
(c) tert-Butyl 1-phenylethyl ether
(d) 1-Phenylethanethiol

PROBLEM tert-Butyl ethers can be prepared by the reaction of an alcohol with 2 -methylpropene in the 18-42 presence of an acid catalyst. Propose a mechanism for this reaction.

PROBLEM Treatment of trans-2-chlorocyclohexanol with NaOH yields 1,2-epoxycyclohexane, but reaction of
18-43 the cis isomer under the same conditions yields cyclohexanone. Propose mechanisms for both reactions, and explain why the different results are obtained.

$\xrightarrow[\mathrm{H}_{2} \mathrm{O}]{\mathrm{NaOH}}$

Reactions of Ethers and Epoxides
PROBLEM Predict the products of the following ether cleavage reactions:

18-44 (a)

(c)

(b)

(d)

PROBLEM How would you carry out the following transformations? More than one step may be required.
18-45 (a)

(b)

(c)

(d)

(e)

PROBLEM What product would you expect from cleavage of tetrahydrofuran with HI ? 18-46


Naming Ethers -
(b)

(c)

(d)

(e)

PROBLEM What product would you expect from cleavage of tetrahydrofuran with HI ? 18-46

PROBLEM Write the mechanism of the hydrolysis of cis-5,6-epoxydecane by reaction with aqueous acid. What 18-47 is the stereochemistry of the product, assuming normal backside $\mathrm{S}_{\mathrm{N}} 2$ attack?

PROBLEM What is the stereochemistry of the product from acid-catalyzed hydrolysis of 18-48 trans-5,6-epoxydecane? How does the product differ from that formed in Problem 18-47?

PROBLEM Acid-catalyzed hydrolysis of a 1,2-epoxycyclohexane produces a trans-diaxial 1,2-diol. What 18-49 product would you expect to obtain from acidic hydrolysis of cis-3-tert-butyl-1,2-epoxycyclohexane? (Recall that the bulky tert-butyl group locks the
cyclohexane ring into a specific conformation.)
PROBLEM Imagine that you have treated ( $2 R, 3 R$ )-2,3-epoxy-3-methylpentane with aqueous acid to carry out a 18-50 ring-opening reaction.


2,3-Epoxy-3-methylpentane (no stereochemistry implied) -
(a) Draw the epoxide, showing stereochemistry.
(b) Draw and name the product, showing stereochemistry. (c) Is the product chiral? Explain.
(d) Is the product optically active? Explain.

PROBLEM Epoxides are reduced by treatment with lithium aluminum hydride to yield alcohols. Propose a
18-51 mechanism for this reaction.

PROBLEM Show the structure and stereochemistry of the alcohol that would result if 1,2-epoxycyclohexane
18-52 were reduced with lithium aluminum deuteride, $\mathrm{LiAlD}_{4}$ (Problem 18-51).


Spectroscopy -
PROBLEM The red fox (Vulpes vulpes) uses a chemical communication system based on scent marks in urine.
18-53 One component of fox urine is a sulfide whose mass spectrum has $\mathrm{M}^{+}=116$. IR spectroscopy shows an intense band at $890 \mathrm{~cm}^{-1}$, and ${ }^{1} \mathrm{H}$ NMR spectroscopy reveals the following peaks:
$1.74 \delta(3 \mathrm{H}$, singlet $) ; 2.11 \delta(3 \mathrm{H}$, singlet $) ; 2.27 \delta(2 \mathrm{H}$, triplet, $J=4.2 \mathrm{~Hz}) ; 2.57 \delta(2 \mathrm{H}$, triplet, $J=$ $4.2 \mathrm{~Hz}) ; 4.73 \delta(2 \mathrm{H}$, broad)

Propose a structure consistent with these data. [Note: $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}$ absorbs at $2.1 \mathrm{\delta}$ ].
PROBLEM Anethole, $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}$, a major constituent of the oil of anise, has the ${ }^{1} \mathrm{H}$ NMR spectrum shown. On
18-54 careful oxidation with $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$, anethole yields $p$-methoxybenzoic acid. What is the structure of anethole? Assign all peaks in the NMR spectrum, and account for the observed splitting patterns.

PROBLEM Propose structures for compounds that have the following ${ }^{1} \mathrm{H}$ NMR spectra:
18-55
(a) $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{~S}(\mathrm{An}-\mathrm{SH}$ proton absorbs near 1.6 ס.)

(b) $\mathrm{C}_{9} \mathrm{H}_{11} \mathrm{BrO}$

(c) $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}_{2}$


General Problems -
PROBLEM Predict the products of the following reactions:


18-56 (a) -
$\xrightarrow{\mathrm{HBr}}$
(b)

?

c)

(d)
?

PROBLEM How would you synthesize anethole (Problem 18-54) from phenol?
18-57
PROBLEM How could you prepare benzyl phenyl ether from benzene and phenol? More than one step is
18-58 required.
PROBLEM Meerwein's reagent, triethyloxonium tetrafluoroborate, is a powerful ethylating agent that converts
18-59 alcohols into ethyl ethers at neutral pH . Show the reaction of Meerwein's reagent with cyclohexanol, and account for the fact that trialkyloxonium salts are much more reactive alkylating agents than alkyl iodides.


$\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{O}^{+} \mathrm{BF}_{4}^{-} \quad$ Meerwein's reagent -
PROBLEM Safrole, a substance isolated from oil of sassafras, is used as a perfumery agent. Propose a synthesis
18-60 of safrole from catechol (1,2-benzenediol).


Safrole -
PROBLEM Grignard reagents react with oxetane, a four-membered cyclic ether, to yield primary alcohols, but
18-61 the reaction is much slower than the corresponding reaction with ethylene oxide. Suggest a reason for the difference in reactivity between oxetane and ethylene oxide.

$$
\square \mathrm{O} \xrightarrow[\text { 2. } \mathrm{H}_{3} \mathrm{O}^{+}]{\text {1. } \mathrm{RMgX}} \mathrm{RCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}
$$

Oxetane
PROBLEM The Zeisel method is an old analytical procedure for determining the number of methoxyl groups
18-62 in a compound. A weighed amount of the compound is heated with concentrated HI, ether cleavage occurs, and the iodomethane product is distilled off and passed into an alcohol solution of $\mathrm{AgNO}_{3}$, where it reacts to form a precipitate of silver iodide. The AgI is then collected and weighed, and the percentage of methoxyl groups in the sample is thereby determined. For example, 1.06 g of vanillin, the material responsible for the characteristic odor of vanilla, yields 1.60 g of AgI. If vanillin has a molecular weight of 152, how many methoxyl groups does it contain?


Safrole -
PROBLEM Disparlure, $\mathrm{C}_{19} \mathrm{H}_{38} \mathrm{O}$, is a sex attractant released by the female spongy moth, Lymantria dispar.
18-63 The ${ }^{1} \mathrm{H}$ NMR spectrum of disparlure shows a large absorption in the alkane region, 1 to $2 \delta$, and a triplet at $2.8 \delta$. Treatment of disparlure, first with aqueous acid and then with $\mathrm{KMnO}_{4}$, yields two carboxylic acids identified as undecanoic acid and 6-methylheptanoic acid. ( $\mathrm{KMnO}_{4}$ cleaves 1,2-diols to yield carboxylic acids.) Neglecting stereochemistry, propose a structure for disparlure. The actual compound is a chiral molecule with $7 R, 8 S$ stereochemistry. Draw disparlure, showing the correct stereochemistry.

PROBLEM How would you synthesize racemic disparlure (Problem 18-63) from compounds having ten or
18-64 fewer carbons?
PROBLEM How would you prepare o-hydroxyphenylacetaldehyde from phenol? More than one step is
18-65 required.

o-Hydroxyphenylacetaldehyde

PROBLEM Identify the reagents a-e in the following scheme:


18-66 -
PROBLEM Propose structures for compounds that have the following ${ }^{1} \mathrm{H}$ NMR spectra:
18-67 (a) $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}_{2}$

(b) $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}$

PROBLEM We saw in Section 17.4 that ketones react with $\mathrm{NaBH}_{4}$ to yield alcohols. We'll also see in Section
18-68 22.3 that ketones react with $\mathrm{Br}_{2}$ to yield $\alpha$-bromo ketones. Perhaps surprisingly, treatment with $\mathrm{NaBH}_{4}$ of the $\alpha$-bromo ketone from acetophenone yields an epoxide rather than a bromo alcohol. Show the structure of the epoxide, and explain its formation.


