Download 5 organic chemistry: functional groups

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M O D U L E
5
ORGANIC CHEMISTRY:
FUNCTIONAL GROUPS
O2.1
Functional Groups
O2.2
Oxidation–Reduction Reactions
O2.3
Alkyl Halides
O2.4
Alcohols and Ethers
O2.5
Aldehydes and Ketones
O2.6
Reactions at the Carbonyl Group
O2.7
Carboxylic Acids and Carboxylate Ions
O2.8
Esters
O2.9
Amines, Alkaloids, and Amides
O2.10 Grignard Reagents
Chemistry in the World Around Us: The Chemistry of Garlic
O2.1 FUNCTIONAL GROUPS
Bromine reacts with 2-butene to form 2,3-dibromobutane.
CH3 CHP CHCH3 ⫹ Br2
Br
A
CH3CHCHCH3
A
Br
It also reacts with 3-methyl-2-pentene to form 2,3-dibromopentane.
CH3
A
CH3 CHPCCH2CH3 ⫹ Br2
Br CH3
A A
CH3CHCCH2CH3
A
Br
Instead of trying to memorize both equations, we can build a general rule that bromine reacts with compounds that contain a CPC double bond to give the product expected from
addition across the double bond. This approach to understanding the chemistry of organic
compounds presumes that certain atoms or groups of atoms known as functional groups
give these compounds their characteristic properties.
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Functional groups focus attention on the important aspects of the structure of a molecule. We don’t have to worry about the differences between the structures of 1-butene
and 2-methyl-2-hexene, for example, when the compounds react with hydrogen bromide.
We can focus on the fact that both compounds are alkenes that add HBr across the CPC
double bond in the direction predicted by Markovnikov’s rule, introduced in Section O1.8.
CH2 P CHCH2CH3 HBr
CH3
A
CH3C PCHCH2CH2CH3 HBr
CH3CHCH2CH3
A
Br
CH3
A
CH3C CH2CH2CH2CH3
A
Br
Some common functional groups are given in Table O2.1.
TABLE 02.1 Common Functional Groups
Functional Group
A
OCO
A
CP C
Cq C
F, Cl, Br, or I
OOH
OOO
ONH2
Name
Example
Alkane
CH3CH2CH3 (propane)
Alkene
Alkyne
Alkyl halide
Alcohol
Ether
Amine
CH3CHPCH2 (propene)
CH3C q CH (propyne)
CH3 Br (methyl bromide)
CH3CH2OH (ethanol)
CH3OCH3 (dimethyl ether)
CH3NH2 (methylamine)
The CPO group plays a particularly important role in organic chemistry. This group
is called a carbonyl, and some of the functional groups based on a carbonyl are shown in
Table O2.2.
TABLE 02.2 Functional Groups That Contain a Carbonyl
Functional Group
O
B
O COH
O
B
O CO
O
B
O CO Cl
O
B
O COOH
O
B
O COOO
O
B
O CONH2
Name
Example
Aldehyde
CH3 CHO (acetaldehyde)
Ketone
CH3COCH3 (acetone)
Acyl chloride
CH3COCl (acetyl chloride)
Carboxylic acid
CH3CO2H (acetic acid)
Ester
CH3CO2CH3 (methyl acetate)
Amide
CH3CONH2 (acetamide)
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Exercise O2.1
Root beer hasn’t tasted the same since the use of sassafras oil as a food additive was outlawed because sassafras oil is 80% safrole, which has been shown to cause cancer in rats
and mice. Identify the functional groups in the structure of safrole.
OOCH2
i
O
Safrole
CH2
G
CHPCH2
Solution
Safrole is an aromatic compound because it contains a benzene ring. It is also an alkene
because it contains a CPC double bond. The most difficult functional group to recognize
in the molecule might be the two ether linkages (OOO).
Exercise O2.2
The following compounds are the active ingredients in over-the-counter drugs used as analgesics (to relieve pain without decreasing sensibility or consciousness), antipyretics (to reduce the body temperature when it is elevated), and/or anti-inflammatory agents (to counteract swelling or inflammation of the joints, skin, and eyes). Identify the functional groups
in each molecule.
O
B
C
CH3
G D G
CH
OH
OH
O
OH
M D
O
C
B
OOCOCH3
Aspirin
(acetylsalicylic acid)
CH3
NH
G D
C
B
Tylenol
O (acetaminophen)
CH3
CH
G D 2
CH
A
Advil
CH3 (ibuprofen)
Solution
All three compounds are aromatic. Aspirin is also a carboxylic acid (OCO2H) and an ester (OCO2CH3). Tylenol is also an alcohol (OOH) and an amide (OCONHO). Ibuprofen contains alkane substituents and a carboxylic acid functional group.
Exercise O2.3
The discovery of penicillin in 1928 marked the beginning of what has been called the
“golden age of chemotherapy,” in which previously life-threatening bacterial infections
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were transformed into little more than a source of discomfort. For those who are allergic
to penicillin, a variety of antibiotics, including tetracycline, are available. Identify the numerous functional groups in the tetracycline molecule.
HO CH3
/∑
H
N(CH3)2
OH
Tetracycline
OH
CPO
OHB
f
OH O NH2
B
O
Solution
The compound contains an aromatic ring fused to three six-membered rings. It is also an
alcohol (with five OOH groups), a ketone (with CPO groups at the bottom of the second
and fourth rings), an amine [the ON(CH3)2 substituent at the top of the fourth ring], and
an amide (the OCONH2 group at the bottom right-hand corner of the fourth ring.)
O2.2 OXIDATION–REDUCTION REACTIONS
Focusing on the functional groups in a molecule allows us to recognize patterns in the behavior of related compounds. Consider what we know about the reaction between sodium
metal and water, for example.
H2(g) 2 Na(aq) 2 OH(aq)
2Na(s) 2 H2O(l)
We can divide the reaction into two half-reactions. One involves the oxidation of sodium
metal to form sodium ions.
Oxidation
Na
Na e
O
S
The other involves the reduction of an H ion in water to form a neutral hydrogen atom
that combines with another hydrogen atom to form an H2 molecule.
Reduction
2H
O
D G
H
2 H 2 e
O OH
2 H 2 SO
Q
2 HT
H2
Once we recognize that water contains an OOH functional group, we can predict what
might happen when sodium metal reacts with an alcohol that contains the same functional
group. Sodium metal should react with methanol (CH3OH), for example, to give H2 gas
and a solution of the Na and CH3O ions dissolved in the alcohol.
2Na(s) 2 CH3OH(l)
H2(g) 2 Na(alc) 2 CH3O(alc)
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Because they involve the transfer of electrons, the reactions between sodium metal and
either water or an alcohol are examples of oxidation–reduction reactions. But what about
the following reaction, in which hydrogen gas reacts with an alkene in the presence of a
transition metal catalyst to form an alkane?
H
G
D
H2
C PC
G
D
H
H
H H
H
Ni
H C
C H
H H
There is no change in the number of valence electrons on any of the atoms in the reaction.
Both before and after the reaction, each carbon atom shares a total of eight valence electrons and each hydrogen atom shares two electrons. Instead of electrons, the reaction
involves the transfer of atoms—in this case, hydrogen atoms. There are so many atomtransfer reactions that chemists developed the concept of oxidation number (see Chapter 5)
to extend the idea of oxidation and reduction to reactions in which electrons aren’t necessarily gained or lost.
Oxidation involves an increase in the oxidation number of an atom. Reduction occurs when the oxidation number of an atom decreases.
During the transformation of ethene into ethane, there is a decrease in the oxidation number of the carbon atom. The reaction therefore involves the reduction of ethene to ethane.
2
H
G
D
H2
C PC
G
D
H
H
H H
H
Ni
H C
C H
H H
3
Reactions in which none of the atoms undergo a change in oxidation number are called
metathesis reactions. Consider the reaction between a carboxylic acid and an amine, for
example,
CH3CO2H CH3NH2 88n CH3CO2 CH3NH3
or the reaction between an alcohol and hydrogen bromide.
CH3CH2OH HBr 88n CH3CH2Br H2O
These are metathesis reactions because there is no change in the oxidation number of any
atom in either reaction.
The oxidation numbers of the carbon atoms in a variety of compounds are given in
Table O2.3. The oxidation numbers can be used to classify organic reactions as either
oxidation–reduction reactions or metathesis reactions.
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TABLE O2.3
Typical Oxidation Numbers of Carbon
Classes of Compounds
Example
Oxidation Number of the
Carbon
Alkane
Alkyllithium
Alkene
Alcohol
Ether
Alkyl halide
Amine
Alkyne
Aldehyde
Carboxylic acid
—
CH4
CH3Li
H2CPCH2
CH3OH
CH3OCH3
CH3Cl
CH3NH2
HCqCH
H2CO
HCO2H
CO2
4
4
2
2
2
2
2
1
0
2
4
Exercise O2.4
Classify the following as either oxidation–reduction or metathesis reactions.
