Solution -
"What is an immediate precursor of the target?" The final step will involve introduction of one of three groups-chlorine, propyl, or sulfonic acid-so we have to consider three possibilities. Of the three, the chlorination of o-propylbenzenesulfonic acid can't be used because the reaction would occur at the wrong position. Similarly, a Friedel-Crafts reaction can't be used as the final step because this reaction doesn't work on sulfonic-acid-substituted (strongly deactivated) benzenes. Thus, the immediate precursor of the desired product is probably $m$-chloropropylbenzene, which can be sulfonated to give a mixture of product isomers that must then be separated.

o-Propylbenzenesulfonic acid
This ring will give the wrong isomer on chlorination.

p-Chlorobenzenesulfonic acid
This deactivated ring will not undergo a Friedel-Crafts reaction.

m-Chloropropylbenzene
This ring will give the desired product on sulfonation.


4-Chloro-2-propylbenzenesulfonic acid -
"What is an immediate precursor of $m$-chloropropylbenzene?" Because the two substituents have a meta relationship, the first substituent placed on the ring must be a meta director so that the second substitution will take place at the proper position. Furthermore, because primary alkyl groups such as propyl can't be introduced directly by Friedel-Crafts alkylation, the precursor of $m$-chloropropylbenzene is probably $m$-chloropropiophenone, which could be catalytically reduced.

"What is an immediate precursor of $m$-chloropropiophenone?" Propiophenone, which could be chlorinated in the meta position.

"What is an immediate precursor of propiophenone?" Benzene, which could undergo Friedel-Crafts acylation with propanoyl chloride and $\mathrm{AlCl}_{3}$.

The final synthesis is a four-step route from benzene:

Planning an organic synthesis has been compared with playing chess. There are no tricks; all that's required is a knowledge of the allowable moves (the organic reactions) and the discipline to plan ahead, carefully evaluating the consequences of each move. Practicing may not be easy, but it's a great way to learn organic chemistry.

PROBLEM How might you synthesize the following substances from benzene?
16-22 (a) $m$-Chloronitrobenzene (b) $m$-Chloroethylbenzene (c) 4-Chloro-1-nitro-2-propylbenzene (d) 3-Bromo-2-methylbenzenesulfonic acid

PROBLEM In planning a synthesis, it's as important to know what not to do as to know what to do. As written,
16-23 the following reaction schemes have flaws in them. What is wrong with each?

(b)


Combinatorial Chemistry -
Traditionally, organic compounds have been synthesized one at a time. This works well for preparing large amounts of a few substances, but it doesn't work so well for preparing small amounts of a great many substances. This latter goal is particularly important in the pharmaceutical industry, where vast numbers of structurally similar compounds must be synthesized and screened to find an optimum drug candidate.

To speed the process of drug discovery, combinatorial chemistry has been developed to prepare what are called combinatorial libraries, in which anywhere from a few dozen to several hundred thousand substances are prepared simultaneously. Among the early successes of combinatorial chemistry is the development of a benzodiazepine library, a class of aromatic compounds commonly used as antianxiety agents.

Benzodiazepine library
( $\mathrm{R}_{\mathbf{1}}-\mathrm{R}_{\mathbf{4}}$ are various organic substituents)

Two main approaches to combinatorial chemistry are used-parallel synthesis and split synthesis. In parallel synthesis, each compound is prepared independently. Typically, a reactant is first linked to the surface of polymer beads, which are then placed into small wells on a 96 -well glass plate. Programmable robotic instruments add different sequences of building blocks to the different wells, thereby making 96 different products. When the reaction sequences are complete, the polymer beads are washed and their products are released.

In split synthesis, the initial reactant is again linked to the surface of polymer beads, which are then divided into several groups. A different building block is added to each group of beads, the different groups are combined, and the reassembled mix is again split to form new groups. Another building block is added to each group, the groups are again combined and redivided, and the process continues. If, for example, the beads are divided into four groups at each step, the number of compounds increases in the progression $4 \rightarrow 16 \rightarrow 64 \rightarrow 256$. After 10 steps, more than 1 million compounds have been prepared (FIGURE 16.23).


Combinatorial Chemistry -
Of course, with so many different final products mixed together, the problem is to identify them. What structure is linked to what bead? Several approaches to this problem have been developed, all of which involve the attachment of encoding labels to each polymer bead to keep track of the chemistry each has undergone. Encoding labels used thus far have included proteins, nucleic acids, halogenated aromatic compounds, and even computer chips.

FIGURE 16.22 Organic chemistry by robot means no spilled flasks!

FIGURE 16.23 The results of split combinatorial synthesis. Assuming that 4 different building blocks are used at each step, 64 compounds result after 3 steps, and more than one million compounds result after 10 steps.


Key Terms -
- acyl group • electrophilic aromatic substitution
- acylation
- Friedel-Crafts reaction
- alkylation
- inductive effect
- benzyne
- nucleophilic aromatic substitution
- combinatorial chemistry
- resonance effect


Summary -
We've continued the coverage of aromatic molecules in this chapter, shifting focus to concentrate on reactions. In particular, we've looked at the relationship between aromatic structure and reactivity, a relationship critical to understanding how numerous biological molecules and pharmaceutical agents are synthesized and why they behave as they do.

An electrophilic aromatic substitution reaction takes place in two steps-initial reaction of an electrophile, $\mathbf{E}^{+}$, with the aromatic ring, followed by loss of $\mathrm{H}^{+}$from the resonance-stabilized carbocation intermediate to regenerate the aromatic ring.

Many variations of the reaction can be carried out, including halogenation, nitration, and sulfonation. Friedel-Crafts alkylation and acylation reactions, which involve reaction of an aromatic ring with carbocation electrophiles, are particularly useful. They are limited, however, by the fact that the aromatic ring must be at least as reactive as a halobenzene. In addition, polyalkylation and carbocation rearrangements often occur in Friedel-Crafts alkylation.

Substituents on the benzene ring affect both the reactivity of the ring toward further substitution and the orientation of that substitution. Groups can be classified as ortho- and para-directing activators, ortho- and para-directing deactivators, or meta-directing deactivators. Substituents influence aromatic rings by a combination of resonance and inductive effects. Resonance effects are transmitted through $\pi$ bonds; inductive effects are transmitted through $\sigma$ bonds.

Halobenzenes undergo nucleophilic aromatic substitution through either of two mechanisms. If the halobenzene has a strongly electron-withdrawing substituent in the ortho or para position, substitution occurs by addition of a nucleophile to the ring, followed by elimination of halide from the intermediate anion. If the halobenzene is not activated by an electron-withdrawing substituent, substitution can occur by elimination of HX to give a benzyne, followed by addition of a nucleophile.


Summary -
The benzylic position of an alkylbenzene can be brominated by reaction with $N$-bromosuccinimide, and the entire side chain can be degraded to a carboxyl group by oxidation with aqueous $\mathrm{KMnO}_{4}$. Aromatic rings can also be reduced to cyclohexanes by hydrogenation over a platinum or rhodium catalyst, and aryl alkyl ketones are reduced to alkylbenzenes by hydrogenation over a platinum catalyst.


Summary of Reactions -
1. Electrophilic aromatic substitution
a. Fluorination (Section 16.2)

b. Bromination (Section 16.1)

c. Chlorination (Section 16.2)

d. Iodination (Section 16.2)

e. Nitration (Section 16.2)

f. Sulfonation (Section 16.2)

g. Friedel-Crafts alkylation (Section 16.3)

h. Friedel-Crafts acylation (Section 16.3)

2. Reduction of aromatic nitro groups (Section 16.2)

3. Nucleophilic aromatic substitution
a. By addition to activated aryl halides (Section 16.6)

b. By formation of benzyne intermediate from unactivated aryl halide (Section 16.7)

4. Oxidation of alkylbenzene side chain (Section 16.8)

5. Benzylic bromination of alkylbenzene side chain (Section 16.8)

6. Catalytic hydrogenation of aromatic ring (Section 16.9)

7. Reduction of aryl alkyl ketones (Section 16.9)


Visualizing Chemistry -
PROBLEM Draw the product from reaction of each of the following substances with (1) $\mathrm{Br}_{2}, \mathrm{FeBr}_{3}$ and (2)
16-24 $\mathrm{CH}_{3} \mathrm{COCl}, \mathrm{AlCl}_{3}$.
(a)

(b)

PROBLEM The following molecular model of a dimethyl-substituted biphenyl represents the lowest-energy
$\mathbf{1 6 - 2 5}$ conformation of the molecule. Why are the two benzene rings tilted at a $63^{\circ}$ angle to each other
rather than being in the same plane so that their $p$ orbitals overlap? Why doesn't complete rotation around the single bond joining the two rings occur?

PROBLEM How would you synthesize the following compound starting from benzene? More than one step is 16-26 needed.

PROBLEM The following compound can't be synthesized using the methods discussed in this chapter. Why 16-27 not?


Mechanisms of Electrophilic Substitutions -
PROBLEM Aromatic iodination can be carried out with a number of reagents, including iodine monochloride,
16-28 ICl. What is the direction of polarization of ICl? Propose a mechanism for the iodination of an aromatic ring with ICl.

PROBLEM The sulfonation of an aromatic ring with $\mathrm{SO}_{3}$ and $\mathrm{H}_{2} \mathrm{SO}_{4}$ is reversible. That is, heating
16-29 benzenesulfonic acid with $\mathrm{H}_{2} \mathrm{SO}_{4}$ yields benzene. Show the mechanism of the desulfonation reaction. What is the electrophile?

PROBLEM The carbocation electrophile in a Friedel-Crafts reaction can be generated by an alternate means
16-30 than reaction of an alkyl chloride with $\mathrm{AlCl}_{3}$. For example, reaction of benzene with 2-methylpropene in the presence of $\mathrm{H}_{3} \mathrm{PO}_{4}$ yields tert-butylbenzene. Propose a mechanism for this reaction.

PROBLEM The $N, N, N$-trimethylammonium group, $-\stackrel{+}{\mathrm{N}}\left(\mathrm{CH}_{3}\right)_{3}$, is one of the few groups that is a meta-directing
16-31 deactivator yet has no electron- withdrawing resonance effect. Explain.

PROBLEM The nitroso group, $-\mathrm{N}=\mathrm{O}$, is one of the few nonhalogens that is an ortho- and para-directing
16-32 deactivator. Explain this behavior by drawing resonance structures of the carbocation intermediates in ortho, meta, and para electrophilic reaction on nitrosobenzene, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}=\mathrm{O}$.

PROBLEM Triphenylmethane can be prepared by reaction of benzene and chloroform in the presence of $\mathrm{AlCl}_{3}$.
16-33 Propose a mechanism for the reaction.

PROBLEM Using resonance structures of the intermediates, explain why bromination of biphenyl occurs at
16-34 ortho and para positions rather than at meta.


Mechanisms of Electrophilic Substitutions -
PROBLEM Using resonance structures of the intermediates, explain why bromination of biphenyl occurs at
16-34 ortho and para positions rather than at meta.

Biphenyl

PROBLEM Benzene and alkyl-substituted benzenes can be hydroxylated by reaction with $\mathrm{H}_{2} \mathrm{O}_{2}$ in the presence
16-35 of an acidic catalyst. What is the structure of the reactive electrophile? Propose a mechanism for the reaction.

Additional Mechanism Practice
PROBLEM Addition of HBr to 1-phenylpropene yields only (1-bromopropyl)benzene. Propose a mechanism for
16-36 the reaction, and explain why none of the other regioisomer is produced.

PROBLEM Hexachlorophene, a substance used in the manufacture of germicidal soaps, is prepared by
16-37 reaction of 2,4,5-trichlorophenol with formaldehyde in the presence of concentrated sulfuric acid. Propose a mechanism for the reaction.

Hexachlorophene
PROBLEM Benzenediazonium carboxylate decomposes when heated to yield $\mathrm{N}_{2}, \mathrm{CO}_{2}$, and a reactive 16-38 substance that can't be isolated. When benzene-diazonium carboxylate is heated in the presence of furan, the following reaction is observed:

What intermediate is involved in this reaction? Propose a mechanism for its formation.

PROBLEM 4-Chloropyridine undergoes reaction with dimethylamine to yield 4-dimethylaminopyridine.
16-39 Propose a mechanism for the reaction.

PROBLEM Propose a mechanism to account for the following reaction:
16-40

PROBLEM In the Gatterman-Koch reaction, a formyl group ( -CHO ) is introduced directly onto a benzene
16-41 ring. For example, reaction of toluene with CO and HCl in the presence of mixed $\mathrm{CuCl} / \mathrm{AlCl}_{3}$ gives $p$-methylbenzaldehyde. Propose a mechanism.


Mechanisms of Electrophilic Substitutions -
PROBLEM Treatment of p-tert-butylphenol with a strong acid such as $\mathrm{H}_{2} \mathrm{SO}_{4}$ yields phenol and
16-42 2-methylpropene. Propose a mechanism.
PROBLEM Benzyl bromide is converted into benzaldehyde by heating in dimethyl sulfoxide. Propose a
16-43 structure for the intermediate, and show the mechanisms of the two steps in the reaction.