Preview of Carbonyl Chemistry -
Carbonyl compounds are everywhere. Most biological molecules contain carbonyl groups, as do most pharmaceutical agents and many of the synthetic chemicals that affect our everyday lives. Citric acid, found in lemons and oranges; acetaminophen, the active ingredient in many over-the-counter headache remedies; and Dacron, the polyester material used in clothing, all contain different kinds of carbonyl groups.

Citric acid
(a carboxylic acid)

Acetaminophen (an amide)

Dacron
(a polyester)

To a great extent, the chemistry of living organisms is the chemistry of carbonyl compounds. Thus, we'll spend the next five chapters discussing the chemistry of the carbonyl group, $\mathbf{C}=\mathbf{O}$ (pronounced car-bo-neel). There are many different kinds of carbonyl compounds and many different reactions, but there are only a few fundamental principles that tie the entire field together. The purpose of this brief preview is not to show details of specific reactions but rather to provide a framework for learning carbonyl-group chemistry. Read through this preview now, and return to it on occasion to remind yourself of the larger picture.


I Kinds of Carbonyl Compounds -
TABLE 18.2 shows some of the many different kinds of carbonyl compounds. All contain an acyl group ( $\mathrm{R}-\mathrm{C}=\mathrm{O}$ ) bonded to another substituent. The R part of the acyl group can be practically any organic part/ structure, and the other substituent to which the acyl group is bonded might be a carbon, hydrogen, oxygen, halogen, nitrogen, or sulfur.

| Name | General formula | Name ending |
| :---: | :---: | :---: |
| Aldehyde | <smiles>[R]C=O</smiles> | -al |
| Ketone | <smiles>[R]C([R])=O</smiles> | -one |
| Carboxylic acid | <smiles>[R]C(=O)O</smiles> | -oic acid |
| Acid halide | <smiles>[R]C([X])=O</smiles> | -yl or -oyl halide |
| Acid anhydride | <smiles>[R]C([R])=O</smiles> | -oic anhydride |
| Acyl phosphate | <smiles>[R]C(=O)OP(=O)([O-])[O-]</smiles> | -yl phosphate |

TABLE 18.2 Some Types of Carbonyl Compounds

| Name | General formula | Name ending |
| :---: | :---: | :---: |
| Ester | <smiles>[R]OC([R])=O</smiles> | -oate |
| Lactone (cyclic ester) | <smiles>CC1(C)CCCOC1=O</smiles> | None |
| Thioester | <smiles>[R]SC([R])=O</smiles> | -thioate |
| Amide | <smiles>[R]C(=O)N(C)C</smiles> | - amide |
| Lactam (cyclic amide) | <smiles>CN1CCCC(C)(C)C1=O</smiles> | None |


I Kinds of Carbonyl Compounds -
It's useful to classify carbonyl compounds into two categories based on the kinds of chemistry they undergo. In one category are aldehydes and ketones; in the other are carboxylic acids and their derivatives. The acyl group in an aldehyde or ketone is bonded to an atom (H or C, respectively) that can't stabilize a negative charge and therefore can't act as a leaving group in a nucleophilic substitution reaction. The acyl group in a carboxylic acid or its derivative, however, is bonded to an atom (oxygen, halogen, sulfur, nitrogen) that can stabilize a negative charge and therefore can act as a leaving group in a nucleophilic substitution reaction.

Aldehyde

Ketone

The $-\mathrm{R}^{\prime}$ and -H in these compounds can't act as leaving groups in nucleophilic substitution reactions.

Carboxylic acid

Amide

Acid halide

Ester

Acid
anhydride

Acyl phosphate

Thioester
The $-\mathrm{OH},-\mathrm{X},-\mathrm{OR}^{\prime},-\mathrm{SR},-\mathrm{NH}_{2}$, $-\mathrm{OCOR}^{\prime}$, and $-\mathrm{OPO}_{3}{ }^{2-}$ in these compounds can act as leaving groups in nucleophilic substitution reactions.


II Nature of the Carbonyl Group -
The carbon-oxygen double bond of a carbonyl group is similar in many respects to the carbon-carbon double bond of an alkene. The carbonyl carbon atom is $s p^{2}$-hybridized and forms three $\sigma$ bonds. The fourth valence electron remains in a carbon $p$ orbital and forms a $\pi$ bond to oxygen by overlapping with an oxygen $p$ orbital.

The oxygen atom also has two nonbonding pairs of electrons, which occupy its remaining two orbitals.

Like alkenes, carbonyl compounds are planar about the double bond and have bond angles of approximately $120^{\circ}$. FIGURE 18.8 shows the structure of acetaldehyde and indicates its bond lengths and angles. As you might expect, the carbon-oxygen double bond is both shorter ( 122 pm versus 143 pm ) and stronger [ $732 \mathrm{~kJ} / \mathrm{mol}$ (175 $\mathrm{kcal} / \mathrm{mol}$ ) versus $385 \mathrm{~kJ} / \mathrm{mol}$ ( $92 \mathrm{kcal} / \mathrm{mol}$ )] than a C-O single bond.

| Bond angle | $\left.\mathbf{(}^{\circ}\right)$ | Bond length | (pm) |
| :--- | :---: | :---: | :---: |
| $\mathrm{H}-\mathrm{C}-\mathrm{C}$ | 118 | $\mathrm{C}=\mathrm{O}$ | 122 |
| $\mathrm{C}-\mathrm{C}=\mathrm{O}$ | 121 | $\mathrm{C}-\mathrm{C}$ | 150 |
| $\mathrm{H}-\mathrm{C}=\mathrm{O}$ | 121 | $\mathrm{OC}-\mathrm{H}$ | 109 |


II Nature of the Carbonyl Group -
FIGURE 18.8 Structure of acetaldehyde.
As indicated by the electrostatic potential map in FIGURE 18.8, the carbon-oxygen double bond is strongly polarized because of the high electronegativity of oxygen relative to carbon. Thus, the carbonyl carbon atom carries a partial positive charge, is an electrophilic (Lewis acidic) site, and reacts with nucleophiles. Conversely, the carbonyl oxygen atom carries a partial negative charge, is a nucleophilic (Lewis basic) site, and reacts with electrophiles. We'll see in the next five chapters that the majority of carbonyl-group reactions can be rationalized by simple polarity arguments.


III General Reactions of Carbonyl Compounds -
Both in the laboratory and in living organisms, most reactions of carbonyl compounds take place by one of four general mechanisms: nucleophilic addition, nucleophilic acyl substitution, alpha substitution, and carbonyl condensation. These mechanisms have many variations, just as alkene electrophilic addition reactions and $\mathrm{S}_{\mathrm{N}} 2$ reactions do, but the variations are much easier to learn when the fundamental features of the mechanisms are made clear. Let's see what the four mechanisms are and what kinds of chemistry carbonyl compounds undergo.


Nucleophilic Addition Reactions of Aldehydes and Ketones (Chapter 19) -
The most common reaction of aldehydes and ketones is the nucleophilic addition reaction, in which a nucleophile, : $\mathrm{Nu}^{-}$, adds to the electrophilic carbon of the carbonyl group. Because the nucleophile uses an electron pair to form a new bond to carbon, two electrons from the carbon-oxygen double bond must move toward the electronegative oxygen atom to give an alkoxide anion. The carbonyl carbon rehybridizes from $s p^{2}$ to $s p^{3}$ during the reaction, and the alkoxide ion product therefore has tetrahedral geometry.