(a) 2 CH3OH H
8n CH3OCH3 H2O
O
O
B
B
8n HCOCH3 H2O
(b) HCOH CH3OH H
(c) CO 2 H2 n CH3OH
(d) CH3Br 2 Li n CH3Li LiBr
Solution
(a) This is a metathesis reaction because there is no change in the oxidation number of
the carbon atoms when an alcohol is converted to an ether.
2 CH3OH 88n CH3OCH3 H2O
2
2
(b) This is a metathesis reaction because there is no change in the oxidation number of any
of the carbon atoms when a carboxylic acid reacts with an alcohol to form an ester.
HCO2H CH3OH 88n HCO2CH3 H2O
2
2
2
2
(c) This is an oxidation–reduction reaction because the carbon atom is reduced from the
2 to the 2 oxidation state when CO combines with H2 to form methanol.
CO 2 H2 88n CH3OH
2
2
(d) This is an oxidation–reduction reaction because the carbon atom is reduced from the
2 to the 4 oxidation state when CH3Br reacts with lithium metal to form CH3Li.
CH3Br 2 Li 88n CH3Li LiBr
2
4
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Because electrons are neither created nor destroyed, oxidation can’t occur in the absence
of reduction, and vice versa. It is often useful, however, to focus attention on one component of the reaction and ask, is that substance oxidized or reduced?
Exercise O2.5
Determine whether the following transformations involve the oxidation or the reduction
of the carbon atom.
H
H C
OH
O
B
H C H
H
O
B
H C H
O
B
H C OH
O
B
H C OH
OPC PO
Solution
Each of the transformations involves the oxidation of the carbon atom. The first reaction
involves oxidation of the carbon atom from the 2 to the 0 oxidation state. In the second
reaction the carbon atom is oxidized to the 2 state, and the third reaction involves oxidation of the carbon atom to the 4 oxidation state.
2
H
H C
OH
0
O
B
H C H
H
0
O
B
H C H
2
O
B
H C OH
O 2
B
H C OH
4
OPCPO
Assigning oxidation numbers to the individual carbon atoms in a complex molecule
can be difficult. Fortunately, there is another way to recognize oxidation–reduction reactions in organic chemistry:
Oxidation occurs when hydrogen atoms are removed from a carbon atom or when
an oxygen atom is added to a carbon atom.
Reduction occurs when hydrogen atoms are added to a carbon atom or when an
oxygen atom is removed from a carbon atom.
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The first reaction in Exercise O2.5 involves oxidation of the carbon atom because
a hydrogen atom is removed from that atom when the alcohol is oxidized to an
aldehyde.
O
B
CH3OH 8n HCH
The second reaction in the exercise is an example of oxidation because an oxygen atom is
added to the carbon atom when an aldehyde is oxidized to a carboxylic acid.
O
B
HC H
O
B
HCOH
Reduction, on the other hand, occurs when hydrogen atoms are added to a carbon atom
or when an oxygen atom is removed from a carbon atom. An alkene is reduced, for example, when it reacts with H2 to form the corresponding alkane.
Ni
CH2PCHCH3 H2 88n
CH3CH2CH3
Figure O2.1 provides a useful guide to the oxidation–reduction reactions of organic
compounds. Each of the arrows in this figure involves a two-electron oxidation of a carbon atom along the path toward carbon dioxide. A line is drawn through the first arrow
because it is impossible to achieve this transformation in a single step.
CH4
CH3OH
–4
–2
O
B
HCH
0
O
B
HCOH
+2
CO2
+4
FIGURE O2.1 The stepwise oxidation
of carbon.
O2.3 ALKYL HALIDES
Imagine that a pair of crystallizing dishes are placed on an overhead projector as shown in
Figure O2.2. An alkene is added to the dish in the upper left corner of the projector, and
an alkane is added to the dish in the upper right corner. A few drops of bromine dissolved
in chloroform (CHCl3) are then added to each of the crystallizing dishes.
CH2 = CH(CH2)3CH3
CH3(CH2)4CH3
FIGURE O2.2 Demonstrating the relative rate of bromination reactions.
The characteristic red-orange color of bromine disappears the instant this reagent is
added to the alkene in the upper left corner as the Br2 molecules add across the CPC double bond in the alkene.
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Br
A
CH2PCH(CH2)3CH3 Br2 88n BrCH2CH(CH2)3CH3
The other crystallizing dish picks up the characteristic color of a dilute solution of bromine
because the reagent doesn’t react with alkanes under normal conditions.
If the crystallizing dish in the upper right corner is moved into the center of the projector, however, the color of the bromine slowly disappears. This can be explained by noting that alkanes react with halogens at high temperatures or in the presence of light to
form alkyl halides, as noted in Section O1.8.
Br
A
CH3(CH2)4CH3 Br2 88n CH3CH(CH2)3CH3 HBr
Light
The light source in an overhead projector is intense enough to initiate the reaction, although the reaction is still much slower than the addition of Br2 to an alkene.
The reaction between an alkane and one of the halogens (F2, Cl2, Br2, or I2) can be
understood by turning to a simpler example.
CH4(g) Cl2(g) 88n CH3Cl(g) HCl(g)
This reaction has the following characteristic properties.
•
•
•
•
•
It doesn’t take place in the dark or at low temperatures.
It occurs in the presence of ultraviolet light or at temperatures above 250ºC.
Once the reaction gets started, it continues after the light is turned off.
The products of the reaction include CH2Cl2 (dichloromethane), CHCl3 (chloroform), and CCl4 (carbon tetrachloride), as well as CH3Cl (chloromethane).
The reaction also produces some C2H6.
These facts are consistent with a chain reaction mechanism that involves three processes:
chain initiation, chain propagation, and chain termination.
Chain Initiation
A Cl2 molecule can dissociate into a pair of chlorine atoms by absorbing energy in the form
of either ultraviolet light or heat.
Cl2 88n 2 Cl
H° 243.4 kJ/molrxn
The chlorine atom produced in the reaction is an example of a free radical—an atom or
molecule that contains one or more unpaired electrons.
Chain Propagation
Free radicals, such as the Cl atom, are extremely reactive. When a chlorine atom collides
with a methane molecule, it can remove a hydrogen atom to form HCl and a CH3
radical.
CH4 Cl 88n CH3 HCl
H° 16 kJ/molrxn
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If the CH3 radical then collides with a Cl2 molecule, it can remove a chlorine atom to form
CH3Cl and a new Cl radical.
CH3 Cl2 88n CH3Cl Cl
H° 87 kJ/molrxn
Because a Cl atom is generated in the second reaction for every Cl atom consumed in
the first, the reaction continues in a chainlike fashion until the radicals involved in the chain
propagation steps are destroyed.
Chain Termination
If a pair of the radicals that keep the chain reaction going collide, they combine in a chain
terminating step. Chain termination can occur in three ways.
2 Cl 88n Cl2
CH3 Cl 88n CH3Cl
2 CH3 88n CH3CH3
H° 243.4 kJ/molrxn
H° 330 kJ/molrxn
H° 350 kJ/molrxn
Because the concentration of the radicals is relatively small, the chain termination reactions are relatively infrequent.
The chain reaction mechanism for free radical reactions explains the observations listed
for the reaction between CH4 and Cl2.
•
The reaction doesn’t occur in the dark or at low temperatures because energy must
be absorbed to generate the free radicals that carry the reaction.
Cl2 88n 2 Cl
H° 243.4 kJ/molrxn
The reaction occurs in the presence of ultraviolet light because a UV photon has
enough energy to dissociate a Cl2 molecule to a pair of Cl atoms. The reaction occurs at high temperatures because Cl2 molecules can dissociate to form Cl atoms
by absorbing thermal energy.
• The reaction continues after the light has been turned off because light is needed
only to generate the Cl atoms that start the reaction. The chain reaction then converts CH4 into CH3Cl without consuming the Cl atoms.
•
CH4 Cl 88n CH3 HCl
CH3 Cl2 88n CH3Cl Cl
•
The reaction doesn’t stop at CH3Cl because the Cl atoms can abstract additional
hydrogen atoms to form CH2Cl2, CHCl3, and eventually CCl4.
CH3Cl Cl 88n CH2Cl HCl
CH2Cl Cl2 88n CH2Cl2 Cl, and so on
•
The formation of C2H6 is a clear indication that the reaction proceeds through a
free radical mechanism. When two CH3 radicals collide, they combine to form an
ethane molecule.
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2 CH3 88n CH3CH3
Free radical halogenation of alkanes provides us with another example of the role of
atom transfer reactions in organic chemistry. The net effect of the reaction is to oxidize a
carbon atom by removing a hydrogen from the atom.