PROBLEM Propose a mechanism for the Smiles rearrangement below.

16-44

PROBLEM Because of their conjugation, azo dyes are highly colored compounds and the major artificial color
16-45 source for textiles and food. Azo dyes are produced by the reaction of aryl diazonium salts with a second aromatic compound. In the product, the aromatic rings are linked by a diazo bridge ( $-\mathrm{N}=\mathrm{N}-$ ). From the reactants provided, propose a structure for each azo dye and draw the electron-pushing mechanism.
(a)

(b)

(c)

$+$

Reactivity and Orientation of Electrophilic Substitutions
PROBLEM Identify each of the following groups as an activator or deactivator and as an o,p-director or 16-46 m -director:
(a) $\stackrel{\gtrless}{<}\left(\mathrm{CH}_{3}\right)_{2}$
(b)

(c)

(d)

PROBLEM Predict the major product(s) of nitration of the following substances. Which react faster than 16-47 benzene, and which slower?
(a) Bromobenzene
(b) Benzonitrile
(c) Benzoic acid
(d) Nitrobenzene
(e) Benzenesulfonic acid (f) ethoxybenzene


Mechanisms of Electrophilic Substitutions -
PROBLEM Rank the compounds in each group according to their reactivity toward electrophilic substitution.
16-48 (a) Chlorobenzene, o-dichlorobenzene, benzene
(b) $p$-Bromonitrobenzene, nitrobenzene, phenol (c) Fluorobenzene, benzaldehyde, $o$-xylene
(d) Benzonitrile, $p$-methylbenzonitrile, $p$-methoxybenzonitrile

PROBLEM Predict the major monoalkylation products you would expect to obtain from reaction of the
16-49 following substances with chloromethane and $\mathrm{AlCl}_{3}$ :
(a) Bromobenzene
(b) $m$-Bromophenol
(c) $p$-Chloroaniline
(d) 2,4-Dichloronitrobenzene
(e) 2,4-Dichlorophenol (f) Benzoic acid
(g) $p$-Methylbenzenesulfonic acid
(h) 2,5-Dibromotoluene

PROBLEM Name and draw the major product(s) of electrophilic chlorination of the following compounds:
16-50 (a) m-Nitrophenol
(b) o-Xylene
(c) p-Nitrobenzoic acid
(d) $p$-Bromobenzenesulfonic acid

PROBLEM Predict the major product(s) you would obtain from sulfonation of the following compounds:
16-51 (a) Fluorobenzene (b) $m$-Bromophenol (c) $m$-Dichlorobenzene (d) 2,4-Dibromophenol
PROBLEM Rank the following aromatic compounds in the expected order of their reactivity toward
16-52 Friedel-Crafts alkylation. Which compounds are unreactive?
(a) Bromobenzene
(b) Toluene
(c) Phenol
(d) Aniline (e) Nitrobenzene
(f) $p$-Bromotoluene

PROBLEM What product(s) would you expect to obtain from the following reactions:
16-53
(a)

(b)

(c)

(d)


Mechanisms of Electrophilic Substitutions -
PROBLEM What product(s) would you expect to obtain from the following reactions:
16-53
(a)

(b)

(c)

(d)

PROBLEM Predict the major product(s) of the following reactions:
16-54 (a)

(c)

Organic Synthesis
PROBLEM How would you synthesize the following substances starting from benzene or phenol? Assume that
16-55 ortho- and para-substitution products can be separated.
(a) o-Bromobenzoic acid
(b) $p$-Methoxytoluene
(c) 2,4,6-Trinitrobenzoic acid
(d) $m$-Bromoaniline

PROBLEM Starting with benzene as your only source of aromatic compounds, how would you synthesize the
16-56 following substances? Assume that you can separate ortho and para isomers if necessary.
(a) p-Chloroacetophenone
(b) $m$-Bromonitrobenzene
(c) $o$-Bromobenzenesulfonic acid
(d) $m$-Chlorobenzenesulfonic acid

PROBLEM Starting with either benzene or toluene, how would you synthesize the following substances:
16-57 Assume that ortho and para isomers can be separated.
(a) 2-Bromo-4-nitrotoluene
(b) 1,3,5-Trinitrobenzene
(c) 2,4,6-Tribromoaniline
(d) m -Fluorobenzoic acid

PROBLEM As written, the following syntheses have flaws. What is wrong with each?
16-58 (a)

(b)

(c)


General Problems -
PROBLEM At what position and on what ring do you expect nitration of 4-bromo-biphenyl to occur? Explain,
16-59 using resonance structures of the potential intermediates.


4-Bromobiphenyl -
PROBLEM Electrophilic substitution on 3-phenylpropanenitrile occurs at the ortho and para positions, but
16-60 reaction with 3 -phenylpropenenitrile occurs at the meta position. Explain, using resonance structures of the intermediates.

3-Phenylpropanenitrile

3-Phenylpropenenitrile

PROBLEM At what position, and on what ring, would you expect the following substances to undergo
16-61 electrophilic substitution?
(a)

(b)

(c)

(d)

PROBLEM At what position, and on what ring, would you expect bromination of benzanilide to occur? Explain
16-62 by drawing resonance structures of the intermediates.


Benzanilide -
PROBLEM Would you expect the Friedel-Crafts reaction of benzene with ( $R$ )-2-chlorobutane to yield optically
16-63 active or racemic product? Explain.

PROBLEM How would you synthesize the following substances starting from benzene?


16-64 (a) -
(b)

(c)

PROBLEM The compound MON-0585 is a nontoxic, biodegradable larvicide that is highly selective against
16-65 mosquito larvae. Synthesize MON-0585 using either benzene or phenol as a source of the aromatic rings.

MON-0585

PROBLEM Phenylboronic acid, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~B}(\mathrm{OH})_{2}$, is nitrated to give $15 \%$ ortho- substitution product and $85 \%$ meta.
16-66 Explain the meta-directing effect of the $-\mathrm{B}(\mathrm{OH})_{2}$ group.

PROBLEM Draw resonance structures of the intermediate carbocations in the bromination of naphthalene,
16-67 and account for the fact that naphthalene undergoes electrophilic substitution at C1 rather than C2.

PROBLEM Propose a mechanism for the reaction of 1-chloroanthraquinone with methoxide ion to give the
16-68 substitution product 1-methoxyanthraquinone. Use curved arrows to show the electron flow in each step.

PROBLEM $p$-Bromotoluene reacts with potassium amide to give a mixture of $m$ - and $p$-methylaniline. Explain. 16-69

PROBLEM Propose a mechanism to account for the reaction of benzene with
16-70 2,2,5,5-tetramethyltetrahydrofuran.

PROBLEM How would you synthesize the following compounds from benzene? Assume that ortho and para
16-71 isomers can be separated.
(a)

(b)

PROBLEM You know the mechanism of HBr addition to alkenes, and you know the effects of various
16-72 substituent groups on aromatic substitution. Use this knowledge to predict which of the following two alkenes reacts faster with HBr. Explain your answer by drawing resonance structures of the carbocation intermediates.

PROBLEM Use your knowledge of directing effects, along with the following data, to deduce the directions of
16-73 the dipole moments in aniline, bromobenzene, and bromoaniline.

$\mu=1.53 \mathrm{D}$


16-64 (a) -
$\mu=1.53 \mathrm{D}$

$\mu=1.52 \mathrm{D}$

$\mu=2.91 \mathrm{D}$

PROBLEM Identify the reagents represented by the letters a-e in the following scheme:
16-74

PROBLEM Phenols (ArOH) are relatively acidic, and the presence of a substituent group on the aromatic ring
16-75 has a large effect. The $p K_{\mathrm{a}}$ of unsubstituted phenol, for example, is 9.89 , while that of $p$-nitrophenol is 7.15. Draw resonance structures of the corresponding phenoxide anions and explain the data.

PROBLEM Would you expect $p$-methylphenol to be more acidic or less acidic than unsubstituted phenol?
16-76 Explain. (See Problem 16-75.)
PROBLEM Predict the product(s) for each of the following reactions. In each case, draw the resonance forms of 16-77 the intermediate to explain the observed regiochemistry.
(a)

(b)

(c)

(d)


Alcohols and Phenols -
FIGURE 17.1 The phenol resveratrol, found in the skin of red grapes, continues to be studied for its potential anti-cancer, antiarthritic, and hypoglycemic properties. (credit: modification of work "Weinreben-Ötlingen" by Pierre Likissas/Wikimedia Commons, CC BY 3.0)


CHAPTER CONTENTS - 17.1 Naming Alcohols and Phenols
17.2 Properties of Alcohols and Phenols
17.3 Preparation of Alcohols: A Review
17.4 Alcohols from Carbonyl Compounds: Reduction
17.5 Alcohols from Carbonyl Compounds: Grignard Reaction
17.6 Reactions of Alcohols
17.7 Oxidation of Alcohols
17.8 Protection of Alcohols
17.9 Phenols and Their Uses
17.10 Reactions of Phenols
17.11 Spectroscopy of Alcohols and Phenols

WHY THIS CHAPTER? Up to this point, we've focused on developing some general ideas of organic reactivity, looking at the chemistry of hydrocarbons and alkyl halides, and examining some of the tools used in structural studies. With that background, it's now time to begin a study of the oxygen-containing functional groups that lie at the heart of organic and biological chemistry. We'll look at alcohols in this chapter and move on to carbonyl compounds in Chapters 19 through 23.

Alcohols and phenols can be thought of as organic derivatives of water in which one of water's hydrogens is replaced by an organic group: $\mathrm{H}-\mathrm{O}-\mathrm{H}$ versus $\mathrm{R}-\mathrm{O}-\mathrm{H}$ and $\mathrm{Ar}-\mathrm{O}-\mathrm{H}$. In practice, the name alcohol is restricted to
compounds that have their - OH group bonded to a saturated, $s p^{3}$-hybridized carbon atom, while compounds with their -OH group bonded to a vinylic, $s p^{2}$-hybridized carbon are called enols. We'll look at enols in Chapter 22.

An alcohol

A phenol


CHAPTER CONTENTS - 17.1 Naming Alcohols and Phenols
An alcohol

A phenol

Alcohols occur widely in nature and have many industrial and pharmaceutical applications. Methanol, for instance, is one of the most important of all industrial chemicals. Historically, methanol was prepared by heating wood in the absence of air and thus came to be called wood alcohol. Today, approximately 173 million tons ( 50 billion gallons) of methanol is manufactured worldwide each year, most of it by catalytic reduction of carbon monoxide with hydrogen gas. Methanol is toxic to humans, causing blindness in small doses ( 15 mL ) and death in larger amounts ( $100-250 \mathrm{~mL}$ ). Industrially, it is used both as a solvent and as a starting material for production of formaldehyde $\left(\mathrm{CH}_{2} \mathrm{O}\right)$ and acetic acid $\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)$.

Ethanol was one of the first organic chemicals to be prepared and purified. Its production by fermentation of grains and sugars has been carried out for perhaps 9000 years, and its purification by distillation goes back at least as far as the 12 th century. Today, approximately 88 million tons ( 26 billion gallons) of ethanol are produced worldwide each year, most of it by fermentation of corn, barley, sorghum, and other plant sources. Almost all of this ethanol is used for bus and automobile fuel.

Ethanol for industrial use as a solvent or chemical intermediate is largely obtained by acid-catalyzed hydration of ethylene at high temperature.

Phenols occur widely throughout nature and also serve as intermediates in the industrial synthesis of products as diverse as adhesives and antiseptics. Phenol itself is a general disinfectant found in coal tar; methyl salicylate is a flavoring agent found in oil of wintergreen; and urushiols are the allergenic constituents of poison oak and poison ivy. Note that the word phenol is the name both of the specific compound (hydroxybenzene) and of the class of compounds.

Phenol (also known as carbolic acid)

Methyl salicylate

Urushiols ( $\mathrm{R}=$ different $\mathrm{C}_{15}$ alkyl and alkenyl chains)


CHAPTER CONTENTS - 17.1 Naming Alcohols and Phenols
Alcohols are classified as primary $\left(1^{\circ}\right)$, secondary $\left(2^{\circ}\right)$, or tertiary $\left(3^{\circ}\right)$, depending on the number of organic groups bonded to the hydroxyl-bearing carbon.


A primary $\left(\mathbf{1}^{\circ}\right)$ alcohol A secondary (2 ${ }^{\circ}$ ) alcohol A tertiary ( $\mathbf{3}^{\circ}$ ) alcohol -
Simple alcohols are named in the IUPAC system as derivatives of the parent alkane, using the suffix -ol.


RULE 1 -
Select the longest carbon chain containing the hydroxyl group, and derive the parent name by replacing the $-e$ ending of the corresponding alkane with $-o l$. The $-e$ is deleted to prevent the occurrence of two adjacent vowels: propanol rather than propaneol, for example.