A carbonyl compound
A tetrahedral intermediate
( $s p^{3}$-hybridized carbon)

Once formed, and depending on the nature of the nucleophile, the tetrahedral alkoxide intermediate can undergo one of two further reactions, as shown in FIGURE 18.9. Often, the tetrahedral alkoxide intermediate is simply protonated by water or acid to form an alcohol product. Alternatively, the tetrahedral intermediate
can be protonated and expel the oxygen to form a new double bond between the carbonyl carbon and the nucleophile. We'll study both processes in detail in Chapter 19.

FIGURE 18.9 The addition reaction of an aldehyde or a ketone with a nucleophile. Depending on the nucleophile, either an alcohol or a compound with a $\mathrm{C}=\mathrm{Nu}$ double bond is formed.


FORMATION OF AN ALCOHOL -
The simplest reaction of a tetrahedral alkoxide intermediate is protonation to yield an alcohol. We've already seen two examples of this kind of process during reduction of aldehydes and ketones with hydride reagents such as $\mathrm{NaBH}_{4}$ and $\mathrm{LiAlH}_{4}$ (Section 17.4) and during Grignard reactions (Section 17.5). During a reduction, the nucleophile that adds to the carbonyl group is a hydride ion, $\mathrm{H}:^{-}$, while during a Grignard reaction, the nucleophile is a carbanion, $\mathrm{R}_{3} \mathrm{C}:^{-}$.

Ketone/ aldehyde

Tetrahedral intermediate

Alcohol


FORMATION OF $\mathbf{C = N u}$ -
The second mode of nucleophilic addition, which often occurs with amine nucleophiles, involves elimination of oxygen and formation of a $\mathrm{C}=\mathrm{Nu}$ double bond. For example, aldehydes and ketones react with primary amines, $\mathrm{RNH}_{2}$, to form imines, $\mathrm{R}_{2} \mathrm{C}=\mathrm{NR}^{\prime}$. These reactions use the same kind of tetrahedral intermediate as that formed during hydride reduction and Grignard reaction, but the initially formed alkoxide ion is not isolated. Instead, it is protonated and then loses water to form an imine, as shown in FIGURE 18.10.

FIGURE 18.10 MECHANISM
Formation of an imine, $\mathrm{R}_{\mathbf{2}} \mathrm{C}=\mathrm{NR}^{\prime}$, by reaction of an amine with an aldehyde or a ketone.
(1) Addition to the ketone or aldehyde carbonyl group by the neutral amine nucleophile gives a dipolar tetrahedral intermediate.

(1)

(2) Transfer of a proton from nitrogen to oxygen then yields an amino alcohol intermediate.

(3) Dehydration of the amino alcohol intermediate gives neutral imine and water as final products.


Nucleophilic Acyl Substitution Reactions of Carboxylic Acid Derivatives (Chapter 21) -
The second fundamental reaction of carbonyl compounds, nucleophilic acyl substitution, is related to the nucleophilic addition reaction just discussed but occurs only with carboxylic acid derivatives rather than with aldehydes and ketones. When the carbonyl group of a carboxylic acid derivative reacts with a nucleophile, addition occurs in the usual way, but the initially formed tetrahedral alkoxide intermediate is not isolated. Because carboxylic acid derivatives have a leaving group bonded to the carbonyl-group carbon, the tetrahedral intermediate can react further by expelling the leaving group and forming a new carbonyl compound:

The net effect of nucleophilic acyl substitution is the replacement of the leaving group by the entering nucleophile. We'll see in Chapter 21, for instance, that acid chlorides are rapidly converted into esters by treatment with alkoxide ions (FIGURE 18.11).


FIGURE 18.11 MECHANISM -
Nucleophilic acyl substitution of an acid chloride with an alkoxide ion yields an ester.

1 Nucleophilic addition of alkoxide ion to an acid chloride yields a tetrahedral intermediate.


© -
(2) An electron pair from oxygen expels chloride ion and yields the substitution product, an ester.


Alpha-Substitution Reactions (Chapter 22) -
The third major reaction of carbonyl compounds, alpha substitution, occurs at the position next to the carbonyl group-the alpha ( $\alpha$ ) position. This reaction results in the substitution of an $\alpha$ hydrogen by an electrophile through the formation of an intermediate enol or enolate ion:

An enol
For reasons that we'll explore in Chapter 22, the presence of a carbonyl group renders the hydrogens on the $\alpha$ carbon acidic. Carbonyl compounds therefore react with strong base to yield enolate ions.

Because they're negatively charged, enolate ions act as nucleophiles and undergo many of the reactions we've already studied. For example, enolates react with primary alkyl halides in the $\mathrm{S}_{\mathrm{N}} 2$ reaction. The nucleophilic enolate ion displaces halide ion, and a new $\mathrm{C}-\mathrm{C}$ bond forms:

The $\mathrm{S}_{\mathrm{N}} 2$ alkylation reaction between an enolate ion and an alkyl halide is a powerful method for making $\mathrm{C}-\mathrm{C}$ bonds, thereby building up larger molecules from smaller precursors. We'll study the alkylation of many kinds of carbonyl compounds in Chapter 22.


Carbonyl Condensation Reactions (Chapter 23) -
The fourth and last fundamental reaction of carbonyl groups, carbonyl condensation, takes place when two carbonyl compounds react with each other. When acetaldehyde is treated with base, for instance, two molecules combine to yield the hydroxy aldehyde product known as aldol (aldehyde + alcohol):

Two acetaldehydes
Aldol
Although carbonyl condensation appears to be different from the three processes already discussed, it's actually quite similar. A carbonyl condensation reaction is simply a combination of a nucleophilic addition step and an $\alpha$-substitution step. The initially formed enolate ion of one acetaldehyde molecule acts as a nucleophile and adds to the carbonyl group of another acetaldehyde molecule, as shown in FIGURE 18.12.


FIGURE 18.12 MECHANISM -
A carbonyl condensation reaction between two molecules of acetaldehyde yields a hydroxy aldehyde product.
(1) Base abstracts an acidic alpha hydrogen from one acetaldehyde molecule, yielding a resonancestabilized enolate ion.

The enolate ion adds as a nucleophile to the carbonyl group of a second acetaldehyde, producing a tetrahedral alkoxide ion.

(2) $\|$

Tetrahedral intermediate
3 The tetrahedral intermediate is protonated by solvent to yield the neutral aldol product and regenerate the base catalyst.


IV Summary -
To a great extent, the chemistry of living organisms is the chemistry of carbonyl compounds. We have not looked at the details of specific carbonyl reactions in this short preview but rather have laid the groundwork for the next five chapters. All the carbonyl-group reactions we'll be studying in Chapters 19 through 23 fall into one of the four fundamental categories discussed in this preview. Knowing where we'll be heading should help you keep matters straight in understanding this most important of all functional groups.

PROBLEM Judging from the following electrostatic potential maps, which kind of carbonyl compound has
18-69 the more electrophilic carbonyl carbon atom, a ketone or an acid chloride? Which has the more nucleophilic carbonyl oxygen atom? Explain.

PROBLEM Predict the product formed by nucleophilic addition of cyanide ion ( $\mathrm{CN}^{-}$) to the carbonyl group of 18-70 acetone, followed by protonation to give an alcohol:

PROBLEM Identify each of the following reactions as a nucleophilic addition, nucleophilic acyl substitution, an 18-71 $\alpha$ substitution, or a carbonyl condensation:
(a)

(b)

(c)


CHAPTER 19 -
Aldehydes and Ketones: Nucleophilic Addition Reactions

FIGURE 19.1 Few flowers are more beautiful or more fragrant than roses. Their perfumed odor is due to several simple organic compounds, including the ketone $\beta$-damascenone. (credit: modification of work "Rosa 'Precious Platinum'" by Bernard Spragg/Wikimedia Commons, CC BY 1.0)


CHAPTER CONTENTS - 19.1 Naming Aldehydes and Ketones
19.2 Preparing Aldehydes and Ketones
19.3 Oxidation of Aldehydes and Ketones
19.4 Nucleophilic Addition Reactions of Aldehydes and Ketones
19.5 Nucleophilic Addition of $\mathrm{H}_{2} \mathrm{O}$ : Hydration
19.6 Nucleophilic Addition of HCN: Cyanohydrin Formation
19.7 Nucleophilic Addition of Hydride and Grignard Reagents: Alcohol Formation
19.8 Nucleophilic Addition of Amines: Imine and Enamine Formation
19.9 Nucleophilic Addition of Hydrazine: The Wolff-Kishner Reaction
19.10 Nucleophilic Addition of Alcohols: Acetal Formation
19.11 Nucleophilic Addition of Phosphorus Ylides: The Wittig Reaction
19.12 Biological Reductions
19.13 Conjugate Nucleophilic Addition to $\alpha, \beta$-Unsaturated Aldehydes and Ketones
19.14 Spectroscopy of Aldehydes and Ketones

WHY THIS CHAPTER? Much of organic chemistry is the chemistry of carbonyl compounds. Aldehydes and ketones, in particular, are intermediates in the synthesis of many pharmaceutical agents, in almost all biological
pathways, and in numerous industrial processes, so an understanding of their properties and reactions is essential. In this chapter, we'll look at some of their most important reactions.

