General Chemistry
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M O D U L E
1 CHEMISTRY OF THE NONMETALS N.1 The Nonmetals N.2 The Chemistry of Hydrogen N.3 The Chemistry of Oxygen Chemistry in the World Around Us: The Chemistry of the Atmosphere N.4 The Chemistry of Sulfur N.5 The Chemistry of Nitrogen N.6 The Chemistry of Phosphorus N.7 The Chemistry of the Halogens N.8 The Chemistry of the Rare Gases N.9 The Inorganic Chemistry of Carbon
More than 75% of the known elements have the characteristic properties of metals (see Figure N.1). They have a metallic luster; they are malleable and ductile; and they conduct heat and electricity. Eight other elements (B, Si, Ge, As, Sb, Te, Po, and At) are best described as semimetals or metalloids. They often look like metals, but they tend to be brittle, and they are more likely to be semiconductors than conductors of electricity. Once the metals and semimetals are removed from the list of known elements, only 17 are left to be classified as nonmetals. Six of these elements belong to the family of rare gases in Group VIIIA, most of which are virtually inert to chemical reactions. Discussions of the chemistry of the nonmetals therefore tend to focus on the following elements: H, C, N, O, F, P, S, Cl, Se, Br, I, and Xe. One way of visualizing the difference between metals, semimetals or metalloids, and nonmetals is the plot of average valence electron energies (AVEE) in Figure N.2. Because the AVEE provides a measure of how tightly an atom holds onto its valence electrons, it can be used to explore the dividing line between the metals and nonmetals. As noted in Section 3.24, elements with AVEE values below 1.06 MJ/mol are metals, whereas those with AVEE values above 1.26 MJ/mol are nonmetals. Elements with AVEE values in the range of 1.06–1.26 MJ/mol have properties between those of the metals and nonmetals and are therefore semimetals or metalloids. 1
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NONMETAL Metals
H
H
He
Nonmetals Li
Be
Na
Mg
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Fr
Ra
Ac
Rf
Db
Sg
Bh
Hs
Mt
Ce
Pr
Nd
Pm
Th
Pa
U
Np
B
C
N
O
F
Ne
Al
Si
P
S
Cl
Ar
Zn
Ga
Ge
As
Se
Br
Kr
Ag
Cd
In
Sn
Sb
Te
I
Xe
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
Semimetals
FIGURE N.1 The elements can be divided into three categories: metals, semimetals, and nonmetals.
Ne F O Ar N
CI S
C
Br Xe Se
P
I
B
As
Si
Te
Ge Sb
Al
Be
Ga Mg
Li
Kr
Sn In
Ca
Na K
Sr Rb
FIGURE N.2 Three-dimensional plot of the average valence electron energies (AVEE) of the main-group elements versus position in the periodic table.
N.1 THE NONMETALS There is a clear pattern in the chemistry of the main-group metals discussed in Chapter 5: The main-group metals are oxidized in all of their chemical reactions. Aluminum, for example, is oxidized by bromine when these elements react to form aluminum bromide. 2 Al 3 Br2 0
Al2Br6
0 Oxidation
3 1
Reduction
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The chemistry of the nonmetals is more interesting because these elements can undergo both oxidation and reduction. Phosphorus, for example, is oxidized when it reacts with more electronegative elements, such as oxygen. P4 5 O2
P4O10
0
5 2
0 Oxidation
Reduction
But it is reduced when it reacts with less electronegative elements, such as calcium. 6 Ca P4 0
2 Ca3P2
0 Oxidation
2 3
Reduction
The behavior of the nonmetals can be summarized as follows. •
Nonmetals tend to oxidize metals. 2 Mg(s) O2(g) 88n 2 MgO(s)
•
Nonmetals with relatively large electronegativities (such as oxygen and chlorine) oxidize substances with which they react. 2 H2S(g) 3 O2(g) 88n 2 SO2(g) 2 H2O(g) PH3(g) 3 Cl2(g) 88n PCl3(l) 3 HCl(g)
•
Nonmetals with relatively small electronegativities (such as carbon and hydrogen) can reduce other substances. Fe2O3(s) 3 C(s) 88n 2 Fe(s) 3 CO(g) CuO(s) H2(g) 88n Cu(s) H2O(g)
N.2 THE CHEMISTRY OF HYDROGEN Hydrogen is the most abundant element in the universe, accounting for 90% of the atoms and 75% of the mass of the universe. But hydrogen is much less abundant on earth. Even when the enormous number of hydrogen atoms in the oceans is included, hydrogen makes up less than 1% of the mass of the planet. The name hydrogen comes from the Greek stems hydro-, “water,” and gennan, “to form or generate.” Thus, hydrogen is literally the “water former.” 2 H2(g) O2(g) 88n 2 H2O(g) Although it is often stated that more compounds contain carbon than any other element, this isn’t necessarily true. Most carbon compounds also contain hydrogen, and hydrogen forms compounds with virtually all the other elements as well.
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Compounds of hydrogen are frequently called hydrides, even though the name hydride literally describes compounds that contain an H ion. There is a regular trend in the formula of the hydrides across a row of the periodic table, as shown in Figure N.3. This trend is so regular that the combining power, or valence, of an element was once defined as the number of hydrogen atoms bound to the element in its hydride.
H2
H2
LiH BeH2
B2H6 CH4 NH3 H2O
NaH MgH2
AlH3 SiH4 PH3 H2S HCl
KH CaH2
GaH3 GeH4 AsH3 H2Se HBr
RbH SrH2
SnH4 SbH3 H2Te
CsH BaH2
PbH4 BiH3 H2Po HAt
HF
HI
FIGURE N.3 Hydrogen combines with every element in the periodic table except those in Group VIIIA. The formulas of the hydrides of the main-group elements are shown here.
Hydrogen has three oxidation states, corresponding to the H ion, a neutral H atom, and the H ion. H 1s0 H 1 s1 H 1s2 Because hydrogen forms compounds with oxidation numbers of both 1 and 1, many periodic tables include the element in both Group IA (with Li, Na, K, Rb, Cs, and Fr) and Group VIIA (with F, Cl, Br, I, and At). There are many reasons for including hydrogen among the elements in Group IA. It forms compounds such as HCl and HNO3 that are analogs of alkali metal compounds such as NaCl and KNO3. Under conditions of very high pressure, hydrogen has the properties of a metal. (It has been argued, for example, that any hydrogen present at the center of the planet Jupiter is present as a metallic solid.) Finally, hydrogen combines with a handful of metals, such as scandium, titanium, chromium, nickel, and palladium, to form materials that behave as if they were alloys of two metals. There are equally valid arguments for placing hydrogen in Group VIIA. It forms compounds such as NaH and CaH2 that are analogs of halogen compounds such as NaF and CaCl2. It also combines with other nonmetals to form covalent compounds such as H2O, CH4, and NH3, the way a nonmetal should. Finally, the element is a gas at room temperature and atmospheric pressure, like other nonmetals such as O2 and N2. It is difficult to decide where hydrogen belongs in the periodic table because of the physical properties of the element. The first ionization energy of hydrogen (1312 kJ/mol), for example, is roughly halfway between the elements with the largest (He 2372 kJ/mol) and smallest (Cs 376 kJ/mol) first ionization energies. Hydrogen also has an electronegativity (EN 2.30) halfway between the extremes of neon, the most electronegative element (EN 4.79), and cesium, the least electronegative (EN 0.66) element. On the basis of electronegativity, it is tempting to classify hydrogen as a semimetal, as shown in Figure N.4.
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Ne F O Ar N
CI S
C
Br Xe Se
P
H
I
B
As
Si
Te
Ge Sb
Al
Be
Ga Mg
Li
Kr
Sn In
Ca
Na K
Sr Rb
FIGURE N.4 Three-dimensional plot of the electronegativities of the main-group elements versus position in the periodic table.
Hydrogen is oxidized by elements that are more electronegative to form compounds in which it has an oxidation number of 1. H2 Cl2 0
2 HCl 11
0 Oxidation
Reduction
Hydrogen is reduced by elements that are less electronegative to form compounds in which its oxidation number is 1. 2 Na H2 0
2 NaH
0 Oxidation
1 1
Reduction
At room temperature, hydrogen is a colorless, odorless gas with a density only onefourteenth the density of air. Small quantities of H2 gas can be prepared in several ways. •
By reacting an active metal with water. 2 Na(s) 2 H2O(l) 88n 2 Na(aq) 2 OH(aq) H2(g)
•
By reacting a less active metal with a strong acid. Zn(s) 2 HCl(aq) 88n Zn2(aq) 2 Cl(aq) H2(g)
•
By reacting an ionic metal hydride with water. NaH(s) H2O(l) 88n Na(aq) OH(aq) H2(g)
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NONMETAL •
By decomposing water into its elements with an electric current. 2 H2O(l) electrolysis 888888n 2 H2(g) O2(g)
Exercise N.1 Use oxidation numbers to determine what is oxidized and what is reduced in the following reactions, which are used to prepare H2 gas. (a) Mg(s) 2 HCl(aq) n Mg2(aq) 2 Cl(aq) H2(g) (b) Ca(s) 2 H2O(l) n Ca2(aq) 2 OH(aq) H2(g) Solution (a) Magnesium metal is oxidized in this reaction and the H ions from hydrochloric acid are reduced. Mg2 2 Cl H2
Mg 2 HCl 0
1
Oxidation
2
0
Reduction
(b) Calcium metal is oxidized in this reaction and the H ions from water are reduced. Ca 2 H2O 0
1
Oxidation
Ca2 2 OH H2 2
0
Reduction
The covalent radius of a neutral hydrogen atom is smaller than any other element. Because small atoms can come very close to each other, they tend to form strong covalent bonds. H2 therefore tends to be unreactive at room temperature. In the presence of a spark, however, a fraction of the H2 molecules dissociate to form hydrogen atoms that are highly reactive. H2(g) 8spark 88n 2 H(g)
H° 435.30 kJ/molrxn
The heat given off when these H atoms react with O2 is enough to catalyze the dissociation of additional H2 molecules. Mixtures of H2 and O2 that are infinitely stable at room temperature therefore explode in the presence of a spark or flame.
N.3 THE CHEMISTRY OF OXYGEN Oxygen is the most abundant element on this planet. The earth’s crust is 46.6% oxygen by mass, the oceans are 86% oxygen by mass, and the atmosphere is 21% oxygen by volume. The name oxygen comes from the Greek stems oxys, “acid,” and gennan, “to form or generate.” Thus, oxygen literally means the “acid former.” The name was introduced by
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Lavoisier, who noticed that compounds rich in oxygen, such as SO2 and P4O10, dissolve in water to give acids. SO2(g) H2O(aq) 88n H2SO3(aq) P4O10(s) 6 H2O(aq) 88n 4 H3PO4(aq) The electron configuration of an oxygen atom—[He] 2s2 2p4—suggests that O atoms can achieve an octet of valence electrons by sharing two pairs of electrons to form an OPO double bond, as shown in Figure N.5. O O
FIGURE N.5 Lewis structure of the O2 molecule.
According to the Lewis structure, all of the electrons in the O2 molecule are paired. The compound should therefore be diamagnetic—it should be repelled by a magnetic field. Experimentally, O2 is found to be paramagnetic—it is attracted to a magnetic field. This can be explained using molecular orbital theory, which predicts that there are two unpaired electrons in the * antibonding molecular orbitals of the O2 molecule. At temperatures below 183°C, O2 condenses to form a liquid with a characteristic light blue color that results from the absorption of light with a wavelength of 630 nm. This absorption isn’t seen in the gas phase and is relatively weak even in the liquid because it requires that three bodies—two O2 molecules and a photon—collide simultaneously, which is a very rare phenomenon, even in the liquid phase.
The Chemistry of Ozone The O2 molecule isn’t the only elemental form of oxygen. In the presence of lightning or another source of a spark, O2 molecules dissociate to form oxygen atoms. spark O2(g) 888 n 2 O(g)
The O atoms can react with O2 molecules to form ozone, O3. O2(g) O(g) 88n O3(g) Ozone is a resonance hybrid of two Lewis structures each of which contains one OPO double bond and one OOO single bond, as shown in Figure N.6. Because the valence electrons on the central atom are distributed toward the corners of a triangle, the O3 molecule is angular or bent, with a bond angle of 116.5°.
O O
O O
O
O
FIGURE N.6 Lewis structures for ozone.
When an element exists in more than one form—such as oxygen (O2) and ozone (O3)— the different forms of the element are called allotropes (from a Greek word meaning “in another manner”). Because they have different structures, allotropes have different chemical and physical properties, as shown in Table N.1.
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TABLE N.1
Properties of Allotropes of Oxygen
Property Melting point Boiling point Density (at 20°C) OOO bond order OOO bond length
Oxygen (O2)
Ozone (O3)
218.75°C 182.96°C 1.331 g/L 2 0.1207 nm
192.5°C 110.5°C 1.998 g/L 1.5 0.1278 nm
Ozone is an unstable compound with a sharp, pungent odor that slowly decomposes to oxygen. 2 O3(g) 88n 3 O2(g) At low concentrations, ozone can be relatively pleasant. (The characteristic clean odor associated with summer thunderstorms is due to the formation of small amounts of O3.) Exposure to O3 at higher concentrations leads to coughing, rapid beating of the heart, chest pain, and general body pain. At concentrations above 1 ppm, ozone is toxic. The most famous characteristic of ozone is its ability to absorb high energy radiation in the ultraviolet portion of the spectrum ( 300 nm), thereby providing a filter that protects us from exposure to high energy ultraviolet radiation emitted by the sun. We can understand the importance of this filter if we think about what happens when radiation from the sun is absorbed by our skin. Electromagnetic radiation in the infrared, visible, and low energy portions of the ultraviolet spectrum carries enough energy to excite an electron into a higher energy orbital. This electron eventually falls back into the orbital from which it was excited, and energy is given off to the surrounding tissue in the form of heat. Anyone who has suffered from a sunburn can appreciate the painful consequences of excessive amounts of this radiation. Radiation in the high energy portion of the ultraviolet spectrum carries enough energy to ionize atoms and molecules. Because living tissue is 70–90% water by weight, the most common ionization reaction involves the loss of an electron from a neutral water molecule to form an H2O ion. H2O 88n H2O e The H2O ions formed in the reaction have an odd number of electrons and are extremely reactive. They can cause permanent damage to the cell tissue and induce processes that eventually result in skin cancer. Relatively small amounts of this radiation can therefore have drastic effects on living tissue. Therefore, the protection by the ozone (O3) layer which prevents high energy radiation from reaching the earth is important to the health of living organisms. In 1974 Molina and Rowland pointed out that chlorofluorocarbons, such as CFCl3 and CF2Cl2, which had been used as refrigerants and as propellants in aerosol cans, were beginning to accumulate in the atmosphere. In the stratosphere, at altitudes of 10 to 50 km above the earth’s surface, chlorofluorocarbons decompose to form Cl atoms and chlorine oxides such as ClO when they absorb sunlight. Cl atoms and ClO molecules also have an unpaired electron in the valence shell of the molecule. As a result, they are unusually
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reactive. In the atmosphere, they react with ozone or with the oxygen atoms that are needed to form ozone. Cl O3 88n ClO O2 ClO O 88n Cl O2 Molina and Rowland postulated that these substances would eventually deplete the ozone shield in the stratosphere, with dangerous implications for biological systems that would be exposed to increased levels of high energy ultraviolet radiation.
Oxygen as an Oxidizing Agent Fluorine is the only element with which oxygen reacts that is more electronegative than oxygen. As a result, oxygen gains electrons in virtually all of its chemical reactions. Each O2 molecule must gain four electrons to satisfy the octets of the two oxygen atoms without sharing electrons. ³¼
¼³
OPO 4 e
þ 2 2 [¼O¼] ³
Oxygen therefore oxidizes metals to form salts in which the oxygen atoms are formally present as O2 ions. Rust forms, for example, when iron reacts with oxygen in the presence of water to give a salt that formally contains the Fe3 and O2 ions, with an average of three water molecules coordinated to each Fe3 ion in the solid. H2O 4 Fe(s) 3 O2(g) 88n 2 Fe2O33 H2O(s)
Oxygen also oxidizes nonmetals, such as carbon, to form covalent compounds in which the oxygen has an oxidation number of 2. C(s) O2(g) 88n CO2(g) Oxygen is the perfect example of an oxidizing agent because it increases the oxidation state of almost any substance with which it reacts. In the course of its reactions, oxygen is reduced. The substances it reacts with are therefore reducing agents.
Exercise N.2 Identify the oxidizing agents and reducing agents in the following reactions. (a) Fe2O3(s) 3 C(s) n 2 Fe(s) 3 CO(g) (b) CH4(g) 2 O2(g) n CO2(g) 2 H2O(g) Solution (a) In this reaction, carbon reduces Fe2O3 to iron metal, which means carbon is the reducing agent. Fe2O3 oxidizes carbon to CO and is therefore the oxidizing agent. Fe2O3 3 C 3
2 Fe 3 CO
0
0 Reduction Oxidation
2
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(b) Oxygen oxidizes CH4 in the reaction, so O2 is the oxidizing agent. The oxygen is reduced by CH4, which means that CH4 is the reducing agent. CH4 2 O2
CO2 2 H2O
4
42
0 Oxidation
2
Reduction
Each year between 75 and 80 quads, or quadrillion (1015) BTU (British thermal units), of energy is consumed in the United States.1 Less than 10% of this energy is provided by nuclear, solar, geothermal, or hydro power. The rest can be traced to a combustion reaction in which a fuel is oxidized by O2. The cars, trucks, and buses that fill our highways are powered by gasoline engines that burn hydrocarbons such as octane, C8H18, 2 C8H18(l) 25 O2(g) 88n 16 CO2(g) 18 H2O(g) or diesel engines that burn larger hydrocarbons such as cetane, C16H34. 2 C16H34(l) 49 O2(g) 88n 32 CO2(g) 34 H2O(g) We heat our homes by burning the methane (CH4) in natural gas, the high molecular weight hydrocarbons in fuel oil, or the hydrocarbons in wood, or by using electricity generated in a power plant that burns either oil or coal. The energy we use to fuel our bodies also comes from combustion reactions. Energy enters our bodies in the form of lipids, proteins, and carbohydrates. These “fuels” are converted into carbohydrates, such as glucose (C6H12O6), which react with oxygen to produce the energy we need to survive. C6H12O6(aq) 6 O2(g) 88n 6 CO2(g) 6 H2O(l) About 65% of the energy given off in the reaction is used to synthesize the ATP (adenosine triphosphate) that fuels biological processes. The remaining 35% is released as the heat that keeps our body temperatures higher than the temperature of the surroundings. There is an ever-growing awareness that the earth contains a finite amount of fossil fuels, such as oil and coal. Nuclear, solar, and geothermal power will be increasingly important sources of energy. But they won’t replace fossil fuels by themselves because they are used to produce electrical energy, which is difficult to store. One possible solution to this problem has been labeled the hydrogen economy. The first step in the hydrogen economy is to use energy from nuclear, solar, or geothermal power to split water into its elements. 2 H2O(l) 88n 2 H2(g) O2(g) The oxygen is then released to the atmosphere, and the hydrogen is either burned as a fuel 2 H2(g) O2(g) 88n 2 H2O(g)
H° 483.64 kJ/molrxn
One BTU is the energy needed to raise the temperature of one pound of water by 1°F; 1 BTU 1.055 kJ.
1
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or used to reduce carbon monoxide to methanol or to gasoline, which can be stored and later burned as a fuel. CO(g) 2 H2(g) 88n CH3OH(l) 8 CO(g) 17 H2(g) 88n C8H18(l) 8 H2O(l)
Peroxides It takes four electrons to reduce an O2 molecule to a pair of O2 ions. If the reaction stops after the O2 molecule has gained only two electrons, the O22 ion is produced. O O 2 e
[ OOO ] 2
The O22 ion has two more electrons than a neutral O2 molecule, which means that the oxygen atoms must share only a single pair of bonding electrons to achieve an octet of valence electrons. The O22 ion is called the peroxide ion because compounds that contain the ion are unusually rich in oxygen. They are not just oxides—they are (hy-)peroxides. The easiest way to prepare a peroxide is to react sodium or barium metal with oxygen. 2 Na(s) O2(g) 88n Na2O2(s) Ba(s) O2(g) 88n BaO2(s) When the peroxides are allowed to react with a strong acid, hydrogen peroxide (H2O2) is produced. BaO2(s) 2 H(aq) 88n Ba2(aq) H2O2(aq) The Lewis structure of hydrogen peroxide contains an OOO single bond.
The electron domain theory predicts that the geometry around each oxygen atom in H2O2 should be bent. But this theory can’t predict whether the four atoms should lie in the same plane or whether the molecule should be visualized as lying in two intersecting planes. The experimentally determined structure of H2O2 is shown in Figure N.7.
H 111.5° 94.8°
H
FIGURE N.7 Geometry of an H2O2 molecule.
The HOOOO bond angle in the molecule is only slightly larger than the angle between a pair of adjacent 2p atomic orbitals on the oxygen atom, and the angle between the planes that form the molecule is slightly larger than the tetrahedral angle. The geometry around each oxygen atom is bent, or angular, with an OOOOH bond angle of 94.8°. In order to keep the nonbonding electrons on the oxygen atoms as far apart as possible, the four atoms in the molecule lie in two planes that intersect at an angle of 111.5°.
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The oxidation number of the oxygen atoms in hydrogen peroxide is 1. H2O2 can therefore act as an oxidizing agent and capture two more electrons to form a pair of hydroxide ions, in which the oxygen has an oxidation number of 2. H2O2 2 e 88n 2 OH Or, it can act as a reducing agent and lose a pair of electrons to form an O2 molecule. H2O2 88n O2 2 H 2 e
Checkpoint Use Lewis structures to explain what happens in the following reactions. H2O2 2 e 88n 2 OH H2O2 88n O2 2 H 2 e Reactions in which a compound simultaneously undergoes both oxidation and reduction are called disproportionation reactions. The products of the disproportionation of H2O2 are oxygen and water. 2 H2O2(aq) 88n O2(g) 2 H2O(l)
Exercise N.3 Use the reactions that describe what happens when H2O2 loses a pair of electrons and what happens when H2O2 picks up a pair of electrons to explain why the disproportionation of hydrogen peroxide gives oxygen and water. 2 H2O2(aq) 88n O2(g) 2 H2O(l) Solution Adding the half-reaction for the oxidation of H2O2 to the half-reaction for the reduction of the compound gives the following results. H2O2 2 e 88n 2 OH H2O2 88n O2 2 H 2 e 2 H2O2 88n O2 2 H 2 OH The H and OH ions produced in the two halves of the reaction combine to form water to give the following overall stoichiometry for the reaction. 2 H2O2 88n O2 2 H2O
The disproportionation of H2O2 is an exothermic reaction. 2 H2O2(aq) 88n O2(g) 2 H2O(l)
H° 189.3 kJ/molrxn
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The reaction is relatively slow, however, in the absence of a catalyst. The principal uses of H2O2 revolve around its oxidizing ability. It is used in dilute (3%) solutions as a disinfectant. In more concentrated solutions (30%), it is used as a bleaching agent for hair, fur, leather, or the wood pulp used to make paper. In very concentrated (40–70%) solutions, H2O2 has been used as rocket fuel because of the ease with which it decomposes to give O2.
Methods of Preparing O2 Small quantities of O2 gas can be prepared in a number of ways. •
By decomposing a dilute solution of hydrogen peroxide with dust or a metal surface as the catalyst. 2 H2O2(aq) 88n O2(g) 2 H2O(l)
•
By reacting hydrogen peroxide with a strong oxidizing agent, such as the permanganate ion, MnO4. 5 H2O2(aq) 2 MnO4(aq) 6 H(aq) 88n 2 Mn2(aq) 5 O2(g) 8 H2O(l)
•
By passing an electric current through water. 2 H2O(l) electrolysis 888888n 2 H2(g) O2(g)
•
By heating potassium chlorate in the presence of a catalyst until it decomposes. MnO2 2 KClO3(s) 888n 2 KCl(s) 3 O2(g)
Checkpoint Which of the following elements or compounds reacts with water to give a solution that could be used to produce O2? (a) Na (b) Na2O (c) Na2O2 (d) NaOH (e) NaCl
Chemistry in the World Around Us The Chemistry of the Atmosphere Although major changes occurred in the atmosphere during the early history of our planet, the chemistry of the earth’s atmosphere has been more or less constant during the time in which the human race evolved. This is no longer true. The amount of methane (CH4) in the atmosphere is increasing at a rate of more than 1% per year. The concentration of carbon dioxide has more than doubled since 1750 and seems to be increasing at an exponential rate (see Figure N.8). In recent years, attention has been focused on one particular change in the atmosphere, the depletion of ozone above the Antarctic continent. For over 25 years, a team of scientists collected data on variations in the amount of O3 at different altitudes above the British Antarctic Survey station at Halley Bay. In 1985, they reported that the O3 concentration declined after the return of solar radiation
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CO2 concentration (ppmv)
340 330 320 310 300 290 280 270 1700
1750
1800
1850 Year
1900
1950
2000
FIGURE N.8 The CO2 concentration in the atmosphere from the time of the industrial revolution to present in units of parts per million by volume (ppmv).
each September. The effect was small when first noticed in the late 1970s, but it reached such high levels by 1984 that the O3 concentration declined by 30% by the end of October of that year. The data from Halley Bay sampled only a small portion of the atmosphere over the Antarctic. Once the notion of ozone depletion was reported, however, scientists were able to retrieve data that had been recorded by satellites over a period of years, which confirmed that the same effect occurred over virtually the entire Antarctic continent. Several features of the ozone depletion were particularly interesting. The drop in the ozone concentration occurred very rapidly, within a period of six weeks each spring. (“Spring” occurs during September and October in the southern hemisphere.) • Although the drop in O3 occurred above the Antarctic, it corresponds to a loss of 3% of the total ozone concentration in the planet’s atmosphere. • The decline in the O3 concentration became more serious each year. By 1989, the O3 concentration during the summer months dropped by 70%. • The decline was temporary. During the winter months, the ozone level built back to normal levels. •
Several possible explanations were available for the ozone holes. One of the most popular was the suggestion by Molina and Rowland that the ozone in the atmosphere could be destroyed by Cl and ClO radicals created when chlorofluorocarbons (CFC’s) in the atmosphere decomposed. The question was: What evidence could be found to either support or refute that hypothesis? The problem is complex, because more than 200 chemical reactions have been included in the models used to explain the chemistry of the atmosphere. It is further complicated by the fact that ClO radicals exist in the atmosphere at concentrations of only about 1 part per trillion by volume (pptv). In the 4 January 1991 issue of Science, James Anderson and co-workers reported data that probed the link between the release of chlorofluorocarbons into the atmosphere and the disappearance of ozone from the stratosphere above the Antarctic each spring. The data were obtained between 23 August and 22 September 1987, using special instruments mounted in high-altitude aircraft. Initial measurements after the plane took off from the tip of Chile (54° S latitude) suggested that the background concentration of ClO radicals was at the threshold of detection: about 1 pptv. As the plane flew toward the south pole, the ClO concentration increased slightly until about 65° S latitude, when it rapidly increased. A plot of the concentration of ClO radicals versus latitude for the 16 September flight is shown in Figure N.9. Once the aircraft reached a latitude of 68° S the abundance of the radicals increased by three orders of magnitude, to a level of approximately 1200 pptv. These data by themselves are suggestive, but Figure N.9 contains more compelling
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16 SEPT
1200
3000 2000
600 O3
1000
ClO 0 0 62 64 66 68 70 72 Latitude (degrees south)
O3 concentration in ppbv
ClO • concentration in pptv
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FIGURE N.9 Variations in the ClO and O3 concentrations in the atmosphere during the September 16, 1987, flight toward the Antarctic continent.
evidence. The plot at the top of Figure N.9 shows that the concentration of O3 dropped by a factor of about 2.5 at virtually the same time that the ClO concentration increased. Anderson and co-workers concluded as follows: “When taken independently, each element in the case contains a segment of the puzzle that in itself is not conclusive. When taken together, however, they provide convincing evidence that the dramatic reduction in O3 over the Antarctic continent would not have occurred had CFC’s not been synthesized and then added to the atmosphere.”2
N.4 THE CHEMISTRY OF SULFUR Because sulfur is directly below oxygen in the periodic table, these elements have similar electron configurations. O S
[He] 2s2 2p4 [Ne] 3s2 3p4
As a result, sulfur forms many compounds that are analogs of oxygen compounds, as shown in Table N.2. The last two examples in Table N.2 show how the prefix thio- can be used to describe compounds in which sulfur replaces an oxygen atom. TABLE N.2 Oxygen Compounds and Their Sulfur Analogs Oxygen Compound
Sulfur Compound
Na2O (sodium oxide) H2O (water) O3 (ozone) CO2 (carbon dioxide) OCN (cyanate) OC(NH2)2 (urea)
Na2S (sodium sulfide) H2S (hydrogen sulfide) SO2 (sulfur dioxide) CS2 (carbon disulfide) SCN (thiocyanate) SC(NH2)2 (thiourea)
There are four principal differences between the chemistry of sulfur and oxygen. OPO double bonds are much stronger than SPS double bonds. SOS single bonds are almost twice as strong as OOO single bonds. • Sulfur is much less electronegative than oxygen. • Sulfur can expand its valence shell to hold more than eight electrons; oxygen cannot. • •
2
J. G. Anderson, D. W. Toohey, and W. H. Brune, Science, 251, 45 (1991).
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These seemingly minor differences have important consequences for the chemistry of these elements.
OX and XP PX Bonds The Effect of Differences in the Strength of XO The radius of a sulfur atom is about 60% larger than that of an oxygen atom. 0.104 nm Covalent radius of sulfur 1.6 Covalent radius of oxygen 0.066 nm As a result, it is harder for sulfur atoms to come close enough together to form double bonds. SPS double bonds are therefore much weaker than OPO double bonds. Double bonds between sulfur and oxygen or carbon atoms can be found in compounds such as SO2 and CS2. But these double bonds are much weaker than the equivalent double bonds to oxygen atoms in O3 or CO2. The CPS double bonds in CS2, for example, are about 65% as strong as the CPO double bonds in CO2. Elemental oxygen consists of O2 molecules in which each atom completes its octet of valence electrons by sharing two pairs of electrons with a single neighboring atom. Because sulfur doesn’t form strong SPS double bonds, elemental sulfur consists of cyclic S8 molecules in which each atom completes its octet by forming single bonds to two different neighboring atoms, as shown in Figure N.10.
S S
S
S
S
S S
S S8
FIGURE N.10 Structure of a cyclic S8 molecule.
S8 molecules can pack to form more than one crystal. The most stable form of sulfur consists of orthorhombic crystals of S8 molecules, which are often found near volcanos. If the orthorhombic crystals are heated until they melt and the molten sulfur is then cooled, an allotrope of sulfur consisting of monoclinic crystals of S8 molecules is formed. The monoclinic crystals slowly transform themselves into the more stable orthorhombic structure over a period of time. The tendency of an element to form bonds to itself is called catenation (from the Latin word catena, “chain”). Because sulfur forms unusually strong SOS single bonds, it is better at catenation than any element except carbon. As a result, the orthorhombic and monoclinic forms of sulfur are not the only allotropes of the element. Allotropes of sulfur also exist that differ in the size of the molecules that form the crystal. Cyclic molecules that contain 6, 7, 8, 10, and 12 sulfur atoms are known. Sulfur melts at 119.25°C to form a yellow liquid that is less viscous than water. If the liquid is heated to 159°C, it turns into a dark red liquid that can’t be poured from its container. The viscosity of the dark red liquid is 2000 times greater than that of molten sulfur because the cyclic S8 molecules open up and link together to form long chains of as many as 100,000 sulfur atoms. When sulfur reacts with an active metal, it can form the sulfide ion, S2. 16 K(s) S8(s) 88n 8 K2S(s)
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This is not the only product that can be obtained, however. A variety of polysulfide ions with a charge of 2 can be produced that differ in the number of sulfur atoms in the chain. 2 K(s) S8(s) 88n K2S2 [K]2[SOS]2 K2S3 [K]2[SOSOS]2 K2S4 [K]2[SOSOSOS]2 K2S5 [K]2[SOSOSOSOS]2 K2S6 [K]2[SOSOSOSOSOS]2 K2S8 [K]2[SOSOSOSOSOSOSOS]2
Exercise N.4 Use the tendency of sulfur to form polysulfide ions to explain why iron has an oxidation number of 2 in iron pyrite, FeS2, one of the most abundant sulfur ores. Solution If the oxidation number of iron in FeS2 is 2, the sulfur must be present as the S22 ion. FeS2 [Fe2][S22] The disulfide ion, S22, is the sulfur analog of the peroxide ion, O22, and has an analogous Lewis structure.
The Effect of Differences in the Electronegativities of Sulfur and Oxygen Because sulfur is much less electronegative than oxygen, it is more likely to form compounds in which it has a positive oxidation number (see Table N.3). TABLE N.3 for Sulfur
Common Oxidation Numbers
Oxidation Number
Examples
2 1 0 1 2 1 22 3 4 5 6
Na2S, H2S Na2S2, H2S2 S8 S2Cl2 S2O32 S4O62 S2O42 SF4, SO2, H2SO3, SO32 S2O62 SF6, SO3, H2SO4, SO42
In theory, sulfur can react with oxygen to form either SO2 or SO3, whose Lewis structures are given in Figure N.11. SO2 is therefore a resonance hybrid of two Lewis structures analogous to the structure of ozone in Figure N.6. SO3 can be thought to result from the donation of a pair of nonbonding electrons on the sulfur atom in SO2 to an empty orbital on a neutral oxygen atom to form a covalent bond.
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S
SO2
O
SO3 O
S O
O
O
O
O
O
S
S
S
O
O
O
O
O
FIGURE N.11 Lewis structures of SO2 and SO3.
In practice, combustion of sulfur compounds gives SO2, regardless of whether sulfur or a compound of sulfur is burned. S8(s) 8 O2(g) 88n 8 SO2(g) CS2(l) 3 O2(g) 88n CO2(g) 2 SO2(g) 3 FeS2(s) 8 O2(g) 88n Fe3O4(s) 6 SO2(g) Although the SO2 formed in the reactions should react with O2 to form SO3, the rate of that reaction is very slow. The rate of the conversion of SO2 into SO3 can be greatly increased by adding an appropriate catalyst. 2O5 /K2O 2 SO2(g) O2(g) V888888 n 2 SO3(g)
Enormous quantities of SO2 are produced by industry each year and then converted to SO3, which can be used to produce sulfuric acid, H2SO4. In theory, sulfuric acid can be made by dissolving SO3 gas in water. SO3(g) H2O(l) 88n H2SO4(aq) In practice, this isn’t convenient. Instead, SO3 is absorbed in 98% H2SO4, where it reacts with the water to form additional H2SO4 molecules. Water is then added, as needed, to keep the concentration of the solution between 96% and 98% H2SO4 by weight. Sulfuric acid is by far the most important industrial chemical. It has even been suggested that there is a correlation between the amount of sulfuric acid a country consumes and its standard of living. Sulfuric acid dissociates in water to give the HSO4 ion, which is known as the hydrogen sulfate, or bisulfate, ion. H2SO4(aq) 88n H(aq) HSO4(aq) Roughly 10% of the hydrogen sulfate ions dissociate further to give the SO42, or sulfate, ion. HSO4(aq) 88n H(aq) SO42(aq) Sulfur dioxide dissolves in water to form sulfurous acid. SO2(g) H2O(l) 88n H2SO3(aq) Sulfurous acid doesn’t dissociate in water to as great an extent as sulfuric acid. Sulfuric acid and sulfurous acid are examples of a class of compounds known as oxyacids because they are literally acids that contain oxygen. Because they are negative ions
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(or anions) that contain oxygen, the SO32 and SO42 ions are known as oxyanions. The Lewis structures of some of the oxides of sulfur that form oxyacids and oxyanions are given in Figure N.12. One of the oxyanions deserves special mention. This ion, which is known as the thiosulfate ion, is formed by the reaction between sulfur and the sulfite (SO32) ion. 8 SO32(aq) S8(s) 88n 8 S2O32(aq) Oxyacids
Oxyanions
O
O
H O
S
O H
O
2–
O
S
O
O
Sulfuric acid, H2SO4
Sulfate, SO42–
S
S
H O
S
O H
O
S
2–
O
O
O
Thiosulfuric acid, H2S2O3
Thiosulfate, S2O32–
O H O
S
2–
O O H
O
O
S
Sulfite, SO32–
Sulfurous acid, H2SO3
2–
O O
S
O S
S
O
S
O
O
Tetrathionate, S4O62– 2–
O H O
S O
O
O O O
S
O H
O
Peroxydisulfuric acid, H2S2O8
O
S O
O O O
S
O
O
Peroxydisulfate, S2O82–
FIGURE N.12 Oxyacids of sulfur and their oxyanions.
Checkpoint Use the following Lewis structures to explain why the S2O32 ion is literally a thiosulfate.
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The Effect of Differences in the Abilities of Sulfur and Oxygen to Expand Their Valence Shell Oxygen reacts with fluorine to form OF2. O2(g) 2 F2(g) 88n 2 OF2(g) The reaction stops at this point because oxygen can hold only eight electrons in its valence shell.
Sulfur, however, reacts with fluorine to form SF4 and SF6 because sulfur can expand its valence shell to hold 10 or even 12 electrons. S8(s) 16 F2(g) 88n 8 SF4(g) S8(s) 24 F2(g) 88n 8 SF6(g) There are 10 valence electrons on the sulfur atom in SF4, so the structure of the molecule is based on a distorted trigonal bipyramid, as shown in Figure N.13. The 12 valence electrons on the central atom in SF6 are distributed toward the corners of an octahedron.
F F
F S
101.6° F
F S
F F
SF4
F
186.9°
F
F
SF6
FIGURE N.13 Structures of SF4 and SF6.
N.5 THE CHEMISTRY OF NITROGEN The chemistry of nitrogen is dominated by the ease with which nitrogen atoms form double and triple bonds. A neutral nitrogen atom contains five valence electrons: 2s2 2p3. A nitrogen atom can therefore achieve an octet of valence electrons by sharing three pairs of electrons with another nitrogen atom.
Because the covalent radius of a nitrogen atom is relatively small, nitrogen atoms come close enough together to form very strong multiple bonds. The NqN triple bond (H°ac 945.41 kJ/molrxn) is almost twice as strong as the OPO double bond (H°ac 498.34 kJ/molrxn). The strength of the NqN triple bond makes the N2 molecule so inert that lithium is one of the few elements with which it reacts at room temperature. 6 Li(s) N2(g) 88n 2 Li3N(s) In spite of the fact that the N2 molecule is unreactive, compounds containing nitrogen exist for virtually every element in the periodic table except those in Group VIIIA (He, Ne,
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Ar, and so on). This can be explained in two ways. First, N2 becomes significantly more reactive as the temperature increases. At high temperatures, nitrogen reacts with hydrogen to form ammonia and with oxygen to form nitrogen oxide. N2(g) 3 H2(g) 88n 2 NH3(g) N2(g) O2(g) 88n 2 NO(g) Second, a number of catalysts found in nature can overcome the inertness of N2 at low temperatures.
The Synthesis of Ammonia It is difficult to imagine a living system that doesn’t contain nitrogen, which is an essential component of the proteins, nucleic acids, vitamins, and hormones that make life possible. Animals pick up the nitrogen they need from the plants or other animals in their diet. Although there is an abundance of N2 in the atmosphere, plants cannot use nitrogen in its elemental form. Plants have to pick up their nitrogen from the soil or absorb it as N2 from the atmosphere. The concentration of nitrogen in the soil is fairly small, so the process by which plants reduce N2 to NH3 (nitrogen fixation) is extremely important. Although 200 million tons of NH3 are produced by nitrogen fixation each year, plants, by themselves, cannot reduce N2 to NH3. The reaction is carried out by blue-green algae and bacteria that are associated with certain plants. The best understood example of nitrogen fixation involves the Rhizobium bacteria found in the root nodules of legumes such as clover, peas, and beans. The bacteria contain a nitrogenase enzyme that is capable of the remarkable feat of reducing N2 from the atmosphere to NH3 at room temperature. Ammonia is made on an industrial scale by a process first developed between 1909 and 1913 by Fritz Haber. In the Haber process, a mixture of N2 and H2 gas at 200 to 300 atm and 400°C to 600°C is passed over a catalyst of finely divided iron metal. Fe N2(g) 3 H2(g) 88n 2 NH3(g)
Almost 20 million tons of NH3 are produced in the United States each year by this process. About 80% of it, worth more than $2 billion, is used to make fertilizers for plants that can’t fix nitrogen from the atmosphere. On the basis of weight, ammonia is the second most important industrial chemical in the United States. (Only sulfuric acid is produced in larger quantities.) Two-thirds of the ammonia used for fertilizers is converted into solids such as ammonium nitrate, NH4NO3; ammonium phosphate, (NH4)3PO4; ammonium sulfate, (NH4)2SO4; and urea, H2NCONH2. The other third is applied directly to the soil as anhydrous (literally, “without water”) ammonia. Ammonia is a gas at room temperature. It can be handled as a liquid when dissolved in water to form an aqueous solution. Alternatively, it can be cooled to temperatures below 33°C, in which case the gas condenses to form the anhydrous liquid, NH3(l).
The Synthesis of Nitric Acid The NH3 produced by the Haber process that isn’t used as fertilizer is burned in oxygen to generate nitrogen oxide. 4 NH3(g) 5 O2(g) 88n 4 NO(g) 6 H2O(g)
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Nitrogen oxide—or nitric oxide, as it was once known—is a colorless gas that reacts rapidly with oxygen to produce nitrogen dioxide, a dark brown gas. 2 NO(g) O2(g) 88n 2 NO2(g) Nitrogen dioxide dissolves in water to give nitric acid and NO, which can be captured and recycled. 3 NO2(g) H2O(l) 88n 2 HNO3(aq) NO(g) Thus, by a three-step process developed by Friedrich Ostwald in 1908, ammonia can be converted into nitric acid. 4 NH3(g) 5 O2(g) 88n 4 NO(g) 6 H2O(g) 2 NO(g) O2(g) 88n 2 NO2(g) 3 NO2(g) H2O(l) 88n 2 HNO3(aq) NO(g) The Haber process for the synthesis of ammonia combined with the Ostwald process for the conversion of ammonia into nitric acid revolutionized the explosives industry. Nitrates have been important explosives ever since Friar Roger Bacon mixed sulfur, saltpeter, and powdered carbon to make gunpowder in 1245. 16 KNO3(s) S8(s) 24 C(s) 88n 8 K2S(s) 24 CO2(g) 8 N2(g) H° 4575 kJ/molrxn Before the Ostwald process the only source of nitrates for use in explosives was naturally occurring minerals such as saltpeter, which is a mixture of NaNO3 and KNO3. Once a dependable supply of nitric acid became available from the Ostwald process, a number of nitrates could be made for use as explosives. Combining NH3 from the Haber process with HNO3 from the Ostwald process, for example, gives ammonium nitrate, which is both an excellent fertilizer and an inexpensive, dependable explosive commonly used in blasting powder. 2 NH4NO3(s) 88n 2 N2(g) O2(g) 4 H2O(g) The destructive power of ammonium nitrate is apparent in photographs of the Alfred P. Murrah Federal Building in Oklahoma City, which was destroyed with a bomb made from ammonium nitrate on April 19, 1995.