RULE 2
Number the alkane chain beginning at the end nearer the hydroxyl group.
RULE 3
Number the substituents according to their position on the chain, and write the name, listing the substituents in alphabetical order and identifying the position to which the -OH is bonded. Note that in naming cis-1,4-cyclohexanediol, the final -e of cyclohexane is not deleted because the next letter, $d$, is not a vowel; that is, cyclohexanediol rather than cyclohexandiol. Also, as with alkenes (Section 7.3), newer IUPAC naming recommendations place the locant immediately before the suffix rather than before the parent.

2-Methyl-2-pentanol
(New: 2-Methylpentan-2-ol) (New: cis-Cyclohexane-1,4-diol)

3-Phenyl-2-butanol

Some simple and widely occurring alcohols have common names that are accepted by IUPAC. For example:


Allyl alcohol (2-propen-1-ol) -
tert-Butyl alcohol (2-methyl-2-propanol)

$$
\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}
$$

Ethylene glycol (1,2-ethanediol)

Glycerol (1,2,3-propanetriol)

Phenols are named as described previously for aromatic compounds according to the rules discussed in Section 15.1, with -phenol used as the parent name rather than -benzene.

PROBLEM Give IUPAC names for the following compounds:
17-1 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM Draw structures corresponding to the following IUPAC names:
17-2 (a) (Z)-2-Ethyl-2-buten-1-ol
(b) 3-Cyclohexen-1-ol
(c) trans-3-Chlorocycloheptanol
(d) 1,4-Pentanediol
(e) 2,6-Dimethylphenol
(f)
-(2-Hydroxyethyl)phenol


Allyl alcohol (2-propen-1-ol) - 17.2 Properties of Alcohols and Phenols
Alcohols and phenols have nearly the same geometry around the oxygen atom as water. The $\mathrm{R}-\mathrm{O}-\mathrm{H}$ bond angle has an approximately tetrahedral value ( $108.5^{\circ}$ in methanol, for instance), and the oxygen atom is $s p^{3}$-hybridized.

Also like water, alcohols and phenols have higher boiling points than might be expected, because of hydrogenbonding (Section 2.12). A positively polarized -OH hydrogen atom from one molecule is attracted to a lone pair of electrons on the electronegative oxygen atom of another molecule, resulting in a weak force that holds the molecules together (FIGURE 17.2). These intermolecular attractions must be overcome for a molecule to break free from the liquid and enter the vapor state, so the boiling temperature is raised. For example, 1-propanol ( $\mathrm{MW}=60$ ), butane $(\mathrm{MW}=58)$, and chloroethane $(\mathrm{MW}=65)$ have similar molecular weights, yet 1-propanol boils at $97^{\circ} \mathrm{C}$, compared with $-0.5^{\circ} \mathrm{C}$ for the alkane and $12.5^{\circ} \mathrm{C}$ for the chloroalkane.

FIGURE 17.2 Hydrogen-bonding in alcohols and phenols. Attraction between a positively polarized -OH hydrogen and a negatively polarized oxygen holds molecules together. The electrostatic potential map of methanol shows the positively polarized -OH hydrogen and the negatively polarized oxygen.

Another similarity with water is that alcohols and phenols are both weakly basic and weakly acidic. As weak bases, they are reversibly protonated by strong acids to yield oxonium ions, $\mathrm{ROH}_{2}{ }^{+}$.

An alcohol
An oxonium ion

$$
\left[\text { or } \mathrm{ArOH}+\mathrm{HX} \rightleftarrows \stackrel{+}{\mathrm{H}}_{2} \mathrm{ArO}^{-}\right]
$$


Allyl alcohol (2-propen-1-ol) - 17.2 Properties of Alcohols and Phenols
As weak acids, they dissociate slightly in dilute aqueous solution by donating a proton to water, generating $\mathrm{H}_{3} \mathrm{O}^{+}$ and an alkoxide ion, $\mathrm{RO}^{-}$, or a phenoxide ion, $\mathrm{ArO}^{-}$.

An alcohol
An alkoxide ion

Recall from the earlier discussion of acidity in Section 2.7 to Section 2.11 that the strength of any acid HA in water can be expressed by an acidity constant, $K_{\mathrm{a}}$.

$$
K_{\mathrm{a}}=\frac{\left[\mathrm{A}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{[\mathrm{HA}]} \quad \mathrm{p} K_{\mathrm{a}}=-\log K_{\mathrm{a}}
$$

Compounds with a smaller $K_{\mathrm{a}}$ and larger $\mathrm{p} K_{\mathrm{a}}$ are less acidic, whereas compounds with a larger $K_{\mathrm{a}}$ and smaller $\mathrm{p} K_{\mathrm{a}}$ are more acidic. As shown in TABLE 17.1, simple alcohols like methanol and ethanol are about as acidic as water, but the more highly substituted tert-butyl alcohol is somewhat weaker. Substituent groups also have a significant effect: 2,2,2-trifluoroethanol is approximately 3700 times stronger than ethanol, for instance. Phenols and thiols, the sulfur analogs of alcohols, are substantially more acidic than water.


Allyl alcohol (2-propen-1-ol) - 17.2 Properties of Alcohols and Phenols
| Compound | $\mathrm{p} K_{\mathrm{a}}$ |  |
| :---: | :---: | :---: |
| $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}$ | 18 | Weaker acid |
| $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ | 16 |  |
| $\mathrm{H}_{2} \mathrm{O}$ | 15.74 |  |
| $\mathrm{CH}_{3} \mathrm{OH}$ | 15.54 |  |
| $\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{OH}$ | 12.43 |  |
| p-Aminophenol | 10.46 |  |
| $\mathrm{CH}_{3} \mathrm{SH}$ | 10.3 |  |
| $p$-Methylphenol | 10.17 |  |
| Phenol | 9.89 |  |
| $p$-Chlorophenol | 9.38 | acid |
| p-Nitrophenol | 7.15 |  |

The effect of alkyl substitution on alcohol acidity is due primarily to solvation of the alkoxide ion formed on acid dissociation. The more readily the alkoxide ion is solvated by water, the more stable it is, the more its formation is energetically favored, and the greater the acidity of the parent alcohol. For example, the oxygen atom of an unhindered alkoxide ion, such as that from methanol, is sterically accessible and is easily solvated by water. The oxygen atom of a hindered alkoxide ion, however, such as that from tert-butyl alcohol, is less easily solvated and is therefore less stable.


Allyl alcohol (2-propen-1-ol) - 17.2 Properties of Alcohols and Phenols
Inductive effects (Section 16.4) are also important in determining alcohol acidities. Electron-withdrawing halogen substituents, for instance, stabilize an alkoxide ion by spreading the charge over a larger volume, thus making the alcohol more acidic. Compare, for instance, the acidities of ethanol ( $\mathrm{p} K_{\mathrm{a}}=16$ ) and 2,2,2-trifluoroethanol ( $\mathrm{p} K_{\mathrm{a}}=12.43$ ), or of tert-butyl alcohol ( $\mathrm{p} K_{\mathrm{a}}=18$ ) and nonafluoro-tert-butyl alcohol $\left(\mathrm{p} K_{\mathrm{a}}=\right.$ 5.4).

Electron-withdrawing groups stabilize the alkoxide ion and lower the $\mathrm{p} K_{\mathrm{a}}$ of the alcohol.

Because alcohols are weak acids, they don't react with weak bases, such as amines or bicarbonate ion, and they only react to a limited extent with metal hydroxides such as NaOH . Alcohols do, however, react with alkali metals and with strong bases such as sodium hydride ( NaH ), sodium amide ( $\mathrm{NaNH}_{2}$ ), and Grignard reagents ( RMgX ). Alkoxides are themselves bases that are frequently used as reagents in organic chemistry. They are named systematically by adding the -ate suffix to the name of the alcohol. Methanol becomes methanolate, for instance.

Phenols are about a million times more acidic than alcohols (TABLE 17.1). They are therefore soluble in dilute aqueous NaOH and can often be separated from a mixture simply by basic extraction into aqueous solution, followed by reacidification.


Allyl alcohol (2-propen-1-ol) - 17.2 Properties of Alcohols and Phenols
Phenols are more acidic than alcohols because the phenoxide anion is resonance-stabilized. Delocalization of the negative charge over the ortho and para positions of the aromatic ring results in increased stability of the phenoxide anion relative to undissociated phenol and in a consequently lower $\Delta G^{\circ}$ for dissociation. FIGURE 17.3 compares electrostatic potential maps of an alkoxide ion $\left(\mathrm{CH}_{3} \mathrm{O}^{-}\right)$with phenoxide ion to show how the negative charge in phenoxide ion is delocalized from oxygen to the ring.

FIGURE 17.3 The resonance-stabilized phenoxide ion is more stable than an alkoxide ion. Electrostatic potential maps show how the negative charge is concentrated on oxygen in the methoxide ion but is spread over the aromatic ring in the phenoxide ion.

Substituted phenols can be either more acidic or less acidic than phenol itself, depending on whether the substituent is electron-withdrawing or electron-donating (Section 16.4). Phenols with an electron-withdrawing substituent are more acidic because these substituents delocalize the negative charge; phenols with an electron-donating substituent are less acidic because these substituents concentrate the charge. The acidifying effect of an electron-withdrawing substituent is particularly noticeable in phenols with a nitro group at the ortho or para position.


(8) WORKED EXAMPLE 17.1 -
Predicting the Relative Acidity of a Substituted Phenol
Is $p$-hydroxybenzaldehyde more acidic or less acidic than phenol?


Strategy -
Identify the substituent on the aromatic ring, and decide whether it is electron-donating or electronwithdrawing. Electron-withdrawing substituents make the phenol more acidic by stabilizing the phenoxide anion, and electron-donating substituents make the phenol less acidic by destabilizing the anion.


Solution -
We saw in Section 16.4 that a carbonyl group is electron-withdrawing. Thus, $p$-hydroxybenzaldehyde is more
acidic $\left(\mathrm{p} K_{\mathrm{a}}=7.9\right)$ than phenol $\left(\mathrm{p} K_{\mathrm{a}}=9.89\right)$.

p-Hydroxybenzaldehyde
( $\mathrm{p} K_{\mathrm{a}}=7.9$ )

PROBLEM The following data for isomeric four-carbon alcohols show that there is a decrease in boiling point
17-3 with increasing substitution of the OH -bearing carbon. How might you account for this trend?

$$
\begin{aligned}
& \text { 1-Butanol, bp } 117.5^{\circ} \mathrm{C} \\
& \text { 2-Butanol, bp } 99.5^{\circ} \mathrm{C} \\
& \text { 2-Methyl-2-propanol, bp } 82.2^{\circ} \mathrm{C}
\end{aligned}
$$

PROBLEM Rank the following substances in order of increasing acidity:
17-4 (a) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH}, \mathrm{HC} \equiv \mathrm{CH},\left(\mathrm{CF}_{3}\right)_{2} \mathrm{CHOH}, \mathrm{CH}_{3} \mathrm{OH}$
(b) Phenol, $p$-methylphenol, $p$-(trifluoromethyl)phenol
(c) Benzyl alcohol, phenol, $p$-hydroxybenzoic acid

PROBLEM $p$-Nitrobenzyl alcohol is more acidic than benzyl alcohol, but p-methoxybenzyl alcohol is less 17-5 acidic. Explain.


Solution - 17.3 Preparation of Alcohols: A Review
Alcohols occupy a central position in organic chemistry. They can be prepared from many other kinds of compounds (alkenes, alkyl halides, ketones, esters, and aldehydes, among others), and they can be transformed into an equally wide assortment of compounds (FIGURE 17.4).

FIGURE 17.4 The central position of alcohols in organic chemistry. Alcohols can be prepared from, and converted into, many other kinds of compounds.

We've already seen several methods of alcohol synthesis:

- Alcohols can be prepared by hydration of alkenes. Because the direct hydration of alkenes with aqueous acid is generally a poor reaction in the laboratory, two indirect methods are commonly used. Hydroboration-oxidation yields the syn, non-Markovnikov hydration product (Section 8.5), whereas oxymercuration-demercuration yields the Markovnikov hydration product (Section 8.4).

- 1,2-Diols can be prepared either by direct hydroxylation of an alkene with $\mathrm{OsO}_{4}$ followed by reduction with $\mathrm{NaHSO}_{3}$ or by acid-catalyzed hydrolysis of an epoxide (Section 8.7). The $\mathrm{OsO}_{4}$ reaction occurs with syn stereochemistry to give a cis diol, and epoxide opening occurs with anti stereochemistry to give a trans diol.

PROBLEM Predict the products of the following reactions:
17-6 (a)

(c)


Solution - 17.4 Alcohols from Carbonyl Compounds: Reduction
The most general method for preparing alcohols, both in the laboratory and in living organisms, is by reduction of a carbonyl compound. Just as reduction of an alkene adds hydrogen to a $\mathrm{C}=\mathrm{C}$ bond to give an alkane (Section 8.6), reduction of a carbonyl compound adds hydrogen to a $\mathrm{C}=\mathrm{O}$ bond to give an alcohol. All kinds of carbonyl compounds can be reduced, including aldehydes, ketones, carboxylic acids, and esters.