Aldehydes (RCHO) and ketones ( $\mathbf{R}_{\mathbf{2}} \mathbf{C O}$ ) are among the most widely occurring of all compounds. In nature, many substances required by living organisms are aldehydes or ketones. The aldehyde pyridoxal phosphate, for instance, is a coenzyme involved in a large number of metabolic reactions; the ketone hydrocortisone is a steroid hormone secreted by the adrenal glands to regulate fat, protein, and carbohydrate metabolism.

Pyridoxal phosphate (PLP)

Hydrocortisone


CHAPTER CONTENTS - 19.1 Naming Aldehydes and Ketones
Pyridoxal phosphate (PLP)

Hydrocortisone

In the chemical industry, simple aldehydes and ketones are produced in large quantities for use as solvents and as starting materials to prepare a host of other compounds. For example, more than 50 million tons per year of formaldehyde, $\mathrm{H}_{2} \mathrm{C}=\mathrm{O}$, is produced worldwide for use in building insulation materials and in the adhesive resins that bind particle board and plywood. Acetone, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}$, is widely used as an industrial solvent, with approximately 8 million tons per year produced worldwide. Formaldehyde is synthesized industrially by catalytic oxidation of methanol, and one method of acetone preparation involves oxidation of 2-propanol produced by hydration of propylene.

Acetone


CHAPTER CONTENTS - 19.1 Naming Aldehydes and Ketones
Aldehydes are named by replacing the terminal -e of the corresponding alkane name with -al. The parent chain must contain the - CHO group, and the - CHO carbon is numbered as carbon 1 . Note in the following examples that the longest chain in 2-ethyl-4-methylpentanal is actually a hexane, but this chain does not include the -CHO group and thus is not the parent.

For cyclic aldehydes in which the -CHO group is directly attached to a ring, the ring name followed by carbaldehyde is used.

Cyclohexanecarbaldehyde

2-Naphthalenecarbaldehyde

A few simple and well-known aldehydes have common names that are recognized by IUPAC. Several that you might encounter are listed in TABLE 19.1.

TABLE 19.1 Common Names of Some Simple Aldehydes

| Formula | Common name | Systematic name |
| :--- | :--- | :--- |
| HCHO | Formaldehyde | Methanal |
| $\mathrm{CH}_{3} \mathrm{CHO}$ | Acetaldehyde | Ethanal |
| $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCHO}$ | Acrolein | Propenal |
| $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCHO}$ | Crotonaldehyde | 2-Butenal |
|  | Benzaldehyde | Benzenecarbaldehyde |

Ketones are named by replacing the terminal -e of the corresponding alkane name with -one. The parent chain is the longest one that includes the ketone group, and the numbering begins at the end nearer the carbonyl carbon. As with alkenes (Section 7.3) and alcohols (Section 17.1), the locant is placed before the parent name using older rules but before the suffix with the newer IUPAC guidelines. For example:

A few ketones are allowed by IUPAC to retain their common names.

Acetophenone

Benzophenone


CHAPTER CONTENTS - 19.1 Naming Aldehydes and Ketones
A few ketones are allowed by IUPAC to retain their common names.

Acetophenone

Benzophenone

When it's necessary to refer to the $\mathrm{R}-\mathrm{C}=\mathrm{O}$ group as a substituent, the name acyl (a-sil) group is used and the name ending $-y l$ is attached. Thus, $-\mathrm{COCH}_{3}$ is an acetyl group, -CHO is a formyl group, -COAr is an aroyl group, and $-\mathrm{COC}_{6} \mathrm{H}_{5}$ is a benzoyl group.

An acyl group

Acetyl

Formyl

Benzoyl

If other functional groups are present and the double-bonded oxygen is considered a substituent on a parent chain, the prefix oxo- is used. For example:

PROBLEM Name the following aldehydes and ketones:
19-1 (a)

(b)

(c)

(d)

(e)

f)

PROBLEM Draw structures corresponding to the following names:
19-2 (a) 3-Methylbutanal (b) 4-Chloro-2-pentanone (c) Phenylacetaldehyde
(d) cis-3-tert-Butylcyclohexanecarbaldehyde
(e) 3-Methyl-3-butenal
(f) 2-(1-Chloroethyl)-5-methylheptanal


Preparing Aldehydes -
Perhaps the best method of aldehyde synthesis is by oxidation of a primary alcohol, as we saw in Section 17.7. The reaction is usually carried out using the Dess-Martin periodinane reagent in dichloromethane solvent at room temperature:

Another method of aldehyde synthesis is one that we'll mention here just briefly and then return to in Section 21.6. Certain carboxylic acid derivatives can be partially reduced to yield aldehydes. The partial reduction of an ester by diisobutylaluminum hydride (DIBAH, or DIBAL-H), for instance, is an important laboratory-scale method of aldehyde synthesis, and mechanistically related processes also occur in biological pathways. The reaction is normally carried out at $-78^{\circ} \mathrm{C}$ (dry-ice temperature) in toluene solution.


Methyl dodecanoate -
Dodecanal (88\%)

PROBLEM How would you prepare pentanal from the following starting materials?
19-3 (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$
(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{3}$
(d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$


Preparing Ketones -
For the most part, methods of ketone synthesis are similar to those for aldehydes. Secondary alcohols are
oxidized by a variety of reagents to give ketones (Section 17.7). The choice of oxidant depends on such factors as reaction scale, cost, and acid or base sensitivity of the alcohol. The Dess-Martin periodinane is a common choice although chromic acid $\left(\mathrm{H}_{2} \mathrm{CrO}_{4}\right)$ was frequently used in older work.

Other methods include the ozonolysis of alkenes in which one of the unsaturated carbon atoms is disubstituted (Section 8.8) and Friedel-Crafts acylation of an aromatic ring with an acid chloride in the presence of $\mathrm{AlCl}_{3}$ catalyst (Section 16.3).

In addition to those methods already discussed, ketones can also be prepared from certain carboxylic acid derivatives, just as aldehydes can. Among the most useful reactions of this sort is that between an acid chloride and a lithium diorganocopper reagent, as we saw in Section 10.7. We'll discuss this reaction in more detail in Section 21.4.

Hexanoyl chloride
2-Heptanone (81\%)
PROBLEM How would you carry out the following reactions? More than one step may be needed.
19-4 (a) 3-Hexyne $\rightarrow 3$-Hexanone (b) Benzene $\rightarrow$ m-Bromoacetophenone
(c) Bromobenzene $\rightarrow$ Acetophenone (d) 1-Methylcyclohexene $\rightarrow$ 2-Methylcyclohexanone


Preparing Ketones - 19.3 Oxidation of Aldehydes and Ketones
Aldehydes are easily oxidized to yield carboxylic acids, but ketones are generally inert toward oxidation. The difference is a consequence of structure: aldehydes have a -CHO proton that can be abstracted during oxidation, but ketones do not.

Many oxidizing agents, including alkaline $\mathrm{KMnO}_{4}$ and hot $\mathrm{HNO}_{3}$, convert aldehydes into carboxylic acids. The oxidation occurs rapidly at room temperature and generally works well.