CH4 Cl2
CH3Cl HCl
4
2
The reaction, however, doesn’t involve the gain or loss of electrons. It occurs by the transfer of a hydrogen atom in one direction and a chlorine atom in the other.
O2.4 ALCOHOLS AND ETHERS
Alcohols contain an OOH group attached to a saturated carbon. The common names for
alcohols are based on the name of the alkyl group.
CH3OH
CH3CH2OH
CH3CHOHCH3
Methyl alcohol
Ethyl alcohol
Isopropyl alcohol
The systematic nomenclature for alcohols adds the ending -ol to the name of the parent
alkane and uses a number to identify the carbon that carries the OOH group. The systematic name for isopropyl alcohol, for example, is 2-propanol.
CH3OH
CH3CH2OH
CH3CHOHCH3
Methanol
Ethanol
2-Propanol
Exercise O2.6
More than 50 organic compounds have been isolated from the oil that gives rise to the
characteristic odor of a rose. One of the most abundant of the compounds is known by the
common name citronellol. Use the systematic nomenclature to name this alcohol, which
has the following structure.
CH3
A
CH
D
G
H2C
CH2
A
A
H2C
CH2O OH
G
CH
B
C
D G
CH3
CH3
Solution
The longest chain of carbon atoms in the compound contains eight atoms.
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CH3
3A
CH 2
4 D
G
H2C
CH2
A
A
H2C 5
CH2O OH
G
1
6 CH
B
C
D7 G
CH3
CH3
8
The longest chain contains the OOH group, which means the compound is named as a derivative of octane. Because it is an alcohol, it would be tempting to name it as an octanol.
But it contains a CPC double bond, which means it must be an octenol. We now have to
indicate that the OOH group is on one end of the chain and the CPC double bond occurs between the sixth and seventh carbon atoms of the chain, which can be done by considering the compound to be a derivative of 6-octen-1-ol. Finally, we have to indicate the
presence of a pair of CH3 groups on the third and seventh carbon atoms. The compound
is therefore given the systematic name 3,7-dimethyl-6-octen-1-ol.
Methanol, or methyl alcohol, is also known as wood alcohol because it was originally
made by heating wood until a liquid distilled. Methanol is highly toxic, and many people
have become blind or have died from drinking it. Ethanol, or ethyl alcohol, is the alcohol
associated with “alcoholic” beverages. It has been made for at least 6000 years by adding
yeast to solutions that are rich in either sugars or starches. The yeast cells obtain energy
from enzyme-catalyzed reactions that convert sugar or starch to ethanol and CO2.
C6H12O6(aq) 88n 2 CH3CH2OH(aq) 2 CO2(g)
When the alcohol reaches a concentration of 10% to 12% by volume, the yeast cells die.
Brandy, rum, gin, and the various whiskeys that have a higher concentration of alcohol are
prepared by distilling the alcohol produced by the fermentation reaction. Ethanol isn’t as
toxic as methanol, but it is still dangerous. Most people are intoxicated at blood-alcohol
levels of about 0.1 gram per 100 mL. An increase in the level of alcohol in the blood to
between 0.4 and 0.6 g/100 mL can lead to coma or death.
The method of choice for determining whether an individual is DUI—driving under
the influence—or DWI—driving while intoxicated—is the Breathalyzer, for which a patent
was issued to R. F. Borkenstein in 1958. The chemistry behind the Breathalyzer is based
on the reaction between alcohol in the breath and the chromate or dichromate ion.
3 CH3CH2OH(g) 2 Cr2O72(aq) 16 H(aq) 88n
3 CH3CO2H(aq) 4 Cr3(aq) 11 H2O(l)
The instrument contains two ampules that hold small samples of potassium dichromate dissolved in sulfuric acid. One of the ampules is used as a reference. The other is opened and
the breath sample to be analyzed is added. If alcohol is present in the breath, it reduces
the yellow-orange Cr2O72 ion to the green Cr3 ion. The extent to which the color balance between the two ampules is disturbed is a direct measure of the amount of alcohol
in the breath sample. Measurements of the alcohol on the breath are then converted into
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estimates of the concentration of the alcohol in the blood by assuming that 2100 mL of air
exhaled from the lungs contains the same amount of alcohol as 1 mL of blood.
Measurements taken with the Breathalyzer are reported in units of percent bloodalcohol concentration (BAC). In most states, a BAC of 0.10% is sufficient for a DUI or
DWI conviction. (This corresponds to a blood-alcohol concentration of 0.10 gram of alcohol per 100 mL of blood.)
Ethanol is oxidized to CO2 and H2O by the alcohol dehydrogenase enzymes in the
body. This reaction gives off 30 kilojoules per gram, which makes ethanol a better source
of energy than carbohydrates (17 kJ/g), and almost as good a source of energy as fat
(38 kJ/g). An ounce of 80-proof liquor can provide as much as 3% of the average daily
caloric intake, and drinking alcohol can contribute to obesity. Many alcoholics are malnourished, however, because of the absence of vitamins in the calories they obtain from
alcoholic beverages.
As a general rule, polar or ionic substances dissolve in polar solvents; nonpolar substances dissolve in nonpolar solvents. As a result, hydrocarbons don’t dissolve in water.
They are often said to be immiscible (literally, “not mixable”) in water. Alcohols, as might
be expected, have properties between the extremes of hydrocarbons and water. When the
hydrocarbon chain is short, the alcohol is soluble in water. There is no limit on the amount
of methanol (CH3OH) and ethanol (CH3CH2OH), for example, that can dissolve in a given
quantity of water. As the hydrocarbon chain becomes longer, the alcohol becomes less soluble in water, as shown in Table O2.4. One end of the alcohol molecules has so much nonpolar character it is said to be hydrophobic (literally, “water-hating”). The other end contains an OOH group that can form hydrogen bonds to neighboring water molecules and
is therefore said to be hydrophilic (literally, “water-loving”). As the hydrocarbon chain becomes longer, the hydrophobic character of the molecule increases, and the solubility of
the alcohol in water gradually decreases until it becomes essentially insoluble in water.
TABLE O2.4
Solubilities of Alcohols in Water
Formula
Name
Solubility in Water (g/100 g)
CH3OH
CH3CH2OH
CH3(CH2)2OH
CH3(CH2)3OH
CH3(CH2)4OH
CH3(CH2)5OH
CH3(CH2)6OH
CH3(CH2)7OH
CH3(CH2)9OH
Methanol
Ethanol
Propanol
Butanol
Pentanol
Hexanol
Heptanol
Octanol
Decanol
Infinitely soluble
Infinitely soluble
Infinitely soluble
9
2.7
0.6
0.18
0.054
Insoluble in water
Alcohols are classified as either primary (1°), secondary (2°), or tertiary (3°) on the
basis of their structures.
CH3CH2OH
OH
A
CH3CHCH3
OH
A
CH3CCH3
A
CH3
1
2
3
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Ethanol is a primary alcohol because there is only one alkyl group attached to the carbon
that carries the OOH substituent. The structure of a primary alcohol can be abbreviated
as RCH2OH, where R stands for an alkyl group. The isopropyl alcohol found in rubbing
alcohol is a secondary alcohol, which has two alkyl groups on the carbon atom with the
OOH substituent (R2CHOH). An example of a tertiary alcohol (R3COH) is tert-butyl (or
t-butyl) alcohol or 2-methyl-2-propanol.
Another class of alcohols are the phenols, in which an OOH group is attached to an
aromatic ring, as shown in Figure O2.3. Phenols are potent disinfectants. When antiseptic
techniques were first introduced in the 1860s by Joseph Lister, it was phenol (or carbolic
acid, as it was then known) that was used. Phenol derivatives, such as o-phenylphenol, are
still used in commercial disinfectants such as Lysol.
OH
OH
o-Phenylphenol
Phenol
FIGURE O2.3 The structures of phenol and
o-phenylphenol.
Water has an unusually high boiling point because of the hydrogen bonds between the
H2O molecules. Alcohols can form similar hydrogen bonds, as shown in Figure O2.4.
CH3
O
H
H ,, O
CH3
FIGURE O2.4 A hydrogen bond between a pair of methanol molecules.
As a result, alcohols have boiling points that are much higher than alkanes with similar
molecular weights. The boiling point of ethanol, for example, is 78.5°C, whereas propane,
with about the same molecular weight, boils at 42.1°C.
Alcohols are Brønsted acids in aqueous solution.
88n H3O(aq) CH3CH2O(aq)
CH3CH2OH(aq) H2O(l) m88
Alcohols therefore react with sodium metal to produce sodium salts of the corresponding
conjugate base, as noted in Section O2.2.