A carbonyl compound -
An alcohol


Reduction of Aldehydes and Ketones -
Aldehydes are easily reduced to give primary alcohols, and ketones are reduced to give secondary alcohols.

Dozens of reagents are used in the laboratory to reduce aldehydes and ketones, depending on the circumstances, but sodium borohydride, $\mathrm{NaBH}_{4}$, is usually chosen because of its safety and ease of handling. Sodium borohydride is a white, crystalline solid that can be weighed in the open atmosphere and used in either water or alcohol solution.


Butanal -
1-Butanol (85\%)
(a $1^{\circ}$ alcohol)


Ketone reduction -
Lithium aluminum hydride, $\mathrm{LiAlH}_{4}$, is another reducing agent often used for reduction of aldehydes and ketones. A grayish powder that is soluble in ether and tetrahydrofuran, $\mathrm{LiAlH}_{4}$ is much more reactive than $\mathrm{NaBH}_{4}$ but also much more dangerous. It reacts violently with water and decomposes explosively when heated above $120^{\circ} \mathrm{C}$.

2-Cyclohexenone
2-Cyclohexenol (94\%)
We'll defer a detailed discussion of these reductions until Chapter 19. For the moment, we'll simply note that they involve the addition of a nucleophilic hydride ion $\left(: \mathrm{H}^{-}\right)$to the positively polarized, electrophilic carbon atom of the carbonyl group. The initial product is an alkoxide ion, which is protonated by addition of $\mathrm{H}_{3} \mathrm{O}^{+}$in a second step to yield the alcohol product.

In living organisms, aldehyde and ketone reductions are carried out by either of the coenzymes NADH (reduced nicotinamide adenine dinucleotide) or NADPH (reduced nicotinamide adenine dinucleotide phosphate). Although these biological "reagents" are much more complex structurally than $\mathrm{NaBH}_{4}$ or $\mathrm{LiAlH}_{4}$, the mechanisms of laboratory and biological reactions are similar. The coenzyme acts as a hydride-ion donor to give an alkoxide anion, and the intermediate anion is then protonated by acid. An example is the reduction of acetoacetyl ACP to $\beta$-hydroxybutyryl ACP, a step in the biological synthesis of fats (FIGURE 17.5). Note that the pro- $R$ hydrogen of NADPH is the one transferred in this example. Enzyme-catalyzed reactions usually occur with high specificity, although it's not usually possible to predict the stereochemical result before the fact.

FIGURE 17.5 The biological reduction of a ketone (acetoacetyl ACP) to an alcohol ( $\beta$-hydroxybutyryl ACP) by NADPH.


Reduction of Carboxylic Acids and Esters -
Carboxylic acids and esters are reduced to give primary alcohols.

A carboxylic acid
An ester
A primary alcohol
These reactions aren't as rapid as the reductions of aldehydes and ketones. $\mathrm{NaBH}_{4}$ reduces esters very slowly and does not reduce carboxylic acids at all. Instead, carboxylic acid and ester reductions are usually carried out with the more reactive reducing agent $\mathrm{LiAlH}_{4}$. All carbonyl groups, including acids, esters, ketones, and aldehydes, are reduced by $\mathrm{LiAlH}_{4}$. Note that one hydrogen atom is delivered to the carbonyl carbon atom during aldehyde and ketone reductions but that two hydrogens become bonded to the former carbonyl carbon during carboxylic acid and ester reductions. We'll defer a discussion of the mechanisms of these reactions until Chapter 21.

Carboxylic acid reduction

Ester reduction

Methyl 2-pentenoate 2-Penten-1-ol (91\%)


Identifying a Reactant, Given the Product -
What carbonyl compounds would you reduce to obtain the following alcohols?
(a)

(b)


Strategy -
Identify the target alcohol as primary, secondary, or tertiary. A primary alcohol can be prepared by reduction of an aldehyde, an ester, or a carboxylic acid; a secondary alcohol can be prepared by reduction of a ketone; and a tertiary alcohol can't be prepared by reduction.


Solution -
(a) The target molecule is a secondary alcohol, which can be prepared only by reduction of a ketone. Either $\mathrm{NaBH}_{4}$ or $\mathrm{LiAlH}_{4}$ can be used.

(b) The target molecule is a primary alcohol, which can be prepared by reduction of an aldehyde, an ester, or a carboxylic acid. $\mathrm{LiAlH}_{4}$ is needed for the ester and carboxylic acid reductions.

PROBLEM What reagent would you use to accomplish each of the following reactions?
17-7 (a)

(b)

PROBLEM What carbonyl compounds give the following alcohols on reduction with $\mathrm{LiAlH}_{4}$ ? Show all 17-8 possibilities.
(a)

(b)

(c)

(d) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{OH}$


Solution - 17.5 Alcohols from Carbonyl Compounds: Grignard Reaction
Grignard reagents (RMgX), prepared by reaction of organohalides with magnesium (Section 10.6), react with carbonyl compounds to yield alcohols in much the same way that hydride reducing agents do. Just as carbonyl reduction involves addition of a hydride ion nucleophile to the $\mathrm{C}=\mathrm{O}$ bond, Grignard reaction involves addition of a carbanion nucleophile ( $\mathrm{R} \mathrm{P}^{-+} \mathrm{MgX}$ ).

$$
\left[\mathrm{R}-\mathrm{X}+\mathrm{Mg} \longrightarrow \underset{\begin{array}{c}
\text { A Grignard } \\
\text { reagent }
\end{array}}{\stackrel{\delta^{-}}{\mathrm{R}-\mathrm{ingX}} \mathrm{\delta}^{+}}\left\{\begin{array}{l}
\mathrm{R}=1^{\circ}, 2^{\circ}, \text { or } 3^{\circ} \text { alkyl, aryl, or vinylic } \\
\mathrm{X}=\mathrm{Cl}, \mathrm{Br}, \mathrm{I}
\end{array}\right]\right.
$$

The nucleophilic addition reaction of Grignard reagents to carbonyl compounds has no direct counterpart in biological chemistry because organomagnesium compounds are too strongly basic to exist in an aqueous medium. Nevertheless, the reaction is worth understanding for two reasons. First, the reaction is an unusually broad and useful method of alcohol synthesis and demonstrates again the relative freedom with which chemists can operate in the laboratory. Second, the reaction does have an indirect biological counterpart, for we'll see in Chapter 23 that the addition of stabilized carbon nucleophiles to carbonyl compounds is used in almost all metabolic pathways as the major process for forming carbon-carbon bonds.

As examples of their addition to carbonyl compounds, Grignard reagents react with formaldehyde, $\mathrm{H}_{2} \mathrm{C}=\mathrm{O}$, to give primary alcohols, with aldehydes to give secondary alcohols, and with ketones to give tertiary alcohols.


Aldehyde reaction -
Ketone reaction

Esters react with Grignard reagents to yield tertiary alcohols in which two of the substituents bonded to the hydroxyl-bearing carbon have come from the Grignard reagent, just as $\mathrm{LiAlH}_{4}$ reduction of an ester adds two hydrogens.

Carboxylic acids don't give addition products when treated directly with Grignard reagents because the acidic carboxyl hydrogen reacts with the basic Grignard reagent to yield a hydrocarbon and the magnesium salt of the acid.

$$
[R-X+M g \quad R-M g X]
$$


A carboxylic acid -
A carboxylic acid salt
Carboxylic acids do, however, react with Grignard reagents to give ketones if they are first treated with $i-\mathrm{Pr}_{2} \mathrm{NMgCl}-\mathrm{LiCl}$, called the turbo-Hauser base, to form a complex and increase the electrophilicity of the carboxylate anion toward nucleophilic addition with a Grignard reagent.

The Grignard reaction, although useful, does have limitations. One major problem is that a Grignard reagent can't be prepared from an organohalide if other reactive functional groups are present in the same molecule. For example, a compound that is both an alkyl halide and a ketone can't form a Grignard reagent because it would react with itself. Similarly, a compound that is both an alkyl halide and a carboxylic acid, alcohol, or amine can't form a Grignard reagent because the acidic $\mathrm{RCO}_{2} \mathrm{H}, \mathrm{ROH}$, or $\mathrm{RNH}_{2}$ hydrogen present in the same molecule would react with the basic Grignard reagent as rapidly as it forms. In general, Grignard reagents can't be prepared from alkyl halides that contain the following functional groups (FG):

As with the reduction of carbonyl compounds discussed in the previous section, we'll defer a detailed treatment of the Grignard reactions until Chapter 19. For the moment, it's sufficient to note that Grignard reagents act as nucleophilic carbanions ( $: \mathrm{R}^{-}$) and that their addition to a carbonyl compound is analogous to the addition of hydride ion. The intermediate is an alkoxide ion, which is protonated by addition of $\mathrm{H}_{3} \mathrm{O}^{+}$in a second step.


Using a Grignard Reaction to Synthesize an Alcohol -
How could you use the addition of a Grignard reagent to a ketone to synthesize 2-phenyl-2-butanol?


Strategy -
Draw the product, and identify the three groups bonded to the alcohol carbon atom. One of the three will have come from the Grignard reagent, and the remaining two will have come from the ketone.


Solution -
2-Phenyl-2-butanol has a methyl group, an ethyl group, and a phenyl group ( $-\mathrm{C}_{6} \mathrm{H}_{5}$ ) attached to the alcohol carbon atom. Thus, the possibilities are addition of ethylmagnesium bromide to acetophenone, addition of methylmagnesium bromide to propiophenone, and addition of phenylmagnesium bromide to 2-butanone.


Using a Grignard Reaction to Synthesize an Alcohol -
How could you use the reaction of a Grignard reagent with a carbonyl compound to synthesize 2-methyl-2-pentanol?


Strategy -
Draw the product, and identify the three groups bonded to the alcohol carbon atom. If the three groups are all different, the starting carbonyl compound must be a ketone. If two of the three groups are identical, the starting carbonyl compound could be either a ketone or an ester.


Solution -
In the present instance, the product is a tertiary alcohol with two methyl groups and one propyl group. Starting from a ketone, the possibilities are addition of methylmagnesium bromide to 2 -pentanone and addition of propylmagnesium bromide to acetone.


Acetone -
Starting from an ester, the only possibility is addition of methylmagnesium bromide to an ester of butanoic acid, such as methyl butanoate.

PROBLEM Show the products obtained from addition of methylmagnesium bromide to the following 17-9 compounds:
(a) Cyclopentanone
(b) Benzophenone (diphenyl ketone)
(c) 3-Hexanone

PROBLEM Use a Grignard reaction to prepare the following alcohols:
17-10 (a) 2-Methyl-2-propanol (b) 1-Methylcyclohexanol (c) 3-Methyl-3-pentanol


(d) 2-Phenyl-2-butanol (e) Benzyl alcohol (f) 4-Methyl-1-pentanol -
PROBLEM Use the reaction of a Grignard reagent with a carbonyl compound to synthesize the following 17-11 compound:


(d) 2-Phenyl-2-butanol (e) Benzyl alcohol (f) 4-Methyl-1-pentanol - 17.6 Reactions of Alcohols
We've already seen several reactions of alcohols-their conversion into alkyl halides and tosylates in Section 10.5 and their dehydration to give alkenes in Section 8.1-albeit without mechanistic details. Let's now look at those details.


Conversion of Alcohols into Alkyl Halides -
Tertiary alcohols react with either HCl or HBr at $0^{\circ} \mathrm{C}$ by an $\mathrm{S}_{\mathrm{N}} 1$ mechanism through a carbocation intermediate. Primary and secondary alcohols are much more resistant to acid, however, and are best converted into halides by treatment with either $\mathrm{SOCl}_{2}$ or $\mathrm{PBr}_{3}$ through an $\mathrm{S}_{\mathrm{N}} 2$ mechanism.

The reaction of a tertiary alcohol with HX takes place by an $\mathrm{S}_{\mathrm{N}} 1$ mechanism when acid protonates the hydroxyl oxygen atom. Water is expelled to generate a carbocation, and the cation reacts with nucleophilic halide ion to give the alkyl halide product.

The reactions of primary and secondary alcohols with $\mathrm{SOCl}_{2}$ and $\mathrm{PBr}_{3}$ take place by $\mathrm{S}_{\mathrm{N}} 2$ mechanisms. Hydroxide ion itself is too poor a leaving group to be displaced by nucleophiles in $\mathrm{S}_{\mathrm{N}} 2$ reactions, but reaction of an alcohol with $\mathrm{SOCl}_{2}$ or $\mathrm{PBr}_{3}$ converts the -OH into a much better leaving group, either a chlorosulfite (-OSOCl) or a dibromophosphite $\left(-\mathrm{OPBr}_{2}\right)$, which is readily expelled by backside nucleophilic substitution.