Intermediate Oxidation Numbers Nitric acid (HNO3) and ammonia (NH3) represent the maximum (5) and minimum (3) oxidation numbers for nitrogen. Nitrogen also forms compounds with every oxidation number between these extremes (see Table N.4).
Negative Oxidation Numbers of Nitrogen besides 3 At about the time that Haber developed the process for making ammonia and Ostwald worked out the process for converting ammonia into nitric acid, Raschig developed a process that used the hypochlorite ion (OCl) to oxidize ammonia to produce hydrazine, N2H4. 2 NH3(aq) OCl(aq) 88n N2H4(aq) Cl(aq) H2O(l)
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TABLE N.4 Common Oxidation Numbers for Nitrogen Oxidation Number
Examples
3 2 1 1 3 0 1 2 3 4 5
NH3, NH4, NH2, Mg3N2 N2H4 NH2OH NaN3, HN3 N2 N2O NO, N2O2 HNO2, NO2, N2O3, NO NO2, N2O4 HNO3, NO3, N2O5
This reaction can be understood by noting that the OCl ion is a two-electron oxidizing agent. Let’s therefore imagine a hypothetical mechanism for the reaction in which the first step is the loss of a pair of nonbonding electrons from an ammonia molecule to form an NH32 ion, as shown in the first step in Figure N.14. A pair of nonbonding electrons from a second NH3 molecule could then be donated into the empty valence shell orbital on the NH32 ion to form an NON bond. In step 3, the product of the reaction in step 2 could then lose a pair of H ions to form a hydrazine molecule. H H N H
Step 1
Step 2
H H N H
Step 3
2+
H + N H H
H H H N N H H H
2+
–2 e–
H H N H
2+
H H H N N H H H H
H N N
H
2+
+ 2 H
H
FIGURE N.14 Hydrazine is prepared by reacting NH3 with a two-electron oxidizing agent.
Hydrazine is a colorless liquid with a faint odor of ammonia that can be collected when the solution is heated until N2H4 distills out of the reaction flask. Many of the physical properties of hydrazine are similar to those of water.
Density Melting Point Boiling Point
H2O
N2H4
1.000 g/cm3 0.00°C 100°C
1.008 g/cm3 1.54°C 113.8°C
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There is a significant difference between the chemical properties of the compounds, however. Hydrazine burns when ignited in air to give nitrogen gas, water vapor, and large amounts of energy. N2H4(l) O2(g) 88n N2(g) 2 H2O(g)
H° 534.3 kJ/molrxn
The principal use of hydrazine is as a rocket fuel. It is second only to liquid hydrogen in terms of the kilograms of thrust produced per kilogram of fuel burned. Hydrazine has several advantages over liquid H2, however. It can be stored at room temperature, whereas liquid hydrogen must be stored at temperatures below 253°C. Hydrazine is also more dense than liquid H2 and therefore requires less storage space. Pure hydrazine is seldom used as a rocket fuel because it freezes at the temperatures encountered in the upper atmosphere. Hydrazine is mixed with N,N-dimethylhydrazine, (CH3)2NNH2, to form a solution that remains a liquid at low temperatures. Mixtures of hydrazine and N,N-dimethylhydrazine were used to fuel the Titan II rockets that carried the Project Gemini spacecraft, and the reaction between hydrazine derivatives and N2O4 is still used to fuel the small rocket engines that enable the space shuttle to maneuver in space. The product of the combustion of hydrazine is unusual. When carbon compounds burn, the carbon is oxidized to CO or CO2. When sulfur compounds burn, SO2 is produced. When hydrazine is burned, the product of the reaction is N2 because of the unusually strong NqN triple bond in the N2 molecule. N2H4(l) O2(g) 88n N2(g) 2 H2O(g)
Positive Oxidation Numbers for Nitrogen: The Nitrogen Halides Fluorine, oxygen, and chlorine are the only elements more electronegative than nitrogen that form compounds with nitrogen. As a result, positive oxidation numbers of nitrogen are found in compounds that contain one or more of these elements. In theory, N2 could react with F2 to form a compound with the formula NF3. In practice, N2 is too inert to undergo the reaction at room temperature. NF3 is made by reacting ammonia with F2 in the presence of a copper metal catalyst. The HF produced in the reaction combines with ammonia to form ammonium fluoride. The overall stoichiometry for the reaction is therefore written as follows. Cu 4 NH3(g) 3 F2(g) 88n NF3(g) 3 NH4F(s)
The Lewis structure of NF3 is analogous to the Lewis structure of NH3, and the molecules have similar shapes. Ammonia reacts with chlorine to form NCl3, which seems at first glance to be closely related to NF3. But there is a significant difference between the compounds. NF3 is essentially inert at room temperature, whereas NCl3 is a shock-sensitive, highly explosive liquid that decomposes to form N2 and Cl2. 2 NCl3(l) 88n N2(g) 3 Cl2(g) Ammonia reacts with iodine to form a solid that is a complex between NI3 and NH3. This material is the subject of a popular, but dangerous, demonstration in which freshly prepared samples of NI3 in ammonia are poured onto filter paper, which is allowed to dry on a ring stand. After the ammonia evaporates, the NH3/NI3 crystals are touched with a feather
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attached to a meter stick, resulting in detonation of the shock-sensitive solid, which decomposes to form a mixture of N2 and I2. 2 NI3(s) 88n N2(g) 3 I2(g)
Positive Oxidation Numbers for Nitrogen: The Nitrogen Oxides Lewis structures for seven oxides of nitrogen with oxidation numbers ranging from 1 to 5 are given in Figure N.15. These compounds all have two things in common: they contain NPO double bonds, and they are less stable than their elements in the gas phase.
N N O
N N O
Dinitrogen oxide, N2O (nitrous oxide)
N O N O
O N
Nitrogen oxide, NO (nitric oxide)
Dinitrogen dioxide, N2O2
O N N O
O
Dinitrogen trioxide, N2O3
O
O
N
O N N
O
O
Nitrogen dioxide, NO2
O
Dinitrogen tetroxide, N2O4
O
O N O
O
N O
Dinitrogen pentoxide, N2O5
FIGURE N.15 The oxides of nitrogen.
Dinitrogen oxide, N2O, which is also known as nitrous oxide, can be prepared by carefully decomposing ammonium nitrate. 170–200°C NH4NO3(s) 888888n N2O(g) 2 H2O(g)
Nitrous oxide is a sweet-smelling, colorless gas best known to nonchemists as “laughing gas.” As early as 1800, Humphry Davy noted that N2O, inhaled in relatively small amounts, produced a state of apparent intoxication often accompanied by either convulsive laughter or crying. When taken in larger doses, nitrous oxide provides fast and efficient relief from pain. N2O was therefore used as the first anesthetic. Because large doses are needed
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to produce anesthesia, and continued exposure to the gas can be fatal, N2O is used today only for relatively short operations. Nitrous oxide has several other interesting properties. First, it is highly soluble in cream; for that reason, it is used as the propellant in whipped cream dispensers. Second, although N2O does not burn by itself, it is better than air at supporting the combustion of other objects. This can be explained by noting that N2O can decompose to form an atmosphere that is one-third O2 by volume, whereas normal air is only 21% oxygen by volume. 2 N2O(g) 88n 2 N2(g) O2(g) For many years, the endings -ous and -ic were used to distinguish between the lowest and highest of a pair of oxidation numbers. N2O is nitrous oxide because the oxidation number of the nitrogen is 1. NO is nitric oxide because the oxidation number of the nitrogen is 2. Enormous quantities of nitrogen oxide, or nitric oxide, are generated each year by the reaction between the N2 and O2 in the atmosphere, catalyzed by a stroke of lightning passing through the atmosphere or by the hot walls of an internal combustion engine. N2(g) O2(g) 88n 2 NO(g) One of the reasons for lowering the compression ratio of automobile engines in recent years is to decrease the temperature of the combustion reaction, thereby decreasing the amount of NO emitted into the atmosphere. NO can be prepared in the laboratory by reacting copper metal with dilute nitric acid. 3 Cu(s) 8 HNO3(aq) 88n 3 Cu(NO3)2(aq) 2 NO(g) 4 H2O(l) The NO molecule contains an odd number of valence electrons. As a result, it is impossible to write a Lewis structure for the molecule in which all of the electrons are paired (see Figure N.15). When NO gas is cooled, pairs of NO molecules combine in a reversible reaction to form a dimer (from Greek, meaning “two parts”), with the formula N2O2, in which all of the valence electrons are paired, as shown in Figure N.15. NO reacts rapidly with O2 to form nitrogen dioxide (once known as nitrogen peroxide), which is a dark brown gas at room temperature. 2 NO(g) O2(g) 88n 2 NO2(g) NO2 can be prepared in the laboratory by heating certain metal nitrates until they decompose. 2 Pb(NO3)2(s) 88n 2 PbO(s) 4 NO2(g) O2(g) It can also be made by reacting copper metal with concentrated nitric acid. Cu(s) 4 HNO3(aq) 88n Cu(NO3)2(aq) 2 NO2(g) 2 H2O(l) NO2 also has an odd number of electrons and therefore contains at least one unpaired electron in its Lewis structures. NO2 dimerizes at low temperatures to form N2O4 molecules, in which all the electrons are paired, as shown in Figure N.15. Mixtures of NO and NO2 combine when cooled to form dinitrogen trioxide, N2O3, which is a blue liquid. The formation of a blue liquid when either NO or NO2 is cooled
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therefore implies the presence of at least a small portion of the other oxide because N2O2 and N2O4 are both colorless. By carefully removing water from concentrated nitric acid at low temperatures with a dehydrating agent we can form dinitrogen pentoxide. 4 HNO3(aq) P4O10(s) 88n 2 N2O5(s) 4 HPO3(s) N2O5 is a colorless solid that decomposes in light or on warming to room temperature. As might be expected, N2O5 dissolves in water to form nitric acid. N2O5(s) H2O(l) 88n 2 HNO3(aq)
N.6 THE CHEMISTRY OF PHOSPHORUS Phosphorus was the first element whose discovery can be traced to a single individual. In 1669, while searching for a way to convert silver into gold, Hennig Brand obtained a white, waxy solid that glowed in the dark and burst spontaneously into flame when exposed to air. Brand made the substance by evaporating the water from urine and allowing the black residue to putrefy for several months. He then mixed the residue with sand, heated the mixture in the presence of a minimum of air, and collected under water the volatile products that distilled out of the reaction flask. Phosphorus forms a number of compounds that are direct analogs of nitrogencontaining compounds. However, the fact that elemental nitrogen is virtually inert at room temperature, whereas elemental phosphorus can burst spontaneously into flame when exposed to air, shows that there are differences between the elements as well. Phosphorus often forms compounds with the same oxidation numbers as the analogous nitrogen compounds, but with different formulas, as shown in Table N.5.
TABLE N.5 Nitrogen and Phosphorus Compounds with the Same Oxidation Numbers but Different Formulas Nitrogen Compound
Phosphorus Compound
Oxidation Number
N2 HNO2 (nitrous acid) N2O3 HNO3 (nitric acid) NaNO3 (sodium nitrate) N2O5
P4 H3PO3 (phosphorous acid) P4O6 H3PO4 (phosphoric acid) Na3PO4 (sodium phosphate) P4O10
0 3 3 5 5 5
The same factors that explain the differences between sulfur and oxygen can be used to explain the differences between phosphorus and nitrogen. • • • •
NqN triple bonds are much stronger than PqP triple bonds. POP single bonds are stronger than NON single bonds. Phosphorus is much less electronegative than nitrogen. Phosphorus can expand its valence shell to hold more than eight electrons, but nitrogen cannot.
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OX and Xq qX Bond Strengths The Effect of Differences in the XO The ratio of the radii of phosphorus and nitrogen atoms is the same as the ratio of the radii of sulfur and oxygen atoms, within experimental error. 0.110 nm Covalent radius of phosphorus 1.6 Covalent radius of nitrogen 0.070 nm As a result, PqP triple bonds are much weaker than NqN triple bonds, for the same reason that SPS double bonds are weaker than OPO double bonds, namely, phosphorus atoms are too big to come close enough together to form strong multiple bonds. Each atom in an N2 molecule completes its octet of valence electrons by sharing three pairs of electrons with a single neighboring atom. Because phosphorus doesn’t form strong multiple bonds with itself, elemental phosphorus consists of tetrahedral P4 molecules in which each atom forms single bonds with three neighboring atoms, as shown in Figure N.16.
P
P
P P
P4, white phosphorus
FIGURE N.16 Tetrahedral P4 molecule.
Phosphorus is a white solid with a waxy appearance, which melts at 44.1°C and boils at 287°C. It is made by reducing calcium phosphate with carbon in the presence of silica (sand) at very high temperatures. 2 Ca3(PO4)2(s) 6 SiO2(s) 10 C(s) 88n 6 CaSiO3(s) P4(s) 10 CO(g) White phosphorus is stored under water because the element spontaneously bursts into flame in the presence of oxygen at temperatures only slightly above room temperature. Although phosphorus is insoluble in water, it is very soluble in carbon disulfide. Solutions of P4 in CS2 are reasonably stable; as soon as the CS2 evaporates, however, the phosphorus bursts into flame. The POPOP bond angle in a tetrahedral P4 molecule is only 60°. This very small angle produces a considerable amount of strain in the P4 molecule, which can be relieved by breaking one of the POP bonds. Phosphorus therefore forms other allotropes by opening up the P4 tetrahedron. When white phosphorus is heated to 300°C, one bond inside each P4 tetrahedron is broken, and the P4 molecules link together to form a polymer (from the Greek words pol, “many,” and meros, “parts”) with the structure shown in Figure N.17. This allotrope of phosphorus is dark red, and its presence in small traces often gives white phosphorus a light yellow color. Red phosphorus is more dense (2.16 g/cm3) than white phosphorus (1.82 g/cm3) and is much less reactive at normal temperatures.
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P
P
P
P
P
P
P
P
29
P
P
P P
Red phosphorus
FIGURE N.17 Portion of the polymeric chain in red phosphorus.
PX and NP PX Double Bonds The Effect of Differences in the Strengths of PP The size of a phosphorus atom also interferes with its ability to form double bonds to other elements such as oxygen, nitrogen, and sulfur. As a result, phosphorus tends to form compounds that contain two POO single bonds, where nitrogen would form an NPO double bond. Nitrogen forms the nitrate (NO3) ion, for example, in which it has an oxidation number of 5. When phosphorus forms an ion with the same oxidation number, it is the phosphate (PO43) ion, as shown in Figure N.18. –
O N O
3–
O O O
O
P O
FIGURE N.18 Lewis structures for the NO3 and PO43 ions.
Similarly, nitrogen forms nitric acid, HNO3, which contains an NPO double bond, whereas phosphorus forms phosphoric acid, H3PO4, which contains POO single bonds, as shown in Figure N.19.
O
O H
H O N O
O
P O H O H
FIGURE N.19 Lewis structures for nitric acid (HNO3) and phosphoric acid (H3PO4).
The Effect of Differences in the Electronegativities of Phosphorus and Nitrogen Because phosphorus is less electronegative than nitrogen, it is more likely to exhibit positive oxidation numbers. The most important oxidation numbers for phosphorus are 3, 3, and 5 (see Table N.6). TABLE N.6 Common Oxidation Numbers of Phosphorus Oxidation Number
Examples
3 3 5
Ca3P2 PF3, P4O6, H3PO3 PF5, P4O10, H3PO4, PO43
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Because it is more electronegative than most metals, phosphorus reacts with metals at elevated temperatures to form phosphides, in which it has an oxidation number of 3. 6 Ca(s) P4(s) 88n 2 Ca3P2(s) The metal phosphides react with water to produce a poisonous, highly reactive, colorless gas known as phosphine (PH3), which has one of the foulest odors the authors have encountered. Ca3P2(s) 6 H2O(l) 88n 2 PH3(g) 3 Ca2(aq) 6 OH(aq) Samples of PH3, the phosphorus analog of ammonia, are often contaminated by traces of P2H4, the phosphorus analog of hydrazine. As if the toxicity and odor of PH3 were not enough, mixtures of PH3 and P2H4 burst spontaneously into flame in the presence of oxygen. Compounds such as Ca3P2 and PH3, in which phosphorus has a negative oxidation number, are far outnumbered by compounds in which the oxidation number of phosphorus is positive. Phosphorus burns in O2 to produce P4O10 in a reaction that gives off extraordinary amounts of energy in the form of heat and light. P4(s) 5 O2(g) 88n P4O10(s)
H° 2984 kJ/molrxn
When phosphorus burns in the presence of a limited amount of O2, P4O6 is produced. P4(s) 3 O2(g) 88n P4O6(s)
H° 1640 kJ/molrxn
P4O6 consists of a tetrahedron in which an oxygen atom has been inserted into each POP bond in the P4 molecule (see Figure N.20). P4O10 has an analogous structure, with an additional oxygen atom bound to each of the four phosphorus atoms.
O
O
P
O
P
O O
O O
O
P
O
P P
P4 O6
P
O
P
O
P
O
O O
O
O
P4 O10
FIGURE N.20 Structures of P4O6 and P4O10.
P4O6 and P4O10 react with water to form phosphorous acid, H3PO3, and phosphoric acid, H3PO4, respectively. P4O6(s) 6 H2O(l) 88n 4 H3PO3(aq) P4O10(s) 6 H2O(l) 88n 4 H3PO4(aq) (P4O10 has such a high affinity for water that it is commonly used as a dehydrating agent.) Phosphorous acid, H3PO3, and phosphoric acid, H3PO4, are examples of a large class of oxyacids of phosphorus. Lewis structures for some of these oxyacids and their related oxyanions are given in Figure N.21.
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Oxyacid
Oxyanion
O
O
H O
P O H
3–
P O
O
O
O
H
3–
Phosphate, PO4
Phosphoric acid, H3PO4
O
2–
O
P O H
H
P O
H
O
O
H
2–
Phosphite, HPO3
Phosphorous acid, H3PO3
O H
H
H O
H –
O
P O P O
O
H
H
O O H
O
H O
O
O
P O P
O P
O
O
O
H
H
H
Triphosphoric acid, H5P3O10
4–
O
P O P O
O
O 4–
Diphosphate, P2O7 (pyrophosphate)
Diphosphoric acid, H4P2O7 (pyrophosphoric acid)
O
P O
Hypophosphite, H2PO2
Hypophosphorous acid, H3PO2
O
–
O
P O H
H
31
O O H
O
O
O
P O P
O P
O
O
5–
O
O 5–
Triphosphate, P3O10
FIGURE N.21 Oxyacids of phosphorus and their oxyanions.
The Effect of Differences in the Abilities of Phosphorus and Nitrogen to Expand Their Valence Shell The reaction between ammonia and fluorine stops at NF3 because nitrogen uses the 2s, 2px, 2py, and 2pz orbitals to hold valence electrons. Nitrogen atoms can therefore hold a maximum of eight valence electrons. Phosphorus, however, can expand the valence shell to hold 10 or more electrons. Thus, phosphorus can react with fluorine to form both PF3 and PF5. Phosphorus can even form the PF6 ion, in which there are 12 valence electrons on the central atom, as shown in Figure N.22.
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F PF6–
F P
F
F
F F PF5
F
F P
F
F F
FIGURE N.22 Structures of PF5 and the PF6 ion.
N.7 THE CHEMISTRY OF THE HALOGENS There are six elements in Group VIIA, the next-to-last column of the periodic table. As expected, these elements have certain properties in common. They all form diatomic molecules (H2, F2, Cl2, Br2, I2, and At2), for example, and they all form negatively charged ions (H, F, Cl, Br, I, and At). When the chemistry of these elements is discussed, hydrogen is separated from the others and astatine is ignored because it is radioactive. (The most stable isotopes of astatine have half-lives of less than a minute. As a result, the largest samples of astatine compounds studied to date have weighed less than 50 ng.) Discussions of the chemistry of the elements in Group VIIA therefore focus on four elements: fluorine, chlorine, bromine, and iodine. These elements are called the halogens (from the Greek words hals, “salt,” and gennan, “to form or generate”) because they are literally the salt formers. None of the halogens can be found in nature in their elemental form. They are invariably found as salts of the halide ions (F, Cl, Br, and I). Fluoride ions are found in minerals such as fluorite (CaF2) and cryolite (Na3AlF6). Chloride ions are found in rock salt (NaCl), in the oceans, which are roughly 2% Cl ion by weight, and in lakes that have a high salt content, such as the Great Salt Lake in Utah, which is 9% Cl ion by weight. Both bromide and iodide ions are found at low concentrations in the oceans, as well as in brine wells in Louisiana, California, and Michigan.
The Halogens in Their Elemental Form Fluorine (F2), a highly toxic, colorless gas, is the most reactive element known—so reactive that asbestos, water, and silicon burst into flame in its presence. It is so reactive it even forms compounds with Kr, Xe, and Rn, elements that were once thought to be inert. Fluorine is such a powerful oxidizing agent that it can coax elements into unusually high oxidation numbers, as in AgF2, PtF6, and IF7. Fluorine is so reactive that it is difficult to find a container in which it can be stored. F2 attacks both glass and quartz, for example, and it causes most metals to burst into flame. Fluorine is handled in equipment built out of certain alloys of copper and nickel. It still reacts with the alloys, but it forms a layer of a fluoride compound on the surface that protects the metal from further reaction. Fluorine is used in the manufacture of Teflon—or poly(tetrafluoroethylene), (C2F4)n— which is used for everything from linings for pots and pans to gaskets that are inert to chemical reactions. Chlorine (Cl2) is a highly toxic gas with a pale yellow-green color. Chlorine is a very strong oxidizing agent, which is used commercially as a bleaching agent and as a disinfectant. It is strong enough to oxidize the dyes that give wood pulp its yellow or brown color, for example, thereby bleaching out this color, and strong enough to destroy bacteria and thereby act as a germicide. Large quantities of chlorine are used each year to make solvents such as carbon tetrachloride (CCl4), chloroform (CHCl3), dichloroethylene (C2H2Cl2), and trichloroethylene (C2HCl3).
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Bromine (Br2) is a reddish-orange liquid with an unpleasant, choking odor. The name of the element, in fact, comes from the Greek stem bromos, “stench.” Bromine is used to prepare flame retardants, fire-extinguishing agents, sedatives, antiknock agents for gasoline, and insecticides. Iodine is an intensely colored solid with an almost metallic luster. The solid is relatively volatile, and it sublimes when heated to form a violet-colored gas. Iodine has been used for many years as a disinfectant in “tincture of iodine.” Iodine compounds are used as catalysts, drugs, and dyes. Silver iodide (AgI) plays an important role in the photographic process and in attempts to make rain by seeding clouds. Iodide is also added to salt to protect against goiter, an iodine deficiency disease characterized by a swelling of the thyroid gland. Some of the chemical and physical properties of the halogens are summarized in Table N.7. TABLE N.7
F2 Cl2 Br2 I2 a
Properties of F2, Cl2, Br2, and I2
Melting Point (°C)
Boiling Point (°C)
219.6 101 7.2 112.9
188.1 34.0 59.5 185.2
Color Colorless Pale green Dark red-brown Dark violet, almost black
First a Ionization Energy (kJ/mol)
Electrona Affinity (kJ/mol)
Ionica Radius (nm)
Density (g/cm3)
1681.0 1251.1 1139.9 1008.4
328.0 348.8 324.6 295.3
0.136 0.181 0.196 0.216
1.513 1.655 3.187 3.960
For the atomic species.
There is a regular increase in many of the properties of the halogens as we proceed down the column from fluorine to iodine, including the melting point, boiling point, intensity of the color of the halogen, the radius of the corresponding halide ion, and the density of the element. On the other hand, there is a regular decrease in the first ionization energy as we go down the column. As a result, there is a regular decrease in the oxidizing strength of the halogens from fluorine to iodine. F2 Cl2 Br2 I2 oxidizing strength
This trend is mirrored by an increase in the reducing strength of the corresponding halides. I Br Cl F reducing strength
Exercise N.5 Use the fact that Cl2 is a stronger oxidizing agent than Br2 to devise a way to prepare elemental bromine from an aqueous solution of the Br ion. Solution When the chemistry of the main-group metals was introduced in Chapter 5, we found that metals can be prepared by reacting one of their salts with a metal that is a stronger reducing agent. Titanium metal, for example, can be prepared by reacting TiCl4 with magnesium metal. TiCl4(l) 2 Mg(s) 88n Ti(s) 2 MgCl2(s)
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Extending this argument to oxidizing agents suggests that we can produce Br2 by reacting a solution that contains the Br ion with something that is an even stronger oxidizing agent than Br2, such as Cl2 dissolved in water. 2 Br(aq) Cl2(aq) 88n Br2(aq) 2 Cl(aq)
Methods of Preparing the Halogens from Their Halides The halogens can be made by reacting a solution of the corresponding halide ion with a substance that is a stronger oxidizing agent than the halogen being prepared. Iodine, for example, can be made by reacting the iodide ion with either bromine or chlorine. 2 I(aq) Br2(aq) 88n I2(aq) 2 Br(aq) Bromine can be prepared by reacting bromide ions with a solution of Cl2 dissolved in water. 2 Br(aq) Cl2(aq) 88n Br2(aq) 2 Cl(aq) To prepare Cl2, we need an unusually strong oxidizing agent, such as manganese dioxide (MnO2) or the permanganate ion (MnO4). 2 Cl(aq) MnO2(aq) 4 H(aq) 88n Cl2(aq) Mn2(aq) 2 H2O(l) The synthesis of fluorine escaped the efforts of chemists for almost 100 years. Part of the problem involved finding an oxidizing agent strong enough to oxidize the F ion to F2. The task of preparing fluorine was made even more difficult by the extraordinary toxicity of both F2 and the hydrogen fluoride (HF) used to make it. The best way of producing strong reducing agents is to pass an electric current through a salt of the metal. Sodium, for example, can be prepared by the electrolysis of molten sodium chloride. electrolysis 2 NaCl(l) 888888 n 2 Na(s) Cl2(g)
The same process can be used to generate strong oxidizing agents, such as F2. Attempts to prepare fluorine by electrolysis, however, were initially unsuccessful. Humphry Davy, who prepared potassium, sodium, barium, strontium, calcium, and magnesium by electrolysis, repeatedly tried to prepare F2 by the electrolysis of fluorite (CaF2), and succeeded only in ruining his health. Joseph Louis Gay-Lussac and Louis Jacques Thenard, who prepared elemental boron for the first time, also tried to prepare fluorine and suffered from very painful exposures to hydrogen fluoride. George and Thomas Knox were badly poisoned during their attempts to make fluorine, and both Paulin Louyet and Jerome Nickles died from fluorine poisoning. Finally, in 1886 Henri Moissan successfully isolated F2 gas from the electrolysis of a mixed salt of KF and HF and noted that crystals of silicon burst into flame when mixed with the gas. Electrolysis of KHF2 is still used to prepare fluorine today. 2 KHF2(s) electrolysis 888888n H2(g) F2(g) 2 KF(s) A schematic drawing of the cell in which KHF2 is electrolyzed to produce F2 gas at the anode and H2 gas at the cathode is shown in Figure N.23.
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35
Anode connection
HF inlet
H2 outlet Cell cover
Carbon anode
Molten solution of KF dissolved in HF Gas separation skirt
Cooling jacket Steel cathode
FIGURE N.23 Schematic drawing of the cell in which KHF2 is electrolyzed.
Common Oxidation Numbers for the Halogens Fluorine is the most electronegative element in the periodic table. As a result, it has an oxidation number of 1 in all its compounds. Because chlorine, bromine, and iodine are less electronegative, it is possible to prepare compounds in which those elements have oxidation numbers of 1, 3, 5, and 7, as shown in Table N.8. TABLE N.8 Common Oxidation Numbers for the Halogens Oxidation Number
Examples
1 0 1 3 5 7
CaF2, HCl, NaBr, AgI F2, Cl2, Br2, I2 HClO, ClF HClO2, ClF3 HClO3, BrF5, BrF6, IF5 HClO4, BrF6, IF7
General Trends in Halogen Chemistry There are several patterns in the chemistry of the halogens. •
• •
The chemistry of fluorine is simplified by the fact that it is the most electronegative element in the periodic table that forms compounds and by the fact that it cannot expand its valence shell to hold more than eight valence electrons. Chlorine, bromine, and iodine can expand their valence shells to hold as many as 14 valence electrons. The chemistry of the halogens is dominated by oxidation–reduction reactions.
The Hydrogen Halides (HX) The hydrogen halides are compounds that contain hydrogen attached to one of the halogens (HF, HCl, HBr, and HI). The compounds are all colorless gases that are soluble in
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water. Up to 512 mL of HCl gas can dissolve in 1 mL of water at 0°C and 1 atm, for example. Each of the hydrogen halides ionizes to at least some extent when it dissolves in water. H2O HCl(g) 88n H(aq) Cl(aq)
Exercise N.6 Explain why it is easy to believe that HCl is an ionic compound. Describe the best evidence that it isn’t an ionic compound. Solution It is easy to believe HCl is an ionic compound because it forms ions when it dissolves in water. H2O HCl(g) 88n H(aq) Cl(aq)
At first glance, the reaction seems to be similar to the reaction that occurs when ionic compounds dissolve in water. H2O NaCl(s) 88n Na(aq) Cl(aq)
The best evidence that HCl isn’t an ionic compound is the fact that it is a gas at room temperature, and ionic compounds are invariably solids at room temperature. Some chemists try to distinguish between the behavior of HCl and NaCl when they dissolve in water as follows. They argue that HCl ionizes when it dissolves in water because ions are created by the reaction. NaCl, on the other hand, dissociates in water because NaCl already consists of Na and Cl ions in the solid.
Several of the hydrogen halides can be prepared directly from the elements. Mixtures of H2 and Cl2, for example, react with explosive violence in the presence of light to form HCl. H2(g) Cl2(g) 88n 2 HCl(g) Because chemists are usually more interested in aqueous solutions of hydrogen halides than they are in the pure gases, the compounds are usually synthesized in water. Aqueous solutions of the hydrogen halides are often called mineral acids because they are literally acids prepared from minerals. Hydrochloric acid is prepared by reacting table salt with sulfuric acid, for example, and hydrofluoric acid is prepared from fluorite and sulfuric acid. 2 NaCl(s) H2SO4(aq) 88n 2 HCl(aq) Na2SO4(aq) CaF2(s) H2SO4(aq) 88n 2 HF(aq) CaSO4(aq) The acids are purified by taking advantage of the ease with which HF and HCl gas boil out of the solutions. The gas given off when one of the solutions is heated is collected and then redissolved in water to give relatively pure samples of the mineral acid.
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The Interhalogen Compounds Interhalogen compounds are formed by reactions between different halogens. All possible interhalogen compounds of the type XY are known. Bromine reacts with chlorine, for example, to give BrCl, which is a gas at room temperature. Br2(l) Cl2(g) 88n 2 BrCl(g) Interhalogen compounds with the general formulas XY3, XY5, and even XY7 are formed when pairs of halogens react. Chlorine reacts with fluorine, for example, to form chlorine trifluoride. Cl2(g) 3 F2(g) 88n 2 ClF3(g) The compounds are easiest to form when Y is fluorine. Iodine is the only halogen that forms an XY7 interhalogen compound, and it does so only with fluorine. ClF3 and BrF5 are extremely reactive compounds. ClF3 is so reactive that wood, asbestos, and even water spontaneously burn in its presence. The compounds are excellent fluorinating agents, which tend to react with each other to form positive ions such as ClF2 and BrF4 and negative ions such as ClF4 and BrF6. 2 BrF5(l) 88n [BrF4][BrF6](s)
Neutral Oxides of the Halogens Under certain conditions, it is possible to isolate neutral oxides of the halogens, such as Cl2O, Cl2O3, ClO2, Cl2O4, Cl2O6, and Cl2O7. Cl2O7, for example, can be obtained by dehydrating perchloric acid, HClO4. The oxides are notoriously unstable compounds that explode when subjected to either thermal or physical shock. Some are so unstable they detonate when warmed to temperatures above 40°C.
Oxyacids of the Halogens and Their Salts Chlorine reacts with the OH ion to form chloride ions and hypochlorite (OCl) ions. Cl2(aq) 2 OH(aq) 88n Cl(aq) OCl(aq) H2O(l) This is a disproportionation reaction in which one-half of the chlorine atoms are oxidized to hypochlorite ions and the other half are reduced to chloride ions. Cl2 2 OH
Cl OCl H2O
0
1
Reduction
1
Oxidation
When the solution is hot, this reaction gives a mixture of the chloride and chlorate (ClO3) ions. 3 Cl2(aq) 6 OH(aq) 88n 5 Cl(aq) ClO3(aq) 3 H2O(l) Under carefully controlled conditions, it is possible to convert a mixture of the chlorate and hypochlorite ions into a solution that contains the chlorite (ClO2) ion. ClO3(aq) ClO(aq) 88n 2 ClO2(aq)
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The last member of this class of compounds, the perchlorate ion (ClO4), is made by electrolyzing solutions of the chlorate ion. The names of the oxyanions of the halogens use the endings -ite and -ate to indicate low and high oxidation numbers and the prefixes hypo- and per- to indicate the very lowest and very highest oxidation numbers, as shown in Table N.9. Each of the ions can be converted into an oxyacid, which is named by replacing the -ite ending with -ous and the -ate ending with -ic. TABLE N.9
Oxyanions and Oxyacids of Chlorine
Oxyanion
Name
Oxyacid
Name
Oxidation State of Chlorine
ClO ClO2 ClO3 ClO4
Hypochlorite Chlorite Chlorate Perchlorate
HClO HOClO HOClO2 HOClO3
Hypochlorous acid Chlorous acid Chloric acid Perchloric acid
1 3 5 7
The hypochlorite (OCl) ion is the active ingredient in liquid bleaches, such as Clorox. Calcium salts of the ion can be found in dry bleaches, such as Clorox 2. Ca(OCl)2 is also the active ingredient in most commercial products used to “chlorinate” swimming pools.
N.8 THE CHEMISTRY OF THE RARE GASES In 1892 Lord Rayleigh noted that nitrogen isolated from air was more dense than nitrogen prepared by decomposing ammonia. William Ramsay attacked this problem by purifying a sample of nitrogen gas to remove any moisture, carbon dioxide, and organic contaminants. He then passed the purified gas over hot magnesium metal, which reacts with nitrogen to form the nitride. 3 Mg(s) N2(s) 88n Mg3N2(s) When he was finished, Ramsay was left with a small residue of gas that occupied roughly 1/80th of the original volume. He excited the gas in an electric discharge tube and found that the resulting emission spectrum contained lines that differed from those of all known gases. After repeated discussions of the results of these experiments, Rayleigh and Ramsay jointly announced the discovery of a new element, which they named argon from the Greek word meaning “lazy one” because the gas refused to react with any element or compound they tested. Argon didn’t fit into any of the known families of elements in the periodic table, but its atomic weight suggested that it might belong to a new group that could be inserted between chlorine and potassium. Shortly after reporting the discovery of argon in 1894, Ramsay found another unreactive gas when he heated a mineral of uranium. The lines in the spectrum of the gas also occurred in the spectrum of the sun, which led Ramsay to name the element helium (from the Greek word helios, “sun”). Experiments with liquid air led Ramsay to a third gas, which he named krypton (“the hidden one”). Experiments with liquid argon led him to a fourth gas, neon (“the new one”), and finally a fifth gas, xenon (“the stranger”). These elements were discovered between 1894 and 1898. Because Moissan had only recently isolated fluorine for the first time and fluorine was the most active of the known
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elements, Ramsay sent a sample of argon to Moissan to see whether it would react with fluorine. It did not. The failure of Moissan’s attempts to react argon with fluorine, coupled with repeated failures by other chemists to get the more abundant of the gases to undergo chemical reaction, eventually led to their being labeled inert gases. The development of the electronic theory of atoms did little to dispel this notion because it was obvious that these gases had very stable electron configurations. As a result, the elements were labeled “inert gases” in almost every textbook and periodic table until about 30 years ago. In 1962 Neil Bartlett found that PtF6 was a strong enough oxidizing agent to remove an electron from an O2 molecule. PtF6(g) O2(g) 88n [O2][PtF6](s) Bartlett realized that the first ionization energy of Xe (1170 kJ/mol) was slightly smaller than the first ionization energy of the O2 molecule (1177 kJ/mol). He therefore predicted that PtF6 might also react with Xe. When he ran the reaction, he isolated the first compound of a Group VIIIA element. Xe(g) PtF6(g) 88n [Xe][PtF6](s) A few months later, workers at the Argonne National Laboratory near Chicago found that Xe reacts with F2 to form XeF4. Since that time, more than 200 compounds of Kr, Xe, and Rn have been isolated. No compounds of the more abundant elements in the group (He, Ne, and Ar) have yet been isolated. However, the fact that elements in the family can undergo chemical reactions has led to the use of the term rare gases rather than inert gases to describe these elements. Compounds of xenon are by far the most numerous of the rare gas compounds. With the exception of XePtF6, rare gas compounds have oxidation numbers of 2, 4, 6, and 8, as shown by the examples cited in Table N.10.
TABLE N.10
Compounds of Xenon and Their Oxidation Numbers
Compound
Oxidation Number
Compound
Oxidation Number
XeF XeF2 Xe2F3 XeF3 XeF4 XeOF2 XeF5 XeF6 Xe2F11
2 2 2 4 4 4 6 6 6
XeO3 XeOF4 XeO2F2 XeO3F XeO4 XeO64 XeO3F2 XeO2F4 XeOF5
6 6 6 6 8 8 8 8 8
There is some controversy over whether the rare gases should be viewed as having the outermost shell of electrons filled (in which case they should be labeled Group VIIIA) or empty (in which case they should be labeled Group 0). The authors believe these elements should be labeled Group VIIIA because they behave as if they contribute eight valence electrons when they form compounds.
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The synthesis of most xenon compounds starts with the reaction between Xe and F2 at high temperatures (250–400°C) to form a mixture of XeF2, XeF4, and XeF6. 3 Xe(g) 6 F2(g) 88n XeF2(s) XeF4(s) XeF6(s) The positively charged XeFn ions are then made by reacting XeF2, XeF4, or XeF6 with either AsF5, SbF5, or BiF5. XeF2(s) SbF5(l) 2 XeF2(s) AsF5(g) XeF4(s) BiF5(s) 2 XeF6(s) AsF5(g)
88n 88n 88n 88n
[XeF][SbF6](s) [Xe2F3][AsF6](s) [XeF3][BiF6](s) [Xe2F11][AsF6](s)
Oxides of xenon, such as XeOF2, XeOF4, XeO2F2, XeO3F2, XeO2F4, XeO3, and XeO4, are prepared by reacting XeF4 or XeF6 with water. The XeO64 ion, for example, is produced when XeF6 dissolves in strong base. 2 XeF6(s) 16 OH(aq) 88n XeO64(aq) Xe(g) O2(g) 12 F(aq) 8 H2O(l) Some xenon compounds are relatively stable. XeF2, XeF4, and XeF6, for example, are stable solids that can be purified by sublimation in a vacuum at 25°C. XeOF4 and Na4XeO6 are also reasonably stable. Others, such as XeO3, XeO4, XeOF2, XeO2F2, XeO3F2, and XeO2F4, are unstable compounds that can decompose violently. The principal use of rare gas compounds at present is as the light-emitting component in lasers. Mixtures of 10% Xe, 89% Ar, and 1% F2, for example, can be “pumped,” or excited, with high energy electrons to form excited XeF molecules, which emit a photon with a wavelength of 354 nm.
N.9 THE INORGANIC CHEMISTRY OF CARBON For more than 200 years, chemists have divided compounds into two categories. Those that were isolated from plants or animals were called organic, while those extracted from ores and minerals were inorganic. Organic chemistry is often defined as the chemistry of carbon. But that definition would include calcium carbonate (CaCO3) and graphite, which more closely resemble inorganic compounds. We will therefore define organic chemistry as the study of compounds such as formic acid (HCO2H), methane (CH4), and vitamin C (C6H8O6) that contain both carbon and hydrogen. This section focuses on inorganic carbon compounds.
Elemental Forms of Carbon: Graphite, Diamond, Coke, and Carbon Black Carbon occurs as a variety of allotropes. There are two crystalline forms—diamond and graphite—and a number of amorphous (noncrystalline) forms, such as charcoal, coke, and carbon black. References to the characteristic hardness of diamond (from the Greek adamas, “invincible”) date back at least 2600 years. It was not until 1797, however, that Smithson Tennant was able to show that diamonds consist solely of carbon. The properties of diamond are remarkable. It is among the least volatile substances known (MP 3550°C, BP 4827°C), it is the hardest substance known, and it expands less on heating than any other material.
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The properties of diamond are a logical consequence of its structure. Carbon, with four valence electrons, forms covalent bonds to four neighboring carbon atoms arranged toward the corners of a tetrahedron, as shown in Figure N.24. Each of the sp3-hybridized atoms is then bound to four other carbon atoms, which form bonds to four other carbon atoms, and so on. As a result, a perfect diamond can be thought of as a single giant molecule. The strength of the individual COC bonds and their arrangement in space give rise to the unusual properties of diamond.
FIGURE N.24 The simplest repeating unit in diamond.
