12
Alcohols and
Phenols
12.1 Structure and Properties of Alcohols
12.2 Acidity of Alcohols and Phenols
12.3 Preparation of Alcohols via
Substitution or Addition
DID YOU EVER WONDER . . .
what causes the hangover associated with
drinking alcohol and whether anything
can be done to prevent a hangover?
12.4 Preparation of Alcohols via Reduction
12.5 Preparation of Diols
12.6 Preparation of Alcohols via
Grignard Reagents
V
12.7 Protection of Alcohols
12.8 Preparation of Phenols
12.9 Reactions of Alcohols:
Substitution and Elimination
12.10 Reactions of Alcohols: Oxidation
12.11 Biological Redox Reactions
12.12 Oxidation of Phenol
g7
/G
et
ty
Im
ag
es
12.13 Synthesis Strategies
m
eisalgia, the medical term for a hangover, refers to the unpleasant physiological effects that result from drinking too much
alcohol. These effects include headache, nausea, vomiting, fatigue,
and a heightened sensitivity to light and noise.
Hangovers are caused by a multitude of factors.
These factors include (but are not limited
to) dehydration caused by the stimulation
of urine production, the loss of vitamin B, and the production of acetaldehyde in the body. Acetaldehyde is a
product of the oxidation of ethanol.
Oxidation is just one of the many
reactions that alcohols undergo. In
this chapter, we will learn about alcohols
and their reactions. Then, we will revisit the topic
of acetaldehyde production in the body, and we will see if
anything can be done to prevent a hangover.
magnetcreative/Getty Images
richcano/Getty Images
530
CHAPTER 12
Alcohols and Phenols
DO YOU REMEMBER?
Before you go on, be sure you understand the following topics.
If necessary, review the suggested sections to prepare for this chapter.
• Brønsted–Lowry Acidity (Sections 3.3–3.4)
• Mechanisms and Curved Arrows
(Sections 6.8–6.11)
• Designating the Configuration of a Chiral
Center (Section 5.3)
• Substitution and Elimination Reactions of
Alcohols (Section 7.10)
Take the DO YOU REMEMBER? QUIZ in the online course to check your understanding.
12.1 Structure and Properties of Alcohols
Alcohols are compounds that possess a hydroxyl group (OH) connected to an sp3-hybridized carbon atom, and are characterized by names ending in “ol”:
OH
OH
Ethanol
Cyclopentanol
A vast number of naturally occurring compounds contain the hydroxyl group. Here are a couple of
examples:
H
H
H
H
HO
H
H
HO
Cholecalciferol (vitamin D3)
Regulates calcium levels and helps
to form and maintain strong bones
Cholesterol
Plays a vital role in the
biosynthesis of many steroids
Phenol is a compound that exhibits an OH group connected directly to a phenyl ring. Substituted
phenols are extremely common in nature and exhibit a wide variety of properties and functions. The
following are just two examples:
O
OH
H
N
H
H
OH
O
OCH3
Phenol
OH
Capsaicin
The compound responsible for the
spicy hot flavor of chili peppers
Tetrahydrocannabinol (THC)
The psychoactive drug
found in marijuana (cannabis)
Nomenclature
Recall that four discrete steps are required to name alkanes, alkenes, and alkynes:
1.
2.
3.
4.
Identify and name the parent.
Identify and name the substituents.
Assign a locant to each substituent.
Assemble the substituents alphabetically.
Alcohols are named using the same four steps and applying the following rules:
• When naming the parent, replace the suffix “e” with “ol” to indicate the presence of a hydroxyl group:
OH
Pentane
Pentanol
12.1 Structure and Properties of Alcohols 531
• When choosing the parent of an alcohol, identify the longest chain that includes the carbon
atom connected to the hydroxyl group.
The parent must include
this carbon atom
Parent = octane
OH
Parent = hexanol
• The position of the hydroxyl group is indicated using a locant. The IUPAC rules published
in 1979 dictate that this locant be placed immediately before the parent, while the IUPAC
recommendations released in 1993 and 2004 allow for the locant to be placed before the suffix “ol.” Both names are acceptable IUPAC names.
OH
1
2
3
3- Pentanol
or
Pentan-3-ol
5
4
• When numbering the parent chain of an alcohol, the hydroxyl group should receive the lowest number possible, despite the presence of alkyl substituents or π bonds.
Correct
Incorrect
OH
OH
4
2
1
5
3
6
4
7
6
7
2
3
1
5
• When a chiral center is present, the configuration must be indicated at the beginning of the
name; for example:
3
2
1
OH
Cl
(R)-2-Chloro-3-phenyl-1-propanol
• Cyclic alcohols are numbered starting at the position bearing the hydroxyl group, so there is
no need to indicate the position of the hydroxyl group; it is understood to be at C–1.
3
OH
Cyclopentanol
2
1
OH
(R)-3,3-Dimethylcyclopentanol
In addition to systematic names, IUPAC nomenclature also recognizes common names for
many alcohols, as seen in the following examples:
OH
OH
OH
n-propyl alcohol
or n-propanol
isopropyl alcohol
or isopropanol
tert-butyl alcohol
or tert-butanol
(systematic name = 1-propanol)
(systematic name = 2-propanol)
(systematic name = 2-methyl-2-propanol)
Notice that the common name for 1-propanol is not propyl alcohol, but rather n-propyl alcohol. When assigning common names, a propyl group is called an n-propyl group to differentiate it
from an isopropyl group. The letter “n” stands for “normal” and indicates that the carbon atoms of
the alkyl group are arranged in a linear chain (not branched) with the hydroxyl group connected to
the end of the chain. Similarly, the common name for 1-butanol is n-butyl alcohol. Note that the
terms n-propyl and n-butyl are used only in common names. In systematic names, the terms propyl
and butyl are used (without the “n”).
Alcohols are designated as primary, secondary, or tertiary, depending on the number of alkyl
groups attached directly to the alpha (α) position (the carbon atom bearing the hydroxyl group).
H
R
α
H
R
OH
R
Primary
α
H
R
OH
R
Secondary
R
α
OH
Tertiary
532 CHAPTER 12 Alcohols and Phenols
The word “phenol” is used to describe a specific compound (hydroxybenzene) but is also used as the
parent name when substituents are attached.
OH
OH
Cl
Phenol
NO2
4-Chloro-2-nitrophenol
SKILLBUILDER
12.1 naming an alcohol
LEARN the skill
Cl
Cl
Assign an IUPAC name for the following alcohol:
OH
SOLUTION
Begin by identifying and naming the parent. Choose the longest chain that includes the
­carbon atom connected to the hydroxyl group and then number the chain to give the
hydroxyl group the lowest number possible.
STEP 1
Identify and name
the parent.
1
2
Cl
8
7
Cl
9
6
4
3
5
OH
3-Nonanol
Then identify the substituents and assign locants.
4,4-Dichloro
STEPS 2 AND 3
Identify the
substituents and
assign locants.
STEP 4
Assemble the
substituents
alphabetically.
1
2
Cl
Cl
9
6
4
3
8
7
5
6-Ethyl
OH
Next, assemble the substituents alphabetically: 4,4-dichloro-6-ethyl-3-nonanol.
Before concluding, we must always check to see if there are any chiral centers. This compound has two chiral centers. Using the skills from Section 5.3, we can assign the R configuration to each chiral center.
STEP 5
Assign the
configuration of any
chiral centers.
Cl
R
Cl
R
OH
Therefore, the complete name is (3R,6R )-4,4-dichloro-6-ethyl-3-nonanol.
PRACTICE the skill 12.1 Provide an IUPAC name for each of the following alcohols:
OH
OH
OH
(a) Br
APPLY the skill
Br
OH
(b)
(c)
(d)
12.2 Safer alternatives to existing insecticides and mosquito repellants might be found in
the essential oils of plants. Phytol is one such isolated compound that has promising pest-­
fighting potential.1 Provide the IUPAC name for phytol, which is both an alcohol and an
12.1 Structure and Properties of Alcohols 533
alkene. A 16-carbon alkane is called hexadecane (6 + 10 = 16), and you might find it useful to
consult the textbook’s Appendix “Nomenclature of Polyfunctional Compounds” for more help!
OH
Phytol
need more PRACTICE? Try Problems 12.29, 12.30a–d,f, 12.31
Commercially Important Alcohols
Methanol (CH3OH) is the simplest alcohol. It is toxic, and ingestion can cause blindness and death,
even in small quantities. Methanol can be obtained from heating wood in the absence of air and has
therefore been called “wood alcohol.” Industrially, methanol is prepared by the reaction between
carbon dioxide (CO2) and hydrogen gas (H2) in the presence of suitable catalysts. Each year, the
United States produces approximately 2 billion gallons of methanol, which is used as a solvent and
as a precursor in the production of other commercially important compounds.
Methanol can also be used as a fuel to power combustion engines. In the second lap of the 1964
Indianapolis 500, a bad accident involving seven cars resulted in a large fire that claimed the lives
of two drivers. That accident prompted the decision that all race cars switch from gasoline-powered
engines to methanol-powered engines, because methanol fires produce no smoke and are more easily extinguished. In 2006, the Indianapolis 500 switched the choice of fuels for the race cars again,
replacing methanol with a mixture of 10% methanol and 90% ethanol.
Ethanol (CH3CH2OH), also called grain alcohol, is obtained from the fermentation of grains
or fruits, a process that has been widely used for thousands of years. Industrially, ethanol is prepared
via the acid-catalyzed hydration of ethylene. Each year, the United States produces more than 10 billion gallons of ethanol, used as a solvent and as a precursor for the production of other commercially
important compounds. Ethanol that is suitable for drinking is highly taxed by most governments.
To avoid these taxes, industrial-grade ethanol is contaminated with small quantities of toxic compounds (such as methanol) that render the mixture unfit for human consumption. The resulting
solution is called “denatured alcohol.”
Isopropanol, also called rubbing alcohol, is prepared industrially via the acid-catalyzed hydration of propylene. Isopropanol has antibacterial properties and is used as a local antiseptic. Sterilizing pads typically contain a solution of isopropanol in water. Isopropanol is also used as an industrial
solvent and as a gasoline additive.
Physical Properties of Alcohols
The physical properties of alcohols are quite different from the physical properties of alkanes or alkyl
halides. For example, compare the boiling points for ethane, chloroethane, and ethanol.
H
H
H
C
C
H
H
H
H
H
H
C
C
H
H
Cl
H
Chloroethane
bp = 12°C
Ethane
bp = –89°C
H
H
C
C
H
H
OH
Ethanol
bp = 78°C
The boiling point of ethanol is much higher than the other two compounds as a result of the
­hydrogen-bonding interactions that occur between molecules of ethanol.
δ+
H
O
R
δ–
δ+
H
O
δ–
R
These interactions are fairly strong intermolecular forces, and they are also critical in understanding how alcohols interact with water. For example, methanol is miscible with water, which
means that methanol can be mixed with water in any proportion (they will never separate into
two layers like a mixture of water and oil). However, not all alcohols are miscible with water. To
534
CHAPTER 12
Alcohols and Phenols
understand why, we must realize that every alcohol has two regions. The hydrophobic region does
not interact well with water, while the hydrophilic region does interact with water via hydrogen
bonding. Figure 12.1 shows the hydrophobic and hydrophilic regions of methanol and n-octanol.
H
FIGURE 12.1
The hydrophobic and hydrophilic
regions of methanol and octanol.
Hydrophobic H
region
C
H
OH
Hydrophobic
region
Hydrophilic
region
OH
Hydrophilic
region
In the case of methanol, the hydrophobic end of the molecule is fairly small. This is true even of ethanol
and n-propanol, but it is not true of n-butanol. The hydrophobic end of n-butanol is large enough to
prevent miscibility. Water can still be mixed with n-butanol, but not in all proportions. In other words,
n-butanol is considered to be soluble in water, rather than miscible. The term soluble means that only a
specific amount of n-butanol can dissolve in a specified volume of water at room temperature.
As the size of the hydrophobic region increases, solubility in water decreases. For example,
n-octanol exhibits extremely low solubility in water at room temperature. Alcohols with more than
eight carbon atoms, such as n-nonanol and n-decanol, are considered to be insoluble in water.
BioLinks Chain Length as a Factor in Drug Design
Potency
Primary alcohols (methanol, ethanol, n-propanol, n-butanol,
etc.) exhibit antibacterial properties. Research indicates that
the antibacterial potency of primary alcohols increases with
increasing molecular weight, and this trend continues up to an
alkyl chain length of eight carbon atoms (n-octanol). Beyond
eight carbon atoms, the potency decreases. That is, n-nonanol is
less potent than n-octanol, and n-dodecanol (12 carbon atoms)
has very little potency.
Studies also show that chain branching decreases the
ability of an alcohol to penetrate cell membranes. Accordingly,
isopropanol is actually less potent as an antibacterial agent than
n-propanol. Nevertheless, isopropanol (rubbing alcohol) is used
because it is less expensive to produce than n-propanol, and the
difference in antibacterial potency does not justify the added
expense of production.
Many other antibacterial agents are specifically designed
with alkyl chains that enable them to penetrate cell membranes.
The design of these agents is optimized by carefully striking
the right balance between the two trends previously discussed.
Various chain lengths are tested to determine optimal potency.
In most cases, the optimal chain length is found to be between
five and nine carbon atoms. Consider, for example, the structure
of resorcinol.
OH
HO
Resorcinol
1
2
3
4
5
6
7
8
9
10 11 12
Alkyl chain length
Two trends explain the observations:
• An alcohol with a larger alkyl chain (hydrophobic region)
exhibits a greater ability to penetrate microbial membranes, which are composed of molecules with hydrophobic
regions. According to this trend, potency should continue to
increase with increased alkyl chain length, even beyond eight
carbon atoms.
• A compound with a larger alkyl chain exhibits lower solubility
in water, decreasing its ability to be transported through aqueous media. This trend explains why the potency of alcohols
decreases steeply as the alkyl chain becomes larger than eight
carbon atoms. A larger alcohol simply cannot reach its destination and therefore has low potency.
The balance between these two trends is achieved for n-octanol,
which has the highest antibacterial potency of the primary
alcohols.
HO
OH
Hexylresorcinol
Resorcinol is a weak antiseptic (antimicrobial agent) used in the
treatment of skin conditions such as eczema and psoriasis. Placing
an alkyl chain on the ring increases its potency as an antiseptic.
Studies indicate that the optimal potency is achieved with a sixcarbon chain length. Hexylresorcinol exhibits bactericidal and
fungicidal properties and is used in many throat lozenges.
CONCEPTUAL CHECKPOINT
12.3 Mandelate esters exhibit spasmolytic activity (they act as muscle relaxants). The nature of the alkyl group (R) greatly affects potency.
Research indicates that the optimal potency is achieved when R is a
nine-carbon chain (a nonyl group). Explain why nonyl mandelate is
more potent than both octyl mandelate and decyl mandelate.
OH
O
O
Mandelate esters
(R = alkyl chain)
R
12.2
Acidity of Alcohols and Phenols
535
12.2 Acidity of Alcohols and Phenols
Acidity of the Hydroxyl Functional Group
As we learned in Chapter 3, the acidity of a compound can be qualitatively evaluated by analyzing
the stability of its conjugate base:
LOOKING BACK
The factors affecting the
stability of a negative
charge were first discussed
in Section 3.4.
R
O
–H+
H
To evaluate the acidity
of this compound...
O
R
...deprotonate...
−
...and assess the stability
of the conjugate base
(an alkoxide ion)
The conjugate base of an alcohol is called an alkoxide ion, and it exhibits a negative charge on an
oxygen atom. A negative charge on an oxygen atom is more stable than a negative charge on a carbon or nitrogen atom but less stable than a negative charge on a halogen, X (Figure 12.2).
LOOKING BACK
Recall that a strong acid
will have a low pKa value.
To review the relationship
between pKa and acidity,
see SkillBuilder 3.2.
Increasing stability
R
FIGURE 12.2
The relative stability of various
anions.
R
−
N
−
O
R
−
X
H
Least
stable
−
Most
stable
Therefore, alcohols are more acidic than amines and alkanes but less acidic than hydrogen halides
(Figure 12.3). The pKa for most alcohols falls in the range of 15–18.