2 Na(s) 2 CH3OH(l) 88n 2 Na(alc) 2 CH3O(alc) H2(g)
The conjugate base of an alcohol is known as an alkoxide.
[Na][CH3O]
[Na][CH3CH2O]
Sodium methoxide
Sodium ethoxide
As we saw in Section O1.8, alcohols can be prepared by adding water to an alkene in
the presence of a strong acid such as concentrated sulfuric acid. The reaction involves
adding an H2O molecule across a CPC double bond. Because the reactions follow
Markovnikov’s rule, the product of the reaction is often a highly substituted 2° or 3°
alcohol.
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FUNCTIONAL GROUP CHEMISTRY
CH3CHP CHCH3 H2O
H2SO4
15
OH
A
CH3CHCH2CH3
Less substituted 1° alcohols can be prepared by substitution reactions that occur when a
primary alkyl halide is allowed to react with the OH ion.
CH3CH2CH2Br OH 88n CH3CH2CH2OH Br
Alcohols (ROH) can be thought of as derivatives of water in which one of the hydrogen atoms has been replaced by an alkyl group. If both of the hydrogen atoms are replaced
by alkyl groups, we get an ether (ROR). These compounds are named by adding the word
ether to the names of the alkyl groups.
CH3CH2OCH2CH3
diethyl ether
Diethyl ether, often known by the generic name “ether,” was once used extensively as an
anesthetic. Because mixtures of diethyl ether and air explode in the presence of a spark,
ether has been replaced by safer anesthetics.
There are important differences between both the physical and chemical properties of
alcohols and ethers. Consider diethyl ether and 1-butanol, for example, which are constitutional isomers with the formula C4H10O.
CH3CH2OCH2CH3
BP 34.5°C
density 0.7138 g/mL
Insoluble in water
CH3CH2CH2CH2OH
BP 117.2°C
density 0.8098 g/mL
Soluble in water
The shapes of the molecules are remarkably similar, as shown in Figure O2.5.
O
OH
FIGURE O2.5 The structures of diethyl ether and 1-butanol.
The fundamental difference between the compounds is the presence of OOH groups in
the alcohol that are missing in the ether. Because hydrogen bonds can’t form between the
molecules in the ether, the boiling point of the compound is more than 80°C lower than
that of the corresponding alcohol. Because there are no hydrogen bonds to organize the
structure of the liquid, the ether is significantly less dense than the corresponding alcohol.
Ethers can act as a hydrogen bond acceptor, as shown in Figure O2.6, but they can’t
act as hydrogen bond donors. As a result, ethers are less soluble in water than alcohols
with the same molecular weight.
H
CH3
O
H
O
CH3
FIGURE O2.6 Water acting as a hydrogen bond donor toward an ether.
The absence of an OOH group in an ether also has important consequences for its
chemical properties. Unlike alcohols, ethers are essentially inert to chemical reactions. They
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don’t react with most oxidizing or reducing agents, and they are stable to most acids and
bases, except at high temperatures. They are therefore frequently used as solvents for chemical reactions.
Compounds that are potential sources of an H ion, or proton, are often described as
being protic. Ethanol, for example, is a protic solvent.
H3O(aq) CH3CH2O(aq)
CH3CH2OH(aq) H2O(l)
Substances that can’t act as a source of a proton are said to be aprotic. Because they don’t
contain an OOH group, ethers are aprotic solvents.
Ethers can be synthesized by splitting out a molecule of water between two alcohols
in the presence of heat and concentrated sulfuric acid.
2 CH3CH2OH
H
CH3CH2OCH2CH3
They can also be formed by reacting a primary alkyl halide with an alkoxide ion.
CH3CH2CH2Br CH3O 88n CH3CH2CH2OCH3 Br
O2.5 ALDEHYDES AND KETONES
The connection between the structures of alkenes and alkanes was established in Section
O1.6, which noted that we can transform an alkene into an alkane by adding an H2 molecule across the CPC double bond.
H
G
D
H2
C PC
G
D
H
H
H H
H
Ni
H C
C H
H H
The driving force behind the reaction is the difference between the strengths of the bonds
that must be broken and the bonds that form in the reaction. In the course of the hydrogenation reaction, a relatively strong HOH bond (435 kJ/mol) and a moderately strong
carbon–carbon bond (~270 kJ/mol) are broken, but two strong COH bonds (439 kJ/mol)
are formed. The reduction of an alkene to an alkane is therefore an exothermic reaction.
What about the addition of an H2 molecule across a CPO double bond?
H O
B
H C C H H2
H
H OH
Pt
H C
C H
H H
Once again, a significant amount of energy has to be invested in the reaction to break the
HOH bond (435 kJ/mol) and the carbon–oxygen bond (~375 kJ/mol). The overall reaction is still exothermic, however, because of the strength of the COH bond (439 kJ/mol)
and the OOH bond (498 kJ/mol) that are formed.
The addition of hydrogen across a CPO double bond raises several important points.
First, and perhaps foremost, it shows the connection between the chemistry of primary alcohols and aldehydes. But it also helps us understand the origin of the term aldehyde. If a
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17
reduction reaction in which H2 is added across a double bond is an example of a hydrogenation reaction, then an oxidation reaction in which an H2 molecule is removed to form
a double bond might be called dehydrogenation. Thus, using the symbol [O] to represent
an oxidizing agent, we see that the product of the oxidation of a primary alcohol is literally an “al-dehyd” or aldehyde. It is an alcohol that has been dehydrogenated.
CH3CH2OH
O
B
CH3CH
[O]
The reaction also illustrates the importance of differentiating between primary, secondary, and tertiary alcohols. Consider the oxidation of isopropyl alcohol, or 2-propanol,
for example.
OH
A
CH3CHCH3
O
B
CH3CCH3
[O]
The product of this particular reaction was originally called aketone, although the name
was eventually softened to azetone and finally acetone. Thus, it is not surprising that any
substance that exhibited chemistry that resembled “aketone” became known as a ketone.
Aldehydes can be formed by oxidizing a primary alcohol; oxidation of a secondary alcohol gives a ketone. What happens when we try to oxidize a tertiary alcohol? The answer is simple: Nothing happens.
OH
A
CH3CCH3
A
CH3
[O]
There aren’t any hydrogen atoms that can be removed from the carbon atom carrying the
OOH group in a 3° alcohol. And any oxidizing agent strong enough to insert an oxygen
atom into a COC bond would oxidize the alcohol all the way to CO2 and H2O.
A variety of oxidizing agents can be used to transform a secondary alcohol to a ketone.
A common reagent for the reaction is some form of chromium(VI)—chromium in the 6
oxidation state—in acidic solution. The reagent can be prepared by adding a salt of the
chromate (CrO42) or dichromate (Cr2O72) ion to sulfuric acid. Or it can be made by
adding chromium trioxide (CrO3) to sulfuric acid. Regardless of how it is prepared, the oxidizing agent in the reactions is chromic acid, H2CrO4.
OH
A
CH3CHCH2CH3
H2CrO4
O
B
CH3CCH2CH3
The choice of oxidizing agents to convert a primary alcohol to an aldehyde is much
more limited. Most reagents that can oxidize the alcohol to an aldehyde carry the reaction
one step further—they oxidize the aldehyde to the corresponding carboxylic acid.
CH3CH2OH
H2CrO4
O
B
CH3CH
O
B
CH3COH
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A weaker oxidizing agent, which is just strong enough to prepare the aldehyde from the
primary alcohol, can be obtained by dissolving the complex that forms between CrO3 and
pyridine, C6H5N, in a solvent such as dichloromethane that doesn’t contain any water.
CH3CH2OH
CrO3 /pyridine
CH2Cl2
O
B
CH3CH
The Nomenclature of Aldehydes and Ketones
The common names of aldehydes are derived from the names of the corresponding carboxylic acids.
O
B
HCOH
O
B
HCH
Formic acid
Formaldehyde
O
B
CH3COH
O
B
CH3CH
Acetic acid
Acetaldehyde
The systematic names for aldehydes are obtained by adding -al to the name of the parent
alkane.
O
B
HCH
O
B
CH3CH
Methanal
Ethanal
The presence of substituents is indicated by numbering the parent alkane chain from the
end of the molecule that carries the OCHO functional group. For example,
O
B
BrCH2CH2CH
3-Bromopropanal
The common name for a ketone is constructed by adding ketone to the names of the
two alkyl groups on the carbon of the CPO double bond, listed in alphabetical order.
O
B
CH3CCH2CH3
Ethyl methyl ketone
The systematic name is obtained by adding -one to the name of the parent alkane and using numbers to indicate the location of the CPO group.