In some ways, the properties of graphite are like those of diamond. Both compounds boil at 4827°C, for example. But graphite is also very different from diamond. Diamond (3.514 g/cm3) is significantly more dense than graphite (2.26 g/cm3). Whereas diamond is the hardest substance known, graphite is one of the softest. Diamond is an excellent insulator, with little or no tendency to carry an electric current. Graphite is such a good conductor of electricity that graphite electrodes are used in electrical cells. The physical properties of graphite can be understood from the structure of the solid shown in Figure N.25. Graphite consists of extended planes of sp2-hybridized carbon atoms in which each carbon is tightly bound to three other carbon atoms. (It takes 477 kJ to break a mole of the bonds within the planes.) The strong bonds between carbon atoms within each plane explain the exceptionally high melting point and boiling point of graphite. The bonds between planes of carbon atoms, however, are relatively weak. (The force of attraction between planes is only 17 kJ/mol.) Because the bonds between planes are weak, it is easy to deform the solid by allowing one plane of atoms to move relative to another. As a result, graphite is soft enough to be used in pencils and as a lubricant in motor oil.
FIGURE N.25 Portion of the structure of extended planes of carbon atoms found in graphite.
The characteristic properties of graphite and diamond might lead you to expect that diamond would be more stable than graphite. This isn’t what is observed experimentally. Graphite at 25°C and 1 atm pressure is slightly more stable than diamond. At very high
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temperatures and pressures, however, diamond becomes more stable than graphite. In 1955 General Electric developed a process to make industrial-grade diamonds by treating graphite with a metal catalyst at temperatures of 2000 to 3000 K and pressures above 125,000 atm. Roughly 40% of industrial-quality diamonds are now synthetic. Although gemquality diamonds can be synthesized, the costs involved are prohibitive. Both diamond and graphite occur as regularly packed crystals. Other forms of carbon are amorphous—they lack a regular structure. Charcoal, carbon black, and coke are all amorphous forms of carbon. Charcoal results from heating wood in the absence of oxygen. To make carbon black, natural gas or other carbon compounds are burned in a limited amount of air to give a thick, black smoke that contains extremely small particles of carbon, which can be collected when the gas is cooled and passed through an electrostatic precipitator. Coke is a more regularly structured material, closer in structure to graphite than either charcoal or carbon black, which is made from coal.
Carbides: Covalent, Ionic, and Interstitial Carbon reacts with less electronegative elements at high temperatures to form compounds known as carbides. When carbon reacts with an element of similar size and electronegativity, a covalent carbide is produced. Silicon carbide, for example, is made by treating silicon dioxide from quartz with an excess of carbon in an electric furnace at 2300 K. SiO2(s) 3 C(s) 88n SiC(s) 2 CO(g) Covalent carbides have properties similar to those of diamond. Both SiC and diamond are inert to chemical reactions, except at very high temperatures; both have very high melting points; and both are among the hardest substances known. SiC was first synthesized by Edward Acheson in 1891. Shortly thereafter, Acheson founded the Carborundum Company to market the material. Then, as now, materials in this class are most commonly used as abrasives. Compounds that contain carbon and one of the more active metals are called ionic carbides. CaO(s) 3 C(s) 88n CaC2(s) CO(g) It is useful to think about ionic carbides as if they contained negatively charged carbon ions: [Ca2][C22] or [Al3]4[C4]3. This model is useful because it explains why these carbides burst into flame when added to water. Ionic carbides such as Al4C3 that formally contain the C4 ion react with water to form methane, which is ignited by the heat given off in the reaction. C4 4 H2O 88n CH4 4 OH The ionic carbides such as CaC2 that formally contain the C22 ion react with water to form acetylene, which is ignited by the heat of reaction. C22 2 H2O 88n C2H2 2 OH At one time, miners’ lamps were fueled by the combustion of acetylene prepared from the reaction of calcium carbide with water. The difference between covalent carbides and ionic carbides can be understood by adding compounds such as SiC, Al4C3, and CaC2 to the bond type triangle introduced in
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Chapter 5. When those compounds are added to Figure 5.10, for example, we find that SiC falls well into the region expected for covalent compounds. CaC2, on the other hand, is clearly an ionic compound. Al4C3 falls on the borderline between ionic and covalent, which is consistent with the fact that the compound is hard—as one would expect for a covalent carbide—and yet reacts with water to form methane—as might be expected for an ionic carbide. Interstitial carbides, such as tungsten carbide (WC), form when carbon combines with a metal that has an intermediate electronegativity and a relatively large atomic radius. In these compounds, the carbon atoms pack in the holes (interstices) between planes of metal atoms. The interstitial carbides, which include TiC, ZrC, and MoC, retain the properties of metals. They act as alloys, rather than as either salts or covalent compounds.
The Oxides of Carbon Although the different forms of carbon are essentially inert at room temperature, they combine with oxygen at high temperatures to produce a mixture of carbon monoxide and carbon dioxide. 2 C(s) O2(g) 88n 2 CO(g) C(s) O2(g) 88n CO2(g)
H° 221.05 kJ/molrxn H° 393.51 kJ/molrxn
CO can also be obtained by reacting red-hot carbon with steam. C(s) H2O(g) 88n CO(g) H2(g) Because the mixture of gases is formed by the reaction of charcoal or coke with water it is often referred to as water gas. It is also known as town gas because it was once made by towns and cities for use as a fuel. Water gas, or town gas, was a common fuel for both home and industrial use before natural gas became readily available. The H2 burns to form water, and the CO is oxidized to CO2. Eventually, as our supply of natural gas is depleted, it will become economical to replace natural gas with other fuels, such as water gas, that can be produced from our abundant supply of coal. Both CO and CO2 are colorless gases. CO boils at 191.5°C, and CO2 sublimes (passes directly from the solid to the gaseous state) at 78.5°C. Although CO has no odor or taste, CO2 has a faint, pungent odor and a distinctly acidic taste. Both are dangerous substances but at very different levels of exposure. Air contaminated with as little as 0.002 gram of CO per liter can be fatal because CO binds tightly to the hemoglobin that carries oxygen through the blood. CO2 is not lethal until the concentration in the air approaches 15%. At that point, it has replaced so much oxygen that a person who attempts to breathe the atmosphere suffocates. The danger of CO2 poisoning is magnified by the fact that CO2 is roughly 1.5 times more dense than the air in our atmosphere. Thus, CO2 can accumulate at the bottom of tanks or wells.
CO2 in the Atmosphere Carbon dioxide influences the temperature of the atmosphere by a phenomenon known as the greenhouse effect. The glass walls and ceilings of a greenhouse absorb some of the lower energy, longer wavelength radiation from sunlight thereby inevitably raising the temperature inside the building. CO2 in the atmosphere does exactly the same thing, it absorbs
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low energy, long wavelength radiation from the Sun that would otherwise be reflected back from the surface of the planet. Thus, CO2 in the atmosphere traps heat. Although there are other factors at work, it is worth noting that Venus, whose atmosphere contains a great deal of CO2, has a surface temperature of roughly 400°C, whereas Mars, with little or no atmosphere, has a surface temperature of 50°C. There are many sources of CO2 in the atmosphere. Over geologic time scales, the largest source has been volcanos. Within the twentieth century, the combustion of petroleum, coal, and natural gas has made a significant contribution to atmospheric levels of CO2 (see Figure N.8). Between 1958 and 1978, the average level of CO2 in the atmosphere increased by 6%, from 315.8 to 334.6 ppm. At one time, the amount of CO2 released to the atmosphere wasn’t a matter for concern because natural processes that removed CO2 from the atmosphere could compensate for the CO2 that entered the atmosphere. The vast majority of the CO2 liberated by volcanic action, for example, was captured by calcium oxide or magnesium oxide to form calcium carbonate or magnesium carbonate. CaO(s) CO2(g) 88n CaCO3(s) MgO(s) CO2(g) 88n MgCO3(s) CaCO3 is found as limestone or marble, or mixed with MgCO3 as dolomite. The amount of CO2 in deposits of carbonate minerals is at least several thousand times larger than the amount in the atmosphere. CO2 also dissolves, to some extent, in water. H2O CO2(g) 88n CO2(aq)
It then reacts with water to form carbonic acid, H2CO3. CO2(aq) H2O(l) 88n H2CO3(aq) As a result of the reactions, the sea contains about 60 times more CO2 than the atmosphere. Can the sea absorb more CO2 from the atmosphere, or is it near its level of saturation? Is the rate at which the sea absorbs CO2 greater than the rate at which we are adding it to the atmosphere? The observed increase in the concentration of CO2 in recent years suggests pessimistic answers to those two questions. A gradual warming of the earth’s atmosphere could result from continued increases in CO2 levels, with adverse effects on the climate and therefore the agriculture of at least the northern hemisphere.
The Chemistry of Carbonates: CO32 and HCO3 Eggshells are almost pure calcium carbonate. CaCO3 can also be found in the shells of many marine organisms and in both limestone and marble. The fact that none of those substances dissolves in water suggests that CaCO3 is normally insoluble in water. Calcium carbonate will dissolve in water saturated with CO2, however, because carbonated water (or carbonic acid) reacts with calcium carbonate to form calcium bicarbonate, which is soluble in water. CaCO3(s) H2CO3(aq) 88n Ca2(aq) 2 HCO3(aq)
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When water rich in carbon dioxide flows through limestone formations, part of the limestone dissolves. If the CO2 escapes from the water, or if some of the water evaporates, solid CaCO3 is redeposited. When this happens as water runs across the roof of a cavern, stalactites, which hang from the roof of the cave, are formed. If the water drops before the carbonate reprecipitates, stalagmites, which grow from the floor of the cave, are formed. The chemistry of carbon dioxide dissolved in water is the basis of the soft drink industry. The first artificially carbonated beverages were introduced in Europe at the end of the nineteenth century. Carbonated soft drinks today consist of carbonated water, a sweetening agent (sugar, saccharin, or aspartame), an acid to impart a sour or tart taste, flavoring agents, coloring agents, and preservatives. As much as 3.5 liters of gaseous CO2 dissolve in a liter of soft drink to provide the characteristic “bite” associated with carbonated beverages. Carbonate chemistry plays an important role in other parts of the food industry as well. Baking soda, or bicarbonate of soda, is sodium bicarbonate, NaHCO3, a weak base, which is used to neutralize the acidity of other ingredients in a recipe. Baking powder is a mixture of baking soda and a weak acid, such as tartaric acid or calcium hydrogen phosphate (CaHPO4). When mixed with water, the acid reacts with the HCO3 ion to form CO2 gas, which causes the dough or batter to rise. HCO3(aq) H(aq) 88n H2CO3(aq) 88n H2O(l) CO2(g) Before commercial baking powders were available, cooks obtained the same effect by mixing roughly a teaspoon of baking soda with a cup of sour milk or buttermilk. The acids that give sour milk and buttermilk their characteristic taste also react with the bicarbonate ion to give CO2.
KEY TERMS Allotrope Anhydrous Carbide Diamagnetic Dimer Disproportionation reaction
Haber process Halide Halogen Hydride Ostwald process Oxidizing agent
Oxyacid Oxyanion Paramagnetic Peroxide Rare gases Reducing agent
PROBLEMS Metals, Nonmetals, and Semimetals 1. List the elements that are nonmetals. Describe where these elements are found in the periodic table. 2. Explain why semimetals (such as B, Si, Ge, As, Sb, Te, Po, and At) exist and describe some of their physical properties. 3. Which member of each of the following pairs of elements is more nonmetallic? (a) As or Bi (b) As or Se (c) As or S (d) As or Ge (e) As or P
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The Chemistry of the Nonmetals 4. Which of the following elements can exist as a triatomic molecule? (a) hydrogen (b) helium (c) sulfur (d) oxygen (e) chlorine 5. Which of the following elements should form compounds with the formulas Na2X, H2X, XO2, and XF6? (a) B (b) C (c) N (d) O (e) S 6. Which of the following elements should form compounds with the formulas XH3, XF3, and Na3XO4? (a) Al (b) Ge (c) As (d) S (e) Cl 7. Which of the following can’t be found in nature? Explain why. (a) MgCl2 (b) CaCO3 (c) F2 (d) Na3AlF6 (e) NaCl
The Role of Nonmetal Elements in Chemical Reactions 8. Explain why more electronegative elements tend to oxidize less electronegative elements. 9. Which of the following ions or molecules can be oxidized? (a) H2SO3 (b) P4 (c) Cl (d) SiO2 (e) PO43 (f) Mg2 10. Which of the following ions or molecules can be reduced? (a) H2O (b) H2SO3 (c) HCl (d) CO2 (e) Mg2 (f) Na
Deciding What Is Oxidized and What Is Reduced 11. For each of the following reactions, identify what is oxidized and what is reduced. (a) Fe2O3(s) 3 CO(g) n 2 Fe(s) 3 CO2(g) (b) H2(g) CO2(g) n H2O(g) CO(g) (c) CH4(g) 2 O2(g) n CO2(g) 2 H2O(g) (d) 2 H2S(g) 3 O2(g) n 2 SO2(g) 2 H2O(g) 12. For each of the following reactions, identify what is oxidized and what is reduced. (a) PH3(g) 3 Cl2(g) n PCl3(g) 3 HCl(g) (b) 2 NO(g) F2(g) n 2 NOF(g) (c) 2 Na(s) 2 NH3(l) n 2 NaNH2(s) H2(g) (d) 3 NO2(g) H2O(l) n 2 HNO3(aq) NO(g) 13. Hydrazine is made by a reaction known as the Raschig process. 2 NH3(aq) NaOCl(aq) 88n N2H4(aq) NaCl(aq) H2O(l) Decide whether this is an oxidation–reduction reaction. If it is, identify the compound oxidized and the compound reduced. 14. The thiosulfate ion, S2O32, is prepared by boiling solutions of sulfur dissolved in sodium sulfite. 8 SO32(aq) S8(s) 88n 8 S2O32(aq) Is this an oxidation–reduction reaction? If it is, identify the compound oxidized and the compound reduced. 15. Chlorine dioxide, ClO2, is used commercially as a bleach or a disinfectant because of its excellent oxidizing ability. ClO2 is prepared by decomposing chlorous acid as follows. 8 HOClO(aq) 88n 6 ClO2(g) Cl2(g) 4 H2O(l)
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16.
17. 18. 19.
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Is this an oxidation–reduction reaction? If it is, identify the oxidizing agent and the reducing agent. Nitric acid sometimes acts as an acid (as a source of the H ion) and sometimes as an oxidizing agent. For each of the following reactions, decide whether HNO3 acts as an acid or as an oxidizing agent. (a) Na2CO3(s) 2 HNO3(aq) n 2 NaNO3(aq) CO2(g) H2O(l) (b) 3 P4(s) 20 HNO3(aq) 8 H2O(l) n 12 H3PO4(aq) 20 NO(g) (c) Al2O3(s) 6 HNO3(aq) n 2 Al(NO3)3(aq) 3 H2O(l) (d) 3 Cu(s) 8 HNO3(aq) n 3 Cu(NO3)2(aq) 2 NO(g) 4 H2O(l) Which of the following would you expect to be the best oxidizing agent? (a) Na (b) H2 (c) N2 (d) P4 (e) O2 Which of the following would you expect to be the best reducing agent? (a) Na (b) F (c) Na (d) Br2 (e) Fe3 For each of the following pairs of elements, determine which is the better reducing agent? (a) P4 or As (b) As or S8 (c) P4 or S8 (d) S8 or Cl2 (e) C or O2
Predicting the Products of Chemical Reactions 20. Predict the products of the (a) Mg(s) N2(g) n (b) Li(s) O2(g) n (c) Br2(l) I(aq) n 21. Predict the products of the (a) SO2(g) H2O(l) n (b) Cl2(g) OH(aq) n (c) CO2(g) H2O(l) n 22. Predict the products of the (a) HCl(g) H2O(l) n (b) P4O10(s) H2O(l) n (c) NO2(g) H2O(l) n 23. Predict the products of the (a) S8(s) O2(g) n (b) Al(s) I2(s) n (c) P4(s) F2(g) n
following reactions.
following reactions.
following reactions.
following reactions.
The Chemistry of Hydrogen 24. Describe three ways of preparing small quantities of H2 in the lab. 25. Explain why it is not a good idea to prepare H2 by reacting sodium metal with a strong acid. 26. Give an example of a compound in which hydrogen has an oxidation number of 1; of 0; of 1. 27. Which of the following reactions produce a compound in which hydrogen has an oxidation number of 1? (a) Li H2 n (b) O2 H2 n (c) S8 H2 n (d) Cl2 H2 n (e) Ca H2 n
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28. Use tables of first ionization energies and electronegativities to explain why it is so difficult to decide whether hydrogen belongs in Group IA or Group VIIA of the periodic table. 29. Which of the following substances can be used as evidence for placing hydrogen in Group IA? Which can be used as evidence for including hydrogen in Group VIIA? (a) CaH2 (b) AlH3 (c) H2S (d) H3PO4 (e) H2 30. The earth’s atmosphere once contained significant amounts of H2. Explain why only traces of H2 are left in the earth’s atmosphere, whereas the atmospheres of other planets—such as Jupiter, Saturn, and Neptune—contain large quantities of H2. 31. Use Lewis structures to explain what happens in the four reactions described in Section N.2 that can be used to prepare small quantities of H2 gas.
The Chemistry of Oxygen and Sulfur 32. Describe three ways of preparing small quantities of O2 in the lab. 33. Describe the relationships among oxygen (O2), the peroxide ion (O22), and the oxide ion (O2). Explain why the number of electrons shared by a pair of oxygen atoms decreases as the oxidation number of the oxygen becomes more negative. 34. Which of the following elements or compounds could eventually produce O2 when it reacts with water? (a) Ba (b) BaO (c) BaO2 (d) Ba(OH)2 (e) BaNO3 35. Explain why the only compounds in which oxygen has a positive oxidation number are compounds, such as OF2, that contain fluorine. 36. Explain why hydrogen peroxide can be either an oxidizing agent or a reducing agent. Describe at least one reaction in which H2O2 oxidizes another substance and one reaction in which it reduces another substance. 37. Write the Lewis structures for ozone, O3, and sulfur dioxide, SO2. Discuss the relationship between the compounds. 38. Explain why elemental oxygen exists as O2 molecules, whereas elemental sulfur forms S8 molecules. 39. Explain why sulfur forms compounds such as SF4 and SF6, when oxygen can only form OF2. 40. Describe the relationship between the thiosulfate and sulfate ions and between the thiocyanate and cyanate ions. Use that relationship to predict the formula of the trithiocarbonate ion. 41. Write the Lewis structures of the following products of the reaction between sodium and sulfur. (a) Na2S (b) Na2S2 (c) Na2S3 (d) Na2S8 42. Explain why sulfur readily forms compounds in the 2, 4, and 6 oxidation states, but only a handful of compounds exist in which oxygen is in a positive oxidation state. 43. Explain why sulfur-containing compounds such as FeS2, CS2, and H2S form SO2 instead of SO3 when they burn. 44. Use Lewis structures to explain why solutions of the SO32 ion react with sulfur to form thiosulfate, S2O32. 45. Use Lewis structures to explain why a two-electron reduction of the S2O62 ion gives SO32. S2O62 2 e 88n 2 SO32
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46. Use Lewis structures to explain why a two-electron oxidation of the S2O32 ion gives the S4O62 ion. 47. Which of the following does not have a reasonable oxidation number for sulfur? (a) Na2S (b) H2S (c) SO32 (d) SO4 (e) SF4 48. Explain why SO2 plays an important role in the phenomenon known as acid rain. 49. Explain why problems with acid rain would be much more severe if sulfur compounds burned to form SO3 instead of SO2.
The Chemistry of Nitrogen and Phosphorus 50. Nitrogen has a reasonable oxidation number in all of the following compounds, and yet one of them is still impossible. Which one is impossible? (a) NF3 (b) NF5 (c) NO3 (d) NO2 (e) NO 51. Earth’s atmosphere contains roughly 4 1016 tons of nitrogen, and yet the biggest problem facing agriculture in the world today is a lack of “nitrogen.” Explain why. 52. Explain why elemental nitrogen is almost inert, but nitrogen compounds such as NH4NO3, NaN3, nitroglycerin, and trinitrotoluene (TNT) form some of the most dangerous explosives. 53. Which of the following oxides of nitrogen are paramagnetic? (a) N2O (b) NO (c) NO2 (d) N2O3 (e) N2O4 (f) N2O5 54. Use Lewis structures to explain what happens in the following reaction. 2 NO O2 88n 2 NO2 55. Use Lewis structures to explain why NO reacts with NO2 to form N2O3 when a mixture of the compounds is cooled. 56. Use the fact that nitrous oxide decomposes to form nitrogen and oxygen to explain why a glowing splint bursts into flame when immersed in a container filled with N2O. 2 N2O(g) 88n 2 N2(g) O2(g) 57. Describe ways of preparing small quantities of each of the following compounds in the laboratory. (a) N2O (b) NO (c) NO2 (d) N2O4 58. Describe how to tell the difference between a flask filled with NO gas and a flask filled with NO2. 59. Lightning catalyzes the reaction between nitrogen and oxygen in the atmosphere to form nitrogen oxide, NO. N2(g) O2(g) 88n 2 NO(g) Explain how lightning acts as one source of acid rain. 60. Which of the following elements or compounds is not involved at some stage in the preparation of nitric acid? (a) O2 (b) N2 (c) NO (d) NO2 (e) H2 61. Explain why phosphorus forms both PCl3 and PCl5 but nitrogen forms only NCl3. 62. Explain why nitrogen is essentially inert at room temperature, but white phosphorus bursts spontaneously into flame when it comes into contact with air. 63. Explain why red phosphorus is much less reactive than white phosphorus.
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64. Explain why nitrogen forms extraordinarily stable N2 molecules at room temperature, but phosphorus forms P2 molecules only at very high temperatures. 65. Explain why nitric acid has the formula HNO3 and phosphoric acid has the formula H3PO4. 66. Write the Lewis structures for phosphoric acid, H3PO4, and phosphorous acid, H3PO3. Explain why phosphoric acid can lose three H ions to form a phosphate ion, PO43, whereas phosphorous acid can lose only two H ions to form the HPO32 ion. 67. Explain why only two of the four hydrogen atoms in H4P2O5 are lost when the oxyacid forms an oxyanion. 68. Describe the role of carbon in the preparation of elemental phosphorus from calcium phosphate. 69. Predict the product of the reaction of phosphorus with excess oxygen and then predict what will happen when the product of the reaction is dissolved in water. 70. Explain why the most common oxidation states of antimony are 3 and 5. 71. Which of the following compounds should not exist? (a) Na3P (b) (NH4)3PO4 (c) PO2 (d) PH3 (e) POCl3
The Chemistry of the Halogens 72. Which of the halogens is the most active, or reactive? Explain why. 73. Describe the difference between halogens and halides. Give examples of each. 74. Fe3 ions can oxidize Br ions to Br2, but they can’t oxidize Cl ions to Cl2. Use this information to determine where the Fe3 ion belongs in the following sequence of decreasing oxidizing strength: F2 Cl2 Br2 I2. 75. HBr can be prepared by reacting PBr3 with water. PBr3(l) 3 H2O(l) 88n 3 HBr(aq) H3PO3(aq) Use this information to explain what happens in the following reaction. P4(s) 6 Br2(s) 12 H2O(l) 88n 12 HBr(aq) 4 H3PO3(aq) 76. Explain why chlorine reacts with fluorine to form ClF3 but not FCl3. 77. Chlorine reacts with base to form the hypochlorite ion. Cl2(aq) 2 OH(aq) 88n Cl(aq) OCl(aq) H2O(l) Use this information to explain why people who make the mistake of mixing Clorox with hydrochloric acid often suffer damage to their lungs from breathing chlorine gas.
The Inorganic Chemistry of Carbon 78. Use the structure of graphite to explain why the bonds between carbon atoms are so strong that it is difficult to boil off individual carbon atoms, yet the material is so soft it can be used as a lubricant. 79. Explain why silicon forms a covalent carbide but calcium forms an ionic carbide. 80. Write balanced equations for the combustion of both CO and H2 that explain why a mixture of the gases can be used as a fuel. 81. Use Lewis structures to explain the following reaction. CO2(g) H2O(l) 88n H2CO3(aq)
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Integrated Problems 82. A recent catalog listed the following prices: $21.40 for 450 grams of sodium, $18.00 for 1 kilogram of zinc, and $52.80 for 250 grams of sodium hydride. Which reagent would be the least expensive source of H2 gas? 83. A solution of hydrogen peroxide in water that is 30% hydrogen peroxide by weight sells for $15.95 per 500 g, and potassium chlorate sells for $12.75 per 500 grams. Is it less expensive to generate oxygen by decomposing H2O2 or KClO3? 84. Write a sequence of reactions for the conversion of elemental nitrogen into nitric acid. Calculate the weight of nitric acid that can be produced from a ton of nitrogen gas. 85. At 1700°C, P4 molecules decompose partially to form P2. P4(g) 88n 2 P2(g) If the average molecular weight of phosphorus at that temperature is 91 g/mol, what fraction of the P4 molecules decompose? 86. Uranium reacts with fluorine to form UF6, which boils at 51°C. The relative rate of diffusion of 235UF6 and 238UF6 in the gas phase was used in the Manhattan project to separate the more abundant 238U isotope from 235U. Predict which substance diffuses more rapidly, and calculate the ratio of the rate of diffusion of the compounds.
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2 TRANSITION METAL CHEMISTRY TM.1 The Transition Metals TM.2 Werner’s Theory of Coordination Complexes TM.3 Typical Coordination Numbers TM.4 The Electron Configurations of Transition Metal Ions TM.5 Oxidation States of the Transition Metals TM.6 Lewis Acid–Lewis Base Approach to Bonding in Complexes TM.7 Typical Ligands TM.8 Coordination Complexes in Nature Chemistry in the World Around Us: Nitrogen Fixation TM.9 Nomenclature of Complexes TM.10 Isomers TM.11 The Valence Bond Approach to Bonding in Complexes TM.12 Crystal Field Theory TM.13 The Spectrochemical Series TM.14 High-Spin versus Low-Spin Octahedral Complexes TM.15 The Colors of Transition Metal Complexes TM.16 Ligand Field Theory
TM.1 THE TRANSITION METALS The elements in the periodic table are often divided into the four categories shown in Figure TM.1: (1) main-group elements, (2) transition metals, (3) lanthanides, and (4) actinides. The main-group elements include the active metals in the two columns on the extreme left of the periodic table and the metals, semimetals, and nonmetals in the six columns on the far right. The transition metals are the metallic elements that serve as a bridge, or transition, between the two sides of the table. The lanthanides and the actinides at the bottom of the table are sometimes known as the inner transition metals because they have atomic numbers that fall between the first and second elements in the last two rows of the transition metals. 1
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Main-group Elements
Transition Metals
Main-group Elements
H
H He
Li Be
B
C
N
O
F Ne
Na Mg
Al Si
P
S
Cl Ar
K Ca Sc Ti Rb Sr
Y
V
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Zr Nb Mo Tc Ru Rh Pb Ag Cd In Sn Sb Te
Cs Ba La Hf Ta W Re Os Ir
I
Xe
Pt Au Hg Tl Pb Bi Po At Rn
Fr Ra Ac Rf Db Sg Bh Hs Mt
Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Actinides
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
FIGURE TM.1 The four categories of elements in the periodic table.
There is some controversy about the classification of the elements on the boundary between the main-group and transition metal elements on the right side of the table (see Figure TM.2). The elements in question are zinc (Zn), cadmium (Cd), and mercury (Hg). The disagreement about whether those elements should be classified as main-group elements or transition metals suggests that the differences between the categories aren’t clear. Transition metals resemble main-group metals in many ways: They look like metals; they are malleable and ductile; they conduct heat and electricity; and they form positive ions. The fact that the two best conductors of electricity are a transition metal (copper) and a main-group metal (aluminum) shows the extent to which the physical properties of maingroup metals and transition metals overlap. Transition Metals
Main-group Elements H He B
C
N
O
F Ne
Al Si
P
S
Cl Ar
Co Ni Cu Zn Ga Ge As Se Br Kr Rh Pb Ag Cd In Sn Sb Te Ir
I
Xe
Pt Au Hg Tl Pb Bi Po At Rn
FIGURE TM.2 There is some debate about whether zinc, cadmium, and mercury should be classified as transition metals or main-group metals.
There are also differences between transition metals and the main-group metals. The transition metals are more electronegative than the main-group metals, for example, and are therefore more likely to form covalent compounds. Another difference between the main-group metals and transition metals can be seen in the formulas of the compounds they form. The main-group metals tend to form salts (such as NaCl, Mg3N2, and CaS) in which there are just enough negative ions to balance the charge on the positive ions. The transition metals form similar compounds [such as FeCl3, HgI2, and Cd(OH)2], but they are more likely than main-group metals to form complexes, such as the FeCl4, HgI42, and Cd(OH)42 ions, that have an excess number of negative ions. A third difference between main-group and transition metal ions is the ease with which they form stable compounds with neutral molecules, such as water and ammonia. Salts of main-group metal ions dissolve in water to form aqueous solutions. H2O NaCl(s) 88n Na(aq) Cl(aq)
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When we let the water evaporate, we get back the original starting material, NaCl(s). Salts of the transition metal ions can display a very different behavior. Chromium(III) chloride, for example, is a violet compound that dissolves in liquid ammonia to form a yellow compound with the formula CrCl36 NH3 that can be isolated when the ammonia is allowed to evaporate. CrCl3(s) 6 NH3(l) 88n CrCl36 NH3(s) The FeCl4 ion and CrCl36 NH3 are called coordination compounds because they contain ions or molecules linked, or coordinated, to a transition metal. They are also known as complex ions or coordination complexes, because they are Lewis acid–base complexes. The ions or molecules that bind to transition metal ions to form complexes are called ligands (from Latin, meaning “to tie or bind”). The number of ligands bound to the transition metal ion is called the coordination number. Although coordination complexes are particularly important in the chemistry of the transition metals, some main-group elements also form complexes. Aluminum, tin, and lead, for example, form complexes such as the AlF63, SnCl42, and PbI42 ions.
TM.2 WERNER’S THEORY OF COORDINATION COMPLEXES Alfred Werner first became interested in coordination complexes in 1892, while preparing lectures for a course on atomic theory. Six months later, he proposed a model for coordination complexes that still serves as the basis for work in the field. Werner then spent the remainder of his life collecting evidence to support his theory. Before we introduce Werner’s model of coordination complexes, let’s look at some of the experimental data available at the turn of the twentieth century. •
Three different cobalt(III) complexes can be isolated when CoCl2 is dissolved in aqueous ammonia and then oxidized by air to the 3 oxidation state. A fourth complex can be made by slightly different techniques. The complexes have different colors and different empirical formulas. CoCl36 CoCl35 CoCl35 CoCl34
•
NH3 NH3H2O NH3 NH3
orange-yellow red purple green
The ammonia in the cobalt(III) complexes is much less reactive than normal. By itself, ammonia reacts rapidly with hydrochloric acid to form ammonium chloride. NH3(aq) HCl(aq) 88n NH4(aq) Cl(aq)
The complexes, however, don’t react with hydrochloric acid, even at 100°C. CoCl36 NH3(aq) HCl(aq) 88n •
A white precipitate of AgCl normally forms when a source of the Ag ion is added to a solution that contains the Cl ion. Ag(aq) Cl(aq) 88n AgCl(s)
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When excess Ag ion is added to solutions of the CoCl36 NH3 and CoCl35 NH3H2O complexes, 3 moles of AgCl are formed for each mole of complex in solution, as might be expected. However, only two of the Cl ions in the CoCl35 NH3 complex and only one of the Cl ions in CoCl34 NH3 can be precipitated with Ag ions. • The results of measurements of the conductivity of aqueous solutions of the complexes suggest that the CoCl36 NH3 and CoCl35 NH3H2O complexes dissociate in water to give a total of four ions. CoCl35 NH3 dissociates to give three ions, however, and CoCl34 NH3 dissociates to give only two ions. Werner explained the observations by differentiating between ions or molecules that were coordinated to the transition metal and those that were present to balance the charge on the metal ion. He assumed that there were always six ions or molecules coordinated to the Co3 ion. Any remaining ions were only used to balance the charge on the complex ion. The formulas of the compounds described above can therefore be written as follows. [Co(NH3)63][Cl]3 [Co(NH3)5(H2O)3][Cl]3 [Co(NH3)5Cl2][Cl]2 [Co(NH3)4Cl2][Cl]
orange-yellow red purple green
Checkpoint Determine the number of chloride ions coordinated to the chromium atom in the Cr(NH3)4(H2O)2Cl3 complex and the number of these ions that are present only to balance the charge on the complex ion.
Exercise TM.1 Describe how Werner’s theory of coordination complexes explains both the number of ions formed when the following compounds dissolve in water and the number of chloride ions that precipitate when the solutions are treated with Ag ions. (a) [Co(NH3)6]Cl3 (b) [Co(NH3)5(H2O)]Cl3 (c) [Co(NH3)5Cl]Cl2 (d) [Co(NH3)4Cl2]Cl Solution The cobalt ion is coordinated to a total of six ligands in each complex. Each complex also has a total of three chloride ions to balance the charge on the Co3 ion. Some of the Cl ions are free to dissociate when the complex dissolves in water. Others are bound to the Co3 ion and neither dissociate nor react with Ag. (a) The three chloride ions in the complex are free to dissociate when it dissolves in water. The complex therefore dissociates in water to give a total of four ions, and all three Cl ions are free to react with Ag ion. H2O [Co(NH3)6]Cl3(s) 88n Co(NH3)63(aq) 3 Cl(aq)
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(b) Once again, the three Cl ions are free to dissociate when the complex dissolves in water, and they precipitate when Ag ions are added to the solution. H2O [Co(NH3)5(H2O)]Cl3(s) 88n Co(NH3)5(H2O)3(aq) 3 Cl(aq)
(c) One of the chloride ions is coordinated to the cobalt in the complex. Thus, only three ions are formed when the compound dissolves in water, and only two Cl ions are free to precipitate with Ag ions. H2O [Co(NH3)5Cl]Cl2(s) 88n Co(NH3)5Cl2(aq) 2 Cl(aq)
(d) Two of the chloride ions are coordinated to the cobalt in the complex. Thus, only two ions are formed when the compound dissolves in water, and only one Cl ion is free to precipitate with Ag ions. H2O [Co(NH3)4Cl2]Cl(s) 88n Co(NH3)4Cl2(aq) Cl(aq)
Werner also assumed that transition metal complexes had definite shapes. According to his theory, the ligands in six-coordinate cobalt(III) complexes are oriented toward the corners of an octahedron, as shown in Figure TM.3.
3+
2+
NH3 NH3
Cl NH3
NH3
Co NH3
NH3 Co
NH3
NH3
NH3 NH3
NH3
Co(NH3)5Cl2+
Co(NH3)63+ orange-yellow
purple
3+
+
OH2 NH3
Cl NH3
NH3
Co NH3
NH3 NH3
Co(NH3)6(H2O)3+ red
NH3 Co
NH3
NH3 Cl
Co(NH3)4Cl2+ green
FIGURE TM.3 Structures of the four cobalt complex ions that played a major role in the development of Werner’s theory.
TM.3 TYPICAL COORDINATION NUMBERS Transition metal complexes have been characterized with coordination numbers that range from 1 to 12, but the most common coordination numbers are 2, 4, and 6. Examples of complexes with these coordination numbers are given in Table TM.1. Note that the charge on the complex is always the sum of the charges on the ions or molecules that form the complex. 88n Cu(NH3)42 Cu2 4 NH3 m88 88n Fe(CN)64 Fe2 6 CN m88
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TABLE TM.1
Examples of Common Coordination Numbers
Metal Ion Ag Ag Ag Cu Cu2 Zn2 Hg2 Co2 Fe2 Fe3 Co3 Ni2 Fe2
Ligand
2 2 2 2 4 4 4 4 6 6 6 6 6
NH3 S2O32 Cl NH3 NH3 CN I SCN H2O H2O NH3 NH3 CN
88n m88 88n m88 88n m88 88n m88 88n m88 88n m88 88n m88 88n m88 88n m88 88n m88 88n m88 88n m88 88n m88
Complex
Coordination Number
Ag(NH3)2 Ag(S2O3)23 AgCl2 Cu(NH3)2 Cu(NH3)42 Zn(CN)42 HgI42 Co(SCN)42 Fe(H2O)62 Fe(H2O)63 Co(NH3)63 Ni(NH3)62 Fe(CN)64
2 2 2 2 4 4 4 4 6 6 6 6 6
Also note that the coordination number of a complex often increases as the charge on the metal ion becomes larger. 88n Cu(NH3)2 Cu 2 NH3 m88 2 88n Cu(NH3)42 Cu 4 NH3 m88
Exercise TM.2 Calculate the charge on the transition metal ion in the following complexes. (a) Na2Co(SCN)4 (b) Ni(NH3)6(NO3)2 (c) K2PtCl6 Solution (a) The complex contains the Na and Co(SCN)42 ions. Each thiocyanate ion (SCN) carries a charge of 1. Since the net charge on the Co(SCN)42 ion is 2, the cobalt ion must carry a charge of 2. (b) The complex contains the Ni(NH3)62 and NO3 ions. Since ammonia is a neutral molecule, the nickel must carry a charge of 2. (c) The complex contains the K and PtCl62 ions. Each chloride ion carries a charge of 1. Since the net charge on the PtCl62 ion is 2, the platinum must carry a charge of 4.
TM.4 THE ELECTRON CONFIGURATIONS OF TRANSITION METAL IONS The electron configurations of ions formed by the main-group metals are discussed in Chapters 3 and 5. Aluminum, for example, loses its three valence electrons when it forms Al3 ions.
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Al [Ne] 3s2 3p1 3 Al [Ne] The relationship between the electron configurations of transition metal elements and their ions is more complex. Consider the chemistry of cobalt, for example, which forms complexes that contain either Co2 or Co3 ions. Before we can predict the electron configuration of the ions we have to answer the question: Which valence electrons are removed from a cobalt atom when the Co2 and Co3 ions are formed? The electron configuration of a neutral cobalt atom is written as follows. Co
[Ar] 4s2 3d7
We might expect cobalt to lose electrons from only the 3d orbitals, but this is not what is observed. The Co2 and Co3 ions have the following electron configurations. Co2 [Ar] 3d7 Co3 [Ar] 3d6 In general, electrons are removed from the valence shell s orbitals before they are removed from valence d orbitals when transition metals are ionized. This raises an interesting question: Why are electrons removed from 4s orbitals before 3d orbitals if they are placed in 4s orbitals before 3d orbitals? There are several ways of answering this question. First, note that the difference between the energies of the 3d and 4s orbitals is very small. For this reason, the electron configurations for chromium and copper differ slightly from what might be expected. Cr [Ar] 4s1 3d5 Cu [Ar] 4s1 3d10 Note also that the relative energies of the 4s and 3d orbitals used when predicting electron configurations only hold true for neutral atoms. When transition metals form positive ions, the 3d orbitals become more stable than the 4s orbitals. As a result, electrons are removed from the 4s orbital before the 3d orbitals.
Exercise TM.3 Predict the electron configuration of the Fe3 ion. Solution We start with the configuration of a neutral iron atom. Fe
[Ar] 4s2 3d6
We then remove three electrons to form the 3 ion. Two of the electrons come from the 4s orbital. The other comes from a 3d orbital. Fe3 [Ar] 3d5
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Because the valence electrons in transition metal ions are concentrated in d orbitals, the ions are often described as having dn configurations. The Co3 and Fe2 ions, for example, are said to have a d 6 configuration. Co3 [Ar] 3d6 Fe2 [Ar] 3d6
TM.5 OXIDATION STATES OF THE TRANSITION METALS Most transition metals exhibit more than one oxidation state. Manganese, for example, forms compounds in every oxidation state from 1 to 7. Some oxidation states, however, are more common than others. The most common oxidation states of the first series of transition metals are given in Table TM.2. Efforts to explain the apparent pattern in Table TM.2 ultimately fail for a combination of reasons. Some of the oxidation states are common because they are relatively stable. Others describe compounds that aren’t necessarily stable but which react slowly. Still others are common only from a historic perspective. TABLE TM.2 Common Oxidation States and d Subshell Electron Configurations of the First Series of Transition Metals Oxidation State 1 2 3 4 5 6 7
Sc
Ti
V
Cr
d3 0
Mn d5
3
d
d 0
Fe d6 d5
Co d7 d6
Ni
Cu
Zn
d8
d10 d9
d10
d3
d
0
d
d0 d0
One point about the oxidation states of transition metals deserves particular attention: Transition metal ions with charges larger than 3 cannot exist in aqueous solution. Consider the following reaction, for example, in which manganese is oxidized from the 2 to the 7 oxidation state. Mn2(aq) 4 H2O(l) 88n MnO4(aq) 8 H(aq) 5 e When the manganese atom is oxidized, it becomes more electronegative. In the 7 oxidation state, the atom is electronegative enough to react with water to form a covalent oxide, MnO4. It is useful to have a way to distinguish between the charge on a transition metal ion and the oxidation state of the transition metal. By convention, symbols such as Mn2 refer to ions that carry a 2 charge. Symbols such as Mn(VII) are used to describe compounds such as KMnO4 in which manganese is in the 7 oxidation state. Mn(VII) isn’t the only example of an oxidation state powerful enough to decompose water. As soon as Mn2 is oxidized to Mn(IV), it reacts with water to form MnO2. A similar phenomenon can be seen in the chemistry of both vanadium and chromium. Vanadium exists in aqueous solutions as the V2 ion. But once it is oxidized to the 4 or 5 oxidation state, it reacts with water to form the VO2 or VO2 ions. The Cr3 ion can be found
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in aqueous solution. But once the ion is oxidized to Cr(VI), it reacts with water to form the CrO42 and Cr2O72 ions.
TM.6 LEWIS ACID–LEWIS BASE APPROACH TO BONDING IN COMPLEXES G. N. Lewis was the first to recognize that the reaction between a transition metal ion and one or more ligands to form a coordination complex was analogous to the reaction between H and OH ions to form water. The reaction between H and OH ions involves the donation of a pair of electrons from the OH ion to the H ion to form a covalent bond. H
OOH
88n HOOOH
The H ion in the reaction acts as an electron pair acceptor. The OH ion, on the other hand, is an electron pair donor. Lewis argued that any ion or molecule that behaves like the H ion should be an acid. Conversely, any ion or molecule that behaves like the OH ion should be a base. A Lewis acid is therefore any ion or molecule that can accept a pair of electrons. A Lewis base is an ion or molecule that can donate a pair of electrons. When Co3 ions react with ammonia, the Co3 ion accepts pairs of nonbonding electrons from six NH3 ligands to form covalent cobalt–nitrogen bonds (see Figure TM.4).
H
H
H
H
N
N
H3N
H H
Co3+
H H
3+
NH3
H
N
H H
NH3 Co
N
H3N
H
N
H N
H
H H
NH3
H NH3
H
FIGURE TM.4 Reaction between Co3 ions and ammonia to form the Co(NH3)63 complex ion.