Hexanal
Hexanoic acid (85\%)
Aldehyde oxidations occur through intermediate 1,1-diols, or hydrates, which are formed by a reversible nucleophilic addition of water to the carbonyl group. Even though it's formed to only a small extent at equilibrium, the hydrate reacts like any typical primary or secondary alcohol and is oxidized to a carbonyl compound (Section 17.7).

An aldehyde

A hydrate

A carboxylic acid

Ketones are inert to most oxidizing agents but undergo a slow cleavage reaction of the $\mathrm{C}-\mathrm{C}$ bond next to the carbonyl group when treated with hot alkaline $\mathrm{KMnO}_{4}$. The reaction is not often used, however, and is mentioned here only for completeness.

Cyclohexanone Hexanedioic acid (79\%)


Preparing Ketones - 19.4 Nucleophilic Addition Reactions of Aldehydes and Ketones
As we saw in the Preview of Carbonyl Chemistry, the most general reaction of aldehydes and ketones is the nucleophilic addition reaction. As shown in FIGURE 19.2, a nucleophile, : $\mathrm{Nu}^{-}$, approaches the carbonyl group from an angle of about $105^{\circ}$ opposite the carbonyl oxygen and forms a bond to the electrophilic $\mathrm{C}-\mathrm{O}$ carbon atom. At the same time, rehybridization of the carbonyl carbon from $s p^{2}$ to $s p^{3}$ occurs, an electron pair from the $\mathrm{C}=\mathrm{O}$ bond moves toward the electronegative oxygen atom, and a tetrahedral alkoxide ion intermediate is produced. Protonation of the alkoxide by addition of acid then gives an alcohol.


FIGURE 19.2 MECHANISM -
A nucleophilic addition reaction to an aldehyde or ketone. The nucleophile approaches the carbonyl group from an angle of approximately $75^{\circ}$ to the plane of the $s p^{2}$ orbitals, the carbonyl carbon rehybridizes from $s p^{2}$ to $s p^{3}$, and an alkoxide ion is formed. Protonation by addition of acid then gives an alcohol.
(1) An electron pair from the nucleophile adds to the electrophilic carbon of the carbonyl group, pushing an electron pair from the $\mathrm{C}=\mathrm{O}$ bond onto oxygen and giving an alkoxide ion intermediate. The carbonyl carbon rehybridizes from $s p^{2}$ to $s p^{3}$.

Aldehyde or ketone

Alkoxide ion
(2) $\mathrm{H}_{3} \mathrm{O}^{+}$

Alcohol

The nucleophile can be either negatively charged (: $\mathrm{Nu}^{-}$) or neutral (: Nu ). If it's neutral, however, it usually carries a hydrogen atom that can subsequently be eliminated, : Nu-H. For example:

Nucleophilic additions to aldehydes and ketones have two general variations, as shown in FIGURE 19.3. In one variation, the tetrahedral intermediate is protonated by water or acid to give an alcohol as the final product. In the second variation, the carbonyl oxygen atom is protonated and then eliminated as $\mathrm{HO}^{-}$or $\mathrm{H}_{2} \mathrm{O}$ to give a product with a $\mathrm{C}=\mathrm{Nu}$ double bond.

FIGURE 19.3 Two general reaction pathways following addition of a nucleophile to an aldehyde or ketone. The top pathway leads to an alcohol product; the bottom pathway leads to a product with a $\mathrm{C}=\mathrm{Nu}$ double bond.


FIGURE 19.2 MECHANISM -
Aldehydes are generally more reactive than ketones in nucleophilic addition reactions for both steric and electronic reasons. Sterically, the presence of only one large substituent bonded to the $\mathrm{C}=\mathrm{O}$ carbon in an aldehyde versus two large substituents in a ketone means that a nucleophile is able to approach an aldehyde more readily. Thus, the transition state leading to the tetrahedral intermediate is less crowded and lower in energy for an aldehyde than for a ketone (FIGURE 19.4).

FIGURE 19.4 Steric hindrance in nucleophilic addition reactions. (a) Nucleophilic addition to an aldehyde is sterically less hindered because only one relatively large substituent is attached to the carbonyl-group carbon. (b) A ketone, however, has two large substituents and is more hindered. The approach of the nucleophile is along the $\mathrm{C}=\mathrm{O}$ bond at an angle of about $75^{\circ}$ to the plane of the carbon $s p^{2}$ orbitals.

Electronically, aldehydes are more reactive than ketones because of the greater polarization of aldehyde carbonyl groups. To see this polarity difference, recall the stability order of carbocations (Section 7.9). A primary carbocation is higher in energy and thus more reactive than a secondary carbocation since it has only one alkyl group inductively stabilizing the positive charge rather than two. In the same way, an aldehyde has only one alkyl group inductively stabilizing the partial positive charge on the carbonyl carbon rather than two, and is a bit more electrophilic, and, therefore, more reactive than a ketone.

Aldehyde
(less stabilization of $\boldsymbol{\delta}+$, more reactive)

$2^{\circ}$ carbocation (more stable, less reactive)

Ketone
(more stabilization of $\boldsymbol{\delta}+$, less reactive)


FIGURE 19.2 MECHANISM -
$2^{\circ}$ carbocation (more stable, less reactive)

Ketone
(more stabilization of $\boldsymbol{\delta}+$, less reactive)

One further comparison: aromatic aldehydes, such as benzaldehyde, are less reactive in nucleophilic addition reactions than aliphatic aldehydes because the electron-donating resonance effect of the aromatic ring makes the carbonyl group less electrophilic. Comparing electrostatic potential maps of formaldehyde and benzaldehyde, for example, shows that the carbonyl carbon atom is less positive (less blue) in the aromatic
aldehyde.

Formaldehyde

Benzaldehyde

PROBLEM Treatment of an aldehyde or ketone with cyanide ion ( ${ }^{-}: \mathrm{C} \equiv \mathrm{N}$ ), followed by protonation of the 19-5 tetrahedral alkoxide ion intermediate, gives a cyanohydrin. Show the structure of the cyanohydrin obtained from cyclohexanone.

PROBLEM $p$-Nitrobenzaldehyde is more reactive toward nucleophilic additions than $p$-methoxybenzaldehyde. 19-6 Explain.


FIGURE 19.2 MECHANISM - 19.5 Nucleophilic Addition of $\mathrm{H}_{2} \mathrm{O}$ : Hydration
Aldehydes and ketones react with water to yield 1,1-diols, or geminal (gem) diols. The hydration reaction is reversible, and a gem diol can eliminate water to regenerate an aldehyde or ketone.

Acetone (99.9\%)
Acetone hydrate (0.1\%)
The position of the equilibrium between a gem diol and an aldehyde or ketone depends on the structure of the carbonyl compound. Equilibrium generally favors the carbonyl compound for steric reasons, but the gem diol is favored for a few simple aldehydes. For example, an aqueous solution of formaldehyde consists of $99.9 \%$ gem diol and $0.1 \%$ aldehyde at equilibrium, whereas an aqueous solution of acetone consists of only about $0.1 \%$ gem diol and 99.9\% ketone.

Formaldehyde (0.1\%) Formaldehyde hydrate (99.9\%)
The hydrate is also favored for $\alpha$-keto carboxylic acids, which have adjacent positively charged carbons that destabilize the keto form and favor the hydrate. Such compounds occur commonly in many biological pathways. Pyruvic acid, which is $60 \%$ hydrate at equilibrium, and $\alpha$-ketoglutaric acid, which is $50 \%$ hydrate, are examples.

The nucleophilic addition of water to an aldehyde or ketone is slow under neutral conditions but is catalyzed by both base and acid. Under basic conditions (FIGURE 19.5a), the nucleophile is negatively charged ( $\mathrm{OH}^{-}$) and uses a pair of its electrons to form a bond to the electrophilic carbon atom of the $\mathrm{C}=\mathrm{O}$ group. At the same time, the $\mathrm{C}=\mathrm{O}$ carbon atom rehybridizes from $s p^{2}$ to $s p^{3}$ and two electrons from the $\mathrm{C}=\mathrm{O} \pi$ bond are pushed onto the oxygen atom, giving an alkoxide ion. Protonation of the alkoxide ion by water then yields a neutral addition product plus regenerated $\mathrm{OH}^{-}$.