Conversion of Alcohols into Tosylates -
Alcohols react with $p$-toluenesulfonyl chloride (tosyl chloride, $p$ - TosCl ) in pyridine solution to yield alkyl
tosylates, ROTos (Section 11.1). Only the $\mathrm{O}-\mathrm{H}$ bond of the alcohol is broken in this reaction; the $\mathrm{C}-\mathrm{O}$ bond remains intact, so no change of configuration occurs if the oxygen is attached to a chirality center. The resultant alkyl tosylates behave much like alkyl halides, undergoing both $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ substitution reactions.

One of the most important reasons for using tosylates in $S_{N} 2$ reactions is stereochemical. The $S_{N} 2$ reaction of an alcohol with an alkyl halide proceeds with two inversions of configuration-one to make the halide from the alcohol and one to substitute the halide-and yields a product with the same stereochemistry as the starting alcohol. The $\mathrm{S}_{\mathrm{N}} 2$ reaction of an alcohol with a tosylate, however, proceeds with only one inversion and yields a product of opposite stereochemistry to the starting alcohol. FIGURE 17.6 shows a series of reactions on the $R$ enantiomer of 2 -octanol that illustrates these stereochemical relationships.

FIGURE 17.6 Stereochemical consequences of $\mathbf{S}_{\mathbf{N}} \mathbf{2}$ reactions on derivatives of ( $\boldsymbol{R}$ )-2-octanol. Substitution through the halide gives a product with the same stereochemistry as the starting alcohol; substitution through the tosylate gives a product with opposite stereochemistry to the starting alcohol.

PROBLEM How would you carry out the following transformation, a step used in the commercial synthesis of 17-12 (S)-ibuprofen?


Dehydration of Alcohols to Yield Alkenes -
A third important reaction of alcohols, both in the laboratory and in biological pathways, is their dehydration to give alkenes. Because of the usefulness of the reaction, a number of methods have been devised for carrying out dehydrations. One method that works particularly well for tertiary alcohols is the acid-catalyzed reaction discussed in Section 8.1. For example, treatment of 1-methylcyclohexanol with warm, aqueous sulfuric acid in a solvent such as tetrahydrofuran results in loss of water and formation of 1-methylcyclohexene.


1-Methylcyclohexanol 1-Methylcyclohexene (91\%) -
Acid-catalyzed dehydrations usually follow Zaitsev's rule (Section 11.7) and yield the more stable alkene as the major product. Thus, 2-methyl-2-butanol gives primarily 2-methyl-2-butene (trisubstituted double bond)
rather than 2-methyl-1-butene (disubstituted double bond).

2-Methyl-2-butanol
2-Methyl-2-butene 2-Methyl-1-butene (trisubstituted) (disubstituted) Major product Minor product
This reaction is an E1 process (Section 11.10) and occurs by the three-step mechanism shown in FIGURE 17.7 on the next page. Protonation of the alcohol oxygen is followed by unimolecular loss of water to generate a carbocation intermediate and final loss of a proton from the neighboring carbon atom to complete the process. As with most E1 reactions, tertiary alcohols react fastest because they lead to stabilized, tertiary carbocation intermediates. Secondary alcohols can be made to react, but the conditions are severe $\left(75 \% \mathrm{H}_{2} \mathrm{SO}_{4}, 100^{\circ} \mathrm{C}\right)$ and sensitive molecules don't survive.

FIGURE 17.7 MECHANISM
Mechanism for the acid-catalyzed dehydration of a tertiary alcohol to yield an
alkene. The process is an E1 reaction and involves a carbocation intermediate.
(1) Two electrons from the oxygen atom bond to $\mathrm{H}^{+}$, yielding a protonated alcohol intermediate.

(2) The carbon-oxygen bond breaks, and the two electrons from the bond stay with oxygen, leaving a carbocation intermediate.

Carbocation
(3) Two electrons from a neighboring carbon-hydrogen bond form the alkene $\pi$ bond, and $\mathrm{H}^{+}$(a proton) is eliminated.


1-Methylcyclohexanol 1-Methylcyclohexene (91\%) -
Carbocation
(3) Two electrons from a neighboring carbon-hydrogen bond form the alkene $\pi$ bond, and $\mathrm{H}^{+}$(a proton) is eliminated.

To circumvent the need for strong acid and allow the dehydration of secondary alcohols in a gentler way, reagents have been developed that are effective under mild, basic conditions. One such reagent, phosphorus oxychloride $\left(\mathrm{POCl}_{3}\right)$ in the basic amine solvent pyridine, is often able to effect the dehydration of secondary and
tertiary alcohols at $0^{\circ} \mathrm{C}$.

1-Methylcyclohexanol
1-Methylcyclohexene (96\%)
Alcohol dehydrations carried out with $\mathrm{POCl}_{3}$ in pyridine take place by an E2 mechanism, as shown in FIGURE 17.8. Because hydroxide ion is a poor leaving group (Section 11.3), direct E 2 elimination of water from an alcohol does not occur. On reaction with $\mathrm{POCl}_{3}$, however, the -OH group is converted into a dichlorophosphate $\left(-\mathrm{OPOCl}_{2}\right)$, which is a good leaving group and is readily eliminated. Pyridine is both the reaction solvent and the base that removes a neighboring proton in the E2 elimination step.

FIGURE 17.8 MECHANISM

Mechanism for the dehydration of secondary and tertiary alcohols by reaction with $\mathrm{POCl}_{3}$ in pyridine. The reaction is an E2 process.
(1) The alcohol hydroxyl group reacts with $\mathrm{POCl}_{3}$ to form a dichlorophosphate intermediate.

2 E2 elimination then occurs by the usual one-step mechanism as the amine base pyridine abstracts a proton from the neighboring carbon at the same time that the
dichlorophosphate group is leaving.


1-Methylcyclohexanol 1-Methylcyclohexene (91\%) -
2 E2 elimination then occurs by the usual one-step mechanism as the amine base pyridine abstracts a proton from the neighboring carbon at the same time that the
dichlorophosphate group is leaving.

As noted in Section 11.11, biological dehydrations are also common and usually occur by an E1cB mechanism on a substrate in which the - OH group is two carbons away from a carbonyl group. One example occurs in the biosynthesis of the aromatic amino acid tyrosine. A base (:B) first abstracts a proton from the carbon adjacent to the carbonyl group, and the anion intermediate then expels the - OH group with simultaneous protonation by an acid (HA) to form water.

PROBLEM What product(s) would you expect from dehydration of the following alcohols with $\mathrm{POCl}_{3}$ in 17-13 pyridine? Indicate the major product in each case.
(a)

(b)

(c)

(d)

(e)


Conversion of Alcohols into Esters -
Alcohols react with carboxylic acids to give esters, a reaction that is common in both the laboratory and living organisms. In the laboratory, the reaction can be carried out in a single step if a strong acid is used as catalyst. More frequently, though, the reactivity of the carboxylic acid is enhanced by first converting it into a carboxylic acid chloride, which then reacts with the alcohol.

In living organisms, a similar process occurs, though a thioester or acyl adenosyl phosphate acts as substrate rather than a carboxylic acid chloride. We'll look at the mechanisms of these reactions in Chapter 21.

An acyl adenosyl phosphate


Conversion of Alcohols into Esters - 17.7 Oxidation of Alcohols
Perhaps the most valuable reaction of alcohols is their oxidation to give carbonyl compounds-the opposite of the reduction of carbonyl compounds to give alcohols. Primary alcohols are oxidized either to aldehydes or carboxylic acids, and secondary alcohols are oxidized to ketones, but tertiary alcohols don't normally react with most oxidizing agents.

Primary alcohol

Secondary alcohol

A ketone

Tertiary alcohol

Primary and secondary alcohols can be oxidized by any of a number of reagents, including $\mathrm{CrO}_{3}$ in aqueous acetic acid and $\mathrm{KMnO}_{4}$ in aqueous NaOH , but chromium-based reagents are rarely used today because of their toxicity and fire danger. Today, primary and secondary alcohols are oxidized to aldehydes and ketones, respectively, using the iodine-containing Dess-Martin periodinane in dichloromethane solution.

Primary alcohols are oxidized to carboxylic acids by heating with $\mathrm{KMnO}_{4}$ in a basic aqueous solution. An aldehyde is involved as an intermediate in the $\mathrm{KMnO}_{4}$ reaction but can't usually be isolated because it is further oxidized too rapidly.

All these oxidations occur by a mechanism that is closely related to the E2 reaction (Section 11.8). In the Dess-Martin oxidation, for instance, the first step involves a substitution reaction between the alcohol and the $\mathrm{I}(\mathrm{V})$ reagent to form a new periodinane intermediate, followed by expulsion of reduced $\mathrm{I}(\mathrm{III})$ as the leaving group. Similarly, when a $\mathrm{Cr}(\mathrm{VI})$ reagent, such as $\mathrm{CrO}_{3}$, is the oxidant, reaction with the alcohol gives a chromate
intermediate followed by expulsion of a reduced $\mathrm{Cr}(\mathrm{VI})$ species.


Conversion of Alcohols into Esters - 17.7 Oxidation of Alcohols
Biological alcohol oxidations are the opposite of biological carbonyl reductions and are facilitated by the coenzymes $\mathrm{NAD}^{+}$and $\mathrm{NADP}^{+}$. A base removes the -OH proton, and the alkoxide ion transfers a hydride ion to the coenzyme. An example is the oxidation of sn-glycerol 3-phosphate to dihydroxyacetone phosphate, a step in the biological metabolism of fats (FIGURE 17.9). Note that addition occurs exclusively on the Re face of the NAD ${ }^{+}$ ring, adding a hydrogen with pro- $R$ stereochemistry.

FIGURE 17.9 The biological oxidation of an alcohol (sn-glycerol 3-phosphate) to give a ketone (dihydroxyacetone phosphate). This mechanism is the exact opposite of the ketone reduction shown previously in FIGURE 17.5.

PROBLEM What alcohols would give the following products on oxidation?
17-14 (a)

(b)

(c)

PROBLEM What products would you expect from oxidation of the following compounds with the Dess-Martin 17-15 periodinane?
(a) 1-Hexanol
(b) 2-Hexanol
(c) Hexanal


Conversion of Alcohols into Esters - 17.8 Protection of Alcohols
It often happens, particularly during the synthesis of complex molecules, that one functional group in a molecule interferes with an intended reaction on another functional group elsewhere in the same molecule. We saw earlier in this chapter, for instance, that a Grignard reagent can't be prepared from an alcohol-containing halidel because the $\mathrm{C}-\mathrm{Mg}$ bond is not compatible with the presence of an acidic -OH group in the same molecule.

When this kind of incompatibility arises, it's sometimes possible to circumvent the problem by protecting the interfering functional group. Protection involves three steps: (1) introducing a protecting group to block the interfering function, (2) carrying out the desired reaction, and (3) removing the protecting group.

One of the more common methods of alcohol protection involves reaction with a chlorotrialkylsilane, $\mathrm{Cl}^{2}-\mathrm{SiR}_{3}$, to yield a trialkylsilyl ether, $\mathrm{R}^{\prime}-\mathrm{O}-\mathrm{SiR}_{3}$. (Chloro-tert-butyldimethylsilane), usually abbreviated either TBS or TBDMS is often used, as is chlorotrimethylsilane (TMS), and the reaction is carried out in the presence of a base, such as triethylamine, to help form the alkoxide anion from the alcohol and to remove the HCl by-product from the reaction.

For example:

The ether-forming step is an $\mathrm{S}_{\mathrm{N}} 2$-like reaction of the alkoxide ion on the silicon atom, with concurrent loss of the leaving chloride anion. Unlike most $\mathrm{S}_{\mathrm{N}} 2$ reactions, though, this reaction takes place at a tertiary center-a trialkyl-substituted silicon atom. The reaction occurs because silicon, a third-row atom, is larger than carbon and forms longer bonds. In addition, three alkyl substituents attached to silicon offer little steric hindrance to ether formation.


Conversion of Alcohols into Esters - 17.8 Protection of Alcohols
Like most other ethers, which we'll study in the next chapter, trialkylsilylethers are relatively unreactive. They have no acidic hydrogens and don't react with oxidizing agents, reducing agents, or Grignard reagents. They do, however, react with aqueous acid or with fluoride ion to regenerate the alcohol. Tetrabutylammonium fluoride is often used.

To solve the problem posed at the beginning of this section, note that it's possible to use an alcohol-containing halide in a Grignard reaction by employing a protection sequence. For example, we can add 3-bromo-1-propanol to acetaldehyde by the route shown in FIGURE 17.10.

Step 1 Protect alcohol:

Step 2a Form Grignard reagent:

Step 2b Do Grignard reaction:

Step 3 Remove protecting group:
FIGURE 17.10 Use of a trialkylsilyl-protected alcohol during a Grignard reaction.
PROBLEM Propose a mechanism for the reaction of cyclohexyl TMS ether with LiF to deprotect the silyl ether. 17-16


Conversion of Alcohols into Esters - 17.9 Phenols and Their Uses
The outbreak of World War I provided a stimulus for the industrial preparation of large amounts of synthetic phenol, which was needed as a raw material to manufacture the explosive, picric acid (2,4,6-trinitrophenol). Today, approximately 132 million tons of phenol is manufactured worldwide each year for use in such products as Bakelite resin and adhesives for binding plywood.