The metal ion is therefore a Lewis acid, and the ligands coordinated to this metal ion are Lewis bases. 88n Co3 6 NH3 Co(NH3)63 m88 Electron pair acceptor Electron pair donor Acid–base complex (Lewis acid) (Lewis base) The Co3 ion is an electron pair acceptor, or Lewis acid, because it has empty valenceshell orbitals that can be used to hold pairs of electrons. To emphasize the empty valence orbitals, we can write the configuration of the Co3 ion as follows. Co3 [Ar] 3d6 4s0 4p0 There is room in the valence shell of the ion for 12 more electrons. (Four electrons can be added to the 3d subshell, two to the 4s orbital, and six to the 4p subshell.) The NH3
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molecule is an electron pair donor, or Lewis base, because it has a pair of nonbonding electrons on the nitrogen atom. According to this model, transition metal ions form coordination complexes because they have empty valence-shell orbitals that can accept pairs of electrons from a Lewis base. Ligands must therefore be Lewis bases—they must contain at least one pair of nonbonding electrons that can be donated to a metal ion.
TM.7 TYPICAL LIGANDS Any ion or molecule with a pair of nonbonding electrons can be a ligand. Many ligands are said to be monodentate (literally, “one-toothed”) because they “bite” the metal in only one place. Typical monodentate ligands are given in Table TM.3. TABLE TM.3
F
Typical Monodentate Ligands
Cl
O H
N H
Br
I
O
O
O C
2
O
O O
C
H
O
H H H 2
O N ]
[ C
S
C
N
S
S
O
O
Other ligands can bind to the metal more than once. Ethylenediamine (en) is a typical bidentate ligand.
H2N
H2C D
CH2
G
Ethylenediamine (en)
NH2
Each end of the molecule contains a pair of nonbonding electrons that can form a covalent bond to a metal ion. Ethylenediamine is also an example of a chelating ligand. The term chelate comes from a Greek stem meaning “claw.” It is used to describe ligands that can grab the metal in two or more places, the way a claw would. A typical ethylenediamine complex is shown in Figure TM.5. +
H H
H
N
CH2
H N
H2C
N Co
H2C N H
CH2
H
H Cl
H Cl Co(en)2Cl2+
FIGURE TM.5 Structure of the Co(en)2Cl2 complex ion.
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A number of multidentate ligands are shown in Table TM.4. Linking ethylenediamine fragments gives tridentate ligands and tetradentate ligands, such as diethylenetriamine (dien) and triethylenetetramine (trien). Adding four OCH2CO2 groups to an ethylenediamine framework gives a hexadentate ligand known as EDTA. This ligand that can single-handedly satisfy the coordination number of six found for so many transition metal ions, as shown in Figure TM.6.
TABLE TM.4
Typical Multidentate Ligands Structure
Name Bidentate Ligands
O O
2
Carbonate
C O
O
2
O C
Oxalate (ox)
C
O
O CH
CH3C
Acetylacetonate (acac)
CCH3
O
O
CH2CH2 H2N
Ethylenediamine (en)
NH2 Tridentate Ligands CH2CH2
CH2CH2
H2N
CH2CH2
CH2CH2
H2N
Diethylenetriamine (dien)
NH2
NH
NH
Tetradentate Ligands CH2CH2 NH2
NH
CH2CO2 N CH2CO2 CH2CO2
Triethylenetetramine (trien)
Nitrilotriacetate (NTA) Hexadentate Ligands
O
O
O
CCH2
CH2C
O
NCH2CH2N
O
CCH2 O
CH2C O
O
Ethylenediaminetetraacetate (EDTA)
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TRANSITION METAL –
O
C O CH2
O C
CH2
O
N CH2
Fe N
O C
CH2
CH2
O
CH2
O C
O [Fe EDTA]–
FIGURE TM.6 EDTA forms very strong complexes with many transition metal ions.
TM.8 COORDINATION COMPLEXES IN NATURE Complex ions play a vital role in the chemistry of living systems, as shown by the three biologically important complexes in Figure TM.7: vitamin B12, chlorophyll a, and the heme found in the proteins hemoglobin and myoglobin that carry oxygen from the lungs to the muscles. As early as 1926, it was known that patients who suffered from pernicious anemia became better when they ate liver. It took more than 20 years, however, to determine that the vitamin B12 in liver was one of the active factors in relieving the anemia. It took another 10 years to determine the structure of vitamin B12, which is a six-coordinate Co3 complex. The Co3 ion is coordinated to four nitrogen atoms that lie in a planar molecule known as a corrin ring. It is also coordinated to a fifth nitrogen atom and to a sixth group, labeled R in Figure TM.7a. The R group can be a CN, NO2, SO32, or OH ion, depending on the source of the vitamin B12. Every carbon atom in our bodies can be traced back to a reaction in which plants use the energy in sunlight to make glucose (C6H12O6) from CO2 and water. light 6 CO2(g) 6 H2O(aq) 88888 88n C6H12O6(aq) 6 O2(g) chlorophyll
The fundamental step in the photosynthesis of glucose is the absorption of sunlight by the chlorophyll in higher plants and algae. Chlorophyll a (see Figure TM.7b) is a complex in which an Mg2 ion is coordinated to four nitrogen atoms in a planar chlorin ring. The chlorin ring in chlorophyll is similar to the corrin ring in vitamin B12, but not identical. The minor differences between the rings adjust their sizes so that each ring is the right size to hold the correct transition metal ion. The coordination complex shown in Figure TM.7c is known as a heme, which contains an Fe2 or Fe3 ion coordinated to four nitrogen atoms in a porphyrin ring. The porphyrin ring in a heme is slightly larger than the corrin ring in vitamin B12 and slightly smaller than the chlorin ring in chlorophyll a, so it is just the right size to hold the Fe2 or Fe3 ion.
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TRANSITION METAL CH2OH
CH2 H
CH3
CH
O H
C
H
CH3
C C
H P O
O
C
N
N
CH3 CH
O– N
CH3
N
C
C
CH2
CH2
C CO
(a)
C
C
CH3
O O (b)
Chlorophyll a CH3
CH2
C H
CH2
CH3
N CH3 H
H2C
CH
CH3
H
C
CH2
CH2CH2CONH2 CH3
C
O
CH2 CH2
C
C
CH2
CH3
N
NH
CH2CH3
CH
N
C
CH
C C
Mg2+ C
HO
C
C
HC
O
H3C
CH
CH
N
CH2CH2CONH2
Co3+
CH2 CH
CH3
N
H2NCOCH2
CH3
N
CH3
CH2 H
H C
CH2
H3C R
CH3
H2NCOCH2
CH2
CH2CONH2
CH
CH2
CH2 CH2CH2CONH2
Vitamin B12
CH2 CH2 CH
CH3
CH2 –
O2CCH2CH2 C H3C
HC
CH2
N Fe
C HC
C
N
C
C C CH3
CH2
C
C
C
CH2
CH2CH2CO2–
H C
N
C
C C H
(c)
CH CH3
C
2+, 3+
N
CH3
C
CH3
CH C C
CH3
C CH
CH2
Heme
FIGURE TM.7 Structures of the biologically important coordination complexes known as vitamin B12, chlorophyll a, and heme.
13
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Chemistry in the World Around Us Nitrogen Fixation Ever since the Haber process was developed, chemists have been fascinated by the contrast between the conditions required to reduce nitrogen to ammonia in the laboratory and the conditions under which this reaction occurs in nature. N2(g) 3 H2(g) 88n 2 NH3(g) In the laboratory, the reaction requires high pressures (200–300 atm) and high temperatures (400–600°C). In nature, it occurs at room temperature and atmospheric pressure. The nitrogenase enzyme that catalyzes the reaction in nature can be separated into two proteins, which contain a total of 34 iron atoms and two molybdenum atoms. One of the proteins contains both Fe and Mo atoms, the other contains only Fe. The Fe protein contains an Fe4S4 unit that binds the ATP that provides the energy that drives the reaction in which nitrogen is reduced to ammonia. The MoFe protein contains four Fe4S4 subunits that can hold one of the electrons necessary to reduce N2 and a pair of MoFe7S8 units. One end of each MoFe7S8 unit is bound to the OSH group on the side chain of a cysteine residue in the protein; the other is bound to an analog of the citrate ion known as homocitrate. Jongsun Kim and Douglas Rees, of the California Institute of Technology, reported X-ray crystal structures of the Fe protein from Azotobacter vinelandii and of the MoFe protein from Clostridium pasteurianum [Science, 257, 1677 (1992)]. Their work suggests that the N2 molecule is bound, side-on, between a pair of iron atoms in the MoF7S8 (homocitrate) complex, as shown in Figure TM.8. Homocitrate Cysteine
-S - Fe - Mo -C -O -N
Histidine
FIGURE TM.8 Model for the binding of N2 to the MoFe protein in the nitrogenase enzyme.
These results explain why chemists have been frustrated for so many years by their inability to understand the role of the molybdenum atoms in the MoFe protein. Contrary to most researchers’ expectations, the N2 molecule isn’t carried by the molybdenum atoms in the protein. It apparently binds to iron atoms in both the industrial catalyst and the nitrogenase enzyme. The primary difference between these catalysts is the ease with which they can pump electrons into the metal atoms that carry the N2 molecule, thereby reducing it to ammonia.
TM.9 NOMENCLATURE OF COMPLEXES The rules for naming chemical compounds are established by nomenclature committees of the International Union of Pure and Applied Chemistry (IUPAC). The IUPAC nomenclature of coordination complexes is based on the following rules.
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Rules for Naming Coordination Complexes • • •
• • • • •
The name of the positive ion is written before the name of the negative ion. The names of the ligands are written before the name of the metal to which they are coordinated. The Greek prefixes mono-, di-, tri-, tetra-, penta-, hexa-, and so on are used to indicate the number of ligands when the ligands are relatively simple. The Greek prefixes bis-, tris-, and tetrakis- are used with more complicated ligands. The names of negative ligands always end in o, as in fluoro (F), chloro (Cl), bromo (Br), iodo (I), oxo (O2), hydroxo (OH), and cyano (CN). A handful of neutral ligands are given common names, such as aquo (H2O), ammine (NH3), and carbonyl (CO). Ligands are listed in the following order: negative ions, neutral molecules, and positive ions. Ligands with the same charge are listed in alphabetical order. The oxidation number of the metal atom is indicated by a Roman numeral in parentheses after the name of the metal atom. The names of complexes with a net negative charge end in -ate. Co(SCN)42, for example, is the tetrathiocyanatocobaltate(II) ion. When the symbol for the metal is derived from its Latin name, -ate is added to the Latin name of the metal. Thus, negatively charged iron complexes are ferrates and negatively charged copper complexes are cuprates.
Exercise TM.4 Name the following coordination complexes. (a) [Cr(NH3)5(H2O)][(NO3)3] (b) [Cr(NH3)4Cl2]Cl Solution (a) The complex contains the Cr(NH3)5(H2O)3 ion and three nitrate ions to balance the charge on the complex ion. It is therefore known as pentaammineaquochromium(III) nitrate (b) Two of the three chloride ions are ligands that are coordinated to the Cr3 ion, and the other is a Cl ion that balances the charge on the Cr(NH3)4Cl2 ion. The compound is therefore known as dichlorotetraamminechromium(III) chloride
Exercise TM.5 Determine the formulas of the following compounds. (a) potassium hexacyanoferrate(II) (b) tris(ethylenediamine)chromium(III) chloride Solution (a) The compound contains enough K ions to balance the charge on the negatively charged ferrate ion. Because the ferrate ion is formed from an Fe2 ion and six CN ligands, it carries a net charge of 4. The formula for the compound is therefore K4Fe(CN)6. (b) The compound contains a Cr3 ion coordinated to three bidentate ethylenediamine or en ligands. It therefore must contain three Cl ions to balance the charge on the complex ion. Thus, the formula for the compound is [Cr(en)3]Cl3.
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TM.10 ISOMERS Cis/Trans Isomers An important test of Werner’s theory of coordination complexes involved the study of coordination complexes that formed isomers (literally, “equal parts”). Isomers are compounds with the same chemical formula that have different structures. There are two isomers of the Co(NH3)4Cl2 complex ion, for example, which differ in the orientation of the two chloride ions around the Co3 ion, as shown in Figure TM.9. In the trans isomer, the chlorides occupy positions across from one another in the octahedron. In the cis isomer, they occupy adjacent positions. The difference between cis and trans isomers can be remembered by noting that the prefix trans is used to describe things that are on opposite sides, as in transatlantic or transcontinental. +
Cl
NH3
H3N
H3N
Co H3N
+
Cl
Cl Co
NH3
H3N
Cl
NH3
NH3 +
cis – [Co(NH3)4Cl2]+
trans – [Co(NH3)4Cl2]
FIGURE TM.9 The trans and cis isomers of the Co(NH3)4Cl2 complex ion.
At the time Werner proposed his theory, only one isomer of the [Co(NH3)4Cl2]Cl complex was known, namely, the green complex described in Section TM.2. Werner predicted that a second isomer should exist, and his discovery in 1907 of a purple compound with the same chemical formula was a key step in convincing scientists who were still skeptical of the model. Cis/trans isomers are also possible in four-coordinate complexes that have a squareplanar geometry. Figure TM.10 shows the structures of the cis and trans isomers of dichlorodiammineplatinum(II). Cl NH3 G D Pt D G H3N Cl
Cl NH3 G D Pt D G Cl NH3
trans-Pt(NH3)2Cl2
cis-Pt(NH3)2Cl2
FIGURE TM.10 The cis/trans isomers of the square-planar Pt(NH3)2Cl2 complex.
The cis isomer of this compound is used as a drug to treat brain tumors, under the trade name Cisplatin. This square-planar complex inserts itself into the grooves in the double helix structure of the DNA in cells, which inhibits the replication of DNA. This slows down the rate at which the tumor grows, which allows the body’s natural defense mechanisms to act on the tumor. Checkpoint Which of the following compounds can form cis/trans isomers? (a) square-planar Rh(CO)Cl3 (c) octahedral Ni(en)2(H2O)22 (b) trigonal-bipyramidal Fe(CO)4(PH3) (d) tetrahedral Ni(CO)3(PH3)
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Chiral Isomers Another form of isomerism can best be understood by considering the difference between gloves and mittens. One glove in each pair fits the left hand, and the other fits the right hand. Mittens usually fit equally well on either hand. To understand why, hold a glove and a mitten in front of a mirror. There is no difference between the mitten shown in Figure TM.11 and its mirror image. Each mitten is therefore said to be superimposable on its mirror image. The same can’t be said of the relationship between a glove and its mirror image. The mirror image of the glove that fits the left hand looks like the glove that fits the right hand, and vice versa.
Mirror
Left-hand glove
Mitten
FIGURE TM.11 Gloves are chiral; mittens are not.
For all practical purposes, the two gloves that form a pair have the same “substituents.” Each glove has four “fingers” and one “thumb.” If you clap them together, you will find even more similarities between the gloves. The thumbs are attached at the same point, significantly below the point where the fingers start. The second “fingers” on both gloves are the longest and the little “fingers” are the shortest. In spite of these similarities, there is a fundamental difference between the gloves that can be observed by trying to place your right hand into a left-hand glove. Like your hands, each glove is the mirror image of the other, and the mirror images are not superimposable. Objects that possess a similar handedness are said to be chiral (literally, “handed”). Gloves are chiral; mittens are not. Feet and shoes are both chiral, but socks are not. The Co(en)33 ion in Figure TM.12 is an example of a chiral molecule, which forms a pair of isomers that are mirror images of each other. The isomers have almost identical physical and chemical properties. They have the same melting point, boiling point, density, and color, for example. They differ only in the way they interact with plane-polarized light. Light consists of electric and magnetic fields that oscillate in all possible directions perpendicular to the path of the light ray. When light is passed through a polarizer, such as a Nicol prism or the lens of polarized sunglasses, the only light that emerges from the polarizer is light whose oscillations are confined to a single plane. In 1813 Jean Baptiste Biot noticed that the plane in which polarized light oscillates was rotated either to the right or the left when plane-polarized light was passed through single crystals of quartz or aqueous solutions of tartaric acid or sugar. Because they seem to interact with light, compounds that can rotate plane-polarized light are said to be optically active. Those that rotate the plane clockwise (to the right) are said to be dextrorotatory (from the Latin word dexter, “right”). Those that rotate the plane counterclockwise (to the left) are called levorotatory (from the Latin word laevus, “left”).
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Mirror
3+
3+ H2C
H2N
CH2
NH2
H2C
CH2 NH2 H2N NH2
CH2 NH2
NH2 H2C H2C
CH2
Co H2N
Co NH2 NH2
H2C CH2 H2N
NH2
H2C
CH2
FIGURE TM.12 The Co(en)33 complex ion is an example of a chiral molecule that exists as a pair of isomers that are mirror images of each other.
All compounds that are chiral are optically active; one isomer is dextrorotatory and the other is levorotatory. Chirality is a property of a molecule that results from its structure. Optical activity is a macroscopic property of a collection of the molecules that arises from the way they interact with light. Checkpoint Which of the following molecules is chiral? (a) octahedral Cr(CO)6 (c) SF4 (see Figure 4.21) (b) tetrahedral Ni(CO)4 (d) octahedral Fe(acac)3
TM.11 THE VALENCE BOND APPROACH TO BONDING IN COMPLEXES The idea that atoms form covalent bonds by sharing pairs of electrons was first proposed by G. N. Lewis in 1902. It was not until 1927, however, that Walter Heitler and Fritz London showed how the sharing of pairs of electrons holds a covalent molecule together. The Heitler–London model of covalent bonds was the basis of the valence bond theory. The last major step in the evolution of the theory was the suggestion that atomic orbitals mix to form hybrid orbitals, such as the sp, sp2, sp3, dsp3, and d2sp3 orbitals shown in Figure TM.13.
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sp3d
sp2
sp3
sp
sp3d 2 or d 2sp3
FIGURE TM.13 The shapes of the sp, sp2, sp3, sp3d, and sp3d2 hybrid orbitals.
It is easy to apply the valence bond theory to some coordination complexes, such as the Co(NH3)63 ion. We start with the electron configuration of the transition metal ion. Co3 [Ar] 3d 6 We then look at the valence-shell orbitals and note that the 4s and 4p orbitals are empty. Co3 [Ar] 3d6 4s0 4p0 Experiments tell us that there are no unpaired electrons in the complex so we begin by concentrating the 3d electrons in the dxy, dxz, and dyz orbitals. This gives the following electron configuration. Co3
hg hg hg 3d
4s
4p
The 3dx 2y2, 3dz2, 4s, 4px, 4py, and 4pz orbitals are then mixed to form a set of empty d2sp3 orbitals that point toward the corners of an octahedron as shown in TM.13. Each of the orbitals can accept a pair of nonbonding electrons from a neutral NH3 molecule to form a complex in which the cobalt atom has a filled shell of valence electrons. Co(NH3)63
hg hg hg
hg hg hg hg hg hg
3d
d 2sp3
Exercise TM.6 Use valence bond theory to describe the Fe(CN)64 complex ion. This complex contains no unpaired electrons. Solution We start with the electron configuration of the transition metal ion. Fe2 [Ar] 3d6 4s0 4p0
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By investing a little energy in the system, we can pair the six 3d electrons, thereby creating empty 3dx 2y2 and 3dz2 orbitals. Fe2
hg hg hg 3d
4s
4p
The orbitals are then mixed with the empty 4s and 4p orbitals to form a set of six d2sp3 hybrid orbitals, which can accept pairs of nonbonding electrons from the CN ligands to form an octahedral complex in which the metal atom has a filled shell of valence electrons. Fe(CN)64
hg hg hg
hg hg hg hg hg hg
3d
d 2sp3
At first glance, some complexes, such as the Ni(NH3)62 ion, seem hard to explain with the valence bond theory. We start, as always, by writing the configuration of the transition metal ion. Ni2 [Ar] 3d8 This configuration creates a problem because there are eight electrons in the 3d orbitals. Even if we invest the energy necessary to pair the 3d electrons, we can’t find enough empty 3d orbitals to use to form a set of d2sp3 hybrids. Ni2
hg hg hg h
h
3d
4s
4p
There is a way around this problem. The five 4d orbitals on nickel are empty, so we can form a set of empty sp3d2 hybrid orbitals by mixing the 4dx 2 -y2, 4dz2, 4s, 4px, 4py, and 4pz orbitals. The hybrid orbitals then accept pairs of nonbonding electrons from six ammonia molecules to form a complex ion. Ni(NH3)62
hg hg hg h 3d
h
hg hg hg hg hg hg sp3d 2
The valence bond theory therefore formally distinguishes between “inner-shell” complexes, which use 3d, 4s, and 4p orbitals to form a set of d2sp3 hybrids, and “outer-shell” complexes, which use 4s, 4p, and 4d orbitals to form sp3d2 hybrid orbitals.
TM.12 CRYSTAL FIELD THEORY At almost exactly the same time that chemists were developing the valence bond model for coordination complexes, physicists such as Hans Bethe, John Van Vleck, and Leslie Orgel were developing an alternative known as crystal field theory. Valence bond theory was used to explain the chemical properties of coordination complexes, such as the fact that the Co3 ion forms a six-coordinate Co(NH3)63 complex. Crystal field theory was developed to explain some of the physical properties of the complexes, such as their color and their behavior in a magnetic field.
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Crystal-field theory can be understood by thinking about the effect of the electrical field of neighboring ions on the energies of the valence orbitals of the transition-metal ions in manganese(II) oxide, MnO, and copper(I) chloride, CuCl.
MnO: Octahedral Crystal Fields Each Mn2 ion in manganese(II) oxide is surrounded by six O2 ions arranged toward the corners of an octahedron, as shown in Figure TM.14. MnO is therefore a model for an octahedral complex in which a transition metal ion is coordinated to six ligands. O2–
O2–
O2– 2+
Mn O2–
O2–
FIGURE TM.14 The octahedral geometry of O2 ions that surround each Mn2 ion in MnO.
2–
O
Repulsion between electrons on the O2 ions and electrons in the 3d orbitals on the metal ion in MnO increases the energy of these orbitals. But this repulsion doesn’t affect the five 3d orbitals the same way. Let’s assume that the six O2 ions that surround each Mn2 ion define an xyz coordinate system. Two of the 3d orbitals (3dx 2y2 and 3dz2) on the Mn2 ion point directly toward the six O2 ions, as shown in Figure TM.15. The other three orbitals (3dxy, 3dxz, and 3dyz) lie between the O2 ions. z
z
x
x y
y
3dx2 – y2
3dz2 z
x y
x
x y
3dxy
z
z
y
3dxz
FIGURE TM.15 The five orbitals in a d subshell.
3dyz
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The energy of the five 3d orbitals increases when the six O2 ions are brought close to the Mn2 ion because of repulsion between the electrons on the O2 ions and the electrons in the d orbitals on the Mn2 ion. The energy of two of these orbitals (3dx 2y2 and 3dz2), however, increases much more than the energy of the other three (3dxy, 3dxz, and 3dyz), as shown in Figure TM.16. The crystal field of the six O2 ions in MnO therefore splits the degeneracy of the five 3d orbitals. Three of the orbitals are now lower in energy than the other two. dx2 – y2
dz2
eg ∆o
dxy
dxz
dyz
t2g
E
Octahedral crystal field
Isolated atom or ion
FIGURE TM.16 The two d orbitals that point toward the ligands in an octahedral complex are higher in energy than the three d orbitals that lie between the ligands.
By convention, the dxy, dxz, and dyz orbitals in an octahedral complex are called the t2g orbitals. The dx 2y2 and dz2 orbitals, on the other hand, are called the eg orbitals. The easiest way to remember this convention is to note that there are three orbitals in the t2g set. t2g eg
dxy, dxz, and dyz dx 2y2 and dz2
The difference between the energies of the t2g and eg orbitals in an octahedral complex is represented by the symbol o. The splitting of the energy of the d orbitals is not trivial; o for the Ti(H2O)63 ion, for example, is 242 kJ/mol. Which is roughly the same as the energy given off when one mole of water is produced by burning a mixture of H2 and O2. The magnitude of the splitting of the t2g and eg orbitals changes from one octahedral complex to another. It depends on the identity of the metal ion, the charge on that ion, and the nature of the ligands coordinated to the metal ion.
CuCl: Tetrahedral Crystal Fields Each Cu ion in copper(I) chloride is surrounded by four Cl ions arranged toward the corners of a tetrahedron, as shown in Figure TM.17. CuCl is therefore a model for a tetrahedral complex in which a transition metal ion is coordinated to four ligands. Cl–
Cu+
Cl–
Cl– Cl–
FIGURE TM.17 The tetrahedral geometry of Cl ions that surround each Cu ion in CuCl.
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Once again, the negative ions in the crystal split the energy of the d atomic orbitals on the transition metal ion. The tetrahedral crystal field splits these orbitals into the same t2g and eg sets of orbitals as does the octahedral crystal field. t2g eg
dxy, dxz, and dyz dx 2y2 and dz2
But the two orbitals in the eg set are now lower in energy than the three orbitals in the t2g set, as shown in Figure TM.18.
dxy
dxz
dx 2 – y2 dz 2 E
Tetrahedral crystal field
dyz
t2g
∆t = –49– ∆o
eg
Isolated atom or ion
FIGURE TM.18 In a tetrahedral complex, the eg orbitals are lower in energy than the t2g orbitals. The difference between the energies of the orbitals is smaller in tetrahedral complexes than in an equivalent octahedral complex.
To understand the splitting of d orbitals in a tetrahedral crystal field, imagine four ligands lying at alternating corners of a cube to form a tetrahedral geometry, as shown in Figure TM.19. The dx 2y2 and dz2 orbitals on the metal ion at the center of the cube lie between the ligands, and the dxy, dxz, and dyz orbitals point toward the ligands. As a result, the splitting observed in a tetrahedral crystal field is the opposite of the splitting in an octahedral complex. z
x
y
FIGURE TM.19 In a tetrahedral complex, dxy, dxz, and dyz orbitals point toward the ligands; the dx 2y2 and dz2 orbitals point between the ligands.
Because a tetrahedral complex has fewer ligands, the magnitude of the splitting is smaller. The difference between the energies of the t2g and eg orbitals in a tetrahedral
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complex (t) is slightly less than half as large as the splitting in analogous octahedral complexes (o). t 4/9 o
Square-Planar Complexes The crystal field theory can be extended to square-planar complexes, such as Pt(NH3)2Cl2. The splitting of the d orbitals in these compounds is shown in Figure TM.20.
dx2 – y2 ∆O dz2 ∼ –2– ∆O
dxy dyz
3
dxz
E
Squareplanar crystal field
Isolated atom or ion
FIGURE TM.20 The splitting of the d orbitals in a square-planar complex.
TM.13 THE SPECTROCHEMICAL SERIES The splitting of the d orbitals in the crystal field model not only depends on the geometry of the complex, it also depends on the nature of the metal ion, the charge on that ion, and the ligands that surround the metal. When the geometry and the ligands are held constant, the splitting decreases in the following order. Pt4 Ir3 Rh3 Co3 Cr3 Fe3 Fe2 Co2 Ni2 Mn2 Strong-field ions
Weak-field ions
Metal ions at one end of the continuum are called strong-field ions, because the splitting due to the crystal field is unusually strong. Ions at the other end are known as weak-field ions. When the geometry and the metal are held constant, the splitting of the d orbitals decreases in the following order. CO CN NO2 NH3 ONCS H2O OH F OSCN Cl Br Strong-field ligands
Weak-field ligands
Ligands that give rise to large differences between the energies of the t2g and eg orbitals are called strong-field ligands. Those at the opposite extreme are known as weak-field ligands. Because they result from studies of the absorption spectra of transition metal complexes, these generalizations are known as the spectrochemical series. The range of values of for a given geometry is remarkably large. The value of o is 100 kJ/mol in the Ni(H2O)62 ion, for example, and 520 kJ/mol in the Rh(CN)63 ion.
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TM.14 HIGH-SPIN VERSUS LOW-SPIN OCTAHEDRAL COMPLEXES Once we know the relative energies of the d orbitals in a transition metal complex, we have to worry about how these orbitals are filled. Orbitals that have the same energy are called degenerate orbitals. Degenerate orbitals are filled according to Hund’s rules. • •
One electron is added to each of the degenerate orbitals in a subshell before a second electron is added to any orbital in the subshell. Electrons are added to a subshell with the same value of the spin quantum number until each orbital in the subshell has at least one electron.
Octahedral transition metal ions with d1, d2, and d3 configurations can therefore be described by the following diagrams. d1
h
d2
h
h
d3
h
h
h
When we try to add a fourth electron, we are faced with a problem. The electron could be used to pair one of the electrons in the lower energy (t2g) set of orbitals or it could be placed in one of the higher energy (eg) orbitals. One of the configurations is called lowspin because it contains only two unpaired electrons. The other is called high-spin because it contains four unpaired electrons with the same spin. The same problem occurs with octahedral d5, d6, and d7 complexes. Low-spin d4
hg h
h
hg hg h
High-spin h h h h h
h
h
h
h
d
5
d
6
hg hg hg
hg h
d
7
h hg hg hg
h h hg hg h
h
h
h
For octahedral d8, d9, and d10 complexes, there is only one way to write satisfactory configurations.
d
8
h h hg hg hg
d
9
hg h hg hg hg
d
10
hg hg hg hg hg
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As a result, we have to differentiate between high-spin and low-spin octahedral complexes only when there are either four, five, six, or seven electrons in the d orbitals. The choice between high-spin and low-spin configurations for octahedral d4, d5, d6, and 7 d complexes is based on by comparing the energy it takes to pair electrons with the energy it takes to excite an electron to the higher energy (eg) orbitals. If it takes less energy to pair the electrons, the complex is low-spin. If it takes less energy to excite the electron, the complex is high-spin. The amount of energy required to pair electrons in the t2g orbitals of an octahedral complex is more or less constant. The amount of energy needed to excite an electron into the higher energy (eg) orbitals, however, depends on the value of o for the complex. As a result, we expect to find low-spin complexes among metal ions and ligands that lie toward the high-field end of the spectrochemical series. High-spin complexes are expected among metal ions and ligands that lie toward the low-field end of the series. ∆ο
hg hg hg Low-spin d
6
h h hg h h High-spin d
∆ο 6
Compounds in which all of the electrons are paired are diamagnetic—they are repelled by both poles of a magnet. Compounds that contain one or more unpaired electrons are paramagnetic—they are attracted to the poles of a magnet. The force of attraction between paramagnetic complexes and a magnetic field is proportional to the number of unpaired electrons in the complex. Thus, we can determine whether a complex is high-spin or lowspin by measuring the strength of the interaction between the complex and a magnetic field.
Exercise TM.6 Explain why the Co(NH3)63 ion is diamagnetic, whereas the CoF63 ion is paramagnetic. Solution Both complexes contain the Co3 ion, which lies toward the strong-field end of the spectrochemical series. NH3 also lies toward the strong-field end of the series. As a result, o for the Co(NH3)63 ion should be relatively large. When the splitting is large, it takes less energy to pair the electrons in the t2g orbitals than it does to excite the electrons to the eg orbitals. The Co(NH3)63 ion is therefore a low-spin d6 complex in which the electrons are all paired, and the ion is diamagnetic. The F ion lies toward the low-field end of the spectrochemical series. As a result, o for the CoF63 ion is much smaller. In this complex, it apparently takes less energy to excite electrons into the eg orbitals than to pair them in the t2g orbitals. As a result, the CoF63 ion is a high-spin d6 complex that contains unpaired electrons and is therefore paramagnetic.
TM.15 THE COLORS OF TRANSITION METAL COMPLEXES One of the joys of working with transition metals is the extraordinary range of colors exhibited by their compounds. By changing the number of ammonia, water, and chloride ligands on a Co3 ion, for example, we can vary the color of the complex over the entire
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visible spectrum, from red to violet, as shown in Section TM.2. We can also vary the color of the complex by keeping the ligand constant and changing the metal ion. The characteristic colors of aqueous solutions of some common transition metal ions are given in Table TM.5. TABLE TM.5 Characteristic Colors of Common Transition Metal Ion Aquo and Other Complexes Ion Cr2 Cr3 Co2 Cu2 Fe3 Mn2
Color Blue Blue-violet Pinkish red Light blue Yellow Very faint pink
Ion
Color
Ni2 Zn2 CrO42 Cr2O72 MnO4
Green Colorless Yellow Orange Deep violet
The color of these complexes also changes with the oxidation state of the metal atom. By reducing the VO2 ion with zinc metal, for example, we can change the color of the solution from yellow to green to blue and then to violet as the vanadium is reduced, one electron at a time, from the 5 to the 2 oxidation state. VO2 VO2 V3 V2
yellow green blue violet
To explain why transition metal complexes are colored, and why the color is so sensitive to changes in the metal, ligand, and oxidation state, we need to understand the physics of color. There are two ways of producing the sensation of color. We can add color where none exists or subtract it from white light. The three primary additive colors are red, green, and blue. When all three are present at the same intensity, we get white light. The three primary subtractive colors are cyan, magenta, and yellow. When those three colors are absorbed with the same intensity, light is absorbed across the entire visible spectrum. If the object absorbs strongly enough, or if the intensity of the light is dim enough, all of the light can be absorbed.
Primary additive colors
Primary subtractive colors
Green
Cyan
Yellow
Blue
Red Magenta
Yellow
Green
Red
Cyan
Magenta Blue
FIGURE TM.21 The primary additive and subtractive colors.
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The additive and subtractive colors are complementary, as shown in Figure TM.21. If we subtract yellow from white light, we get a mixture of cyan and magenta, and the light looks blue. If we subtract blue—a mixture of cyan and magenta—from white light, the light looks yellow. The Cu(NH3)42 complex ion has a blue color because it absorbs light in the yellow portion of the spectrum. The CrO42 ion appears yellow because it absorbs blue light. Table TM.6 describes what happens when light in different portions of the visible spectrum is absorbed.
TABLE TM.6 Relationship between the Color of Transition Metal Complexes and the Wavelength of Light Absorbed Wavelength Absorbed (nm)
Color of Light Absorbed
Color of Complex
410 430 480 500 530 560 580 610 680
Violet Indigo Blue Blue-green Green Lemon yellow Yellow Orange Red
Lemon yellow Yellow Orange Red Purple Violet Indigo Blue Blue-green
Light is absorbed when it carries just enough energy to excite an electron from one orbital to another. Compounds therefore absorb light when the difference between the energy of an orbital that contains an electron and the energy of an empty orbital corresponds to a wavelength in the narrow band of the electromagnetic spectrum that is visible to the naked eye. How much energy is associated with the absorption of a typical photon, such as a photon with a wavelength of 480 nanometers? Since we know the wavelength of the photon, we can calculate its frequency. c 2.998 108 m/s 6.25 1014 s1 480 109 m We can then calculate the energy of the photon from its frequency and Planck’s constant. E h (6.626 1034 Js)(6.25 1014 s1) 4.14 1019 J But this is the energy associated with only a single photon. Multiplying by Avogadro’s number gives us the energy in units of kilojoules per mole. 6.022 1023 photons 4.14 1019 J 249 kJ/mol 1 mol 1 photon Repeating the calculation for all frequencies in the visible portion of the spectrum reveals that the energy associated with a photon of light ranges from 160 to 300 kJ/mol. If we refer back to the discussion of the spectrochemical series in Section TM.13, we find that
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these energies fall in the range of values of . Anything that changes the difference between the energies of the t2g and eg orbitals in a transition metal complex therefore influences the color of the light absorbed by the complex. Thus, it isn’t surprising that the color of the complex is sensitive to a variety of factors, including the geometry of the complex, the identity of the metal, the nature of the ligand, and the oxidation state of the metal.
TM.16 LIGAND FIELD THEORY The valence bond model described in Section TM.11 and the crystal field theory described in Section TM.12 each explain some aspects of the chemistry of the transition metals, but neither model is good at predicting all of the properties of transition metal complexes. A third model, based on molecular orbital theory, was therefore developed that is known as ligand field theory. Ligand field theory is more powerful than either the valence bond or crystal field theories. Unfortunately, it is also more abstract. The ligand field model for an octahedral transition metal complex such as the Co(NH3)63 ion assumes that the 3d, 4s, and 4p orbitals on the metal overlap with one orbital on each of the six ligands to form a total of 15 molecular orbitals, as shown in Figure TM.22. Six of the orbitals are bonding molecular orbitals, whose energies are much lower than those of the original atomic orbitals. Another six are antibonding molecular orbitals, whose energies are higher than those of the original atomic orbitals. Three are best described as nonbonding molecular orbitals, because they have essentially the same energy as the 3d atomic orbitals on the metal.
4p
4s ∆O
E
3d Metal orbitals
Ligand orbitals
FIGURE TM.22 The molecular orbitals in this ligand field diagram for Co(NH3)63 were generated by allowing the valence-shell 3d, 4s, and 4p orbitals on the transition metal to overlap with an orbital on each of the six ligands that contribute pairs of nonbonding electrons to form an octahedral complex.
Ligand field theory enables the 3d, 4s, and 4p orbitals on the metal to overlap with orbitals on the ligand to form the octahedral covalent bond skeleton that holds the complex together. At the same time, the model generates a set of five orbitals in the center of the
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diagram that are split into t2g and eg subshells, as predicted by the crystal field theory. As a result, we don’t have to worry about “inner-shell” versus “outer-shell” metal complexes. In effect, we can use the 3d orbitals in two different ways. We can use them to form the covalent bond skeleton and then use them again to form the orbitals that hold the electrons that were originally in the 3d orbitals of the transition metal.
KEY TERMS Actinide Bidentate ligand Chelating ligand Chiral Cis Complex ion Coordination compound Coordination number Crystal field theory Dextrorotatory
Diamagnetic Electron pair acceptor/donor High-spin Isomers Lanthanide Levorotatory Lewis acid/base Ligand Ligand field theory
Low-spin Main-group elements Monodentate Optically active Paramagnetic Spectrochemical series Trans Transition metals Valence bond theory
PROBLEMS Transition Metals and Coordination Complexes 1. Identify the subshell of atomic orbitals filled among the transition metals in the fifth row of the periodic table. 2. Describe some of the ways in which transition metals differ from main-group metals such as aluminum, tin, and lead. Describe ways in which they are similar. 3. Use the electron configurations of Zn2, Cd2, and Hg2 to explain why those ions often behave as if they were main-group metals. 4. Use the electronegativities of cobalt and nitrogen to predict whether the CoON bond in the Co(NH3)63 ion is best described as ionic, polar covalent, or covalent. 5. Define the terms coordination number and ligand.
Werner’s Model of Coordination Complexes 6. Werner wrote the formula of one of his coordination complexes as CoCl36 NH3. Today, we write that compound as [Co(NH3)6]Cl3 to indicate the presence of Co(NH3)63 and Cl ions. Write modern formulas for the compounds Werner described as CoCl35 NH3, CoCl34 NH3, and CoCl35 NH3H2O. 7. What is the charge on the cobalt ion in the following complex [Co(NH3)6]Cl3? 8. Explain why aqueous solutions of CoCl36 NH3 conduct electricity better than aqueous solutions of CoCl34 NH3. 9. Explain why Ag ions precipitate three chloride ions from an aqueous solution of CoCl36 NH3 but only one chloride ion from an aqueous solution of CoCl34 NH3. 10. Predict the number of Cl ions that could precipitate from an aqueous solution of the Co(en)2Cl2 complex.
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11. Predict the number of ions formed when the NiSO44 NH32 H2O complex dissociates in water.
Typical Coordination Numbers 12. Use the examples in Table TM.1 to identify at least one factor that influences the coordination number of a transition metal ion. 13. Determine both the coordination number and the charge on the transition metal ion in each of the following complexes. (a) CuF42 (b) Cr(CO)6 (c) Fe(CN)64 (d) Pt(NH3)2Cl2 14. Determine both the coordination number and the charge on the transition metal ion in each of the following complexes. (a) Co(SCN)42 (b) Fe(acac)3 (c) Ni(en)2(H2O)22 (d) Co(NH3)5(H2O)3
The Electron Configuration of Transition Metal Ions 15. Write the electron configuration of the following transition metal ions. (a) V2 (b) Cr2 (c) Mn2 (d) Fe2 (e) Ni2 16. Explain why the Co2 ion can be described as a d7 ion. 17. Which of the following ions can be described as d5? (a) Cr2 (b) Mn2 (c) Fe3 (d) Co3 (e) Cu 18. Explain the difference between the symbols Cr3 and Cr(VI). 19. Which of the following is not an example of a d0 transition metal complex? (a) TiO2 (b) VO2 (c) Cr2O72 (d) MnO4 20. Explain why manganese becomes more electronegative when it is oxidized from Mn2 to Mn(VII). 21. Use electronegativities to predict whether the MnOO bond in MnO4 is best described as covalent or ionic.