FIGURE 19.2 MECHANISM - 19.5 Nucleophilic Addition of $\mathrm{H}_{2} \mathrm{O}$ : Hydration
Under acidic conditions (FIGURE 19.5b), the carbonyl oxygen atom is first protonated by $\mathrm{H}_{3} \mathrm{O}^{+}$to make the carbonyl group more strongly electrophilic. A neutral nucleophile, $\mathrm{H}_{2} \mathrm{O}$, then uses a pair of electrons to bond to the carbon atom of the $\mathrm{C}=\mathrm{O}$ group, and two electrons from the $\mathrm{C}=\mathrm{O} \pi$ bond move onto the oxygen atom. The positive charge on oxygen is thereby neutralized, while the nucleophile gains a positive charge. Finally, deprotonation by water gives the neutral addition product and regenerates the $\mathrm{H}_{3} \mathrm{O}^{+}$catalyst.

Note the key difference between the base-catalyzed and acid-catalyzed reactions. The base-catalyzed reaction takes place rapidly because water is converted into hydroxide ion, a much better nucleophile. The acidcatalyzed reaction takes place rapidly because the carbonyl compound is converted by protonation into a much better electrophile.


FIGURE 19.5 MECHANISM -
The mechanism for a nucleophilic addition reaction of aldehydes and ketones under both basic and acidic conditions. (a) Under basic conditions, a negatively charged nucleophile adds to the carbonyl group to give an alkoxide ion intermediate, which is subsequently protonated. (b) Under acidic conditions, protonation of the carbonyl group occurs first, followed by addition of a neutral nucleophile and subsequent deprotonation.


(a) Basic conditions -
(1) The negatively charged nucleophile $\mathrm{OH}^{-}$adds to the electrophilic carbon and pushes $\pi$ electrons from the $\mathrm{C}=0$ bond onto oxygen, giving an alkoxide ion.
(2) The alkoxide ion is protonated by water to give the neutral hydrate as the addition product and regenerating $\mathrm{OH}^{-}$.

Hydrate (gem diol)
(b) Acidic conditions
(1) The carbonyl oxygen is protonated by acid $\mathrm{H}_{3} \mathrm{O}^{+}$, making the carbon more strongly electrophilic.
oxygen. The oxygen becomes neutral, and the nucleophile gains the + charge.
(3) Water deprotonates the intermediate, giving the neutral hydrate addition product and regenerating the acid catalyst $\mathrm{H}_{3} \mathrm{O}^{+}$.

Hydrate (gem diol)

The hydration reaction just described is typical of what happens when an aldehyde or ketone is treated with a nucleophile of the type $\mathrm{H}-\mathrm{Y}$, where the Y atom is electronegative and can stabilize a negative charge (oxygen, halogen, or sulfur, for instance). In such reactions, the nucleophilic addition is reversible, with the equilibrium generally favoring the carbonyl reactant rather than the tetrahedral addition product. In other words, treatment of an aldehyde or ketone with $\mathrm{CH}_{3} \mathrm{OH}, \mathrm{H}_{2} \mathrm{O}, \mathrm{HCl}, \mathrm{HBr}$, or $\mathrm{H}_{2} \mathrm{SO}_{4}$ does not normally lead to a stable alcohol addition product.

PROBLEM When dissolved in water, trichloroacetaldehyde exists primarily as its hydrate, called chloral 19-7 hydrate. Show the structure of chloral hydrate.


(a) Basic conditions -
PROBLEM When dissolved in water, trichloroacetaldehyde exists primarily as its hydrate, called chloral 19-7 hydrate. Show the structure of chloral hydrate.

PROBLEM The oxygen in water is primarily (99.8\%) ${ }^{16} \mathrm{O}$, but water enriched with the heavy isotope ${ }^{18} \mathrm{O}$ is 19-8 also available. When an aldehyde or ketone is dissolved in ${ }^{18} \mathrm{O}$-enriched water, the isotopic label becomes incorporated into the carbonyl group. Explain.

$$
\mathrm{R}_{2} \mathrm{C}=\mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{R}_{2} \mathrm{C}=\mathrm{O}+\mathrm{H}_{2} \mathrm{O} \quad \text { where } \mathrm{O}={ }^{18} \mathrm{O}
$$


(a) Basic conditions - 19.6 Nucleophilic Addition of HCN: Cyanohydrin Formation
Aldehydes and unhindered ketones undergo a nucleophilic addition reaction with HCN to yield cyanohydrins, $\mathrm{RCH}(\mathrm{OH}) \mathrm{C} \equiv \mathrm{N}$. Studies carried out in the early 1900s by Arthur Lapworth showed that cyanohydrin formation is reversible and base-catalyzed. Reaction occurs slowly when pure HCN is used but rapidly when a small amount of base is added to generate the nucleophilic cyanide ion, $\mathrm{CN}^{-}$. Addition of $\mathrm{CN}^{-}$takes place by a typical nucleophilic addition pathway, yielding a tetrahedral intermediate that is protonated by HCN to give cyanohydrin product plus regenerated $\mathrm{CN}^{-}$.

Cyanohydrin formation is somewhat unusual because it is one of the few examples of the addition of a protic acid ( $\mathrm{H}-\mathrm{Y}$ ) to a carbonyl group. As noted in the previous section, protic acids such as $\mathrm{H}_{2} \mathrm{O}, \mathrm{HBr}, \mathrm{HCl}$, and $\mathrm{H}_{2} \mathrm{SO}_{4}$ don't normally yield carbonyl addition products because the equilibrium constants are unfavorable. With HCN, however, equilibrium favors the cyanohydrin adduct.

Cyanohydrin formation is useful because of the further chemistry that can be carried out on the product. For example, a nitrile ( $\mathrm{R}-\mathrm{C} \equiv \mathrm{N}$ ) can be reduced with $\mathrm{LiAlH}_{4}$ to yield a primary amine $\left(\mathrm{RCH}_{2} \mathrm{NH}_{2}\right)$ and can be hydrolyzed by hot aqueous acid to yield a carboxylic acid. Thus, cyanohydrin formation provides a method for transforming an aldehyde or ketone into a different functional group.

PROBLEM Cyclohexanone forms a cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not. 19-9 Explain.


Addition of Hydride Reagents: Reduction -
We saw in Section 17.4 that the most common method for preparing alcohols, both in the laboratory and in living organisms, is by the reduction of carbonyl compounds. Aldehydes are reduced with sodium borohydride $\left(\mathrm{NaBH}_{4}\right)$ to give primary alcohols, and ketones are reduced similarly to give secondary alcohols.

Carbonyl reduction occurs by a typical nucleophilic addition mechanism under basic conditions, as shown earlier in FIGURE 19.5a. Although the details of carbonyl-group reductions are complex, $\mathrm{LiAlH}_{4}$ and $\mathrm{NaBH}_{4}$ act as if they were donors of hydride ion nucleophile, $: \mathrm{H}^{-}$, and the initially formed alkoxide ion intermediate is then protonated by addition of aqueous acid. The reaction is effectively irreversible because the reverse process would require expulsion of a very poor leaving group.


FIGURE 19.6 MECHANISM -
Mechanism of the Grignard reaction. Complexation of the carbonyl oxygen with the Lewis acid $\mathrm{Mg}^{2+}$ and subsequent nucleophilic addition of a carbanion to an aldehyde or ketone is followed by protonation of the alkoxide intermediate to yield an alcohol.
(1) The Lewis acid $\mathrm{Mg}^{2+}$ first forms an acid-base complex with the basic oxygen atom of the aldehyde or ketone, thereby making the carbonyl group a better acceptor.
(2) Nucleophilic addition of an alkyl group : $\mathrm{R}^{-}$to the aldehyde or ketone produces a tetrahedral magnesium alkoxide intermediate...
(2)

A tetrahedral intermediate
(3) . . . which undergoes hydrolysis when water is added in a separate step. The final product is a neutral alcohol.
(3) $\mathrm{H}_{2} \mathrm{O}$

An alcohol


Addition of Grignard Reagents, RMgX -
Just as aldehydes and ketones undergo nucleophilic addition with hydride ion to give alcohols, they undergo a similar addition with Grignard reagent nucleophiles, $\mathrm{R}:^{-+} \mathrm{MgX}$. Aldehydes give secondary alcohols on reaction with Grignard reagents in ether solution, and ketones give tertiary alcohols.