Increasing acidity
FIGURE 12.3
The relative acidity of
alkanes, amines, alcohols,
and hydrogen halides.
R
H
R
pKa between
45 and 50
NH2
R
pKa between
35 and 40
X
OH
pKa between
15 and 18
H
pKa between
–10 and 3
Reagents for Deprotonating an Alcohol
There are two common ways to deprotonate an alcohol, forming an alkoxide ion.
1. A strong base can be used to deprotonate the alcohol. A commonly used base is sodium hydride
(NaH), because hydride (H−) deprotonates the alcohol to generate hydrogen gas, which bubbles out of solution:
H
O
Ethanol
−
+
Na
+
H
O
Sodium hydride
−
+
+
Na
Sodium ethoxide
H2
Hydrogen gas
2. Alternatively, an alcohol can be deprotonated upon treatment with Li, Na, or K. These metals
react with the alcohol to liberate hydrogen gas, producing the alkoxide ion.
O
H
Na
−
+
O Na
+
1
2
H2
CONCEPTUAL CHECKPOINT
12.4
Draw the alkoxide ion that is formed in each of the following reactions:
OH
Na
(a)
?
OH
OH
(b)
NaH
?
OH
Li
(c)
?
NaH
(d)
?
536
CHAPTER 12
Alcohols and Phenols
Reagents for Protonating an Alkoxide Ion
In this chapter and throughout the rest of this text, we will explore many reactions that generate
alkoxide ions as products. Whenever the product of a reaction is an alkoxide ion, it can be readily
converted into its corresponding alcohol upon treatment with an acid, such as H3O+:
H
R O
−
H
O
+
H
R O
An alkoxide ion
H
An alcohol
H3O+ can be prepared by diluting a strong acid, such as HCl or H2SO4, in water, although there
are other ways to prepare even milder aqueous acidic conditions, as we will see in upcoming chapters. It should be noted that most alkoxide ions can be protonated even with weak acids, such as
water (H2O), although it is generally more efficient to use H3O+. Throughout the remainder of this
text, we will use H3O+ as the proton source for protonating an alkoxide ion.
CONCEPTUAL CHECKPOINT
12.5 Draw the alcohol that is formed when each of the following alkoxide ions is treated with
H3O+, and in each case, show a mechanism for formation of the alcohol.
(a)
O
−
(b)
O
O
−
−
O
(c)
Factors Affecting the Acidity of Alcohols and Phenols
How can we predict which, of a number of alcohols, is more acidic? In this section, we will explore
three factors for comparing the acidity of alcohols.
1. Resonance. One of the most significant factors affecting the acidity of alcohols and phenols is
resonance. As a striking example, compare the pKa values of cyclohexanol and phenol:
OH
OH
Cyclohexanol
(pKa = 18)
Phenol
(pKa = 10)
When phenol is deprotonated, the conjugate base is stabilized by resonance.
O
−
O
O
O
−
O
−
−
−
This resonance-stabilized anion is called a phenolate, or phenoxide, ion. Resonance stabilization of the phenoxide ion explains why phenol is eight orders of magnitude (100,000,000
times) more acidic than cyclohexanol. As a result, phenol does not need to be deprotonated
with a very strong base like sodium hydride. Instead, it can be deprotonated by hydroxide.
O
H
(pKa = 10.0)
+
+
Na
−
O
OH
−
+
Na
+
H2O
(pKa = 15.7)
12.2 Acidity of Alcohols and Phenols 537
The acidity of phenols is one of the ways in which phenols differ from alcohols. Later in this
chapter and again in Chapter 18, we will see other reasons why phenols belong to a class of
their own.
2. Induction. Another factor in comparing the acidity of alcohols is induction. As an example,
compare the pKa values of ethanol and trichloroethanol.
Cl
OH
OH
Cl
Ethanol
(pKa = 16.0)
LOOKING BACK
For a review of inductive
effects, see Section 3.4.
Cl
Trichloroethanol
(pKa = 12.2)
Trichloroethanol is four orders of magnitude (10,000 times) more acidic than ethanol, because
the conjugate base of trichloroethanol is stabilized by the electron-withdrawing effects of the
nearby chlorine atoms.
3. Solvation effects. To explore the effect of alkyl branching, compare the acidity of ethanol and
tert-butanol.
OH
OH
Ethanol
(pKa = 16.0)
tert-Butanol
(pKa = 18.0)
The pKa values indicate that tert-butanol is less acidic than ethanol, by two orders of m
­ agnitude.
This difference in acidity is best explained by a steric effect. The ethoxide ion is not sterically
hindered and is therefore more solvated (stabilized) by the solvent, while tert-butoxide is sterically hindered and is less solvated (Figure 12.4). The conjugate base of tert-butanol is less stabilized than the conjugate base of ethanol, rendering tert-butanol less acidic.
nt
nt
lve
So
O
Solvent
−
Ethoxide
So
Solv
O
tert-Butoxide
lve
nt
SKILLBUILDER
12.2 comparing the acidity of alcohols
LEARN the skill
Identify which of the following compounds is expected to be more acidic:
OH
O
Compound A
OH
Solvent
−
ent
FIGURE 12.4
An ethoxide ion is stabilized by
the solvent to a greater extent
than tert-butoxide is stabilized by
the solvent.
S
ve
ol
O
Compound B
So
lve
nt
538 CHAPTER 12 Alcohols and Phenols
SOLUTION
Begin by drawing the conjugate base of each and then compare the stability of those conjugate bases.
−
O
−
O
O
O
Conjugate base
of compound B
Conjugate base
of compound A
The conjugate base of compound B is not resonance stabilized, but the conjugate base of
compound A is resonance stabilized.
Conjugate base of compound A
−
O
O
O
O
O
O
−
−
As a result, the conjugate base of compound A will be more stable than the conjugate base
of compound B. Therefore, compound A will be more acidic.
We expect compound B to have a pKa somewhere in the range of 15–18 (the range
expected for alcohols). The pKa of compound A will be more difficult to predict. However,
we expect that it will be lower (more acidic) than a regular alcohol. In other words, the pKa
value will be lower than 15.
PRACTICE the skill 12.6 For each of the following pairs of alcohols, identify the one that is more acidic and
explain your choice:
Cl
Cl
OH
OH
OH
Cl
OH
(a)
Cl
(b)
Cl
O
OH
OH
(c)
APPLY the skill
OH
(d)
OH
12.7 Flavonoids are a class of naturally occurring compounds with promising medicinal
uses due to their anti-inflammatory, anti-cancer and anti-HIV properties. Much of this biological activity is due to the radical scavenging ability of flavonoids, which can be affected by
the acidity of their hydroxyl groups.2 Naringenin, the major flavonoid found in grapefruit, has
three acidic hydroxyl groups. pKa values are shown pK = 7.05
pKa2 = 8.84
a1
H
for two of these hydroxyl groups (highlighted). ProHO
O
OH
vide an explanation for the substantial difference in
acidity between these two hydroxyl groups.
OH
need more PRACTICE? Try Problems 12.32, 12.63
O
Naringenin
12.3 Preparation of Alcohols via Substitution or Addition
Substitution Reactions
As we saw in Chapter 7, alcohols can be prepared by substitution reactions in which a leaving group
is replaced by a hydroxyl group.
R
X
R
OH
12.4
Preparation of Alcohols via Reduction
539
A primary substrate will require SN2 conditions (a strong nucleophile), while a tertiary substrate will
require SN1 conditions (a weak nucleophile).
Primary:
Cl
+
NaOH
SN2
OH
+
O
SN1
OH
H
+
NaCl
+
HCl
Tertiary:
Cl
LOOKING BACK
H
With a secondary substrate, neither SN2 nor SN1 is particularly effective for preparing a secondary
alcohol. Under SN1 conditions, the reaction is generally too slow, while S N2 conditions (use of
hydroxide as the nucleophile) will generally favor elimination over substitution.
For a review of the factors
that affect substitution vs.
elimination, see Section 7.9.
Addition Reactions
In Chapter 8, we learned several addition reactions that produce alcohols.
OH
Dilute H2SO4
Acid-catalyzed hydration
(Section 8.6)
OH
1) Hg(OAc)2, H2O
2) NaBH4
Oxymercuration-demercuration
(Section 8.7)
1) BH3 THF
2) H2O2, NaOH
OH
Hydroboration-oxidation
(Section 8.8)
Acid-catalyzed hydration proceeds with Markovnikov addition (Section 8.6). That is, the hydroxyl
group is installed at the more substituted position. It is a useful method if the substrate is not susceptible to carbocation rearrangements (Section 6.11). In a case where the substrate can possibly
rearrange, oxymercuration-demercuration can be employed. This approach also proceeds via Markovnikov addition, but it does not involve carbocation rearrangements. Hydroboration-oxidation is
used to achieve an anti-Markovnikov addition of water.
CONCEPTUAL CHECKPOINT
12.8
Identify the reagents that you would use to accomplish each of the following transformations:
OH
OH
Br
(a)
(b)
OH
Br
(c)
OH
(d)
12.9
OH
(f )
(e)
HO
Identify reagents that can be used to achieve each of the following transformations:
(a) To convert 1-hexene into a primary alcohol
(b) To convert 3,3-dimethyl-1-hexene into a secondary alcohol
(c) To convert 2-methyl-1-hexene into a tertiary alcohol
12.4 Preparation of Alcohols via Reduction
In this section, we will explore a new method for preparing alcohols. This method involves a change
in oxidation state, so let’s spend a few moments to understand oxidation states and their relationship with formal charges.
540 CHAPTER 12 Alcohols and Phenols
Oxidation States
An oxidation state refers to a method of electron bookkeeping. Chapter 1 developed a ­different
method for electron bookkeeping called formal charge. To calculate the formal charge on an atom,
we treat all bonds to that atom as if they were purely covalent, and we break them ­homolytically. For
calculating oxidation states, we will take another extreme approach. We treat all bonds as if they
were purely ionic, and we break them heterolytically, giving each pair of electrons to the more electronegative atom in each case. Formal charges and ­oxidation states therefore represent two extreme
methods of electron bookkeeping. In Figure 12.5, the formal charge of the carbon atom is zero,
because we count four electrons on the central carbon atom, which is equivalent to the number of
valence e­ lectrons a carbon atom is supposed to have. In contrast, the same carbon atom has an oxidation state of −2, because we count six electrons on the carbon atom, which is two more electrons
than it is supposed to have.
OH
Treat all bonds as covalent
and break them homolytically
H
C
Treat all bonds as ionic
and break them heterolytically
H
H
Formal charge
Oxidation state
OH
OH
H
FIGURE 12.5
Two different electron
bookkeeping methods: formal
charge vs. oxidation state.
C
−
H
H+
+
C −
−
+H
−
H
Four
electrons
Six
electrons
+
H
A carbon atom with four bonds will always have no formal charge, but its oxidation state can range
anywhere from −4 to +4.
Methane
Methanol
Formaldehyde
Formic
acid
Carbon
dioxide
OH
O
O
O
H
H
C
H
H
C
H
H
–4
–2
H
H
H
OH
H
+2
0
C
O
+4
A reaction involving an increase in the oxidation state of a carbon atom is called an oxidation. For
example, when ­methanol is converted into formaldehyde, we say that methanol was oxidized. In
contrast, a reaction involving a decrease in the oxidation state of a carbon atom is called a reduction. For example, when formaldehyde is converted into methanol, we say that formaldehyde was
reduced. Let’s get some practice identifying oxidations and reductions.
SKILLBUILDER
12.3 identifying oxidation and reduction reactions
LEARN the skill
In the following transformation, identify whether the compound has been oxidized, reduced,
or neither:
O
RO
OR
12.4 Preparation of Alcohols via Reduction 541
SOLUTION
Focus on the carbon atom where a change has occurred and determine whether the oxidation state has changed as a result of the transformation. Let’s begin with the starting material:
STEP 1
Determine the
oxidation state of the
starting material.
O
H3C
C
O
Break all bonds heterolytically,
except for C — C bonds
CH3
H3C
C
CH3
Each ​CO​bond is broken heterolytically, giving all four electrons of the C
​ O​double bond
to the oxygen atom. Each ​CC​bond cannot be broken heterolytically, because C and C
have the same electronegativity. For each ​CC​bond, just divide the electrons between
the two carbon atoms, breaking the bond homolytically. This leaves a total of two electrons
on the central carbon atom. Compare this number with the number of valence electrons
that a carbon atom is supposed to have (four). The carbon atom in this example is missing
two electrons. Therefore, it has an oxidation state of +2.
Now analyze the carbon atom in the product. The
OR
RO
RO OR
same result is obtained: an oxidation state of +2. This
C
C
should make sense, because the reaction has simply H C
CH3
3
H 3C
CH3
exchanged one C
​ O​double bond for two C
​ O​ single
bonds. The oxidation state of the carbon atom has not
Two electrons
changed during the reaction, and therefore the starting
material has been neither oxidized nor reduced.
Two electrons
STEP 2
Determine the
oxidation state of the
product.
STEP 3
Determine if there
has been a change in
oxidation state.
PRACTICE the skill 12.10 In each of the following transformations, identify whether the starting material
has been oxidized, reduced, or neither. Try to determine the answer without calculating
­oxidation states and then use the calculations to see if your intuition was correct.
O
Cl
O
Cl
(a)
(b)
H
OH
OH
APPLY the skill
(c)
O
OH
OH
O
O
O
O
(d)
O
O
O
OH
Cl
(e)
(f )
12.11 Compound 1 was converted into compound 2 and both compounds were evaluated for their cytotoxicity (ability to kill cells). It was observed that compound 2 was four to
ten times less cytotoxic than compound 1.3 Determine whether the ring has been oxidized
or reduced in this reaction (or you may find that it has been neither oxidized nor reduced).
Try to determine the answer first by intuition and then use calculations to see if your intuition was correct.
R
R
OCOCH3
Na2S2O4
H3CO
O
O
H3CO
OH
O
OH
1
2
O
H
need more PRACTICE? Try Problems 12.49, 12.50, 12.74
Reducing Agents
The conversion of a ketone (or aldehyde) to an alcohol is a reduction.
OH
O
Reduction
O
R=
O
542
CHAPTER 12
Alcohols and Phenols
The reaction requires a reducing agent, which is itself oxidized as a result of the reaction. In this section,
we will explore three reducing agents that can be used to convert a ketone or aldehyde to an alcohol:
1. In Chapter 8, we learned that an alkene can undergo hydrogenation in the presence of a metal
catalyst such as platinum, palladium, or nickel. A similar reaction can occur for ketones or
aldehydes, although more forcing conditions are generally required (higher temperature and
pressure), so this method is rarely used by synthetic chemists.
O
OH
H2
Pt, Pd, or Ni
95%
2. Sodium borohydride (NaBH4) is a common reducing agent that can be used to reduce ketones
or aldehydes.
O
+
Na
H
OH
NaBH4
H
EtOH, MeOH, or H2O
−
B
H
H
Sodium borohydride
(NaBH4 )
90%
Sodium borohydride functions as a source of hydride (H:−) and the solvent functions as the
source of a proton (H+). The solvent can be ethanol, methanol, or water. The precise mechanism of action has been heavily investigated and is somewhat complex. Nevertheless, Mechanism 12.1 presents a simplified version that will be sufficient for our purposes. The first step
involves the transfer of hydride to the carbonyl group (the CO bond), and the second step
is a proton transfer.
MECHANISM 12.1 REDUCTION OF A KETONE OR ALDEHYDE WITH NaBH4
Nucleophilic attack
Proton transfer
O
O
−
R
R
NaBH4 delivers H −
to the carbonyl group
H
H
−
B
R
H
R
H
OH
O
H
CH3
The resulting
alkoxide ion
is then protonated
by the solvent
to form an alcohol
R
H
R
H
Hydride (H:−), by itself, is not a good nucleophile because it is not polarizable. As a result, the
reaction above cannot be achieved by using NaH (sodium hydride). NaH only functions as
a base, not as a nucleophile. But NaBH4 does function as a nucleophile. Specifically, NaBH4
functions as a delivery agent of nucleophilic H−.