Lewis Acid–Lewis Base Approach to Bonding in Complexes 22. Describe what to look for when deciding whether an ion or molecule is a Lewis acid. 23. Explain how Lewis acids, such as the Co3 ion, pick up Lewis bases, or ligands, to form coordination complexes. 24. Which of the following are Lewis acids? (a) Fe3 (b) BF3 (c) H2 (d) Ag (e) Cu2 25. Which of the following are Lewis bases, and therefore potential ligands? (a) CO (b) O2 (c) Cl (d) N2 (e) NH3 26. Which of the following are Lewis bases, and therefore potential ligands? (a) CN (b) SCN (c) CO32 (d) NO (e) S2O32
Typical Ligands 27. Define the terms monodentate, bidentate, tridentate, and tetradentate. Give an example of each category of ligands. 28. Use Lewis structures to explain why carbon monoxide could act as a bridge between a pair of transition metals, but it can’t be a chelating ligand that coordinates to the same metal twice. 29. Draw the structures of the following coordination complexes. (a) Fe(acac)3 (b) Co(en)33 (c) Fe(EDTA) (d) Fe(CN)64
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Nomenclature of Complexes 30. Name the following complexes. (a) Cu(NH3)42 (b) Mn(H2O)62 (c) Fe(CN)64 (d) Ni(en)32 (e) Cr(acac)3 31. Name the following complexes. (a) Pt(NH3)2Cl2 (b) Ni(CO)4 (c) Co(en)33 32. Name the following complexes. (a) Na3[Co(NO2)6] (b) Na2[Zn(CN)4] (c) [Co(NH3)4Cl2]Cl (d) [Ag(NH3)2]Cl 33. Write the formulas for the following compounds. (a) hexamminechromium(III) chloride (b) chloropentamminechromium(III) chloride (c) trisethylenediamminecobalt(III) chloride (d) potassium tetranitritodiamminecobaltate(III)
Isomers 34. Which of the following octahedral complexes can form cis/trans isomers? (a) Co(NH3)63 (b) Co(NH3)5Cl2 (c) Co(NH3)5(H2O)3 (d) Co(NH3)4Cl2 (e) Co(NH3)4(H2O)23 35. Predict the structures of the cis/trans isomers of Ni(en)2(H2O)22. 36. The octahedral Mo(PH3)3(CO)3 complex can exist as a pair of isomers. Predict the structures of the compounds. 37. Which of the following square-planar complexes can form cis/trans isomers? (a) Cu(NH3)42 (b) Pt(NH3)2Cl2 (c) RhCl3(CO) (d) IrCl(CO)(PH3)2 38. Explain why square-planar complexes with the generic formula MX2Y2 can form cis/trans isomers, but tetrahedral complexes with the same generic formula cannot. 39. Explain why square-planar complexes with the generic formula MX3Y can’t form cis/trans isomers. 40. Use the fact that Rh(CO)(H)(PH3)2 forms cis/trans isomers to predict whether the geometry around the transition metal is square planar or tetrahedral. 41. Compounds are optically active when the mirror image of the compound cannot be superimposed on itself. Draw the mirror images of the following complex ions and determine which of the ions are chiral. (a) Cu(NH3)42 (a square-planar complex) (b) Co(NH3)62 (an octahedral complex) (c) Ag(NH3)2 (a linear complex) (d) Cr(en)33 (an octahedral complex) 42. Explain why neither of the cis/trans isomers of Pt(NH3)2Cl2 is chiral. 43. Explain why the cis isomer of Ni(en)2(H2O)22 is chiral, but the trans isomer is not. 44. Explain why the cis isomer of Ni(en)2(H2O)22 is chiral, but the cis isomer of Ni(NH3)4(H2O)22 is not. 45. Which of the following octahedral complexes are chiral? (a) Cr(acac)3 (b) Cr(C2O4)33 (c) Cr(CN)63 (d) Cr(CO)4(NH3)2
The Valence Bond Approach to Bonding in Complexes 46. Describe the difference between the valence bond model for the Co(NH3)63 complex ion and the valence bond model for the Ni(NH3)62 complex ion. 47. Apply the valence bond model of bonding in transition metal complexes to Ni(CO)4, Fe(CO)5, and Cr(CO)6. (Hint: Assume that the valence electrons on transition metals in complexes in which the metal is in the zero oxidation state are concentrated in the d orbitals.)
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48. Use the results of the previous problem to explain why the transition metal is sp3 hybridized in Ni(CO)4, dsp3 hybridized in Fe(CO)5, and d2sp3 hybridized in Cr(CO)6. 49. Use valence bond theory to explain the 2 charge on the Fe(CO)42 ion. Predict the charge on the equivalent Co(CO)4x ion. Use acid–base chemistry to predict the charge on the HFe(CO)4x ion. 50. Use Lewis structures to explain what happens in the following reaction. 88n Cr2O72(aq) H2O(l) 2 CrO42(aq) 2 H(aq) m88 Predict the charge on the product of the following reaction. Explain why this transitionmetal compound is a gas at room temperature. 88n Mn2O7x(l) H2O(l) MnO42(aq) 2 H(aq) m88 51. Apply the valence bond model of bonding in transition metal complexes to the Zn(NH3)42 and Fe(H2O)63 complex ions. 52. Use the assumption that transition metals often pick up enough ligands to fill their valence shell to predict the charge on the Mn(CO)5x ion. 53. Use the assumption that transition metals often pick up enough ligands to fill their valence shell to predict the charge on the HgI4x ion. 54. Use the assumption that transition metals often pick up enough ligands to fill their valence shell to predict the coordination number of the Cd2 ion in the Cd(OH)x2 ion. 55. Explain why Zn2 ions form both Zn(CN)42 and Zn(NH3)42 ions.
Crystal Field Theory 56. Describe what happens to the energies of the 3d atomic orbitals in an octahedral crystal field. 57. Describe what happens to the energies of the 3d atomic orbitals in a tetrahedral crystal field. 58. Which of the 3d atomic orbitals in an octahedral crystal field belong to the t2g set of orbitals? Which belong to the eg set? 59. In an octahedral field what do the orbitals in a t2g set have in common? What do the orbitals in the eg set have in common? 60. The 3d orbitals are split into t2g and eg sets in both octahedral and tetrahedral crystal fields. Is there any difference between the orbitals that go into the t2g set in octahedral and in tetrahedral crystal fields? 61. Explain why t for a tetrahedral complex is much smaller than o for the analogous octahedral complex. 62. Use the splitting of the 3d atomic orbitals in an octahedral crystal field to explain the stability of the oxidation states corresponding to d3 and d6 electron configurations in the Cr(NH3)63 and Fe(CN)64 complex ions. 63. The difference between the energies of the t2g and eg sets of atomic orbitals in an octahedral or tetrahedral crystal field depends on both the metal atom and the ligands that form the complex. Which of the following metal ions would give the largest difference? (a) Rh3 (b) Cr3 (c) Fe3 (d) Co2 (e) Mn2 64. Which of the following ligands would give the largest value of o? (a) CN (b) NH3 (c) H2O (d) OH (e) F
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High-Spin versus Low-Spin Complexes Describe the difference between a high-spin and a low-spin d6 complex. What factors determine whether a complex is high-spin or low-spin? Explain why the Mn(H2O)62 ion is a high-spin complex. One of the Fe(H2O)62 and Fe(CN)64 complex ions is high-spin and the other is lowspin. Which is which? 69. Use the relative magnitudes of o and t to explain why there are no low-spin tetrahedral complexes. 70. Compare the positions of the Co2 with Co3 and Fe2 with Fe3 ions in the spectrochemical series. Explain why the value of generally increases with the charge on the transition metal ion. 71. Compare the positions of the Co3, Rh3, and Ir3 ions in the spectrochemical series. What happens to the value of as we go down a column among the transition metals? 65. 66. 67. 68.
The Colors of Transition Metal Complexes 72. Describe the characteristic colors of aqueous solutions of the following transition metal ions. (a) Cu2 (b) Fe3 (c) Ni2 (d) CrO42 (e) MnO4 73. Explain why so many of the pigments used in oil paints, such as vermilion (HgS), cadmium red (CdS), cobalt yellow [K3Co(NO2)6], chrome yellow (PbCrO4), Prussian blue (Fe4[Fe(CN)6]3), and cobalt blue (CoOAl2O3), contain transition metal ions. 74. Explain why Cu(NH3)42 complexes have a deep-blue color if they don’t absorb blue light. What light do they absorb? 75. CrO42 ions are bright yellow. In what portion of the visible spectrum do the ions absorb light? 76. When CrO42 reacts with acid to form Cr2O72 ions, the color shifts from bright yellow to orange. Does this mean that the light absorbed shifts toward a higher or a lower frequency? 77. Ni2 forms a complex with the dimethylglyoxime (DMG) ligand that absorbs light in the blue-green portion of the spectrum. What is the color of the Ni(DMG)2 complex? 78. Explain why a white piece of paper looks as if it has a faint pink color to a person who has been working for several hours at a computer terminal that has a green screen.
Ligand Field Theory 79. Describe how ligand field theory eliminates the difference between inner-shell complexes, such as the Co(NH3)63 ion, and outer-shell complexes, such as the Ni(NH3)62 ion. 80. Explain how ligand field theory allows the valence-shell d orbitals on the transition metal to be used simultaneously to form the skeleton structure of the complex and to hold the electrons that were originally in the d orbitals on the transition metal.
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M O D U L E
3 COMPLEX ION EQUILIBRIA COM.1 COM.2 COM.3 COM.4 COM.5 COM.6 COM.7
Complex Ions The Stepwise Formation of Complex Ions Complex Dissociation Equilibrium Constants Approximate Complex Ion Calculations Using Complex Ion Equilibria to Dissolve an Insoluble Salt A Qualitative View of Combined Equilibria A Quantitative View of Combined Equilibria
COM.1 COMPLEX IONS The basic assumption behind the discussion of solubility equilibria is the idea that salts dissociate into their ions when they dissolve in water. Copper sulfate, for example, dissociates into the Cu2 and SO42 ions in water. HO
2 88n CuSO4(s) m88 Cu2(aq) SO42(aq)
If we add 2 M NH3 to the solution, the first thing we notice is the formation of a light blue, almost bluish-white, precipitate. This can be explained by combining what we know about acid–base and solubility equilibria. Ammonia acts as a base toward water to form a mixture of the ammonium and hydroxide ions. 88n NH4(aq) OH(aq) NH3(aq) H2O(l) m88
Kb 1.8 105
The OH ions formed in the reaction combine with Cu2 ions in the solution to form an insoluble Cu(OH)2 precipitate. 88n Cu(OH)2(s) Cu2(aq) 2 OH(aq) m88
Ksp 2.2 1020
In theory, the OH ion concentration should increase when more ammonia is added to the solution. As a result, more Cu(OH)2 should precipitate from the solution. At first, this is exactly what happens. In the presence of excess ammonia, however, the Cu(OH)2 precipitate dissolves, and the solution turns deep blue. This raises an important question: “Why does the Cu(OH)2 precipitate that forms in dilute ammonia dissolve in excess ammonia?” The first step toward answering the question involves writing the electron configuration of copper metal and its Cu2 ion. 1
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[Ar] 4s1 3d10 [Ar] 3d9
Cu Cu2
It is sometimes useful to think about the electron configuration of the Cu2 ion in terms of the entire set of valence-shell orbitals. In addition to the nine electrons in the 3d subshell, the Cu2 ion has an empty 4s orbital and a set of three empty 4p orbitals. Cu2
[Ar] 4s0 3d9 4p0
The empty 4s and 4p orbitals on the Cu2 ion are used to pick up pairs of nonbonding electrons from four NH3 molecules to form a Cu(NH3)42 ion, as shown in Figure COM.1. H
H
H N
H
N
Cu2+ N
NH3
H H
H
2
NH3
H
H
Cu
NH3
NH3
N H H
H
FIGURE COM.1 The Cu2 ion has four empty valence-shell orbitals that can accept a pair of nonbonding electrons from an NH3 molecule to form a covalent CuON bond.
G. N. Lewis was the first to recognize the similarity between this reaction and the acid–base reaction in which an H ion combines with an OH ion to form a water molecule. O OH H SO Q
O OH HO O Q
Both reactions involve the transfer of a pair of nonbonding electrons from one atom to an empty orbital on another atom to form a covalent bond. Both reactions can therefore be interpreted in terms of an electron pair acceptor combining with an electron pair donor. Lewis suggested that we could expand our definition of acids by assuming that an acid is any substance that acts like the H ion to accept a pair of nonbonding electrons. A Lewis acid is therefore an electron pair acceptor. A Lewis base, on the other hand, is any substance that acts like the OH ion to donate a pair of nonbonding electrons. A Lewis base is therefore an electron pair donor. Checkpoint Use Lewis structures to predict which of the following can act as an electron pair donor, or Lewis base. (a) H2O (b) CO (c) O2 (d) H (e) BH3 The product of the reaction of a Lewis acid with a Lewis base is an acid–base complex. When the Cu2 ion reacts with four NH3 molecules, the product of the reaction is called a complex ion. Cu2(aq) Electron pair acceptor (Lewis acid)
4 NH3(aq) Electron pair donor (Lewis base)
88n Cu(NH3)42(aq) m88 Acid–base complex or complex ion
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3
Any atom or molecule with at least one empty valence-shell orbital can be a Lewis acid. Any atom or molecule that contains at least one pair of nonbonding electrons is a Lewis base. All of the substances whose Lewis structures are shown in Figure COM.2, for example, can act as Lewis bases to form complex ions. O H
N
O
H H H
H –
O H
C
N
–
O
F
C O
–
–
Cl
O
Br
C O
–
I
–
2–
O –
S
C
N
S
C
S
N
C
N
S
S
O
O FIGURE COM.2 Lewis structures of some potential Lewis bases.
COM.2 THE STEPWISE FORMATION OF COMPLEX IONS When a transition metal ion binds one or more Lewis bases to form an acid–base complex, it picks up the Lewis bases one at a time. The Ag ion, for example, combines with NH3 in a two-step reaction. It first picks up one NH3 molecule to form a one-coordinate complex. 88n Ag(NH3)(aq) Ag(aq) NH3(aq) m88 This intermediate then picks up a second NH3 molecule in a separate step. 88n Ag(NH3)2(aq) Ag(NH3)(aq) NH3(aq) m88 It is possible to write equilibrium constant expressions for each step in these complex ion formation reactions. The complex formation equilibrium constant (Kf) expressions for the two steps in the formation of the Ag(NH3)2 complex ion are written as follows. [Ag(NH3)] Kf1 [Ag][NH3] [Ag(NH3)2] Kf2 [Ag(NH3)][NH3] The difference between Kf1 and Kf2 for the complexes between Ag and ammonia is only a factor of 4. Ag(aq) NH3(aq) Ag(NH3)(aq) NH3(aq)
88n Ag(NH3)(aq) m88 88n Ag(NH3)2(aq) m88
Kf1 1.7 103 Kf2 6.5 103
This means that most of the Ag ions that pick up an NH3 molecule to form the onecoordinate Ag(NH3) complex ion are likely to pick up a second NH3 molecule to form
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the two-coordinate Ag(NH3)2 complex ion. Table COM.1 summarizes the concentrations of the Ag, Ag(NH3), and Ag(NH3)2 ions over a range of NH3 concentrations. TABLE COM.1 Effect of Changes in the NH3 Concentration on the Fraction of Silver Present as the Ag, Ag(NH3), and Ag(NH3)2 Ions [NH3] (M)
Ag (%)
Ag(NH3) (%)
Ag(NH3)2 (%)
106 105 104 103 102 101
99.8 98.2 78.1 7.3 0.09 0.0009
0.2 1.7 13.3 12.4 1.5 0.2
0.001 0.1 8.6 80.4 98.4 99.8
Data from Table COM.1 are plotted in Figure COM.3. Essentially all of the silver is present as the Ag ion at very low concentrations of NH3. As the NH3 concentration increases, the dominant species soon becomes the two-coordinate Ag(NH3)2 ion. Even at NH3 concentrations as small as 0.0010 M, most of the silver is present as the Ag(NH3)2 ion.
Proportion of total silver ion concentration
100%
80% [Ag+]
[Ag(NH3)2+]
60%
40%
20% [Ag(NH3)+] 0 10–6
10–5
10–4
10–3
10–2
10–1
[NH3] (M)
FIGURE COM.3 The effect of changes in the NH3 concentration on the proportion of the total silver ion concentration present as Ag, Ag(NH3), and Ag(NH3)2 ions.
The concentration of the one-coordinate Ag(NH3) intermediate is never very large. Either the NH3 concentration is so small that most of the silver is present as the Ag ion, or it is large enough that essentially all of the silver is present as the two-coordinate Ag(NH3)2 complex ion. If the only important components of the equilibrium are the free Ag ion (at low NH3 concentrations) and the two-coordinate Ag(NH3)2 complex ion (at moderate to high NH3 concentrations), we can collapse the individual steps in the reaction into the following overall equation. 88n Ag(NH3)2(aq) Ag(aq) 2 NH3(aq) m88
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The overall complex formation equilibrium constant expression for the reaction is written as follows. [Ag(NH3)2] Kf [Ag][NH3]2 This expression is equal to the product of the equilibrium constant expressions for the individual steps in the reaction. [Ag(NH3)2] [Ag(NH3)2] [Ag(NH3)] [Ag(NH3)][NH3] [Ag][NH3]2 [Ag][NH3] The overall complex formation equilibrium constant is therefore equal to the product of the Kf values for the individual steps. Kf Kf1 Kf2 1.1 107 Overall complex formation equilibrium constants for common complex ions can be found in Appendix B.1.
Exercise COM.1 Calculate the complex formation equilibrium constant for the two-coordinate Fe(SCN)2 complex ion from the following data. Fe3(aq) SCN(aq) Fe(SCN)2(aq) SCN(aq)
88n Fe(SCN)2(aq) m88 88n Fe(SCN)2(aq) m88
Kf1 890 Kf2 2.6
Solution The difference between the stepwise formation equilibrium constants for the complex is relatively small. [Fe(SCN)2] Kf1 890 [Fe3][SCN] [Fe(SCN)2] Kf2 2.6 [Fe(SCN)2][SCN] Solutions of these complex ions are therefore best described in terms of an overall complex formation equilibrium when there is an excess of the ligand. 88n Fe(SCN)2(aq) Fe3(aq) 2 SCN(aq) m88 The equilibrium constant expression for the overall reaction is equal to the product of the expressions for the individual steps in the reaction. [Fe(SCN)2] [Fe(SCN)2] [Fe(SCN)2] [Fe(SCN)2][SCN] [Fe3][SCN]2 [Fe3][SCN]
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The overall equilibrium constant is therefore the product of Kf1 and Kf2. Kf Kf1 Kf2 2.3 103
COM.3 COMPLEX DISSOCIATION EQUILIBRIUM CONSTANTS Complex ions can also be described in terms of complex dissociation equilibria. We can start by assuming, for example, that most of the silver ions in an aqueous solution of ammonia are present as the two-coordinate Ag(NH3)2 complex ion. We then assume that some of the ions dissociate to form Ag(NH3) complex ions and then eventually Ag ions. 88n Ag(NH3)(aq) NH3(aq) Ag(NH3)2(aq) m88 88n Ag(aq) NH3(aq) Ag(NH3)(aq) m88 A complex dissociation equilibrium constant (Kd) expression can be written for each of the reactions. [Ag(NH3)][NH3] Kd1 [Ag(NH3)2] [Ag][NH3] Kd2 [Ag(NH3)] Alternatively, the individual steps in the reaction can be collapsed into an overall equation, which can be described by an overall equilibrium constant expression. 88n Ag(aq) 2 NH3(aq) Ag(NH3)2(aq) m88 [Ag][NH3]2 Kd [Ag(NH3)2]
Exercise COM.2 Calculate the complex dissociation equilibrium constant for the Cu(NH3)42 ion from the value of Kf for the complex. [For Cu(NH3)42, Kf 2.1 1013.] Solution The equilibrium expression for the complex formation reaction is written as follows. [Cu(NH3)42] Kf [Cu2][NH3]4 The equilibrium constant expression for the complex dissociation reaction is nothing more or less than the inverse of the complex formation equilibrium expression. [Cu2][NH3]4 Kd [Cu(NH3)42]
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7
The value of Kd is therefore equal to the inverse of Kf. 1 Kd 4.8 1014 Kf
COM.4 APPROXIMATE COMPLEX ION CALCULATIONS Complex formation equilibria provide another example of the general rule that it is useful to begin equilibrium calculations by comparing the reaction quotient for the initial conditions with the equilibrium constant for the reaction. Consider, for example, the following question: What fraction of the total iron(III) concentration is present as the Fe 3 ion in a solution that was initially 0.10 M Fe 3 and 1.0 M SCN? [For Fe(SCN)2, Kf 2.3 10 3.] The initial conditions for the reaction can be summarized as follows. Initial Initial
Fe3+ + SCN– Fe(SCN)2+ Fe(SCN)2+ + SCN– Fe(SCN)2+
Initial: [Fe3+] = 0.10 M [SCN–] = 1.0 M
∆(SCN–) = 2∆(Fe3+) as we form the Fe(SCN)2+ complex
88n Fe(SCN)2(aq) Fe3(aq) 2 SCN(aq) m88 0.10 M 1.0 M 0
Kf 2.3 103
The initial value of the reaction quotient is therefore equal to zero. (Fe(SCN)2) (0) 3 Qf 3 2 0 HO CT A A A CH3 H H 3°
2°
1°
The activation energy for the chain propagation steps in free radical bromination reactions is significantly larger than the activation energy for the steps during chlorination. As a result, free radical bromination reactions are more selective than chlorination reactions. Bromination reactions are far more likely to give the product predicted from the relative stability of the free radical intermediate. Bromination of 2-methylpropane, for example, gives almost exclusively 2-bromo-2-methylpropane, not the statistically more likely 1-bromo-2-methylpropane.
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ORGANIC REACTION MECHANISMS
CH3 A CH3CHCH3 Br2
CH3 A CH3CCH3 HBr A Br 99%
O3.8 BIMOLECULAR NUCLEOPHILIC SUBSTITUTION OR SN2 REACTIONS Most of our knowledge of the mechanisms of chemical reactions has come from the study of the factors that influence the rate of these reactions. The type of reaction that has been studied more than any other involves attack by a nucleophile on a saturated carbon atom. Consider the following reaction, for example, which converts an alkyl bromide into an alcohol. CH3Br OH 88n CH3OH Br In the course of the reaction, one nucleophile (the OH ion) is substituted for another (the Br ion). This is therefore a nucleophilic substitution reaction. The rate of the reaction is first-order in both CH3Br and the OH ion, and secondorder overall. Rate k(CH3Br)(OH)
O
#
Br
HO O C
H
{H H
#
HO
C O Br H{ H
H A HO C { H
#
H
O
In the 1930s, Sir Christopher Ingold proposed a mechanism for the reaction in which both the alkyl halide and the hydroxyl ion are involved in the rate-limiting or slowest step of the reaction. The OH ion attacks the “backside” of the CH3Br molecule. (It attacks the carbon atom at a point directly opposite the Br substituent or leaving group.) When this happens, a pair of nonbonding electrons on the OH ion are used to form a covalent bond to the carbon atom at the same time that the carbon–bromine is broken, as shown in Figure O3.2. Because the rate-limiting step in the reaction involves both the CH3Br and OH molecules, it is called a bimolecular nucleophilic substitution reaction, or SN2 reaction.
Br
FIGURE O3.2 The mechanism for nucleophilic attack on CH3Br.
The most important point to remember about the mechanism of SN2 reactions is that they occur in a single step. The species in the middle of Figure O3.2 is known as a transition state. If you envision the reaction as an endless series of snapshots that capture the infinitesimally small changes which occur as one bond forms and the other bond breaks, the transition state is the snapshot in the series that has the highest energy—and is therefore the least stable. The transition state has an infinitesimally short lifetime, on the order of 1012 second. In the course of an SN2 reaction, the other three substituents on the carbon atom are “flipped” from one side of the atom to the other. This inevitably leads to inversion of the
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configuration at a stereocenter. Consider the following reaction, for example, in which cis1-bromo-3-methylcyclopentane is converted to trans-3-methylcyclopentanol. CH3
Br
CH3
Br OH
OH
Or consider the reaction in which the R isomer of 2-bromobutane is transformed into the S isomer of 2-butanol. CH3CH2
HO
CH3 H
CH2CH3 C
Br
HO
C
CH3 H
Br
O3.9 UNIMOLECULAR NUCLEOPHILIC SUBSTITUTION OR SN1 REACTIONS The kinetics of nucleophilic substitution reactions have been studied in greater detail than any other type of reaction because they don’t always proceed through the same mechanism. Consider the reaction between the OH ion and t-butyl bromide, for example. (CH3)3CBr OH 88n (CH3)3COH Br The rate of the reaction depends only on the concentration of the alkyl bromide. (Adding more OH ion to the solution has no effect on the rate of reaction.) Rate k((CH3)3CBr) Ingold and co-workers argued that this rate law is consistent with a mechanism in which the rate-limiting or slowest step involves the breaking of the carbon–bromine bond to form a pair of ions. As one might expect, the pair of electrons in the COBr bond end up on the more electronegative bromine atom. CH3
CH3 Rate-limiting step
CH3
C CH3
Br
CH3
C Br CH3
Because the bromine atom has formally gained an electron from the carbon atom, it is now a negatively charged Br ion. Because the carbon atom has formally lost an electron, it is now a “carbocation.” The first step in the mechanism is a relatively slow reaction because the activation energy for this step is roughly 80 kJ/molrxn. If the reaction is done in water, the next step is extremely fast. The (CH3)3C ion is a Lewis acid because it has an empty orbital that can be used to accept a pair of electrons. Water, on the other hand, is a reasonably good Lewis base. A Lewis acid-base reaction therefore rapidly occurs in which a pair of nonbonding electrons on a water molecule are donated to the carbocation to form a covalent COO bond.
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CH3
CH3
C H2O
CH3
CH3
CH3
OH2
C CH3
The product of the reaction is a stronger acid than water. As a result, it transfers a proton to water. CH3 CH3
C
CH3
OH2 H2O
CH3
C
CH3
OH H3O
CH3
Because the slowest step of the reaction only involves t-butyl bromide, the overall rate of reaction depends only on the concentration of that species. The reaction is therefore a unimolecular nucleophilic substitution reaction, or SN1 reaction. The central carbon atom in the t-butyl carbocation formed in the first step of the reaction is planar, as shown in Figure O3.3. This means that water can attack the carbocation in the second step with equal probability from either side of the carbon atom. This has no effect on the products of the reaction, because the starting material is not optically active. But what would happen if we started with an optically active halide, such as Obromobutane?
{
CH3 O C#
CH3
CH3
FIGURE O3.3 The geometry of the (CH3)3C ion.
Regardless of whether we start with the R or S isomer of 2-bromobutane, we get the same intermediate when the COBr bond breaks. CH3CH2 C H CH3
Br
CH2CH3 A C CH3 H
Br
CH2CH3 Br
C
H CH3
The intermediate formed in the first step in the SN1 mechanism is therefore achiral. Mixtures of equal quantities of the / or R/S stereoisomers of a compound are said to be racemic. The term traces back to the Latin word racemus, which means “a cluster of grapes.” Just as there is an equal probability of finding grapes on either side of the stem in a cluster of grapes, there is an equal probability of finding the R and S enantiomers in a racemic mixture. SN1 reactions are therefore said to proceed with racemization. If we start with a pure sample of (R)-2-bromobutane, for example, we expect the product of the SN1 reaction with the OH ion to be a racemic mixture of the two enantiomers of 2-butanol. We are now ready to address a pair of important questions. First, why does CH3Br react with the OH ion by the SN2 mechanism if (CH3)3CBr does not? The SN2 mechanism requires direct attack by the OH ion on the carbon atom that carries the COBr bond. It is much easier for the OH ion to get past the small hydrogen atoms in CH3Br than it is for the ion to get past the bulkier CH3 groups in (CH3)3CBr.
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H HO
H H
C
Br
H A HO , C ,Br
21
H HO
H H
C
H H
Br
Thus, SN2 reactions at the 1° carbon atom in CH3Br are much faster than the analogous reaction at the 3° carbon atom in (CH3)3CBr. Why, then, does (CH3)3CBr react with the OH ion by the SN1 mechanism if CH3Br does not? The SN1 reaction proceeds through a carbocation intermediate, and the stability of such ions decreases in the following order. CH3 CH3 CH3 H A A A A CH3 O C > CH3 O C > HO C > HO C A A A A CH3 H H H Organic chemists explain this by noting that alkyl groups are slightly “electron releasing.” They can donate electron density to a neighboring group. This tends to delocalize the charge over a larger volume of the molecule, which stabilizes the carbocation. When we encountered a similar phenomenon in the chemistry of free radicals in Section O3.7 we noted that 3° radicals are roughly 30 kJ/mol more stable than 1° radicals. In this case, the difference is much larger. A 3° carbocation is 340 kJ/mol more stable than a 1° carbocation!3 As a result, it is much easier for (CH3)3CBr to form a carbocation intermediate than it is for CH3Br to undergo a similar reaction. In theory, both starting materials could undergo both reaction mechanisms. But the rate of SN2 reactions for CH3Br are much faster than the corresponding SN1 reactions, whereas the rate of SN1 reactions for (CH3)3CBr are very much faster than SN2 reactions.
O3.10 ELIMINATION REACTIONS Why do we need to worry about whether a nucleophilic substitution reaction occurs by an SN1 or SN2 mechanism? At first glance, it would appear that the same product is obtained regardless of the mechanism of the reaction. Consider the following substitution reaction, for example. Br A CH3CHCH2CH3 CH3O
OCH3 A CH3CHCH2CH3 Br
The only apparent difference between the two mechanisms is the stereochemistry of the product. If the reaction proceeds through an SN2 mechanism, it gives inversion of configuration—conversion of an R starting material into an S product, or vice versa. If the reaction proceeds through a carbocation intermediate via an SN1 mechanism, we get a racemic mixture. The importance of understanding the mechanism of nucleophilic substitution reactions can best be appreciated by studying the distribution of products of the example given above.
3
J. E. Bartmess, Mass Spectrometry Review, 8, 297–343 (1989).
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When 2-bromobutane is allowed to react with the methoxide ion in methanol, less than half of the starting material is converted into methyl isopropyl ether; the rest is transformed into 2-propene. Br A CH3CHCH2CH3 CH3O
OCH3 A CH3CHCH2CH3 CH2 P CHCH2CH3 ca. 40%
ca. 60%
The reaction that produces the alkene involves the loss of an HBr molecule to form a CPC double bond. It is therefore an example of an elimination reaction. Starting materials that are likely to undergo a bimolecular SN2 reaction undergo elimination reactions by a bimolecular elimination mechanism, or E2 reaction. This is a onestep reaction in which the nucleophile attacks a COH bond on the carbon atom adjacent to the site of SN2 reaction. CH3O
Br A HO CH2CHCH2CH3
CH3OH
CH2 P CHCH2CH3 Br
Starting materials that are likely to undergo a unimolecular SN1 reaction undergo elimination reactions by a unimolecular elimination mechanism, or E1 reaction. As might be expected, the rate-limiting step is the formation of the carbocation.
Rate-limiting step
CH3 A CH3 O C OBr A CH3
CH3 A CH3 O C Br A CH3
The solvent then acts as a base, removing an H ion from one of the alkyl groups adjacent to the carbocation. The electrons in the COH bond that is broken are donated to the empty orbital on the carbocation to form a double bond. CH3OH H CH3 A A CH2 O C A CH3
CH3 A SCH2 O C A CH3
CH3 D CH2 P C G CH3
O3.11 SUBSTITUTION VERSUS ELIMINATION REACTIONS There are three ways of pushing the reaction between an alkyl halide and a nucleophile toward elimination instead of substitution. •
Start with a highly substituted substrate, which is more likely to undergo elimination. Only 10% of a primary alkyl bromide undergoes elimination to form an alkene, for example, when it reacts with an alkoxide ion dissolved in alcohol. The vast majority of the starting material goes on to form the product expected for an SN2 reaction.
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ORGANIC REACTION MECHANISMS CH3O
CH3CH2CH2Br
CH3OH
23
CH3CH2CH2OCH3
More than half of a secondary alkyl bromide undergoes elimination under the same conditions, as we have already seen. Br A CH3CHCH2CH3 CH3O
OCH3 A CH3CHCH2CH3 CH2 P CHCH2CH3
When the starting material is a tertiary alkyl halide, more than 90% of the product is formed by an E1 elimination reaction. CH3 A CH3 O C OBr A CH3 •
CH3 D CH2 P C G CH3
CH3O CH3OH
Use a very strong base as the nucleophile. When we use a relatively weak base, such as ethyl alcohol, only about 20% of t-butyl bromide undergoes elimination. CH3 A CH3 O C OBr A CH3
CH3CH2OH
CH3 CH3 A D CH3 O C O OCH2CH3 CH2 P C G A CH3 CH3 ~80%
~20%
In the presence of the ethoxide ion, which is a much stronger base, the product of the reaction is predominantly the alkene. CH3 A CH3 O C OBr A CH3
CH3CH2O CH3CH2OH
CH3 D CH2 P C G CH3 ~95%
•
Increase the temperature at which the reaction is run. Because both E1 and E2 reactions lead to an increase in the number of particles in the system, they are associated with a positive entropy term. Thus, increasing the temperature of the reaction makes the overall free energy of reaction more negative, and the reaction becomes more favorable.
Summary of Substitution/Elimination Reactions •
Methyl halides and primary alkyl halides such as CH3CH2Br undergo nucleophilic substitution reactions. CH3CH2Br CN
CH3CH2CN Br
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Secondary alkyl halides undergo SN2 reactions when handled gently—at low temperatures and with moderately strong nucleophiles. Br A CH3CHCH3 HS
•
At high temperatures, or in the presence of a strong base, secondary alkyl halides undergo E2 elimination reactions. Br A CH3CHCH3
•
SH A CH3CHCH3 Br
(CH3)3CO
CH2 P CHCH3
heat
Tertiary alkyl halides undergo a combination of SN1 and E1 reactions. If the reaction is kept cool, and the nucleophile is a relatively weak base, it is possible to get nucleophilic substitution. At high temperatures, or with strong bases, elimination reactions predominate. CH3 A CH3 O C OBr A CH3
CH3O
CH3 D CH2 P C G CH3
KEY TERMS Acetal Acid Acid dissociation equilibrium constant, Ka Alkali Base Bimolecular nucleophilic substitution reaction Brønsted acid
Brønsted base Diol E1 reaction E2 reaction Elimination reaction Hemiacetal Heterolytic Homolytic Leaving group
Lewis acid Lewis base Nucleophilic substitution reaction SN1 reaction SN2 reaction Transition state Unimolecular nucleophilic substitution reaction
PROBLEMS Acids and Bases 1. Predict the products of the reaction between ethanol (CH3CH2OH) and sodium hydride (NaH). 2. Write the Lewis structure for the conjugate base of ethanol (CH3CH2OH). 3. Which of the following is the conjugate acid of ethanol (CH3CH2OH)? (a) CH3CH2OH (b) CH3CH2O (c) CH3CH2OH2 (d) CH3CH2 (e) H3O 4. Some beginning chemistry students get confused when acetic acid is described as a weak acid because they see four hydrogen atoms in a CH3CO2H molecule. Explain why only one of the H atoms dissociates when acetic acid is dissolved in water.
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5. Which of the following are Lewis acids but not a Brønsted acid? (a) H (b) NH4 (c) BF3 (d) CH3CH2OH (e) Mg2 6. Arrange the following compounds in order of increasing basicity. (a) NH3 (b) NH2 (c) NH4 (d) N3 7. Identify the Brønsted acids in the following reaction. HCqCH CH3Li 88n HCqC CH4 Li 8. Arrange the following hydrocarbons in order of increasing acidity. (a) C2H2 (b) C2H4 (c) C2H6 9. Which of the following reagents is a strong enough base to generate the ethoxide ion (CH3CH2O) from ethanol (CH3CH2OH)? (a) NaOH (b) NaH (c) NaNH2 (d) CH3MgBr (e) HCqCNa 10. Which of the following acid–base reactions should occur as written? (a) CH3CH2OH NaOH n Na CH3CH2O H2O (b) CH4 NaNH2 n CH3 Na NH3 (c) CH3Li HCqCH n CH4 HCqCLi (d) CH3CH2OH HS n CH3CH2O H2S
Attack at a Carbonyl 11. Use Lewis structures and the concept of oxidation–reduction reactions to explain why it takes two moles of lithium metal to reduce a mole of methyl bromide to form methyllithium. CH3Br 2 Li 88n CH3Li LiBr 12. Explain why ethers, but not alcohols, are used as solvents for reactions that involve Grignard or alkyllithium reagents. 13. Explain why the bonding of a Lewis acid at the oxygen atom of a carbonyl group increases the rate at which nucleophilic attack occurs at the carbon atom. 14. Predict the products of the following reactions. O B Li CH3CCH2CH3
A
H
B
15. Predict the products of the following reactions. O B C O CH3 CH3MgBr
A
H
B
16. Design a two-step reaction sequence using either a Grignard or alkyllithium reagent that could be used to produce the following compound. OH A C O CH2CH3 A CH3
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The Mechanism of Reduction Reactions 17. Both H2 on a metal catalyst and LiAlH4, can be used to reduce a carbonyl to an alcohol. Explain why H2 on a metal catalyst can be used to reduce a CPC double bond but not LiAlH4. 18. Explain why the yield of reactions that try to reduce a carbonyl with LiAlH4 in ethanol is effectively zero. 19. Reduction of a CPC double bond with H2 gas on a metal catalyst gives a product in which the two hydrogen atoms are added to the same side of the double bond. What does this tell us about the relative rates of the first and second steps in the reaction? 20. Why is it important to “poison” the metal catalyst before trying to reduce an alkyne to an alkene with H2? 21. The reagent that attacks the carbon atom is the same when either LiAlH4 or NaBH4 is used to reduce a CPO double bond. (In each case, it is the H ion.) What does the fact that LiAlH4 is significantly more reactive than NaBH4 tell us about the relative acidity of AlH3 and BH3 as Lewis acids? 22. Predict the product of the following reaction. O CH3 B G CHO C O CH3 D CH3
1. LiAlH4 in ether 2. H3O
23. Predict the product of the following reaction. CH3C q CCH3
H2 Pd on CaCO3
Nucleophilic Attack by Water or an Alcohol 24. Explain why the rate of nucleophilic attack on a carbonyl by either water or an alcohol is relatively slow in the absence of an acid or base catalyst. 25. Identify the starting materials that would give the following product. OH A C O OCH2CH3 A CH3 26. Identify the starting materials that would give the following acetal, which has a delightful odor of ‘Bing’ cherries. OCH2CH3 A CH A OCH2CH3 27. Maltose, or malt sugar, is an important component of the barley malt used to brew beer. This disaccharide is formed when the OOH group on C-4 of one -D-glucopyranose forms an acetal by reacting with C-1 of a second -D-glucopyranose residue. Draw the structure of the compound.
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27
Addition/Elimination Reactions 28. Explain why acyl chlorides are more reactive than carboxylic acid esters toward addition/elimination reactions. 29. Use the fact that the best leaving groups for addition/elimination reactions are weak bases to explain why attack on a carbonyl by the H ion or a source of the CH3 ion are not reversible reactions. 30. Which of the following would be the best leaving group? (a) NH2 (pKb 19) (b) CH3O (pKb 4) (c) HS (pKb 7) (d) CH3CO2 (pKb 9.3)
Free Radical Reactions 31. Draw the structures of all possible products of the free radical chlorination of 2-methylbutane. Predict the relative abundance of the products if the reaction is so fast that it is not selective. 32. Draw the structures of all possible products of the free radical bromination of methylcyclobutane. Predict the relative abundance of the products if the reaction is equally likely to occur at each of the hydrogen atoms. CH3 Br2
33. Assume that the reaction in the previous problem is slow enough to give almost exclusively the product that would be formed from the most stable free radical intermediate. Predict the structure of the product. 34. Calculate the value of H° for the reaction in which H2, F2, Cl2, Br2, and I2 dissociate to form free radicals.
Nucleophilic Substitution Reactions 35. Which of the following starting materials is most likely to undergo an SN2 reaction with the OCH3 ion?
Br A
Br B
Br C
36. Predict the products of the following reactions. (a) CH3CH2CH2I OH n (b) (CH3)2CHBr I n (c) CH3Br CN n (d) CH3CH2I HS n 37. The relative rates of nucleophilic substitution reactions often decrease in the order CH3 OH Cl. Is this consistent with what we know about the relative strengths of the ions as Brønsted bases? 38. The relative rates of nucleophilic substitution reactions often decrease in the order I Br Cl F. Is this consistent with what we know about the relative acidities of HI (Ka 3 109), HBr (Ka 1 109), HCl (Ka 106), and HF (Ka 7.2 104)?
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Or does it suggest that there is a difference between trends in nucleophilicity and basicity as we go down a column of the periodic table? (In other words, the X ion becomes less basic and more nucleophilic as we go down the column.) 39. Write the mechanism for the following reaction. CH3 A CH3 O C O OCH3 Br A CH3
CH3 A CH3 O C OBr OCH3 A CH3
40. Identify a set of starting materials that would give the following compound as the product of a nucleophilic substitution reaction. CH3CH2NH2 41. Assume that you had a choice of two reagents to react with the following starting material, CH3OH and the CH3O ion. CH3 A C OBr A H Which nucleophile would be more likely to give a racemized product?
Elimination Reactions 42. Write the mechanism for the following reaction. CH3 A CH3 O C OBr OCH3 A CH3
CH3 D CH2 P C G CH3
43. Which of the following would be the most likely to undergo an E1 elimination reaction with a very strong base, such as the (CH3)3CO ion?