Phenol was manufactured for many years by the Dow process, in which chlorobenzene reacts with NaOH at high temperature and pressure (Section 16.7). Now, however, an alternative synthesis uses isopropylbenzene, commonly called cumene. Cumene reacts with air at high temperature by benzylic oxidation through a radical mechanism to form cumene hydroperoxide, which is converted into phenol and acetone by treatment with acid. This is a particularly efficient process because two valuable chemicals are prepared at the same time.

As shown in FIGURE 17.11, the reaction occurs by protonation of oxygen followed by shift of the phenyl group from carbon to oxygen with simultaneous loss of water. Replacement of the water then yields an intermediate called a hemiacetal-a compound that contains one -OR group and one -OH group bonded to the same carbon atom-which breaks down to phenol and acetone.


FIGURE 17.11 MECHANISM -
Mechanism for the formation of phenol by acid-catalyzed rearrangement of cumene hydroperoxide.

Protonation of the hydroperoxy group
on the terminal oxygen atom gives an oxonium ion...
(1) $\downarrow \mathrm{H}_{3} \mathrm{O}^{+}$

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

(3) Nucleophilic addition of water to the carbocation yields another oxonium ion...

(4)... which rearranges by a proton shift from one oxygen to another.
(4) $\downarrow$

(5) Elimination of phenol gives acetone as co-product and regenerates the acid catalyst.

$$
\text { (5 } \|-\mathrm{H}_{3} \mathrm{O}^{+}
$$

In addition to its use in making resins and adhesives, phenol is also the starting material for the synthesis of chlorinated phenols and the food preservatives BHT (butylated hydroxytoluene) and BHA (butylated hydroxyanisole). Pentachlorophenol, a widely used wood preservative, is prepared by reaction of phenol with excess $\mathrm{Cl}_{2}$. The herbicide 2,4-D (2,4-dichlorophenoxyacetic acid) is prepared from 2,4-dichlorophenol, and the hospital antiseptic agent hexachlorophene is prepared from 2,4,5-trichlorophenol.

Pentachlorophenol (wood preservative)

2,4-Dichlorophenoxyacetic acid, 2,4-D (herbicide)

Hexachlorophene (antiseptic)

The food preservative BHT is prepared by Friedel-Crafts alkylation of $p$-methylphenol ( $p$-cresol) with 2 -methylpropene in the presence of acid; BHA is prepared similarly by alkylation of $p$-methoxyphenol.

BHT

BHA


FIGURE 17.11 MECHANISM -
BHT

BHA

PROBLEM Show the mechanism for the reaction of $p$-methylphenol with 2-methylpropene and $\mathrm{H}_{3} \mathrm{PO}_{4}$ catalyst 17-17 to yield the food additive BHT.


Electrophilic Aromatic Substitution Reactions -
The hydroxyl group is a strongly activating, ortho- and para-directing substituent in electrophilic aromatic substitution reactions (Section 16.4). As a result, phenols are highly reactive substrates for electrophilic halogenation, nitration, sulfonation, and Friedel-Crafts reactions.

$+$


Oxidation of Phenols: Quinones -
Phenols don't undergo oxidation in the same way as alcohols because they don't have a hydrogen atom on the hydroxyl-bearing carbon. Instead, oxidation of a phenol yields a 2,5-cyclohexadiene-1,4-dione, or quinone. Many different oxidizing agents will accomplish the transformation, but potassium nitrosodisulfonate [ $\left(\mathrm{KSO}_{3}\right)_{2} \mathrm{NO}$ ], called Fremy's salt, is often used.

Quinones are a valuable class of compounds because of their oxidation-reduction, or redox, properties. They can be easily reduced to hydroquinones (p-dihydroxybenzenes) by reagents such as $\mathrm{NaBH}_{4}$ and $\mathrm{SnCl}_{2}$, and hydroquinones can be easily reoxidized back to quinones.

Benzoquinone
Hydroquinone
The redox properties of quinones are crucial to the functioning of living cells, where compounds called ubiquinones act as biochemical oxidizing agents to mediate the electron-transfer processes involved in energy production. Ubiquinones, also called coenzymes $Q$, are components of the cells in all aerobic organisms, from the simplest bacterium to humans. They are so named because of their ubiquitous occurrence throughout nature.

Ubiquinones ( $\boldsymbol{n}=\mathbf{1 - 1 0}$ )
Ubiquinones function within the mitochondria of cells to mediate the respiration process in which electrons are transported from the biological reducing agent NADH to molecular oxygen. Through a complex series of steps, the ultimate result is a cycle whereby NADH is oxidized to $\mathrm{NAD}^{+}, \mathrm{O}_{2}$ is reduced to water, and energy is produced. Ubiquinone acts only as an intermediary and is itself unchanged.


Step 2 -
Net change: NADH $+\frac{1}{2} \mathrm{O}_{2}+\mathrm{H}^{+} \longrightarrow \mathrm{NAD}^{+}+\mathrm{H}_{2} \mathrm{O}$


Infrared Spectroscopy -
Alcohols have a strong C-O stretching absorption near $1050 \mathrm{~cm}^{-1}$ and a characteristic $\mathrm{O}-\mathrm{H}$ stretching absorption at 3300 to $3600 \mathrm{~cm}^{-1}$. The exact position of the $\mathrm{O}-\mathrm{H}$ stretch depends on the extent of hydrogenbonding in the molecule. Unassociated alcohols show a fairly sharp absorption near $3600 \mathrm{~cm}^{-1}$, whereas hydrogen-bonded alcohols show a broader absorption in the 3300 to $3400 \mathrm{~cm}^{-1}$ range. The hydrogen-bonded hydroxyl absorption appears at $3350 \mathrm{~cm}^{-1}$ in the IR spectrum of cyclohexanol (FIGURE 17.12).

FIGURE 17.12 IR spectrum of cyclohexanol. Characteristic $\mathrm{O}-\mathrm{H}$ and $\mathrm{C}-\mathrm{O}$ stretching absorptions are indicated.
Phenols also show a characteristic broad IR absorption at $3500 \mathrm{~cm}^{-1}$ due to the -OH group, as well as the usual 1500 and $1600 \mathrm{~cm}^{-1}$ aromatic bands (FIGURE 17.13). In phenol itself, monosubstituted aromatic-ring peaks are visible at 690 and $760 \mathrm{~cm}^{-1}$.

PROBLEM Assume that you need to prepare 5-cholesten-3-one from cholesterol. How could you use IR
17-18 spectroscopy to tell whether the reaction was successful? What differences would you look for in the IR spectra of starting material and product?

Cholesterol

5-Cholestene-3-one


Nuclear Magnetic Resonance Spectroscopy -
Carbon atoms bonded to electron-withdrawing - OH groups are deshielded and absorb at a lower field in the ${ }^{13} \mathrm{C}$ NMR spectrum than do typical alkane carbons. Most alcohol carbon absorptions fall in the range 50 to $80 \delta$, as shown in the following drawing for cyclohexanol:

Alcohols also show characteristic absorptions in the ${ }^{1} \mathrm{H}$ NMR spectrum. Hydrogens on the oxygen-bearing carbon atom are deshielded by the electron-withdrawing effect of the nearby oxygen, and their absorptions
occur in the range 3.4 to $4.5 \delta$. Spin-spin splitting, however, is not usually observed between the $\mathrm{O}-\mathrm{H}$ proton of an alcohol and the neighboring protons on carbon. Most samples contain small amounts of acidic impurities, which catalyze an exchange of the $\mathrm{O}-\mathrm{H}$ proton on a timescale so rapid that the effect of spin-spin splitting is removed. It's often possible to take advantage of this rapid proton exchange to identify the position of the $\mathrm{O}-\mathrm{H}$ absorption. If a small amount of deuterated water, $\mathrm{D}_{2} \mathrm{O}$, is added to an NMR sample tube, the $\mathrm{O}-\mathrm{H}$ proton is rapidly exchanged for deuterium and the hydroxyl absorption disappears from the spectrum.

Typical spin-spin splitting is observed between protons on the oxygen-bearing carbon and other neighbors. For example, the signal of the two $-\mathrm{CH}_{2} \mathrm{O}$ - protons in 1-propanol is split into a triplet by coupling with the neighboring $-\mathrm{CH}_{2}-$ protons (FIGURE 17.14).


Nuclear Magnetic Resonance Spectroscopy -
FIGURE 17.14 ${ }^{\mathbf{1}} \mathbf{H}$ NMR spectrum of 1-propanol. The protons on the oxygen-bearing carbon are split into a triplet at $3.58 \delta$.
Phenols, like all aromatic compounds, show ${ }^{1} \mathrm{H}$ NMR absorptions near 7 to $8 \delta$, the expected position for aromatic-ring protons (Section 15.7). In addition, phenol $0-H$ protons absorb at 3 to $8 \delta$. In neither case are these absorptions uniquely diagnostic for phenols, since other kinds of protons absorb in the same range.

PROBLEM When the ${ }^{1} \mathrm{H}$ NMR spectrum of an alcohol is run in dimethyl sulfoxide (DMSO) solvent rather than
17-19 in chloroform, exchange of the $\mathrm{O}-\mathrm{H}$ proton is slow and spin-spin splitting is seen between the $\mathrm{O}-\mathrm{H}$ proton and $\mathrm{C}-\mathrm{H}$ protons on the adjacent carbon. What spin multiplicities would you expect for the hydroxyl protons in the following alcohols?
(a) 2-Methyl-2-propanol
(b) Cyclohexanol
(c) Ethanol
(d) 2-Propanol
(e) Cholesterol
(f) 1-Methylcyclohexanol


Mass Spectrometry -
As noted in Section 12.3, alcohols undergo fragmentation in the mass spectrometer by two characteristic pathways, alpha cleavage and dehydration. In the alpha-cleavage pathway, a $\mathrm{C}-\mathrm{C}$ bond nearest the hydroxyl group is broken, yielding a neutral radical plus a resonance-stabilized, oxygen-containing cation.

In the dehydration pathway, water is eliminated, yielding an alkene radical cation.

Both fragmentation modes are apparent in the mass spectrum of 1-butanol (FIGURE 17.15). The peak at $\mathrm{m} / \mathrm{z}=$ 56 is due to loss of water from the molecular ion, and the peak at $m / z=31$ is due to an alpha cleavage.

FIGURE 17.15 Mass spectrum of 1-butanol ( $\mathbf{M}^{+}=\mathbf{7 4}$ ). Dehydration gives a peak at $m / z=56$, and fragmentation by alpha cleavage gives a peak at $m / z=31$.


Ethanol: Chemical, Drug, and Poison -
The production of ethanol by fermentation of grains and sugars is one of the oldest known organic reactions, going back at least 8000 years in the Middle East and perhaps as many as 9000 years in China. Fermentation is carried out by adding yeast to an aqueous sugar solution, where enzymes break down carbohydrates into ethanol and $\mathrm{CO}_{2}$. As noted in the chapter introduction, approximately 26 billion gallons of ethanol is produced each year by fermentation of corn and sugar, with essentially the entire amount used to make bus and automobile fuel.

$$
\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \xrightarrow{\text { Yeast }} 2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+2 \mathrm{CO}_{2}
$$


A carbohydrate -
Ethanol is classified medically as a central nervous system (CNS) depressant. Its effects-that is, being drunk-resemble the human response to anesthetics. There is an initial excitability and increase in sociable behavior, but this results from depression of inhibition rather than from stimulation. At a blood alcohol concentration of $0.1 \%$ to $0.3 \%$, motor coordination is affected, accompanied by loss of balance, slurred speech, and amnesia. When blood alcohol concentration rises to between $0.3 \%$ and $0.4 \%$, nausea and loss of consciousness occur. Above $0.6 \%$, spontaneous respiration and cardiovascular regulation are affected, ultimately leading to death. The $\mathrm{LD}_{50}$ of ethanol is $10.6 \mathrm{~g} / \mathrm{kg}$ (Chapter 1 Chemistry Matters).

The passage of ethanol through the body begins with its absorption in the stomach and small intestine, followed by rapid distribution to all body fluids and organs. In the pituitary gland, ethanol inhibits the production of a hormone that regulates urine flow, causing increased urine production and dehydration. In the stomach, ethanol stimulates production of acid. Throughout the body, ethanol causes blood vessels to dilate, resulting in flushing of the skin and a sensation of warmth as blood moves into capillaries beneath the surface. The result is not a warming of the body, but an increased loss of heat at the surface.