When nucleophilic H− attacks a carbonyl group, an important change in geometry occurs.
Prior to the attack, the carbon atom of the carbonyl group is sp2-hybridized and has trigonal
planar geometry. But as a result of the attack, this carbon atom becomes sp3-hybridized, with
tetrahedral geometry. This change is especially relevant when an unsymmetrical ketone (a ketone
with two different R groups) is reduced with sodium borohydride. In such a case, a new chiral
center is created, and a pair of stereoisomers is obtained. Consider the following example:
O
NaBH4
MeOH
OH
OH
+
12.4
Preparation of Alcohols via Reduction
543
In this case, reduction of the ketone gives a racemic mixture of enantiomers, because the
hydride nucleophile can attack either face of the planar carbonyl group with equal likelihood:
O
−
−
Nuc
Nuc
3. Lithium aluminum hydride (LiAlH4) is another common reducing agent, and its structure is
very similar to NaBH4.
+
Na
H
−
H
B
+
Li
H
H
H
−
Al
H
H
H
Sodium borohydride
(NaBH4)
Lithium aluminum hydride
(LiAIH4)
Lithium aluminum hydride, commonly abbreviated as LAH, is also a delivery agent of H −,
but it is a much stronger reagent. It reacts violently with protic solvents (such as water), and
therefore, a protic solvent cannot be present together with LiAlH4 in the reaction flask. First the
ketone or aldehyde is treated with LiAlH4, and then, in a separate step (called workup), the proton source is added to the reaction flask. For the workup step, H3O+ can be used as the proton
source, although H2O can also be used as a proton source:
O
1) LiAIH4
H
2) H3O+ (or H2O)
OH
86%
Notice that LiAlH4 and the workup step are listed as two separate steps. When LiAlH4 is used
to reduce a ketone or aldehyde, the workup step must be shown as a separate step, because
LiAlH4 cannot be in the reaction flask together with H3O+(or H2O). The mechanism of action
has been heavily investigated and is somewhat complex. A simplified version is presented here
(Mechanism 12.2). Note the similarity between Mechanism 12.1 and Mechanism 12.2.
MECHANISM 12.2 REDUCTION OF A KETONE OR ALDEHYDE WITH LiAlH4
Nucleophilic attack
Proton transfer
O
O
R
R
H
H
−
Al
−
LiAlH4 delivers H
to the carbonyl group
H
O
+
OH
H
H
R
R
H
H
−
R
As a workup step,
acid is introduced
R
into the reaction flask,
and the alkoxide ion is protonated
to give an alcohol
H
H
This simplified mechanism does not take into account many important observations, such as
the role of the lithium cation (Li+). However, a full treatment of the mechanism of hydride
reducing agents is beyond the scope of this text, and this simplified version will suffice.
When an unsymmetrical ketone is reduced with LiAlH4, a pair of stereoisomers is obtained
(just as we saw for reduction with NaBH4), because the hydride nucleophile can attack either
face of the planar carbonyl group:
O
1) LiAlH4
2) H3O+
OH
OH
+
544
CHAPTER 12
Alcohols and Phenols
Both NaBH4 and LiAlH4 will reduce ketones or aldehydes. These hydride delivery agents offer
one significant advantage over catalytic hydrogenation in that they can selectively reduce a carbonyl
group in the presence of a CC bond. Consider the following example:
O
NaBH4
OH
OH
MeOH
H2
1) LiAlH4
Pt, Pd, or Ni
2) H3O+
90–95%
90–95%
When treated with a hydride reducing agent, only the carbonyl group is reduced. In contrast, catalytic hydrogenation will also reduce the CC bond under the conditions required to reduce the
carbonyl group (high temperature and pressure). For this reason, hydride reducing agents such as
NaBH4 and LiAlH4 are generally preferred over catalytic hydrogenation. Many other hydride reducing reagents are commercially available. Some are even more reactive than LiAlH4 and others are
even milder than NaBH4. Many of these reagents are derivatives of NaBH4 and LiAlH4.
Li
R
R
Al
−
+
H
Na
R
−
R
B
R
+
H
R
Each R group can be an alkyl group, a cyano group, an alkoxy group, or any one of a number of
groups. By carefully choosing the three R groups to be either electron donating or electron withdrawing, it is possible to modify the reactivity of the hydride reagent. Hundreds of different hydride
reducing agents are available, and each has its own selectivity and advantages. For now, we will simply focus our attention on the differences between LiAlH4 and NaBH4.
We mentioned that LiAlH4 is much more reactive than NaBH4. For example, LiAlH4 will
reduce a carboxylic acid or an ester to produce an alcohol, but NaBH4 will not.
BY THE WAY
Although NaBH4 does not
reduce esters under mild
conditions, it is observed that
NaBH4 will sometimes reduce
esters when more forcing
conditions are employed,
such as high temperature.
O
1) Excess LiAIH4
R
2) H3O+
OH
R
OH
R
OH
A carboxylic acid
O
1) Excess LiAIH4
R
2) H3O+
OMe
+
MeOH
An ester
Reduction of the ester involves the transfer of hydride two times, as seen in Mechanism 12.3.
MECHANISM 12.3 REDUCTION OF AN ESTER WITH LiAlH4
Loss of a
leaving group
Nucleophilic attack
Nucleophilic attack
Proton transfer
H
O
O
OMe
R
H
H
−
Al
H
H
LiAlH4 delivers
H−
to the
carbonyl group
R
−
O
OMe
H
Expulsion of
an alkoxide
ion causes the
carbonyl group to
re-form
O
H
R
H
H
−
Al
H
H
LiAlH4 delivers
H−
to the
carbonyl group
R
−
H
H
O
+
OH
H
H
R
As a workup step,
acid is introduced
into the reaction flask,
and the alkoxide ion is protonated
to give an alcohol
H
H
12.4 Preparation of Alcohols via Reduction 545
LiAlH4 delivers hydride to the carbonyl group, but then loss of a leaving group causes the c­ arbonyl
group to re-form. In the presence of LiAlH4, the newly formed carbonyl group can be attacked
again by hydride. The leaving group in the second step of the mechanism is a methoxide ion, which
is generally not a good leaving group. For example, methoxide does not function as a leaving group
in E2 or SN2 reactions. The reason that it can function as a leaving group in this case stems from the
nature of the intermediate after the first attack of hydride. That i­ntermediate is high in energy and
already exhibits a negatively charged oxygen atom. The intermediate is therefore capable of ejecting
methoxide, because such a step is not uphill in energy.
O
−
R
O
OMe
H
R
H
−
+
OMe
High-energy intermediate
In Chapter 20, we will discuss this reaction in greater detail as well as a mechanism for the reaction
of a carboxylic acid with LiAlH4.
SKILLBUILDER
12.4 drawing a mechanism and predicting the products of hydride reductions
LEARN the skill
Draw a mechanism and predict the product for the following reaction:
O
1) LiAIH4
2) H3O+
?
SOLUTION
LiAlH4 is a hydride reducing agent, and the starting material is a ketone. When a hydride reducing agent reacts with a ketone or aldehyde, we can draw a mechanism that consists of two
steps: nucleophilic attack, followed by proton transfer. In the first step, draw the structure of
LiAlH4 and show a hydride being delivered to the carbonyl group. Do not simply draw H:− as
the reagent, because H:− is not nucleophilic by itself. It must be delivered by the delivery agent
(LiAlH4). Draw the complete structure of LiAlH4 (showing all bonds) and then draw a curved
arrow that shows the electrons coming from one A
​ lH​bond and attacking the carbonyl group.
STEPS 1 AND 2
Draw the two curved
arrows that show the
delivery of a hydride
and then draw the
resulting alkoxide ion.
O
O
LiAlH4 delivers a hydride
to the carbonyl group
H
H
H
Al
−
−
CH3
+ Enantiomer
H
H
Notice that a chiral center is created during this reduction process, so we expect the alkoxide ion to be produced as a racemic mixture of enantiomers. Then, after the reaction is
complete, acid is introduced into the reaction flask as a workup step. During this step, the
alkoxide ion is protonated to give an alcohol.
STEPS 3
Draw the two curved
arrows showing the
alkoxide ion being
protonated by the
proton source.
O
−
H
OH
H
CH3
+ Enantiomer
H
O
+
H
H
CH3
+ Enantiomer
546 CHAPTER 12 Alcohols and Phenols
The product is a racemic mixture of a secondary alcohol, which is what we expect from
reduction of a ketone. The enantiomeric products can also be drawn like this:
OH
OH
+
PRACTICE the skill 12.12 Draw a mechanism and predict the major product(s) for each reaction:
O
H
?
1) LiAIH4
2) H3O+
(a)
O
(c)
1) LiAIH4
2) H3O+
(b)
O
?
NaBH4
MeOH
APPLY the skill
O
1) xs LiAIH4
2) H3O+
O
(d)
?
?
12.13 Ambuic acid has been isolated from fungi that grow on economically important
plants found in the rain forest. Ambuic acid is active against several other fungi that are
harmful to these plants, indicating that it may provide chemical protection. During a synthesis of ambuic acid, compound 1 was reduced to compound 2 using NaBH4.4
O
OH
NaBH4
RO
O
OH
OH
1
CH3OH
RO
RO =
O Si
HO
O
OH
Me
OH
Many
steps
t-Bu
O
OH
O
2
Ambuic acid
CO2H
Me
(a) Draw a mechanism for the conversion of 1 to 2.
(b) Since compound 1 is chiral, reduction of the carbonyl group leads to diastereomeric
products (rather than enantiomers), and it is observed that one diastereomer
(compound 2) is favored. Based on the configuration of the newly formed alcohol,
determine whether the preference is for hydride to be delivered to the top face or to
the bottom face of compound 1.
(Note: The source of this preference is beyond the scope of our current discussion.)
need more PRACTICE? Try Problems 12.44, 12.45c, 12.46e,f, 12.51, 12.55, 12.72
12.5 Preparation of Diols
Diols are compounds with two hydroxyl groups, and the following additional rules are used to name them:
1. The positions of both hydroxyl groups are identified with numbers placed before the parent.
2. The suffix “diol” is added to the end of the name:
OH
OH
HO
OH
1,5-Hexanediol
1,3-Propanediol
Notice that an “e” appears in between the parent and the suffix. In a regular alcohol, the “e” is dropped (i.e.,
propanol or hexanol). A few simple diols have common names that are accepted by IUPAC nomenclature.
OH
HO
OH
Ethylene glycol
HO
Propylene glycol
The term glycol indicates the presence of two hydroxyl groups.
12.6
Preparation of Alcohols via Grignard Reagents
547
Diols can be prepared from diketones via reduction using any of the reducing agents that we have seen.
H2 / Pd
O
O
OH
OH
1) LiAIH4 (two equiv.)
2) H3O+
NaBH4 (two equiv.)
MeOH
Alternatively, diols can be made via dihydroxylation of an alkene. In Chapter 8, we explored reagents
for achieving either syn or anti dihydroxylation.
OH
1) RCO3H
Anti dihydroxylation
(Section 8.11)
2) H3O+
OH
OH
catalytic OsO4
Syn dihydroxylation
(Section 8.12)
NMO
OH
WorldLinks Antifreeze
Figure8Photos/Getty
Figure8Photos/Getty Images
Images
Automobiles are powered by internal combustion engines, and
various parts of the engines can become very hot. To prevent
damage caused by overheating, a liquid coolant is used
to transfer some of the heat away from the sensitive
engine parts. In most climates, pure water is
unsuitable as a coolant because it can freeze
if outdoor temperatures drop below 0ºC. To
prevent freezing, a coolant called antifreeze
is used. Antifreeze is a solution of water
and other compounds that significantly
lower the freezing point of the mixture.
Ethylene glycol and propylene glycol are
commonly used for this purpose.
OH
HO
OH
Ethylene glycol
HO
Propylene glycol
12.6 Preparation of Alcohols via Grignard Reagents
In this section, we will discuss formation of alcohols using Grignard reagents. A Grignard reagent is
formed by the reaction between an alkyl halide and magnesium.
R
X
Mg
R
Mg
X
Grignard reagent
These reagents are named after the French chemist Victor Grignard, who demonstrated their utility
in preparing alcohols. For his achievements, he was awarded the 1912 Nobel Prize in Chemistry.
Below are a couple of specific examples of Grignard reagents.
MgBr
Br
Mg
Cl
Mg
MgCl
548
CHAPTER 12
Alcohols and Phenols
A Grignard reagent is characterized by the presence of a CMg bond. Carbon is more electronegative
than magnesium, so the carbon atom withdraws electron density from magnesium via induction. This
gives rise to a partial negative charge (δ−) on the carbon atom. In fact, the difference in electronegativity
between C and Mg is so large that the bond can be treated as ionic.
δ+
δ– MgBr
−
+
MgBr
Grignard reagents are carbon nucleophiles capable of attacking a wide range of electrophiles, including the carbonyl group of ketones or aldehydes (Mechanism 12.4).
MECHANISM 12.4 THE REACTION BETWEEN A GRIGNARD
REAGENT AND A KETONE OR ALDEHYDE
Nucleophilic attack
Proton transfer
H
O
O
R
R
−
R
R
The Grignard reagent acts
as a nucleophile and
attacks the carbonyl group
H
O
OH
H
+
As a workup step,
an acid is introduced
into the reaction flask,
and the alkoxide ion is protonated
to give an alcohol
R
−
R
R
R
R
The product is an alcohol, and the mechanism presented here is similar to the mechanism that was
presented for reduction via hydride reagents (LiAlH4 or NaBH4). In fact, the reaction here is also a
reduction, but it involves introduction of an R group.
O
OH
1) RMgX
2) H3O+ (or H2O)
R
Notice that the proton source (H3O+ or H2O) is added to the reaction mixture in a separate step
(similar to reduction with LiAlH4). Even a weak acid, such as H2O, cannot be present together with
the Grignard reagent, because the Grignard reagent is also a strong base and will deprotonate water.
−
R
+
MgX
+
H
O
H
(pKa = 15.7)
R H
+
HOMgX
(pKa ∼ 50)
The difference in pKa values is so vast that the reaction is essentially irreversible. Every water molecule
present in the reaction flask will destroy one molecule of Grignard reagent. Grignard reagents will even
react with the moisture in the air, so care must be taken to use conditions that scrupulously avoid the
presence of water. After the Grignard reagent has attacked the ketone, an acid can be introduced into the
reaction flask to protonate the alkoxide. This protonation step (also called workup) can be accomplished
by introducing even a weak acid, such as H2O, into the reaction flask. Throughout the remainder of this
text, we will always show H3O+ as the source of protons for the workup of a Grignard reaction.
A Grignard reagent will react with a ketone or an aldehyde to produce an alcohol.
O
H3C
OH
1) CH3MgBr
2) H3O+
O
1) CH3MgBr
H
2) H3O+
OH
H
(Racemic)
CH3
12.6
Preparation of Alcohols via Grignard Reagents
549
Notice that the second example involves formation of a chiral center, and a racemic mixture of enantiomers is obtained. As we explained earlier (Section 12.4), the nucleophile can attack either face of
the carbonyl group with equal likelihood, giving rise to a racemic mixture.
CONCEPTUAL CHECKPOINT
12.14
Predict the major product(s) for each of the following reactions:
O
?
1) MeMgBr
2) H3O+
(a)
O
?
1) PhMgBr
2) H3O+
H
(b)
Br
?
1) Mg
O
O
(c)
1)
?
MgBr
2) H3O+
2)
H
H
3) H3O+
(d)
Grignard reagents also react with esters to produce alcohols, with introduction of two R groups.
O
OH
1) Excess CH3MgBr
+
CH3
CH3
2) H3O+
OMe
MeOH
The mechanism presented for this process (Mechanism 12.5) is similar to the mechanism presented
for reduction of an ester with LiAlH4 (Mechanism 12.3).