Br
Br
A
B
Br C
44. Predict the product of the following elimination reaction. (CH3)3CO Br 45. Elimination reactions can often give two different products, depending on which carbon atom adjacent to the COX bond is attacked. As a rule, the dominant product of
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the reactions is the most highly substituted alkene. Predict the product of the following elimination reaction. CH3 CH3 G D CHO CH (CH3)3CO D G CH3 Br
Substitution versus Elimination Reactions 46. Assume that 2-bromopropane undergoes a combination of E2 and SN2 reactions when it reacts with the CH3O ion in methanol. Predict the products of the reactions. 47. Predict the most likely product or products of the following reactions. Br (a)
(b)
(c)
CH3OH
Br
Br
CH3NH2
CH3O heat
48. Predict whether the following reactions are more likely to undergo elimination or substitution. Identify the mechanism of the dominant reaction (E1 versus E2; SN1 versus SN2). I
(a)
Br
A (b) O C OBr A
NH2
H2O
(c) Br
49. Explain why the following reaction will not give the indicated product. Predict the product that would form. CH3 CH3 A A CH3 O C OBr CH3 O C OO A A CH3 CH3
CH3 CH3 A A CH3 O C OO OO A CH3 A CH3 CH 3
50. Explain why the following reaction will not give the indicated product. Predict the product that would form. H2O
Br
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Integrated Problems 51. The following Lewis structures were drawn correctly by an organic chemistry student who forgot to indicate whether the molecules carry a positive or negative charge. Correct the work by specifying whether each molecule is negatively charged, positively charged, or neutral. H A O OH HO C O O A A H H
H H A A HO C ONOH A A H H
H A HO C O C qNS A H
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M O D U L E
7 POLYMER CHEMISTRY P.1 Polymers P.2 Definition of Terms P.3 Elastomers P.4 Free Radical Polymerization Reactions P.5 Ionic Polymerization Reactions P.6 Coordination Polymerization P.7 Addition Polymers P.8 Condensation Polymers P.9 Properties of Polymers Chemistry in the World Around Us: The Search for Synthetic Fibers
P.1 POLYMERS Imagine that one evening you decide to go bowling. Wearing a pair of jeans, a bowling shirt, and a lightweight jacket, you walk into the local bowling alley. After you change into an appropriate pair of shoes, you pick up a ball, stride to the line, smoothly release the ball, watch as it rolls down the lane until it collides with a set of wooden pins, and then use a pencil to record your score on a sheet of paper. The fabric of your cotton jeans, your polyester shirt, your nylon jacket, and your leather shoes have something in common with the rubber bowling ball, the polyurethane coating on the bowling lane, the wooden pins, the mixture of graphite and clay in the “lead” pencil, and the sheet of paper on which you wrote your score. Each of these substances is a polymer. In 1833, Jöns Jakob Berzelius suggested that compounds with the same molecular formula but different structures should be called isomers (literally, “equal parts”). He then proposed the term polymer (literally, “many parts”) to describe compounds that had the same empirical formula but different molecular weights. Ethylene (C2H4) and butene (C4H8) are compounds that Berzelius would classify as polymers. Each compound has the same empirical formula (CH2) but they have different molecular weights. Acetylene (C2H2) and benzene (C6H6) are another example of compounds Berzelius would call polymers. The term polymer eventually came to mean compounds such as cellulose and natural rubber that have unusually large molecular weights, in the range of 10,000 to 100,000 or more grams per mole. These molecules are so large they are often called macromolecules— literally, molecules large enough to be seen with the naked eye. A perfect diamond, for example, can be thought of as a single molecule containing an array of COC bonds arranged 1
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toward the corners of a tetrahedron around each carbon atom in the crystal. The plastic case that enclosed one of the radios that your grandparents listened to can be thought of as a single molecule. So can a 14-pound bowling ball. The first explanation of why polymer molecules are so heavy was offered by Hermann Staudinger in 1920. Staudinger argued that polymers contain long chains of relatively simple repeating units, or monomers. Natural rubber, for example, is a polymer that contains large numbers of OCH2C(CH3)PCHCH2O units. The number of monomers in a polymer can differ from one chain to the next. Rubber, for example, is a mixture of polymer chains whose mass differs by a factor of 10 or more, but whose average molecular weight is 100,000 grams per mole. The term rubber was first used in 1770 by Joseph Priestley to describe the gum from a South American tree that could be used to “rub out” pencil marks. Because natural rubber is tacky, strong-smelling, perishable, too soft when warm, and too hard when cold, it had only limited uses. Nathaniel Hayward was the first to note that rubber loses some of its sticky properties when treated with sulfur. It was Charles Goodyear, however, who accidentally dropped a mixture of rubber and sulfur onto a hot stove and discovered “vulcanized” rubber, which is stable over a wide range of temperatures and is far more durable than natural rubber. Cellulose is another example of a polymer that contains many copies of a simple repeating unit: C6H10O5. Wood is about 50% cellulose by weight; cotton is almost 90% cellulose. Cellulose is used to make paper from wood pulp and cloth from cotton. In the last hundred years, cellulose has also served as the starting material for the synthesis of the first plastics— cellulose nitrate and celluloid—and the first synthetic fibers—Chardonnet silk, or rayon. Each OC6H10O5O repeating unit in cellulose contains three OOH groups that can react with nitric acid to form nitrate esters known as cellulose nitrate. In 1869 John Wesley Hyatt found that mixtures of cellulose nitrate and camphor dissolve in alcohol to produce a plastic substance he named celluloid. Cellulose nitrate, or celluloid, was used as a substitute for ivory in the manufacture of a variety of items ranging from billiard balls to movie film. Because it is extremely flammable, cellulose nitrate has been replaced by other plastics for almost all uses except Ping-Pong balls. No other plastic has been found that has quite the same “bounce” as celluloid. The cellulose from wood pulp contains too many impurities to be used to make fibers. It can be purified, however, by dissolving the polymer in a mixture of NaOH and carbon disulfide (CS2). When the viscous solution is forced through tiny holes in a nozzle into an acid bath, the cellulose fiber is regenerated. When this process was introduced in 1885, the product was the first synthetic fiber. It was originally called Chardonnet silk, but soon become known as rayon. A similar process is still used to make a thin film of regenerated cellulose known as cellophane.
P.2 DEFINITIONS OF TERMS Linear, Branched, and Cross-linked Polymers The term polymer is used to describe compounds with relatively large molecular weights formed by linking together many small monomers. Polyethylene, for example, is formed by polymerizing ethylene molecules. n CH2PCH2
[...(CH2CH2)n...]
Ethylene
Polyethylene
Polyethylene is called a linear or straight-chain polymer because it consists of a long string of carbon–carbon bonds. Those terms are misleading, however, because the geometry
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3
around each carbon atom is tetrahedral and the chain is neither linear nor straight, as shown in Figure P.1.
FIGURE P.1 Small portion of a straight-chain polyethylene molecule.
As the polymer chain grows, it folds back on itself in a random fashion to form structures such as the one shown in Figure P.2.
Linear polymer
FIGURE P.2 Random structure formed by a straight-chain polymer as it folds back on itself.
Polymers with branches at irregular intervals along the polymer chain are called branched polymers (see Figure P.3). The branches make it difficult for the polymer molecules to pack in a regular array, and therefore make the polymer less crystalline. Crosslinked polymers contain branches that connect polymer chains, as shown in Figure P.4. At first, adding cross-links between polymer chains makes the polymer more elastic. The vulcanization of rubber, for example, results from the introduction of short chains of sulfur atoms that link the polymer chains in natural rubber. As the number of cross-links increases, the polymer becomes more rigid.
Branched polymer
FIGURE P.3 Branched polymers contain short side chains that extend from the backbone of the polymer.
Cross-linked polymer
FIGURE P.4 Cross-linked polymers have branches that connect chains.
The decision to classify a polymer as branched or cross-linked is based on the extent to which the side chains on the polymer backbone link adjacent polymer chains. The easiest way to distinguish between these categories is to study the effect of various solvents on the polymer. Branched polymers are often soluble in one or more solvents because it is possible to separate the polymer chains. Cross-linked polymers are insoluble in all solvents because the polymer chains are tied together by strong covalent bonds.
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Linear and branched polymers form a class of materials known as thermoplastics. These materials flow when heated and can be molded into a variety of shapes which they retain when they cool. Heavy cross-linking produces materials known as thermoset plastics. Once the cross-links form, the polymers take on a shape that cannot be changed without destroying the plastic. The polypropylene used in the plastic chairs that fill so many classrooms is a thermoplastic; as you lean back on the chair you can feel it give. The plastic case in which early radios were placed is an example of a thermoset plastic; it had a tendency to shatter rather than bend if the radio was dropped on the floor.
Exercise P.1 Polyethylene can be obtained in two different forms. High-density polyethylene (0.96 g/cm3) is a linear polymer. Low-density polyethylene (0.92 g/cm3) is a branched polymer with short side chains on 3% of the atoms along the polymer chain. Explain how the structures of these polymers give rise to the difference in their densities. Solution Linear polymers are more regular than branched polymers. Linear polymers can therefore pack more tightly, with less wasted space. As a result, linear polymers are slightly more dense than branched polymers.
Homopolymers and Copolymers Polyethylene is an example of a homopolymer that is formed by polymerizing a single monomer. Copolymers are formed by polymerizing more than one monomer. Ethylene (CH2PCH2) and propylene (CH2PCHCH3) can be copolymerized, for example, to produce a polymer that has two kinds of repeating units.
x CH2 PCH2 y CH2 PCHCH3
CH3 A ...(CH2CH2)x CH2CH y...
Copolymers are classified on the basis of the way monomers are arranged along the polymer chain, as shown in Figure P.5. Random copolymers contain repeating units arranged in a purely random fashion. Regular copolymers contain a sequence of regularly alternating repeating units. The repeating units in block copolymers occur in blocks of different lengths. Graft copolymers have a chain of one repeating unit grafted onto the backbone of another. [...OAOBOBOAOAOAOBOAOBOBO...] Random copolymer
[...OAOBOAOBOAOBOAOBOAOBO...] Regular copolymer
[...OAOAOAOAOBOBOBOBOAOAOBOBOBOBOBOBOAOAOA...] Block copolymer
...OAOAOAOAOAOAOAOAOAOAO... A BOBOBOBOBO... Graft copolymer
FIGURE P.5 Random, regular, block, and graft copolymers.
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Tacticity Polymers with regular substituents on the polymer chain possess a property known as tacticity (from the Latin word tacticus, “fit for arranging”). Tacticity results from the different ways in which the substituents can be arranged on the polymer backbone (see Figure P.6). When the substituents are arranged in an irregular, random fashion, the polymer is atactic (literally, “no arrangement”). When the substituents are all on the same side of the chain, the polymer is isotactic (literally, “the same arrangement”). If the substituents alternate regularly from one side of the chain to the other, the polymer is syndiotactic.
CH3 CH3 CH3 A A A ... OCH2OCHOCH2OCHOCH2OCHOCH2OCHOCH2OCHO... A A CH3 CH3 Atactic polypropylene
... OCH2OCHOCH2OCHOCH2OCHOCH2OCHOCH2OCHO...
Syndiotactic polystyrene
... OCH2OCHOCH2OCHOCH2OCHOCH2OCHOCH2OCHO... A A A A A Cl Cl Cl Cl Cl Isotactic poly(vinyl chloride)
FIGURE P.6 Atactic, isotactic, and syndiotactic polymers.
Exercise P.2 Atactic polypropylene is a soft, rubbery material with no commercial value. The isotactic polymer is a rigid substance with an excellent resistance to mechanical stress. Explain the difference between the physical properties of the two forms of polypropylene. Solution Isotactic polypropylene is easier to pack in a regular fashion than the atactic form of the polymer. As a result, isotactic polypropylene is more crystalline, which makes the solid more rigid.
Addition versus Condensation Polymers Polyethylene, polypropylene, and poly(vinyl chloride) are addition polymers formed by adding monomers to a growing polymer chain. Addition polymers can be recognized by noting that the repeating unit always has the same formula as the monomer from which the polymer is formed.
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n CH2PCH2
[...(CH2CH2)n...]
Polyethylene
n CH2PCHCH3
CH3 A ... CH2CH n...
Polypropylene
n CH2PCHCl
Cl A ... CH2CH n...
Poly(vinyl chloride)
To condense means to make something more dense, or compact. Polymers formed when a small molecule condenses out during the polymerization reaction are therefore called condensation polymers. Silicone, for example, is a condensation polymer formed by polymerizing (CH3)2Si(OH)2. Each time a monomer is added to the polymer chain, a molecule of water is condensed out, as shown in Figure P.7. Note that the repeating unit in a condensation polymer is inevitably smaller than the monomer from which it is made. CH3 CH3 CH3 A A A ... HOOOSiOOOH HOOOSiOOOH HOOOSiOOOH ... A A A CH3 CH3 CH3 H2O
CH3 CH3 CH3 A A A ...OOSiOOOSiOOOSiOO... A A A CH3 CH3 CH3 Silicone
FIGURE P.7 Silicone is a condensation polymer produced by eliminating a molecule of water each time a bond between monomers is formed.
Exercise P.3 Classify the products of the following reactions as either addition or condensation polymers. (a) poly(methyl methacrylate), sold as Lucite or Plexiglas CH3 A n CH2 P CCO2CH3
CH3 A ... CH2C A CO2CH3
n...
(b) nylon 6
n H2N(CH2)5CCl
O
P
P
O
... HN(CH2)5C n... n HCl
Solution (a) Poly(methyl methacrylate) is an addition polymer because every atom in the monomer ends up in the repeating unit of the polymer.
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(b) Nylon 6 is a condensation polymer because a molecule of HCl is eliminated each time a monomer is added to the chain.
P.3 ELASTOMERS Elastomers are polymers that have the characteristic properties of rubber—they are both flexible and elastic. To be elastic, a polymer must meet the following criteria. • • • •
It must contain long, flexible molecules that are coiled in the natural state and that can be stretched without breaking, as shown in Figure P.8. It must contain a few cross-links between polymer chains so that one chain does not slip past another when the substance is stretched. It cannot contain too many cross-links, or else it would be too rigid to be stretched. The force of attraction between chains must be relatively small, so that the polymer can curl back into its coiled shape after it has been stretched.
Stretch
Relax
FIGURE P.8 The effect of stretching and relaxing the cross-linked chains in an elastomer.
We can understand these requirements by taking a closer look at the chemistry of natural rubber, which is a polymer of a C5H8 hydrocarbon known as isoprene. CH3 A n CH2 P CCHP CH2
CH3 A ... CH2C P CHCH2 n...
Isoprene
Natural rubber
The double bonds in natural rubber are all in the cis form. The force of attraction between polymer chains is relatively small, so the polymer can curl back into its original shape after the molecules have been oriented by stretching. By adding sulfur to natural rubber it is possible to introduce a small number of cross-links between polymer chains that hold the chains together when the polymer is stretched. At first glance, it might seem easy to make synthetic rubber. All we have to do is find a suitable catalyst that can polymerize isoprene. The task is made more difficult by the fact that the cis isomer of isoprene rearranges into the trans isomer during polymerization, and the trans isomer of polyisoprene, which is known as gutta percha, is not elastic. It is therefore important to control the geometry around the CPC double bond during polymerization to make sure that as few of the bonds as possible are converted to the trans geometry. Until recently, this wasn’t possible, and other approaches to making synthetic rubber were necessary. The first solution to the problem involved polymerizing 2-chloro-1,3-butadiene, or “chloroprene,” to form the first major synthetic rubber, neoprene. Cl A n CH2 P CCHP CH2
Cl A ... CH2C P CHCH2 n...
Chloroprene
Neoprene
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This approach is still used to produce a copolymer of 75% butadiene and 25% styrene known as styrene–butadiene rubber (SBR). Roughly 40% of the rubber used in the world today is SBR; another 35% is natural rubber that has been treated with sulfur. The effect of cross-linking on elastomers can be demonstrated with a pair of rubber balls available from Flinn Scientific. One of the balls is a polybutadiene rubber that contains an unusually large amount of sulfur. Because the polymer chains are extensively crosslinked, this ball dissipates very little energy in the form of heat when it bounces. It is therefore extremely resilient when bounced on the floor. The other ball is a styrene–butadiene copolymer with much less cross-linking. When dropped on the floor, the ball seems to “die.” This copolymer is used in applications where an energy-absorbing medium is desired, such as automobile tires which must absorb some of the energy associated with the bumps we encounter as we drive down the highway.
P.4 FREE RADICAL POLYMERIZATION REACTIONS It isn’t difficult to form addition polymers from monomers containing CPC double bonds; many of these compounds polymerize spontaneously unless polymerization is actively inhibited. One of the problems with early techniques for refining gasoline, for example, was the polymerization of alkene components when the gasoline was stored. Even with modern gasolines, deposits of “gunk” can form when a car or motorcycle is stored for extended periods without draining the gas tank. The simplest way to catalyze the polymerization reaction that leads to an addition polymer is to add a source of a free radical to the monomer. The term free radical is used to describe a family of very reactive, short-lived components of a reaction that contain one or more unpaired electrons. In the presence of a free radical, addition polymers form by a chain reaction mechanism that contains chain initiation, chain propagation, and chain termination steps.
Chain Initiation A source of free radicals is needed to initiate the chain reaction. The free radicals are usually produced by decomposing a peroxide such as di-tert-butyl peroxide or benzoyl peroxide, shown below. In the presence of either heat or light, the peroxides decompose to form a pair of free radicals that contain an unpaired electron. CH3 CH3 A A CH3 O C O O O OO CO CH3 A A CH3 CH3
O
O C O O OOO C O
P
O
P
P
O
CH3 A 2 CH3 O C O O A CH3
2
O CO O
Chain Propagation The free radical produced in the chain initiation step adds to an alkene to form a new free radical.
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CH3 A CH3 O C O O ⫹ CH2 P CH2 A CH3
9
CH3 A CH3 O C O O O CH2O CH2 A CH3
The product of this reaction can then add additional monomers in a chain reaction. CH3 A CH3 O C O O O CH2O CH2 ⫹ CH2 P CH2 A CH3
CH3 A CH3 O C O O O CH2O CH2 O CH2O CH2 A CH3
Chain Termination Whenever pairs of radicals combine to form a covalent bond, the chain reactions carried by these radicals are terminated. CH3 CH3 A A CH3 O C O O O (CH2CH2)n ⫹ (CH2CH2)m O O O C O CH3 A A CH3 CH3 CH3 CH3 A A CH3OC O O O (CH2CH2)n⫹m O O O C O CH3 A A CH3 CH3
The Formation of Branched Polymers We might expect the product of the free radical polymerization of ethylene to be a straightchain polymer. As the chain grows, however, it begins to fold back on itself. This allows an intramolecular reaction to occur in which the site at which polymerization occurs is transferred from the end of the chain to a carbon atom along the backbone. CH2H CH2 E CH2 A CH (CH3)3C O O O CH2 H E 2 CH2
CH2H CH3 E CH2 A CH (CH3)3C O O O CH H E 2 CH2
When this happens, branches are introduced onto the polymer chain. Free radical polymerization of ethylene produces a polymer that contains branches on between 1% and 5% of the carbon atoms. Of these branches, 10% contain two carbon atoms, 50% contain four carbon atoms, and 40% are longer side chains.
P.5 IONIC POLYMERIZATION REACTIONS Addition polymers can also be made by chain reactions that proceed through intermediates that carry either a negative or positive charge.
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Anionic Polymerization When the chain reaction is initiated and carried by negatively charged intermediates, the reaction is known as anionic polymerization. Like free radical polymerizations, these chain reactions take place via chain initiation, chain propagation, and chain termination steps. The reaction is initiated by a Grignard reagent or alkyllithium reagent, which can be thought of as a source of a negatively charged CH3 or CH3CH2 ion. H H A A HO C O C OLi A A H H
H H A A HO C O C A A H H
Li
The CH3 or CH3CH2 ion from one of these metal alkyls can attack an alkene to form a carbon–carbon bond. H H A A HO C O C A A H H
H H D G CPC D G H X
H H H H A A A A HO C O C O C O C A A A A H H H X
The product of the chain initiation reaction is a new carbanion that can attack another alkene in a chain propagation step. H H H H A A A A HO C O C O C O C A A A A H H H H
H H D G CPC D G H X
H H H H H H A A A A A A HO C O C O C O C O C O C A A A A A A H H H X H X
The chain reaction is terminated when the carbanion reacts with traces of water in the solvent in which the reaction is run. CH3CH2(CH2CH)nCH2CH A A X X
H2O
CH3CH2(CH2CH)nCH2CH2 OH A A X X
Cationic Polymerization The intermediate that carries the chain reaction during polymerization can also be a positive ion, or cation. In this case, the cationic polymerization reaction is initiated by adding a strong acid to an alkene to form a carbocation. CH2P CH H A X
CH3CH A X
The ion produced in this reaction adds monomers to produce a growing polymer chain. CH3CH CH2P CH A A X X
CH3CHCH2CH A A X X
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The chain reaction is terminated when the carbonium ion reacts with water that contaminates the solvent in which the polymerization is run. CH3CH(CH2CH)nCH2CH H2O A A A X X X
CH3CH(CH2CH)nCH2CHOH H A A A X X X
Advantages of Free-Radical versus Ionic Polymerization The initiation step of ionic polymerization reactions has a much smaller activation energy than the equivalent step for free radical polymerizations. As a result, ionic polymerization reactions are relatively insensitive to temperature, and they can be run at temperatures as low as 70°C. Ionic polymerization therefore tends to produce a more regular polymer, with less branching along the backbone, and more controlled tacticity. Because the intermediates involved in ionic polymerization reactions can’t combine with one another, chain termination occurs only when the growing chain reacts with impurities or with reagents that can be specifically added to control the rate of chain growth. It is therefore easier to control the average molecular weight of the product of ionic polymerization reactions. Ionic polymerizations are more difficult to carry out on an industrial scale than free radical polymerizations. Ionic polymerization is therefore only used for monomers that don’t polymerize by the free radical mechanism or to prepare polymers with a regular structure.
P.6 COORDINATION POLYMERIZATION In 1963 Karl Ziegler and Giulio Natta received the Nobel Prize in chemistry for their discovery of coordination compound catalysts for addition polymerization reactions. Ziegler–Natta catalysts provide the opportunity to control both the linearity and tacticity of the polymer. Free radical polymerization of ethylene produces a low-density, branched polymer with side chains of one to five carbon atoms on up to 3% of the atoms along the polymer chain. Ziegler–Natta catalysts produce a more linear polymer, which is more rigid, with a higher density and a higher tensile strength. Polypropylene produced by free radical reactions, for example, is a soft, rubbery, atactic polymer with no commercial value. Ziegler–Natta catalysts provide an isotactic polypropylene, which is harder, tougher, and more crystalline. A typical Ziegler–Natta catalyst can be produced by mixing solutions of titanium(IV) chloride (TiCl4) and triethylaluminum [Al(CH2CH3)3] dissolved in a hydrocarbon solvent from which both oxygen and water have been rigorously excluded. The product of the reaction is an insoluble olive-colored complex in which the titanium has been reduced to the Ti(III) oxidation state. The catalyst formed in the reaction can be described as coordinately unsaturated because there is an open coordination site on the titanium atom. This allows an alkene to act as a Lewis base toward the titanium atom, donating a pair of electrons to form a transition metal complex. H H G D C B C D G H H
CH CH3 G D 2 Ti D G
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The alkene is then inserted into a TiOCH2CH3 bond to form a growing polymer chain and a site at which another alkene can bond. H H G D C B C D G H H
CH CH3 G D 2 Ti D G
CH CH2CH2CH3 G D 2 Ti D G
Thus, the titanium atom provides a template on which a linear polymer with carefully controlled stereochemistry can grow.
P.7 ADDITION POLYMERS Addition polymers such as polyethylene, polypropylene, poly(vinyl chloride), and polystyrene are linear or branched polymers with little or no cross-linking. As a result, they are thermoplastic materials, which flow easily when heated and can be molded into a variety of shapes. The structures, names, and trade names of some common addition polymers are given in Table P.1.
Polyethylene Low-density polyethylene (LDPE) is produced by free-radical polymerization at high temperatures (200°C) and high pressures (above 1000 atm). The high-density polymer (HDPE) is obtained using Ziegler–Natta catalysis at temperatures below 100°C and pressures less than 100 atm. More polyethylene is produced each year than any other plastic. About 7800 million pounds of low-density and 4400 million pounds of high-density polyethylene were sold in 1980. Polyethylene has no taste or odor and is lightweight, nontoxic, and relatively inexpensive. It is used as a film for packaging food, clothing, and hardware. Most commercial trash bags, sandwich bags, and plastic wrapping are made from polyethylene films. Polyethylene is also used for everything from seat covers to milk bottles, pails, pans, and dishes.
Polypropylene The isotactic polypropylene from Ziegler–Natta-catalyzed polymerization is a rigid, thermally stable polymer with an excellent resistance to stress, cracking, and chemical reaction. Although it costs more per pound than polyethylene, it is much stronger. Thus, bottles made from polypropylene can be thinner, contain less polymer, and cost less than conventional polyethylene products. Polypropylene’s most important impact on today’s college student takes the form of the plastic stackable chairs that abound on college campuses.
Poly(tetrafluoroethylene) Tetrafluoroethylene (CF2PCF2) is a gas that boils at 76°C and is therefore stored in cylinders at high pressure. In 1938 Roy Plunkett received a cylinder of tetrafluoroethylene that didn’t deliver as much gas as it should have. Instead of returning the cylinder, he cut it open with a hacksaw and discovered a white, waxy powder that was the first polytetrafluoroethylene polymer. After considerable effort, a less fortuitous route to the polymer was discovered, and polytetrafluoroethylene, or Teflon, became commercially available.
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Teflon is a remarkable substance. It has the best resistance to chemical attack of any polymer, and it can be used at any temperature between 73°C and 260°C with little effect on its properties. It also has a very low coefficient of friction. (In simpler terms, it has a waxy or slippery touch.) Even materials as “sticky” as crude rubber, adhesives, bread dough, and candy won’t stick to a Teflon-coated surface. Teflon is so slippery that it has even been sprayed on plants, so that insects that might prey on the plants fall off.
TABLE P.I
Common Addition Polymers
Structure
Chemical Name
Trade Name or Common Name
(OCH2OCH2O)n (OCF2OCF2O)n (OCH2OCHO)n A CH3 CH3 A (OCH2OCO)n A CH3 (OCH2OCHO)n
polyethylene poly(tetrafluoroethylene) polypropylene
Teflon Herculon
polyisobutylene
butyl rubber
(OCH2OCHO)n A CN (OCH2OCHO)n A Cl (OCH2OCHO)n A CO2CH3 CH3 A (OCH2OCO)n A CO2CH3 H H A A (OCH2OCPCOCH2O)n Cl A (OCH2OCPCHOCH2O)n H CH3 A A (OCH2OCPCOCH2O)n H A (OCH2OCPCOCH2O)n A CH3
polyacrylonitrile
Orlon
poly(vinyl chloride)
PVC
polystyrene
poly(methyl acrylate)
poly(methyl methacrylate)
Plexiglas, Lucite
polybutadiene polychloroprene
neoprene
poly(cis-1,4-isoprene)
natural rubber
poly(trans-1,4-isoprene)
gutta percha
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Poly(vinyl Chloride) and Poly(vinylidene Chloride) Chlorine is one of the top ten industrial chemicals in the United States—more than 20 billion pounds are produced annually. About 20% of the chlorine is used to make vinyl chloride (CH2PCHCl) for the production of poly(vinyl chloride), or PVC. The chlorine substituents on the polymer chain make PVC more fire-resistant than polyethylene or polypropylene. They also increase the force of attraction between polymer chains, which increases the hardness of the plastic. The properties of PVC can be varied over a wide range by adding plasticizers, stabilizers, fillers, and dyes, making PVC one of the most versatile plastics. A copolymer of vinyl chloride (CH2PCHCl) and vinylidene chloride (CH2PCCl2) is sold under the trade name Saran. The same increase in the force of attraction between polymer chains that makes PVC harder than polyethylene gives thin films of Saran a tendency to “cling.”
Acrylics Acrylic acid is the common name for 2-propenoic acid: CH2PCHCO2H. Acrylic fibers such as Orlon are made by polymerizing a derivative of acrylic acid known as acrylonitrile.
n CH2PCHCN
CN A ... CH2CH n...
Polyacrylonitrile
Other acrylic polymers are formed by polymerizing an ester of 2-propenoic acid, such as methyl acrylate. O B n CH2PCHCOCH3
CO2CH3 A ... CH2CH n...
Poly(methyl acrylate)
One of the most important acrylic polymers is poly(methyl methacrylate), or PMMA, which is sold under the trade names Lucite and Plexiglas. O B n CH2 P CCOCH3 A CH3
CO2CH3 A ... CH2C n... A CH3
Poly(methyl methacrylate), PMMA
PMMA is a lightweight, crystal-clear, glasslike polymer used in airplane windows, taillight lenses, and light fixtures. Because it is hard, stable to sunlight, and extremely durable, PMMA is also used to make the reflectors embedded between lanes of interstate highways. The unusual transparency of PMMA makes the polymer ideal for hard contact lenses. Unfortunately, PMMA is impermeable to oxygen and water. Oxygen must therefore be transported to the cornea of the eye in the tears and then passed under the contact lens each time the eye blinks. Soft plastic lenses that pass both oxygen and water are made by using ethylene glycol dimethacrylate to cross-link poly(2-hydroxyethyl methacrylate). CO2CH2CH2OH A ... CH2C n... A CH3
Poly(2-hydroxyethyl methacrylate)
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O O B B CH2 PCCOCH2CH2OCC P CH2 A A CH3 CH3
15
Ethylene glycol dimethacrylate
An interesting polymer can be prepared by copolymerizing a mixture of acrylic acid and the sodium salt of acrylic acid. The product of the reaction has the following structure. CO2H A ... CH2CH
x
CO2Na A CH2CH y...
Sodium polyacrylate
The difference between the Na+ ion concentration inside the polymer network and in the solution in which the polymer is immersed generates an osmotic pressure that draws water into the polymer. The amount of liquid that can be absorbed depends on the ionic strength of the solution—the total concentration of positive and negative ions in the solution. The polymer can absorb 800 times its own weight of distilled water, but only 300 times its weight of tap water. Because the ionic strength of urine is equivalent to a 0.1 M NaCl solution, the superabsorbent polymer, which can be found in disposable diapers, can absorb up to 60 times its weight of urine.
P.8 CONDENSATION POLYMERS The first plastic (celluloid) and the first artificial fiber (rayon) were produced from cellulose, as noted in Section P.1. The first truly synthetic plastic was Bakelite, developed by Leo Baekland between 1905 and 1914. The synthesis of Bakelite starts with the reaction between formaldehyde (H2CO) and phenol (C6H5OH) to form a mixture of ortho- and para-substituted phenols. At temperatures above 100°C, the phenols condense to form a polymer in which the aromatic rings are bridged by either OCH2OCH2O or OCH2O linkages. The cross-linking in the polymer is so extensive that it is a thermoset plastic. Once a piece is formed, any attempt to change the shape of the plastic is doomed to failure. Research started by Wallace Carothers and co-workers at Du Pont in the 1920s and 1930s eventually led to the discovery of the families of condensation polymers known as polyamides and polyesters. The polyamides were obtained by reacting a diacyl chloride with a diamine. O O B B n H2N(CH2)xNH2 n ClC(CH2)yCCl Diamine
Diacyl chloride
O O B B ... NH(CH2)xNHC(CH2)yC n... 2n HCl Polyamide
The polyesters were made by reacting the diacyl chloride with a dialcohol. O O B B n HO(CH2)xOH n ClC(CH2)yCCl Dialcohol
Diacyl chloride
O O B B ... O(C H2)xOC(CH2)yC n... 2n HCl Polyester
While studying polyesters, Julian Hill found that he could wind a small amount of the polymer on the end of a stirring rod and draw it slowly out of solution as a silky fiber. One day, when Carothers wasn’t in the lab, Hill and his colleagues tried to see how long a fiber
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they could make by stretching a sample of the polymer as they ran down the hall. They soon realized that this playful exercise had oriented the polymer molecules in two dimensions and produced a new material with superior properties. They then tried the same thing with one of the polyamides and produced a sample of what became the first synthetic fiber: nylon. The polymerization process can be demonstrated by carefully pouring a solution of hexamethylenediamine in water on top of a solution of adipoyl chloride in CH2Cl2. O O B B n H2N(CH2 )6NH2 n ClC(CH2)4CCl Hexamethylene diamine
O O B B ... NH(CH2)6NHC(CH2)4C n ... 2n HCl
Adipoyl chloride
Nylon 6,6
A thin film of polymer forms at the interface between the two phases. By grasping the film with a pair of tweezers, we can draw a continuous string of nylon from the solution. The product of the reaction is known as Nylon 6,6 because the polymer is formed from a diamine that has six carbon atoms and a derivative of a dicarboxylic acid that has six carbon atoms. The effect of pulling on the polymer with the tweezers is much like that of stretching an elastomer—the polymer molecules become oriented in two dimensions. Why don’t the polymer molecules return to their original shape when we stop pulling? Section P.3 suggested that polymers are elastic when there is no strong force of attraction between the polymer chains. Polyamides and polyesters form strong hydrogen bonds between the polymer chains that keep the polymer molecules oriented, as shown in Figure P.9.
H A CON B O Y H A NOC B O Y H A NOC B O Y
FIGURE P.9 The hydrogen bonds that form between the polymer chains when polyamides and polyesters are stretched help to keep the chains oriented in a two-dimensional fiber.
Exercise P.4 A synthetic fiber known as Nylon 6 has the following structure. O B ... NH(CH2)5 C n ... Explain how the polymer is made.
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Solution The polyamide must be made from a monomer that contains both an ONH2 and a OCOCl functional group. The polymer is therefore made by the following condensation reaction. O B n H2N(CH2)5 CCl
O B ... NH(CH2)5 C n ... ⫹ n HCl
The first polyester fibers were produced by reacting ethylene glycol and either terephthalic acid or one of its esters to give poly(ethylene terephthalate). This polymer is still used to make thin films (Mylar) and textile fibers (Dacron and Fortrel). O B n HOCH2CH2OH ⫹ n CH3OC
O B ... OCH2CH2OC
O B COCH3
O B C n ... ⫹ 2 n CH3OH
Phosgene (COCl2) reacts with alcohols to form esters that are analogous to those formed when acyl chlorides react with alcohols. O B ClCCl ⫹ 2 HOR
O B ROCOR ⫹ 2 HCl
The product of the reaction is called a carbonate ester because it is the diester of carbonic acid, H2CO3. Polycarbonates are produced when one of the esters reacts with an appropriate alcohol, as shown in Figure P.10. The polycarbonate shown in Figure P.10 is known as Lexan. It has a very high resistance to impact and is used in safety glass, bulletproof windows, and motorcycle helmets.
O B OOCOO
n
... O
⫹ n HO
CH3 A C A CH3
CH3 A C A CH3
O B OOC n... ⫹ 2n
OH
OH
FIGURE P.10 Polycarbonates are condensation polymers formed by reacting a diester of carbonic acid with a dialcohol.
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The structures and names of some common condensation polymers are given in Table P.2. TABLE P.2 Common Condensation Polymers Trade Name or Common Name
Structure Polyamides O O B B (ONHO(CH2)6ONHOCO(CH2)4OCO)n O O B B (ONHO(CH2)6ONHOCO(CH2)3OCO)n O B (ONHO(CH2)5OCO)n O O B B (ONH CH2 NHOCO(CH2)10OCO)n
Nylon 6, 6 Nylon 6, 10 Nylon 6 Qiana
Polyaramides O B CO)n
(ONH
Kevlar Polyesters
O B (OOOCH2CH2OOOC
O B CO)n
O B CH2OOOC
(OOOCH2
Dacron, Mylar O B CO)n
Kodel
Polycarbonates CH3 A C A CH3
(OO
O B OOCO)n
Lexan Silicones
CH3 A (OOOSiO)n A CH3
silicone rubber
P.9 PROPERTIES OF POLYMERS The following variables can be controlled when producing a polymer. • • • • •
The The The The The
monomer polymerized or the monomers copolymerized. reagent used to initiate the polymerization reaction. identity and amount of the reagent used to cross-link the polymer chains. temperature and pressure at which the polymerization occurs. solvent in which the monomer is polymerized.
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The way the polymer is collected, which can produce either a more or less random alignment of the polymer chains or a fabric in which the chains are aligned in one direction.
Changing one or more of these parameters can affect the linearity of the polymer, its average molecular weight, the tacticity of side chains on the polymer backbone, and the density of the product. It is also possible to change the properties of a polymer by adding either stabilizers or plasticizers. Stabilizers are used to increase the ability of a plastic to resist oxidation, to make it less sensitive to either heat or light, or to make the material flame retardant. Plasticizers increase the flexibility of a plastic by acting as a lubricant, decreasing the friction between molecules as one polymer chain moves past another. They also increase the amount of empty space—the so-called free volume—within the polymer by opening up space between the polymer chains to increase the ease with which the chain ends, the side chains, and the main chain can move. The result of all of these manipulations can be a polymer as strong as Kevlar, which is used to make bulletproof vests, or a material as easy to rip as a piece of paper. It can be as hard as a bowling ball or as soft as a piece of tissue paper. It can be as brittle as the disposable polystyrene glasses used at parties or as elastic as a Styrofoam coffee cup. The following list describes some of the important properties of a polymer. Heat capacity/heat conductivity: The extent to which the plastic or polymer acts as an effective insulator against the flow of heat. (The polystyrene in disposable plastic glasses isn’t a very good insulator. However, blowing air through styrene while it is being polymerized gives the Styrofoam used for disposable coffee cups, which is a much better insulator.) Thermal expansion: The extent to which the polymer expands or contracts when heated or cooled. (Silicone is often used to seal glass windows to their frames because it has a very low coefficient of thermal expansion.) Thermal expansion is also concerned with the question of whether the polymer expands or contracts by the same amount in all directions. (Polymers are usually anisotropic. They contain strong covalent bonds along the polymer chain and much weaker dispersive forces between the polymer chains. As a result, polymers can expand by differing amounts in different directions.) Crystallinity: The extent to which the polymer chains are arranged in a regular structure instead of a random fashion. (Some polymers, such as Silly Putty and Play Dough, are too amorphous and lack the rigidity needed to make a useful product. Polymers that are too crystalline often are also too brittle.) Permeability: The tendency of a polymer to pass extraneous materials. (Polyethylene is used to wrap foods because it is 4000 times less permeable to oxygen than polystyrene.) Elastic modulus: The force it takes to stretch the plastic in one direction. Tensile strength: The strength of the plastic (i.e., the force that must be applied in one direction to stretch the plastic until it breaks). Resilience: The ability of the plastic to resist abrasion and wear. Refractive index: The extent to which the plastic affects light as it passes through the polymer. (Does it pass light the way PMMA does, or does it absorb light like PVC?) Resistance to electric current: Is the material an insulator, like most polymers, or does it conduct an electric current? (There is a growing interest in conducting polymers,
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which can be charged and discharged, and photoconducting polymers that can pick up an electric charge when exposed to light.)
Chemistry in the World Around Us The Search for Synthetic Fibers Synthetic fibers can be traced back to 1885, when the French Count Hilaire de Chardonnet received a patent for a synthetic silk. The first step in making Chardonnet silk, as it was known, involved dissolving cellulose from wood pulp in nitric acid to form cellulose nitrate. Anyone who has watched what happens when a flame is held close to a PingPong ball (see Section P.1) should understand that cellulose nitrate, by itself, is far too flammable to be used as a fiber. The next step in the production of Chardonnet silk therefore involved decomposing cellulose nitrate back to cellulose under conditions that generated a continuous fiber. This was achieved by extruding a viscous solution of cellulose nitrate in alcohol through small holes into water. In 1903, a slightly different process was patented in the United Kingdom. It involved dissolving cellulose from wood pulp in a mixture of caustic soda and carbon disulfide (CS2). The resulting viscous material was then extruded through small holes into a solution of sulfuric acid to form a fiber that was sold under the trade name of Rayon. By 1980, the demand for cellulose-based fibers had reached the point where more than 3,250,000 metric tonnes were produced each year. More than 100 years after the development of the first synthetic fiber, some people take comfort from the fact that the most important fiber is still cotton, which accounts for almost 50% of the total world production of textile fibers. Advances in synthetic fibers, however, have generated materials that not only compete with, but surpass, natural fibers. Until recently, the standard against which all fibers were compared as potential insulators was the soft, fluffy down obtained from ducks and geese. Down is simultaneously lightweight and an excellent thermal insulator. Unfortunately, it also readily absorbs water, and, when wet, loses much of its ability to act as an insulator. Thus, even if the supply of down were plentiful—which it is not—and even if down were inexpensive—which it is not—there would be a potential demand for a synthetic fiber that had the insulating properties of down but did not “wet.” Several years ago, a synthetic insulator known as primaloft was prepared under a contract issued by the U.S. Army Research, Development and Engineering Center at Natick [Chemical and Engineering News, Oct. 16, 1989, p. 25]. Scanning electron microscopy has shown that natural down is a mixture of relatively large-diameter fibers, which make it stiff, and very thin fibers that ensure the presence of many small pockets of air. The result is a lightweight but stiff material that is a good insulator. Primaloft achieves the same effect by combining a small number of large-diameter polyester fibers with many more small fibers. The small fibers have a diameter of 7 m, roughly one-fourth the diameter of a human hair. Primaloft is just as “warm” as down and even has the same feel when enclosed in a jacket or parka. More importantly, primaloft absorbs much less water when wet, and therefore it doesn’t lose its insulating capacity in the rain. Furthermore, it is considerably less expensive than down. A variety of esoteric measurements are done to compare different textile fibers. Questions that are asked include the following: Does the fabric swell when the fibers absorb water? Does the fabric have a tendency to build up static electric charge? Is the
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fiber even, or does the yarn vary in diameter as it is spun? Does the fiber shrink on exposure to water? How well does the fiber “breathe?” (Is it permeable to air?) What is the bursting strength of the fiber? What is its tear strength? How well does it resist snag? How well does it stand up to abrasion and other forms of wear? How well does it conduct heat? (Is it a good insulator, like natural wool?) How well does it pass moisture in the form of water vapor? Is it water repellent, or, at least, can it be made water repellent? How well does it stretch? How well does it accept various dyes? Is it colorfast, once dyed? And, of course, perhaps as important as any other property—is it flammable? Research in the 1990s will continue the search for new fabrics, such as those recently formulated by the Hoechst Celanese Corporation for athletic clothes that have the remarkable ability to pass water vapor but not liquid water through the fabric. These fabrics allow perspiration to evaporate and still provide the protection against rain or snow expected for water-repellant fabrics.
KEY TERMS Acrylic Addition polymer Anionic polymerization Atactic Block copolymer Branched polymer Cationic polymerization Condensation polymer Coordination polymerization Copolymer Cross-linked polymer Elastomer
Free radical Free radical polymerization Graft copolymer Homopolymer Isomer Isotactic Linear polymer Macromolecule Monomer Nylon Polyamide Polycarbonate
Polyester Polyethylene Polymer Polypropylene Poly(vinyl chloride), PVC Random copolymer Regular copolymer Straight-chain polymer Syndiotactic Tacticity Teflon Ziegler–Natta catalyst
PROBLEMS Polymers 1. Use Berzelius’ definition of a polymer to sort the following compounds into groups of polymers. (a) formaldehyde, H2CO (b) ethylene, C2H4 (c) glyceraldehyde, C3H6O3 (d) 2-butene, C4H8 (e) cyclohexane, C6H12 (f ) glucose, C6H12O6 2. Which of the following substances are polymers? (a) Teflon (b) propane (c) acetic acid (d) poly(vinyl chloride) (e) polyethylene
Definition of Terms 3. Describe the difference between linear, branched, and cross-linked polymers.