O
B
CH3CCH2CH3
2-Butanone
Common Aldehydes and Ketones
Formaldehyde has a sharp, somewhat unpleasant odor. The aromatic aldehydes in Figure
O2.7, on the other hand, have a very pleasant “flavor.” Benzaldehyde has the characteristic
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19
odor of almonds, vanillin is responsible for the flavor of vanilla, and cinnamaldehyde makes
an important contribution to the flavor of cinnamon.
O
B
CH
Benzaldehyde
O
B
CH
HO
CH3O
Vanillin
O
B
CH
Cinnamaldehyde
FIGURE O2.7 Aromatic aldehydes with characteristic
odors.
O2.6 REACTIONS AT THE CARBONYL GROUP
It is somewhat misleading to write the carbonyl group as a covalent CPO double bond.
The difference between the electronegativities of carbon and oxygen is large enough to
make the CPO bond moderately polar. As a result, the carbonyl group is best described
as a hybrid of the following resonance structures.
O
S
OS
SO
A
C
D G
O
B
C
D G
We can represent the polar nature of the hybrid by indicating the presence of a slight negative charge on the oxygen () and a slight positive charge () on the carbon of the
CPO double bond.
O
O
O
B
C
D G
O
S
Reagents that attack the electron-rich end of the CPO bond are called electrophiles
(literally, “lovers of electrons”). Electrophiles include ions (such as H and Fe3) and neutral molecules (such as AlCl3 and BF3) that are Lewis acids, or electron pair acceptors.
Reagents that attack the electron-poor end of the bond are nucleophiles (literally,
“lovers of nuclei”). Nucleophiles are Lewis bases (such as NH3 or the OHion).
Electrophiles
O
B (H, Fe3, BF , etc.)
3
C
D G
Nucleophiles
(OH, NH3, etc.)
The polarity of the CPO double bond can be used to explain the reactions of carbonyl
compounds. Aldehydes and ketones react with a source of the hydride ion (H) because
the H ion is a Lewis base, or nucleophile, that attacks the end of the CPO bond.
When this happens, the two valence electrons on the H ion form a covalent bond to the
carbon atom. Since carbon is tetravalent, one pair of electrons in the CPO bond is displaced onto the oxygen to form an intermediate with a negative charge on the oxygen atom.
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OS
SO
A
CH3OC OCH3
A
H
O
S
20
EQA
O
B
C
D G
CH3
CH3
HS
The alkoxide ion can then remove an H ion from water to form an alcohol.
O OH
SO
A
CH3OC OCH3 OH
A
H
OS
SO
A
CH3OC OCH3 H2O
A
H
Common sources of the H ion include lithium aluminum hydride (LiAlH4) and sodium
borohydride (NaBH4). Both compounds are ionic.
[Li][AlH4]
[Na][BH4]
LiAlH4:
NaBH4:
The aluminum hydride (AlH4) and borohydride (BH4) ions act as if they were complexes between an H ion, acting as a Lewis base, and neutral AlH3 or BH3 molecules,
acting as a Lewis acid.
H
A
HOAlOH
A
H
HS
H
A
AlOH
A
H
LiAlH4 is such a good source of the H ion that it reacts with the H ions in water or
other protic solvents to form H2 gas. The first step in the reduction of a carbonyl with
LiAlH4 is therefore carried out using an ether as the solvent. The product of the hydride
reduction reaction is then allowed to react with water in a second step to form the corresponding alcohol.
O
B
CH3CH2CH
1. LiAlH4
in ether
2. H2O
CH3CH2CH2OH
NaBH4 is less reactive toward protic solvents, which means that borohydride reductions
are usually done in a single step, using an alcohol as the solvent.
O
OH
B
A
NaBH4
CH3CH2CCH3 CH CH OH CH3CH2CHCH3
3
2
O2.7 CARBOXYLIC ACIDS AND CARBOXYLATE IONS
When one of the substituents on a carbonyl group is an OH group, the compound is a carboxylic acid with the generic formula RCO2H. These compounds are acids, as the name
suggests, which form carboxylate ions (RCO2) by the loss of an H ion.
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O
J
O
CH3 C
G
OH
21
O
J
O
CH3 C
H
G O
The carboxylate ion formed in the reaction is a hybrid of two resonance structures.
O
J
CH3O C
G O
O
D
CH3O C
M
O
Resonance delocalizes the negative charge in the carboxylate ion, which makes the ion
more stable than the alkoxide ion formed when an alcohol loses an H ion. By increasing
the stability of the conjugate base, resonance increases the acidity of the acid that forms
the base. Carboxylic acids are therefore much stronger acids than the analogous alcohols.
The Ka value for a typical carboxylic acid is about 105, whereas alcohols have Ka values
only 1016.
O
J
CH3C
G
OH
CH3OH
O
J
CH3C
H
G O
Ka 1.8 105
CH3O H
Ka 8 1016
Carboxylic acids were among the first organic compounds to be discovered. As a result, they have well-established common names that are often derived from the Latin stems
of their sources in nature. Formic acid (Latin formica, “an ant”) and acetic acid (Latin acetum, “vinegar”) were first obtained by distilling ants and vinegar, respectively. Butyric acid
(Latin butyrum, “butter”) is found in rancid butter, and caproic, caprylic, and capric acids
(Latin caper, “goat”) are all obtained from goat fat. A list of common carboxylic acids is
given in Table O2.5.
The systematic nomenclature of carboxylic acids is easy to understand. The ending
-oic acid is added to the name of the parent alkane to indicate the presence of the OCO2H
functional group.
HCO2H
CH3CO2H
CH3CH2CO2H
Methanoic acid
Ethanoic acid
Propanoic acid
Unfortunately, because of the long history of carboxylic acids and their importance in biology and biochemistry, you are more likely to encounter these compounds by their common names.
Formic acid and acetic acid have sharp, pungent odors. As the length of the alkyl chain
increases, the odor of carboxylic acids becomes more unpleasant. Butyric acid, for example, is found in sweat, and the odor of rancid meat is due to carboxylic acids released as
the meat spoils.
The solubility data in Table O2.5 show that carboxylic acids also become less soluble
in water as the length of the alkyl chain increases. The OCO2H end of the molecule is polar and therefore soluble in water. As the alkyl chain gets longer, the molecule becomes
more nonpolar and less soluble in water.
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TABLE O2.5
Common Carboxylic Acids
Common Name
Solubility in H2O
(g/100 mL)
Formula
Formic acid
Acetic acid
Propionic acid
Butyric acid
Caproic acid
Caprylic acid
Capric acid
Lauric acid
Myristic acid
Palmitic acid
Stearic acid
Saturated carboxylic acids and fatty acids
HCO2H
CH3CO2H
CH3CH2CO2H
CH3(CH2)2CO2H
CH3(CH2)4CO2H
CH3(CH2)6CO2H
CH3(CH2)8CO2H
CH3(CH2)10CO2H
CH3(CH2)12CO2H
CH3(CH2)14CO2H
CH3(CH2)16CO2H
Palmitoleic acid
Oleic acid
Linoleic acid
Linolenic acid
Unsaturated fatty acids
CH3(CH2)5CHPCH(CH2)7CO2H
CH3(CH2)7CHPCH(CH2)7CO2H
CH3(CH2)4CHPCHCH2CHPCH(CH2)7CO2H
CH3CH2CHPCHCH2CHPCHCH2CHPCH(CH2)7CO2H
0.968
0.068
0.015
0.0055
0.0020
0.00072
0.00029
—
—
—
—
Compounds that contain two OCO2H functional groups are known as dicarboxylic
acids. A number of dicarboxylic acids (see Table O2.6) can be isolated from natural sources.
Tartaric acid, for example, is a by-product of the fermentation of wine, and succinic, fumaric, malic, and oxaloacetic acids are intermediates in the metabolic pathway used to oxidize sugars to CO2 and H2O.
TABLE 02.6 Common Dicarboxylic Acids
Structure
Common Name
HO2CCO2H
HO2CCH2CO2H
HO2CCH2CH2CO2H
HO 2C
CO H
G
D 2
C PC
D
G
H
H
H
HO 2C
G
D
C PC
D
G
H
CO2 H
Oxalic acid
Malonic acid
Succinic acid
Maleic acid
Structure
OH
A
HO2CCH2CHCO2H
OH OH
A A
HO2CCHCHCO2H
O
B
HO2CCH2CCO2H
Common Name
Malic acid
Tartaric acid
Oxaloacetic acid
Fumaric acid
Several tricarboxylic acids also play an important role in the metabolism of sugar. The
most important example of this class of compounds is the citric acid that gives so many
fruit juices their characteristic acidity.
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CH2CO2H
A
HOC OCO2H
A
CH2CO2H
23
Citric acid
O2.8 ESTERS
Carboxylic acids (OCO2H) can react with alcohols (ROH) in the presence of either acid
or base to form esters (OCO2R). Acetic acid, for example, reacts with ethanol to form
ethyl acetate.