MECHANISM 12.5 THE REACTION BETWEEN A GRIGNARD REAGENT AND AN ESTER
Loss of a
leaving group
Nucleophilic attack
O
O
OMe
R
−
R
The Grignard
reagent acts as a
nucleophile and
attacks the
carbonyl group
R
−
Proton transfer
O
O
OMe
R
Nucleophilic attack
Expulsion of
an alkoxide
ion causes the
carbonyl group to
re-form
R
R
−
R
H
R
R
The Grignard
reagent acts as a
nucleophile and
attacks the
carbonyl group
H
−
O
+
OH
H
As a workup step,
an acid is introduced
into the reaction flask,
and the alkoxide ion is
protonated to give an
alcohol
R
R
R
R
We will explore this reaction and others like it in much more detail in Chapter 20. A Grignard
reagent will not attack the carbonyl group of a carboxylic acid. Instead, the Grignard reagent will
simply function as a base to deprotonate the carboxylic acid.
−
R
+
O
+
MgX
+
H
O
C
R
R
+
H
MgX
O
O
C
−
R
In other words, a Grignard reagent is incompatible with a carboxylic acid. For similar reasons, it is
not possible to form a Grignard reagent in the presence of even mildly acidic protons, such as the
proton of a hydroxyl group.
Not compatible
OH
Mg
Br
OH
−
+
MgBr
Cannot form this Grignard reagent
550 CHAPTER 12 Alcohols and Phenols
This Grignard reagent cannot be formed, as it will simply attack itself to produce an alkoxide. In
Section 12.7, we will learn how to circumvent this problem. But first, let’s get some practice preparing alcohols via Grignard reactions.
SKILLBUILDER
12.5 preparing an alcohol via a grignard reaction
LEARN the skill
Show how you would use a Grignard reaction to prepare the following compound:
OH
SOLUTION
First identify the carbon atom attached directly to the hydroxyl group (the α position):
OH
STEP 1
Identify the three
groups connected to
the α position.
Next, identify all groups connected to this position. In this case, the α position is connected
to a phenyl group, a methyl group, and an ethyl group:
OH
ethyl
methyl
phenyl
STEP 2
Disconnect one of the
groups connected to
the α position.
We can make a disconnection at a bond between the α position and any one of these three
groups, which gives us three possible disconnections to consider. As usual, for each possible
disconnection, we must work backward to a suitable nucleophile and electrophile. Below, we
consider a disconnection between the α carbon and one of the three groups (the ethyl group):
was an
electrophile
(C=O)
O
OH
What starting materials
are needed?
δ+
was a
nucleophile
(RMgBr)
BrMg
δ–
+
Electrophile
Nucleophile
The α carbon in the target molecule has an attached OH group, providing a clue that it used to be
a carbonyl carbon. The carbonyl group was the electrophilic component in the desired C​​C bondforming reaction, and a Grignard reagent served as the nucleophilic carbon atom. The same logic
can be applied for the other two possible disconnections, shown in the following retrosyntheses:
OH
δ–
O
MgBr
+
Nucleophile
OH
δ+
Electrophile
O
δ+
Electrophile
+
CH3MgBr
δ–
Nucleophile
12.6 Preparation of Alcohols via Grignard Reagents 551
Our retrosynthetic analysis has revealed three possible retrosyntheses, and we can now
draw the forward process for each of these three possible ways of preparing the target
molecule. In each case, we would need to start with a ketone in which two of the groups are
already present and we introduce the third group as a Grignard reagent. Here are all three
possibilities:
O
STEP 3
Show how each group
could have been
installed via a Grignard
reaction.
1) MeMgBr
2) H3O+
Et
Ph
O
Ph
OH
1) EtMgBr
Me
2) H3O+
Et
1) PhMgBr
2) H3O+
Ph
Et
Me
O
Me
Notice that there are three perfectly correct answers in this case. Indeed, synthesis problems often have more than one correct answer.
PRACTICE the skill 12.15 Show how a Grignard reaction can be used to make each of the following compounds, starting with an aldehyde or ketone starting material:
OH
OH
OH
(a)
(b)
(c)
OH
OH
(d)
(e)
(f )
OH
12.16 Two of the compounds from Problem 12.15 can be prepared from the reaction
between a Grignard reagent and an ester. Identify those two compounds and explain why
the other four compounds cannot be prepared from an ester.
APPLY the skill
12.17 Extracts from the flowering plants of Asteraceae have been used to produce medicine, rubber, and pesticides. (+)-Carissone, a compound isolated from
such extracts, has been found to have antibacterial properties. During a synthesis of
(+)-carissone, a Grignard reaction was used to convert compound 1 into compound 2,
as shown below.5
?
HO
CO2Me
1
Oxidation
(Section 12.10)
HO
OH
2
O
OH
(+)-Carissone
(a) What Grignard reagent is needed for the transformation shown?
(b) Typically, two equivalents of a Grignard reagent are required when reacting with an
ester, but in this case a minimum of three equivalents are required. Explain why a third
equivalent is required in this case. (Hint: What reaction happens first when a Grignard
reagent is added to the given starting material? Consider all of the functional groups
that are present.)
(c) Draw a mechanism that uses all three equivalents of the Grignard reagent and
generates the diol product, compound 2 (after neutralization with an aqueous workup).
need more PRACTICE? Try Problems 12.37, 12.39b, 12.52, 12.54, 12.56, 12.58, 12.65b,h,n,o,q, 12.70
552 CHAPTER 12 Alcohols and Phenols
12.7 Protection of Alcohols
Consider the following hypothetical Grignard reaction:
O
1)
Mg
Br
BrMg
OH
OH
Hypothetical Grignard reagent
(cannot be formed)
R
HO
R
2) H3O+
R
OH
R
To achieve this transformation, the required Grignard reagent would have to bear an OH group.
But we saw in Section 12.6 that it is not possible to form this Grignard reagent, because the strongly
basic Grignard reagent is incompatible with the weakly acidic hydroxyl group. To circumvent this
problem, we employ a three-step process.
1. Protect the hydroxyl group by replacing its proton with a protecting group, which is a group
that lacks acidic protons and will not react with a Grignard reagent (we will soon see an example of a protecting group).
2. Form the Grignard reagent and perform the desired Grignard reaction.
3. Remove the protecting group. This step regenerates the original OH group:
Br
HO
OH
R
Grignard
reaction
Protect
OH group
Br
Remove
protecting group
HO
O
O
R
Protecting
group
OH
R
R
Protecting
group
The protecting group enables us to perform the desired Grignard reaction. There are many protecting groups commonly used by chemists. The following is an example of a protecting group, called a
trimethylsilyl group, abbreviated TMS (and highlighted below):
Me
Br
OH
Br
O
Si
Br
Me
OTMS
Me
Trimethylsilyl group
(a protecting group)
Unlike a hydroxyl group, this protecting group is compatible with a Grignard reagent because it has
no acidic protons. In other words, the OH group is “protected” from undergoing a proton transfer
reaction after being converted to an OTMS group.
The OH group of an alcohol is converted into an OTMS group upon treatment with tri­
methylsilyl chloride, abbreviated TMSCl.
Me
R
O
H
Me
Si
Cl
Me
(TMSCl)
Et3N
Et
R
O
TMS
+
+
N
Et
Et
H
Cl
−
This reaction is believed to proceed via an SN2-like process (called SN2-Si), in which the hydroxyl
group functions as a nucleophile to attack the silicon atom and a chloride ion is expelled as a leaving
group. A base, such as triethylamine, is then used to remove the proton connected to the oxygen
atom. Notice that the first step involves an SN2-like process occurring at a tertiary substrate. This
should seem surprising because in Chapter 7 we learned that a nucleophile cannot effectively attack
a sterically hindered substrate. This case is different because the electrophilic center is a silicon atom
rather than a carbon atom. Bonds to silicon atoms are typically much longer than bonds to carbon
atoms, and this longer bond length opens up the back side for attack. To visualize this, compare
space-filling models of tert-butyl chloride and trimethylsilyl chloride (Figure 12.6).
12.8
tert-Butyl chloride
Trimethylsilyl chloride
Me
Me
FIGURE 12.6
Space-filling models of tertbutyl chloride and trimethylsilyl
chloride. The latter is less
sterically hindered and is
susceptible to attack by a
nucleophile.
Me
Cl
C
553
Preparation of Phenols
Me
Si
Me
Attack is too sterically hindered
Cl
Me
Attack is less sterically hindered
After the desired Grignard reaction has been performed, the trimethylsilyl group can be removed
easily with either H3O+ or fluoride ion.
+
R
O
H3O
TMS
or F −
R
OH
A commonly used source of fluoride ion is tetrabutylammonium fluoride (TBAF).
F
N
+
−
Tetrabutylammonium fluoride (TBAF)
The overall process is shown below.
HO
Br
OH
Grignard reaction
1) Mg
TMSCl
Et3N
Br
OTMS
2)
Protect OH group
O
OH
R R
HO
OTMS
R R
R
R
3) H3O+
TBAF
Remove
protecting group
CONCEPTUAL CHECKPOINT
12.18 Each of the following transformations involves a Grignard reaction, and requires the use of
a protecting group. Identify reagents that can be used to achieve each transformation:
OH
HO
Br
OH
Br
HO
HO
(a)
HO
OH
(b)
12.8 Preparation of Phenols
Phenol is prepared industrially via a multistep process involving the formation and oxidation of
cumene.
OOH
OH
O
H3O+
O2
+
H3PO4
Benzene
Cumene
Cumene
hydroperoxide
Phenol
Acetone
554 CHAPTER 12 Alcohols and Phenols
A by-product of this process is acetone, which is also a commercially important compound. Over
two million tons of phenol are produced each year in the United States. Phenol is used as a ­precursor
in the synthesis of a wide variety of pharmaceuticals and other commercially ­useful compounds,
including synthetic polymers (discussed in Chapter 27), adhesives for ­plywood, and antioxidant
food additives (BHT and BHA, discussed in Chapter 10).
OH
OH
OH
O
O
Butylated hydroxytoluene
(BHT)
Butylated hydroxyanisole
(BHA)
12.9 Reactions of Alcohols: Substitution and Elimination
SN1 Reactions with Alcohols
As seen in Section 7.10, tertiary alcohols will undergo a substitution reaction when treated with a
hydrogen halide.
OH
X
HX
+
H2O
This reaction proceeds via an SN1 mechanism.
Proton transfer
Nucleophilic attack
Loss of a leaving group
H
OH
H
O
Br
+
– H2O
H
−
Br
Br
+
Recall that an SN1 mechanism has two core steps (loss of leaving group and nucleophilic attack).
When the starting material is an alcohol, the mechanism must begin with an additional step, in
which the hydroxyl group is protonated. This reaction proceeds via a carbocation intermediate and
is therefore most appropriate for tertiary alcohols. When dealing with a primary or secondary alcohol, an SN2 pathway is required in order to convert an alcohol into an alkyl halide.
SN2 Reactions with Alcohols
Primary and secondary alcohols will undergo substitution reactions with a variety of reagents, all
of which proceed via an SN2 process. In this section, we will explore three such reactions that all
employ an SN2 process.
1. Primary alcohols will react with HBr via an SN2 process.
Proton transfer
Nucleophilic attack +
loss of a leaving group
H
OH
H
O
Br
+
H
Br
−
Br
SN2
+
H2 O
The hydroxyl group is first protonated, converting it into an excellent leaving group, followed
by an SN2 process. This reaction works well for HBr but does not work well for HCl. To replace
the hydroxyl group with chloride, ZnCl2 can be used as a catalyst.
OH
HCl
ZnCl2
Cl
12.9
Reactions of Alcohols: Substitution and Elimination
555
Because of its ionic nature, the use of ZnCl2 is limited to alcohols that are water-soluble. For
such alcohols (Section 12.1), ZnCl2 converts the hydroxyl group into a better leaving group,
thereby enabling the subsequent SN2 process.
Nuc attack
on a Lewis acid
Nuc attack +
loss of LG
−
ZnCl2
OH
ZnCl2
O
−
+
Cl
Cl
H
SN2
2. As seen in Section 7.12, an alcohol can be converted into a tosylate, followed by nucleophilic
attack.
OH
OTs
Bad
leaving group
X
−
TsCl
py
X
(SN2)
Good
leaving group
Using tosyl chloride and pyridine, the hydroxyl group is converted into a tosylate group (an excellent leaving group), which is susceptible to an SN2 process. Notice the stereochemical outcome of
the previous reaction. The configuration of the chiral center is not inverted during formation of
the tosylate, but it is inverted during the SN2 process. The net result is inversion of configuration.
OH
Br
1) TsCl, py
2) NaBr
R
S
3. Primary and secondary alcohols can react with SOCl2 or PBr3 via an SN2 process.
SOCl2
py
Cl
PBr3
Br
OH
These two reactions are believed to proceed via similar mechanisms. In the reaction with SOCl2,
the first few steps convert a bad leaving group into a good leaving group, then the halide attacks
in an SN2 process (as shown in Mechanism 12.6, although the exact mechanism with SOCl2
depends on the experimental conditions).
MECHANISM 12.6 THE REACTION BETWEEN SOCl2 AND AN ALCOHOL
Bad
leaving group
OH
Cl
SOCl2
py
SO2
+
Cl
S
Cl
O
Cl
H
+
O
Cl
H
Cl
S
+
O
−
O
−
N
S
O
Cl
O
S
Cl
O
Good
leaving group
+
Cl
−
556
CHAPTER 12
Alcohols and Phenols
The reaction mechanism presented for PBr 3 (Mechanism 12.7) has similar characteristics,
including conversion of the hydroxyl group into a better leaving group followed by nucleophilic
attack.
MECHANISM 12.7 THE REACTION BETWEEN PBr3 AND AN ALCOHOL
Br
P
OH
Br
H
Br
O
Bad
leaving group
+
P
Br
+
Br
Br
SN 2
−
+
HOPBr2
Br
Good
leaving group
Notice the similarity among all of the SN2 processes that we have seen in this section. All
involve the conversion of the hydroxyl group into a better leaving group followed by nucleophilic attack. If any of these reactions occurs at a chiral center, inversion of configuration is to be
expected.
SKILLBUILDER
12.6 proposing reagents for the conversion of an alcohol into an alKyl halide
LEARN the skill
Identify the reagents you would use to accomplish the following transformation:
Cl
OH
STEP 1
Analyze the substrate.
STEP 2
Analyze the
stereochemical
outcome and
determine whether SN2
is required.
STEP 3
Identify reagents that
achieve the process
determined in step 2.
SOLUTION
This transformation is a substitution process, so we must decide whether to use SN1 conditions or SN2 conditions. The key factor is the substrate. The substrate here is secondary,
indicating that an SN2 process is necessary. Furthermore, the transformation requires inversion of configuration, which can only be accomplished via SN2 (an SN1 process would give
racemization, as we saw in Section 7.8). In addition, the substrate would likely undergo a
carbocation rearrangement if subjected to SN1 conditions. For all of these reasons, an SN2
process is required.
The OH group is a very poor leaving group, so treating the alcohol with chloride will not
achieve the desired SN2 reaction. However, if the OH group is first converted into a tosylate
(a good leaving group), then the desired SN2 process can be accomplished by treating the
tosylate with chloride. Alternatively, the alcohol can be treated with thionyl chloride (SOCl2)
and pyridine.
OH
Cl
1) TsCI, py
2) NaCI
SOCl2, py
In this case, HCl and ZnCl2 cannot be used, because the alcohol is not expected to be
water soluble (it has eight carbon atoms).
12.9
Reactions of Alcohols: Substitution and Elimination
557
PRACTICE the skill 12.19 Identify reagents that you could use to accomplish each of the following
transformations:
Br
OH
(a)
(b)
APPLY the skill
Cl
OH
(c)
(e)
Br
OH
OH
(f )
Cl
Br
OH
(d)
Br
OH
12.20 Tunicates are marine animals that are called “sea squirts” because when they are
taken out of water, they tend to contract and expel seawater. Lepadiformine is a cytotoxic
agent (toxic to cells) isolated from a marine tunicate. During a recent synthesis of lepadiformine, the investigators observed the formation of an interesting by-product (3) while
treating diol 1 with a reagent similar in function to PBr3.6
HO
R
PBr3
R
dibromide +
1
OH
O
2
3
(a) Draw a structure for the desired product (2).