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4. Explain the following observations. (a) Linear polymers, such as the polyethylene in garbage bags, tear when stretched. (b) Lightly cross-linked polymers, such as rubber, return to their original shape when stretched. (c) Highly cross-linked polymers, such as Bakelite, break when “stretched.” 5. Describe the difference between a homopolymer (such as polystyrene) and a copolymer (such as styrene–butadiene rubber). 6. Describe the difference between random, regular, block, and graft copolymers. Give an example of the sequence of monomer units that would be found in a short length of each of the polymers. 7. Describe the differences in the structures of atactic, isotactic, and syndiotactic forms of poly(vinyl chloride). Which of those would you expect to be the easiest to make? Which might be the most useful? 8. Calculate the range of molecular weights of polyethylene molecules in a sample in which individual chains contain between 500 and 50,000 (OCH2CH2O) units. 9. Individual chains in polystyrene polymers typically weigh between 200,000 and 300,000 amu. If the formula for the monomer is C6H5CHPCH2 and the polymer is an example of an addition polymer, how many monomers does the typical chain contain? 10. Calculate the average molecular weight of the polymer chains in the cellulose nitrate in Ping-Pong balls if the formula for the monomer is C6H9NO7 and the average polymer chain contains 1500 monomers. 11. Explain why it is possible to measure the average molecular weight of linear polymers such as polyethylene, but not cross-linked polymers such as those found in rubber. 12. Thermoplastic polymers flow when heated and can be molded into shapes they retain on cooling. Explain how increasing the amount of cross-linking between polymer chains can transform a thermoplastic polymer into a rigid thermoset polymer. 13. Describe the difference between addition and condensation polymers. Give an example of each. 14. Classify the following as either addition or condensation polymers. (a) polyethylene, [...(CH2CH2)n...] (b) poly(vinyl chloride), [...(CH2CHCl)n...] (c) Nylon 6,6, [...(CO(CH2)4CONH(CH2)6NH)n...] (d) polyester, [...(OCH2CH2OCOC6H4CO)n...] (e) silicone, [...(OSi(CH3)2)n...] (f ) Plexiglas, [...(CH2C(CH3)(CO2CH3))n...]
Elastomers 15. What are the characteristic properties of an elastomer? 16. Explain why elastomers must contain long, flexible molecules that are coiled in the natural state. Explain why polymers have to have some cross-links to be elastomers but can’t have too many. 17. Natural rubber becomes too soft when heated and too hard when cooled to be of much use. Explain what happens when natural rubber is treated with sulfur that turns the material into a commercially useful product.
Free Radical Polymerization Reactions 18. Describe what happens during the chain initiation, chain propagation, and chain termination steps when propylene (CH3CHPCH2) is polymerized by a free radical mechanism.
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19. Write the Lewis structures of the intermediates in the free radical polymerization of vinyl chloride, CH2PCHCl.
Ionic and Coordination Polymerization Reactions 20. Describe what happens during the chain initiation, chain propagation, and chain termination steps in the anionic polymerization of vinyl chloride (CH2PCHCl) when methyllithium (CH3Li) is used as the chain initiator. Write the Lewis structures of all intermediates. 21. Describe what happens during the chain initiation, chain propagation, and chain termination steps in the cationic polymerization of vinylidene chloride (CH2PCCl2) when hydrobromic acid (HBr) is used to initiate the reaction. Write the Lewis structures of all intermediates. 22. A typical Ziegler–Natta catalyst consists of a mixture of TiCl3 and Al(C2H5)3. Explain how the formation of a complex between a molecule of ethylene and the titanium atom in the catalyst can be thought of as an example of a Lewis acid-base reaction.
Addition Polymers 23. Polypropylene can be made in both atactic and isotactic forms. The atactic form is soft and rubbery, with no commercial value. The isotactic form is much more crystalline— it is hard enough, for example, to be used for furniture. Explain the difference between the physical properties of the two forms of the polymer. 24. Polyethylene can be made in both high-density and low-density forms. One of the polymers has a linear structure; the other is branched, with short side chains of up to five carbon atoms attached to the polymer backbone. Which structure would you expect to give the denser polymer? Which structure would give the more crystalline polymer? 25. Teflon, (CF2CF2)n, is a waxy polymer to which practically nothing sticks. It is also the most chemically inert of all polymers. Describe at least five ways in which a polymer with these properties can be used. 26. Use the concept of van der Waals forces to explain why plastic wrap made from Saran clings to itself, whereas plastic wrap made from polyethylene does not.
Condensation Polymers 27. Predict the formula of the repeating unit in the condensation polymers formed by the following reactions. O O B B HCl ⫹ ... (a) Dacron: HOCH2CH2OH ⫹ ClC CCl O O B B HCl ⫹ ... (b) Nylon 6,6: H2N(CH2 ) 6NH2 ⫹ ClC(CH2 ) 4CCl CH3OH ⫹ ... (c) polycarbonate: HOCH2CH2OH ⫹ (CH3O)2C P O O B HCl ⫹ ... CCl (d) polyaramide: H2N (e) silicone rubber: (CH3)2Si(OH)2 HCl ⫹ ... 28. Explain the difference between the reactions used to prepare Nylon 6,6 and Nylon 6.
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8 BIOCHEMISTRY BIO.1 Biochemistry BIO.2 The Amino Acids BIO.3 The Acid–Base Chemistry of Amino Acids BIO.4 The Chemistry of Hemoglobin and Myoglobin BIO.5 Peptides and Proteins BIO.6 The Structure of Proteins BIO.7 Carbohydrates: The Monosaccharides BIO.8 Carbohydrates: The Disaccharides and Polysaccharides BIO.9 Lipids Chemistry in the World Around Us: The Search for New Drugs BIO.10 Nucleic Acids BIO.11 Protein Biosynthesis
BIO.1 BIOCHEMISTRY Biochemistry is the study of the chemistry of living organisms. Biomolecules are often sorted into four categories: (1) peptides and proteins, (2) carbohydrates, (3) nucleic acids, and (4) lipids. The term protein comes from the Greek word proteios, which means “of first importance.” Biochemistry, for many years, was almost synonymous with the study of proteins because these compounds serve the broadest array of functions of any class of biomolecules, including the following: Structure: The actin and myosin in muscles, the collagen in skin and bone, and the keratins in hair, horn, and hoof are examples of proteins whose primary function is to produce the structure of the organism. Catalysis: Most chemical reactions in living systems are catalyzed by enzymes, which are proteins. Control: Many proteins regulate or control biological activity. Insulin, for example, controls the rate at which sugar is taken into the cell. Energy: Some proteins, including the casein in milk and the albumin in eggs, are used primarily to store food energy. Transport: O2 is carried through the bloodstream by hemoglobin. Other proteins transport sugars, amino acids, and ions across cell membranes. 1
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Protection: The first line of defense against viruses and bacteria are the antibodies produced by the immune system, which are based on proteins. The term carbohydrate reflects the fact that many of the compounds in that category have the empirical formula CH2O—they are literally “hydrates of carbon.” Carbohydrates are the primary source of food energy for most living systems. They include simple sugars such as glucose (C6H12O6) and sucrose (C12H22O11) as well as polymers of these sugars such as starch, glycogen, and cellulose. Carbohydrates are produced from CO2 and H2O during photosynthesis and are therefore the end products of the process by which plants capture the energy in sunlight. The name nucleic acid was originally given to a class of relatively strong acids that were found in the nuclei of cells. As monomers, nucleic acids such as adenosine triphosphate (ATP) are involved in the process by which cells capture food energy and make it available to fuel the processes that keep cells alive. As polymers, they store and process the information that allows the organism to grow and eventually reproduce. Compounds are classified as proteins, carbohydrates, and nucleic acids on the basis of similarities in their structures. Lipids, on the other hand, are defined on the basis of their physical properties. Any molecule in a biological system that is soluble in nonpolar solvents is classified as a lipid (from the Greek word lipos, “fat”). The lipid known as cholesterol, for example, is virtually insoluble in water, but it is soluble in a variety of nonpolar solvents—including the nonpolar region between the inner and outer surfaces of a cell membrane.
BIO.2 THE AMINO ACIDS Proteins are formed by polymerizing monomers that are known as amino acids because they contain an amine (ONH2) and a carboxylic acid (OCO2H) functional group. With only one exception,1 the amino acids used to synthesize proteins are primary amines with the following generic formula. R A H2NCHCO2H An amino acid
These compounds are known as -amino acids because the ONH2 group is on the carbon atom next to the OCO2H group, the so-called carbon atom of the carboxylic acid. The chemistry of amino acids is complicated by the fact that the ONH2 group is a base and the OCO2H group is an acid. In aqueous solution, an H ion is therefore transferred from one end of the molecule to the other to form a zwitterion (from a German word meaning “mongrel ion,” or hybrid ion). R R A A H2NCHCO2H 88n H3NCHCO2 Zwitterions are simultaneously electrically charged and electrically neutral. They contain positive and negative charges, but the net charge on the molecule is zero. 1
The sole exception is the amino acid proline, which is a secondary amine.
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More than 300 amino acids are listed in the Practical Handbook of Biochemistry and Molecular Biology,2 but only the 20 amino acids in Table BIO.1 are used to synthesize proteins. Most of the amino acids differ only in the nature of the R substituent. The standard amino acids are therefore classified on the basis of the R groups. Amino acids with nonpolar substituents are said to be hydrophobic (“water-hating”). Amino acids with polar R groups that form hydrogen bonds to water are classified as hydrophilic (“water-loving”). The remaining amino acids have substituents that carry either negative or positive charges in aqueous solution at neutral pH and are therefore strongly hydrophilic.
Exercise BIO.1 Use the structures of the following amino acids in Table BIO.1 to classify the compounds as either nonpolar/hydrophobic, polar/hydrophilic, negatively charged/hydrophilic, or positively charged/hydrophilic. (a) valine, R OCH(CH3)2 (b) serine, R OCH2OH (c) aspartic acid, R OCH2CO2 (d) lysine, R O(CH2)4NH3 Solution (a) Valine is a nonpolar, hydrophobic amino acid. (b) Serine is a polar, hydrophilic amino acid. (c) Aspartic acid is a hydrophilic amino acid with a negatively charged R group. (d) Lysine is a hydrophilic amino acid with a positively charged R group. With the exception of glycine, the common amino acids all contain at least one chiral carbon atom. The amino acids therefore exist as pairs of stereoisomers. The structures of the D and L isomers of alanine are shown in Figure BIO.1. Although D amino acids can be found in nature, only the L isomers are used to form proteins. The D isomers are most often found attached to the cell walls of bacteria and in antibiotics that attack bacteria. The presence of the D isomers protects the bacteria from enzymes the host organism uses to protect itself from bacterial infection by hydrolyzing the proteins in the bacterial cell wall.
D-Alanine
L-Alanine
#
CH3 A C { { CO2 H NH3
#
CH3 A C { { H NH3 CO2
FIGURE BIO.1 The alanine.
D
and
L
stereoisomers of the amino acid
A few biologically important derivatives of the standard amino acids are shown in Figure BIO.2. Anyone who has used an “antihistamine” to alleviate the symptoms of exposure to an allergen can appreciate the role that histamine—a decarboxylated derivative of histidine—plays in mediating the body’s response to allergic reactions. L-DOPA, which is a derivative of tyrosine, has been used to treat Parkinson’s disease. The compound received notoriety in the film Awakening, which documented its use as a treatment for other neurological disorders. Thyroxine, which is an iodinated ether of tyrosine, is a hormone that acts on the thyroid gland to stimulate the rate of metabolism. 2
CRC Press, Boca Raton, Florida, 1989.
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TABLE BIO.1
The 20 Standard Amino Acids
Name
Structure (at neutral pH)
Name
Structure (at neutral pH)
Nonpolar (Hydrophobic) R Groups H A Glycine (Gly) H3NOCHOCO2
Polar (Hydrophilic) R Groups CH2OH A H3N OCHOCO2 Serine (Ser)
CH3 A H3N OCHOCO2
OH D G CH A H3NOCHOCO2
Alanine (Ala)
CH3
CH3
D
CH3
Valine (Val)
G CH A H3NOCHOCO2
Leucine (Leu)
CH D 3 G CH A CH2 A H3NOCHOCO2
Threonine (Thr)
OH A
CH3
CH3
Isoleucine (Ile)
G CH A H3NOCHOCO2 H2C H2C
Methionine (Met)
Phenylalanine (Phe)
H
H
CH CO2 H
CH3 A S A CH2 A CH2 A H3NOCHOCO2
CH2 A H3NOCHOCO2
NOH
Tryptophan (Trp)
Cysteine (Cys)
CH2SH A H3N OCHOCO2
Asparagine (Asn)
O B C ONH2 A CH2 A H3NOCHOCO2
Glutamine (Gln)
O B C ONH2 A CH2 A CH2 A H3NOCHOCO2
CH2 N
Proline (Pro)
CH2CH3 D
Tyrosine (Tyr)
A CH2 A H3NOCHOCO2
A CH2 A H3NOCHOCO2
Negatively Charged R Groups
Aspartic acid (Asp)
CO2 A CH2 A H3NOCHOCO2
Glutamic acid (Glu)
CO2 A CH2 A CH2 A H3NOCHOCO2
(continued)
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Name
Structure (at neutral pH)
Name
Positively Charged R Groups
Structure (at neutral pH)
Positively Charged R Groups NH3 A CH2 A CH2 A CH2 A CH2 A H3NOCHOCO2 Lysine (Lys)
B
NH2 A C NH2 A NH A CH2 A CH2 A CH2 A H3NOCHOCO2
Arginine (Arg)
5
H
N
D
H H
N
CH2 H A H3NOCHOCO2
Histidine (His)
N Histamine
HO
A
CH2CH2NH3
A
A
N A H
A A
CO2
CH2CH
A HO
A I
NH3
A
O
A
A
CO2
A A
HO
A
IA
L-DOPA
CH2CH
Thyroxine
NH3
FIGURE BIO.2 Three biologically important derivatives of amino acids.
BIO.3 THE ACID–BASE CHEMISTRY OF AMINO ACIDS Acetic acid and ammonia often play an important role in the discussion of the chemistry of acids and bases. One of the compounds is a weak acid; the other is a weak base. 88n CH3CO2 H3O CH3CO2H H2O m88 88n NH4 OH NH3 H2O m88
Ka 1.8 105 Kb 1.8 105
Thus, it is not surprising that an H ion is transferred from one end of the molecule to the other when an amino acid dissolves in water. R R A A H2NCHCO2H 88n H3NCHCO2 The zwitterion is the dominant species in aqueous solutions at physiological pH (pH 7).
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The zwitterion can undergo acid–base reactions, however, if we add either a strong acid or a strong base to the solution. Imagine what would happen if we add a strong acid to a neutral solution of an amino acid in water. In the presence of a strong acid, the OCO2 end of the molecule picks up an H ion to form a molecule with a net positive charge. R R A A H H3NCHCO2 88n H3NCHCO2H In the presence of a strong base, the ONH3 end of the molecule loses an H ion to form a molecule with a net negative charge. R R A A OH H3NCHCO2 88n H2NCHCO2 Figure BIO.3 shows what happens to the pH of an acidic solution of glycine when the amino acid is titrated with a strong base, such as NaOH.
12 +
10 pK2
O
+
H
H3NCH
2 CO
O
–
CO NCH 2 H2
–
+
8
pI
O – + H+
pH 6
+
pK1 2
O H 2C NC 3 +
H
0
0.5
OH
H
2 CO
4 NC H3
1.0
1.5
2.
H+ ions dissociated/molecule
FIGURE BIO.3 Titration curve for the titration of glycine with a strong base.
To understand the titration curve, let’s start with the equation that describes the acid dissociation equilibrium constant expression for an acid, HA. [H3O][A] Ka [HA] Let’s now rearrange the Ka expression, [HA] [H3O] Ka [A]
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take the log to the base 10 of both sides of the equation, [HA] log[H3O] log Ka log [A] and then multiply both sides of the equation by 1. [HA] log[H3O] log Ka log [A] By definition, the term on the left side of the equation is the pH of the solution, and the first term on the right side is the pKa of the acid. [HA] pH pKa log [A] The negative sign on the right side of this equation is often viewed as “inconvenient.” The derivation therefore continues by taking advantage of the following feature of logarithmic mathematics [HA] [A] log log [A] [HA] to give the following form of the equation. [A] pH pKa log [HA] This equation is known as the Henderson–Hasselbalch equation, and it can be used to calculate the pH of the solution at any point in the titration curve in Figure BIO.3. The following occurs as we go from left to right across the titration curve. •
•
•
•
•
The pH initially increases as we add base to the solution because the base deprotonates some of the positively charged H3NCH2CO2H ions that were present in the strongly acidic solution. The pH then levels off because we form a buffer solution in which we have reasonable concentrations of both an acid, H3NCH2CO2H, and its conjugate base, H3NCH2CO2. When virtually all of the H3NCH2CO2H molecules have been deprotonated, we no longer have a buffer solution and the pH rises rapidly when more NaOH is added to the solution. The pH then levels off as some of the neutral H3NCH2CO2 molecules lose protons to form negatively charged H2NCH2CO2 ions. When the ions are formed, we once again get a buffer solution in which the pH remains relatively constant until essentially all of the H3NCH2CO2 molecules have been converted into H2NCH2CO2 ions. At this point, the pH rises rapidly until it reaches the value observed for a strong base.
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The pH titration curve tells us the volume of base required to titrate the positively charged H3NCH2CO2H molecule to the H3NCH2CO2 zwitterion. If we only add half as much base, only half of the positive ions would be titrated to zwitterions. In other words, the concentration of the H3NCH2CO2H and H3NCH2CO2 ions would be the same. Or, using the symbolism in the Henderson–Hasselbalch equation: [HA] [A] Because the concentrations of the ions are the same, the logarithm of the ratio of their concentrations is zero. [A] log 0 [HA] Thus, at this particular point in the titration curve, the Henderson–Hasselbalch equation gives the following equality. pH pKa We can therefore determine the pKa of an acid by measuring the pH of a solution in which the acid has been half-titrated. Because there are two titratable groups in glycine, we get two points at which the amino acid is half-titrated. The first occurs when half of the positive H3NCH2CO2H ions have been converted to neutral H3NCH2CO2 molecules. The second occurs when half of the H3NCH2CO2 zwitterions have been converted to negatively charged H2NCH2CO2 ions. The following results are obtained when this technique is applied to glycine. H3NCH2CO2 pKa 9.78 pKa 2.35 Let’s compare these values with the pKa values of acetic acid and the ammonium ion. CH3CO2H NH4
pKa 4.74 pKa 9.24
The acid–base properties of the -amino group in an amino acid are very similar to the properties of ammonia and the ammonium ion. The -amine, however, has a significant effect on the acidity of the carboxylic acid. The -amine increases the value of Ka for the carboxylic acid by a factor of about 100. The inductive effect of the -amine can only be felt at the -CO2H group. If we look at the chemistry of glutamic acid, for example, the OCO2H group on the R substituent has an acidity that is close to that of acetic acid. CO2 A CH2 A CH2 A
pKa 4.07
H3NCHCO2 pKa 9.47
pKa 2.10
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When we titrate an amino acid from the low end of the pH scale (pH 1) to the high end (pH 13), we start with an ion that has a net positive charge and end up with an ion that has a net negative charge. OH
H3NCHRCO2H 88n H2NCHRCO2 pH 7
pH 7
Somewhere between these extremes, we have to find a situation in which the vast majority of the amino acids are present as the zwitterion—with no net electric charge. This point is called the isoelectric point (pI) of the amino acid. For simple amino acids, in which the R group doesn’t contain any titratable groups, the isoelectric point can be calculated by averaging the pKa values for the -carboxylic acid and -amino groups. Glycine, for example, has a pI of about 6. pI
2.35 9.78 6.07 2
At pH 6, more than 99.98% of the glycine molecules in the solution are present as the neutral H3NCH2CO2 zwitterion. When calculating the pI of an amino acid that has a titratable group on the R side chain, it is useful to start by writing the structure of the amino acid at physiological pH (pH 7). Lysine, for example, could be represented by the following diagram. NH3 pKa 10.79 A (CH2)4 A H3NCHCO2 pKa 9.18
pKa 2.16
At physiological pH, lysine has a net positive charge. Thus, we have to increase the pH of the solution to remove positive charge in order to reach the isoelectric point. The pI for lysine is simply the average of the pKa values of the two ONH3 groups. pI
9.18 10.79 ≈ 9.99 2
At pH 10, all of the carboxylic acid groups are present as OCO2 ions, and the total population of the ONH3 groups is equal to one. Thus, the net charge on the molecule at that pH is zero. If we apply the same technique, given above, to the pKa data for glutamic acid, we get a pI of 3.09. The three amino acids in this section therefore have very different pI values. Glutamic acid (R OCH2CH2CO2) Glycine (R OH) Lysine (R OCH2CH2CH2CH2NH3)
pI 3.09 pI 6.07 pI 9.99
Thus, it isn’t surprising that a common technique for separating amino acids (or the proteins they form) involves placing a mixture in the center of a gel and then applying a strong voltage across the gel. This technique, which is known as gel electrophoresis, is based on
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the fact that amino acids or proteins that carry a net positive charge at the pH at which the separation is done will move toward the negative electrode, whereas those with a net negative charge will move toward the positive electrode.
BIO.4 THE CHEMISTRY OF HEMOGLOBIN AND MYOGLOBIN At one time or another, everyone has experienced the momentary sensation of having to stop, to “catch one’s breath,” until enough O2 can be absorbed by the lungs and transported through the bloodstream. Imagine what life would be like if we had to rely only on our lungs and the water in our blood to transport oxygen through our bodies. O2 is only marginally soluble (0.0001 M) in blood plasma at physiological pH. If we had to rely on the oxygen that dissolved in blood as our source of oxygen, we would get roughly 1% of the oxygen to which we are accustomed. (Consider what life would be like if the amount of oxygen you received was equivalent to only one breath every 5 min, instead of one breath every 3 s.) The evolution of forms of life even as complex as an earthworm required the development of a mechanism to actively transport oxygen through the system. Our bloodstream contains about 150 g/L of the protein known as hemoglobin (Hb), which is so effective as an oxygen carrier that the concentration of O2 in the bloodstream reaches 0.01 M—the same concentration as in air. Once the Hb–O2 complex reaches the tissue that consumes oxygen, the O2 molecules are transferred to another protein—myoglobin (Mb)—which transports oxygen through the muscle tissue. The site at which oxygen binds to both hemoglobin and myoglobin is the heme shown in Figure BIO.4. At the center of the heme is an Fe(II) atom. Four of the six coordination sites around the atom are occupied by nitrogen atoms from a planar porphyrin ring. The O B
B
CH2 B CH
CH3
N
A
A
N
A
A
A
H3C
CH3 A
CH A
A
H2C
B
O
A CH2
A CH2
CH2
A
C
COO
N HN
B
A
A CH2 A O C N H C A H
A
CH2
N
D
A
OOC
A
A
H3C
A
N
A
A
A
Fe(II)
FIGURE BIO.4 The heme group in hemoglobin.
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fifth coordination site is occupied by a nitrogen atom from a histidine side chain on one of the amino acids in the protein. The last coordination site is available to bind an O2 molecule. The heme is therefore the oxygen-carrying portion of the hemoglobin and myoglobin molecules. This raises the question, What is the function of the globular protein or “globin” portion of the molecules? The structure of sperm whale myoglobin, shown in Figure BIO.5, suggests that the oxygen-carrying heme group is buried inside the protein portion of the molecule, which keeps pairs of heme groups from coming too close together. This is important, because the proteins need to bind O2 reversibly and the Fe(II) heme, by itself, cannot do this. When there is no globin to protect the heme, it reacts with oxygen to form an oxidized Fe(III) atom instead of an Fe(II)–O2 complex.
FIGURE BIO.5 Schematic drawing of the structure of sperm whale myoglobin. The heme group is in a pocket formed by the protein in the top left of the diagram.
Hemoglobin consists of four protein chains, each about the size of a myoglobin molecule, which fold to give a structure that looks very similar to myoglobin. Thus, hemoglobin has four separate heme groups that can bind a molecule of O2. Even though the distance between the iron atoms of adjacent hemes in hemoglobin is very large—between 250 and 370 nm—the act of binding an O2 molecule at one of the four hemes in hemoglobin leads to a significant increase in the affinity for O2 binding at the other hemes. The cooperative interaction between different binding sites makes hemoglobin an unusually good oxygen-transport protein because it enables the molecule to pick up as much oxygen as possible once the partial pressure of the gas reaches a particular threshold level, and then give off as much oxygen as possible when the partial pressure of O2 drops significantly below the threshold level. The hemes are much too far apart to interact directly. However, changes that occur in the structure of the globin that surrounds a heme when it picks up an O2 molecule are mechanically transmitted to the other globins in the protein. These changes carry the signal that facilitates the gain or loss of an O2 molecule by the other hemes. Drawings of the structures of proteins often convey the impression of a fixed, rigid structure, in which the side chains of individual amino acid residues are locked into position. Nothing could be further from the truth. The changes that occur in the structure of hemoglobin when oxygen binds to the hemes are so large that crystals of deoxygenated hemoglobin shatter when exposed to oxygen. Further evidence for the flexibility of proteins can be obtained by noting that there is no path in the crystal structures of myoglobin and hemoglobin along which an O2 molecule can travel to reach the heme group. The fact that
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these proteins reversibly bind oxygen suggests that they must undergo simple changes in their conformation—changes that have been called breathing motions—that open up and then close down the pathway along which an O2 molecule travels as it enters the protein. Computer simulations of the motion within proteins suggests that the interior of a protein has a significant “fluidity,” with groups moving within the protein by as much as 20 nm.
BIO.5 PEPTIDES AND PROTEINS Myoglobin and hemoglobin are important examples of the class of compounds known as proteins, which are linear polymers of between 40 and 10,000 (or more) amino acids. The average molecular weight of an amino acid is about 110 amu. As a result, a modestly sized protein with only 300 amino acids has a molecular weight of 33,000 g/mol, and very large proteins can have molecular weights as high as 1,000,000 g/mol. Proteins are formed by joining the OCO2H end of one amino acid with the ONH2 end of another to form an amide. The OCONHO bond between amino acids is known as a peptide bond because relatively short polymers of amino acids are known as peptides. R O R O B B H3NCHCO H3NCHCO
R O R O B B H3NCHCNHCHCO A dipeptide
The same OCONHO bond forms the backbone of both proteins and synthetic fibers such as nylon. This raises an interesting question: How do we explain the enormous range of structures and functions of proteins when nylon has such regular properties? Nylon has a regular structure that repeats monotonously from one end of the polymer to the other because the monomers from which it is made are symmetrical. The two ends of an amino acid, on the other hand, are different. Each monomer has both an ONH2 head and a OCO2H tail. Thus, four different dipeptides can be formed from only two amino acids. Aspartic acid (Asp) and phenylalanine (Phe), for example, can give two symmetrical dipeptides—Phe-Phe and Asp-Asp—and two unsymmetrical dipeptides—Phe-Asp and Asp-Phe—as shown in Figure BIO.6. When the full range of amino acids is considered, it is possible to make 400 (202) different dipeptides, 64 million (206) different hexapeptides, and 1052 (2040) different proteins that contain only 40 amino acids.
A
CO A 2
CO A 2
CH A 2 O B
CH A 2
CH A 2 O B
CH A 2 CH
A
NH
A
CO A 2
CO A 2
A
CH A 2 O B
CH A 2
CH A 2 O B
CH A 2 NH
CH
A
C
A
CH
A
H3N
A
CO2
A
Phe-Asp
CH
A
NH
A
C
A
A
A
CH
CO2
Asp-Asp
Phe-Phe
H3N
C
A
CH
A
H3N
A
CO2
A
CH
A
NH
A
C
A
CH
A
A
H3N
A
CO2
Asp-Phe
FIGURE BIO.6 The four dipeptides that can be formed by phenylalanine and aspartic acid.
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Differences between the structures of even closely related dipeptides such as Asp-Phe and Phe-Asp give rise to significant differences in their properties. The methyl ester of Asp-Phe, for example, has a very sweet taste and is sold as an artificial sweetener under the name aspartame.
CO2 A CH2 CH2 O A A B H3NOCHOCONHOCHOCO2CH3 Aspartame
The ester of the dipeptide with the opposite arrangement of amino acids, Phe-Asp, does not taste sweet and has no commercial value. As the length of the polymer chain increases and the number of possible combinations of R groups increases, polymer chains with an almost infinite variety of structures and properties are produced. In recent years, a group of naturally occurring peptides that mimic painkilling drugs such as morphine has been discovered in human brain cells. These enkephalins (from Greek and meaning “in the head”) hold the promise of a synthetic painkiller that is both safe and nonaddictive. One of the enkephalins is a pentapeptide that contains four different amino acids—tyrosine (Tyr), glycine (Gly), phenylalanine (Phe), and methionine (Met). The first step in describing the structure of the peptide is to list the amino acids in the order in which they are found on the peptide chain: Tyr-Gly-Gly-Phe-Met. We then have to identify the amino acid at the OCO2H end of the chain and the amino acid at the ONH2 end. By convention, proteins are listed from the N-terminal amino acid residue toward the C-terminal end. The structure of the enkephalin is shown in Figure BIO.7. OH A
CH A 3 S A
A CH A 2 O B
CH A 2
CH A 2 O B
CH A 2
Tyr-Gly-Gly-Phe-Met
CH
A
NH
A
C
A
CH
A
NH
A
C
A
CH2
A
NH
A
C
A
CH2
A
NH
A
N-terminal amino acid
C
A
CH
A
A
H3N
O B
O B
A
CO2
C-terminal amino acid
FIGURE BIO.7 The structure of the naturally occurring pentapeptide Tyr-Gly-Gly-Phe-Met that binds to the same sites in the brain as synthetic painkillers such as morphine.
The pentapeptide in Figure BIO.7 illustrates the perils that face anyone who tries to synthesize peptides or proteins from amino acids. In order to make the enkephalin in large quantities, we would have to overcome the following problems. •
Peptide bonds are not easy to form. In theory, a carboxylic acid could react with an amine to form an amide. O O B B ... OCOOH H2NO ... 88n ...OCONHO... H2O
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In practice, carboxylic acids are more likely to react with amines in a simple acid–base reaction to form a salt. O O B B ...OCOOH H2NO... 88n ... OCOOJ H3NO... We therefore have to find a way to force the reaction to form the amide. Forming a peptide bond is also an uphill process (G° 17 kJ/molrxn), so a way must be found to drive the reaction forward. • Because the sequence of amino acids is important, they must be added to the chain one at a time, in a carefully controlled fashion. Thus, a significant entropy factor must be overcome during the synthesis of peptides or proteins. • The R groups of certain amino acids must be protected during polymerization so that no reactions take place on the side chains. •
In 1984 R. B. Merrifield received the Nobel Prize in chemistry for developing an automated approach to the synthesis of peptides. The first step involves attaching the amino acid that will become the C-terminal residue to an inert, insoluble polystyrene resin. Amino acids are then incorporated, one at a time, by coupling them onto the growing peptide chain. Because the product of each step in the reaction is a solid, it can be easily collected, washed, and purified before the next step in the reaction. The Merrifield synthesis uses a dehydrating agent known as dicyclohexylcarbodiimide (DCC) to drive the reaction that forms the peptide bond. To prevent reactions at the wrong site, appropriate blocking groups are added to reactive sites on the side chains of the amino acids before they are polymerized. If we use the symbol B to indicate an appropriate blocking group, the synthesis of a dipeptide can be represented by the following equation. B B A A R O R O B A B A DCC BONHOCHOCOOH H2NOCHOCOresin 8n B B A A O O R R A B B A BONHOCHOCONHOCHOCOresin The blocking group on the N-terminal end of the dipeptide is then removed, and a third blocked amino acid residue is added to give a tripeptide. The process of adding one amino acid at a time is continued until the polypeptide or protein synthesis is complete. The polypeptide or protein chain is then removed by reacting the resin with HBr in a suitable solvent. When it was first introduced, the peptide synthesis process was automated on an apparatus that required about 4 hours to add an amino acid residue to the peptide chain. Thus, insulin could be synthesized in approximately 8 days, while ribonuclease, with 124 amino acids, required more than a month. The beauty of the Merrifield synthesis is the yield of each step, which is essentially 99%. The synthesis of ribonuclease, for example, took 369 chemical reactions and 11,931 automated steps, and yet still had an overall yield of 17%.
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BIO.6 THE STRUCTURE OF PROTEINS The Primary Structure of Proteins The primary structure of a protein is nothing more than the sequence of amino acids, read off one at a time, as if printed on ticker tape. Insulin obtained from cows, for example, consists of two chains (A and B) with the primary structures shown in Figure BIO.8. There is more to the structure of a protein, however, than the sequence of amino acids. The polypeptide chain folds back on itself to form a secondary structure. Interactions between amino acid side chains then produce a tertiary structure. For some proteins, such as hemoglobin, interactions between individual polypeptide chains give rise to a quaternary structure. S OS A chain Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-Ser-Leu-Tyr-Gln-Leu-Glu-Asn-Tyr-Cys-Asn S
S OS
S
B chain
Phe-Val-Asn-Gln-His-Leu-Cys-Gly-Ser-His-Leu-Val-Glu-Ala-Leu-Tyr-Leu-Val-Cys-Gly Glu Arg Gly Phe Ala-Lys-Pro-Thr-Tyr-Phe
N-terminal amino acid
C-terminal amino acid
FIGURE BIO.8 Insulin is a relatively small protein that consists of two chains held together by covalent SOS bonds between side chains of cysteine residues.
The Secondary Structure of Proteins
O S
The peptide bond is a resonance hybrid of the two Lewis structures shown below. The Lewis structure on the left implies that the geometry around the carbon atom is trigonal planar and that the carbon atom and its three nearest neighbors lie in the same plane. The Lewis structure on the right suggests a trigonal planar geometry for the nitrogen atom as well. Because the peptide bond is a hybrid of the resonance forms, the six atoms involved all lie in the same plane. O B C E H EC ... O N ...C A H
OS SO A C ... C E E N ...C A H
Since the NOH and CPO bonds are relatively polar, hydrogen bonds form between adjacent peptide chains. G
D NOH, , OPC H G D G D C C D G D G CPO , ,HON R H D G R
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The fact that the six atoms in the peptide bond must lie in the same plane limits the number of ways in which a polypeptide can be arranged in space. By building models, Linus Pauling and Robert Corey discovered two ways in which a polypeptide chain could maximize the hydrogen bonds between peptides. In one of the structures, the chain forms the helix shown in Figure BIO.9. The other structure is the -pleated sheet in Figure BIO.10.
Hydrogen Oxygen Nitrogen Carbon R group
FIGURE BIO.9 The hydrogen bonds between adjacent peptide bonds allow polypeptides to form a right-handed helix. One turn along the helix contains 3.6 amino acid residues.
Hydrogen Oxygen Nitrogen Carbon R group
FIGURE BIO.10 The polypeptide chain can fold back on itself to form a structure that looks something liked a pleated sheet. Two different -pleated sheet structures are found in nature that differ in whether the adjacent polypeptide chains run in the same or opposite directions. This drawing shows the antiparallel structure found in silk.
The Tertiary Structure of Proteins Most proteins have structures that lie between the extremes of ideal helixes and -pleated sheets because other factors influence the way proteins fold to form three-dimensional structures. Particular attention must be paid to interactions between the side chains of the amino acids that form the backbone of the protein. Figure BIO.11 shows four ways in which the amino acid side chains can interact to form the tertiary structure of a protein. Disulfide (SOS) linkages: If the folding of a protein brings two cysteine residues together, the two OSH side chains can be oxidized to form a covalent SOS bond. These disulfide bonds cross-link the polypeptide chain. Hydrogen bonding: In addition to the hydrogen bonds between atoms involved in peptide bonds that give rise to the secondary structure of the protein, hydrogen bonds can form between amino acid side chains.
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4 CH3
CH2 1
CH3
CH CH3
S
CH2CH3 CH3 CH
S CH2 O H2C
OH
O–
+
N H3
C
(CH2)4
– O2
CH2
C
H2C 2 3
FIGURE BIO.11 Four factors are responsible for the tertiary structure of proteins: (1) disulfide linkages, (2) hydrogen bonding, (3) electrostatic interactions, and (4) hydrophobic interactions.
Ionic bonding: The structure of a protein can be stabilized by the force of attraction between amino acid side chains of opposite charge, such as the ONH3 side chain of Lys and the OCO2 side chain of Asp. Hydrophobic interactions: Proteins often fold so that the hydrophobic side chains of the amino acids Gly, Ala, Val, Leu, Ile, Pro, Met, Phe, and Trp are buried within the protein, where they can interact to form hydrophobic pockets. These hydrophobic interactions stabilize the structure of the protein. Human hair is composed primarily of proteins known as the -keratins that are about 14% cysteine. Hair curls as it grows because of the disulfide (SOS) links between cysteine residues on adjacent protein molecules. The first step in changing the way hair curls involves shaping the hair to our satisfaction and then locking it into place with curlers. The hair is then treated with a mild reducing agent that reduces the SOS bonds to pairs of OSH groups. This relaxes the proteins in the hair, allowing them to pick up the structure dictated by the curlers. The OSH side chains on cysteine residues that are now adjacent to each other are then oxidized by the O2 in air. New SOS linkages form, locking the hair permanently in place (at least until new hair grows). The -keratins are divided into two categories, “hard” and “soft,” on the basis of the amount of cysteine they contain. The -keratins in skin are soft because they contain relatively small amounts of sulfur, and disulfide cross-links are uncommon. Although hair is classified as a hard keratin, horn and hoof, which contain even more sulfur, are much less pliable because of the extensive disulfide cross-links that form.
The Quaternary Structure of Proteins As we have seen, hemoglobin is the protein that carries O2 through the bloodstream to the muscles. This protein consists of four polypeptide chains—two chains that contain 141 amino acids and two chains that contain 146 amino acids. Hemoglobin is therefore an example of a protein that has a quaternary structure. It consists of four polymer chains that must be assembled to form the complete protein. The polymer chains in a quaternary protein are not linked by covalent bonds such as the SOS bonds that hold together the polypeptide chains in insulin. The primary force of attraction between the and chains in hemoglobin is the result of interactions between hydrophobic substituents on the polymer chains. In other quaternary proteins, hydrogen bonding or ionic interactions between amino acid side chains on the surfaces of adjacent polymer chains also contribute to the process by which the polymer chains are held together.
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The Denaturation of Proteins Proteins are fragile molecules that are remarkably sensitive to changes in structure. The replacement of a polar Glu residue by a nonpolar Val at the sixth position on the chains of hemoglobin, for example, gives rise to the disease known as sickle-cell anemia. The introduction of a hydrophobic Val residue at that position changes the quaternary structure of hemoglobin. The “sticky,” nonpolar side chain on a valine residue at that position causes hemoglobin molecules to cluster together in an abnormal fashion, interfering with their function as oxygen-carrying proteins. Sickle-cell anemia is the result of a change in the way the protein is assembled from amino acids. The structure of a protein can also be changed after it has been made. Anything that causes a protein to leave its normal, or natural, structure is said to denature the protein. Factors that can lead to denaturation include the following. • • •
• •
Heating, which disrupts the secondary and tertiary structure of the protein. (The changes we observe when we fry an egg result from denaturation caused by heating.) Changes in pH that interfere with ionic bonding between amino acid side chains. Detergents, which make nonpolar amino acid side chains soluble and thereby destroy the hydrophobic interactions that give rise to the tertiary and quaternary structure of the protein. Oxidizing or reducing agents that either create or destroy SOS bonds. Reagents such as urea (H2NCONH2) that disrupt the hydrogen bonds which form the secondary structure of the protein.
BIO.7 CARBOHYDRATES: THE MONOSACCHARIDES The term carbohydrate was originally used to describe compounds that were literally “hydrates of carbon” because they had the empirical formula CH2O. In recent years, carbohydrates have been classified on the basis of their structures, not their formulas. They are now defined as polyhydroxy aldehydes and ketones. Among the compounds that belong to the family are cellulose, starch, glycogen, and most sugars. There are three classes of carbohydrates: monosaccharides, disaccharides, and polysaccharides. The monosaccharides are white, crystalline solids that contain a single aldehyde or ketone functional group. They are subdivided into two classes—aldoses and ketoses— on the basis of whether they are aldehydes or ketones. They are also classified as a triose, tetrose, pentose, hexose, or heptose on the basis of whether they contain three, four, five, six, or seven carbon atoms. With only one exception,3 the monosaccharides are optically active compounds. Although both D and L isomers are possible, most of the monosaccharides found in nature are in the D configuration. Structures for the D and L isomers of the simplest aldose, glyceraldehyde, are shown below.
CHO A C D ( OH HOCH2 H D -Glyceraldehyde
CHO A C D ( H HOCH2 OH L-Glyceraldehyde
3
Dihydroxyacetone is the only carbohydrate that is not optically active. O B HOCH2CCH2OH
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The structures of many monosaccharides were first determined by Emil Fischer in the 1880s and 1890s and are still written according to a convention he developed. The Fischer projection represents what the molecule would look like if its three-dimensional structure were projected onto a piece of paper. By convention, Fischer projections are written vertically, with the aldehyde or ketone at the top. The OOH group on the second-to-last carbon atom is written on the right side of the skeleton structure for the D isomer and on the left for the L isomer. Fischer projections for the two isomers of glyceraldehyde are shown below. CHO A HOCOOH A CH2OH
CHO A HOOCOH A CH2OH
D-Glyceraldehyde
L-Glyceraldehyde
The Fischer projections can be obtained from the skeleton structures shown above by imaging what would happen if you placed a model of each isomer on an overhead projector so that the CHO and CH2OH groups rested on the glass and then looked at the images of the models that would be projected on a screen. Fischer projections for some of the more common monosaccharides are given in Figure BIO.12. Aldoses
Ketoses O B
C A
A
C A
H OH
A
A
H
C A
A
OH
A
H
HO
O
A
OH
B
C A
O
A
A
H
C A
C A
A
CH2OH D-Mannose
C A H
A
OH
CH A 2OH CH A 2OH B
OH
A
A
C A
H
A
A
CH2OH D-Galactose
C A
H
A
H
A
OH
C A
H
A
H
A
HO
H
C A
A
A
C A
H
A
C A
HO
A
A
H
C A
C A
OH
A
HO
A
CH2OH D-Glucose
HO
C A
O B H
A
OH
H
A
OH
C A
A
A
C A
H
A
C A
A
CH2OH D-Arabinose
A
H
C A
OH
A
OH
A
H
C A
O B H
A
OH
A
C A
HO
H
A
A
CH2OH D-Xylose
C A
A
H
A
OH
C A
H
A
H
A
HO
H
H
A
A
C A
C A
C A A
OH
C A
A
CH2OH D-Ribose
C A
A
H
H
A
OH
C A A
HO
O B A
H
OH
A
A
C A
OH
A
A
H
C A C A
H
A
H
A
H
A
C A
O B
A
O B
OH
CH2OH
CH2OH
D-Ribulose
D-Fructose
FIGURE BIO.12 Fischer projections for some of the common monosaccharides.