O
B
CH3COH CH3CH2OH
O
B
CH3COCH2CH3 H2O
H
This isn’t an efficient way of preparing an ester, however, because the equilibrium constant
for the reaction is relatively small (Kc 3). Chemists therefore tend to synthesize esters
from carboxylic acids in a two-step process. They start by reacting the acid with a chlorinating agent such as thionyl chloride (SOCl2) to form the corresponding acyl chloride.
O
B
CH3COH SOCl2
O
B
CH3CCl SO2 HCl
They then react the acyl chloride with an alcohol in the presence of base to form the ester.
O
B
CH3CCl CH3CH2OH
B
O
B
CH3COCH2CH3 BHCl
The base absorbs the HCl given off in the reaction, thereby driving it to completion.
As might be expected, esters are named as if they were derivatives of a carboxylic acid
and an alcohol. The ending -ate or -oate is added to the name of the parent carboxylic acid,
and the alcohol is identified using the “alkyl alcohol” convention. The following ester, for
example, can be named as a derivative of acetic acid (CH3CO2H) and ethyl alcohol
(CH3CH2OH).
O
B
CH3COCH2CH3
Ethyl acetate
Or it can be named as a derivative of ethanoic acid (CH3CO2H) and ethyl alcohol
(CH3CH2OH).
O
B
CH3COCH2CH3
Ethyl ethanoate
The term ester is commonly used to describe the product of the reaction of any strong
acid with an alcohol. Sulfuric acid, for example, reacts with methanol to form a diester
known as dimethyl sulfate.
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O
A
CH3O S OCH3 2 H2O
A
O
O
A
HO S OH 2 CH3OH
A
O
Phosphoric acid reacts with alcohols to form triesters such as trimethyl phosphate.
O
A
CH3O P OCH3 3 H2O
A
OCH3
O
A
HO P OH 3 CH3OH
A
OH
Compounds that contain the OCO2R functional group might therefore best be called carboxylic acid esters, to indicate the acid from which they are formed.
Carboxylic acid esters with low molecular weights are colorless, volatile liquids that often have a pleasant odor. They are important components of both natural and synthetic
flavors (see Figure O2.8).
B
CH3O
C
O
O
OH
E
CH2
CH3
CH2
Methyl salicylate
(oil of wintergreen)
CH3
CH2
CH3
CH2
B
O
CH
CH3
Isoamyl butyrate
(chocolate)
O
CH2
C
CH2
C
B
O
O
CH2
CH3
CH3
C
B
O
CH2
CH3
CH2
CH
CH3
Isoamyl acetate
(apple)
Ethyl butyrate
(pineapple)
FIGURE O2.8 Typical carboxylic acid esters with pleasant odors or flavors.
O2.9 AMINES, ALKALOIDS, AND AMIDES
Amines are derivatives of ammonia in which one or more hydrogen atoms are replaced by
alkyl groups. We indicate the degree of substitution by labeling the amine as either primary (RNH2), secondary (R2NH), or tertiary (R3N). The common names of the compounds
are derived from the names of the alkyl groups.
(CH3)2CHNH2
CH3
A
CH3CH2NH
CH3
A
CH3 NCH3
Isopropylamine
Ethylmethylamine
Trimethylamine
The systematic names of primary amines are derived from the name of the parent alkane
by adding the prefix amino- and a number specifying the carbon that carries the ONH2
group.
CHP CH
NH2
D
G
D
CH2O CH
CH3
G
CH3
5-Amino-2-hexene
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25
The chemistry of amines mirrors the chemistry of ammonia. Amines are weak bases
that pick up a proton to form ammonium salts. Trimethylamine, for example, reacts with
acid to form the trimethylammonium ion.
CH3
A
CH3ONS H
A
CH3
CH3
A
CH3ONOH
A
CH3
The salts are more soluble in water than the corresponding amines, and the reaction can
be used to dissolve otherwise insoluble amines in aqueous solution.
The difference between amines and their ammonium salts plays an important role in
both over-the-counter and illicit drugs. Cocaine, for example, is commonly sold as the hydrochloride salt, which is a white, crystalline solid. By extracting the solid into ether, it is
possible to obtain the “free base.” A glance at the side panel of almost any over-thecounter medicine will provide examples of ammonium salts of amines that are used to ensure that the drug dissolves in water (see Figure O2.9).
CH3
CH3O
NH2Cl
CH3
CH
N
Dextromethorphan
hydrobromide
(cough suppressant)
H
Br
CH3
HO
CH
Pseudoephedrine
hydrochloride
(decongestant)
FIGURE O2.9 Over-the-counter drugs sold as amine hydrochloride or hydrobromide salts.
Amines that are isolated from natural sources, especially from plants, are known as alkaloids. They include poisons such as nicotine, coniine, and strychnine (shown in Figure
O2.10). Nicotine has a pleasant, invigorating effect when taken in minuscule quantities but
is extremely toxic in larger amounts. Coniine is the active ingredient in hemlock, a poison
that has been used since the time of Socrates. Strychnine is another toxic alkaloid that has
been a popular poison in murder mysteries.
N
N
CH3
Nicotine
N
H
Coniine
O
N
O
Strychnine
N
FIGURE O2.10 The alkaloids are complex amines that can
be isolated from plants.
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The alkaloids also include a number of drugs, such as morphine, quinine, and cocaine.
Morphine is obtained from poppies; quinine can be found in the bark of the chinchona
tree; and cocaine is isolated from coca leaves. This family of compounds also includes synthetic analogs of naturally occurring alkaloids, such as heroin and LSD (see Figure O2.11).
To illustrate the importance of minor changes in the structure of a molecule, three
amines with similar structures are shown in Figure O2.12. One of the compounds is caffeine, which is the pleasantly addictive substance that makes a cup of coffee an important
part of the day for so many people. Another is theobromine, which is the pleasantly addictive substance that draws so many people to chocolate. The third compound is theophylline, which is a prescription drug used as a bronchodilator for people with asthma.
O
B
CH3COO
HOH
Morphine
Heroin
O
O
ECH3
N
NOCH3
E
HO
E
CH3CO
B
O
CH3H
N
CHPCH2
O
Cocaine
CO2CH3
O
HO
P
H
OC O
CH3OH
O
H E
CH
A
N
Quinine
N
CH3CH2H
E
CH3CH2
O
B
NOCO
NH
Lysergic acid
diethylamide
(LSD)
N
A
CH3
FIGURE O2.11 The structures of some alkaloids used as drugs.
O
CH3H
N
O
N
N
A
CH3
Caffeine
N
CH3
f
O
HH
N
O
N
A
CH3
CH3
f
N
N
Theobromine
O
H
i
N
N
A
CH3
N
CH3H
N
O
Theophylline
FIGURE O2.12 The caffeine in coffee, the theobromine in chocolate, and the active ingredient in a common bronchodilator are central nervous system stimulants with very
similar structures.
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It is tempting to assume that carboxylic acids will react with amines to form the class
of compounds known as amides. In practice, when aqueous solutions of carboxylic acids
and amines are mixed all we get is an acid–base reaction.
O
B
CH3COH CH3NH2
O
B
CH3COCH3NH3
The best way to prepare an amide is to react the appropriate acyl chloride with an amine.
O
B
CH3CCl 2 CH3NH2
O
B
CH3CNHCH3 CH3NH3Cl
Excess amine is used to drive the reaction to completion by absorbing the HCl given off
in the reaction.
O2.10 GRIGNARD REAGENTS
So far, we have built a small repertoire of reactions that can be used to convert one functional group to another. We have briefly discussed converting alkenes to alkanes; alkanes
to alkyl halides; alkyl halides to alcohols; alcohols to ethers, aldehydes, and ketones; and
aldehydes to carboxylic acids. We have also shown how carboxylic acids can be converted
to esters and amides. We have yet to encounter a reaction, however, that addresses a basic
question: How do we make COC bonds? One answer resulted from the work that Francois
Auguste Victor Grignard started as part of his Ph.D. research at the turn of the twentieth
century.
Grignard noted that alkyl halides react with magnesium metal in diethyl ether (Et2O)
to form compounds that contain a metal–carbon bond. Methyl bromide, for example, forms
methylmagnesium bromide.
Et O
2
CH3Br Mg 88n
CH3MgBr
Because carbon is considerably more electronegative than magnesium, the metal–carbon
bond in the product has a significant amount of ionic character. Grignard reagents such as
CH3MgBr are best thought of as hybrids of ionic and covalent Lewis structures.
CH3OMgOBr
[CH3S][Mg 2][Br]
Grignard reagents are therefore a potential source of carbanions (literally, “anions of
carbon”). The Lewis structure of the CH3 ion suggests that carbanions can be Lewis bases,
or electron pair donors.