(b) Propose a plausible mechanism for formation of by-product 3 using PBr3 as the reagent. Explain
how your mechanism provides a justification for the stereochemical outcome.
need more PRACTICE? Try Problems 12.34a–c, 12.62f, 12.65p
BioLinks Drug Metabolism
Drug metabolism refers to the set of reactions by an organism, where drugs are converted into
other compounds that are either used by the body or excreted. Our bodies dispose of the
medications we ingest by a variety of metabolic pathways.
One common metabolic pathway is called glucuronic acid conjugation, or simply
glucuronidation. This process is very similar to all of the SN2 processes that we investigated in
Section 12.9. Specifically, a bad leaving group (hydroxyl) is first converted into a good leaving
group, followed by nucleophilic attack. Glucuronidation exhibits the same two key steps.
1. Formation of UDPGA (uridine-5’-diphospho-α-D-glucuronic acid) from glucose involves
conversion of a bad leaving group into a good leaving group.
OH
OH
O
O
HO
HO
OH
D-Glucose
O
HO
HO
OH
Bad LG
OH
O
UDPGA
O
P
O
−
O
O
O
P
O
−
O
N
HO
OH
Good LG (called UDP)
UDPGA is a compound with a very good leaving group. This large leaving group is called UDP
(highlighted above). In this transformation, one of the hydroxyl groups in glucose has been
converted into a good leaving group.
H
N
O
O
558
CHAPTER 12
Alcohols and Phenols
2. Next, an SN2 process occurs in which the drug being metabolized (such as an alcohol) attacks
UDPGA, expelling the good leaving group.
OH
OH
O
O
O
HO
HO
OH
UDP-glucuronyltransferase
O UDP
O
HO
HO
– H+
ROH
O
R
OH
β-Glucuronide
This SN2 process requires an enzyme (a biological catalyst) called UDP-glucuronyl transferase.
The reaction proceeds via inversion of configuration (as expected of an SN2 process) to
produce a β-glucuronide, which is highly water soluble and is readily excreted in the urine.
Many functional groups undergo glucuronidation, but alcohols and phenols are the most
common classes of compounds that undergo this metabolic pathway. For example, morphine,
acetaminophen, and chloramphenicol are all metabolized via glucuronidation:
N
H
N
HO
O
OH
H
H
OH
Morphine
An opiate analgesic
used to treat severe pain
OH
O
Cl
O2N
HO
Acetaminophen
An analgesic (pain-relieving)
and antipyretic (fever-reducing) agent,
sold under the trade name Tylenol
N
H
Cl
O
Chloramphenicol
An antibiotic used in eye drops
to treat bacterial conjuctivitis
In each of these three compounds, the highlighted hydroxyl group attacks UDPGA. Glucuronidation
is the main metabolic pathway by which these drugs are eliminated from the body.
E1 and E2 Reactions with Alcohols
Recall from Section 7.10 that alcohols undergo elimination reactions in acidic conditions.
conc. H2SO4
OH
+
heat
H2O
For tertiary alcohols, this transformation follows an E1 mechanism:
Proton transfer
Proton transfer
Loss of LG
H
H
OH
O
+
H
H
O+
–H2O
+
H
H
H
O
H
Recall that the two core steps of an E1 mechanism are loss of a leaving group followed by a proton
transfer. However, when the starting material is an alcohol, the mechanism must begin with an additional step, in which the hydroxyl group is protonated.
Recall that elimination generally favors the more substituted alkene.
OH
conc. H2SO4
+
heat
Major
Minor
This transformation can also be accomplished via an E2 pathway if the hydroxyl group is first converted into a better leaving group, such as a tosylate. A strong base can then be employed to accomplish an E2 reaction.
OH
TsCl
py
OTs
NaOEt
(E2)
12.10
Reactions of Alcohols: Oxidation
559
As seen in Section 7.10, when more than one alkene product is possible, an E2 process will generally
favor formation of the more substituted alkene (unless a sterically hindered base is used), and no
carbocation rearrangements are observed in E2 processes.
CONCEPTUAL CHECKPOINT
12.21
Predict the major and minor products for each of the following transformations:
OH
(a)
conc. H2SO4
heat
?
?
1) TsCl, py
2) NaOEt
OH
(b)
12.10 Reactions of Alcohols: Oxidation
In Section 12.4, we saw that alcohols can be formed via a reduction process. In this section, we will
explore the reverse process, called oxidation, which involves an increase in oxidation state.
Oxidation
OH
H
O
Reduction
The outcome of an oxidation process depends on whether the starting alcohol is primary, secondary,
or tertiary. Let’s first consider the oxidation of a primary alcohol.
O
OH
O
[O]
[O]
R
H
A primary alcohol
LOOKING AHEAD
For an exception to this
general rule, see Section
19.11, where we will learn
about a special oxidizing
reagent that can oxidize a
ketone to form an ester.
H
R
H
R
An aldehyde
OH
A carboxylic acid
Notice that a primary alcohol has two protons at the α position (the carbon atom bearing the
hydroxyl group). As a result, primary alcohols can be oxidized twice. The first oxidation produces an
aldehyde, and then oxidation of the aldehyde produces a carboxylic acid.
Secondary alcohols only have one proton at the α position so they can only be oxidized once,
forming a ketone.
OH
O
[O]
R
R
H
R
A secondary alcohol
R
A ketone
Generally speaking, the ketone is not further oxidized.
Tertiary alcohols do not have any protons at the α position, and as a result, they generally do
not undergo oxidation:
OH
[O]
R
R
R
No reaction
A large number of reagents are available for oxidizing primary and secondary alcohols. A common
oxidizing reagent is chromic acid (H2CrO4), which can be formed either from chromium trioxide
(CrO3) or from sodium dichromate (Na2Cr2O7) in aqueous acidic solution.
O
O
Cr
H3O+
Acetone
O
O
Chromium trioxide
HO Cr OH
+
Na
O
O
O
−
−
O Cr O Cr O
O
O
Sodium dichromate
+
Na
H2SO4
H2O
Chromic acid
560
CHAPTER 12
Alcohols and Phenols
Chromic acid oxidations are believed to proceed via a mechanism comprised of two stages (Mechanism 12.8). The first stage involves formation of a chromate ester, and the second stage is an E2
process to form a carbon-oxygen π bond (rather than a carbon-carbon π bond).
MECHANISM 12.8 OXIDATION OF AN ALCOHOL WITH CHROMIC ACID
STAGE 1
O
R
H
OH
+
HO Cr OH
R
Fast and
reversible
O
R
H
O
O
+
Cr OH
R
H2 O
O
Chromate ester
STAGE 2
O
R
H
O
R
Cr OH
H
R
O
H
O
E2
+
+
H3O
+
−
HCrO3
R
O
Chromate ester
When a primary alcohol is oxidized with chromic acid, a carboxylic acid is obtained. It is generally
difficult to control the reaction to produce the aldehyde.
OH
R
O
Na2Cr2O7
H2SO4 , H2O
An alcohol
R
O
H
R
OH
A carboxylic acid
An aldehyde
Difficult to isolate
In order to produce the aldehyde without further oxidation, it is necessary to use a more selective
oxidizing reagent—one that will react with the alcohol but will not react with the aldehyde. One
such reagent is pyridinium chlorochromate (PCC), typically dissolved in methylene chloride
(CH2Cl2):
+
O
N
PCC
R
OH
R
CH2Cl2
H
H
−
CrO3Cl
Pyridinium chlorochromate
(PCC)
Under these conditions, the alcohol is oxidized to give an aldehyde which is not further oxidized to
a carboxylic acid. When a secondary alcohol is treated with a chromium oxidizing agent (whether
chromic acid or PCC), the alcohol is oxidized to give a ketone:
Na2Cr2O7
H2SO4, H2O
OH
R
R
PCC
CH2Cl2
O
R
R
Chromium-based oxidations have largely been replaced with alternative oxidation techniques
that are more green (less harmful to the environment). Chromium-based oxidations produce byproducts including chromium salts that are hazardous and require special disposal. Two alternative
oxidation methods, called the Swern oxidation and the Dess–Martin oxidation, will now be
discussed briefly.
12.10
561
Reactions of Alcohols: Oxidation
The Swern oxidation uses dimethyl sulfoxide (DMSO) and oxalyl chloride (COCl)2, as
shown:
OH
R
O
1) DMSO, (COCl)2
2) Et3N
R
R
R
The alcohol is added to a mixture of DMSO and oxalyl chloride at low temperature (often –78°C),
followed by treatment with a base, such as triethylamine, Et3N. The reaction is believed to proceed
via Mechanism 12.9, which can be broken down into two stages. In the first stage, DMSO reacts
with oxalyl chloride and is thereby converted into a chlorodimethylsulfonium ion (highlighted).
This ion is believed to function as the active oxidizing agent, as seen in stage 2 of the mechanism.
Notice that the last step of the process is an intramolecular elimination reaction. This is the step in
which the carbon atom undergoes oxidation to give a ketone.
MECHANISM 12.9 SWERN OXIDATION
STAGE 1
Nucleophilic attack
Cl
Cl
H3C
+S
O
Cl
−
H3C
DMSO
O
DMSO
functions as
a nucleophile
and attacks
oxalyl chloride
Substitution
(Nuc attack + Loss of LG)
Loss of a
leaving group
O
O
H3C
+S
H3C
O
O
Cl
O
H3C
−
CH3
S
O
A
+
chloride
H3C
ion is expelled
as a leaving group
Cl
O
+
Cl
−
Cl
A chloride ion
functions
as a nucleophile,
resulting in a
substitution process
Cl
+
−
+ CO
S+
CH3
+ CO2
Chlorodimethyl
sulfonium ion
STAGE 2
Substitution
(Nuc attack + Loss of LG)
H3C
H3C
S
OH
R
C
H3C
+
Cl
+
CH3
+O
H
S
C
R
The alcohol
functions as
R
R
a nucleophile
A 2° alcohol and attacks the
chlorodimethylsulfonium ion,
ejecting chloride as a leaving group
H
Proton
transfer
Et3N
O
C
R
Triethyl
amine
R
functions as a base,
and removes a proton
H
Proton
transfer
CH3
S
+
H
C
H
H
H
Et3N
O
C
R
Triethyl
amine
R
functions as a base,
and removes a proton
CH3
Intramolecular elimination
(Proton transfer + Loss of LG)
S
H
+
H
−
C
CH3
O
C
H
+
S
An intraCH3
R
R
molecular
DMS
elimination A ketone
process gives the
(by-product)
oxidation product (a ketone)
and DMS as a by-product
The Swern oxidation can also be used to convert primary alcohols into aldehydes:
OH
R
O
1) DMSO, (COCl)2
H
H
A 1° alcohol
2) Et3N
R
H
An aldehyde
Under these conditions, the aldehyde is not further oxidized to a carboxylic acid. The by-products
of the Swern oxidation are volatile (low-boiling), so they can be easily separated from the reaction
mixture.
562 CHAPTER 12 Alcohols and Phenols
Another alternative to chromium-based oxidations is called the Dess–Martin periodinane
(DMP) oxidation. Much like the Swern oxidation, DMP-based oxidations will also convert primary
alcohols into aldehydes, and secondary alcohols are converted into ketones:
OH
R
H
DMP
CH2Cl2
H
A 1° alcohol
O
R
H
AcO OAc
OAc
An aldehyde
DMP =
OH
R
R
H
A 2° alcohol
DMP
CH2Cl2
O
OAc =
O
O
O
R
O
R
Dess–Martin Periodinane
A ketone
DMP-based oxidations employ nonacidic conditions (neutral pH) and can also be achieved at room
temperature. These features offer advantages over chromium-based oxidations, which require acidic
conditions and high temperature. In many cases, acidic conditions must be avoided in order to prevent undesired side reactions.
DMP-based oxidations are believed to proceed via the following type of periodinane
intermediate:
R
AcO
OH
R
H
H
H
O
H
O
DMP
CH2Cl2
OAc
O
O
+
O
O
+
2 AcOH
H
R
O
O
Periodinane intermediate
By-products
Both the Swern and DMP-based oxidations are “greener” alternatives to chromium-based oxidations, yet both processes have very poor atom efficiency. Also, the Swern oxidation produces a very
foul-odored by-product (dimethyl sulfide, DMS), and DMP must be handled with care, because it
is shock sensitive (explosive).
SKILLBUILDER
12.7 predicting the products of an oxidation reaction
LEARN the skill
Predict the major organic product of the following reaction:
OH
CrO3
H3O+, acetone
?
SOLUTION
STEP 1
Identify whether the
alcohol is primary or
secondary.
STEP 2
If the alcohol is
primary, consider
whether an aldehyde
or a carboxylic acid is
obtained.
In order to predict the product, identify whether the alcohol is primary or secondary. In this
case, the alcohol is primary, and therefore, we must decide whether the process will produce an aldehyde or a carboxylic acid:
O
O
OH
[O]
H
An aldehyde
[O]
OH
A carboxylic acid
The product is determined by the choice of reagent. Chromium trioxide in acidic conditions forms chromic acid, which will oxidize the alcohol to give the carboxylic acid.
12.11 Biological Redox Reactions 563
In order to stop at the aldehyde, we could use PCC, or a Swern oxidation, or a DMPbased oxidation.
O
STEP 3
Draw the product.
OH
OH
CrO3
H3O+, acetone
PRACTICE the skill 12.22 Predict the major organic product for each of the following reactions:
OH
OH
(a)
?
Na2Cr2O7
H2SO4 , H2O
?
Na2Cr2O7
H2SO4 , H2O
(b)
H
(c)
xs CrO3
OH
H3O+
acetone
O
?
OH
(d)
1) DMSO, (COCI)2
2) Et3N
APPLY the skill
?
OH
(e)
HO
?
PCC
CH2Cl2
O
(f )
DMP
CH2Cl2
H
?
12.23 Galantamine, a natural product produced by the amaryllis family of flowering plants,
has been used in the early treatment of Alzheimer’s disease. In a synthesis of g
­ alantamine
and related compounds,7 compound 1 was converted into compound 2. Propose an
­efficient synthesis for the conversion of 1 to 2.
MeO
O
H
MeO
O
O
MeO
H
O
O
H
OH
O
O
O
1
H
N
2
Me
Galantamine
need more PRACTICE? Try Problems 12.34e–f, 12.36, 12.46, 12.53
12.11 Biological Redox Reactions
BY THE WAY
NADH is much less reactive
than NaBH4 or LiAlH4 and
requires a catalyst to do its
job. Nature’s catalysts are
called enzymes (Section 25.8).
In this chapter, we have seen several reducing agents and several oxidizing reagents used by chemists
in the laboratory. Nature employs its own reducing and oxidizing agents, although they are generally much more complex in structure and much more selective in their reactivity. Indeed, biological
reactions generally occur enantioselectively, affording a single enantiomer, rather than a racemic
mixture. One important biological reducing agent is NADH, which is a molecule comprised of
many parts, shown in Figure 12.7.
O
HO
A sugar
(ribose) HO
O
O
P
O−
N
FIGURE 12.7
The structure of NADH, a
biological reducing agent.
Reactive center
Diphosphate linkage
N
A base
N
(adenine)
N
NH2
H
O
O
P
H
O
O
O−
OH
OH
A sugar
(ribose)
N
NH2
O
564 CHAPTER 12 Alcohols and Phenols
The reactive center in NADH (highlighted in orange) functions as a hydride delivery agent (very
much like NaBH4 or LiAlH4) and can reduce ketones or aldehydes to form alcohols. NADH acts as
a reducing agent, and in the process, it is oxidized. The oxidized form is called NAD+.
H
R
H
N
NH2
R
+
+
O
H
+
Base
H
Enzymatic
catalysis
R
+
±
NH2
N
R
Ketone
O
+ H
R
OH
+ Base
R
O
Alcohol
NAD+
NADH
The reverse process can also occur. That is, NAD+ can act as an oxidizing agent by accepting a
hydride from an alcohol, and in the process, NAD+ is reduced to give NADH.
H
+
N
R
NH2
R
O
+
H
H
O
+
Base
H
Enzymatic
catalysis
N
R
R
Alcohol
NAD+
BY THE WAY
To remember which is the
reducing agent and which is
the oxidizing agent, simply
remember that NADH has
an H at the end of its name,
because it can function as a
delivery agent of hydride. So
NADH is a reducing agent.