Exercise BIO.2 Glucose and fructose have the same formula: C6H12O6. Glucose is the sugar with the highest concentration in the bloodstream; fructose is found in fruit and honey. Use the Fischer projections in Figure BIO.12 to explain the difference between the structures of the compounds. Predict what an enzyme would have to do to convert glucose to fructose, or vice versa. Solution Both compounds have the same structure for the third, fourth, fifth, and sixth carbon atoms. One compound, however, is an aldehyde (glucose), while the other is a ketone (fructose). Glucose could be converted to fructose by simultaneously reducing the first carbon atom from an aldehyde to a primary alcohol and oxidizing the second carbon from a secondary alcohol to a ketone. Fructose can be converted to glucose, on the other hand, by simultaneously oxidizing the first carbon atom and reducing the second. If the carbon chain is long enough, the alcohol at one end of a monosaccharide can attack the carbonyl group at the other end to form a cyclic compound. When a six-membered
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ring is formed, the product of the reaction is called a pyranose (see Figure BIO.13). When a five-membered ring is formed, it is called a furanose (see Figure BIO.14). There are two possible structures for the pyranose and furanose forms of a monosaccharide, which are called the and anomers. These anomers are isomers that differ in the orientation of the H and OH on the first carbon. H OH H O B OH
A
H
A
OH
A
C A
H
A
C A
A
H
C A
A
H
C A
A
HO
A
H
A
C A
O
HO
OH
H
HO
-D-Glucopyranose
OH
H H
HO
H OH H O
HO
CH2OH
OH
HO
-D-Glucopyranose
OH
H H
H
FIGURE BIO.13 The and anomers of D-glucopyranose.
CH2OH H O
CH2OH OH -D-Fructofuranose
CH A 2OH H
A
OH
A
C A
O
A
C A
A
H
C A
A
H
A
HO
B
C A
OH
OH
H H
OH
CH2OH H
CH2OH
O
OH OH -D-Fructofuranose
CH2OH
H OH
H
FIGURE BIO.14 The and anomers of fructofuranose.
The reactions that lead to the formation of a pyranose or a furanose are reversible. Thus, it doesn’t matter whether we start with a pure sample of -D-glucopyranose or -Dglucopyranose. Within minutes, the anomers are interconverted to give an equilibrium mixture that is 63.6% of the anomer and 36.4% of the anomer. The 21 preference for the anomer can be understood by comparing the structures of the molecules shown in Figure BIO.13. In the anomer, all of the bulky OOH and OCH2OH substituents lie more or less within the plane of the six-membered ring. In the anomer, one of the OOH groups is perpendicular to the plane of the six-membered ring, in a region where it feels strong repulsive forces from the hydrogen atoms that lie in similar positions around the ring. As a result, the anomer is slightly more stable than the anomer.
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BIO.8 CARBOHYDRATES: THE DISACCHARIDES AND POLYSACCHARIDES Disaccharides are formed by condensing a pair of monosaccharides. The structures of three important disaccharides with the formula C12H22O11 are shown in Figure BIO.15. Maltose, or malt sugar, which forms when starch breaks down, is an important component of the barley malt used to brew beer. Lactose, or milk sugar, is a disaccharide found in milk. Very young children have a special enzyme known as lactase that helps digest lactose. As they grow older, many people lose the ability to digest lactose and cannot tolerate milk or milk products. Because human milk has twice as much lactose as milk from cows, young children who develop lactose intolerance while they are being breast-fed are switched to cows’ milk or a synthetic formula based on sucrose.
OH H CH2
H
OH
O H
HO
CH2
H
HO H H
OH
H
O OH
HO H H
OH OH CH2
Maltose
O
OH H
OH H
H O
H
CH2
H
O
O
HO H H
Lactose
OH
HO
OH H
H H
OH H
OH H CH2 HO
H O H
HO H H
OH
OH
Sucrose
H
O
CH2OH H
OH O
CH2OH
H
FIGURE BIO.15 The structures of three common disaccharides.
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The substance most people refer to as “sugar” is the disaccharide sucrose, which is extracted from either sugar cane or beets. Sucrose is the sweetest of the disaccharides. It is roughly three times as sweet as maltose and six times as sweet as lactose. In recent years, sucrose has been replaced in many commercial products by corn syrup, which is obtained when the polysaccharides in cornstarch are broken down. Corn syrup is primarily glucose, which is only about 70% as sweet as sucrose. Fructose, however, is about two and one-half times as sweet as glucose. A commercial process has therefore been developed that uses an isomerase enzyme to convert about half of the glucose in corn syrup into fructose (see Exercise BIO.2). The resulting high-fructose corn sweetener is just as sweet as sucrose and has found extensive use in soft drinks. The monosaccharides and disaccharides represent only a small fraction of the total amount of carbohydrates in the natural world. The great bulk of the carbohydrates in nature are present as polysaccharides, which have relatively large molecular weights. The polysaccharides serve two principal functions. They are used by both plants and animals to store glucose as a source of future food energy, and they provide some of the mechanical structure of cells. Very few forms of life receive a constant supply of energy from their environment. In order to survive, plant and animal cells have had to develop a way of storing energy during times of plenty in order to survive the times of shortage that follow. Plants store food energy as polysaccharides known as starch. There are two basic kinds of starch: amylose and amylopectin. Amylose is found in algae and other lower forms of plants. It is a linear polymer of glucose residues whose structure can be predicted by adding -D-glucopyranose rings to the structure of maltose. Amylopectin is the dominant form of starch in the higher plants. It is a branched polymer of about 6000 glucose residues with branches on 1 in every 24 glucose rings. A small portion of the structure of amylose is shown in Figure BIO.16. HOCH2 HO HO
O HOCH2 OH
O O
HO
HOCH2 OH
O
n
O
HO
Amylose n 1000–6000
OH OH
FIGURE BIO.16 The structure of amylose formed by linking -D-glucopyranose rings.
The polysaccharide that animals use for the short-term storage of food energy is known as glycogen. Glycogen has almost the same structure as amylopectin, with two minor differences. The glycogen molecule is roughly twice as large as amylopectin, and it has roughly twice as many branches. There is an advantage to branched polysaccharides such as amylopectin and glycogen. During times of shortage, enzymes attack one end of the polymer chain and cut off glucose molecules, one at a time. The more branches, the more points there are at which the enzyme attacks the polysaccharide. Thus, a highly branched polysaccharide is better suited for the rapid release of glucose than is a linear polymer.
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Polysaccharides are also used to form the walls of plant and bacterial cells. Cells that do not have a cell wall often break open in solutions whose salt concentrations are either too low (hypotonic) or too high (hypertonic). If the ionic strength of the solution is much smaller than that of the cell, osmotic pressure forces water into the cell to bring the system into balance, which causes the cell to burst. If the ionic strength of the solution is too high, osmotic pressure forces water out of the cell, and the cell breaks open as it shrinks. The cell wall provides the mechanical strength that helps protect plant cells that live in freshwater ponds (too little salt) or seawater (too much salt) from osmotic shock. The cell wall also provides the mechanical strength that allows plant cells to support the weight of other cells. The most abundant structural polysaccharide is cellulose. There is so much cellulose in the cell walls of plants that it is the most abundant of all biological molecules. Cellulose is a linear polymer of glucose residues, with a structure that resembles amylose more closely than amylopectin, as shown in Figure BIO.17. The difference between cellulose and amylose can be seen by comparing Figures BIO.16 and BIO.17. Cellulose is formed by linking -glucopyranose rings, instead of the -glucopyranose rings in starch and glycogen. HOCH2 HO HO
O OH
(
HOCH2 O HO
O OH
(
O
n
HOCH2 HO
O OH
OH
Cellulose n = 5000–10,000
FIGURE BIO.17 The structure of cellulose formed by linking -D-glucopyranose rings.
The OOH substituent that serves as the primary link between -glucopyranose rings in starch and glycogen is perpendicular to the plane of the six-membered ring. As a result, the glucopyranose rings in those carbohydrates form a structure that resembles the stairs of a staircase. The OOH substituent that links the -glucopyranose rings in cellulose lies in the plane of the six-membered ring. The cellulose molecule therefore stretches out in a linear fashion. This makes it easier for strong hydrogen bonds to form between the OOH groups of adjacent molecules. This, in turn, gives cellulose the rigidity required for it to serve as a source of the mechanical structure of plant cells. Cellulose and starch provide an excellent example of the link between the structure and function of biomolecules. At the turn of the twentieth century, Emil Fischer suggested that the structure of an enzyme is matched to the substance on which it acts, in much the same way that a lock and key are matched. Thus, the amylase enzymes in saliva that break down the linkages between glucose molecules in starch cannot act on the linkages in cellulose. Most animals cannot digest cellulose because they don’t have an enzyme that can cleave linkages between glucose molecules. Cellulose in their diet therefore serves only as fiber, or roughage. The digestive tracts of some animals, such as cows, horses, sheep, and goats, contain bacteria that have enzymes that cleave the linkages, so those animals can digest cellulose.
Exercise BIO.3 Termites provide an example of the symbiotic relationship between bacteria and higher organisms. Termites cannot digest the cellulose in the wood they eat, but their digestive tracts are infested with bacteria that can. Propose a simple way of ridding a house from termites, without killing other insects that might be beneficial.
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Solution Killing termites isn’t difficult—there are a number of poisons that do an excellent job. Unfortunately, the compounds can also poison other insects or even small animals or children. The simplest way of getting rid of termites is to treat the wood with a poison that kills the bacteria that digest the cellulose in wood, so that the termites starve to death.
For many years, biochemists considered carbohydrates to be dull, inert compounds that filled the space between the exciting molecules in the cell—the proteins. Carbohydrates were impurities to be removed when “purifying” a protein. Biochemists now recognize that most proteins are actually glycoproteins, in which carbohydrates are covalently linked to the protein chain. Glycoproteins play a particularly important role in the formation of the rigid cell walls that surround bacterial cells.
BIO.9 LIPIDS Any biomolecule that dissolves in relatively nonpolar solvents—such as chloroform (CHCl3), benzene (C6H6), and diethyl ether (CH3CH2OCH2CH3)—is classified as a lipid (from the Greek word lipos, “fat”). Because they are soluble in nonpolar solvents, lipids are often insoluble—or only marginally soluble—in water, and they often feel oily or greasy to the touch.
Neutral Fats and Oils Long-chain carboxylic acids such as stearic acid [CH3(CH2)16CO2H] are called fatty acids because they can be isolated from animal fats. The fatty acids are subdivided into two categories on the basis of whether they contain CPC double bonds: saturated fatty acids and unsaturated fatty acids. The common names of carboxylic acids trace back to Latin or Greek stems that indicate a natural source of the acid. The destructive distillation of ants, for example, produces formic acid (from the Latin word formica, “ant”). Vinegar is a 5–6% solution of acetic acid in water. Acetic acid therefore takes its common name from the Latin term for vinegar: acetum. The next acid, as we build up the hydrocarbon chain, is propionic acid, which takes its name from the Greek stems protos and pion. The name literally means “first fat,” because it is the simplest carboxylic acid that can be isolated from fat. The next member of the family is butyric acid, from the Latin word butyrum, or “butter,” because it can be obtained from rancid butter. The fifth carboxylic acid is known as valeric acid, because it can be obtained from plants in the genus Valeriana, which are perennial herbs. HCO2H CH3CO2H CH3CH2CO2H CH3CH2CH2CO2H CH3CH2CH2CH2CO2H
Formic acid Acetic acid Propionic acid Butyric acid Valeric acid
From this point on, common carboxylic acids tend to have an even number of carbon atoms. The next three derivatives are all given names from the Latin term for goat, caper. The carboxylic acids with 12, 14, 16 and 18 carbon atoms are named from the Latin stem for the bay tree, laurel; the genus for the spice nutmeg, Myristica; the Latin stem for the palm tree, palma; and the Greek stem for the tallow used to make candles, stear.
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CH3(CH2)4CO2H CH3(CH2)6CO2H CH3(CH2)8CO2H CH3(CH2)10CO2H CH3(CH2)12CO2H CH3(CH2)14CO2H CH3(CH2)16CO2H
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Caproic acid Caprylic acid Capric acid Lauric acid Myristic acid Palmitic acid Stearic acid
The very small carboxylic acids have a sharp odor. (Formic acid has an odor that is even sharper than acetic acid.) By the time the hydrocarbon chain has grown to a total of four carbon atoms, the odor of the compounds has taken a significant turn for the worse. (Butyric acid is the source of the characteristic odor of rancid butter and spoiled meat.) As the length of the hydrocarbon chain increases further, the odor of the acid changes once again—this time, becoming more pleasant. There are four important unsaturated fatty acids. One of them, a derivative of palmitic acid, is known as palmitoleic acid. CH3(CH2)5
(CH2)7CO2H D G CPC G D H H
Palmitoleic acid
The other three are derivatives of stearic acid. The first has a single CPC double bond in the center of the fatty acid chain, and is known as oleic acid. CH3(CH2)7 (CH2)7CO2H G D CPC G D H H
Oleic acid
The second, which is known as linoleic acid, has another CPC double bond in the nonpolar half of the fatty acid chain. CH 2 (CH2)7CO2H D G D G C PC C PC D D G G H HH H
CH3(CH2)4
Linoleic acid
The third—linolenic acid—has one more CPC double bond in the same half of the fatty acid chain. CH 2 (CH2)7CO2H CH 2 D D G D G G C PC C PC C PC C D D D G G G H HH H HH
CH3CH2
Linolenic acid
There are several regularities in the chemistry of the unsaturated fatty acids. First, they contain cis double bonds. Second, the double bonds are always isolated from each other by a CH2 group. So much attention is paid to the structures of the fatty acids in discussions of the chemistry of lipids that it is easy to miss an important point: Free fatty acids are seldom found in nature. They are usually tied up with alcohols or amines to form esters (RCO2R) or
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amides (RCONHR). The most abundant lipids are the triesters formed when a glycerol molecule reacts with three fatty acids, as shown in Figure BIO.18. These lipids have been known by a variety of names, including fat, neutral fat, glyceride, triglyceride, and triacylglyceride. P
O
P
P
CH2OC(CH2)12CH3 O A A CH3(CH2)12COCH A O A CH2OC(CH2)12CH3
Trimyristin
FIGURE BIO.18 The triester trimyristin can be isolated from nutmeg.
Most animal fats are complex mixtures of different triglycerides. As the percentage of unsaturated fatty acids in the fats increases, the melting point of the triglyceride decreases until it eventually becomes an oil at room temperature. Beef fat, which is one-third unsaturated fatty acids, is a solid. Olive oil, which is roughly 80% unsaturated, is a liquid. The effect of unsaturated fatty acids on the melting point of a triglyceride can be understood by recognizing that the cis CPC double bond introduces a rigid 30° bend in the hydrocarbon chain, as shown in Figure BIO.19.
FIGURE BIO.19 A phospholipid containing one saturated (bottom) and one unsaturated (top) fatty acid.
This bend or “kink” increases the average distance between triglyceride molecules, which decreases the van der Waals interactions between neighboring molecules. Thus, the introduction of unsaturated fatty acids into a triglyceride increases the fluidity of the lipid. Table BIO.2 compares the relative abundances of the common fatty acids in a typical animal fat (butter) and a vegetable oil (olive oil). TABLE BIO.2 Relative Abundance of Fatty Acids in a Typical Fat and a Typical Oil Fatty Acid
Butter
Butyric Caproic Caprylic Capric Lauric Myristic Palmitic Stearic Palmitoleic Oleic Linoleic Linolenic
3–4% 1–2% 1% 2–3% 2–5% 8–15% 25–29% 9–12% 4–6% 18–33% 2–4% 1%
Olive Oil
1% 5–15% 1–4% 67–84% 8–12%
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Fats and oils are used by living cells for only one purpose—to store energy. They are a far more efficient storage system than glycogen or starch because they give off between two and three times as much energy when they are burned. (The metabolism of glycogen releases 15.7 kilojoules per gram of carbohydrate consumed, whereas the metabolism of lipids gives approximately 40 kJ/g.) This explains why the seeds of many plants are relatively rich in oils, which provide the energy the seed needs to grow until the leaves can begin to produce energy by photosynthesis. The average human contains enough fat (21% of the body weight for men, 26% for women) to provide the energy they need to survive for up to 3 months. But there is only enough glycogen stored in the human body at any time to provide enough energy for one day. Thus, glycogen is only used for the short-term storage of food energy. In “times of plenty,” the body stores energy in the form of fat to compensate for “times of shortage” to come.
Checkpoint Several hundred people were poisoned in Spain a few years ago when an unscrupulous merchant diluted olive oil with diesel oil (see Section O1.11). Explain both the similarities and differences between the structures of the components of the two “oils.”
Polar Lipids Fats and oils are neutral compounds. When one of the fatty acids in a triglyceride is replaced by a phosphate group, a phospholipid is obtained that has two nonpolar hydrophobic tails and a charged hydrophilic head.
O B CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOOCH2 A CH3CH2CH2CH2CH2CH2CH2CH2CHPCHCH2CH2CH2CH2CH2CH2CH2COOOCH O B B A O CH2OOOPOO A O
A variety of biochemically important molecules can be obtained by forming a second ester linkage to the phosphate group. These phosphate diesters are often called phosphatides, and they contain an alcohol at the position labeled with an X in the following structure.
CH3(CH2)14CO2CH2 A O CH3(CH2)14CO2CH B A CH2OOOPOX A O
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The most important phosphatides contain the following X groups. X OCH2CH2N(CH3)3 X OCH2CH2NH3
Phosphatidylcholine Phosphatidylethanolamine
Note that the phosphatidylcholines and phosphatidylethanolamines are zwitterions because they simultaneously carry both positive and negative charges within a molecule that has no net charge. The phosphatidylcholines are also known as lecithins, while the phosphatidylethanolamines are known as cephalins. The phosphatidylcholines and phosphatidylethanolamines are described as amphipathic (literally, “both paths”) because they contain a polar, hydrophilic head and a pair of nonpolar, hydrophobic tails. The compounds can therefore spontaneously associate in aqueous solutions to form a bilayer in which the molecules are oriented so that the nonpolar tails of adjacent phospholipids form a hydrophobic pocket and the polar heads point toward the water that surrounds both sides of the bilayer. If we symbolize the hydrophilic head by a small circle and the hydrophobic tails by a pair of wavy lines, we can see that the bilayer is the first step toward the formation of a membrane (see Figure BIO.20). Hydrophilic areas
Hydrophilic heads
Hydrophobic areas
Hydrophobic tails
Phospholipid bilayer
Hydrophilic areas
Integral protein
FIGURE BIO.20 Schematic drawing of the phospholipid bilayer that forms the membranes that separate cells from each other and compartmentalize each cell.
The best model for the structure of cell membranes involves a bilayer of amphipathic lipids approximately 8 nm wide into which various proteins are embedded. It is a common mistake to assume that the membrane is static. It is not. There is a considerable amount of mobility or flexibility in the hydrocarbon tails of the lipid molecules. However, the strong hydrophobic character of the region between the inner and outer surface of the membrane resists the passage of highly charged or polar intermediates across the membrane.
Chemistry in the World Around Us The Search for New Drugs In 1763, the Reverend Edmund Stone took the first step toward the discovery of one of the most commonly used medicines when he noted that the bark of the English willow was an effective treatment for patients suffering from a fever. Stone explained the effect
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of willow bark by noting that “many natural maladies carry their cures along with them, or their remedies lie not far from their causes.” Thus, he argued, the English willow grows in the same moist regions where one was likely to catch the fever treated with its bark. It took 50 years before the active ingredient in willow bark was isolated and named salicin, from the Latin name for the willow (Salix alba). Another 50 years elapsed before a large-scale synthesis for the compound was available. By that time, the compound was known as salicylic acid because saturated solutions in water are highly acidic (pH 2.4). By the end of the nineteenth century, salicylic acid was being used to treat rheumatic fever, gout, and arthritis. Many patients treated with the drug complained of chronic stomach irritation because of its acidity and the large doses required (6–8 g/day). Because his father was one of these patients, Felix Hoffman searched the chemical literature for a less acidic derivative of salicylic acid. In 1898, Hoffman reported that the acetyl ester of salicylic acid was simultaneously more effective and easier to tolerate than the parent compound. He named the compound aspirin, taking the prefix a- from the name of the acetyl group and spirin from the German name of the parent compound spirsäure. The existence of a drug that reduced both pain and fever initiated a search for other compounds that could achieve the same result. Although based on trial and error, this search inevitably produced a variety of substances, such as those in Figure BIO.21, that are analgesics, antipyretics, and/or anti-inflammatory agents. Analgesics relieve pain without decreasing sensibility or consciousness. Antipyretics reduce the body temperature when it is elevated. Anti-inflammatory agents counteract swelling or inflammation of the joints, skin, and eyes. CH3
CO2H
OH
CH
NH
CH2
CPO
CH
CO2H O
P
CO2H OH
Salicylic acid
OCCH3
Acetylsalicylic acid (aspirin)
CH3 Acetaminophen (Tylenol)
CH3
CH3 Ibuprofen (Motrin, Advil)
FIGURE BIO.21 The structures of some common analgesics, antipyretics, and anti-inflammatory agents.
Although the use of aspirin has been widespread since the nineteenth century, the mechanism for its action was first described in 1971 [J. R. Vane, Nature, 231(25), 232–235 (1971)]. Vane noted that injury to tissue was often followed by the release of a group of hormones known as the prostaglandins, which have widespread physiological effects at very low concentrations. The prostaglandins regulate blood pressure, mediate the inflammatory response of the joints, induce the process by which blood clots, regulate the sleep/wake cycle, and, when appropriate, induce labor. Vane suggested that aspirin and other nonsteroidal anti-inflammatory drugs (or NSAIDs) inhibit the enzyme that starts the process by which prostaglandins such as PGE2 and PGF2 are synthesized from the 20-carbon unsaturated fatty acid known as arachidonic
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acid shown in Figure BIO.22. The steroidal anti-inflammatory drugs (such as hydrocortisone) achieve a similar effect by inhibiting the enzyme that releases arachidonic acid into the cell. CO2H
Arachidonic acid
O
HO CO2H
HO
OH PGE2
CO2H HO
OH PGF2
FIGURE BIO.22 The structures of the PGE2 and PGF2 prostaglandins synthesized from arachidonic acid.
Now that they are beginning to understand the mechanism by which drugs operate, medicinal chemists can approach the design of drugs by a rational process. One paper described progress toward the design of a drug to treat the debilitating diseases caused by protozoan parasites that afflict millions of people in Latin America, Africa, and Asia [W. N. Hunter et al., Journal of Molecular Biology, 227, 322–333 (1992)]. The potential target for the drug is an enzyme—trypanothione reductase (TR)—that protects the parasite from oxidative damage from the immune system of its mammalian host. Mammalian cells use a similar enzyme, known as glutathione reductase (GR), to protect against damage from oxidation reactions. Hunter and co-workers found that the human GR enzyme has a smaller, more positively charged active site compared to the TR enzyme in the parasite. The structural information in this study can now be used to rationally modify a substrate of the enzymes until it possesses the following characteristics. The substrate must be too large to bind to the GR enzyme in humans. The substrate must have a high affinity for binding to the TR enzyme in the parasite. • The substrate must inhibit the activity of the TR enzyme, thereby allowing the immune system of the mammalian host to attack and eventually destroy the parasite. • •
BIO.10 NUCLEIC ACIDS Nucleic acids, which are relatively strong acids found in the nuclei of cells, were first isolated in 1869. The nucleic acids include polymers with molecular weights as high as 100,000,000 grams per mole. They can be broken down, or digested, to form monomers known as nucleotides. Each nucleotide contains three units: a sugar, an amine, and a phosphate, as shown in Figure BIO.23. Nucleic acids are divided into classes on the basis of the sugar used to form the nucleotides. Ribonucleic acid (RNA) is built on a -D-ribofuranose ring. Deoxyribonucleic acid (DNA) contains a modified ribofuranose in which the OOH group on the second carbon atom has been removed, as shown in Figure BIO.24.
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NH2 N
O B O O O P O O CH2 A O
Phosphate
Amine
N
O O
Sugar
FIGURE BIO.23 The structure of the nucleotide known as cytidine monophosphate.
OH 5
5
HO OCH2 H O 4 C C H 3 OH
HO OCH2 H O 4 C C H 3 OH
Amine H
C
1
C
2
H OH
-D-Ribofuranose found in RNA
Amine H C 2
C
1
H
H
-D-Deoxyribofuranose found in DNA
FIGURE BIO.24 The structures of the carbohydrates on which DNA and RNA are built.
The amines that form nucleic acids fall into two categories: purines and pyrimidines. There are three pyrimidines—cytosine, thymine, and uracil—and two purines—adenine and guanine—as shown in Figure BIO.25. DNA and RNA each contain four nucleotides. Both contain the same purines—adenine and guanine—and both also contain the pyrimidine cytosine. But the fourth nucleotide in DNA is thymine, whereas RNA uses uracil to complete its quartet of nucleotides. NH2
O
O
H N O
CH3
N N
H N
N
O
H
N
O
H
Cytosine
H
Thymine
Uracil
PYRIMIDINES
O
NH2 H
N
N N H
N
N
N
H2N H
Adenine
N H
N H
Guanine PURINES
FIGURE BIO.25 The two classes of amine substituents in nucleic acids.
The carbon atoms in the sugar at the center of a nucleotide are numbered from 1′ to 5′. The OOH group on the 3′ carbon of one nucleotide can react with the phosphate
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attached to the 5′ carbon of another to form a dinucleotide held together by phosphate ester bonds. As the chain continues to grow, it becomes a polynucleotide. A short segment of a DNA chain is shown in Figure BIO.26. Reading from the 5′ end of the chain to the 3′ end, this DNA segment contains the following sequence of amine substituents: adenine (A), cytosine (C), guanine (G), and thymine (T). For many years, the role of nucleic acids in living systems was unknown. In 1944 Oswald Avery presented evidence that nucleic acids were involved in the storage and transfer of the genetic information needed for the synthesis of proteins. This suggestion was actively opposed by many of his contemporaries, who believed that the structure of the nucleic acids was too regular—and therefore too dull—to carry the information that codes for the thousands of different proteins a cell needs to survive. In retrospect, the first clue about how nucleic acids function was obtained by Erwin Chargaff, who found that DNA always contains the same amounts of certain pairs of bases. There is always just as much adenine as thymine, for example, and just as much guanine as cytosine. In 1954, James Watson and Francis Crick proposed a structure for DNA that explained how DNA could be used to store genetic information. Their structure consisted of two polynucleotide chains running in opposite directions that were linked by hydrogen bonds between a specific purine (A or G) on one strand and a specific pyrimidine (C or T) NH2
5 end
O A OO P OO O CH2 A O
N
N
A
N
N H
NH2 O A N O O P O O A O O CH2 O N
C
O O A N O O P O O A O O CH2 H2N N
N G
N H
O O A N O O P O O A N O O CH2 O
3 end
OH
FIGURE BIO.26 Short segment of a single strand of DNA.
CH3 T
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H bonds Sugar– phosphate chain
H CH3
N
† †
N
O,H N H
N N
N N
O
H
Thymine (T)
Adenine (A)
H N H ,O N O Cytosine (C)
†
N
Sugar– phosphate chain
H
N N
N N
H H bonds
N H
H Guanine (G)
FIGURE BIO.27 The two strands in DNA are held together by hydrogen bonds between specific pairs of purine and pyrimidine bases. The hydrogen bonds between A and T and between G and C are shown.
on the other, as shown in Figure BIO.27. The strands form a helix that is not quite as tightly coiled as the helix Pauling and Corey proposed for proteins. The helix structure must be able to explain two processes. There must be some way to make perfect copies of the DNA that can be handed down to future generations (replication). There also must be some way to decode the information on the DNA chain (transcription) and to translate the information into a sequence of amino acids in a protein (translation). Replication is easy to understand. According to Watson and Crick, an adenine on one strand of DNA is always paired with a guanine on the other, and a cytosine is always paired with a thymine. The two strands of DNA therefore complement each other perfectly; the sequence of nucleotides on one strand can always be predicted from the sequence on the other. Replication occurs when the two strands of the parent DNA molecule separate and both strands are copied simultaneously. Thus, one strand from the parent DNA is present in each of the daughter molecules produced when a cell divides.
BIO.11 PROTEIN BIOSYNTHESIS The information that tells a cell how to build the proteins it needs to survive is coded in the structure of the DNA in the nucleus of that cell. The code can’t be based on a one-toone match between nucleotides and amino acids because there are only four nucleotides but there are 20 amino acids that must be coded. If the nucleotides are grouped in threes, however, there are 64 possible triplets, or codons, which is more than enough combinations to code for the 20 amino acids. To understand how proteins are made, we have to divide the decoding process into two steps. DNA only stores the genetic information; it isn’t involved in the process by which the information is used. The first step in protein biosynthesis therefore has to involve transcribing the information in the DNA structure into a useful form. In a separate step, this information can be translated into a sequence of amino acids.
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Transcription Before the information in DNA can be decoded, a small portion of the DNA double helix must be uncoiled. A strand of RNA is then synthesized that is a complementary copy of one strand of the DNA. Assume that the section of the DNA that is copied has the following sequence of nucleotides, starting from the 3 end. 3 T-A-C-A-A-G-C-A-G-T-T-G-G-T-C-G-T-G... 5
DNA
When we predict the sequence of nucleotides in the RNA complement, we have to remember that RNA uses U where T would be found in DNA. We also have to remember that base pairing occurs between two chains that run in opposite directions. The RNA complement of the DNA should therefore be written as follows. 3 T-A-C-A-A-G-C-A-G-T-T-G-G-T-C-G-T-G... 5 A-U-G-U-U-C-G-U-C-A-A-C-C-A-G-C-A-C...
5 3
DNA m-RNA
Since the RNA strand contains the message that was coded in the DNA, it is called messenger RNA, or mRNA.
Translation The messenger RNA now binds to a ribosome, where the message is translated into a sequence of amino acids. The amino acids that are incorporated into the protein being synthesized are carried by relatively small RNA molecules known as transfer RNA, or tRNA. There are at least 60 tRNAs, which differ slightly in their structures, in each cell. At one end of each tRNA is a specific sequence of three nucleotides that can bind to the messenger RNA. At the other end is a specific amino acid. Thus, each three-nucleotide segment of the messenger RNA molecule codes for the incorporation of a particular amino acid. The relationship between the triplets, or codons, on the mRNA and the amino acids is shown in Table BIO.3.
Exercise BIO.4 Assume that the DNA chain that codes for the synthesis of a particular protein contains the triplet A-G-T (reading from the 3′ to the 5′ end). Predict the sequence of nucleotides in the triplet, or codon, that would be built in the messenger RNA constructed on the DNA template. Then predict the amino acid that would be incorporated at this point in the protein. Solution The messenger RNA synthesized when the DNA chain is read is a complement of the DNA. Every time a G appears, a C is placed on the complementary chain, and vice versa. Each time a T appears in the DNA message, an A is incorporated into the mRNA chain. When the DNA contains an A, the complementary RNA chain contains a U. DNA mRNA
3 5
...A-G-T... ...U-C-A...
5 3
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C
A
G
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The Genetic Code Second Position
U
C
A
G
Third Position (3′ End)
Phe Phe Leu Leu Leu Leu Leu Leu Ile Ile Ile Met Val Val Val Val
Ser Ser Ser Ser Pro Pro Pro Pro Thr Thr Thr Thr Ala Ala Ala Ala
Tyr Tyr —a —a His His Gln Gln Asn Asn Lys Lys Asp Asp Glu Glu
Cys Cys —a Trp Arg Arg Arg Arg Ser Ser Arg Arg Gly Gly Gly Gly
U C A G U C A G U C A G U C A G
a
There are three triplets that code for termination of the polypeptide chain: UAA, UGA, and UAG.
According to Table BIO.3, the mRNA chain is read from the 5′ to the 3′ end. We therefore look up the U-C-A codon in Table BIO.3 and find that the triplet codes for the amino acid serine.
The signal to start making a polypeptide chain in simple, prokaryotic cells4 is the triplet AUG, which codes for the amino acid methionine (Met). The synthesis of every protein in these cells therefore starts with a Met residue at the N-terminal end of the polypeptide chain. After the tRNA that carries Met binds to the start signal on the messenger RNA, a tRNA carrying the second amino acid binds to the next codon. A dipeptide is synthesized when the Met residue is transferred from the first tRNA to the amino acid on the second tRNA. If the DNA described in this section were translated, the dipeptide would be Met-Phe (reading from the N-terminal to the C-terminal amino acid). The messenger RNA now moves through the ribosome, and a tRNA carrying the third amino acid (Val) binds to the next codon. The dipeptide is then transferred to the amino acid on the third tRNA to form a tripeptide. This sequence of steps continues until one of three codons is encountered: UAA, UGA, or UAG. These codons give the signal for terminating the synthesis of the polypeptide chain, and the chain is cleaved from the last tRNA residue. The sequence of DNA described in this section would produce the following sequence of amino acids. 4
Prokaryotic cells do not have a cell nucleus. Cells that have a nucleus are called eukaryotic.
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Met-Phe-Val-Asn-Gln-His- ... This polypeptide is not necessarily an active protein. All proteins in prokaryotic cells start with Met when synthesized, but not all proteins have Met first in their active form. It is often necessary to clip off the Met after the polypeptide has been synthesized to give a protein with a different N-terminal amino acid. Modifications to the polypeptide often have to be made before an active protein is formed. Insulin, for example, consists of two polypeptide chains connected by disulfide linkages. In theory, it would be possible to make the chains one at a time and then try to assemble them to make the final protein. Nature, however, has been more subtle. The polypeptide chain that is synthesized contains a total of 81 amino acids. All of the disulfide bonds that will be present in insulin are present in that chain. The protein is made when a sequence of 30 amino acids is clipped out of the middle of the polypeptide chain.
KEY TERMS Amino acid Amphipathic Anomer Biochemistry Breathing motion Carbohydrate Codon Cooperative interaction Denaturation Deoxyribonucleic acid (DNA) Disaccharide Disulfide Fatty acid Gel electrophoresis Glycoprotein Heme
Hemoglobin Henderson–Hasselbalch equation Hydrophobic interaction Isoelectric point Lipid Membrane Messenger RNA (mRNA) Monosaccharide Myoglobin Nucleic acid Nucleotide Peptide Phospholipid Polynucleotide Polysaccharide
Porphyrin Primary structure Protein Protein biosynthesis Purine Pyrimidine Quaternary structure Replication Ribonucleic acid (RNA) Secondary structure Tertiary structure Transcription Transfer RNA (tRNA) Translation Triglyceride Zwitterion
PROBLEMS The Amino Acids 1. Explain why amino acids such as glycine (H2NCH2CO2H) exist in aqueous solution as zwitterions (H3NCH2CO2). 2. Which of the following forms of the amino acid glycine is present in strongly acidic solutions? In strongly basic solutions? (a) H3NCH2CO2H (b) H3NCH2CO2 (c) H2NCH2CO2 3. Write the formula for lysine as it would exist in aqueous solution at pH 1. 4. Write the formula for serine as it would exist at pH 1 and pH 13.
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5. Which of the following amino acids contain side chains that cannot contribute to forming a hydrophobic pocket? (a) alanine (b) isoleucine (c) methionine (d) serine (e) lysine 6. The common amino acids in proteins have the generic formula H2NCHRCO2H. Classify the following amino acids as either hydrophilic or hydrophobic. (a) alanine, R OCH3 (b) glutamic acid, R OCH2CH2CO2 (c) arginine, R OCH2CH2CH2NHC(NH2)PNH2 (d) methionine, R OCH2CH2SCH3 (e) threonine, R OCH(CH3)OH 7. Use examples to explain why amino acids with an R group that is either positively charged or negatively charged at neutral pH are hydrophilic. 8. The three titratable groups in the following fully protonated amino acid are labeled I, II, and III. Arrange the groups in order of increasing pKa. I
CO2H A CH2 A H3NCHCO2H II
9. 10.
11.
12.
III
(a) I II III (b) I III II (c) II I III (d) II III I (e) III I II (f) III II I The pI for tyrosine is 5.7. What would be the charge on the average tyrosine molecule when the pH of an aqueous solution was 3? The pKa values for the -carboxylic acid and the -amino groups in arginine are 1.82 and 8.99, respectively. The pKa for the titratable CPNH2 side chain in the amino acid is 12.48. Calculate the pI for arginine. The pKa values for the -carboxylic acid and the -amino groups in cysteine are 1.92 and 10.78, respectively. The pKa for the titratable OSH side chain in the amino acid is 8.33. Calculate the pI for cysteine. Which of the following amino acids would be the most likely to have an isoelectric point near pH 3? (a) Ala (b) Asp (c) Lys (d) Ser (e) Gln
Peptides and Proteins 13. How do amino acids, peptides, and proteins differ? 14. How many dipeptides can be formed when alanine, R OCH3, is condensed with cysteine, R OCH2SH? What is the difference between the dipeptides? 15. Describe the tripeptides that can form when alanine, R OCH3, is condensed with cysteine, R OCH2SH. 16. We can describe peptide chains by listing the amino acids from the N-terminal to the C-terminal residue. Write the amino acid sequences for all the tetrapeptides containing four different amino acids that can be formed from the amino acids Ala, Lys, Ser, and Tyr. 17. Write the Lewis structure for the following tripeptide: Met-Thr-Glu.
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The Structure of Proteins 18. Explain why proteins, polysaccharides, and nucleic acids can all be classified as condensation polymers. 19. Describe how the primary, secondary, tertiary, and quaternary structures of a protein differ. 20. Describe the four factors that give rise to the tertiary structure of a protein. Give an example of each. 21. Which of the following amino acids has the fewest interactions that contribute to the formation of the tertiary structure of a protein? (a) Gly (b) Glu (c) Cys (d) Met (e) Pro 22. List the amino acids that can form hydrogen bonds in the tertiary structure of a protein. 23. List the amino acids that can form ionic bonds in the tertiary structure of a protein. 24. List the amino acids that can form the hydrophobic pockets in the tertiary structure of a protein. 25. Explain how each of the following can denature a protein. (a) increasing the temperature (b) changing the pH (c) adding a detergent (d) adding an oxidizing or reducing agent (e) adding compounds, such as urea, (H2N)2CPO, that form strong hydrogen bonds
Carbohydrates 26. Which of the following compounds satisfy the literal definition of the term carbohydrate? (a) glucose, C6H12O6 (b) sucrose, C12H22O11 (c) cellulose, (C6H10O5)n (d) glyceraldehyde, HOCH2CHOHCHO (e) dihydroxyacetone, (HOCH2)2CPO (f) ribose, C5H10O5 27. Use examples to explain the difference between an aldose and a hexose and between a triose and a tetrose. 28. Describe the difference between -D-glucopyranose and -D-glucopyranose. Describe the difference between -D-glucopyranose and -D-glucofuranose. 29. Which of the following compounds is not an aldose?
CH2OH A HOC O OH A HOC O OH A CH2OH Erythrose
CHO A HOC OOH A HOOC OH A HOC O OH A HOC O OH A CH2OH
CHO A HOC OOH A HO OC OH A HO OC O H A HOC OOH A CH2OH
CH2OH A HOC OOH A HOOC OH A HOC OOH A HOC OOH A CH2OH
Glucose
Galactose
Glucitol
30. How do maltose, lactose, and sucrose differ? 31. How do glycogen, amylose, amylopectin, and cellulose differ? 32. Explain why humans are able to digest glycogen, amylose, and amylopectin but not cellulose. What would enable us to digest cellulose?
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33. The following formulas are the Fischer projections for D-erythrose and D-threose. Draw the Fischer projections for the L isomers of these carbohydrates. CH2OH A HOC OOH A HOC OOH A CH2OH
CH2OH A HOOC OH A HOC OOH A CH2OH
D-Erythrose
D-Threose
Lipids 34. Explain why lipids, such as fats and oils, are not soluble in water but are soluble in solvents such as chloroform, CHCl3, benzene, C6H6, and diethyl ether, (CH3CH2)2O. 35. In what ways are corn oil and olive oil similar to petroleum oil? What is the difference between these oils that allows us to digest the first two but not the third? 36. Draw the structure of the triglyceride formed when stearic acid combines with glycerol. 37. What is the difference between the structures of the fatty acids in fats and oils? How does this difference cause one of these to be a solid and the other a liquid at room temperature? 38. Calculate the average oxidation states of the carbon atoms in a typical carbohydrate, such as glucose, C6H12O6, and a typical fatty acid, such as palmitic acid, C16H32O2. Use the oxidation states to explain why fatty acids give off much more energy than carbohydrates when burned. 39. Glucose, C6H12O6, gives off 2870 kJ/mol when burned, and palmitic acid, C16H32O2, gives off 9790 kJ/mol when burned. Use the data to estimate the energy released per gram when carbohydrates and lipids are burned. 40. Which of the following carboxylic acids is least likely to be found in olive oil? (a) butyric acid (b) palmitic acid (c) oleic acid 41. The starting materials for the synthesis of phospholipids are the phosphatidic acids. Draw the structure of the phosphatidic acid that would be produced by esterifying glycerol with two molecules of stearic acid (C17H35CO2H) and one molecule of phosphoric acid. 42. Explain why phospholipids, such as the phosphatidylcholines, can be described as phosphodiesters. 43. Explain why phospholipids, such as the phosphatidylcholines (lecithins), are such important components of the structure of biological membranes. 44. Imagine a lipid bilayer formed by phosphatidylcholine. Which part(s) of the molecule would form the hydrophilic surface of the bilayer?
Nucleic Acids 45. Draw the structure of one of the nucleotides found in nucleic acids. Show how the structures of the monomers that form DNA differ from those that form RNA. 46. Classify the five common nucleic acids—adenine, cytosine, guanine, thymine, and uracil—as either purines or pyrimidines. 47. Explain why studies of nucleic acids invariably find roughly equivalent amounts of purine and pyrimidine nucleotides. Explain why nucleic acids that contain more adenine than guanine also contain more thymine than cytosine.
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48. Which of the following equations are correct statements of the results of the work of Edwin Chargaff, which provided the basis for Watson and Crick’s discovery of the structure of DNA? (a) (%G %A) (%C %T) (b) %G %A (c) %C %T (d) %G %C (e) %A %T 49. Draw the structures of the side chains on nucleic acids and show how hydrogen bonds form between adenine and thymine and between cytosine and guanine. 50. Describe the difference between replication, transcription, and translation.
Protein Biosynthesis 51. Describe the difference between the roles played by mRNA and tRNA in protein biosynthesis. 52. Assume that the following sequence of nucleotides on one strand of a DNA molecule is used as the template to code for a protein. 3′
T-A-C-A-A-G-C-A-G-T-T-G-G-T-C-G-T-G-
5′
Write the sequence of nucleotides, reading from the 5′ to the 3′ end, in the messenger RNA produced when this strand of DNA is transcribed. Describe the polypeptide, starting from the N-terminal end, that would be produced during protein biosynthesis.
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