FIGURE 17.16 The Harger Drunkometer was the first breath analyzer, introduced in 1938 to help convict drunk drivers. (credit: "Unidentified police officer with a Harger "Drunkometer" breathalyzer" by Florida Memory, State Library and Archives of Florida/Wikimedia Commons, Public Doman)

Ethanol metabolism occurs mainly in the liver and proceeds by oxidation in two steps, first to acetaldehyde $\left(\mathrm{CH}_{3} \mathrm{CHO}\right)$ and then to acetic acid $\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)$. When continuously present in the body, ethanol and acetaldehyde are toxic, leading to the devastating physical and metabolic deterioration seen in people with chronic alcohol use disorder. The liver usually suffers the worst damage since it is the major site of alcohol metabolism.


A carbohydrate -
Approximately 17,000 people are killed each year in the United States in alcohol-related automobile accidents. Thus, all 50 states have made it illegal to drive with a blood alcohol concentration (BAC) above $0.08 \%$.

Fortunately, simple tests have been devised for measuring blood alcohol concentration. The original breath analyzer test measured alcohol concentration in expired air by the color change occurring when the brightorange oxidizing agent potassium dichromate $\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right)$ reduced to blue-green chromium(III). Current consumer devices use a conductivity sensor, and tests used by law-enforcement agencies use IR spectroscopy to measure blood-alcohol levels in expired air. Just breathe into the machine, and let the spectrum tell the tale.


Key Terms -
- alcohol
- phenoxide ion $\mathrm{ArO}^{-}$
- alkoxide ion, $\mathrm{RO}^{-}$
- protecting group
- hydroquinone
- quinone
- phenol


Summary -
In previous chapters, we focused on developing general ideas of organic reactivity, looking at the chemistry of hydrocarbons and alkyl halides and seeing some of the tools used in structural studies. With that accomplished, we have now begun to study the oxygen-containing functional groups that lie at the heart of organic and biological chemistry.

Alcohols are among the most versatile of all organic compounds. They occur widely in nature, are important industrially, and have an unusually rich chemistry. The most widely used methods of alcohol synthesis start with carbonyl compounds. Aldehydes, esters, and carboxylic acids are reduced by reaction with $\mathrm{LiAlH}_{4}$ to give primary alcohols $\left(\mathrm{RCH}_{2} \mathrm{OH}\right)$; ketones are reduced to yield secondary alcohols ( $\mathrm{R}_{2} \mathrm{CHOH}$ ).

Alcohols are also prepared by reaction of carbonyl compounds with Grignard reagents, RMgX. Addition of a Grignard reagent to formaldehyde yields a primary alcohol, addition to an aldehyde yields a secondary alcohol, and addition to a ketone or an ester yields a tertiary alcohol. The Grignard reaction is limited by the fact that Grignard reagents can't be prepared from alkyl halides that contain reactive functional groups in the same molecule. This problem can sometimes be avoided by protecting the interfering functional group. Alcohols are often protected by formation of trialkylsilyl ethers.

Alcohols undergo many reactions and can be converted into many other functional groups. They can be dehydrated to give alkenes by treatment with $\mathrm{POCl}_{3}$ and can be transformed into alkyl halides by treatment with $\mathrm{PBr}_{3}$ or $\mathrm{SOCl}_{2}$. Furthermore, alcohols are weakly acidic ( $\mathrm{p} K_{\mathrm{a}} \approx 16-18$ ) and react with strong bases and with alkali metals to form alkoxide anions, which are used frequently in organic synthesis. Perhaps the most important reaction of alcohols is their oxidation to carbonyl compounds. Primary alcohols yield either aldehydes or carboxylic acids, secondary alcohols yield ketones, but tertiary alcohols are not normally oxidized.


Summary -
Phenols are aromatic counterparts of alcohols but are more acidic ( $\mathrm{p} K_{\mathrm{a}} \approx 10$ ) because their corresponding phenoxide anions are resonance stabilized by delocalization of the negative charge into the aromatic ring. Substitution of the aromatic ring by an electron-withdrawing group increases phenol acidity, and substitution by an electron-donating group decreases acidity. Phenols can be oxidized to quinones, and quinones can be reduced back to hydroquinones.


Summary of Reactions -
1. Synthesis of alcohols
a. Reduction of carbonyl compounds (Section 17.4)
(1) Aldehydes

Primary alcohol
(2) Ketones


Secondary alcohol -
(3) Esters

Primary alcohol
(4) Carboxylic acids

Primary alcohol
b. Grignard addition to carbonyl compounds (Section 17.5)
(1) Formaldehyde

Primary alcohol
(2) Aldehydes

Secondary alcohol
(3) Ketones

Tertiary alcohol
(4) Esters

Tertiary alcohol
2. Reactions of alcohols
a. Dehydration (Section 17.6)
(1) Tertiary alcohols

(2) Secondary and tertiary alcohols
b. Oxidation (Section 17.7)
(1) Primary alcohols

Carboxylic acid
(2) Secondary alcohols

Ketone
3. Oxidation of phenols to quinones (Section 17.10)


Visualizing Chemistry -
PROBLEM Give IUPAC names for the following compounds:


17-20 (a) -
(c)

(b)

(d)

PROBLEM Draw the structure of the carbonyl compound(s) from which each of the following alcohols might
17-21 have been prepared, and show the products you would obtain by treatment of each alcohol with (1) Na metal, (2) $\mathrm{SOCl}_{2}$, and (3) Dess-Martin periodinane.
(a)

(b)

PROBLEM Predict the product from reaction of the following substance (reddish brown $=\mathrm{Br}$ ) with:


17-22 -
(a) $\mathrm{PBr}_{3}$
(b) Aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$
(c) $\mathrm{SOCl}_{2}$
(d) Dess-Martin periodinane (e) $\mathrm{Br}_{2}, \mathrm{FeBr}_{3}$

PROBLEM Predict the product from reaction of the following substance with:
17-23

(a) $\mathrm{NaBH}_{4}$; then $\mathrm{H}_{3} \mathrm{O}^{+}$(b) $\mathrm{LiAlH}_{4}$; then $\mathrm{H}_{3} \mathrm{O}^{+}$(c) $2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}$; then $\mathrm{H}_{3} \mathrm{O}^{+}$

PROBLEM Name and assign $R$ or $S$ stereochemistry to the product(s) you would obtain by reaction of the
17-24 following substance with ethylmagnesium bromide. Is the product chiral? Is it optically active? Explain.


Mechanism Problems -
PROBLEM Evidence for the intermediate carbocations in the acid-catalyzed dehydration of alcohols comes
17-25 from the observation that rearrangements sometimes occur. Propose a mechanism to account for the formation of 2,3-dimethyl-2-butene from 3,3-dimethyl-2-butanol.

PROBLEM Acid-catalyzed dehydration of 2,2-dimethylcyclohexanol yields a mixture of
17-26 1,2-dimethylcyclohexene and isopropylidenecyclopentane. Propose a mechanism to account for the formation of both products.


Isopropylidenecyclopentane -
PROBLEM Epoxides react with Grignard reagents to yield alcohols. Propose a mechanism.
17-27

PROBLEM Treatment of the following epoxide with aqueous acid produces a carbocation intermediate that
$\mathbf{1 7 - 2 8}$ reacts with water to give a diol product. Show the structure of the carbocation, and propose a mechanism for the second step.

PROBLEM Reduction of 2-butanone with $\mathrm{NaBH}_{4}$ yields 2-butanol. Is the product chiral? Is it optically active?
17-29 Explain.
PROBLEM The conversion of $3^{\circ}$ alcohols into $3^{\circ}$ alkyl halides under acidic conditions involves two cationic
17-30 intermediates. For each reaction, draw the complete mechanism using curved arrows.
(a)

(b)

(c)

PROBLEM Identify the type of substitution mechanism $\left(\mathrm{S}_{\mathrm{N}} 1, \mathrm{~S}_{\mathrm{N}} 2\right)$ involved in the conversion of the following
17-31 alcohols into the corresponding alkyl halide.
(a)

(b)

(c)

PROBLEM The conversion of $3^{\circ}$ alcohols into alkenes under acidic conditions involves two cationic 17-32 intermediates. For each reaction, draw the complete mechanism using curved arrows.
(a)

(b)

(c)

PROBLEM For each reaction, write the mechanism using curved arrows for the conversion of the alcohol into
17-33 the corresponding alkene with $\mathrm{POCl}_{3}$. In each case, explain the regiochemistry of the elimination.
(a)

(b)

(c)

PROBLEM The trimethylsilyl (TMS) protecting group is one of several silicon protecting groups for alcohols.
17-34 For each reaction, draw the mechanism for the protection of ( $R$ )-3-bromo-1-butanol with the following silyl chlorides, using triethylamine as the base:
(a) tert-butyldimethylsilyl chloride (TBS-Cl)
(b) triisopropylsilyl chloride (TIPS-Cl)
(c) triethylsilyl chloride (TES-Cl)


Isopropylidenecyclopentane -
PROBLEM When the following alcohol is treated with $\mathrm{POCl}_{3}$ and pyridine, the expected elimination product
17-35 is formed. However, when the same alcohol is treated with $\mathrm{H}_{2} \mathrm{SO}_{4}$, the elimination product is 1,2-dimethylcyclopentene. Propose a mechanism for each pathway to account for these differences.

PROBLEM Phenols generally have lower $\mathrm{p} K_{\mathrm{a}}$ 's than alcohols because of resonance stabilization with the
17-36 aromatic ring. Draw all of the resonance contributors for the following phenolate ions.
(a)

(b)

(c)

Naming Alcohols
PROBLEM Give IUPAC names for the following compounds:
17-37 (a)

(b)

(c)

(d)

(e)

PROBLEM Draw and name the eight isomeric alcohols with formula $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}$. 17-38

PROBLEM Draw structures corresponding to the following IUPAC names:
17-39 (a) Trans-3-Chlorocycloheptanol
(b) 2-Ethyl-2-buten-1-ol
(c) o-(2-Hydroxyethyl)phenol
(d) 3-Methyl-1-phenyl-1-butanol

PROBLEM Bombykol, the sex pheromone secreted by the female silkworm moth has the formula $\mathrm{C}_{16} \mathrm{H}_{28} \mathrm{O}$
17-40 and the systematic name ( $10 E, 12 Z$ )-10,12-hexadecadien-1-ol. Draw bombykol, showing the correct geometry for the two double bonds.

PROBLEM Carvacrol is a naturally occurring substance isolated from oregano, thyme, and marjoram. What is 17-41 its IUPAC name?

Carvacrol

Synthesizing Alcohols
PROBLEM What Grignard reagent and what carbonyl compound might you start with to prepare the following
17-42 alcohols:
(a)

(b)

(c)

(d)

(e)

(f)


Isopropylidenecyclopentane -
(b)

(c)

(d)

(e)

(f)

PROBLEM What carbonyl compounds would you reduce to prepare the following alcohols: List all possibilities.
17-43 (a)

(b)

(c)

PROBLEM What carbonyl compounds might you start with to prepare the following compounds by Grignard
17-44 reaction? List all possibilities.
(a) 2-Methyl-2-propanol
(b) 1-Ethylcyclohexanol
(c) 3-Phenyl-3-pentanol
(d) 2-Phenyl-2-pentanol
(e)

(f)

PROBLEM How would you synthesize the following alcohols, starting with benzene and other alcohols of six or 17-45 fewer carbons as your only organic reagents?
(a)

(b)

(c)

(d)

Reactions of Alcohols
PROBLEM What products would you obtain from reaction of 1-pentanol with the following reagents:
17-46
(a) $\mathrm{PBr}_{3}$
(b) $\mathrm{SOCl}_{2}$
(c) Dess-Martin periodinane

PROBLEM How would you prepare the following compounds from 2-phenylethanol: More than one step may 17-47 be required.
(a) Styrene $\left(\mathrm{PhCH}=\mathrm{CH}_{2}\right)$ (b) Phenylacetaldehyde $\left(\mathrm{PhCH}_{2} \mathrm{CHO}\right)$
(c) Phenylacetic acid $\left(\mathrm{PhCH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)$ (d) Benzoic acid (e) Ethylbenzene (f) Benzaldehyde
(g) 1-Phenylethanol (h) 1-Bromo-2-phenylethane

PROBLEM How would you prepare the following compounds from 1-phenylethanol: More than one step may 17-48 be required.
(a) Acetophenone $\left(\mathrm{PhCOCH}_{3}\right)$
(b) Benzyl alcohol
(c) m-Bromobenzoic acid
(d) 2-Phenyl-2-propanol


Isopropylidenecyclopentane -
PROBLEM How would you prepare the following substances from cyclopentanol: More than one step may be 17-49 required.
(a) Cyclopentanone
(b) Cyclopentene
(c) 1-Methylcyclopentanol
(d) trans-2-Methylcyclopentanol

PROBLEM What products would you expect to obtain from reaction of 1-methylcyclohexanol with the 17-50 following reagents?
(a) HBr (b)
(b) NaH (c) $\mathrm{H}_{2} \mathrm{SO}_{4}$


Spectroscopy -
PROBLEM The following ${ }^{1} \mathrm{H}$ NMR spectrum is that of an alcohol, $\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}$. Propose a structure.