FIGURE 12.8
NADH and NAD+ play an
important role in the citric acid
cycle as well as in ATP synthesis.
R
H
NH2
O
+
O
+
H
+
Base
R
Ketone
NADH
NAD+ and NADH are present in all living cells and function in a wide variety of redox reactions.
NADH is a reducing agent, and NAD+ is an oxidizing agent. Two important biological processes
demonstrate the critical role of NADH and NAD+ in biological systems (Figure 12.8). The citric
acid cycle is part of the process by which food is metabolized. This process involves the conversion
of NAD+ to NADH. Another important biological process is the conversion of ADP (adenosine
diphosphate) to ATP (adenosine triphosphate), called ATP synthesis, which involves the conversion
of NADH to NAD+. ATP is a molecule that is higher in energy than ADP, and the conversion of
ADP to ATP is the way that our bodies store the energy that is obtained from metabolism of the
food we eat. The energy stored in ATP can be released by conversion back into ADP, and that stored
energy is used to power all of our life functions.
Citric
acid
cycle
NAD+
ADP
NADH
ATP
The citric acid cycle and ATP synthesis are linked into one story. That story can be summarized
like this: Energy from the sun is absorbed by vegetation and is used to convert CO2 molecules
into larger organic compounds (a process called carbon dioxide fixing). These organic compounds
have bonds (such as ​CC​ and ​CH​) that are higher in energy than the ​CO​bonds in CO2.
Therefore, energy will be released if the organic compounds are converted back into CO2 (this
is precisely why the combustion of gasoline releases energy). Humans eat the vegetation (or we
eat the animals that ate the vegetation), and our bodies utilize a series of chemical reactions that
break down those large molecules into CO2, thereby releasing energy. That energy is once again
captured and stored in the form of high-energy ATP molecules, which are then used to provide
energy for our biological processes. In this way, our bodies are ultimately powered by energy from
the sun.
12.12
Oxidation of Phenol
565
BioLinks Biological Oxidation of Methanol and Ethanol
Methanol is oxidized in our bodies by NAD+.
NAD+
OH
H
H
H
NADH
Alcohol
dehydrogenase
NAD+
O
H2O,
Aldehyde
dehydrogenase
H
H
NADH
Formaldehyde
Methanol
O
OH
H
Formic acid
The enzyme that catalyzes this oxidation process is called
alcohol dehydrogenase. Methanol is oxidized twice. The first
oxidation produces formaldehyde (a toxic compound), while the
second oxidation produces formic acid. Formic acid is highly
toxic, even in small quantities. A buildup of formic acid in the
eyes leads to blindness, and a buildup of formic acid in other
organs leads to organ failure and death.
A methanol overdose is typically treated by administering
ethanol to the patient as soon after methanol ingestion as
possible. Ethanol binds to the alcohol-metabolizing enzymes
more tightly than methanol, resulting in a decreased rate of
methanol oxidation and allowing most of the methanol to be
excreted unmetabolized. On the downside, the dose of ethanol
required to prevent methanol metabolism can cause major
intoxication and acidosis. A newer drug, fomepizole, can inhibit
methanol metabolism without the adverse effects of ethanol, but
is under patent and very expensive. It is not commonly stocked
by most hospital pharmacies.
The oxidation of ethanol produces acetic acid instead of
formic acid, and acetic acid is not toxic.
NAD+
OH
H
H
Ethanol
NADH
Alcohol
dehydrogenase
NAD+
O
H
Acetaldehyde
NADH
H2O,
Aldehyde
dehydrogenase
mg7/Getty Images
richcano/Getty Images
O
OH
Acetic acid
Ethanol is a primary alcohol and is oxidized twice. The first
oxidation produces acetaldehyde, while the second oxidation
produces acetic acid. Acetic acid can be used by the body for a
variety of functions, but acetaldehyde is less useful. When a person
drinks large quantities of ethanol (binge drinking), the concentration
of acetaldehyde temporarily builds up. A high concentration of
acetaldehyde causes nausea, vomiting, and other nasty symptoms.
We mentioned in the chapter opener that the effects of a
hangover are caused by a variety of factors. The impact of some
of those factors, such as dehydration, can be reduced by drinking
a glass of water in between drinks. But other factors, such as the
buildup of acetaldehyde, are the unavoidable consequences of
binge drinking. The only way to avoid high concentrations of
acetaldehyde is to drink small quantities of alcohol over a long
period of time. Binge drinking will always produce the unpleasant
effects of a hangover. There are many products on the market that
claim to prevent hangovers, but there is little scientific evidence
that any of these products are effective. The only way to prevent a
hangover is to drink responsibly. That is, drink small quantities of
alcohol over a long period of time, together with plenty of water.
In addition to the unpleasant effects of a hangover, binge
drinking can also potentially cause a host of more serious health
issues, such as an irregular heart rhythm or acute pancreatitis
(inflammation of the pancreas), both of which can be life
threatening.
12.12 Oxidation of Phenol
In Sections 12.10 and 12.11, we discussed reactions that achieve the oxidation of primary and secondary alcohols. In this section, we consider the oxidation of phenol. Based on our discussion thus
far, we might expect that phenol would not readily undergo oxidation because it lacks a proton at
the α position, much like a tertiary alcohol.
OH
OH
No alpha proton
α
α
R
R
R
3° alcohol
Phenol
No alpha proton
566 CHAPTER 12 Alcohols and Phenols
Nevertheless, phenol is observed to undergo oxidation even more readily than primary and secondary alcohols. The product is benzoquinone.
OH
O
Na2Cr2O7
H2SO4, H2O
Phenol
O
Benzoquinone
Benzoquinone is important because it is readily reduced to give hydroquinone:
Reduction
O
O
Oxidation
Benzoquinone
OH
OH
Hydroquinone
The reversibility of this process is critical for cellular respiration, a process by which molecular
­oxygen is used to convert food into CO2 and water, with the release of energy. Key players in this
process are a group of compounds called quinones:
O
O
CH3
CH2CH
O
CCH2 H
n
n = 6–10
O
Quinones
These compounds are often called ubiquinones because they are ubiquitous in nature—that is, they
are found in all cells. The redox properties of quinones are utilized to convert molecular oxygen
into water. The initial step of this process is the reduction of a quinone to a hydroquinone, as
shown here:
O
Initial step:
OH
NADH
O
+
O
R
O
A quinone
NAD+
O
H+
O
R
OH
A hydroquinone
The hydroquinone is then oxidized back to quinone, which involves the transfer of protons across a
biological membrane. This flow of protons creates an electric potential, which can be viewed as the
biological analog of charging a battery. The charged biological battery is used to drive the synthesis
of ATP from ADP and phosphate, via a process catalyzed by the enzyme ATP synthase. During this
process, protons are ultimately combined with oxygen to make water. The conversion of molecular
oxygen into water is a critical step in the breakdown of food molecules to release the energy stored
in their chemical bonds.
The idea that ATP synthesis could be driven by a battery-like force was first proposed by British
biochemist Peter Mitchell in 1961 and was initially very controversial. But more and more experimental data supported his hypothesis, and it is now the accepted explanation for how most (but not
all) ATP is synthesized in aerobic (air-breathing) organisms. Mitchell received the Nobel prize in
Chemistry in 1978 for elucidating this mechanism.
12.13 Synthesis Strategies 567
12.13 Synthesis Strategies
Recall from Chapter 11 that there are two issues to consider when proposing a synthesis:
1. A change in the carbon skeleton.
2. A change in the functional group.
In this chapter, we have learned valuable skills for dealing with both of these issues. Let’s focus on
them one at a time.
Functional Group Interconversion
In Chapter 11, we saw how to interconvert triple bonds, double bonds, and single bonds
(Figure 12.9).
FIGURE 12.9
A map showing the reactions
that enable the interconversion
between alkanes, alkenes, and
alkynes.
H2
H2
Lindlar's cat.
Pt
R
R
R
1) Br2/CCl4
2) xs NaNH2
1) Br2, hv
3) H2O
2) NaOEt
In this chapter, we learned how to interconvert between ketones and secondary alcohols. This level
of control will be very important later, as we will see many reactions involving compounds containing carbonyl groups.
OH
Na2Cr2O7
H2SO4, H2O
O
1) LiAIH4
2) H3O+
Primary alcohols and aldehydes can also be interconverted. Oxidation with PCC (or DMP, or Swern
conditions) will convert an alcohol into an aldehyde.
OH
PCC
CH2Cl2
O
1) LiAIH4
2) H3O+
Let’s now combine some important reactions we have seen in the previous few chapters into one
map that will help us organize reactions that interconvert functional groups. We will focus on the six
functional groups highlighted in Figure 12.10. These six functional groups have been separated into
categories based on oxidation state. Conversion of a functional group from one category to another
category is either a reduction or an oxidation. For example, converting an alkane into an alkyl halide
is an oxidation reaction. In contrast, interconverting functional groups within a category is neither
oxidation nor reduction. For example, conversion of an alcohol into an alkyl halide does not involve
a change in oxidation state.
568 CHAPTER 12 Alcohols and Phenols
High
oxidation state
Low
oxidation state
R
Alkynes
FIGURE 12.10
A diagram showing six different
functional groups and their
relative oxidation states. The
interconversion between alkenes,
alcohols, and alkyl halides does
not constitute oxidation or
reduction.
Reduction
O
R
Oxidation
R
R
Ketones or
aldehydes
Alcohols
Alkanes
Reduction
X
OH
R
Oxidation
Alkenes
R
Alkyl halides
We have seen many reagents that allow us to interconvert functional groups, both within a category and across categories (oxidation/reduction). Those reagents are shown in Figure 12.11. It is
worth your time to study Figure 12.11 carefully, as it summarizes many of the key reactions that we
have seen until now. Note that each functional group can be converted into any one of the other
functional groups. Some conversions can be done in one step, such as the conversion of an alcohol
into a ketone. Other conversions require multiple steps, such as the conversion of an alkane into a
ketone. Having a clear mental picture of this chart will be extremely helpful when solving synthesis
problems.
Alkynes
H2, Lindlar’s catalyst
C
1) Br2
C
Alkenes
or Na, NH3
C
Alkanes
H2
C
Pt, Pd or Ni
C
C
2) xs NaNH2
3) H2O
H2SO4, H2O
HgSO4
or
1) R2BH
2) H2O2, NaOH
Ketones/
aldehydes
conc. H2SO4, heat
or
1) TsCl/pyr
2) NaOMe
1) LiAIH4, 2) H3O+
or NaBH4, MeOH
Alcohols
OH
O
C
H3O+
or
1) BH3 • THF
2) H2O2, NaOH
or
1) Hg(OAc)2, H2O
2) NaBH4, NaOH
H2CrO4 or PCC
or DMP or Swern
C
HX or SOCl2/py
or PBr3
or 1) TsCl/py
2) NaX
NaOH (SN2)
HX
or
HBr, ROOR
NaOMe
or
t-BuOK
Alkyl
halides
Br
Br2, hν
C
or H2O (SN1)
FIGURE 12.11
A map of reactions for interconversion between alkanes, alkenes, alkynes, alkyl halides, alcohols, and
ketones/aldehydes.
In Chapter 11, we summarized the transformations we had learned to that point as a “Synthetic
Toolbox Review” in Table 11.1. In each new chapter, we will continue to add new tools to our synthetic toolbox. In this chapter, we’ve learned new methods for preparing alcohols (by the reduction
of ketones or aldehydes) and for preparing ketones, aldehydes, and carboxylic acids (by the oxidation
of alcohols). Table 12.1 illustrates how these new methods fit into our existing synthetic toolbox.
12.13 Synthesis Strategies 569
TABLE 12.1 an update to our existing synthetic toolbox (table 11.1) that incorporates
reduction and oxidation methods
functional
possible starting material to
group in
make target molecule by
target molecule
functional group interconversion
example
H3O+
OH
Alkene
1) BH3·THF
OH
2) H2O2, NaOH
Alcohol
Alkyl Halide
NaOH
Br
OH
O
Ketone/Aldehyde
1) LiAlH4
2) H3O+
H
OH
1) O3
Alkene
O
2) DMS
O
H2O, H2SO4
HgSO4
Ketone
or
Aldehyde
(Markovnikov)
Alkyne
O
1) R2BH
2) H2O2, NaOH
(anti-Markovnikov)
OH
Alcohol
Na2Cr2O7
O
H2SO4, H2O
O
1) O3
Alkyne
Carboxylic
acid
O
Alcohol
OH
SKILLBUILDER
12.8 converting functional groups
LEARN the skill
OH
2) H2O
Propose an efficient synthesis for the following transformation:
O
Na2Cr2O7
H2SO4, H2O
OH
570 CHAPTER 12 Alcohols and Phenols
SOLUTION
In this example, the reactant and product have the same carbon skeleton and differ only in the
identity of the functional group. The reactant has a carbon-carbon double bond, while the product has a carbon-oxygen double bond (a carbonyl group). It is always preferable to achieve a
transformation in the fewest number of steps possible, so we first consider whether there is a single step that will achieve this transformation. As seen in Figure 12.11, we have not yet discussed
a one-step method for achieving the desired transformation. Therefore, to plan the multistep synthesis, we can try to identify the structure of an intermediate compound that will provide a reasonable pathway for the required transformation. This compound can serve as a stepping stone
in the transformation of the given starting material to the desired product (the target molecule):
?
O
?
Intermediate
compound
The retrosynthetic analysis of the target molecule asks, “What reaction have we seen that produces a ketone product?” This analysis reveals two potential starting materials for the last step
of the synthesis. We have seen that a ketone can be made from an alcohol or from an alkyne:
O
OH
or
Both strategies are reasonable, because both the alcohol and the alkyne shown above can
be prepared from the given alkene starting material:
OH
O
Therefore, there are (at least) two perfectly acceptable answers to this problem.
1) H3O+
2) PCC or Na2Cr2O7,
H2SO4, H2O
O
1) Br2
2) xs NaNH2
3) H2O
4) H2SO4, H2O, HgSO4
PRACTICE the skill 12.24 Propose an efficient synthesis for each of the following transformations:
OH
(a)
O
OH
(b)
(c)
H
O
O
H
(d)
OH
APPLY the skill
(e)
(f )
12.25 Lithium borohydride (LiBH4) is a useful reducing agent, as it is more selective than
LiAlH4 but less selective than NaBH4. Specifically, LiBH4 will reduce esters, but it will not
reduce amides. This selectivity is frequently employed by synthetic organic chemists. For
12.13 Synthesis Strategies 571
example, during a synthesis of (−)-croalbinecine, a natural product with cytotoxic properties, compound 1 was converted to compound 3, as shown below.8 Draw the structures of
2 and 3.
CO2Et
Ph
HO
1) Excess LiBH4
N
2) H2O
O
1
1) Excess NaH,
excess PhCH2Br,
DMF
2) H2, Lindlar's catalyst
2
3
OCH3
need more PRACTICE? Try Problems 12.38, 12.46, 12.61, 12.64, 12.66, 12.67
​CC​Bond Formation
In this chapter, we learned a new way to form ​CC​bonds using the reaction between a Grignard
reagent and a ketone or aldehyde.
O
H3C
OH
1) CH3MgBr
2) H3O+
OH
O
1) CH3MgBr
H
2) H3O+
H
CH3
We also saw that an ester will react with two equivalents of a Grignard reagent to form an alcohol. In
this reaction, two new ​CC​bonds have been formed.
OH
O
1) Excess CH3MgBr
OMe
CH3
CH3
2) H3O+
+
MeOH
The ability to form a new ​CC​bond represents an extremely important synthetic tool. By modifying carbon skeletons, chemists can build complex molecules, such as biologically active, natural
products. Thus far, we have seen a variety of methods for extending carbon chains. These methods
are summarized in Table 12.2.