H
A
HO CS
A
H
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Grignard reagents such as methylmagnesium bromide are therefore sources of a nucleophile that can attack the end of the CPO double bond in aldehydes and ketones.
CH3
CH3S
CH3
A
CH3O C O O
A
CH3
G
CPO
D
CH3
If we treat the product of the reaction with water, we get a tertiary alcohol.
CH3
A
CH3OC OO H2O
A
CH3
CH3
A
CH3OC OOH OH
A
CH3
If we wanted to make a secondary alcohol, we could add the Grignard reagent to an aldehyde, instead of a ketone.
O
B
CH3CH
1. CH 3MgBr
2. H2O
OH
A
CH3CHCH3
By reacting a Grignard reagent with formaldehyde we can add a single carbon atom to
form a primary alcohol.
CH3
A
CH3CCH2MgBr H2CPO
A
CH3
CH3
A
CH3CCH2CH2O
A
CH3
H2O
CH3
A
CH3CCH2CH2OH
A
CH3
The alcohol can then be oxidized to the corresponding aldehyde.
CH3
A
CH3CCH2CH2OH
A
CH3
CrO3 /pyridine
CH2Cl2
CH3 O
B
A
CH3CCH2CH
A
CH3
The Grignard reagent therefore provides us with a way of performing the following overall transformation.
CH3
A
CH3CCH2MgBr
A
CH3
1. H2CO
2. H2O
3. CrO3 /pyridine
in CH2Cl2
CH3 O
B
A
CH3CCH2CH
A
CH3
A single carbon atom can also be added if the Grignard reagent is allowed to react
with CO2 to form a carboxylic acid.
CH3
A
CH3CCH2MgBr CO2
A
CH3
CH3 O
A
B
CH3CCH2CO
A
CH3
H2O
CH3 O
A
B
CH3CCH2 COH OH A
CH3
Perhaps the most important aspect of the chemistry of Grignard reagents is the ease
with which the reaction allows us to couple alkyl chains. Isopropylmagnesium bromide, for
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29
example, can be used to graft an isopropyl group onto the hydrocarbon chain of an appropriate ketone, as shown in Figure O2.13.
A
CH3CH2CCH3
A
CH
D G
CH3
CH3
H2O
H
A
CH3CH2CCH3
A
CH
D G
CH3
CH3
O
B
G
CHMgBr
CH
CH
CCH
3
2
3
E
CH3
CH3
FIGURE O2.13 Grignard reagents provide a powerful way to introduce alkyl groups onto a
hydrocarbon chain.
Chemistry in the World Around Us
The Chemistry of Garlic
The volatile materials that can be distilled from plants were named essential oils by
Paracelsus in the sixteenth century because they were thought to be the quintessence
(literally, the “fifth essence,” or vital principle) responsible for the odor and flavor of the
plants from which they were isolated.
The Egyptians extracted essential oils from fragrant herbs more than 5000 years ago
by pressing the herb or by extracting the fragrant material with olive or palm oil. Some
essential oils are still obtained by pressing. Others are extracted into a nonpolar solvent.
The most common method for isolating essential oils, however, is steam distillation. The
ground botanical is immersed in water which is heated to boiling, or boiling water is allowed to pass through a sample of the ground botanical. The oil and water vapor pass
into a condenser, where the oil separates from the water vapor.
The function of the essential oils isn’t fully understood. Some act as attractants for
the insects involved in pollination. Most are either bacteriostats (which stop the growth
of bacteria) or bactericides (which kill bacteria). In some cases, they can be a source of
metabolic energy. In other cases, they appear to be waste products of plant metabolism.
The essential oils are mixtures of up to 200 organic compounds, many of which are
either terpenes (with 10 carbon atoms) or sesquiterpenes (with 15 carbon atoms). Although the three components shown in Figure O2.14 represent almost 60% of the mass
of a sample of rose oil, 50 other components of that essential oil have been identified.
CH2OH
CPO
A
H
CH2OH
Citronellol
(3-7-dimethyl6-octen-1-ol)
Geraniol
(E-3,7-dimethyl2,6-octadien-1-ol)
Neral
(Z-3,7-dimethyl2,6-octadienal)
FIGURE O2.14 The three most abundant components of rose oil.
Garlic, onions, and mustard seed differ from most other sources of essential oils.
In each case, the fragrance producing part of the plant must be crushed before the
volatile components are released. For more than 100 years, chemists have known
that the principal component of the oil that distills from garlic is diallyl disulfide
[F. W. Semmler, Archiv der Pharmazie, 230, 434–448 (1892)].
CH2PCHCH2 SSCH2CHPCH2
Diallyl disulfide
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Only recently, however, have chemists explained how the compound is produced when a
clove of garlic is crushed [E. Block, Angewandte Chemie, International Edition in English, 31, 1101–1264 (1992)].
Before garlic is crushed, the intact cell contains S–2-propenyl-L-cysteine S-oxide—or
alliin—which can be found in the cell cytoplasm.
O
NH3
B
A
CH2PCHCH2SCH2CHCO2
Alliin
Within the cell there are vacuoles that contain an enzyme known as alliinase. When the
cell is crushed, the enzyme is released. The enzyme transforms the natural product alliin
into an intermediate that reacts with itself to form a compound known as allicin.
O
B
CH2PCHCH2SSCH2CHPCH2
Allicin
Allicin has been described as an odoriferous, unstable antibacterial substance that polymerizes easily and must be stored at low temperatures. When heated, it breaks down to
give a variety of compounds, including the diallyl disulfide obtained when oil of garlic is
distilled from the raw material.
An alliinase enzyme can also be found in onions, where it converts an isomer of alliin known as S-(E)-1-propenyl-L-cysteine S-oxide into propanethial S-oxide.
CH3CH2CHPSOO
Propanethial S-oxide
The product of the reaction is known as the lacrimator factor of onion because it is the
substance primarily responsible for the tears generated when onions are cut.
A great deal of progress has been made in recent years in identifying the various
organosulfur compounds formed when garlic and onion are cut and in understanding the
process by which the compounds are formed. The structures of some of the principal
organosulfur compounds associated with garlic are shown in Figure O2.15. In spite of
the progress made so far, much still has to be learned about the compounds that can be
isolated from extracts of the genus Allium, which includes both garlic and onion.
O
S
S
H
SOH
S
S
S
O
O
S
S
S
O
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
FIGURE O2.15 Some of the organosulfur
compounds associated with garlic.
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31
KEY TERMS
Acyl chloride
Alcohol
Aldehyde
Alkaloid
Alkoxide
Amide
Amine
Aprotic
Carbanion
Carbonyl
Carboxylate ion
Carboxylic acid
Carboxylic acid ester
Chain initiation
Chain propagation
Chain reaction
Chain termination
Dicarboxylic acid
Electrophile
Ester
Ether
Free radical
Functional group
Grignard reagent
Hydrophilic
Hydrophobic
Ketone
Nucleophile
Oxidation number
Protic
PROBLEMS
Functional Groups
1. Give examples of compounds that contain each of the following functional groups.
(a) an alcohol (b) an aldehyde (c) an amine (d) an amide (e) an alkyl halide
(f) an alkene (g) an alkyne
2. Describe the difference between the members of each of the following pairs of functional groups.
(a) an alcohol and an alkoxide ion
(b) an alcohol and an aldehyde
(c) an amine and an amide
(d) an ether and an ester
(e) an aldehyde and a ketone
3. Classify the following compounds as an alcohol, an aldehyde, an ether, or a ketone.
O
O
B
B
(a) CH3CH2CH (b) CH3CH2OH (c) CH3CH2OCH2CH3 (d) CH3CH2CCH3
4. Classify the following compounds as primary, secondary, or tertiary alcohols.
CH3
CH3
OH
A
A
A
(a) CH3CH2OH (b) CH3CH2CHOH (c) CH3CH2COH (d) CH3CH2CHCH2CH3
A
CH3
5. Cortisone is an adrenocortical steroid that is used as an anti-inflammatory agent.
Use the fact that carbon is tetravalent to determine the molecular formula of the compound from the following stick structure. Identify the functional groups present in the
molecule.
O
M
CH3
A
J
O
CH2OH
A
CPO
CH3A
A O OH
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6. Piperine {(E,E)-1-[5-(1,3-benzodioxol-5-yl)-1-oxo-2,4-pentadienyl]piperidine} can be
extracted from black pepper. Identify the functional groups in the structure of the
compound.
O
N
O
B
O
7. PGE2 is a member of the family of compounds known as prostaglandins, which have
very significant physiological effects in even small quantities. They can affect blood
pressure, heart rate, body temperature, blood clotting, and conception. Some induce
inflammation; others relieve it. Aspirin, which is both an anti-inflammatory and an antipyretic (fever-reducing) drug, acts by blocking the synthesis of prostaglandins. Identify the functional groups in the structure of the PGE2 molecule and calculate its molecular formula.