As shown in FIGURE 19.6, a Grignard reaction begins with an acid-base complexation of $\mathrm{Mg}^{2+}$ to the carbonyl oxygen atom of the aldehyde or ketone, thereby making the carbonyl group a better electrophile. Nucleophilic addition of R:- then produces a tetrahedral magnesium alkoxide intermediate, and protonation by addition of water or dilute aqueous acid in a separate step yields the neutral alcohol. Like reduction, Grignard additions are effectively irreversible because a carbanion is too poor a leaving group to be expelled in a reversal step.


Addition of Grignard Reagents, RMgX - 19.8 Nucleophilic Addition of Amines: Imine and Enamine Formation
Primary amines, $\mathrm{RNH}_{2}$, add to aldehydes and ketones to yield imines, $\mathrm{R}_{2} \mathrm{C}=\mathrm{NR}$. Secondary amines, $\mathrm{R}_{2} \mathrm{NH}$, add similarly to yield enamines, $\mathrm{R}_{2} \mathrm{~N}-\mathrm{CR}=\mathrm{CR}_{2}$ (ene + amine $=$ unsaturated amine).

Imines are particularly common as intermediates in biological pathways, where they are often called Schiff bases. The amino acid alanine, for instance, is metabolized in the body by reaction with the aldehyde pyridoxal phosphate (PLP), a derivative of vitamin $B_{6}$, to yield a Schiff base that is further degraded.

Imine formation and enamine formation seem different because one leads to a product with a $\mathrm{C}=\mathrm{N}$ bond and the other leads to a product with a $\mathrm{C}=\mathrm{C}$ bond. Actually, though, the reactions are quite similar. Both are typical examples of nucleophilic addition reactions in which water is eliminated from the initially formed tetrahedral intermediate and a new $\mathrm{C}=\mathrm{Nu}$ double bond is formed.

Imines are formed in a reversible, acid-catalyzed process (FIGURE 19.7) that begins with nucleophilic addition of the primary amine to the carbonyl group, followed by transfer of a proton from nitrogen to oxygen to yield a neutral amino alcohol, or carbinolamine. Protonation of the carbinolamine oxygen by an acid catalyst then converts the -OH into a better leaving group $\left(-\mathrm{OH}_{2}{ }^{+}\right)$, and E1-like loss of water produces an iminium ion. Loss of a proton from nitrogen gives the final product and regenerates the acid catalyst.


FIGURE 19.7 MECHANISM -
Mechanism of imine formation by reaction of an aldehyde or ketone with a primary amine. The key step is the initial nucleophilic addition to yield a carbinolamine intermediate, which then loses water to give the imine.

1 Nucleophilic attack on the ketone or aldehyde by the lone-pair electrons of an amine leads to a dipolar tetrahedral intermediate.

(2) A proton is then transferred from nitrogen to oxygen, yielding a neutral carbinolamine.
(2) $\downarrow$ Proton transfer

Carbinolamine
3 Acid catalyst protonates the hydroxyl oxygen.
(3) $\downarrow \mathrm{H}_{3} \mathrm{O}^{+}$

4 The nitrogen lone-pair electrons expel water, giving an iminium ion.

Iminium ion
Loss of $\mathrm{H}^{+}$from nitrogen then gives the neutral imine product.
(5)

Imine

Imine formation with such reagents as hydroxylamine and 2,4-dinitrophenylhydrazine is sometimes useful because the products of these reactions-oximes and 2,4-dinitrophenylhydrazones (2,4-DNPS), respectively-are often crystalline and easy to handle. Such crystalline derivatives are occasionally prepared as a means of purifying and characterizing liquid ketones or aldehydes.


2,4-Dinitrophenylhydrazone -
Reaction of an aldehyde or ketone with a secondary amine, $\mathrm{R}_{2} \mathrm{NH}$, rather than a primary amine yields an enamine. As shown in FIGURE 19.8, the process is identical to imine formation up to the iminium ion stage, but at this point there is no proton on nitrogen that can be lost to form a neutral imine product. Instead, a proton is lost from the neighboring carbon (the $\alpha$ carbon), yielding an enamine.


FIGURE 19.8 MECHANISM -
Mechanism for enamine formation by reaction of an aldehyde or ketone with a secondary amine, $\mathbf{R}_{\mathbf{2}} \mathbf{N H}$. The iminium ion intermediate formed in step 3 has no hydrogen attached to N and so must lose $\mathrm{H}^{+}$from the carbon two atoms away.

1. Nucleophilic addition of a secondary amine to the ketone or aldehyde, followed by proton transfer from (1) $\downarrow \mathrm{R}_{2} \mathrm{NH}$ nitrogen to oxygen, yields an intermediate carbinolamine in the normal way.

2 Protonation of the hydroxyl by acid catalyst converts it into a better (2 $\downarrow{ } \mathrm{H}_{3} \mathrm{O}^{+}$ leaving group.

(3) Elimination of water by the lone-pair electrons on nitrogen then yields an intermediate iminium ion.

$$
\text { (3 } \downarrow \|_{-\mathrm{H}_{2} \mathrm{O}}
$$

4 Loss of a proton from the alpha carbon atom yields the enamine product and regenerates the acid catalyst.
(4) $\downarrow$

Enamine

Imine and enamine formation are slow at both high pH and low pH but reach a maximum rate at a weakly acidic pH around 4 to 5 . For example, the profile of pH versus rate shown in FIGURE 19.9 for the reaction between acetone and hydroxylamine, $\mathrm{NH}_{2} \mathrm{OH}$, indicates that the maximum reaction rate occurs at pH 4.5 .

We can explain the observed pH dependence of imine formation by looking at the individual steps in the mechanism. As indicated in FIGURE 19.8, an acid catalyst is required in step 3 to protonate the intermediate carbinolamine, thereby converting the - OH into a better leaving group. Thus, reaction will be slow if not enough acid is present (that is, at high pH ). On the other hand, if too much acid is present (low pH ), the basic amine nucleophile is completely protonated, so the initial nucleophilic addition step can't occur.


FIGURE 19.8 MECHANISM -
FIGURE 19.9 Dependence on $\mathbf{p H}$ of the rate of reaction between acetone and hydroxylamine: $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}+\mathrm{NH}_{2} \mathrm{OH} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{NOH}+\mathrm{H}_{2} \mathrm{O}$.

Evidently, a pH of 4.5 represents a compromise between the need for some acid to catalyze the rate-limiting dehydration step but not too much acid so as to avoid complete protonation of the amine. Each nucleophilic addition reaction has its own requirements, and conditions must be optimized to obtain maximum reaction rates.


Predicting the Product of Reaction between a Ketone and an Amine -
Show the products you would obtain by acid-catalyzed reaction of 3-pentanone with methylamine, $\mathrm{CH}_{3} \mathrm{NH}_{2}$, and with dimethylamine, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}$.


Strategy -
An aldehyde or ketone reacts with a primary amine, $\mathrm{RNH}_{2}$, to yield an imine, in which the carbonyl oxygen atom has been replaced by the $=\mathrm{N}-\mathrm{R}$ group of the amine. Reaction of the same aldehyde or ketone with a secondary amine, $\mathrm{R}_{2} \mathrm{NH}$, yields an enamine, in which the oxygen atom has been replaced by the $-\mathrm{NR}_{2}$ group of the amine and the double bond has moved to a position between the former carbonyl carbon and the neighboring carbon.


Solution -
An enamine

PROBLEM Show the products you would obtain by acid-catalyzed reaction of cyclohexanone with ethylamine, 19-10 $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}$, and with diethylamine, $\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{NH}$.