TABLE 12.2
summary of carbon-carbon bond-forming reactions
possible starting material to
functional group in
make target molecule via
target molecule
carbon-carbon bond formation
Alkyne
Acetylene or
Terminal Alkyne
example
(new C​​C bond highlighted in red)
1) NaNH2
2) Me
O
Ketone/Aldehyde
Alcohol
1) EtMgBr
H
O
Ester
OH
2) H3O+
1) MeMgBr (xs)
OEt
CH3
2) H3O+
OH
CH3
CH3
572
CHAPTER 12
Alcohols and Phenols
A Grignard reaction accomplishes more than just CC bond formation; it also reduces the
carbonyl group to an alcohol in the process. Now consider the following two steps: Suppose we perform a Grignard reaction with an aldehyde and then oxidize the resulting alcohol to give a ketone.
OH
O
H
O
[O]
Attack with
Grignard
R
R
The net result of these two steps is to convert an aldehyde into a ketone. This is very helpful, because
we have not yet seen a direct way to convert aldehydes into ketones.
O
O
1) RMgBr
2) H3O+
H
3) Na2Cr2O7,
H2SO4, H2O
R
CONCEPTUAL CHECKPOINT
12.26
Identify reagents that you could use to accomplish each of the following transformations:
O
(a)
O
O
H
O
H
(b)
Functional Group Transformations and CC Bond Formation
We will now combine the two skills covered in this section and work on synthesis problems that
involve both functional group interconversion and CC bond formation. Recall from Chapter 11
that it is very helpful to approach a synthesis problem working forward as well as working backward
(retrosynthetic analysis). We will use all of these skills in the following SkillBuilder.
SKILLBUILDER
12.9 proposing a synthesis
LEARN the skill
Propose an efficient synthesis for the following transformation:
Br
OH
SOLUTION
Always approach a synthesis problem by initially asking two questions.
1. Is there a change in the carbon skeleton? Yes, the carbon skeleton is increasing in size
by two carbon atoms.
2. Is there a change in the functional groups? Yes, the starting material is an alcohol, and
the product is an alkyl bromide.
To begin a retrosynthetic analysis, we examine the target molecule and ask, “What starting
material could be used to produce an alkyl bromide?” We have seen that an alkyl bromide
can be made either from an alkene or from an alcohol:
Br
OH
or
12.13 Synthesis Strategies 573
Let’s first consider the route involving the alkene intermediate. It is always critical to analyze
any proposed strategy and make sure that the regiochemistry and stereochemistry of each
step are correct. This particular strategy does not have any issues of stereochemistry (there
are no chiral centers or stereoisomeric alkenes). However, the regiochemistry is problematic.
There is a serious flaw with the addition of HBr to the alkene. Specifically, it is not possible to
control the regiochemical outcome of that reaction. That step will likely produce a mixture of
two possible alkyl bromides (2-bromopentane and 3-bromopentane). This represents a critical flaw. We cannot propose a synthesis in which we have no control over the regiochemistry
of one of the steps. All of our steps should lead to the desired product as the major product.
The alkene is not a viable option for preparing the desired target molecule, so let’s turn
our attention to the other possibility—making the alkyl bromide from an alcohol starting
material. This transformation is a reasonable one, and the alcohol intermediate is suitable
for the required C​​C bond disconnection as we continue our retrosynthetic analysis. After
identifying the α carbon, a disconnection is made to give a suitable electrophile (a three-­
carbon aldehyde) and nucleophile (a two-carbon Grignard reagent):
OH
O
δ+ H
+
BrMg
δ–
Electrophile
Nucleophile
To complete the synthesis, we just need to bridge the gap between the given starting material and the required aldehyde:
Br
OH
?
PBr3
O
OH
1) EtMgBr
H
2) H3O+
Conversion of an alcohol to an aldehyde is an oxidation reaction that can be accomplished
with PCC, or with DMP, or with Swern conditions. Our answer can be summarized in the
­following way:
OH
Br
1) PCC, CH2Cl2
2) EtMgBr
3) H3O+
4) PBr3
PRACTICE the skill 12.27 Propose an efficient synthesis for each of the following transformations:
OH
OH
(b)
(a)
O
OH
O
O
H
(c)
(d)
OH
O
(e)
APPLY the skill
Cl
Cl
(f )
Ph
H
12.28 In a study of the chemical components of salted and
pickled prunes, hexyl butanoate was identified as one of 181
detected compounds.9 Propose a synthesis of this compound
using acetylene as your only source of carbon atoms. (Hint:
See SkillBuilder 7.8.)
need more PRACTICE? Try Problems 12.36, 12.37, 12.39, 12.43, 12.65, 12.68, 12.71
Ph
Ph
Ph
O
O
Hexyl butanoate
574
CHAPTER 12
Alcohols and Phenols
REVIEW OF REACTIONS
Preparation of Alkoxides
NaH
or Na
ROH
RO
−
SN2 Reactions with Alcohols
HBr
+
Na
OH
Preparation of Alcohols via Reduction
NaBH4 , MeOH
O
H
R
PBr3
H
R
SOCl2
py
H
1) LiAIH4
2) H3O+
NaBH4 , MeOH
H2
Pt, Pd, or Ni
R
S
R
OH
R
Cl
OH
O
R
S
R
H
O
H2
Pt, Pd, or Ni
HCl
ZnCl2
R
1) LiAIH4
E1 and E2 Reactions with Alcohols
2) H3O+
OH
O
1) Excess LiAIH4
R
2) H3O+
OH
OH
1) Excess LiAIH4
OMe
2) H3O+
R
+
OH
TsCl
py
OH
R
R
O
Na2Cr2O7
H2SO4 , H2O
R
Secondary
alcohol
O
OH
R
OMe
1) Excess RMgX
2) H3O+
R
R
R
+
MeOH
Na2Cr2O7
H2SO4 , H2O
R
Primary
alcohol
OH
Protection and Deprotection of Alcohols
R
TMSCl
Et3N
R
OH
R
OTMS
TBAF
Primary
alcohol
R
OH
R
HX
R
OH
Carboxylic
acid
O
PCC, CH2Cl2
or DMP, CH2Cl2
or 1) DMSO, (COCI)2
2) Et3N
R
H
Aldehyde
OH
O
Na2Cr2O7
H2SO4 , H2O
SN1 Reactions with Alcohols
R
R
Ketone
R
OH
O
R
R
X
R
H2O
NaOEt
OTs
Oxidation of Alcohols and Phenols
OH
1) RMgX
2) H3O+
+
MeOH
Preparation of Alcohols via Grignard Reagents
O
conc. H2SO4
Heat
OH
R
O
R
Br
1) TsCl, py
2) NaBr
+
H2O
Phenol
O
Benzoquinone
Review of Concepts and Vocabulary
575
REVIEW OF CONCEPTS AND VOCABULARY
SECTION 12.1
SECTION 12.7
• Alcohols are compounds that contain a hydroxyl group (OH)
• Protecting groups, such as the trimethylsilyl group, can be
connected to an sp3-hybridized carbon atom.
• When naming an alcohol, the parent is the longest chain containing the hydroxyl group.
• All alcohols possess a hydrophilic region and a hydrophobic region. Small alcohols (methanol, ethanol, n-propanol)
are miscible with water. A substance is said to be soluble
in water when only a certain amount of the substance will
dissolve in a specified volume of water at room temperature.
n-Butanol is soluble in water.
used to circumvent the problem of Grignard incompatibility
and can be easily removed after the desired Grignard reaction has been performed.
SECTION 12.8
• Phenol, also called hydroxybenzene, is used as a precursor in
the synthesis of a wide variety of pharmaceuticals and other
commercially useful compounds.
SECTION 12.9
SECTION 12.2
• Tertiary alcohols will undergo an SN1 reaction when treated
• The conjugate base of an alcohol is called an alkoxide ion.
• The pKa for most alcohols falls in the range of 15–18.
• Alcohols are commonly deprotonated with either sodium
• Primary and secondary alcohols can undergo an SN2 process
hydride (NaH) or an alkali metal (Na, Li, or K).
• Several factors determine the relative acidity of alcohols,
including resonance, induction, and solvating effects.
• The conjugate base of phenol is called a phenolate, or
phenoxide ion.
SECTION 12.3
• When preparing an alcohol via a substitution reaction, pri-
mary substrates require SN2 conditions, while tertiary substrates require SN1 conditions.
• Addition reactions that will produce alcohols include acidcatalyzed hydration, oxymercuration-demercuration, and
hydroboration-oxidation.
• Alcohols can be formed by treating a carbonyl group (CO
SECTION 12.4
bond) with a reducing agent. The resulting reaction involves
a decrease in oxidation state and is called reduction.
• LiAlH4 is more reactive than NaBH4. For example, LiAlH4 will
reduce carboxylic acids and esters, while NaBH4 will not.
with a hydrogen halide.
when treated with HX, SOCl2, or PBr3 or when the hydroxyl group
is converted into a tosylate group followed by nucleophilic attack.
• Tertiary alcohols undergo E1 elimination when treated with
sulfuric acid.
• For an E2 process, the hydroxyl group must first be converted
into a tosylate or an alkyl halide.
SECTION 12.10
• Primary alcohols can undergo oxidation twice to give a carboxylic acid.
• Secondary alcohols are oxidized only once to give a ketone.
• Tertiary alcohols do not undergo oxidation.
• A common oxidizing reagent is chromic acid (H2CrO4), which
can be formed either from chromium trioxide (CrO3) or from
sodium dichromate (Na2Cr2O7) in aqueous acidic solution.
• PCC can be used to convert a primary alcohol into an
aldehyde. Alternatively, primary alcohols can be converted
into aldehydes with a Swern oxidation or a DMP-based
oxidation.
SECTION 12.11
• NADH is a biological reducing agent that functions as a
• Diols are compounds with two hydroxyl groups.
• Diols can be prepared from diketones via reduction using a
hydride delivery agent (very much like NaBH4 or LiAlH4),
while NAD+ is an oxidizing agent.
• NADH and NAD+ play critical roles in biological systems.
Examples include the citric acid cycle and ATP synthesis.
• Diols can also be made via syn dihydroxylation or anti dihy-
SECTION 12.12
SECTION 12.5
reducing agent.
droxylation of an alkene.
• A reaction involving an increase in the oxidation state of a
carbon atom is called an oxidation.
• Phenols undergo oxidation to quinones. Quinones are
SECTION 12.6
SECTION 12.13
• Grignard reagents are carbon nucleophiles that are capable
of attacking a wide range of electrophiles, including the carbonyl group of ketones or aldehydes, to produce an alcohol.
• Grignard reagents also react with esters to produce alcohols
with introduction of two R groups.
biologically important because their redox properties play a
significant role in cellular respiration.
• There are two key issues to consider when proposing a
synthesis:
1. A change in the carbon skeleton.
2. A change in the functional group.
576 CHAPTER 12 Alcohols and Phenols
SKILLBUILDER REVIEW
12.1 NAMING AN ALCOHOL
STEP 1 Choose the longest
chain containing the OH group
and number the chain starting
from the end closest to the
OH group.
Cl
1
2
4,4-Dichloro
9
Cl
6
4
3
8
7
Cl
STEPS 2 AND 3 Identify the
substituents and assign locants.
2
1
5
Cl
OH
Cl
9
STEP 5 Assign the
configuration of any
chiral center.
Cl
Cl
6
4
3
8
7
STEP 4 Assemble the
substituents alphabetically.
5
R
R
OH
OH
6-Ethyl
OH
Cl
4,4-Dichloro6-ethyl-3-nonanol
3-Nonanol
(3R,6R)-4,4-dichloro6-ethyl-3-nonanol
Try Problems 12.1, 12.2, 12.29, 12.30a–d,f, 12.31
12.2 COMPARING THE ACIDITY OF ALCOHOLS
Look for resonance effects;
for example:
OH
Look for inductive effects;
for example:
OH
Look for solvating effects;
for example:
Cl
OH
Cl
OH
OH
Cl
OH
More acidic
More acidic
More acidic
Try Problems 12.6, 12.7, 12.32, 12.63
12.3 IDENTIFYING OXIDATION AND REDUCTION REACTIONS
EXAMPLE Determine
whether the starting
material has been oxidized,
reduced, or neither.
STEP 1 Determine the oxidation
state of the starting material.
Break all bonds heterolytically,
except for C C bonds.
O
O
STEP 2 Determine the oxidation
state of the product. Break all
bonds heterolytically, except
for C C bonds.
RO
H3C
RO
OR
CH3
H3C
Increase = oxidation
Decrease = reduction
No change = neither
OR
+2
C
C
STEP 3 Determine if there has
been a change in oxidation state.
+2
This example is neither
an oxidation nor a reduction
CH3
Two electrons,
but carbon should have four.
This carbon is missing
two electrons
Two electrons,
but carbon should have four.
This carbon is missing
two electrons
Oxidation state = +2
Oxidation state = +2
Try Problems 12.10, 12.11, 12.49, 12.50, 12.74
12.4 DRAWING A MECHANISM AND PREDICTING THE PRODUCTS OF HYDRIDE REDUCTIONS
STEP 2 Draw the alkoxide
intermediate.
STEP 1 Draw the complete structure
of LiAlH4, and draw two curved arrows
that show the delivery of hydride to
the carbonyl group.
H
O
H
Al
H
−
H
O
STEP 3 Draw two curved arrows
showing the alkoxide intermediate
being protonated by the proton
source.
−
H
H
CH3
+ Enantiomer
H
O
+
OH
H
H
CH3
+ Enantiomer
Try Problems 12.12, 12.13, 12.44, 12.45c, 12.46e,f, 12.51, 12.55, 12.72
SkillBuilder Review 577
12.5 PREPARING AN ALCOHOL VIA A GRIGNARD REACTION
STEP 1 Identify the α
position.
STEP 2 Identify the
three groups connected
to the α position.
Ph
O
1) MeMgBr
Et
Ph
OH
OH
STEP 3 Show how each group could have been
installed via a Grignard reaction.
O
Et
Me
Ph
2) H3O+
OH
1) EtMgBr
Me
2) H3O
+
Ph
O
Me
Et
Et
Me
1) PhMgBr
2) H3O+
Try Problems 12.15–12.17, 12.37, 12.39b, 12.52, 12.54, 12.56, 12.58, 12.65b,h,n,o,q, 12.70
12.6 PROPOSING REAGENTS FOR THE CONVERSION OF AN ALCOHOL INTO AN ALKYL HALIDE
EXAMPLE Identify
the necessary reagents.
STEP 1 Analyze
the substrate:
= SN2
Secondary = SN2
Tertiary = SN1
Primary
OH
STEP 2 Analyze
the stereochemistry:
inversion = SN2.
STEP 3 Reaction must occur via SN2
so use reagents that favor SN2.
OH
OH
Cl
Substrate is
secondary
OH
1) TsCl, py
2) NaCl
Cl
Cl
SOCl2
py
Try Problems 12.19, 12.20, 12.34a–c, 12.62f, 12.65p
12.7 PREDICTING THE PRODUCTS OF AN OXIDATION REACTION
STEP 1 Identify whether
the alcohol is primary or
secondary.
EXAMPLE
H
OH
H
OH
CrO3
H3O+
acetone
?
Primary
STEP 2 A primary alcohol can be oxidized
either to an aldehyde or to a carboxylic acid,
depending on the reagents.
O
STEP 3 Analyze the reagents. Chromic
acid is used to form the carboxylic acid.
PCC is used to form the aldehyde.
Alternatively, the aldehyde can be obtained
with a Swern or DMP-based oxidation.
O
O
H
OH
An aldehyde
A carboxylic acid
OH
Try Problems 12.22, 12.23, 12.34e–f, 12.36, 12.46, 12.53
578
CHAPTER 12
Alcohols and Phenols
12.8 CONVERTING FUNCTIONAL GROUPS
Alkynes
H2, Lindlar’s catalyst
C
1) Br2
Alkenes
or Na, NH3
C
C
Alkanes
H2
C
Pt, Pd or Ni
C
C
2) xs NaNH2
3) H2O
H2SO4, H2O
HgSO4
or
1) R2BH
2) H2O2, NaOH
conc. H2SO4, heat
or
1) TsCl/pyr
2) NaOMe
1) LiAIH4, 2) H3O+
Ketones/
aldehydes
H3O+
or
1) BH3 • THF
2) H2O2, NaOH
or
1) Hg(OAc)2, H2O
2) NaBH4, NaOH
HX or SOCl2/py
or PBr3
Alcohols
or NaBH4, MeOH
C
H2CrO4 or PCC
C
Alkyl
halides
or 1) TsCl/py
2) NaX
OH
O
Br2, hν
Br
NaOH (SN2)
or DMP or Swern
NaOMe
or
t-BuOK
HX
or
HBr, ROOR
C
or H2O (SN1)
Try Problems 12.24, 12.25, 12.38, 12.46, 12.61, 12.64, 12.66, 12.67
12.9 PROPOSING A SYNTHESIS
STEP 1 Is there a
change in the
carbon skeleton?