O
CO2H
OH
OH
8. Pseudoephedrine hydrochloride is the active ingredient in a variety of decongestants,
including Sudafed. Use the structure of the compound shown in Figure O2.9 to identify the functional groups in the molecule.
9. Cocaine was once used in Coca Cola. Quinine is still added to tonic water. Use the
structures of the alkaloids shown in Figure O2.11 to identify the functional groups in
the compounds.
OXIDATION–REDUCTION REACTIONS
10. Arrange the following substances in order of increasing oxidation number of the carbon atom.
(a) C (b) HCHO (c) HCO2H (d) CO (e) CO2 (f) CH4 (g) CH3OH
11. Classify the following reactions as examples of either metathesis or oxidation–reduction
reactions.
(a) CH4 Cl2 n CH3Cl HCl
(b) CH3OH n HCHO
(c) HCHO n HCO2H
(d) CH3OH HBr n CH3Br H2O
(e) (CH3)2CO n (CH3)2CHOH
12. Classify the following transformations as examples of either oxidation or reduction.
(a) CH3CH2OH n CH3CHO
(b) CH3CHO n CH3CO2H
(c) (CH3)2CPO n (CH3)2CHOH
(d) CH3CHPCHCH3 n CH3CH2CH2CH3
(e) (CH3)2CHCqCH n (CH3)2CHCH2CH3
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33
13. Which of the following compounds can be oxidized to form an aldehyde?
(a) CH3CH2OH (b) CH3CHOHCH3 (c) CH3OCH3 (d) (CH3)2CPO
14. Which of the following compounds should be the most difficult to oxidize?
(a) CH3CH2OH (b) (CH3)2CHOH (c) (CH3)3COH (d) CH3CHO
15. Which of the following compounds can be prepared by reducing CH3CHO?
(a) CH3CH3 (b) CH3CH2OH (c) CH3CO2H (d) CH3CH2CO2H
16. Which of the following compounds can be prepared by oxidizing CH3CHO?
(a) CH3CH3 (b) CH3CH2OH (c) CH3CO2H (d) CH3CH2CO2H
17. Predict the product of the oxidation of 2-methyl-3-pentanol.
18. Predict the product of the reduction of 2-methyl-2-pentene with hydrogen gas over a
nickel metal catalyst.
Alkyl Halides
19. Use Lewis structures to describe the free radical chain reaction mechanism involved
in the bromination of methane to form methyl bromide. Clearly label the chain initiation, chain propagation, and chain termination steps.
20. Use Lewis structures to describe the difference between a CH3 ion, a CH3 radical,
and a CH3 ion. Which of these Lewis structures describes a carbanion? Which describes
a carbocation?
21. How many different products could be formed by the free radical chlorination of pentane? If attack at the different hydrogen atoms in the compound were more or less
random, what would be the relative abundance of the different isomers of chloropentane formed in the reaction?
22. Consider the reaction between a Cl atom and a pentane molecule. Classify the different intermediates that could be produced in the reaction as either primary, secondary,
or tertiary free radicals.
Alcohols and Ethers
23. Describe the differences between the structures of water, methyl alcohol, and dimethyl
ether.
24. Draw the structures of the seven constitutional isomers that have the formula C4H10O.
Classify the isomers as either alcohols or ethers.
25. Ethyl alcohol and dimethyl ether have the same chemical formula, C2H6O. Explain
why one of the compounds reacts rapidly with sodium metal but the other does not.
26. Predict the product of the dehydration of ethyl alcohol with sulfuric acid at 130°C.
27. Use examples to explain why it is possible to oxidize either a primary or secondary alcohol, but not a tertiary alcohol.
28. Draw the structure of o-phenylphenol, the active ingredient in Lysol.
29. Explain why alcohols are Brønsted acids in water.
30. Write the structures of the conjugate bases formed when the following alcohols act as
a Brønsted acid.
(a) CH3CH2OH (ethyl alcohol) (b) C6H5OH (phenol)
(c) (CH3)2CHOH (isopropyl alcohol)
31. Use the fact that there are no hydrogen bonds between ether molecules to explain why
diethyl ether (C4H10O) boils at 34.5°C, whereas its constitutional isomer—
1-butanol (C4H10O)—boils at 118°C.
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32. Assign a systematic name to the following alcohols.
OH
OH
A
B
33. Assign a systematic name to the compound known by the common name menthol.
OH
Aldehydes and Ketones
34. At which end of a carbonyl group will each of the following substances react?
(a) H (b) H (c) OH (d) NH3 (e) BF3
35. Explain why mild oxidation of a primary alcohol gives an aldehyde, whereas oxidation
of a secondary alcohol gives a ketone.
36. Explain why strong oxidizing agents can’t be used to convert a primary alcohol to an
aldehyde.
37. Assign systematic names to the following compounds.
P
O
P
O
H
A
B
38. Assign a systematic name to the compound known by the common name geranial.
O
Carboxylic Acids, Carboxylate Ions, and Carboxylic Acid Esters
39. Explain the difference between a carboxylic acid, a carboxylate ion, and a carboxylic
acid ester.
40. What major differences between carboxylic acids (such as butyric acid,
CH3CH2CH2CO2H) and esters (such as ethyl butyrate, CH3CH2CH2CO2CH2CH3)
help explain why butyric acid gives rise to the odor of rotten meat but ethyl butyrate
gives rise to the pleasant odor of pineapple?
41. Explain why fats, which are esters of long-chain carboxylic acids, are insoluble in water. Explain what happens during saponification (see Chapter 8) that makes the resulting compounds marginally soluble in water.
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42. Explain the difference between CH3CH2OH and CH3CO2H that makes one of the
COOH bonds in the compounds more than 10 orders of magnitude more acidic than
the other when values of Ka for the compounds are compared.
43. Use the structures shown in Figure O2.11 to propose a series of reactions that could
be used to convert morphine into heroin.
Amines, Alkaloids, and Amides
44. Classify each of the compounds in Figures O2.10 and O2.11 as either a primary amine,
a secondary amine, a tertiary amine, and/or an amide.
45. Explain why reacting a complex amine, such as pseudoephedrine, with an acid makes
the compounds more soluble in water.
46. Draw the structure of caffeine and label each of the nitrogen atoms in the compound
as either an amine or an amide.
Grignard Reagents
47. Use Lewis structures to decide whether the organic component of a Grignard reagent
is a nucleophile (Lewis base) or an electrophile (Lewis acid). Predict which end of a
CPO bond this portion of the Grignard reagent will attack.
48. If CH3CH2MgBr can be thought of as containing the CH3CH2, Mg2, and Br ions,
what would be the product of the reaction between the Grignard reagent and dilute
hydrochloric acid?
49. Explain why Grignard reagents are prepared in an aprotic solvent, such as an ether,
rather than a protic solvent such as an alcohol.
50. Write a step-by-step mechanism for the reaction between CH3MgBr and H2CPO to
form CH3CH2OH.
51. Predict the product of the reaction between (CH3)2CHMgBr and 2-pentanone.
O
B
52. If CH3CH2CCH3 is used as the starting material, show how a tertiary alcohol can be
synthesized by a Grignard reaction.
53. Select from the reagents given below to design a sequence of reactions that would lead
to the following compound.
OH
A
CH3CHCHCH3
A
CH3
(a) (CH3)2CHMgBr (b) CH3CHO (c) CH3MgBr (d) CH3COCH2CH3
(e) CH3COCH3 (f) CH3CH2MgBr (g) CH3COCH3
Qualitative Organic Analysis
54.
55.
56.
57.
Describe a way of determining whether a compound is an alkane or an alkene.
Describe a way of determining whether a compound is a carboxylic acid or an ester.
Describe a way of determining whether a compound is an alcohol or an ether.
Describe a way of determining whether a compound is a primary or a tertiary alcohol.
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58.
59.
60.
61.
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FUNCTIONAL GROUP CHEMISTRY
Describe
Describe
Describe
Describe
a way of determining whether a compound is an alcohol or an alkyl halide.
a way of determining whether a compound is a carboxylic acid or an amide.
a way of determining whether a compound is an alcohol or an amine.
a way of determining whether a compound is an aldehyde or ketone.
Integrated Problems
62. Identify the Brønsted acids and the Brønsted bases in the following reaction.
CH3CqCH NH2 88n CH3CqC NH3
63. Succinic acid plays an important role in the Krebs cycle, malic acid (apple acid) is found
in apples, and tartaric acid ( fruit acid) is found in many fruits. Which of these dicarboxylic acids is chiral? See Table O2.6.