PROBLEM Imine formation is reversible. Show all the steps involved in the acid-catalyzed reaction of an imine 19-11 with water (hydrolysis) to yield an aldehyde or ketone plus primary amine.

PROBLEM Draw the following molecule as a skeletal structure, and show how it can be prepared from a ketone 19-12
and an amine.


Solution - 19.9 Nucleophilic Addition of Hydrazine: The Wolff-Kishner Reaction
A useful variant of the imine-forming reaction just discussed involves the treatment of an aldehyde or ketone with hydrazine, $\mathrm{H}_{2} \mathrm{~N}-\mathrm{NH}_{2}$, in the presence of KOH . Called the Wolff-Kishner reaction, the process is a useful and general method for converting an aldehyde or ketone into an alkane, $\mathrm{R}_{2} \mathrm{C}=\mathrm{O} \rightarrow \mathrm{R}_{2} \mathrm{CH}_{2}$.

As shown in FIGURE 19.10, the Wolff-Kishner reaction involves formation of a hydrazone intermediate, $\mathrm{R}_{2} \mathrm{C}=\mathrm{N}-\mathrm{NH}_{2}$, followed by base-catalyzed double-bond migration, loss of $\mathrm{N}_{2}$ gas to give a carbanion, and protonation to give the alkane product. The double-bond migration takes place when a base removes one of the weakly acidic NH protons in step 2 to generate a hydrazone anion, which has an allylic resonance structure that places the double bond between nitrogens and the negative charge on carbon. Reprotonation then occurs on carbon to generate the double-bond rearrangement product. The next step-loss of nitrogen and formation of an alkyl anion-is driven by the large thermodynamic stability of the $\mathrm{N}_{2}$ molecule.


FIGURE 19.10 MECHANISM -
Mechanism for the Wolff-Kishner reduction of an aldehyde or ketone to yield an alkane.

(1) Reaction of the aldehyde or ketone with hydrazine yields a hydrazone in the normal way.

2 Base abstracts a weakly acidic $\mathrm{N}-\mathrm{H}$ proton, yielding a hydrazone anion. This anion has a resonance form that places the negative charge on carbon and the double bond between nitrogens.
(3) Protonation of the hydrazone anion takes place on carbon to yield a neutral intermediate.
(4) Deprotonation of the remaining weakly acidic N-H occurs with simultaneous loss of nitrogen to give a carbanion...
5. . . which is protonated to give the alkane product

(1) $\downarrow \downarrow$

(5) $\mathrm{H}_{2} \mathrm{O}$

Note that the Wolff-Kishner reduction accomplishes the same overall transformation as the catalytic hydrogenation of an acylbenzene to yield an alkylbenzene (Section 16.10). The Wolff-Kishner reduction is more general and more useful than catalytic hydrogenation, however, because it works well with both alkyl and aryl ketones.

PROBLEM Show how you could prepare the following compounds from 4-methyl-3-penten-2-one, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=$ 19-13 $\mathrm{CHCOCH}_{3}$.
(a)

(b)

(c)


FIGURE 19.10 MECHANISM - 19.10 Nucleophilic Addition of Alcohols: Acetal Formation
Aldehydes and ketones react reversibly with 2 equivalents of an alcohol in the presence of an acid catalyst to yield acetals, $\mathbf{R}_{\mathbf{2}} \mathbf{C}\left(\mathbf{O R}^{\prime}\right)_{\mathbf{2}}$, which are often called ketals if derived from a ketone. Cyclohexanone, for instance, reacts with methanol in the presence of HCl to give the corresponding dimethyl acetal.

Acetal formation is similar to the hydration reaction discussed in Section 19.5. Like water, alcohols are weak nucleophiles that add to aldehydes and ketones slowly under neutral conditions. Under acidic conditions, however, the reactivity of the carbonyl group is increased by protonation, so addition of an alcohol occurs rapidly.

A neutral carbonyl group is moderately electrophilic because of the polarity of the $\mathrm{C}-\mathrm{O}$ bond.

A protonated carbonyl group is strongly electrophilic because of the positive charge on carbon.

As shown in FIGURE 19.11, nucleophilic addition of an alcohol to the carbonyl group initially yields a hydroxy ether called a hemiacetal, analogous to the gem diol formed by addition of water. Hemiacetals are formed reversibly, with equilibrium normally favoring the carbonyl compound. In the presence of acid, however, a further reaction occurs. Protonation of the - OH group, followed by an E1-like loss of water, leads to an oxonium ion, $\mathrm{R}_{2} \mathrm{C}=\mathrm{O}^{+} \mathrm{R}$, which undergoes a second nucleophilic addition of alcohol to yield the protonated acetal. Loss of a proton completes the reaction.


FIGURE 19.11 MECHANISM -
Mechanism of acid-catalyzed acetal formation by reaction of an aldehyde or ketone with an alcohol.

1 Protonation of the carbonyl oxygen strongly polarizes the carbonyl group and...

2 . . . activates the carbonyl group for nucleophilic attack by oxygen lone-pair electrons from the alcohol.

3 Loss of a proton yields a neutral hemiacetal tetrahedral intermediate.

(1)

(3)

Hemiacetal

(4) Protonation of the hemiacetal hydroxyl converts it into a good leaving group.

(5) Dehydration yields an intermediate oxonium ion.
(6) Addition of a second equivalent of alcohol gives a protonated acetal.
(5)

(6) $\| R-\ddot{o}-H$

Loss of a proton yields the neutral acetal product.

Acetal

Because all the steps in acetal formation are reversible, the reaction can be driven either forward (from carbonyl
compound to acetal) or backward (from acetal to carbonyl compound), depending on the conditions. The forward reaction is favored by conditions that remove water from the medium and thus drive the equilibrium to the right. In practice, this is often done by distilling off water as it forms. The reverse reaction is favored by treating the acetal with a large excess of aqueous acid to drive the equilibrium to the left.

Acetals are useful because they can act as protecting groups for aldehydes and ketones in the same way that trimethylsilyl ethers act as protecting groups for alcohols (Section 17.8). As we saw previously, it sometimes happens that one functional group interferes with intended chemistry elsewhere in a complex molecule. For example, if we wanted to reduce only the ester group of ethyl 4 -oxopentanoate, the ketone would interfere. Treatment of the starting keto ester with $\mathrm{LiAlH}_{4}$ would reduce both the keto group and the ester group to give a diol product.


FIGURE 19.11 MECHANISM -
By protecting the keto group as an acetal, however, the problem can be circumvented. Like other ethers, acetals are unreactive to bases, hydride reducing agents, Grignard reagents, and catalytic hydrogenation conditions, but they are cleaved by acid. Thus, we can accomplish the selective reduction of the ester group in ethyl 4 -oxopentanoate by first converting the keto group to an acetal, then reducing the ester with $\mathrm{LiAlH}_{4}$, and then removing the acetal by treatment with aqueous acid. (In practice, it's often convenient to use 1 equivalent of a diol such as ethylene glycol as the alcohol to form a cyclic acetal. The mechanism of cyclic acetal formation using 1 equivalent of ethylene glycol is exactly the same as that using 2 equivalents of methanol or other monoalcohols.)

Acetal and hemiacetal groups are particularly common in carbohydrate chemistry. Glucose, for instance, is a polyhydroxy aldehyde that undergoes an internal nucleophilic addition reaction and exists primarily as a cyclic hemiacetal.


Predicting the Product of Reaction between a Ketone and an Alcohol -
Show the structure of the acetal you would obtain by acid-catalyzed reaction of 2-pentanone with


Strategy -
Acid-catalyzed reaction of an aldehyde or ketone with 2 equivalents of a monoalcohol or 1 equivalent of a diol yields an acetal, in which the carbonyl oxygen atom is replaced by two -OR groups from the alcohol.


Solution -
PROBLEM Show all the steps in the acid-catalyzed formation of a cyclic acetal from ethylene glycol and an 19-14 aldehyde or ketone.

PROBLEM Identify the carbonyl compound and the alcohol that were used to prepare the following acetal:
19-15