17-51 -
PROBLEM Propose structures for alcohols that have the following ${ }^{1}$ H NMR spectra:
17-52
(a) $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}$

(b) $\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}$

PROBLEM Propose a structure consistent with the following spectral data for a compound $\mathrm{C}_{8} \mathrm{H}_{18} \mathrm{O}_{2}$ :
17-53

> IR: $3350 \mathrm{~cm}^{-1}$
> ${ }^{1} \mathrm{H}$ NMR: $1.24 \delta(12 \mathrm{H}$, singlet $) ; 1.56 \delta(4 \mathrm{H}$, singlet $) ; 1.95 \delta(2 \mathrm{H}$, singlet $)$

PROBLEM The ${ }^{1} \mathrm{H}$ NMR spectrum shown is that of 3-methyl-3-buten-1-ol. Assign all the observed resonance
17-54 peaks to specific protons, and account for the splitting patterns.


17-51 -
PROBLEM A compound of unknown structure gave the following spectroscopic data:
17-55
Mass spectrum: $\mathrm{M}^{+}=88.1$
IR: $3600 \mathrm{~cm}^{-1}$
${ }^{1} \mathrm{H}$ NMR: $1.4 \delta(2 \mathrm{H}$, quartet, $J=7 \mathrm{~Hz}) ; 1.2 \delta(6 \mathrm{H}$, singlet); $1.0 \delta(1 \mathrm{H}$, singlet); $0.9 \delta(3 \mathrm{H}$, triplet, $J=7 \mathrm{~Hz})$
${ }^{13}$ C NMR: 74, 35, 27, $25 \delta$
(a) Assuming that the compound contains C and H but may or may not contain O , give three possible molecular formulas.
(b) How many hydrogens does the compound contain?
(c) What functional group(s) does the compound contain?
(d) How many carbons does the compound contain?
(e) What is the molecular formula of the compound?
(f) What is the structure of the compound?
(g) Assign peaks in the molecule's ${ }^{1} \mathrm{H}$ NMR spectrum corresponding to specific protons.

PROBLEM Propose a structure for a compound $\mathrm{C}_{15} \mathrm{H}_{24} \mathrm{O}$ that has the following ${ }^{1} \mathrm{H}$ NMR spectrum. The peak 17-56 marked by an asterisk disappears when $\mathrm{D}_{2} \mathrm{O}$ is added to the sample.


General Problems -
PROBLEM How would you carry out the following transformations?
17-57 (a)

(b)

(c)

PROBLEM Benzoquinone is an excellent dienophile in the Diels-Alder reaction. What product would you
17-58 expect from reaction of benzoquinone with 1 equivalent of 1,3-butadiene? From reaction with 2 equivalents of 1,3 -butadiene?

PROBLEM Rank the following substituted phenols in order of increasing acidity, and explain your answer:
17-59

PROBLEM Benzyl chloride can be converted into benzaldehyde by treatment with nitromethane and base.
17-60 The reaction involves initial conversion of nitromethane into its anion, followed by $S_{N} 2$ reaction of the anion with benzyl chloride and subsequent E2 reaction. Write the mechanism in detail, using curved arrows to indicate the electron flow in each step.

PROBLEM Reaction of (S)-3-methyl-2-pentanone with methylmagnesium bromide followed by acidification
17-61 yields 2,3-dimethyl-2-pentanol. What is the stereochemistry of the product? Is the product optically active?

3-Methyl-2-pentanone

PROBLEM Testosterone is one of the most important male steroid hormones. When testosterone is dehydrated
17-62 by treatment with acid, rearrangement occurs to yield the product shown. Propose a mechanism to account for this reaction.

PROBLEM Starting from testosterone (Problem 17-62), how would you prepare the following substances?

17-63 (a)

(c)

(b)

(d)

PROBLEM $p$-Nitrophenol and 2,6-dimethyl-4-nitrophenol both have $\mathrm{p} K_{\mathrm{a}}=7.15$, but 17-64 3,5-dimethyl-4-nitrophenol has $\mathrm{p} K_{\mathrm{a}}=8.25$. Why is 3,5-dimethyl-4-nitrophenol so much less acidic?

$\mathrm{p} K_{\mathrm{a}}=7.15$

$\mathrm{p} K_{\mathrm{a}}=7.15$


General Problems -
$\mathrm{p} K_{\mathrm{a}}=7.15$

$\mathrm{p} K_{\mathrm{a}}=7.15$

$\mathrm{p} K_{\mathrm{a}}=8.25$

PROBLEM Compound $\mathbf{A}, \mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}$, undergoes reaction with dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$ at $25{ }^{\circ} \mathrm{C}$ to yield a mixture of two 17-65 alkenes, $\mathrm{C}_{10} \mathrm{H}_{16}$. The major alkene product, $\mathbf{B}$, gives only cyclopentanone after ozone treatment followed by reduction with zinc in acetic acid. Write the reactions involved, and identify $\mathbf{A}$ and $\mathbf{B}$.

PROBLEM Compound $\mathbf{A}, \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}$, is one of the basic building blocks of nature. All steroids and many other 17-66 naturally occurring compounds are built from compound $\mathbf{A}$. Spectroscopic analysis of $\mathbf{A}$ yields the following information:

IR: $3400 \mathrm{~cm}^{-1} ; 1640 \mathrm{~cm}^{-1}$
${ }^{1} \mathrm{H}$ NMR: $1.63 \delta(3 \mathrm{H}$, singlet $) ; 1.70 \delta(3 \mathrm{H}$, singlet $) ; 3.83 \delta(1 \mathrm{H}$, broad singlet); $4.15 \delta(2 \mathrm{H}$, doublet, $J=7 \mathrm{~Hz}) ; 5.70 \delta(1 \mathrm{H}$, triplet, $J=7 \mathrm{~Hz})$
(a) How many double bonds and/or rings does $\mathbf{A}$ have?
(b) From the IR spectrum, what is the identity of the oxygen-containing functional group?
(c) What kinds of hydrogens are responsible for the NMR absorptions listed?
(d) Propose a structure for $\mathbf{A}$.


General Problems -
PROBLEM Dehydration of trans-2-methylcyclopentanol with $\mathrm{POCl}_{3}$ in pyridine yields predominantly 17-67 3-methylcyclopentene. Is the stereochemistry of this dehydration syn or anti?

PROBLEM 2,3-Dimethyl-2,3-butanediol has the common name pinacol. On heating with aqueous acid, pinacol
17-68 rearranges to pinacolone, 3,3-dimethyl-2-butanone. Suggest a mechanism for this reaction.

PROBLEM As a rule, axial alcohols oxidize somewhat faster than equatorial alcohols. Which would you expect
17-69 to oxidize faster, cis-4-tert-butylcyclohexanol or trans-4-tert-butylcyclohexanol? Draw the more stable chair conformation of each molecule.

PROBLEM Propose a synthesis of bicyclohexylidene, starting from cyclohexanone as the only source of carbon.
17-70


Bicyclohexylidene -
PROBLEM A problem often encountered in the oxidation of primary alcohols to carboxylic acids is that esters
17-71 are sometimes produced as by-products. For example, oxidation of ethanol yields acetic acid and ethyl acetate:

Propose a mechanism to account for the formation of ethyl acetate. Take into account the reversible reaction between aldehydes and alcohols:

PROBLEM Identify the reagents a-f in the following scheme:
17-72

PROBLEM Galactose, a constituent of the disaccharide lactose found in dairy products, is metabolized by a
17-73 pathway that includes the isomerization of UDP-galactose to UDP-glucose, where UDP = uridylyl diphosphate. The enzyme responsible for the transformation uses $\mathrm{NAD}^{+}$as cofactor. Propose a mechanism.

PROBLEM Propose structures for alcohols that have the following ${ }^{1} \mathrm{H}$ NMR spectra:
17-74 (a) $\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}$

(b) $\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2}$

PROBLEM Compound $\mathbf{A}, \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}$, has the IR and ${ }^{1} \mathrm{H}$ NMR spectra shown. Propose a structure consistent with
17-75 the observed spectra, and label each peak in the NMR spectrum. Note that the absorption at $5.5 \delta$ disappears when $\mathrm{D}_{2} \mathrm{O}$ is added.


CHAPTER 18 <br> Ethers and Epoxides; Thiols and Sulfides -
FIGURE 18.1 The appalling and unforgettable odor of skunks is due to a mixture of several simple thiols. (credit: modification of work "Striped Skunk" by Tom Friedel/Wikimedia Commons, CC BY 3.0)


CHAPTER CONTENTS -
18.1 Names and Properties of Ethers
18.2 Preparing Ethers
18.3 Reactions of Ethers: Acidic Cleavage
18.4 Cyclic Ethers: Epoxides
18.5 Reactions of Epoxides: Ring-Opening
18.6 Crown Ethers
18.7 Thiols and Sulfides
18.8 Spectroscopy of Ethers

WHY THIS CHAPTER? This chapter finishes the coverage of functional groups with $\mathrm{C}-\mathrm{O}$ and $\mathrm{C}-\mathrm{S}$ single bonds that was begun in the chapter on Alcohols and Phenols. We'll focus primarily on ethers and take only a brief look at thiols and sulfides before going on to an extensive coverage of compounds with $\mathrm{C}=\mathrm{O}$ double bonds in Chapters 19 through 23.

Ethers ( $\mathbf{R} \mathbf{- 0}-\mathbf{R}^{\prime}$ ), like the alcohols we saw in the preceding chapter, are organic derivatives of water, but they have two organic groups bonded to the same oxygen atom rather than one. The organic groups might be alkyl, aryl, or vinylic, and the oxygen atom might be in an open chain or a ring.

Perhaps the most well-known ether is diethyl ether, which has a long history of medicinal use as an anesthetic and industrial use as a solvent. Other useful ethers include anisole, a pleasant-smelling aromatic ether used in perfumery, and tetrahydrofuran (THF), a cyclic ether often used as a solvent.

$\mathrm{CH}_{3} \mathrm{CH}_{2}, \mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{3}$

Diethyl ether

Anisole (methyl phenyl ether)

Tetrahydrofuran

Thiols ( $\mathrm{R}-\mathrm{S}-\mathrm{H}$ ) and sulfides ( $\mathrm{R}-\mathrm{S}-\mathrm{R}^{\prime}$ ) are sulfur analogs of alcohols and ethers, respectively. Both functional groups are found in various biomolecules, although not as commonly as their oxygen-containing relatives.


CHAPTER CONTENTS - 18.1 Names and Properties of Ethers
Simple ethers with no other functional groups are named by identifying the two organic substituents and adding the word ether.

Isopropyl methyl ether

Ethyl phenyl ether

If other functional groups are present, the ether part is considered an alkoxy substituent. For example:

$p$-Dimethoxybenzene

4-tert-Butoxy-1-cyclohexene

Like alcohols, ethers have nearly the same geometry as water. The $\mathrm{R}-\mathrm{O}-\mathrm{R}$ bonds have an approximately tetrahedral bond angle ( $112^{\circ}$ in dimethyl ether), and the oxygen atom is $s p^{3}$-hybridized.

The electronegative oxygen atom gives ethers a slight dipole moment, and the boiling points of ethers are often slightly higher than the boiling points of comparable alkanes. TABLE 18.1 compares the boiling points of some common ethers and their corresponding hydrocarbons.

TABLE 18.1 Comparison of Boiling Points of Ethers and Hydrocarbons

| Ether | Boiling point ${ }^{\circ} \mathrm{C}$ | Hydrocarbon | Boiling point ${ }^{\circ} \mathrm{C}$ |
| :--- | :---: | :--- | :---: |
| $\mathrm{CH}_{3} \mathrm{OCH}_{3}$ | -25 | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | -45 |
| $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}$ | 34.6 | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ | 36 |

TABLE 18.1 Comparison of Boiling Points of Ethers and Hydrocarbons

| Ether | Boiling point ${ }^{\circ} \mathrm{C}$ | Hydrocarbon | Boiling point ${ }^{\circ} \mathrm{C}$ |
| :---: | :---: | :---: | :---: |
|  | 65 | 49 |  |
|  |  |  |  |


CHAPTER CONTENTS - 18.1 Names and Properties of Ethers
Ethers are relatively stable and unreactive in many respects, but some ethers react slowly with the oxygen in air to give peroxides, compounds that contain an $\mathrm{O}-\mathrm{O}$ bond. The peroxides from low-molecular-weight ethers such as diisopropyl ether and tetrahydrofuran are explosive and extremely dangerous, even in small amounts. Ethers are very useful as solvents in the laboratory, but they must always be used cautiously and should not be stored for long periods of time.

PROBLEM Name the following ethers:

18-1 (a)

(e) $\underset{\substack{\mathrm{CH} \\ \mathrm{CH}_{3} \mathrm{CHCH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}}}{\text { (e) }}$
(b)

(f) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{OCH}=\mathrm{CH}_{2}$
(c)

(d)