Keep track of all
the C C bond-forming
reactions that you
have learned
until now.
STEP 2 Is there a change
in the functional groups?
The chart from the previous
SkillBuilder summarizes
many of the important
functional group
interconversions that we
have seen.
STEP 3 After proposing a
synthesis, analyze your answer
with the following two questions:
Is the regiochemical outcome
of each step correct?
Is the stereochemical outcome
of each step correct?
MORE TIPS Remember that the
desired product should be the major
product of your proposed synthesis.
Always think backwards (retrosynthetic
analysis) as well as forwards and then
try to bridge the gap.
Most synthesis problems will have multiple
correct answers. Do not feel that you have
to find the “one” correct answer.
Try Problems 12.27, 12.28, 12.36, 12.37, 12.39, 12.43, 12.65, 12.68, 12.71
PRACTICE PROBLEMS
12.29 Assign an IUPAC name for each of the following compounds:
HO
12.32 Rank each set of alcohols below in order of increasing acidity,
and explain your reasoning.
OH
Cl
(b)
(a)
OH
(c)
• Included in Answers section
Br
Cl
OH
Cl
Cl
Cl
Cl
(a)
OH
(b)
OH
Cl
Cl
OH
OH
(d)
12.30 Draw the structure of each compound:
(a) cis-1,2-Cyclohexanediol
Cl
OH
(b) Isobutyl alcohol
(c) 2,4,6-Trinitrophenol
(d) (R)-2,2-Dimethyl-3-heptanol
(e) Ethylene glycol
(f ) (S)-2-Methyl-1-butanol
12.31 • Draw and name all constitutionally isomeric alcohols with the
molecular formula C4H10O.
OH
OH
Draw resonance structures for each of the following anions:
−
O
(a)
OH
NO2
(c)
12.33
OH
O
(b)
−
−
O
(c)
Practice Problems
12.34 Predict the major product of the reaction between 1-butanol and:
(a) PBr3
(b) SOCl2, py
(c) HCl, ZnCl2
(d) DMP, CH2Cl2
(e) Na2Cr2O7, H2SO4, H2O
(f ) Li
(g) NaH
(h) TMSCl, Et3N
(i) TsCl, pyridine
(j) Na
12.40
Propose a mechanism for the following reaction:
O
Br
12.35 Acid-catalyzed hydration of 3,3-dimethyl-1-butene produces
2,3-dimethyl-2-butanol. Show a mechanism for this reaction.
12.42 Consider the following sequence of reactions and identify the
structures of compounds A, B, and C:
12.36 • Starting with 1-butanol, show the reagents you would use to
prepare each of the following compounds:
O
H
(a)
OH
Mg
Compound A
(C6H11Br)
O
1)
2) H3O+
(c)
Conc. H2SO4
O
(d)
Heat
OH
OH
OH
(a)
(b)
OH
12.44 Predict the product and draw a mechanism for each of the following reactions:
O
O
OH
OH
(a)
12.43 Using 2-propanol as your only source of carbon, show how you
would prepare 2-methyl-2-pentanol.
(c)
12.38 Each of the following alcohols can be prepared via reduction
of a ketone or aldehyde. In each case, identify the aldehyde or ketone
that would be required.
1) LiAIH4
2) H3O+
(a)
12.45
(b)
OH
SOCl2
py
(a)
O
Cl
OH
(b)
+
OH
OH
OH
OH
H
O
OH
(c)
(d)
OH
O
O
H
(e)
O
OH
OH
OH
O
(b)
O
(b)
CH3OH
O
O
H
Br
PBr3
12.46 Identify the reagents you would use to accomplish each of the
following transformations:
(a)
(c)
MeOH
(b)
2) H3O+
O
(c)
H
(a)
?
NaBH4
1) Excess LiAIH4
(c)
O
?
Draw a mechanism for each of the following reactions:
O
12.39 • What reagents would you use to accomplish each of the following transformations?
OH
(f )
12.47 • Predict the major products for each of the following synthetic
sequences:
(d)
OH
1) O3
2) DMS
OH
(a)
3) Excess LiAIH4
4) H3O+
1) O3
2) DMS
(f )
Compound C
(e)
12.37 • Using a Grignard reaction, show how you could prepare each
of the following alcohols:
(e)
Compound B
OH
OH
(b)
NaH
OH
12.41 Acid-catalyzed hydration of 1-methylcyclohexene yields a mixture
of two alcohols. The major product does not react with chromic acid, while
the minor product will undergo chromic acid oxidation. Identify the major
and minor products, and explain their reactivity toward chromic acid.
(k) Potassium tert-butoxide
O
579
O
(b)
3) Excess LiAIH4
4) H3O+
?
?
580
CHAPTER 12
O
Alcohols and Phenols
1) EtMgBr
2) H3O+
?
3) Na2Cr2O7, H2SO4, H2O
4) EtMgBr
5) H3O+
H
(c)
1) LiAIH4
2) H3O+
1) H3O+
2) Na2Cr2O7, H2SO4, H2O
?
3) PhMgBr
4) H3O+
(e)
OH
12.50 In the following reaction, determine whether or not the alkyne has
been oxidized or reduced. Using the answer from the previous problem,
try to determine the answer without calculating oxidation states and then
use the calculations to see if your intuition was correct.
H
(d)
H
H3O+
?
3) TsCl, pyridine
O
(Hint: Evaluate each carbon atom separately and then look at the net
change for the alkene as a whole.)
O
H2SO4 , H2O
HgSO4
12.48 Provide a mechanism for each of the following transformations:
O
12.51 Draw a mechanism and predict the major product of the following reaction:
OH
1) MeMgBr
O
2) H3O+
(a)
O
1) Excess LiAIH4
O
2) H3O+
O
?
1) Excess MeMgBr
2) H3O+
(b)
OH
HO
12.49 In Chapter 8, we learned about addition of water across a π
bond. Identify whether the alkene has been oxidized or reduced (or if it
has been neither oxidized nor reduced).
12.52 Draw a mechanism and predict the product for the following
reaction. In this case, H3O+ must be used as a proton source instead of
water. Explain why.
O
O
?
1) xs MeMgBr
2) H3O+
ACS-STYLE PROBLEMS (Multiple Choice)
Problems 12.53–12.61 follow the style of the ACS organic
chemistry exam. For each of these problems, there will be
one correct answer and three distractors.
12.55 • What is the major product of the given reaction?
O
OCH3
12.53 • Of the following, which are the best reagents to perform the
given transformation?
OH
OH
?
O
O
(c) PCC, CH2Cl2
(d) KMnO4, NaOH, cold
12.54 • Which of the following is NOT an efficient synthesis of
2-methyl-2-pentanol?
O
2) H3O
(a)
O
O
(d)
+
1) xs CH3MgBr
OCH3
(c)
O
1) CH3MgBr
2) H3O+
1) xs CH3MgBr
OH
2) H3O+
(b)
O
OCH3
(b)
O
OH
O
(b) CrO3, H3O+, acetone
OH
H
?
xs NaBH4
CH3OH
O
(a)
H
(a) Na2Cr2O7, H2SO4, H2O
O
OH
(c)
OH
(d)
12.56 • A logical retrosynthesis of the given target molecule leads to
which Grignard reagent?
Ph
OH
1) CH3CH2CH2MgBr
2) H3O+
BrMg
(a)
Ph
OH
(b)
MgBr
(d) PhMgBr
MgBr
O
(c) Ph
Integrated Problems
12.57 • Predict the major product of the following sequence of reactions.
OH
2) Mg
3) CH3OH
OCH3
(a)
12.60 • Predict the major product of the following reaction.
?
1) PBr3
O
O
CN
(c)
NaCN
acetone
S
O
OH
(b)
O
(d)
S
O
(a)
?
1)
(b)
CN
(c)
2) H3O+
(a)
O
O
(b) H
H
H
(c)
(d)
12.61 • Which reagent (or reagents) will efficiently achieve the following transformation?
HO
O
O
?
O
12.58 • What is the missing reagent in the following synthesis?
MgBr
581
?
OH
OMe
(d)
12.59 • Predict the major product of the following sequence of reactions.
1) O3
2) DMS
?
3) LiAlH4
4) H3O+
(a) HO
OH
(a)
(c)
OH
(b) HO
TsCl, py
SOCl2, py
t-BuOK
(b)
NaOMe
(d)
PBr3
NaOEt
conc. H2SO4
heat
O
(c) O
O
(d)
INTEGRATED PROBLEMS
12.62 Consider the following sequence of transformations and identify suitable reagents for a–h:
H
a
NO2
NO2
b
2-Nitrophenol
c
OH
H
e
Br
f
OH
g
O
H
h
12.63 Consider the structures of 2-nitrophenol and 3-nitrophenol.
These compounds have very different pKa values. Predict which one
has the lower pKa and explain why. (Hint: In order to solve this problem,
you must draw the structure of each nitro group.)
3-Nitrophenol
12.64 Identify reagents that can be used to accomplish each of the
transformations shown here:
O
d
HO
OH
O
OH
O
OH
OH
Br
582
CHAPTER 12
Alcohols and Phenols
12.65 Propose an efficient synthesis for each of the following
transformations:
O
12.67 Using any reagents of your choosing, show how you would
convert tert-butyl alcohol into 2-methyl-1-propanol.
O
H
12.66 Show at least two different methods for preparing 1-methylcyclohexene from 1-methylcyclohexanol.
12.68 Using starting materials and reagents that have no more than
two carbon atoms, identify a method for preparing each of the following
target molecules:
(a)
O
H
OH
OH
(b)
(b)
(a)
O
Cl
(c)
OH
H
O
(c)
12.69
OH
Propose a mechanism for the following transformation:
Cl
(d)
(d)
O
O
O
2) H3O+
O
(e)
1) Excess LiAlH4
HO
OH
O
OH
12.70
(f )
Propose a mechanism for the following transformation:
O
O
(g)
OH
1) Excess MeMgBr
HO
2) H3O+
O
OH
O
(h)
12.71 Show reagents that can be used to achieve the following
transformation:
OH
(i)
O
OH
H
(j)
O
12.72
Propose a mechanism for the following transformation:
(k)
O
O
O
1) Excess LiAIH4
2) H3O+
O
CH3OH
12.73 A carbocation is resonance-stabilized when it is adjacent to an
oxygen atom:
(l)
O
HO
+
O
(m)
O
+
O
Resonance-stabilized
(n)
O
OH
(o)
HO
O
(p)
H
OH
O
conc. H2SO4
Heat
Br
OH
(q)
Such a carbocation is even more stable than a tertiary carbocation. Using
this information, propose a mechanism for the following transformation
exhibited by a diol. This reaction is called a pinacol rearrangement:
12.74 Determine whether the pinacol rearrangement, shown in the
previous problem, is a reduction, an oxidation, or neither a reduction
nor an oxidation.
583
Integrated Problems
12.75 (S)-Gizzerosine is an amino acid that is believed to be responsible for a serious disease called “black vomit” in chickens. However, the
same compound can potentially be used as a drug for the treatment of
osteoporosis and acid buildup in the stomach. Provide a two-step synthesis for achieving the following transformation, which was employed
during a synthesis of (S)-gizzerosine:10
O
H
N
H
Proton NMR
3
1
2
2
O
2 steps
HO
OCH3
O
12.80 Propose a structure for a compound with the molecular formula
C3H8O that exhibits the following 1H NMR and 13C NMR spectra:
OCH3
O
Boc
Boc =
N
H
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
Chemical shift (ppm)
Boc
O
Carbon 13 NMR
12.76 Briarellin E (compound 3) is produced by coral in the Caribbean
and Mediterranean seas and belongs to a larger family of marine natural products being investigated as potential anticancer agents. During
a synthesis of 3, compound 1 was treated with excess TBAF to give
compound 2.11 Draw the structure of compound 2.
H
Si O
H
H
TBAF
O
HO
OH
H
2
O
O
Si O
1
25.7
64.2
70
60
50
40
10.0
30
20
10
12.81 Propose two possible structures for a compound with the
molecular formula C5H12O that exhibits the following 13C NMR and
IR spectra:
H
O
HO
Carbon 13 NMR
O
3
29.1
9.5
73.8
12.77 The compound duryne was one of several structurally related
compounds isolated from a marine sponge.12 Propose an efficient synthesis of duryne starting with any compounds containing eleven or fewer
carbon atoms.
OH
0
Chemical shift (ppm)
100
90
70
80
60
50
40
30
20
10
0
Chemical shift (ppm)
100
OH
80
(CH2)9
Duryne
12.78 Estragole is an insect repellant that has been isolated from the
leaves of the Clausena anisata tree.13 Propose a synthesis of estragole
starting from 4-methylphenol.
% Transmittance
(CH2)9
60
40
20
O
Estragole
0
4000
3500
3000
2500
2000
1500
1000
Wavenumber
(cm–1)
Problems 12.79–12.82 are intended for students who have
already covered spectroscopy (Chapters 14 and 15).
12.79 Propose a structure for a compound with the molecular formula
C10H14O that exhibits the following 1H NMR spectrum:
12.82 Propose a structure for a compound with the molecular formula
C8H10O that exhibits the following 1H NMR spectrum:
Proton NMR
Proton NMR
2
9
2
2
2
5
1
1
8
8
7
6
5
4
Chemical shift (ppm)
3
2
1
7
6
5
4
Chemical shift (ppm)
3
2
1
584
CHAPTER 12
Alcohols and Phenols
CHALLENGE PROBLEMS
12.83 The following two isomeric ketones were among the 68 compounds isolated from the steam-distilled essential oil of fresh and airdried marijuana buds.14 Propose a separate synthesis for each of these
two compounds using only disubstituted alkenes containing four carbon atoms as starting materials.
O
O
HO
H
O
TsCl
TBAF
K
4
3 pyridine
O
NaOMe
O
1
R
Si
R
O
R
O
H
O
12.84 As part of a study of the male sex pheromone of a Brazilian
bug that feeds on the leaves and fruit of tomato plants, the compound
5,9,12,16-tetramethyleicosane was synthesized.15 The synthesis of this
alkane was reported, and it used compounds A–C.
H O
H
A
O
O
O
O
O
Me
B
H
H
O
O
E
D
C
H
Me
H
H
H
Me
O
O
H
O
I
H
G
F
H
H
H O
Me
Me
K
J
H
O
H
O H
H
2
(Brevetoxin B)
O
(a) Draw the structures of 3 and 4.
Br
A
B
C
(b) Draw a mechanism for the conversion of 4 to 5 upon treatment
with NaOMe.
O
A key step in the synthesis is shown below, in which a mixture of E and
Z isomers is obtained:
OH
5
HO
Me
Me
5,9,12,16-Tetramethyleicosane
H
HBr
Br
(a) Draw a plausible mechanism for this process.
(b) Propose an efficient synthesis of 5,9,12,16-tetramethyleicosane
using A, B, and C as your only sources of carbon.
12.85 Brevetoxin B (compound 2) is produced by Ptychodiscus
brevis, a marine organism responsible for red tides. Brevetoxin B is a
potent neurotoxin, as a result of its ability to bind to sodium channels
and force them to remain open. Its biological action, together with its
complex architecture, has made it an attractive synthetic target. During
K. C. Nicolaou’s synthesis of brevetoxin B, the K ring was assembled via
the construction of compound 1. The cis relationship of the hydroxyl
group and the silicon-based protected group in 1 was established via
the successful conversion of a small amount of 1 into 5, as shown.16
Note that the conversion of 3 to 4 involves the selective tosylation of
the primary hydroxyl group:
(c) Explain why the successful conversion of 1 to 5 verifies the cis
relationship between the hydroxyl group and the silyl-protected
oxygen group in 1.
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